CHAPTER 2 The correspondence between the problem set in this fifth edition versus the problem set in the 4’th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod).

2.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitational field ? How much mass can a force of 1 N support ?

Solution: ma = 0 = F = F – mg F = mg = 2 · 9.80665 = 19.613 N F = mg => m = F/g = 1 / 9.80665 = 0.102 kg 2.2 A model car rolls down an incline with a slope so the gravitational “pull” in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg. Find the acceleration.

Solution: ma = F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2 2.3 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. If the total car and driver mass is 1075 kg. Find the necessary force.

Solution: ma = F ; a = dV / dt = (60 · 1000) / (3600 · 5) = 3.33 m/s2 Fnet = ma = 1075 · 3.333 = 3583 N 2.4 A washing machine has 2 kg of clothes spinning at a rate that generates an acceleration of 24 m/s2. What is the force needed to hold the clothes?

Solution: F = ma = 2 kg · 24 m/s2 = 48 N 2.5 A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed of 75 km/h. What are the force and total time required?

2.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What force is needed and what is the final velocity?

Solution: a = dV / dt => dV = a dt => V = a t = 3 · 10 = 30 m/s V = 30 m/s ; F = ma = 950 · 3 = 2850 N 2.7 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration?

Solution: ma = F a = F / m m = msteel + mpropane = 15 + (1.75 · 44.094) = 92.165 kg a = 2000 / 92.165 = 21.7 m/s2 2.8 A rope hangs over a pulley with the two equally long ends down. On one end you attach a mass of 5 kg and on the other end you attach 10 kg. Assuming standard gravitation and no friction in the pulley what is the acceleration of the 10 kg mass when released?

Solution: Do the equation of motion for the mass m2 along the downwards direction, in that case the mass m1 moves up (i.e. has -a for the acceleration) m2 a = m2 g – m1 g – m1a (m1 + m2 ) a = (m2 – m1 )g This is net force in motion direction a = (10 – 5) g / (10 + 5) = g / 3 = 3.27 m/s2 g 2 1

2.9 A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force.

2.10 On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading?

Solution: Moon gravitation is: g = gearth/6 m m m Beam Balance Reading is 5 kg This is mass comparison Spring Balance Reading is in kg units length F g 5 Reading will be kg 6 This is force comparison 2.11 One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500- L tank. Find the specific volume on both a mass and mole basis (v and v ).

Solution: v = V/m = 0.5/1 = 0.5 m3/kg V v = V/n = = Mv = 32 · 0.5 = 16 m3/kmol m/M 2.12 A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall (average) specific volume.

Solution: mair = V = air ( Vtot – mgranite / ) = 1.15 [ 5 – (900 / 2400) ] = 1.15 · 4.625 = 5.32 kg v = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg 2.13 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m3. If the system is decelerated with 6 m/s2 what is the needed force?

2.14 A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 1500 kPa.

Solution: Force balance: F› = PA = F? = P A + mpg ; P = 1 bar = 100 kPa 00 A = ( /4) D2 = ( /4) · 0.1252 = 0.01227 m2 mp = (P-P )A/g = ( 1500 – 100 ) · 1000 · 0.01227 / 9.80665 = 1752 kg 0

2.15 A barometer to measure absolute pressure shows a mercury column height of 725 mm. The temperature is such that the density of the mercury is 13550 kg/m3. Find the ambient pressure.

Solution: Hg : l = 725 mm = 0.725 m; = 13550 kg/m3 P = g l = 13550 · 9.80665 · 0.725 · 10-3 = 96.34 kPa

2.16 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gun- powder is burned a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder (cannon) is pointing horizontally?

Solution: ma = F = P1 · A – P0 · A a = (P – P ) · A / m = ( 7000 – 101 ) [ ( 0.152 /4 )/5 ] = 24.38 m/s2 10 2.17 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction.

2.18 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg resting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston?

Solution: Force balance: F› = F? = PA = mpg + P A 0 P = P + mpg/A = 100 kPa + (100 · 9.80665) / (0.01 · 1000) 0 = 100 kPa + 98.07 = 198 kPa 2.19 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700 kg of a car?

Solution: F? = ma = mg = 740 · 9.80665 = 7256.9 N Force balance: F› = ( P – P0 ) A = F? => P = P0 + F? / A A = D2 (1 / 4) = 0.031416 m2 P = 101 + 7256.9 / (0.031416 · 1000) = 332 kPa 2.20 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel.

Solution: Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P = 1250 + 96 = 1346 kPa 0

2.21 The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure is 97 kPa. If a U-tube with mercury, density 13550 kg/m3, is attached to the tank to measure the vacuum, what column height difference would it show?

2.22 A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring and the outside atmospheric pressure of 100 kPa. The spring exerts no force on the piston when it is at the bottom of the cylinder and for the state shown, the pressure is 400 kPa with volume 0.4 L. The valve is opened to let some air in, causing the piston to rise 2 cm. Find the new pressure.

Solution: A linear spring has a force linear proportional to displacement. F = k x, so the equilibrium pressure then varies linearly with volume: P = a + bV, with an intersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume (V -> 0) when there is no spring force F = PA = PoA + mpg and the initial state. Piston area = A = ( /4) · 0.12 = 0.00785 m2 P

P P2 400 106.2 1 mpg 5Ê· Ê9.80665 a = P + = 100 + 0 Ap 0.00785 2 = 106.2 kPa intersect for zero volume.

V = 0.4 + 0.00785 · 20 = 0.557 L 2 dP P =P + V 2 1 dV V (400-106.2) = 400 + (0.557 – 0.4) 0 0.4 0.557 0.4Ê-Ê0 = 515.3 kPa

2.23 A U-tube manometer filled with water, density 1000 kg/m3, shows a height difference of 25 cm. What is the gauge pressure? If the right branch is tilted to make an angle of 30° with the horizontal, as shown in Fig. P2.23, what should the length of the column in the tilted tube be relative to the U-tube?

Solution: H 30° P = F/A = mg/A = V g/A = h g = 0.25 · 1000 · 9.807 = 2452.5 Pa h = 2.45 kPa

2.24 The difference in height between the columns of a manometer is 200 mm with a fluid of density 900 kg/m3. What is the pressure difference? What is the height difference if the same pressure difference is measured using mercury, density 13600 kg/ m3, as manometer fluid?

Solution: P = gh = 900 · 9.807 · 0.2 = 1765.26 Pa = 1.77 kPa 11 900 hhg = P/ ( hg g) = ( 1 gh1) / ( hg g) = · 0.2 = 0.0132 m= 13.2 mm 13600 2.25 Two reservoirs, A and B, open to the atmosphere, are connected with a mercury manometer. Reservoir A is moved up/down so the two top surfaces are level at h3 as shown in Fig. P2.25. Assuming that you know , Hg and measure the A 12 3 B Solution: Balance forces on each side: P + g(h – h ) + Hggh = P + g(h – h ) + Hggh 0A32 20B31 1 h Ê-Êh h Ê-Êh 32 21 = + Hg B A h Ê-Êh h Ê-Êh ? 3 1? ? 3 1? 2.26 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3, the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tall with diameter 4m. What is the total force from the bottom of each tank to the water and what is the pressure at the bottom of each tank?

2.27 The density of mercury changes approximately linearly with temperature as Hg = 13595 – 2.5 T kg/ m3 T in Celsius so the same pressure difference will result in a manometer reading that is influenced by temperature. If a pressure difference of 100 kPa is measured in the summer at 35°C and in the winter at -15°C, what is the difference in column height between the two measurements?

Solution: P = gh h = P/ g ; su = 13507.5 ; w = 13632.5 hsu = 100· 103/(13507.5 · 9.807) = 0.7549 m hw = 100· 103/(13632.5 · 9.807) = 0.7480 m h = hsu – hw = 0.0069 m = 6.9 mm 2.28 Liquid water with density is filled on top of a thin piston in a cylinder with cross- sectional area A and total height H. Air is let in under the piston so it pushes up, spilling the water over the edge. Deduce the formula for the air pressure as a function of the piston elevation from the bottom, h.

Solution: P0 h Force balance Piston: F› = F? P PA = P0A + mH2Og P = P0 + mH2Og /A HP 0

h, V P = P + (H-h) g air 0

2.29 A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of piston motion towards the gas. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure.

2.30 A piece of experimental apparatus is located where g = 9.5 m/s2 and the temperature is 5°C. An air flow inside the apparatus is determined by measuring the pressure drop across an orifice with a mercury manometer (see Problem 2.27 for density) showing a height difference of 200 mm. What is the pressure drop in kPa?

Solution: P = gh ; = 13600 Hg P = 13600 · 9.5 · 0.2 = 25840 Pa = 25.84 kPa 2.31 Repeat the previous problem if the flow inside the apparatus is liquid water, 1000 kg/m3, instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns.

Solution: Balance forces in the manometer: (H – h ) – (H – h ) = h = h – h P1 P 2 1 Hg 1 2 2 ··

H h2 P A + h gA + (H – h )gA 1 H2O 1 Hg 1 h 1 = P A + h gA + (H – h )gA 2 H2O 2 Hg 2 P – P = (h – h )g + (h – h )g 1 2 H2O 2 1 Hg 1 2 P – P = h g – h g = 13600 · 0.2 · 9.5 – 1000 · 0.2 · 9.5 1 2 Hg Hg H2O Hg = 25840 – 1900 = 23940 Pa = 23.94 kPa 2.32 Two piston/cylinder arrangements, A and B, have their gas chambers connected by a pipe. Cross-sectional areas are A = 75 cm2 and A = 25 cm2 with the piston AB mass in A being mA = 25 kg. Outside pressure is 100 kPa and standard gravitation. Find the mass mB so that none of the pistons have to rest on the bottom.

2.33 Two hydraulic piston/cylinders are of same size and setup as in Problem 2.32, but with neglible piston masses. A single point force of 250 N presses down on piston A. Find the needed extra force on piston B so that none of the pistons have to move.

Solution: No motion in connecting pipe: P = P & Forces on pistons balance AB A = 75 cm2 ; A = 25 cm2 AB P = P0 + F / A = P = P0 + F / A A AAB BB F = F A / A = 250 · 25 / 75 = 83.33 N BABA 2.34 At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the ocean and you later climb a hill up to 250 m elevation. Assume the density of water is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do you feel at each place?

Solution: P = gh Pocean 0 P = 1025 · 100 + 1000 · 9.81 · 15 =P + = 2.4965 · 105 Pa = 250 kPa Phill = P – P = 1025 · 100 – 1.18 · 9.81 · 250 0 = 0.99606 · 105 Pa = 99.61 kPa

2.35 In the city water tower, water is pumped up to a level 25 m above ground in a pressurized tank with air at 125 kPa over the water surface. This is illustrated in Fig. P2.35. Assuming the water density is 1000 kg/m3 and standard gravity, find the pressure required to pump more water in at ground level.

2.36 Two cylinders are connected by a piston as shown in Fig. P2.36. Cylinder A is used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and there is standard gravity. What is the gas pressure in cylinder B?

Solution: Force balance for the piston: P A + mpg + P (A – A ) = P A BB 0ABAA A = ( /4)0.12 = 0.00785 m2; A = ( /4)0.0252 = 0.000491 m2 AB P A = P A – mpg – P (A – A ) = 500· 0.00785 – (25 · 9.807/1000) BBAA 0AB – 100 (0.00785 – 0.000491) = 2.944 kN P = 2.944/0.000491 = 5996 kPa = 6.0 MPa B

2.37 Two cylinders are filled with liquid water, = 1000 kg/m3, and connected by a line with a closed valve. A has 100 kg and B has 500 kg of water, their cross-sectional areas are A = 0.1 m2 and A = 0.25 m2 and the height h is 1 m. Find the pressure AB on each side of the valve. The valve is opened and water flows to an equilibrium. Find the final pressure at the valve location.

Solution: A = vH2OmA A A A = 1 m V = m / = 0.1 = A h => h A

B = vH2OmB B B B B = 2 m V = m / = 0.5 = A h => h P = P + g(h +H) = 101325 + 1000 · 9.81 · 3 = 130 755 Pa VB 0 B P = P + gh = 101325 + 1000 · 9.81 · 1 = 111 135 Pa VA 0 A Equilibrium: same height over valve in both h A Ê+Ê(h +H)A AA B B Vtot = V + V = h A + (h – H)A h = = 2.43 m A B 2 A 2 B 2 A Ê+ÊA AB P = P + gh = 101.325 + (1000 · 9.81 · 2.43)/1000 = 125.2 kPa V2 0 2 2.38 Using the freezing and boiling point temperatures for water in both Celsius and Fahrenheit scales, develop a conversion formula between the scales. Find the conversion formula between Kelvin and Rankine temperature scales.

Solution: TFreezing = 0 oC = 32 F; TBoiling = 100 oC = 212 F T = 100 oC = 180 F ToC = (TF – 32)/1.8 or TF = 1.8 ToC + 32 For the absolute K & R scales both are zero at absolute zero.

English Unit Problems 2.39E A 2500-lbm car moving at 15 mi/h is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/h. What are the force and total time required?

Solution: dV a= = V/ t t= V/a dt t = (50 -15) · 1609.34 · 3.28084/(3600 · 15) = 3.42 sec F = ma = 2500 · 15 / 32.174 lbf= 1165 lbf 2.40E Two pound moles of diatomic oxygen gas are enclosed in a 20-lbm steel container. A force of 2000 lbf now accelerates this system. What is the acceleration?

Solution: mO2 = nO2MO2 = 2 · 32 = 64 lbm mtot = mO2 + msteel = 64 + 20 = 84 lbm a = Fgc = (2000 · 32.174) / 84 = 766 ft/s2 mtot 2.41E A bucket of concrete of total mass 400 lbm is raised by a crane with an acceleration of 6 ft/s2 relative to the ground at a location where the local gravitational acceleration is 31 ft/s2. Find the required force.

Solution: F = ma = Fup – mg Fup = ma + mg = 400 · ( 6 + 31 ) / 32.174 = 460 lbf 2.42E One pound-mass of diatomic oxygen (O molecular weight 32) is contained in a 2 100-gal tank. Find the specific volume on both a mass and mole basis (v and v ).

2.43E A 30-lbm steel gas tank holds 10 ft3 of liquid gasoline, having a density of 50 lbm/ft3. What force is needed to accelerate this combined system at a rate of 15 ft/s2?

Solution: m = mtank + mgasoline = 30 + 10 · 50 = 530 lbm ma F = = (530 · 15) / 32.174 = 247.1 lbf g C

2.44E A differential pressure gauge mounted on a vessel shows 185 lbf/in.2 and a local barometer gives atmospheric pressure as 0.96 atm. Find the absolute pressure inside the vessel.

Solution: P = Pgauge + P0 = 185 + 0.96 · 14.696 = 199.1 lbf/in2

3 2.45E A U-tube manometer filled with water, density 62.3 lbm/ft , shows a height difference of 10 in. What is the gauge pressure? If the right branch is tilted to make an angle of 30° with the horizontal, as shown in Fig. P2.23, what should the length of the column in the tilted tube be relative to the U-tube?

Solution: H 30° P = F/A = mg/Ag = h g/g CC = [(10/12) · 62.3 · 32.174] / 32.174 ·144 h = Pgauge = 0.36 lbf/in2

h = H · sin 30° H = h/sin 30° = 2h = 20 in = 0.833 ft 2 2.46E A piston/cylinder with cross-sectional area of 0.1 ft has a piston mass of 200 lbm resting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressure of 1 atm, what should the water pressure be to lift the piston?

2.47E The density of mercury changes approximately linearly with temperature as 3 = 851.5 – 0.086 T lbm/ft T in degrees Fahrenheit Hg so the same pressure difference will result in a manometer reading that is 2 influenced by temperature. If a pressure difference of 14.7 lbf/in. is measured in the summer at 95 F and in the winter at 5 F, what is the difference in column height between the two measurements?

Solution: P = gh/g h = Pg / g cc 33 = 843.33 lbm/ft ; = 851.07 lbm/ft su w 14.7°·°144·°°32.174 h = = 2.51 ft = 30.12 in su 843.33°·°32.174 14.7°·°144·°°32.174 h = = 2.487 ft = 29.84 in w 851.07°·°32.174 h = h – h = 0.023 ft = 0.28 in su w

2 2.48E A piston, m = 10 lbm, is fitted in a cylinder, A = 2.5 in. , that contains a gas. The p2 setup is in a centrifuge that creates an acceleration of 75 ft/s . Assuming standard atmospheric pressure outside the cylinder, find the gas pressure.

Solution: P 0 F? = F› = P A + m g = PA 0p 10°·°75 g P = P + m g/Ag = 14.696 + 0 p c 2.5°·°32.174 2 = 14.696 + 9.324 = 24.02 lbf/in

2.49E At the beach, atmospheric pressure is 1025 mbar. You dive 30 ft down in the ocean and you later climb a hill up to 300 ft elevation. Assume the density of water is about 62.3 lbm/ft3 and the density of air is 0.0735 lbm/ft3. What pressure do you feel at each place?

CHAPTER 3 The SI set of problems are revised from the 4th edition as: New Old New Old New Old 1 new 21 new 41 33 2 new 22 13 mod 42 34 3 new 23 16 mod 43 35 4 new 24 17 44 36 mod 5 new 25 new 45 37 mod 6 new 26 18 mod 46 38 mod 7 7 mod 27 19 d.mod 47 39 8 3 28 20 e.mod 48 40 9 2 29 21 a.b.mod 49 41 10 4 30 22 b.mod 50 42 mod 11 5 31 23 51 43 12 new 32 24 52 44 13 6 33 26 53 45 14 8 mod 34 27 mod 54 46 15 10 35 14 55 47 16 11 36 28 56 48 17 12 37 29 mod 57 49 18 new 38 30 mod 58 50 19 15 mod 39 31 59 51 20 new 40 32 mod 60 52 The english unit problem set is revised from the 4th edition as: New Old New Old New Old 61 new 69 61 77 69 62 53 70 62 mod 78 70 63 55 71 63 79 71 64 56 72 64 80 72 65 new 73 65 81 new 66 new 74 66 82 74 67 59 mod 75 67 83 75 mod 68 60 76 68 mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem.

3.1 Water at 27°C can exist in different phases dependent upon the pressure. Give the approximate pressure range in kPa for water being in each one of the three phases vapor, liquid or solid.

Solution: The phases can be seen in Fig. 3.6, a sketch ln P T =27 °C = 300 L CR.P.

From Fig. 3.6: S V PVL » 4 · 10 -3 MPa = 4 kPa, PLS = 103 MPa T P < 4 kPa VAPOR P > 1000 MPa SOLID(ICE) 0.004 MPa < P < 1000 MPa LIQUID

3.2 Find the lowest temperature at which it is possible to have water in the liquid phase. At what pressure must the liquid exist?

Solution: ln P There is no liquid at lower temperatures than on the fusion line, see Fig. 3.6, saturated ice III to liquid phase boundary is at T » 263K » – 10°C and P » 2100 MPa S lowest T liquid L CR.P.

V T 3.3 If density of ice is 920 kg/m3, find the pressure at the bottom of a 1000 m thick ice cap on the north pole. What is the melting temperature at that pressure?

3.4 A substance is at 2 MPa, 17°C in a rigid tank. Using only the critical properties can the phase of the mass be determined if the substance is nitrogen, water or propane ?

Solution: Find state relative to critical point properties which are: a) Nitrogen N2 : 3.39 MPa 126.2 K b) Water H2O : 22.12 MPa 647.3 K c) Propane C3H8 : 4.25 MPa 369.8 K Ê State is at 17 °C = 290 K and 2 MPa < Pc ln P for all cases: N2 : T >> Tc Superheated vapor P < Pc H2O : T << Tc ; P << Pc C3H8 : T < Tc ; P < Pc you cannot say c a b Vapor T

3.5 A cylinder fitted with a frictionless piston contains butane at 25°C, 500 kPa. Can the butane reasonably be assumed to behave as an ideal gas at this state ?

Solution Butane 25°C, 500 kPa, Table A.2: Tc = 425 K; Pc = 3.8 MPa Tr = ( 25 + 273 ) / 425 = 0.701; Pr = 0.5/3.8 = 0.13 Look at generalized chart in Figure D.1 Actual Pr > Pr, sat => liquid!! not a gas 3.6 A 1-m3 tank is filled with a gas at room temperature 20°C and pressure 100 kPa. How much mass is there if the gas is a) air, b) neon or c) propane ?

Solution: Table A.2 T= 20 °C = 293.15 K ; P = 100 kPa << Pc for all Air : T >> TC,N2; TC,O2 = 154.6 K so ideal gas; R= 0.287 Neon : T >> Tc = 44.4 K so ideal gas; R = 0.41195 Propane: T < Tc = 370 K, but P << Pc = 4.25 MPa so gas R = 0.18855

3.7 A cylinder has a thick piston initially held by a pin as shown in Fig. P3.7. The cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K. The metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101 kPa. The pin is now removed, allowing the piston to move and after a while the gas returns to ambient temperature. Is the piston against the stops?

Solution: Piston mp = Ap · l · piston = 8000 kg/m3 mpg ApÊ· Ê0.1Ê· Ê9.807Ê· Ê8000 PextÊonÊCO = P + = 101 + = 108.8 kPa 2 0 Ap ApÊ· Ê1000 Pin released, as P > Pfloat piston moves up, T = To & if piston at stops, 12 then V = V · 150 / 100 21 100 P = P · V / V = 200 · = 133 kPa > Pext 2 1 1 2 150 piston is at stops, and P = 133 kPa 2

3.8 A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 25°C. To what pressure should it be charged if there Solution: Assume CO is an ideal gas table A.5: P = mRT/V 2

3.9 A 1-m3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in Fig. P3.9. The valve is opened and air flows into the tank until the pressure reaches 5 MPa, at which point the valve is closed and the temperature inside is 450K. b. The tank eventually cools to room temperature, 300 K. What is the pressure inside the tank then?

Solution: P V 1000Ê· Ê1 1 m = = = 8.711 kg air1 RT 0.287Ê· Ê400 1 P V 5000Ê· Ê1 2 m = = = 38.715 kg air2 RT 0.287Ê· Ê450 2 Process 2 ? 3 is constant V, constant mass cooling to T 3 P = P · (T /T ) = 5000 · (300/450) = 3.33 MPa 3232

3.10 A hollow metal sphere of 150-mm inside diameter is weighed on a precision beam balance when evacuated and again after being filled to 875 kPa with an unknown gas. The difference in mass is 0.0025 kg, and the temperature is 25°C. What is the gas, assuming it is a pure substance listed in Table A.5 ?

3.11 A piston/cylinder arrangement, shown in Fig. P3.11, contains air at 250 kPa, 300°C. The 50-kg piston has a diameter of 0.1 m and initially pushes against the stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools as heat is b. How far has the piston dropped when the temperature reaches ambient? Solution: Piston Ap = 4 · 0.12 = 0.00785 m2 Balance forces when piston floats: mpg 50Ê· Ê9.807 Pfloat = Po + Ap = 100 + 0.00785Ê· Ê1000 P 2 = 162.5 kPa = P = P 23 To find temperature at 2 assume ideal gas: P2 162.5 T = T · = 573.15 · = 372.5 K 2 1 P 250 1 P 1 3 2 V Vstop

b) Process 2 -> 3 is constant pressure as piston floats to T3 = To = 293.15 K V = V = Ap · H = 0.00785 · 0.25 = 0.00196 m3 = 1.96 L 21 T3 293.15 Ideal gas and P = P => V = V · = 1.96 · = 1.54 L 2 3 3 2 T 372.5 2 H = (V2 -V3)/A = (1.96-1.54) · 0.001/0.00785 = 0.053 m = 5.3 cm

3.13 A vacuum pump is used to evacuate a chamber where some specimens are dried at 50°C. The pump rate of volume displacement is 0.5 m3/s with an inlet pressure of 0.1 kPa and temperature 50°C. How much water vapor has been removed over a 30- min period?

Solution: Use ideal gas P << lowest P in steam tables. R is from table A.5 m = m t with mass flow rate as: m= V/v = PV/RT (ideal gas) . 0.1Ê· Ê0.5Ê· Ê30· 60 m = PV t/RT = = 0.603 kg (0.46152Ê· Ê323.15) 3.14 An initially deflated and flat balloon is connected by a valve to a 12 m3 storage tank containing helium gas at 2 MPa and ambient temperature, 20°C. The valve is opened and the balloon is inflated at constant pressure, Po = 100 kPa, equal to ambient pressure, until it becomes spherical at D = 1 m. If the balloon is larger 1 than this, the balloon material is stretched giving a pressure inside as D1 D1 P = P + C 1Ê-Ê 0 ? D?D The balloon is inflated to a final diameter of 4 m, at which point the pressure inside is 400 kPa. The temperature remains constant at 20°C. What is the maximum pressure inside the balloon at any time during this inflation process? What is the pressure inside the helium storage tank at this time?

3.15 The helium balloon described in Problem 3.14 is released into the atmosphere and rises to an elevation of 5000 m, with a local ambient pressure of Po = 50 kPa and temperature of -20°C. What is then the diameter of the balloon?

Solution: Balloon of Problem 3.14, where now after filling D = 4 m, we have : m1 = P1V1/RT1 = 400 ( /6) 43 /2.077 · 293.15 = 22.015 kg P1 = 400 = 100 + C(1 – 0.25)0.25 => C = 1600 For final state we have : P0 = 50 kPa, T2 = T0 = -20°C = 253.15 K State 2: T2 and on process line for balloon, i.e. the P-V relation: P = 50 + 1600 ( D*Ê-1 – D*Ê-2 ), D* = D/D ; V = ( /6) D3 1 P2V2 = m R T2 = 22.015 · 2.077 · 253.15 = 11575 or PD+3 = 11575 · 6/ = 22107 substitute P into the P-V relation 22107 D*Ê-3 = 50 + 1600 ( D*Ê-1 – D*Ê-2 ) Divide by 1600 13.8169 D*Ê-3 – 0.03125 – D*Ê-1 + D*Ê-2 = 0 Multiply by D*3 13.8169 – 0.03125 D*Ê3 – D*ÊÊ2 + D*ÊÊ1 = 0 Qubic equation. By trial and error D* = 3.98 so D = D*D = 3.98 m 1

3.16 A cylinder is fitted with a 10-cm-diameter piston that is restrained by a linear spring (force proportional to distance) as shown in Fig. P3.16. The spring force constant is 80 kN/m and the piston initially rests on the stops, with a cylinder volume of 1 L. The valve to the air line is opened and the piston begins to rise when the cylinder pressure is 150 kPa. When the valve is closed, the cylinder volume is 1.5 L and the Solution: Fs = ks x = ks V/Ap ; V1 = 1 L = 0.001 m3, Ap = 4 0.12 = 0.007854 m2 State 2: V = 1.5 L = 0.0015 m3; T = 80°C = 353.15 K 33 The pressure varies linearly with volume seen from a force balance as: PAp = P Ap + mp g + ks(V – V0)/Ap 0 Between the states 1 and 2 only volume varies so: P

ks(V3-V2) 80· 103(0.0015Ê-Ê0.001) P = P + = 150 + 3 2 Ap2 0.0078542Ê· Ê1000 = 798.5 kPa P3V3 798.5Ê· Ê0.0015 m = = = 0.012 kg RT3 0.287Ê· Ê353.15 3

3.17 Air in a tire is initially at -10°C, 190 kPa. After driving awhile, the temperature goes up to 10°C. Find the new pressure. You must make one assumption on your own. Solution: Assume constant volume and that air is an ideal gas P = P · T /T = 190 · 283.15/263.15 = 204.4 kPa 2121

3.18 A substance is at 2 MPa, 17°C in a 0.25-m3 rigid tank. Estimate the mass from the compressibility factor if the substance is a) air, b) butane or c) propane. Solution: Figure D.1 for compressibility Z and table A.2 for critical properties.

Nitrogen Pr = 2/3.39 = 0.59; Tr = 290/126.2 = 2.3; Z » 0.98 m = PV/ZRT = 2000 · 0.25/(0.98 · 0.2968 · 290) = 5.928 kg Butane Pr = 2/3.80 = 0.526; Tr = 290/425.2 = 0.682; Z » 0.085 m = PV/ZRT = 2000 · 0.25/(0.085 · 0.14304 · 290) = 141.8 kg Propane Pr = 2/4.25 = 0.47; Tr = 290/369.8 = 0.784; Z » 0.08 m = PV/ZRT = 2000 · 0.25/(0.08 · 0.18855 · 290) = 114.3 kg Za Tr= 2.0 Tr = 0.7 Tr = 0.7 cb 0.1 1 ln Pr

3.19 Argon is kept in a rigid 5 m3 tank at -30°C, 3 MPa. Determine the mass using the compressibility factor. What is the error (%) if the ideal gas model is used?

3.20 A bottle with a volume of 0.1 m3 contains butane with a quality of 75% and a temperature of 300 K. Estimate the total butane mass in the bottle using the Solution: m = V/v so find v given T1 and x as : v = v + x vfg f Tr = 300/425.2 = 0.705 => Fig. D.1 Zf » 0.02; Zg » 0.9 P = Psat = Prsat · Pc = 0.1· 3.80 · 1000 = 380 kPa vf = ZfRT/P = 0.02 · 0.14304 · 300/380 = 0.00226 m3/kg vg = ZgRT/P = 0.9 · 0.14304 · 300/380 = 0.1016 m3/kg v = 0.00226 + 0.75 · (0.1016 – 0.00226) = 0.076765 m3/kg m = 0.1/0.076765 = 1.303 kg

3.21 A mass of 2 kg of acetylene is in a 0.045 m3 rigid container at a pressure of 4.3 MPa. Use the generalized charts to estimate the temperature. (This becomes trial Solution: Table A.2, A.5: Pr = 4.3/6.14 = 0.70; Tc = 308.3; R= 0.3193 v = V/m = 0.045/2 = 0.0225 m3/kg State given by (P, v) v = ZRT/P Since Z is a function of the state Fig. D.1 and thus T, we have trial and error. Try sat. vapor at Pr = 0.7 => Fig. D.1: Zg = 0.59; Tr = 0.94 vg = 0.59 · 0.3193 · 0.94 · 308.3/4300 = 0.0127 too small Tr = 1 => Z = 0.7 => v = 0.7 · 0.3193 · 1 · 308.3/4300 = 0.016 Tr = 1.2 => Z = 0.86 => v = 0.86 · 0.3193 · 1.2 · 308.3/4300 = 0.0236 Interpolate to get: Tr » 1.17 T » 361 K

3.22 Is it reasonable to assume that at the given states the substance behaves as an ideal Solution: a) Oxygen, O at 30°C, 3 MPa Ideal Gas ( T » Tc = 155 K from A.2) 2

d) R-134a at 30°C, 3 MPa NO compressed liquid P > Psat (B.5.1) e) R-134a at 30°C, 100 kPa Ideal Gas P is low < Psat (B.5.1) ln P c, d e a, b Vapor T

3.23 Determine whether water at each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor.

Solution: All states start in table B.1.1 (if T given) or B.1.2 (if P given) a. 10 MPa, 0.003 m3/kg b. 1 MPa, 190°C : T > Tsat = 179.91oC so it is superheated vapor c. 200°C, 0.1 m3/kg: v < vg = 0.12736 m3/kg, so it is two-phase d. 10 kPa, 10°C : P > Pg = 1.2276 kPa so compressed liquid e. 130°C, 200 kPa: P < Pg = 270.1 kPa so superheated vapor f. 70°C, 1 m3/kg vf = 0.001023; vg = 5.042 m3/kg, so mixture of liquid and vapor PT C.P. C.P.

States shown are placed relative to the two-phase region, not to each other.

3.24 Determine whether refrigerant R-22 in each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor.

Solution: All cases are seen in Table B.4.1 a. 50°C, 0.05 m3/kg superheated vapor, v > vg =0.01167 at 50°C b. 1.0 MPa, 20°C compressed liquid, P > Pg = 909.9 kPa at 20°C c. 0.1 MPa, 0.1 m3/kg mixture liq. & vapor, vf < v < vg at 0.1 MPa d. 50°C, 0.3 m3/kg superheated vapor, v > vg = 0.01167 at 50°C e -20°C, 200 kPa superheated vapor, P < Pg = 244.8 kPa at -20°C f. 2 MPa, 0.012 m3/kg superheated vapor, v > vg = 0.01132 at 2 MPa P States shown are placed relative to the two-phase region, not to each other.

ba d cT e c ad b e v v f 3.25 Verify the accuracy of the ideal gas model when it is used to calculate specific volume for saturated water vapor as shown in Fig. 3.9. Do the calculation for 10 kPa and 1 MPa.

Solution: a. H O T = 275°C P = 5 MPa Table B.1.1 or B.1.2 2 Psat = 5.94 MPa => superheated vapor v = 0.04141 m3/kg b. H O T = -2°C P = 100 kPa Table B.1.5 2 Psat = 0.518 kPa =>compressed solid v vi = 0.0010904 m3/kg c. CO T = 267°C P = 0.5 MPa Table A.5 2 RT 0.18892Ê· Ê540 3 sup. vap. assume ideal gas v = = = 0.204 m /kg P 500 d. Air T = 20°C P = 200 kPa Table A.5 RT 0.287Ê· Ê293 3 sup. vap. assume ideal gas v = = = 0.420 m /kg P 200 e. NH T = 170°C P = 600 kPa Table B.2.2 3 T > Tc => sup. vap. v = (0.34699 + 0.36389)/2 = 0.3554m3/kg P C.P. T C.P. c, d, e aa States shown are c, d, e placed relative to the two-phase region, not T bv bv

e. NH3 T = 20°C P = 100 kPa => Table B.2.1 Psat = 847.5 kPa sup. vap. B.2.2 v = 1.4153 m3/kg P C.P. T C.P. d States shown are b P = const. a, c, e placed relative to the d two-phase region, not b a, c, e T to each other.

v v Solution: a. H2O T = 120°C v = 0.5 m3/kg < vg Table B.1.1 sat. liq. + vap. P = 198.5 kPa, x = (0.5 - 0.00106)/0.8908 = 0.56 b. H O P = 100 kPa v = 1.8 m3/kg Table B.1.2 v > vg 2 sup. vap., interpolate in Table B.1.3 1.8Ê-Ê1.694 T = (150 – 99.62) + 99.62 = 121.65 °C 1.93636Ê-Ê1.694 c. H O T = 263 K v = 200 m3/kg Table B.1.5 2 sat. solid + vap., P = 0.26 kPa, x = (200-0.001)/466.756 = 0.4285 d. Ne P = 750 kPa v = 0.2 m3/kg; Table A.5 Pv 750Ê· Ê0.2 ideal gas, T = = = 364.1 K R 0.41195 e. NH T = 20°C v = 0.1 m3/kg Table B.2.1 3 sat. liq. + vap. , P = 857.5 kPa, x = (0.1-0.00164)/0.14758 = 0.666 d b a, e States shown are placed relative to the two-phase region, not to each other.T

Solution: a. R-22 T = 10°C v = 0.01 m3/kg Table B.4.1 sat. liq. + vap. P = 680.7 kPa, x = (0.01-0.0008)/0.03391 = 0.2713 b. H2O T = 350°C v = 0.2 m3/kg Table B.1.1 v > vg sup. vap. P 1.40 MPa, x = undefined c. CO2 T = 800 K P = 200 kPa Table A.5 RT 0.18892Ê· Ê800 ideal gas v = = = 0.756 m3/kg P 200 d. N2 T = 200 K P = 100 kPa Table B.6.2 T > Tc sup. vap. v = 0.592 m3/kg e. CH T = 190 K x = 0.75 Table B.7.1 P = 4520 kPa 4 sat. liq + vap. v = 0.00497 + x · 0.003 = 0.00722 m3/kg

c, d b a, e States shown are placed relative to the two-phase region, not to each other.T

v T C.P. c, d a, e b v 3.32 Give the phase and the missing properties of P, T, v and x. These may be a little more difficult if the appendix tables are used instead of the software.

c) H2O T = 60°C, v = 0.001016 m3/kg: Table B.1.1 v < vf = 0.001017 => compr. liq. see Table B.1.4 v = 0.001015 at 5 MPa so P 0.5(5000 + 19.9) = 2.51 MPa d) NH3 T = 30°C, P = 60 kPa : Table B.2.1 P < Psat => sup. vapor interpolate in Table B.2.2 v = 2.94578 + (60-50)(1.95906-2.94578)/(75-50) = 2.551 m3/kg v is not linearly proportional to P (more like 1/P) so the computer table gives a more accurate value of 2.45 e) R-134a v = 0.005m3/kg , x = 0.5: sat. liq. + vap. Table B.5.1 v = (1-x) vf + x vg => vf + vg = 0.01 m3/kg An iterpolation gives: T 68.7°C, P = 2.06 MPa

a c b, e d States shown are placed relative to the two-phase region, not to each other.T

3.34 What is the percent error in pressure if the ideal gas model is used to represent the behavior of superheated vapor R-22 at 50°C, 0.03082 m3/kg? What if the generalized compressibility chart, Fig. D.1, is used instead (iterations needed)?

Solution: Real gas behavior: P = 900 kPa from Table B.4.2 _ Ideal gas constant: R = R/M = 8.31451/86.47 = 0.096155 P = RT/v = 0.096155 · (273.15 + 50) / 0.03082 = 1008 kPa which is 12% too high Generalized chart Fig D.1 and critical properties from A.2: Tr = 323.2/363.3 = 0.875; Pc = 4970 kPa Assume P = 900 kPa => Pr = 0.181 => Z 0.905 v = ZRT/P = 0.905 · 0.096155 · 323.15 / 900 = 0.03125 too high Assume P = 950 kPa => Pr = 0.191 => Z 0.9 v = ZRT/P = 0.9 · 0.096155 · 323.15 / 950 = 0.029473 too low 0.03082Ê-Ê0.029437 P 900 + ( 950 – 900 ) · = 938 kPa 4.2 % high 0.03125Ê-Ê0.029437 3.35 Determine the mass of methane gas stored in a 2 m3 tank at -30°C, 3 MPa. Estimate the percent error in the mass determination if the ideal gas model is used. Solution: The methane Table B.7.2 linear interpolation between 225 and 250 K.

3.37 A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C. The vessel is now heated. If a safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 200°C?

Solution: Process: v = V/m = constant v = 1/2 = 0.5 m3/kg 2-phase 1 200°C, 0.5 m3/kg seen in Table B.1.3 to be between 400 and 500 kPa so interpolate 0.5-0.53422 P 400 + · (500-400) Ê0.42492-0.53422 = 431.3 kPa T 500 kPa 400 kPa 100 C

3.39 A 400-m3 storage tank is being constructed to hold LNG, liquified natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 90% liquid and 10% vapor, by volume, at 100 kPa, what mass of LNG (kg) will the tank hold? What is the quality in the tank?

Solution: Vliq 0.9Ê· Ê400 Vvap 0.1Ê· Ê400 mliq = = = 152542 kg; mvap = = = 69.9 kg vf 0.00236 vg 0.5726 mtot = 152 612 kg, x = mvap / mtot = 4.58· 10-4 (If you use computer table, vf 0.002366, vg 0.5567)

3.40 A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5°C per hour due to a failure in the refrigeration system. How long time will it take before the methane becomes single phase and what is the pressure then?

Solution: Use Table B.7.1 Assume rigid tank v = const = v = 0.002439 + 0.25· 0.30367 = 0.078366 1

All single phase when v = vg => T 145 K t = /5° C (145 – 120 ) / 5 = 5 hours P = Psat= 824 kPa 3.41 Saturated liquid water at 60°C is put under pressure to decrease the volume by 1% keeping the temperature constant. To what pressure should it be compressed?

Solution: H O T = 60°C , x = 0.0; Table B.1.1 2 v = 0.99 · vfÊ(60°C) = 0.99· 0.001017 = 0.0010068 m3/kg Between 20 & 30 MPa in Table B.1.4, P 23.8 MPa 3.42 Saturated water vapor at 60°C has its pressure decreased to increase the volume by 10% keeping the temperature constant. To what pressure should it be Solution: From initial state: v = 1.10 · vg = 1.1 · 7.6707 = 8.4378 m3/kg Interpolate at 60°C between saturated (P = 19.94 kPa) and superheated vapor P = 10 kPa in Tables B.1.1 and B.1.3 P 19.941 + (8.4378 – 7.6707)(10-19.941)/(15.3345-7.6707) = 18.9 kPa Comment: T,v P = 18 kPa (software) v is not linear in P, more like 1/P, so the linear interpolation in P is not very accurate.

3.43 A boiler feed pump delivers 0.05 m3/s of water at 240°C, 20 MPa. What is the mass flowrate (kg/s)? What would be the percent error if the properties of saturated liquid at 240°C were used in the calculation? What if the properties of saturated liquid at Solution: At 240°C, 20 MPa: v = 0.001205 m3/kg (from B.1.4) m = V/v = 0.05/0.001205 = 41.5 kg/s vfÊ(240°C) = 0.001229 m = 40.68 kg/s error 2% vfÊ(20ÊMPa) = 0.002036 m = 24.56 kg/s error 41% 3.44 A glass jar is filled with saturated water at 500 kPa, quality 25%, and a tight lid is put on. Now it is cooled to -10°C. What is the mass fraction of solid at this Solution: Constant volume and mass v = v From Table B.1.2 and B.1.5: 12 v = 0.001093 + 0.25 · 0.3738 = 0.094543 = v = 0.0010891 + x · 446.756 1 22 x = 0.0002 mass fraction vapor 2

xsolid =1-x = 0.9998 or 99.98 % 2 3.45 A cylinder/piston arrangement contains water at 105°C, 85% quality with a volume of 1 L. The system is heated, causing the piston to rise and encounter a linear spring as shown in Fig. P3.45. At this point the volume is 1.5 L, piston diameter is 150 mm, and the spring constant is 100 N/mm. The heating continues, so the piston compresses the spring. What is the cylinder temperature when the Solution: P = 120.8 kPa, v = vf + x vfg = 0.001047 + 0.85*1.41831 = 1.20661 11 m = V / v = 0.001 = 8.288· 10-4 kg 1 1 1.20661 200 v = v (V / V ) = 1.20661· 1.5 = 1.8099 2121 & P = P = 120.8 kPa ( T = 203.5°C ) 12 P = P + (ks/Ap2) m(v3-v2) linear spring P

3.46 Saturated (liquid + vapor) ammonia at 60°C is contained in a rigid steel tank. It is used in an experiment, where it should pass through the critical point when the system is heated. What should the initial mass fraction of liquid be?

Solution: Process: Constant mass and volume, v = C From table B.2.1: v = v = 0.004255 = 0.001834 + x · 0.04697 12 1 => x = 0.01515 1 liquid = 1 – x = 0.948 1 T Crit. point

60 C 1 v 3.47 For a certain experiment, R-22 vapor is contained in a sealed glass tube at 20°C. It is desired to know the pressure at this condition, but there is no means of measuring it, since the tube is sealed. However, if the tube is cooled to -20°C small droplets of liquid are observed on the glass walls. What is the initial pressure?

Solution: R-22 fixed volume (V) & mass (m) at 20°C cool to -20°C ~ sat. vapor

T v = const = v = 0.092843 m3/kg P gÊatÊ-20°C 1 T -20oC 2 T 20 oC State 1: 20°C, 0.092843 m3/kg 1

3.48 A steel tank contains 6 kg of propane (liquid + vapor) at 20°C with a volume of 0.015 m3. The tank is now slowly heated. Will the liquid level inside eventually rise to the top or drop to the bottom of the tank? What if the initial mass is 1 kg instead of 6 kg?

Solution: Constant volume and mass v = v = V/m = 0.0025 m3/kg 21

C.P. vc = 0.203/44.094 = 0.004604 > v1 T

level rises to top v1 = 0.015 > vc If m = 1 kg 20°C level falls v V c

3.49 A cylinder containing ammonia is fitted with a piston restrained by an external force that is proportional to cylinder volume squared. Initial conditions are 10°C, 90% quality and a volume of 5 L. A valve on the cylinder is opened and additional ammonia flows into the cylinder until the mass inside has doubled. If at this point the pressure is 1.2 MPa, what is the final temperature?

3.52 Ammonia in a piston/cylinder arrangement is at 700 kPa, 80°C. It is now cooled at constant pressure to saturated vapor (state 2) at which point the piston is locked with a pin. The cooling continues to -10°C (state 3). Show the processes 1 to 2 and 2 to 3 Solution: PT

2 80 700 1 14 290 3 -10 v 1

2 3 v 3.53 A piston/cylinder arrangement is loaded with a linear spring and the outside 3 atmosphere. It contains water at 5 MPa, 400°C with the volume being 0.1 m . If the piston is at the bottom, the spring exerts a force such that P = 200 kPa. The lift system now cools until the pressure reaches 1200 kPa. Find the mass of water, the 22

5000 1200 200 P 1: Table B.1.3 v = 0.05781 1 1 m = V/v = 0.1/0.05781 = 1.73 kg 1

3.54 Water in a piston/cylinder is at 90°C, 100 kPa, and the piston loading is such that pressure is proportional to volume, P = CV. Heat is now added until the temperature reaches 200°C. Find the final pressure and also the quality if in the two-phase region. Solution: Final state: 200°C , on process line P = CV P 2 State 1: Table B.1.1: v1 = 0.001036 m3/kg P = P v /v from process equation 2 121 Check state 2 in Table B.1.1 1 v vg(T2) = 0.12736; Pg(T2) = 1.5538 MPa If v = vg(T ) P = 12.3 MPa > Pg not OK 222 If sat. P = Pg(T ) = 1553.8 kPa v = 0.0161 m3kg < vg sat. OK, 22 2 P = 1553.8 kPa, x = (0.0161 - 0.001156) / 0.1262 = 0.118 22 3.55 A spring-loaded piston/cylinder contains water at 500°C, 3 MPa. The setup is such that pressure is proportional to volume, P = CV. It is now cooled until the water becomes saturated vapor. Sketch the P-v diagram and find the final Solution: P = Cv C = P /v = 3000/0.11619 = 25820 11 P State 2: x = 1 & P = Cv (on process line) 1 222 Trial & error on T2sat or P2sat: 2 at 2 MPa vg = 0.09963 C = 20074 2.5 MPa vg = 0.07998 C = 31258 2.25 MPa vg = 0.08875 C = 25352 v Interpolate to get right C P = 2270 kPa 2

3 3.57 A sealed rigid vessel of 2 m contains a saturated mixture of liquid and vapor R- 134a at 10°C. If it is heated to 50°C, the liquid phase disappears. Find the pressure at 50°C and the initial mass of the liquid.

Solution: P State 2 is saturated vapor, from table B.5.1 P = Psat(50°C) = 1.318 MPa 2 2 State 1: same specific volume as state 2 v = v = 0.015124 m3/kg 12 1 v = 0.000794 + x · 0.048658 11 v x = 0.2945 1

m = V/v = 2/0.015124 = 132.24 kg; mliq = (1 – x )m = 93.295 kg 11 3.58 Two tanks are connected as shown in Fig. P3.58, both containing water. Tank A is at 33 200 kPa, v = 0.5 m /kg, V = 1 m and tank B contains 3.5 kg at 0.5 MPa, 400°C. A The valve is now opened and the two come to a uniform state. Find the final specific volume.

3.59 A tank contains 2 kg of nitrogen at 100 K with a quality of 50%. Through a volume flowmeter and valve, 0.5 kg is now removed while the temperature remains constant. Find the final state inside the tank and the volume of nitrogen removed if the valve/meter is located at a. The top of the tank b. The bottom of the tank Solution m = m – 0.5 = 1.5 kg 21 v = 0.001452 + x · 0.029764 = 0.016334 11 Vtank = m1v1 = 0.0327 m3 = Vtank/m2 = 0.0218 < vg v (T) 2 0.0218-0.001452 x = = 0.6836 2 0.031216-0.001452 Top: flow out is sat. vap. v = 0.031216 g

V = m v = 0.0156 m3 out out g Bottom: flow out is sat. liq. v = 0.001452 f

V = m v = 0.000726 m3 out out f 3.60 Consider two tanks, A and B, connected by a valve, as shown in Fig. P3.60. Each has a volume of 200 L and tank A has R-12 at 25°C, 10% liquid and 90% vapor by volume, while tank B is evacuated. The valve is now opened and saturated vapor flows from A to B until the pressure in B has reached that in A, at which point the valve is closed. This process occurs slowly such that all temperatures stay at 25°C throughout the process. How much has the quality changed in tank A during the process?

English Unit Problems 3.61E A substance is at 300 lbf/in.2, 65 F in a rigid tank. Using only the critical properties can the phase of the mass be determined if the substance is nitrogen, water or propane?

Solution: Find state relative to the critical point properties, table C.1 Nitrogen 492 lbf/in.2 227.2 R Water 3208 lbf/in.2 1165.1 R Propane 616 lbf/in.2 665.6 R P < Pc for all and T = 65 F =65 + 459.67 = 525 R N T >> Tc Yes gas and P < Pc 2 H O T << Tc P << Pc so you cannot say 2 C3H8 T < Tc P < Pc you cannot say 3.62E A cylindrical gas tank 3 ft long, inside diameter of 8 in., is evacuated and then filled with carbon dioxide gas at 77 F. To what pressure should it be charged if there should be 2.6 lbm of carbon dioxide?

Solution: Assume CO is an ideal gas table C.4: P = mRT/V 2

Vcyl = A · L = 4 (8)2 · 3 · 12 = 1809.6 in3 2.6Ê· Ê35.1Ê· Ê(77Ê+Ê459.67)Ê· Ê12 P = = 324.8 lbf/in2 1809.6 3.63E A vacuum pump is used to evacuate a chamber where some specimens are dried at 120 F. The pump rate of volume displacement is 900 ft3/min with an inlet pressure of 1 mm Hg and temperature 120 F. How much water vapor has been removed over a 30-min period?

3.64E A cylinder is fitted with a 4-in.-diameter piston that is restrained by a linear spring (force proportional to distance) as shown in Fig. P3.16. The spring force constant is 400 lbf/in. and the piston initially rests on the stops, with a cylinder volume of 60 in.3. The valve to the air line is opened and the piston begins to rise when the cylinder pressure is 22 lbf/in.2. When the valve is closed, the cylinder volume is 90 in.3 and the temperature is 180 F. What mass of air is inside the cylinder?

Solution: V = V = 60 in3; Ap = · 42 = 12.566 in2 12 4 P = 22 lbf/in2 ; V = 90 in3 , T = 180°F = 639.7 R P 233 3 ks(V3-V2) Linear spring: P = P + 3 2 Ap2 2 = 22 + 400 (90-60) = 98 lbf/in2 1 12.5662 v

P3V3 98Ê· Ê90 m = = = 0.02154 lbm RT3 12Ê· Ê53.34Ê· Ê639.7 3.65E A substance is at 70 F, 300 lbf/in.2 in a 10 ft3 tank. Estimate the mass from the compressibility chart if the substance is a) air, b) butane or c) propane. Solution: Use Fig. D.1 for compressibility Z and table C.1 for critical properties m =PV/ZRT = 300 · 144 · 10 / 530 Z R = 815.09 / Z R = 815.09 / Z R Air use nitrogen 492 lbf/in.2; 227.2 R Pr = 0.61; Tr = 2.33; Z = 0.98 m =PV/ZRT = 815.09 / Z R = 815.09/(0.98· 55.15) = 15.08 lbm Butane 551 lbf/in.2; 765.4 R Pr = 0.544; Tr = 0.692; Z = 0.09 m =PV/ZRT = 815.09 / Z R = 815.09 /(0.09· 26.58) = 340.7 lbm Propane 616 lbf/in.2; 665.6 R Pr = 0.487; Tr = 0.796; Z = 0.08 m =PV/ZRT = 815.09 / Z R = 815.09 / (0.08· 35.04) = 290.8 lbm 3.66E Determine the mass of an ethane gas stored in a 25 ft3 tank at 250 F, 440 lbf/in.2 using the compressibility chart. Estimate the error (%) if the ideal gas model is used.

3.67E Argon is kept in a rigid 100 ft3 tank at -30 F, 450 lbf/in.2. Determine the mass using the compressibility factor. What is the error (%) if the ideal gas model is used?

Tr = ( 460 – 30 ) / 271.4 = 1.58, Pr = 450/706 = 0.64 Z = 0.95 m = PV/ZRT = 450 · 144 · 100 / (0.95 · 38.68 · 430) = 410 lbm Ideal gas Z = 1 m = PV/RT = 390 lbm 5% error 3.68E Determine whether water at each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor.

3.72E What is the percent error in specific volume if the ideal gas model is used to represent the behavior of superheated ammonia at 100 F, 80 lbf/in.2? What if the generalized compressibility chart, Fig. D.1, is used instead?

Solution: Ammonia Table C.9.2: v = 4.186 ft3/lbm Ideal gas v = RT = 90.72Ê· Ê559.7 = 4.4076 ft3/lbm 5.3% error P 80Ê· Ê144 Generalized compressibility chart and Table C.1 Tr = 559.7/729.9 = 0.767, Pr = 80/1646 = 0.0486 => Z 0.96 v = ZRT/P = 0.96 · 4.4076 = 4.231 ft3/lbm 1.0% error 3.73E A water storage tank contains liquid and vapor in equilibrium at 220 F. The distance from the bottom of the tank to the liquid level is 25 ft. What is the absolute pressure at the bottom of the tank?

Solution: Table C.8.1: vf = 0.01677 ft3/lbm gÊl 32.174Ê· Ê25 2 P = = = 10.35 lbf/in gcvf 32.174Ê· Ê0.01677Ê· Ê144 3.74E A sealed rigid vessel has volume of 35 ft3 and contains 2 lbm of water at 200 F. The vessel is now heated. If a safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 400 F?

Solution: Process: v = V/m = constant = v = 17.5 ft3/lbm 1 Table C.8.2: 400 F, 17.5 ft3/lbm between 20 & 40 lbf/in2 P 32.4 lbf/in2 (28.97 by software)

3.75E Saturated liquid water at 200 F is put under pressure to decrease the volume by 1%, keeping the temperature constant. To what pressure should it be compressed?

3.76E Saturated water vapor at 200 F has its pressure decreased to increase the volume by 10%, keeping the temperature constant. To what pressure should it be expanded?

Solution: v = 1.1 · vg = 1.1 · 33.63 = 36.993 ft3/lbm Interpolate between sat. at 200 F and sup. vapor in Table C.8.2 at 200 F, 10 lbf/in2 P 10.54 lbf/in2 3.77E A boiler feed pump delivers 100 ft3/min of water at 400 F, 3000 lbf/in.2. What is the mass flowrate (lbm/s)? What would be the percent error if the properties of saturated liquid at 400 F were used in the calculation? What if the properties of 2 Solution: Table C.8.4: v = 0.0183 ft3/lbm . V 100 m = = = 91.07 lbm/s v 60Ê· Ê0.018334 vfÊ(400ÊF) = 0.01864 m = 89.41 error 1.8% vfÊ(3000Êlbf/in2) = 0.03475 m = 47.96 error 47%

3.79E A steel tank contains 14 lbm of propane (liquid + vapor) at 70 F with a volume of 0.25 ft3. The tank is now slowly heated. Will the liquid level inside eventually rise to the top or drop to the bottom of the tank? What if the initial mass is 2 lbm Solution: P Constant volume and mass v = v = V/m = 0.25/14 = 0.01786 21 vc = 3.2/44.097 = 0.07256 ft3/lbm v < vc so eventually sat. liquid 2 level rises If v = v = 0.25/2 = 0.125 > vc 21 v Now sat. vap. is reached so level drops 3.80E A pressure cooker (closed tank) contains water at 200 F with the liquid volume being 1/10 of the vapor volume. It is heated until the pressure reaches 300 lbf/in.2. Find the final temperature. Has the final state more or less vapor than the initial state?

3.81E Two tanks are connected together as shown in Fig. P3.58, both containing water. Tank A is at 30 lbf/in.2, v = 8 ft3/lbm, V = 40 ft3 and tank B contains 8 lbm at 80 lbf/in. 2, 750 F. The valve is now opened and the two come to a uniform state. Find the final specific volume.

Solution: m = V /v = 40/8 = 5 lbm A AA Table C.8.2: v = (8.561 + 9.322)/2 = 8.9415 B V = m v = 8 · 8.9415 = 71.532 ft3 B BB Final state: mtot = m + m = 5 + 8 = 13 lbm AB Vtot = V + V = 111.532 ft3 AB v2 = Vtot/mtot = 111.532/13 = 8.579 ft3/lbm

3.82E A spring-loaded piston/cylinder contains water at 900 F, 450 lbf/in.2. The setup is such that pressure is proportional to volume, P = CV. It is now cooled until the water becomes saturated vapor. Find the final pressure.

3.83E Refrigerant-22 in a piston/cylinder arrangement is initially at 120 F, x = 1. It is then expanded in a process so that P = Cv-1 to a pressure of 30 lbf/in.2. Find the final temperature and specific volume.

4.1 A piston of mass 2 kg is lowered 0.5 m in the standard gravitational field. Find the required force and work involved in the process.

Solution: F = ma = 2 · 9.80665 = 19.61 N W = F dx = F dx = F x = 19.61 · 0.5 = 9.805 J

4.2 An escalator raises a 100 kg bucket of sand 10 m in 1 minute. Determine the total amount of work done and the instantaneous rate of work during the process.

Solution: W = F dx = F dx = F x = 100 · 9.80665 · 10 = 9807 J W = W / t = 9807 / 60 = 163 W

4.3 A linear spring, F = k (x – x ), with spring constant k = 500 N/m, is stretched so s until it is 100 mm longer. Find the required force and work input.

Solution: F = ks(x – x ) = 500 · 0.1 = 50 N 0 W = F dx = ? Êks(xÊ-Êx )d(xÊ-Êx ) = ks(x – x )2/2 000 = 500 · 0.12/2 = 2.5 J

n 4.4 A nonlinear spring has the force versus displacement relation of F = kns(x – xo) . If the spring end is moved to x from the relaxed state, determine the formula for 1 the required work.

4.5 A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant R-134a vapor at 1000 kPa, 140°C. The setup is cooled at constant pressure until the R-134a reaches a quality of 25%. Calculate the work done in the process. Solution: Constant pressure process boundary work. State properties from Table B.5.2 State 1: v = 0.03150 m3/kg , State 2: v = 0.000871 + 0.25 · 0.01956 = 0.00576 m3/kg Interpolated to be at 1000 kPa, numbers at 1017 kPa could have been used in which case: v = 0.00566 W12 = P dV = P (V2-V1) = mP (v2-v1) = 5 · 1000(0.00576 – 0.03150) = -128.7 kJ

4.6 A piston/cylinder arrangement shown in Fig. P4.6 initially contains air at 150 kPa, 400°C. The setup is allowed to cool to the ambient temperature of 20°C. a. Is the piston resting on the stops in the final state? What is the final b. What is the specific work done by the air during this process?

Solution: P1 = 150 kPa, T1 = 400°C = 673.2 K P T2 = T0 = 20°C = 293.2 K 1a 1 P For all states air behave as an ideal gas. 1 a) If piston at stops at 2, V2 = V1/2 P2 2 V and pressure less than Plift = P1 V1 T2 293.2 P2 = P1 · · = 150 · 2 · = 130.7 kPa < P1 V2 T1 673.2 Piston is resting on stops.

4.7 The refrigerant R-22 is contained in a piston/cylinder as shown in Fig. P4.7, where the volume is 11 L when the piston hits the stops. The initial state is -30°C, 150 kPa with a volume of 10 L. This system is brought indoors and warms up to 15°C. Solution:

Initially piston floats, V < Vstop so the P piston moves at constant Pext = P1 until 2 it reaches the stops or 15°C, whichever is first. P 1 1 1a a) From Table B.4.2: v1 = 0.1487, V 0.010 V stop m = V/v = = 0.06725 kg 0.1487 0.011 v2 = V/m = = 0.16357 m3/kg => T = -9 °C & T2 = 15°C 0.06725 1a Since T2 > T1a then it follows that P2 > P1 and the piston is againts stop.

W12 = Pext dV = Pext(V2 – V1) = 150(0.011 – 0.010) = 0.15 kJ

4.8 Consider a mass going through a polytropic process where pressure is directly proportional to volume (n = – 1). The process start with P = 0, V = 0 and ends with P = 600 kPa, V = 0.01 m3.The physical setup could be as in Problem 2.22. Find the Solution: The setup has a pressure that varies linear with volume going through the initial and the final state points. The work is the area below the process curve.

4.9 A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3. 3 Stops in the cylinder restricts the enclosed volume to 0.5 m , similar to the setup in Problem 4.7. The water is now heated to 200°C. Find the final pressure, volume and the work done by the water.

Solution: Initially the piston floats so the equilibrium P lift pressure is 200 kPa 2 1: 200 kPa, v1= 0.1/50 = 0.002 m3/kg, 1 2: 200 °C, ON LINE P 1 1a V Check state 1a: vstop = 0.5/50 = 0.01 => Table B.1.2: 200 kPa , vf < vstop < vg V stop State 1a is two phase at 200 kPa and Tstop » 120.2 °C so as T2 > Tstop the state is higher up in the P-V diagram with v = vstop < vg = 0.127 (at 200°C) 2 State 2 two phase => P2 = Psat(T2) = 1.554 MPa, V2 = Vstop = 0.5 m3 W = Wstop = 200 (0.5 – 0.1) = 80 kJ 121 4.10 A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. Initially the piston floats, similar to the setup in Problem 4.7, with a maximum enclosed volume of 0.002 m3 if the piston touches the stops. Now heat is added so a final pressure of 600 kPa is reached. Find the final volume and the work in the process. Solution: Take CV as the water which is a control Table B.1.1: 20°C => P = 2.34 kPa sat State 1: Compressed liquid P 1 v = vf(20) = 0.001002 m3/kg State 1a: v = 0.002 m3/kg , 300 kPa stop P 2 1 1a V V stop

4.11 A piston/cylinder contains butane, C4H10, at 300°C, 100 kPa with a volume of 0.02 m3. The gas is now compressed slowly in an isothermal process to 300 kPa. a. Show that it is reasonable to assume that butane behaves as an ideal gas during b. Determine the work done by the butane during the process.

Solution: T 573.15 P 100 a) Tr1 = Tc = 425.2 = 1.35; Pr1 = Pc = = 0.026 3800 From the generalized chart in figure D.1 Z1 = 0.99 T 573.15 P 300 Tr2 = = = 1.35; Pr2 = = = 0.079 Tc 425.2 Pc 3800 From the generalized chart in figure D.1 Z2 = 0.98 b) Ideal gas T = constant PV = mRT = constant P1 100 W = ? PdV = P1V1 ln = 100 · 0.02 · ln = -2.2 kJ P 300 2

4.12 The piston/cylinder shown in Fig. P4.12 contains carbon dioxide at 300 kPa, 100°C with a volume of 0.2 m3. Mass is added at such a rate that the gas 1.2 compresses according to the relation PV = constant to a final temperature of 200°C. Determine the work done during the process.

4.13 Air in a spring loaded piston/cylinder has a pressure that is linear with volume, P = A + BV. With an initial state of P = 150 kPa, V = 1 L and a final state of 800 kPa and volume 1.5 L it is similar to the setup in Problem 3.16. Find the work done by the air.

Solution: Knowing the process equation: P = A + BV giving a linear variation of pressure versus volume the straight line in the P-V diagram is fixed by the two points as state 1 and state 2. The work as the integral of PdV equals the area under the process curve in the P-V diagram.

1 0 P State 1: P = 150 kPa V = 1 L = 0.001 m3 11 2 State 2: P = 800 kPa V = 1.5 L = 0.0015 m3 22 Process: P = A + BV linear in V V Ê2 P Ê+ÊP W = ? ÊPdV = ( )(V 12 -V) 12 221 Ê1 1 = (150 + 800)(1.5 – 1)· 0.001 = 0.2375 kJ 2

4.14 A gas initially at 1 MPa, 500°C is contained in a piston and cylinder arrangement 3 with an initial volume of 0.1 m . The gas is then slowly expanded according to the relation PV = constant until a final pressure of 100 kPa is reached. Determine the work for this process.

Solution: By knowing the process and the states 1 and 2 we can find the relation between the pressure and the volume so the work integral can be performed.

Process: PV = C V2 = P1V1/P2 = 1000 · 0.1/100 = 1 m3

3 4.15 Consider a two-part process with an expansion from 0.1 to 0.2 m at a constant 3 pressure of 150 kPa followed by an expansion from 0.2 to 0.4 m with a linearly rising pressure from 150 kPa ending at 300 kPa. Show the process in a P-V Solution: By knowing the pressure versus volume variation the work is found.

300 150 12 1W3 = 1W2 + 2W3 P 3 Ê2 Ê3 = ? ÊPdV + ? ÊPdV Ê1 Ê2 = P1 (V2 – V1) V 1 + (P2 + P3)(V3-V2) 0.1 0.2 0.4 2 1 W = 150 (0.2-1.0) + (150 + 300) (0.4 – 0.2) = 15 + 45 = 60 kJ 2

4.17 The gas space above the water in a closed storage tank contains nitrogen at 25°C, 100 kPa. Total tank volume is 4 m3, and there is 500 kg of water at 25°C. An additional 500 kg water is now forced into the tank. Assuming constant temperature throughout, find the final pressure of the nitrogen and the work done on the nitrogen in this process.

Solution: The water is compressed liquid and in the process the pressure goes up so the water stays as liquid. Incompressible so the specific volume does not change. The nitrogen is an ideal gas and thus highly compressible.

State 1: VH2OÊ1 = 500 · 0.001003 = 0.5015 m3 VN2Ê1 = 4.0 – 0.5015 = 3.4985 m3 State 2: VN2Ê2 = 4.0 – 2 · 0.5015 = 2.997 m3 IdealÊGas 3.4985 TÊ=Êconst PN2Ê2 = 100 · 2.997 = 116.7 kPa Constant temperature gives P = mRT/V i.e. pressure inverse in V Ê2 W12ÊbyÊN2 = ? Ê PN2dVN2 = P1V1 ln(V2/V1 ) Ê1 2.997 = 100 · 3.4985 · ln = -54.1 kJ 3.4985

4.18 A steam radiator in a room at 25°C has saturated water vapor at 110 kPa flowing through it, when the inlet and exit valves are closed. What is the pressure and the o quality of the water, when it has cooled to 25 C? How much work is done?

4.19 A balloon behaves such that the pressure inside is proportional to the diameter squared. It contains 2 kg of ammonia at 0°C, 60% quality. The balloon and Considering the ammonia as a control mass, find the amount of work done in the process.

Solution: Process : P D2, with V D3 this implies P D2 V2/3 so PVÊ-2/3 = constant, which is a polytropic process, n = -2/3 From table B.2.1: V = mv = 2(0.001566 + 0.6 · 0.28783) = 0.3485 m3 11 P2 3/2 600 3/2 V = V ( ) = 0.3485( ) = 0.5758 m3 2 1 P 429.3 1 Ê2 P V Ê-ÊP V 2211 W = ? ÊPdV = (Equation 4.4) 12 1Ê-Ên Ê1 600Ê· Ê0.5758Ê-Ê429.3Ê· Ê0.3485 = = 117.5 kJ 1Ê-Ê(-2/3)

4.20 Consider a piston cylinder with 0.5 kg of R-134a as saturated vapor at -10°C. It is now compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Find the final volume and temperature, and determine the work done during the process.

4.21 A cylinder having an initial volume of 3 m3 contains 0.1 kg of water at 40°C. The water is then compressed in an isothermal quasi-equilibrium process until it has a quality of 50%. Calculate the work done in the process. Assume the water vapor Solution: C.V. Water 33 1 1 0.1 GÊ ) v = V /m = = 30 m /kg ( > v T P G P1 Tbl B.1.1 => PGÊ = 7.384 kPa very low 40 oC so H2O ~ ideal gas from 1-2 321 vG 19.52 P1 = PGÊ = 7.384 · = 4.8 kPa v 30 1

V2 = mv2 = 0.1 · 19.52 = 1.952 m3 v ÊÊ2 V2 1.952 W12 = ? 1 1 V1 3 T = C: ÊPdV = P V ln = 4.8 · 3.0 · ln = -6.19 kJ 1

v3 = 0.001008 + 0.5 · 19.519 = 9.7605 => V3 = mv3 = 0.976 m3 ÊÊ3 P = C = Pg: W23 = ? ÊPdV = PgÊ(V3-V2) = 7.384(0.976 – 1.952) = -7.21 kJ 2

Total work: W13 = -6.19 – 7.21 = -13.4 kJ 4.22 Consider the nonequilibrium process described in Problem 3.7. Determine the Solution: If piston floats or moves: P = Plift = Po + hg = 101.3 + 8000*0.1*9.807 / 1000 = 108.8 kPa V = V 1 50 / 100 = ( /4) 0.12 · 0.1· 1.5 = 0.000785· 1.5 = 0.0011775 m3 2 1· For max volume we must have P > Plift so check using ideal gas and constant T process: P2 = P1 V1/ V2 = 200/1.5 = 133 kPa and piston is at stops.

4.23 Two kilograms of water is contained in a piston/cylinder (Fig. P4.23) with a massless piston loaded with a linear spring and the outside atmosphere. Initially the spring force is zero and P = Po = 100 kPa with a volume of 0.2 m3. If the piston just hits 1 the upper stops the volume is 0.8 m3 and T = 600°C. Heat is now added until the pressure reaches 1.2 MPa. Find the final temperature, show the P–V diagram and find Solution: P 2 State 1: v1 = V/m = 0.2 / 2 = 0.1 m3/kg ,

, 3 3 Process: 1 ? 2 ? 3 or 1 ? 3′ State at stops: 2 or 2′ 2

P v2 = Vstop/m = 0.4 m3/kg & T2 = 600°C 11 Table B.1.3 Pstop = 1 MPa < P3 V V V stop since Pstop < P3 the process is as 1 ? 2 ? 3 1

State 3: P3 = 1.2 MPa, v3 = v2 = 0.4 m3/kg T3 770°C 11 W13 = W12 + W23 = 2(P1 2)(V2 1 2(100 + 1000)(0.8 – 0.2) +P -V)+0= = 330 kJ 4.24 A piston/cylinder (Fig. P4.24) contains 1 kg of water at 20°C with a volume of 0.1 m3. Initially the piston rests on some stops with the top surface open to the atmosphere, Po and a mass so a water pressure of 400 kPa will lift it. To what temperature should the water be heated to lift the piston? If it is heated to 12 Solution: (a) State to reach lift pressure of P P = 400 kPa, v = V/m = 0.1 m3/kg 1a 2 Table B.1.2: vf < v < vg = 0.4625 P 2 => T = T sat = 143.63°C (b) State 2 is saturated vapor at 400 kPa P 1 1 V since state a is two-phase.

4.25 Assume the same system as in the previous problem, but let the piston be locked with a pin. If the water is heated to saturated vapor find the final temperature, volume and the work, 1W2.

Solution: Constant mass and constant volume process State 2: x2 = 1, v2 = v1 = V1/m = 0.1 m3/kg vg(T) = 0.1 Table B.1.1 => T2 212.5°C

V2 = V1 = 0.1 m3, 1W2 = ? ÊPdV = 0 P 2 1 line V 4.26 A piston cylinder setup similar to Problem 4.24 contains 0.1 kg saturated liquid and vapor water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 500 kPa will float it. The water is heated to 300°C. Find the final pressure, 12 Solution:

4.27 A 400-L tank, A (see Fig. P4.27) contains argon gas at 250 kPa, 30°C. Cylinder B, having a frictionless piston of such mass that a pressure of 150 kPa will float it, is initially empty. The valve is opened and argon flows into B and eventually reaches a uniform state of 150 kPa, 30°C throughout. What is the work done by the argon?

Solution: Take C.V. as all the argon in both A and B. Boundary movement work done in cylinder B against constant external pressure of 150 kPa. Argon is an ideal gas, so write out that the mass and temperature at state 1 and 2 are the same P V = m RT = m RT = P ( V + V ) A1 A A A1 A 2 2 A B2 250Ê· Ê0.4 => V = – 0.4 = 0.2667 m3 B2 150 Ê2 W = ? ÊPextdV = Pext(V – V ) = 150 (0.2667 – 0) = 40 kJ 12 B2 B1 Ê1

4.28 Air at 200 kPa, 30°C is contained in a cylinder/piston arrangement with initial volume 0.1 m3 . The inside pressure balances ambient pressure of 100 kPa plus an externally imposed force that is proportional to V0.5. Now heat is transferred to the system to a final pressure of 225 kPa. Find the final temperature and the work done in Solution: C.V. Air. This is a control mass. Use initial state and process to find T2 P = P + CV1/2; 200 = 100 + C(0.1)1/2, C = 316.23 => 10 225 = 100 + CV 1/2 V = 0.156 m3 22 P1V1 P2V2 = mRT2 = T2 T1 T2 = (P2V2 / P1V1) T1 = 225 · 0.156 · 303.15 / (200 · 0.1) = 532 K = 258.9°C W12 = P dV = (P0 + CV1/2) dV = P0 (V2 – V1) + C · 2 · (V23/2 – V13/2) 3

4.29 A spring-loaded piston/cylinder arrangement contains R-134a at 20°C, 24% quality with a volume 50 L. The setup is heated and thus expands, moving the piston. It is noted that when the last drop of liquid disappears the temperature is 40°C. The heating is stopped when T = 130°C. Verify the final pressure is about 1200 kPa by iteration and find the work done in the process.

Solution: P 3 P 2 P 1 P State 1: Table B.5.1 => 3 v1 = 0.000817 + 0.24*0.03524 = 0.009274 P1 = 572.8 kPa, 2

4.30 A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then further cooled to 20°C, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V.

Solution: P 2000 1555 857 W 3 13 = ? ÊPdV » ( 2 1 2000Ê+Ê1555 1555Ê+Ê857 = 1(0.08313 – 0.10571) + 1(0.07543 – 0.08313) 22 = -49.4 kJ State 1: (T, P) Table B.2.2 1 v1 = 0.10571 o 2 o 40 C P2 = 1555 kPa, v2 = 0.08313 3o 20 C State 3: (T, x) P3 = 857 kPa, v3 = (0.001638+0.14922)/2 = 0.07543 v

P1Ê+ÊP2 P2Ê+ÊP3 )m(v2 – v1) + ( )m(v3 – v2) 2

4.31 A vertical cylinder (Fig. P4.31) has a 90-kg piston locked with a pin trapping 10 L of R-22 at 10°C, 90% quality inside. Atmospheric pressure is 100 kPa, and the cylinder cross-sectional area is 0.006 m2. The pin is removed, allowing the piston to move and come to rest with a final temperature of 10°C for the R-22. Find the final pressure, final volume and the work done by the R-22.

Solution: State 1: (T,x) from table B.4.1 v1 = 0.0008 + 0.9 · 0.03391 = 0.03132 m = V1/v1 = 0.010/0.03132 = 0.319 kg Force balance on piston gives the equilibrium pressure 90Ê· Ê9.807 P2 = P0 + mPg/ AP = 100 + = 247 kPa 0.006Ê· Ê1000 State 2: (T,P) interpolate V2 = mv2 = 0.319 · 0.10565 = 0.0337 m3 = 33.7 L R-22

4.32 A piston/cylinder has 1 kg of R-134a at state 1 with 110°C, 600 kPa, and is then brought to saturated vapor, state 2, by cooling while the piston is locked with a pin. Now the piston is balanced with an additional constant force and the pin is removed. The cooling continues to a state 3 where the R-134a is saturated liquid. Show the processes in a P-V diagram and find the work in each of the two steps, 1 to 2 and 2 to 3.

Solution : P Properties from table B.5.1 and 5.2 State 1: (T,P) => v = 0.04943 State 2 given by fixed volume and x2 = 1.0 1

State 2: v2 = v1 = vg => T = 10 C State 3 reached at constant P (F = constant) 2

3 Final state 3: v3 = vf = 0.000794 V

Since no volume change from 1 to 2 => 1W2 = 0 2W3 = P dV = P(V3 -V2) = mP(v3 -v2) Constant pressure = 415.8(0.000794-0.04943) 1 = -20.22 kJ

4.33 Consider the process described in Problem 3.49. With the ammonia as a control mass, determine the boundary work during the process.

Solution : P 700 2 80 1 14 290 3 -10

v T 2 3 1 v Ê2 1W3 = 1W2 + 2W3 = ? ÊPdV = P1(V2 – V1) = mP1(v2 – v1) Ê1 Since constant volume from 2 to 3, see P-v diagram. From table B.2 v1 = 0.2367, P1 = 700 kPa, v2 = vg = 0.1815 m3/kg 1w3 = P1(v2- v1) = 700 · (0.1815 – 0.2367) = -38.64 kJ/kg

Solution : 5000 1200 200 P 1: 5 MPa, 400°C v1= 0.05781 1 m = V/v1 = 0.1/0.05781 = 1.73 kg Straight line: P = Pa + Cv 2 P2Ê-ÊPa v2 = v1 = 0.01204 m3/kg P1Ê-ÊPa v v2 < vg(1200 kPa) so two-phase T2 = 188°C 0 ? 0.05781 x2 = (v2 - 0.001139)/0.1622 = 0.0672 a

The P-V coordinates for the two states are then: P1 = 5 MPa, V1 = 0.1 m3, P2 = 1200 kPa, V2 = mv2 = 0.02083 m 3

Solution : P Process equation: P = Cv 1 Tabel B.1.3: C = P /v = 3000/0.11619 = 25820 11 2 State 2: x = 1 & P = Cv (on process line) 222 Trial & error on T2sat or P2sat: v at 2 MPa vg = 0.09963 C = P/vg = 20074 2.5 MPa vg = 0.07998 C = P/vg = 31258 2.25 MPa vg = 0.08875 C = P/vg = 25352 Now interpolate to match the right slope C: P2 = 2270 kPa, v2 = P2/C = 2270/25820 = 0.0879 m3/kg P is linear in V so the work becomes (area in P-v diagram) 1 1w2 = P dv = (P1 + P2)(v2 – v1) 2 1 = (3000+2270)(0.0879 – 0.11619) = – 74.5 kJ/kg 2

4.38 A spherical elastic balloon initially containing 5 kg ammonia as saturated vapor at 20°C is connected by a valve to a 3-m3 evacuated tank. The balloon is made such that the pressure inside is proportional to the diameter. The valve is now opened, allowing ammonia to flow into the tank until the pressure in the balloon has dropped to 600 kPa, at which point the valve is closed. The final temperature in both the balloon and the tank is 20°C. Determine a. The final pressure in the tank b. The work done by the ammonia

Solution : D Balloon State 1: (T, x) and size m1 = 5 kg Table B.2.1: v1 = 0.14922 m3/kg, P1 = 857 kPa 3 V1 = m1v1 = 0.7461 m3 = D1 D1 = 1.125 m 6 Tank state 1: V = 3 m3 ; m1 = 0 V

Process in the balloon: P = K1D = K2 VÊ1/3 PVÊ-1/3 = constant Final state 2: Balloon has P2 = 600 kPa and T2 = 20°C Table B.2.2: v2 = 0.22154 m3 D2 P2 600 From process equation: = = D2 = 0.7876 m D1 P1 857 V2 = 6 D23 = 0.2558 m => m2 = V2/v2 = 1.155 kg in balloon 3

4.39 A 0.5-m-long steel rod with a 1-cm diameter is stretched in a tensile test. What is the required work to obtain a relative strain of 0.1%? The modulus of elasticity of steel is 2 · 108 kPa.

Solution : AEL0 -W12 = (e)2, A = (0.01)2 = 78.54· 10-6 m2 24 78.54· 10-6Ê· Ê2· 108Ê· Ê0.5 -W12 = (10-3)2 = 3.93 J 2

4.42 For the magnetic substance described in Problem 4.41, determine the work done in a process at constant magnetic field intensity (temperature varies), instead of one at constant temperature.

Solution : Assume M = cH /T For H = constant (and neglecting volume change) W = µoH d(VM) = µoHÊ2Vc d 1 ?T? H 2 Vc Ø1 1 Ø or W12 = µo ? Ê-Ê œ ºT2 T1ß

4.43 A battery is well insulated while being charged by 12.3 V at a current of 6 A. Take the battery as a control mass and find the instantaneous rate of work and the total work done over 4 hours..

Solution : Battery thermally insulated Q = 0 For constant voltage E and current i, Power = E i = 12.3 · 6 = 73.8 W [Units V*A = W] W = power dt = power t

4.44 Two springs with same spring constant are installed in a massless piston/cylinder with the outside air at 100 kPa. If the piston is at the bottom, both springs are relaxed and the second spring comes in contact with the piston at V = 2 m3. The cylinder (Fig. P4.44) contains ammonia initially at -2°C, x = 0.13, V = 1 m3, which is then heated until the pressure finally reaches 1200 kPa. At what pressure will the piston touch the second spring? Find the final temperature and the total work done by the ammonia.

Solution : P State 1: P = 399.7 kPa Table B.2.1 3 v = 0.00156 + 0.13·0.3106 = 0.0419 1 2 At bottom state 0: 0 m3, 100 kPa P 0 W W State 2: V = 2 m3 and on line 0-1-2 12 23 V 0

012V 3 Slope of line 0-1-2: P/ V = (P – P )/ V = (399.7-100)/1 = 299.7 kPa/ m3 10 P2 = P1 + (V2 – V1) P/ V = 399.7 + (2-1)·299.7 = 699.4 kPa State 3: Last line segment has twice the slope.

4.45 Consider the process of inflating a helium balloon, as described in Problem 3.14. For a control volume that consists of the space inside the balloon, determine the work done during the overall process.

Solution : Inflation at constant P = P0 = 100 kPa to D1 = 1 m, then P = P0 + C ( D*Ê-1 – D*Ê-2 ), D* = D / D1, to D2 = 4 m, P2 = 400 kPa, from which we find the constant C as: 400 = 100 + C[ (1/4) – (1/4)2 ] => C = 1600 The volumes are: V = D3 => V = 0.5236 m3; V = 33.51 m3 612 Ê2 WCV = P0(V1 – 0)+ ? ÊPdV Ê1 Ê2 = P0(V1 – 0) + P0(V2 – V1) + ? ÊC(D*Ê-1Ê-ÊD*Ê-2)dV Ê1 V = D3, dV = D2 dD = D 3ÊD*Ê2 dD* 6 2 21 ÊÊÊÊÊD2*=4 WCV = P0V2 + 3CV1 ? ÊÊÊÊÊ(D*-1)dD* ÊÊÊÊÊÊD1*=1

4.46 A cylinder (Fig. P4.46), A = 7.012 cm2, has two pistons mounted, the upper cyl one, mp1 = 100 kg, initially resting on the stops. The lower piston, mp2 = 0 kg, has 2 kg water below it, with a spring in vacuum connecting the two pistons. The spring force is zero when the lower piston stands at the bottom, and when the lower piston hits the stops the volume is 0.3 m3. The water, initially at 50 kPa, V a. Find the initial temperature and the pressure that will lift the upper piston. b. Find the final T, P, v and the work done by the water.

Solution : State 1: P1, v1 = V1/m = 0.00103 m3/kg T1 = 81.33°C Force balance on the combined set of pistons and spring.

(mp1+mp2)g 100Ê· Ê9.807 Plift = P0 + = 101.325 + = 1500 kPa Acyl 7.012· 10-4· 103 P To place the process line in the P-v diagram 1a: Plift & line from (0,0) to state 1: 1a 2 v1Plift 0.00103Ê· Ê1500 v1a = P1 = 50 = 0.0309 m3 P 2a /kg lift

0v vv vv Vstop 0.3 1 1a 2 stop

v2a = m = 2 = 0.15 m3/kg check saturated vapor state at Plift vg(Plift) = 0.13177 m3/kg v2a > vg(Plift) so state 2a is superheated vapor.

2 4.47 The sun shines on a 150 m road surface so it is at 45°C. Below the 5 cm thick asphalt, average conductivity of 0.06 W/m K, is a layer of compacted rubbles at a temperature of 15°C. Find the rate of heat transfer to the rubbles.

Solution : . T 45-15 Q = k A = 0.06 · 150 · = 5400W x 0.05

4.48 A pot of steel, conductivity 50 W/m K, with a 5 mm thick bottom is filled with 15°C liquid water. The pot has a diameter of 20 cm and is now placed on an electric stove that delivers 250 W as heat transfer. Find the temperature on the outer pot bottom surface assuming the inner surface is at 15°C.

Solution : Steady conduction through the bottom of the steel pot. Assume the inside Q=kA =Qx/k x

T = 250 · 0.005/(50 · · 0.22) = 0.796 4 T = 15 + 0.796 15.8 °C

4.49 A water-heater is covered up with insulation boards over a total surface area of 3 2 m . The inside board surface is at 75°C and the outside surface is at 20°C and the board material has a conductivity of 0.08 W/m K. How thick a board should it be to limit the heat transfer loss to 200 W ?

4.50 You drive a car on a winter day with the atmospheric air at -15°C and you keep the outside front windshield surface temperature at +2°C by blowing hot air on the inside surface. If the windshield is 0.5 m2 and the outside convection coefficient is 250 W/m2K find the rate of energy loos through the front windshield.

Solution : The heat transfer from the inside must match the loss on the outer surface Q conv = h A = 250 · 0.5 · (2-(-15)) = 250 · 0.5 · 17 = 2125 W This is a substantial amount of power.

4.51 A large condenser (heat exchanger) in a power plant must transfer a total of 100 MW from steam running in a pipe to sea water being pumped through the heat exchanger. Assume the wall separating the steam and seawater is 4 mm of steel, conductivity 50 W/m K and that a maximum of 5°C difference between the two fluids is allowed in the design. Find the required minimum area for the heat transfer neglecting any convective heat transfer in the flows.

Solution : Q = k A = Q x/k x A = 100 · 106 · 0.004/(50 · 5) = 1600 m2

4.52 The black grille on the back of a refrigerator has a surface temperature of 35°C with a total surface area of 1 m2. Heat transfer to the room air at 20°C takes place with an average convective heat transfer coefficient of 15 W/m2 K. How much energy can be removed during 15 minutes of operation?

4.53 Due to a faulty door contact the small light bulb (25 W) inside a refrigerator is kept on and limited insulation lets 50 W of energy from the outside seep into the refrigerated space. How much of a temperature difference to the ambient at 20°C must the refrigerator have in its heat exchanger with an area of 1 m2 and an average heat transfer coefficient of 15 W/m2 K to reject the leaks of energy.

Solution : Q tot = 25 + 50 = 75 W to go out Q = hA T = 15 · 1 · T = 75 T = Q / hA = 75/(15· 1) = 5 °C OR T must be at least 25 °C

4.54 The brake shoe and steel drum on a car continuously absorbs 25 W as the car slows down. Assume a total outside surface area of 0.1 m2 with a convective heat transfer coefficient of 10 W/m2 K to the air at 20°C. How hot does the outside brake and drum surface become when steady conditions are reached?

Solution : Q = hA = Q / hA T = ( BRAKE – 20 ) = 25/(10 · 0.1) = 25 °C TBRAKE = 20 + 25 = 45 °C

4.55 A wall surface on a house is at 30°C with an emissivity of = 0.7. The surrounding ambient to the house is at 15°C, average emissivity of 0.9. Find the rate of radiation energy from each of those surfaces per unit area.

4.56 A log of burning wood in the fireplace has a surface temperature of 450°C. Assume the emissivity is 1 (perfect black body) and find the radiant emission of energy per unit surface area.

Solution : . 4 –8 4 Q /A = 1 · T = 5.67 · 10 · ( 273.15 + 450) = 15505 W/m2 = 15.5 kW/m2

4.57 A radiant heat lamp is a rod, 0.5 m long and 0.5 cm in diameter, through which 400 W of electric energy is deposited. Assume the surface has an emissivity of 0.9 and neglect incoming radiation. What will the rod surface temperature be ?

Solution : Q = AT4 = Q rad el T4 = Q / A = 400 / (0.9 · 5.67 · 10 –8 · 0.5 · · 0.005) el = 9.9803 · 1011 K4 T 1000 K OR 725 °C

4.58 Consider a window-mounted air conditioning unit used in the summer to cool incoming air. Examine the system boundaries for rates of work and heat transfer, Solution : Air-conditioner unit, steady operation with no change of temperature of AC unit. – electrical work (power) input operates unit, +Q rate of heat transfer from the room, a larger -Q rate of heat transfer (sum of the other two energy rates) out to the outside air.

4.59 Consider a hot-air heating system for a home. Examine the following systems for a) The combustion chamber and combustion gas side of the heat transfer area Fuel and air enter, warm products of the combustion exit, large -Q to the air in b) The furnace as a whole, including the hot- and cold-air ducts and chimney Fuel and air enter, warm products exit through the chimney, cool air into the cold air return duct, warm air exit hot-air duct to heat the house. Small heat transfer losses from furnace, chimney and ductwork to the house.

4.60 Consider a household refrigerator that has just been filled up with room- temperature food. Define a control volume (mass) and examine its boundaries for a. Immediately after the food is placed in the refrigerator b. After a long period of time has elapsed and the food is cold a) short term.: -Q from warm food to cold refrigerator air. Food cools. II. C.V. refrigerator space, not food, not refrigerator system a) short term: +Q from the warm food, +Q from heat leak from room into cold space. -Q (sum of both) to refrigeration system. If not equal the refrigerator space initially warms slightly and then cools down to preset T. b) long term: small -Q heat leak balanced by -Q to refrigeration system. Note: For refrigeration system CV any Q in from refrigerator space plus electrical W input to operate system, sum of which is Q rejected to the room.

4.61 A room is heated with an electric space heater on a winter day. Examine the following control volumes, regarding heat transfer and work , including sign. Electrical work (power) input, and equal (after system warm up) Q out to the b) Room Q input from the heater balances Q loss to the outside, for steady (no c) The space heater and the room together Electrical work input balances Q loss to the outside, for steady operation.

English Unit Problems 4.62E A cylinder fitted with a frictionless piston contains 10 lbm of superheated refrigerant R-134a vapor at 100 lbf/in.2, 300 F. The setup is cooled at constant pressure until the water reaches a quality of 25%. Calculate the work done in the process.

Solution: v1 = 0.76629; v2 = 0.013331 + 0.25 · 0.46652 = 0.12996 Ê2 W12 = ? ÊPdV = P(V2 – V1) = mP(v2 – v1) Ê1 144 = 10 · 100 · · (0.12996 – 0.76629) = -117.78 Btu 778

4.63E An escalator raises a 200 lbm bucket of sand 30 ft in 1 minute. Determine the total amount of work done and the instantaneous rate of work during the process.

Solution: W = ? Fdx = F? dx = F x = 200 · 30 = 6000 ft lbf = (6000/778) Btu = 7.71 Btu W = W / t = 7.71 / 60 = 0.129 Btu/s

s o), with spring constant ks = 4.64E A linear spring, F = k (x – x 35 lbf/ft, is stretched until it is 2.5 in. longer. Find the required force and work input.

4.65E The piston/cylinder shown in Fig. P4.12 contains carbon dioxide at 50 lbf/in.2, 200 F with a volume of 5 ft3. Mass is added at such a rate that the gas compresses according to the relation PV1.2 = constant to a final temperature of 350 F. Determine the work done during the process.

Solution: From Eq. 4.4 for PVn = const ( n =/ 1 ) Ê2 P2V2Ê-ÊP1V1 W12 = ? ÊPdV = Assuming ideal gas, PV = mRT 1Ê-Ên Ê1 mR(T2Ê-ÊT1) P1V1 50Ê· Ê144Ê· Ê5 W12 = , But mR = = = 0.07014 1Ê-Ên T1 659.7Ê· Ê778 0.07014(809.7Ê-Ê659.7) W12 = = -52.605 Btu 1Ê-Ê1.2

4.66E Consider a mass going through a polytropic process where pressure is directly proportional to volume (n = – 1). The process start with P = 0, V = 0 and ends with P = 90 lbf/in.2, V = 0.4 ft3.The physical setup could be as in Problem 2.22. Find the boundary work done by the mass.

4.67E The gas space above the water in a closed storage tank contains nitrogen at 80 F, 15 lbf/in.2. Total tank volume is 150 ft3 and there is 1000 lbm of water at 80 F. An additional 1000 lbm water is now forced into the tank. Assuming constant temperature throughout, find the final pressure of the nitrogen and the work done on the nitrogen in this process.

Solution: Water is compressed liquid, so it is incompressible VH2OÊ1 = mv1 = 1000 · 0.016073 = 16.073 ft3 VN2Ê1 = Vtank – VH2OÊ1 = 150 – 16.073 = 133.93 ft3 VN2Ê2 = Vtank – VH2OÊ2 = 150 – 32.146 = 117.85 ft3 N2 is an ideal gas so PN 2 = PN 1 · VN 1/VN 2 = 15 · 133.93 = 17.046 lbf/in2 2Ê 2Ê 2Ê 2Ê 117.85 V2 15· 144· 133.93 117.85 W12 = ? PdV = P1V1 ln = ln = -47.5 Btu V1 778 133.93

4.68E A steam radiator in a room at 75 F has saturated water vapor at 16 lbf/in.2 flowing through it, when the inlet and exit valves are closed. What is the pressure and the quality of the water, when it has cooled to 75F? How much work is done?

Solution: T 75°F P 1: x1 = 1, P1 = 16 lbf/in2 1 1 v1 = vg1= 24.754 ft3/lbm 2: T2 = 75°F, v2 = v1 = 24.754 2 v P2 = PgÊ2 = 0.43 lbf/in2 v2 = 24.754 = 0.01606 + x2(739.584 – 0.01606)

4.69E Consider a two-part process with an expansion from 3 to 6 ft3 at a constant pressure of 20 lbf/in.2 followed by an expansion from 6 to 12 ft3 with a linearly rising pressure from 20 lbf/in.2 ending at 40 lbf/in.2. Show the process in a P-V Solution: By knowing the pressure versus volume variation the work is found.

40 20 12 P 1W3 = 1W2 + 2W3 3 Ê2 Ê3 = ? ÊPdV + ? ÊPdV Ê1 Ê2 = P1 (V2 – V1) V 1 3 6 12 + (P2 + P3)(V3-V2) 2 1 2 = (34560/778) = 44.42 Btu 4.70E A cylinder having an initial volume of 100 ft3 contains 0.2 lbm of water at 100 F. The water is then compressed in an isothermal quasi-equilibrium process until it has a quality of 50%. Calculate the work done in the process assuming water vapor is an ideal gas.

4.71E Air at 30 lbf/in.2, 85 F is contained in a cylinder/piston arrangement with initial volume 3.5 ft3. The inside pressure balances ambient pressure of 14.7 lbf/in.2 plus an externally imposed force that is proportional to V0.5. Now heat is transferred to the system to a final pressure of 40 lbf/in.2. Find the final temperature and the work done in the process.

Solution: C.V. Air. This is a control mass. Use initial state and process to find T2 P = P + CV1/2; 30 = 14.7 + C(3.5)1/2, C = 8.1782 => 10 40 = 14.7 + CV 1/2 V = [ (40 – 14.7)/8.1782 ]2 = 9.57 ft3 22 P2V2 = mRT2 = P1V1 T2 / T1 T2 = (P2V2 / P1V1) T1 = 40 · 9.57 *545 / (30 · 3.5) = 1987 R W = P dV = (P + CV1/2) dV 12 0 = P (V – V ) + C 2 · (V 3/2 – V 3/2) 0 2 1 ·3 2 1 2· 8.1782 3/2 3/2 144 = [14.7 (9.57 – 3.5) + (0.957 – 3.5 ) ] = 10.85 Btu 3 778

4.72E A cylinder containing 2 lbm of ammonia has an externally loaded piston. Initially the ammonia is at 280 lbf/in.2, 360 F and is now cooled to saturated vapor at 105 F, and then further cooled to 65 F, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V. Solution: P

Solution: State 1: P = 274.6 lbf/in2 v = 0.1924 ft3/lbm 11

Process: Pv = C = P1v1 = P2v2 1w2 = ? Pdv = C v-1 dv= C ln v2 1 vP P = 30 lbf/in2; v = 1 1 = 0.1924 · 274.6 / 30 = 1.761 ft3 State 2: /lbm 2 2P 2

v2 P1 274.6 1w2 = P1v1 ln v1 = P1v1 ln P2 = 274.6 · 0.1924 · 144 ln 30 = 16845 ft·lbf/lbm = 21.65 Btu/lbm

4.76E A 1-ft-long steel rod with a 0.5-in. diameter is stretched in a tensile test. What is the required work to obtain a relative strain of 0.1%? The modulus of elasticity of steel is 30 · 106 lbf/in.2.

Solution: AEL0 -W12 = (e)2, A = (0.5)2 = in2 2 4 16 -W12 = 1 ( ) 30· 106 · 1 · (10-3)2 = 2.94 ft·lbf 2 16

4.77E The sun shines on a 1500 ft2 road surface so it is at 115 F. Below the 2 inch thick asphalt, average conductivity of 0.035 Btu/h ft F, is a layer of compacted rubbles at a temperature of 60 F. Find the rate of heat transfer to the rubbles.

4.78E A water-heater is covered up with insulation boards over a total surface area of 30 ft2. The inside board surface is at 175 F and the outside surface is at 70 F and the board material has a conductivity of 0.05 Btu/h ft F. How thick a board should it be to limit the heat transfer loss to 720 Btu/h ?

Solution: .T Q = k · A · x = kA / O x x = 0.05 · 30 (175 -70)/720 = 0.219 ft = 2.6 in 4.79E The black grille on the back of a refrigerator has a surface temperature of 95 F with a total surface area of 10 ft2. Heat transfer to the room air at 70 F takes place with an average convective heat transfer coefficient of 3 Btu/h ft2 R. How much energy can be removed during 15 minutes of operation?

CHAPTER 5 The correspondence between the new problem set and the previous 4th edition chapter 5 problem set.

The problems that are labeled advanced are: New Old New Old New Old 91 21 95 50 99 73 92 38 96 new 100 82 93 47 97 new 101 new 94 49 98 new

5.1 A hydraulic hoist raises a 1750 kg car 1.8 m in an auto repair shop. The hydraulic pump has a constant pressure of 800 kPa on its piston. What is the increase in potential energy of the car and how much volume should the pump displace to deliver that No change in kinetic or internal energy of the car, neglect hoist mass. E2 – E1 = PE2 – PE1 = mg (Z2 – Z1) = 1750 x 9.80665 x 1.8 = 30891 J The increase in potential energy is work into car from pump at constant P. W = E2 – E1 = P dV = P V V = (E2 – E1) / P = 30891/(800 x 1000) = 0.0386 m3

5.2 Airplane takeoff from an aircraft carrier is assisted by a steam driven piston/cylinder with an average pressure of 750 kPa. A 3500 kg airplane should be accelerated from zero to a speed of 30 m/s with 25% of the energy coming from the steam piston. Find E – E = m (1/2) (V 2 – 0) = 3500 x (1/2) x 302 = 1575000 J = 1575 kJ 21 2 W = P dv = Pavg V = 0.25 (E2 – E1) = 0.25 x 1575 = 393.75 kJ V = 393.75/750 = 0.525 m3

5.3 Solve Problem 5.2, but assume the steam pressure in the cylinder starts at 1000 kPa, dropping linearly with volume to reach 100 kPa at the end of the process. Solution: C.V. Airplane.

P E2 – E1 = m (1/2) (V22 – 0) 1000 = 3500 x (1/2) x 302 = 1575000 J = 1575 kJ 100 W = P dv = (1/2)(Pbeg + Pend) V = 0.25 x 1575 = 393.75 kJ V = 393.75/[ (1/2)(1000 + 100) ] = 0.716 m3 1

5.4 A piston motion moves a 25 kg hammerhead vertically down 1 m from rest to a velocity of 50 m/s in a stamping machine. What is the change in total energy of the Solution: C.V. Hammerhead The hammerhead does not change internal energy i.e. same P,T E2 – E1 = m(u2 – u1) + m((1/2)V2 2 – 0) + mg (h2 – 0) = 0 + 25 x (1/2) x 502 + 25 x 9.80665 x (-1) = 31250 – 245.17 = 31005 J = 31 kJ

5.5 A 25 kg piston is above a gas in a long vertical cylinder. Now the piston is released from rest and accelerates up in the cylinder reaching the end 5 m higher at a velocity of 25 m/s. The gas pressure drops during the process so the average is 600 kPa with an outside atmosphere at 100 kPa. Neglect the change in gas kinetic and potential energy, Solution: C.V. Piston (E2 – E1)PIST. = m(u2 – u1) + m[(1/2)V2 2 – 0] + mg (h2 – 0) = 0 +25 x (1/2) x 252 + 25 x9.80665 x 5 = 7812.5 + 1225.8 = 9038.3 J = 9.038 kJ Energy equation for the piston is: E2 – E1 = Wgas – Watm = Pavg Vgas – Po Vgas (remark Vatm = – Vgas so the two work terms are of opposite sign) Vgas = 9.038/(600 – 100) = 0.018 m3

V P o H Pavg g P 1 2

5.6 Find the missing properties a. H O T = 250°C, v = 0.02 m3/kg P = ? u = ? 2 b. N T = 277°C, P = 0.5 MPa x = ? h = ? 2 c. H O T = -2°C, P = 100 kPa u = ? v = ? 2 d. R-134a P = 200 kPa, v = 0.12 m3/kg u = ? T = ? e. NH T = 65°C, P = 600 kPa u = ? v = ? 3 Solution: a) Table B.1.1 vf < v < vg P = Psat = 3973 kPa x= (v - vf)/ vfg = (0.02 – 0.001251)/0.04887 = 0.383 u = uf + x ufg = 1080.37 + 0.38365 x 1522.0 = 1664.28 kJ/kg Table C.7 h = [(1/2) x 5911 + (1/2) x 8894]/28.013 = 264.25 kJ/kg c) Table B.1.1 : T < Ttriple point => B.1.5: P > Psat so compressed solid u ui = -337.62 kJ/kg v vi = 1.09×10-3 m3/kg approximate compressed solid with saturated solid properties at same T. T ~ 32.5°C = 30 + (40 – 30) x (0.12 – 0.11889)/(0.12335 – 0.11889) u = h – Pv = 429.07 – 200 x 0.12 = 405.07 kJ/kg e) Table B.2.1 P < Psat => superheated vapor B.2.2: v = 0.5*0.25981 + 0.5 x 0.26888 = 0.2645 m3/kg u = h – Pv = 1594.65 – 600 x 0.2645 = 1435.95 kJ/kg

b d, e States shown are placed relative to the two-phase region, not to each other.aT cv Tb P = const.

5.7 Find the missing properties and give the phase of the substance 2 2 3 Solution: a) Table B.1.1: uf < u < ug => 2-phase mixture of liquid and vapor x = (u – uf)/ ufg = (2390 – 376.82)/2117.7 = 0.9506 v = vf + x vfg = 0.001036 + 0.9506 x 2.35953 = 2.244 m3/kg h = hf + x hfg = 376.96 + 0.9506 x 2283.19 = 2547.4 kJ/kg

b) Table B.1.2: u < uf so compressed liquid B.1.4, x = undefined 1200-1121.03 T 260 + (280 - 260) · = 275.8°C 1220.9-1121.03 1200-1121.03 v = 0.001265 + 0.000057 · = 0.0013096 m3/kg 1220.9-1121.03

c) Table B.3.1: P > Psat => x = undef, compr. liquid Approximate as saturated liquid at same T, h = hf = 31.45 kJ/kg

d) Table B.5.1: h > hg => x = undef, superheated vapor B.5.2, find it at given T between 1400 kPa and 1600 kPa to match h: 430-434.08 v 0.01503 + (0.01239-0.01503) · = 0.01269 m3/kg 429.32-434.08

e) Table B.2.1: P < Psat => x = undef, superheated vapor, from B.2.2: v = 1.4153 m3/kg ; u = h – Pv = 1516.1 – 100 · 1.4153 = 1374.6 kJ/kg

States shown are placed relative to the two-phase region, not to each other.

5.8 Find the missing properties and give the phase of the substance 3 2 2 23 3 4 Solution:

a) Table B.1.1: vf < v < vg => L+V mix, P = 198.5 kPa, x = (0.5 – 0.00106)/0.8908 = 0.56, u = 503.48 + 0.56 · 2025.76 = 1637.9 kJ/kg

b) Table B.1.4: compressed liquid, v = 0.001039 m3/kg, u = 416.1 kJ/kg

c) Table A.2: T >> Tcrit. => sup. vapor, ideal gas, R from Table A.5 v = RT/P = 0.18892 x 800/200 = 0.7557 m3/kg 22806 Table A.8: u = h – Pv = – 200 x 0.7557 = 367 kJ/kg 44.01

d) Table B.2.1: v > vg => sup. vapor, x = undefined 0.1Ê-Ê0.10539 B.2.2: P = 1600 + 200 · = 1685 kPa 0.09267-0.10539

e) Table B.7.1: L+V mix, v = 0.00497 + 0.75 · 0.003 = 0.00722 m3/kg u = 69.1 + 0.75 x 67.01 = 119.36 kJ/kg

P b c ad States shown are placed relative to the two-phase region, not e T

5.9 Find the missing properties among (P, T, v, u, h) together with x if applicable and give a. R-22 T = 10°C, u = 200 kJ/kg b. H O T = 350°C, h = 3150 kJ/kg 2 c. R-12 P = 600 kPa, h = 230 kJ/kg d. R-134aT = 40°C, u = 407 kJ/kg e. NH T = 20°C, v = 0.1 m3/kg 3 Solution: a) Table B.4.1: u < ug => L+V mixture, P = 680.7 kPa x = (200 – 55.92)/173.87 = 0.8287, v = 0.0008 + 0.8287 · 0.03391 = 0.0289 m3/kg, h = 56.46 + 0.8287 · 196.96 = 219.7 kJ/kg b) Table B.1.1: h > hg => superheated vapor follow 350°C in B.1.3 P ~ 1375 kPa, v = 0.204 m3/kg, u = 2869.5 kJ/kg c) Table B.3.1: h > hg = > sup. vapor, T = 69.7°C, v = 0.03624 m3/kg, u = 208.25 kJ/kg d) Table B.5.1: u > ug => sup. vap., calculate u at some P to end with P = 500 kPa, v = 0.04656 m3/kg, h = 430.72 kJ/kg e) Table B.2.1: v < vg => L+V mixture, P = 857.22 kPa x = (0.1 – 0.001638)/0.14758 = 0.666 h = 274.3 + 0.666 · 1185.9 = 1064.1 kJ/kg u = h – Pv = 978.38 kJ/kg ( = 272.89 + 0.666 · 1059.3)

States shown are placed relative to the two-phase region, not to each other.

5.10 A 100-L rigid tank contains nitrogen (N2) at 900 K, 6 MPa. The tank is now cooled to C.V.: Nitrogen in tank. m = m ; m(u – u ) = Q – W 21211212 Process: V = constant, v = v = V/m => W = 0/ 21 12 Table B.6.2: State 1: v = 0.045514 => m = V/v = 2.197 kg 11 u = h – P v = 963.59 – 6000 x 0.045514 = 690.506 1 1 11 State 2: 100 K, v = v = V/m, look in table B.6.2 at 100 K 21 500 kPa: v= 0.05306 ; h = 94.46, 600 kPa: v = 0.042709, h = 91.4 so a linear interpolation gives: P = 572.9 kPa, h = 92.265 kJ/kg, 22 u = h – P v = 92.265 – 572.9 · 0.04551 = 66.19 kJ/kg 2 2 22 Q = m(u – u ) = 2.197 (66.19 – 690.506) = -1372 kJ 12 21

5.11 Water in a 150-L closed, rigid tank is at 100°C, 90% quality. The tank is then cooled to -10°C. Calculate the heat transfer during the process.

Solution: C.V.: Water in tank. m = m ; m(u – u ) = Q – W 21211212 Process: V = constant, v = v , W = 0 2112 State 1: v = 0.001044 + 0.9 x 1.6719 = 1.5057 m3/kg 1 u = 418.94 + 0.9 x 2087.6 = 2297.8 kJ/kg 1 State 2: T , v = v mix of sat. solid + vap. Table B.1.5 221 v = 1.5057 = 0.0010891 + x x 466.7 => x = 0.003224 222 u = -354.09 + 0.003224 x 2715.5 = -345.34 kJ/kg 2 m = V/v = 0.15/1.5057 = 0.09962 kg 1 Q = m(u – u ) = 0.09962(-345.34 – 2297.8) = -263.3 kJ 12 21

5.12 A cylinder fitted with a frictionless piston contains 2 kg of superheated refrigerant R- 134a vapor at 350 kPa, 100°C. The cylinder is now cooled so the R-134a remains at constant pressure until it reaches a quality of 75%. Calculate the heat transfer in the process.

Solution: C.V.: R-134a m = m = m; m(u – u ) = Q – W 21 211212 Process: P = const. W = ? PdV = P V = P(V – V ) = Pm(v – v ) 12 21 21

PT 1 21 2 V V

5.13 A test cylinder with constant volume of 0.1 L contains water at the critical point. It now cools down to room temperature of 20°C. Calculate the heat transfer from the Solution: C.V.: Water m = m = m ; m(u – u ) = Q – W P 1 21 211212 Process: Constant volume v = v 21 Properties from Table B.1.1 State 1: v = vc = 0.003155 u = 2029.6 11 m = V/v = 0.0317 kg 1 2

v State 2: T , v = v = 0.001002 + x x 57.79 221 2 x = 3.7×10-5, u = 83.95 + x x 2319 = 84.04 222 Constant volume => W = 0/ 12 Q = m(u – u ) = 0.0317(84.04 – 2029.6) = -61.7 kJ 12 21

5.14 Ammonia at 0°C, quality 60% is contained in a rigid 200-L tank. The tank and ammonia is now heated to a final pressure of 1 MPa. Determine the heat transfer for the process.

Solution: C.V.: NH 3 m = m = m ; m(u – u ) = Q – W 21 211212 Process: Constant volume v = v & W = 0/ 2112 State 1: Table B.2.1 v = 0.001566 + x x 0.28783 = 0.17426 m3/kg 11 u = 179.69 + 0.6 x 1138.3 = 862.67 kJ/kg 1 P

5.15 A 10-L rigid tank contains R-22 at -10°C, 80% quality. A 10-A electric current (from a 6-V battery) is passed through a resistor inside the tank for 10 min, after which the R-22 temperature is 40°C. What was the heat transfer to or from the tank during this Solution:

m = m = m ; m(u – u ) = Q – W 21 211212 Process: Constant V v = v 21 => no boundary work, but electrical work State 1 from table B.4.1 P

2 1 V v = 0.000759 + 0.8 · 0.06458 = 0.05242 m3/kg 1 u = 32.74 + 0.8 · 190.25 = 184.9 kJ/kg 1 m = V/v = 0.010/0.05242 = 0.1908 kg 21 P = 500 + 100 · (0.05242-0.05636)/(0.04628-0.05636) = 535 kPa, 2

5.16 A piston/cylinder arrangement contains 1 kg of water, shown in Fig. P5.16. The piston is spring loaded and initially rests on some stops. A pressure of 300 kPa will just float the piston and, at a volume of 1.5 m3, a pressure of 500 kPa will balance the piston. The initial state of the water is 100 kPa with a volume of 0.5 m3. Heat is now added b. Find the work and heat transfer in the process and plot the P–V diagram. P C.V.: H O 2 500 Cont.: m = m = 1 kg 400 2 1 300 2 Energy: m(u2 – u1) = 1Q2 – 1W2 Straight line (linear spring): 100 1 From (0.5, 300) to (1.5, 500) 0.5 1.0 1.5 V The initial pressure can not lift the piston. 1: 100 kPa, v = V /m = 0.5 m3/kg < v ; => T = 99.6°C 111 g 1 x = (0.5 – 0.001043)/1.69296 = 0.2947 1

u = 417.33 + 0.2947 · 2088.72 = 1032.9 kJ/kg 1 2: 400 kPa and on line, see figure => V = 1.0 m3 , v = V /m = 1.0 m3/kg 2 221 Superheated vapor Table B.1.2: T = 595°C, u = 3292 kJ/kg 22 1 W = P dV = AREA = (300 + 400)(1.0 – 0.5) = 175 kJ 12 2 Q = m(u -u ) + W = 1(3292 – 1032.9) + 175 = 2434 kJ 122112

5.17 A closed steel bottle contains ammonia at -20°C, x = 20% and the volume is 0.05 m3. It has a safety valve that opens at a pressure of 1.4 MPa. By accident, the bottle is heated until the safety valve opens. Find the temperature and heat transfer when the C.V.: NH : m = m = m ; m(u – u ) = Q – W 3 21 211212 Process: constant volume process W = 0 P 12 State 1: v = 0.001504 + 0.2 · 0.62184 = 0.1259 1 => m = V/v = 0.05/0.1259 = 0.397 kg 1

u = 88.76 + 0.2 · 1210.7 = 330.9 kJ/kg 1 State 2: P , v = v => superheated vapor 221 2

5.19 Consider the same setup and initial conditions as in the previous problem. Assuming See 5.18 solution up to *, then: W = P (V – V ) ; V ‡ V if > then P = P piston floats 1 2 B1 2 1 2 A 2 B1 Energy: m u + W = m u + m u = m u + P V – P V 2 2 1 2 A1 A1 B1 B1 2 2 2 2 B1 1 mh=mu+mu+PV 2 2 A1 A1 B1 B1 B1 1 = 0.5903 · 2506.1 + 0.9695 · 2965.5 + 300 · 2 = 4954 kJ h = 4954/1.5598 = 3176.3 Table B.1.2 : v = 0.95717 > va = 0.641 22 (P , h ) T = 352°C and V = 1.56 · 0.9572 = 1.493 m3 B1 2 2 2 W = 300 (1.493 – 2) = -152.1 kJ 12 5.20 A vertical cylinder fitted with a piston contains 5 kg of R-22 at 10°C, shown in Fig. P5.20. Heat is transferred to the system, causing the piston to rise until it reaches a set of stops at which point the volume has doubled. Additional heat is transferred until the temperature inside reaches 50°C, at which point the pressure inside the cylinder is 1.3 Solution: C.V. R-22. Control mass goes through process: 1 -> 2 -> 3 As piston floats pressure is constant (1 -> 2) and the P volume is constant for the second part (2 -> 3) So we have: v = v = 2 x v 321 State 3: Table B.4.2 (P,T) v = 0.02015 3

u = h – Pv = 274.39 – 1300 · 0.02015 = 248.2 kJ/kg 3 So we can then determine state 1 and 2 Table B.4.1: 3

5.21 A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3. Stops in the cylinder are placed to restrict the enclosed volume to 0.5 m3 similar to Fig. P5.20. The water is now heated until the piston reaches the stops. Find the necessary heat Solution: C.V. H O m = constant P 2 m(e – e ) = m(u – u ) = Q – W 21 211212 Process : P = constant (forces on piston constant) W = P dV = P (V – V ) 12 121 12 Q = m(u – u ) + P (V – V ) = m(h – h ) V 122112121 Properties from Table B.1.1 0.1 0.5 State 1 : v = 0.1/50 = 0.002 m3/kg => 2-phase 1 x = (0.002 – 0.001061)/0.88467 = 0.001061 h = 504.68 + 0.001061 x 2201.96 = 507.02 kJ/kg State 2 : v = 0.5/50 = 0.01 m3/kg also 2-phase same P 2 x = (0.01 – 0.001061)/0.88467 = 0.01010 2 h = 504.68 + 0.01010 x 2201.96 = 526.92 kJ/kg 2 Q = 50 x (526.92 – 507.02) = 995 kJ 12 [ W = P (V – V ) = 200 x (0.5 – 0.1) = 80 kJ ] 12121

5.22 Ten kilograms of water in a piston/cylinder with constant pressure is at 450°C and a volume of 0.633 m3. It is now cooled to 20°C. Show the P–v diagram and find the Solution: P Constant pressure W = mP(v -v ) 12 21

2 Q = m(u – u ) + W = m(h – h ) 112211221 Properties from Table B.1.3 and B.1.4 State 1: v = 0.633/10 = 0.0633 m3/kg 1 P = 5 Mpa, h = 3316.2 kJ/kg 11 State 2: 5 MPa, 20°C v = 0.0009995 2 v h = 88.65 kJ/kg 2

Solution: Take CV as the water. Properties from table B.1 m = m = m ; m(u – u ) = Q – W 21 211212 State 1: Compressed liq. v = vf (20) = 0.001002, u = uf = 83.94 State 2: Since P > P then v = vstop = 0.002 and P = 600 kPa lift For the given P : vf < v < vg so 2-phase T = Tsat = 158.85 °C v = 0.002 = 0.001101 + x x (0.3157-0.001101) => x = 0.002858 u = 669.88 + 0.002858 x1897.5 = 675.3 kJ/kg Work is done while piston moves at P = constant = 300 kPa so we get lift 1W2 = P dV = m P (v2 -v1) = 1×300(0.002 – 0.001002) = 0.299 kJ lift Heat transfer is found from energy equation 1Q2 = m(u2 – u1) + 1W2 = 1(675.3 – 83.94) + 0.299 = 591.66 kJ Solution: m = m = m ; m(u – u ) = Q – W 21 211212

State 1: 20 C, v1 = V/m = 0.1/1 = 0.1 m3/kg P x = (0.1 – 0.001002)/57.789 = 0.001713 1a 2 u1 = 83.94 + 0.001713 · 2318.98 = 87.92 kJ/kg P2 To find state 2 check on state 1a: P 1 1 P = 400 kPa, v = v1 = 0.1 m3/kg Table B.1.2: vf < v < vg = 0.4625 V

v = v = 0.4625 m3/kg , V = m v = 0.4625 m3 u2 = ug= 2553.6 kJ/kg 2g22, Pressure is constant as volume increase beyond initial volume.

5.25 A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 a. Find the final temperature and plot the P-v diagram for the process. b. Calculate the work and heat transfer for the process.

Solution: m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 State 1: Compr. liq., use sat. liq. same T, Table B.1.1 P 2 v = vf (20) = 0.001002, u = uf = 83.94 kJ/kg P State 2: v = V/m = 0.1/1 = 0.1 m3/kg and P = 3 MPa 1 => Sup. vapor T = 400 C ; u = 2932.7 kJ/kg P 2

1 v Work is done while piston moves at linearly varying pressure, so we get 1W2 = P dV = Pavg(V2 -V1) = 0.5x(300+3000)(0.1 – 0.001) = 163.35 kJ Heat transfer is found from energy equation 1Q2 = m(u2 – u1) + 1W2 = 1 x (2932.7 – 83.94) + 163.35 = 3012 kJ

5.26 An insulated cylinder fitted with a piston contains R-12 at 25°C with a quality of 90% and a volume of 45 L. The piston is allowed to move, and the R-12 expands until it exists as saturated vapor. During this process the R-12 does 7.0 kJ of work against the piston. Determine the final temperature, assuming the process is adiabatic. Solution: Take CV as the R-12. m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 State 1: (T, x) Tabel B.3.1 => v = 0.000763 + 0.9 x 0.02609 = 0.024244 m3/kg 1 m = V /v = 0.045/0.024244 = 1.856 kg 11 u = 59.21 + 0.9 x 121.03 = 168.137 kJ/kg 1

5.27 Two kilograms of nitrogen at 100 K, x = 0.5 is heated in a constant pressure process to 300 K in a piston/cylinder arrangement. Find the initial and final volumes and the total Solution: m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 State 1: Table B.6.1 v = 0.001452 + 0.5 · 0.02975 = 0.01633 m3/kg, V = 0.0327 m3 11 h = -73.20 + 0.5 · 160.68 = 7.14 kJ/kg 1 State 2: P = 779.2 kPa , 300 K => sup. vapor interpolate in Table B.6.2 v = 0.14824 + (0.11115-0.14824)· 179.2/200 = 0.115 m3/kg, V = 0.23 m3 22 h = 310.06 + (309.62-310.06) · 179.2/200 = 309.66 kJ/kg 2

Process: P = const. W = ? PdV = Pm(v – v ) 12 21 Q = m(u – u ) + W = m(h – h ) = 2 x (309.66 – 7.14) = 605 kJ 12211221

5.28 A piston/cylinder arrangement has the piston loaded with outside atmospheric pressure and the piston mass to a pressure of 150 kPa, shown in Fig. P5.28. It contains water at -2°C, which is then heated until the water becomes saturated vapor. Find the final Solution: Continuity: m = m , Energy: u – u = q – w 21 211212 ÊÊ2Ê Process: P = const. = P , => w = ? ÊPÊdv = P (v – v ) 1 12 121 Ê1 State 1: T , P => Table B.1.5 compressed solid, take as saturated solid. 11 v = 1.09×10-3 m3/kg, u = -337.62 kJ/kg 11 State 2: x = 1, P = P = 150 kPa due to process => Table B.1.2 21

5.29 Consider the system shown in Fig. P5.29. Tank A has a volume of 100 L and contains saturated vapor R-134a at 30°C. When the valve is cracked open, R-134a flows slowly into cylinder B. The piston mass requires a pressure of 200 kPa in cylinder B to raise the piston. The process ends when the pressure in tank A has fallen to 200 kPa. During this process heat is exchanged with the surroundings such that the R-134a always remains at Solution: m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 State 1: 30°C, x = 1. Table B.5.1: v = 0.02671 m3/kg, u = 394.48 kJ/kg 11 m = V/v = 0.1 / 0.02671 = 3.744 kg 1

State 2: 30°C, 200 kPa superheated vapor Table B.5.2 v = 0.11889 m3/kg, u = 426.87 – 200 x 0.11889 = 403.09 kJ/kg 22 Work done in B against constant external force (equilibrium P in cyl. B) W = Pext dV = Pextm(v – v ) = 200×3.744x(0.11889 – 0.02671) = 69.02 kJ 12 21 Q = m(u – u ) + W = 3.744 x(403.09 – 394.48) + 69.02 = 101.26 kJ 122112

5.30 A spherical balloon contains 2 kg of R-22 at 0°C, 30% quality. This system is heated until the pressure in the balloon reaches 600 kPa. For this process, it can be assumed that the pressure in the balloon is directly proportional to the balloon diameter. How does pressure vary with volume and what is the heat transfer for the process? m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 State 1: 0°C, x = 0.3. Table B.4.1 gives P = 497.6 kPa 1

v = 0.000778 + 0.3 · 0.04636 = 0.014686 m3/kg 1 u = 44.2 + 0.3 x 182.3 = 98.9 kJ/kg 1

5.31 A piston held by a pin in an insulated cylinder, shown in Fig. P5.31, contains 2 kg water at 100°C, quality 98%. The piston has a mass of 102 kg, with cross-sectional area of 100 cm2, and the ambient pressure is 100 kPa. The pin is released, which allows the piston to move. Determine the final state of the water, assuming the process 102ÊxÊ9.807 P = Pext = P0 + mpg/A = 100 + = 200 kPa 2 100×10-4ÊxÊ103 1 2 ? PextdV = Pext 2 1 W = m(v – v ) q = 0/ = u – u + P v – P v = h – u – P v 12 2 1 22 21 2 1 21 h = u + P v = 2464.8 + 200 x 1.6395 = 2792.7 kJ/kg 2 1 21 State 2: (P2 , h2) Table B.1.3 => T2 161.75°C

5.32 A piston/cylinder arrangement has a linear spring and the outside atmosphere acting on the piston, shown in Fig. P5.32. It contains water at 3 MPa, 400°C with the volume being 0.1 m3. If the piston is at the bottom, the spring exerts a force such that a pressure of 200 kPa inside is required to balance the forces. The system now cools until the pressure reaches 1 MPa. Find the heat transfer for the process.

Solution: P 1 State 1: Table B.1.3 3 MPa v = 0.09936 m3/kg, u = 2932.8 kJ/kg 11 2 m = V/v = 0.1/0.09936 = 1.006 kg 1 MPa 1 200 kPa P = P0 + (P1 – P0)v/v1 0vv 21

5.33 A vertical piston/cylinder has a linear spring mounted as shown in Fig. P5.32. The spring is mounted so at zero cylinder volume a balancing pressure inside is 100 kPa. The cylinder contains 0.5 kg of water at 125°C, 70% quality. Heat is now transferred to the water until the cylinder pressure reaches 300 kPa. How much work is done by the water during this process and what is the heat transfer?

Solution: Energy eq.: m(u2 – u1) + = 1Q2 – 1W2 Process: Linear spring => P = P0 + C(V – 0) Po = 100 kPa Fspr = 0 at Vo = 0, Q to P2 = 300 kPa State 1: Two phase table B.1.1, P1 = PgÊ125C = 232.1 kPa v1 = 0.001065 + 0.7(0.77059 – 0.001065) = 0.53973

P u1 = 534.7 + 0.7 · 2009.9 = 1931.6 kJ/kg 300 V1 = mv1 = 0.27 m3 232 232.1-100 100 300 = 232.1 + (V2 – 0.27) 0.27-0 V 67.9 = 489.259V2 – 132.1 0 0.27 V2 => V2 = 0.4088 m3

5.34 Two heavily insulated tanks are connected by a valve, as shown in Fig. P5.34. Tank A contains 0.6 kg of water at 300 kPa, 300°C. Tank B has a volume of 300 L and contains water at 600 kPa, 80% quality. The valve is opened, and the two tanks eventually come to a uniform state. Assuming the process to be adiabatic, show the final state (u,v) is two-phase and iterate on final pressure to match required internal energy.

Solution: C.V.: Both tanks m = m + m ; m u – m u – m u = Q – W = 0/ 2 A1 B1 2 2 A1 A1 B1 B1 1 2 1 2 State 1A: Table B.1.3 v = 0.8753 m3/kg, u = 2806.7 kJ/kg A1 A1 State 1B: Table B.1.2 v = 0.001101 + 0.8 · 0.31457 = 0.25278 m3/kg B1 u = 669.88 + 0.8 · 1897.52 = 2187.9 kJ/kg B1 m = V /v = 0.3/0.25278 = 1.187 kg B1 B B1 Continuity eq.: m = m + m = 1.787 kg 2 A1 B1 m u = 0.6 x 2806.7 + 1.187 x 2187.9 = 4281 kJ u = 2395.67 kJ/kg 22 2 v = Vtot/m = (0.6 x 0.8753 + 0.3)/1.787 = 0.462 m3/kg 22

22 state is two-phase Trial & error v = vf + xvfg ; u = uf + xufg v2Ê-Êvf 2 = 2395.67 = uf + vfg ufg u

Compute RHS for a guessed pressure: T 400 kPa 400 kPa u=2553 1.7 kPa u=2396 sat vap 0.46 77.9 v

P = 350 kPa: RHS = 583.93 + [(0.462-0.001079)/0.52317]*1964.98 = 2315.1 P = 375 kPa: RHS = 594.38 + [(0.462-0.001081)/0.49029]*1956.93 = 2434.1 Interpolate to match correct u: P 367 kPa 2

5.35 A piston/cylinder contains 1 kg of ammonia at 20°C with a volume of 0.1 m3, shown in Fig. P5.35. Initially the piston rests on some stops with the top surface open to the atmosphere, Po, so a pressure of 1400 kPa is required to lift it. To what temperature should the ammonia be heated to lift the piston? If it is heated to saturated vapor find Solution: m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 State 1: 20°C; v = 0.10 < vg x = (0.1 – 0.001638)/0.14758 = 0.6665 11 u = uf + x ufg = 272.89 + 0.6665 x1059.3 = 978.9 11 P Process: Piston starts to lift at state 1a (Plift, v ) 1 1a State 1a: 1400 kPa, v1 Table B.2.2 (sup.vap.) 1400 1200 2 0.1Ê–Ê0.09942 Ta = 50 + (60 – 50) = 51.2 °C 857 0.10423Ê–Ê0.09942 1 v State 2: x = 1.0, v = v => V = mv = 0.1 m3 1

T = 30 + (0.1 – 0.11049) x 5/(0.09397 – 0.11049) = 33.2 °C 2 u = 1338.7; W = 0; q = u – u = 359.8 kJ/kg 2 12 1221 5.36 A cylinder/piston arrangement contains 5 kg of water at 100°C with x = 20% and the piston, mP = 75 kg, resting on some stops, similar to Fig. P5.35. The outside pressure is 100 kPa, and the cylinder area is A = 24.5 cm2. Heat is now added until the water reaches a saturated vapor state. Find the initial volume, final pressure, work, and heat Continuty: m2 = m1 = m ; Energy: m(u2 – u1) = 1Q2 – 1W2 Process: V = constant if P < Plift otherwise P = Plift see P-v diagram. 75ÊxÊ9.807 P3 = P2 = Plift = P0 + mp g / Ap = 100 + = 400 kPa 0.00245ÊxÊ1000 State 1: (T,x) Table B.1.1 P v1 = 0.001044 + 0.2 · 1.6719 2 3 143 C V1 = mv1 = 5 x 0.3354 = 1.677 m 3

u1 = 418.91 + 0.2 · 2087.58 100 C 1v = 836.4 kJ/kg State 3: (P, x = 1) Table B.1.1 => v3 = 0.4625 > v1, u3 = 2553.6 kJ/kg 1W3 = 2W3 = Pextm(v3 – v2) = 400 x 5(0.46246 – 0.3354) = 254.1 kJ 1Q3 = 5 (2553.6 – 836.4) + 254.1 = 8840 kJ HO 2 m p

5.37 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.37. Room A is at 200 kPa, v = 0.5 m3/kg, VA = 1 m3, and room B contains 3.5 kg at 0.5 MPa, 400°C. The membrane now ruptures and heat transfer takes place so the water comes to a uniform state at 100°C. Find the heat transfer during the m=m+m;mu-mu-mu=Q-W 2 A1 B1 2 2 A1 A1 B1 B1 1 2 1 2 State 1A: (P, v) Table B.1.2, m = V /v = 1/0.5 = 2 kg A1 A A1 x = (0.5 – 0.001061)/0.88467 = 0.564 A1 u = 504.47 + 0.564 · 2025.02 = 1646.6 kJ/kg A1 State 1B: Table B.1.3, v = 0.6173, u = 2963.2, V = m v = 2.16 m3 B1 B1 B B1 B1 Process constant total volume: Vtot = VA + VB = 3.16 m3 and 1W2 = 0/ m2 = mA1 + mB1 = 5.5 kg => v2 = Vtot/m2 = 0.5746 m3/kg State 2: T , v Table B.1.1 x = (0.5746 – 0.001044)/1.67185 = 0.343 , 22 2 u = 418.91 + 0.343*2087.58= 1134.95 kJ/kg 2 Q = m u – m u – m u = -7421 kJ 1 2 2 2 A1 A1 B1 B1

5.38 Two tanks are connected by a valve and line as shown in Fig. P5.38. The volumes are both 1 m3 with R-134a at 20°C, quality 15% in A and tank B is evacuated. The valve is opened and saturated vapor flows from A into B until the pressures become equal. The process occurs slowly enough that all temperatures stay at 20°C during the process. Find C.V.: A + B State 1A: vA1 = 0.000817 + 0.15 x 0.03524 = 0.006103 uA1 = 227.03 + 0.15 x 162.16 = 251.35 mA1 = VA/vA1 = 163.854 kg m2 = mA1 ; V2 = VA + VB = 2 m ; v2 = V2/m2 = 0.012206 m /kg 33

5.39 Consider the same system as in the previous problem. Let the valve be opened and transfer enough heat to both tanks so all the liquid disappears. Find the necessary heat State 1A: vA1 = 0.000817 + 0.15 x 0.03524 = 0.006103 uA1 = 227.03 + 0.15 x 162.16 = 251.35 mA1 = VA/vA1 = 163.854 kg m = m ; V = V + V = 2 m3 ; v = V /m = 0.012206 m3/kg 2 A1 2 A B 2 2 2 State 2: x = 100%, v = 0.012206 22 T = 55 + 5 x (0.012206 – 0.01316)/(0.01146 – 0.01316) = 57.8°C 2 u = 406.01 + 0.56 x (407.85 – 406.01) = 407.04 kJ/kg 2 Q = m (u – u ) = 163.854 x (407.04 – 251.35) = 25510 kJ 1 2 2 2 A1

5.40 A cylinder having a piston restrained by a linear spring contains 0.5 kg of saturated vapor water at 120°C, as shown in Fig. P5.40. Heat is transferred to the water, causing the piston to rise, and with a spring constant of 15 kN/m, piston cross-sectional area 0.05 m2, the presure varies linearly with volume until a final pressure of 500 kPa is reached. Find the final temperature in the cylinder and the heat transfer for the process. Continuty: m2 = m1 = m ; Energy: m(u2 – u1) = 1Q2 – 1W2 State 1: (T, x) Table B.1.1 => v1 = 0.89186 m3/kg, u1 = 2529.2 kJ/kg ksm 15ÊxÊ0.5 Process: P2 = P1 + (v2 – v1) = 198.5 + (v2 – 0.89186) Ap2 (0.05)2 => v2 = 0.89186 + (500 – 198.5) · (0.052/7.5) = 0.9924 m /kg 3

5.41 A water-filled reactor with volume of 1 m3 is at 20 MPa, 360°C and placed inside a containment room as shown in Fig. P5.41. The room is well insulated and initially evacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room volume so the final pressure does not exceed 200 kPa. Solution: C.V.: Containment room and reactor.

Mass: m = m = Vreactor/v = 1/0.001823 = 548.5 kg 21 1 Energy: m(u – u ) = Q – W = 0 – 0 = 0 u = u = 1702.8 kJ/kg 211212 21 State 2: P = 200 kPa, u < ug => Two-phase Table B.1.2 22 x = (u – uf)/ ufg = (1702.8 – 504.47)/2025.02 = 0.59176 22 v = 0.001061 + 0.59176 x 0.88467 = 0.52457 m3/kg 2

V = m v = 548.5 x0.52457 = 287.7 m3 2 22 5.42 Assume the same setup as the previous problem, but the room has a volume of 100 m3. Show that the final state is two-phase and find the final pressure by trial and error. C.V.: Containment room and reactor.

Mass: m = m = Vreactor/v = 1/0.001823 = 548.5 kg 21 1 Energy: m(u2 – u1) = 1Q2 – 1W2 = 0 – 0 = 0 u2 = u1 = 1702.8 Total volume and mass => v2 = Vroom/m2 = 0.1823 m3/kg State 2: u2 , v2 Table B.1.1 see Fig. T

0.184 · 1897.52 = 1762.9 state is two-phase (notice u2 << ug Trial & error v = vf + xvfg ; u = uf + xufg v2Ê-Êvf u2 = 1702.8 = ufÊ + ufg vfg Compute RHS for a guessed pressure P : 2

0.1823-0.001101 P = 600 kPa: RHS = 669.88 + 2 0.31457 0.1823-0.001097 P = 550 kPa: RHS = 655.30 + 2 0.34159· 1909.17 = 1668.1 Linear interpolation to match u = 1702.8 gives P2 568.5 kPa 1060 kPa u=25851060 kPa

5.43 Refrigerant-12 is contained in a piston/cylinder arrangement at 2 MPa, 150°C with a massless piston against the stops, at which point V = 0.5 m3. The side above the piston is connected by an open valve to an air line at 10°C, 450 kPa, shown in Fig. P5.43. The whole setup now cools to the surrounding temperature of 10°C. Find the heat transfer and show the process in a P–v diagram.

Continuity: m = constant, Energy: m(u – u ) = Q – W 211212 Process: F? = F› = P·A = PairA + Fstop ; if V < Vstop Fstop = 0/ State 1: v = 0.01265 m3/kg, u = 277 - 2000*0.01265 = 252.1 kJ/kg 11 m = V/v = 39.523 kg 2

v2 vf = 0.000733 V2 = 0.02897; u2 = uf – = 45.06

P 2 MPa T = 10 450 kPa 1 2 150 ~73 v 11.96 10 T 2 P = 2 MPa

5.44 A 10-m high open cylinder, Acyl = 0.1 m2, contains 20°C water above and 2 kg of 20°C water below a 198.5-kg thin insulated floating piston, shown in Fig. P5.44. Assume standard g, Po. Now heat is added to the water below the piston so that it expands, pushing the piston up, causing the water on top to spill over the edge. This process continues until the piston reaches the top of the cylinder. Find the final state of the water below the piston (T, P, v) and the heat added during the process.

Solution: AA = mp Bg + P0 Piston force balance at initial state: F› = F? = P g + m A State 1A,B: Comp. Liq. v vf = 0.001002 m3/kg; u1A = 83.95 kJ/kg VA1 = mAvA1 = 0.002 m3; mtot = Vtot/v = 1/0.001002 = 998 kg mass above the piston mB1 = mtot – mA = 996 kg (198.5+996)*9.807 PA1 = P0 + (mp + mB)g/A = 101.325 + Ê0.1*1000 = 218.5 kPa mpg State 2A: PA2 = P0 + A = 120.82 kPa ; vA2 = Vtot/ mA= 0.5 m3/kg xA2 = (0.5 – 0.001047)/1.4183 = 0.352 ; T2 = 105°C uA2 = 440.0 + 0.352 · 2072.34 = 1169.5 kJ/kg

Continuity eq. in A: mA2 = mA1 Energy: mA(u2 – u1) = 1Q2 – 1W2 Process: P linear in V as mB is linear with V 1 1W2 = ? PdV = (218.5 + 120.82)(1 – 0.002) 2 = 169.32 kJ P 1

5.45 A rigid container has two rooms filled with water, each 1 m3 separated by a wall. Room A has P = 200 kPa with a quality x = 0.80. Room B has P = 2 MPa and T = 400°C. The partition wall is removed and the water comes to a uniform state which after a while due to heat transfer has a temperature of 200°C. Find the final pressure Continuity: m – m – m = 0 ; V = V + V = 2 m3 2 A1 B1 2 A B Energy: m u – m u – m u = Q – W = Q 2 2 A1 A1 A1 A1 1 2 1 2 1 2 State 1A: Table B.1.2 u = 504.47 + 0.8 · 2025.02 = 2124.47, A1 v = 0.001061 + 0.8 · 0.88467 = 0.70877 m = 1/v = 1.411 kg A1 A1 A1 State 1B: u = 2945.2, v = 0.1512 m = 1/v = 6.614 kg B1 B1 B1 B1 State 2: T , v = V /m = 2/(1.411 + 6.614) = 0.24924 m3/kg 2222 Table B.1.3 superheated vapor. 800 kPa < P < 1 MPa 2 0.24924-0.2608 P 800 + · 200 = 842 kPa u 2628.8 kJ/kg 2 0.20596-0.2608 2 Q = 8.025 x 2628.8 – 1.411 x 2124.47 – 6.614 x 2945.2 = - 1381 kJ 12

Solution: P State 1: v1 = V/m = 0.2 / 2 = 0.1 m3/kg , 2 , 3 x = 0.058 , u = 539.45 kJ/kg 11 3 Process: 1 ? 2 ? 3 or 1 ? 3′ 2 State at stops: 2 or 2′ P v2 = Vstop/m = 0.4 m3/kg & T2 = 600°C 11 Table B.1.3 Pstop = 1 MPa < P3 V V V since Pstop < P3 the process is as 1 ? 2 ? 3 stop 1

State 3: P3 = 1.2 MPa, v3 = v2 = 0.4 m3/kg T3 770°C; u = 3603.5 kJ/kg 3

11 W13 = W12 + W23 = (P1 + P2)(V2 – V1) + 0 = (100 + 1000)(0.8 – 0.2) 22 = 330 kJ Q = m(u – u ) + W = 2 x (3603.5 -539.45) + 330 = 6458 kJ 133113

5.48 Consider the piston/cylinder arrangement shown in Fig. P5.48. A frictionless piston is free to move between two sets of stops. When the piston rests on the lower stops, the enclosed volume is 400 L. When the piston reaches the upper stops, the volume is 600 L. The cylinder initially contains water at 100 kPa, 20% quality. It is heated until the water eventually exists as saturated vapor. The mass of the piston requires 300 kPa pressure to move it against the outside ambient pressure. Determine the final pressure in the cylinder, the heat transfer and the work for the overall process. 0.4 State 1: v = 0.001043 + 0.2×1.693 = 0.33964; m = V /v = = 1.178 kg 1 1 1 0.33964 u1 = 417.36 + 0.2 x 2088.7 = 835.1 kJ/kg 0.6 State 3: v3 = = 0.5095 < vG = 0.6058 at P3 = 300 kPa 1.178 Piston does reach upper stops.

5.49 Calculate the heat transfer for the process described in Problem 4.30. A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then further cooled to 20°C, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V.

Solution: Continuity: m = constant, Energy: m(u – u ) = Q – W 311313

5.50 A cylinder fitted with a frictionless piston that is restrained by a linear spring contains R-22 at 20°C, quality 60% with a volume of 8 L, shown in Fig. P5.50. The piston cross-sectional area is 0.04 m2, and the spring constant is 500 kN/m. A total of 62 kJ of heat is now added to the R-22. Verify that the final pressure is around 1600 kPa and Solution: Continuity: m = constant, Energy: m(u – u ) = Q – W 211212 R-22 State 1: 20°C ; x1 = 0.6 ; Table B.4.1 => P1 = 910 kPa v1 = 0.000824 – 0.6 x 0.02518 = 0.01593 m3/kg Q u1 = 67.92 + 0.6 x 164.92 = 166.87 kJ/kg V1 = 8 L => m = V1/v1 = 0.008/0.01593 = 0.502 kg Process: P linear in V with Ap = 0.04 m2 , ks = 500 kN/m to match P in kPa as: ks P2 = P1 + (V2 – V1) P Ap2 500 1600 2 = 910 + (mv2 – 0.008) 2 (0.04)2 ks /A 1 1W2 = ? Pdv = (1/2) (P2 + P1) (V2 – V1) 910 62 = 1Q2 = m(u2 – u1) + 1W2 V

Now we want to find the final P such that the process gives the stated Q of 62 kJ. Assume P2 = 1600 kPa then the final state is 1600Ê-Ê910 V2 = 0.008 + 500 x 0.042 = 0.0102 m 3

v2 = 0.0102/0.502 = 0.02033 T2Ê=Ê106.4°CÊÊh2Ê=Ê318.1 State 2: At P , v Table B.4.2 ? u2Ê=Êh2Ê-ÊP2v2Ê=Ê285.5 22

910Ê+Ê1600 1W2 = (0.0102 – 0.008) = 2.76 kJ ?2? 1Q2 = 0.502 x (285.5 – 166.87) + 2.76 = 62.3 kJ = 62 OK

5.51 A 1-L capsule of water at 700 kPa, 150°C is placed in a larger insulated and otherwise evacuated vessel. The capsule breaks and its contents fill the entire volume. If the final Continuity: m2 = m1 = m = V/v1 = 0.916 kg Process: expansion with 1Q2 = 0/ , 1W2 = 0/ Energy: m(u2 – u1) = 1Q2 – 1W2 = 0/ u2 = u1 State 1: v1 vf = 0.001091 m3/kg; u1 uf = 631.66 kJ/kg State 2: P2 , u2 x2 = (631.66 – 444.16)/2069.3 = 0.09061 v2 = 0.001048 + 0.09061 · 1.37385 = 0.1255 m3/kg V2 = mv2 = 0.916 x 0.1255 = 0.115 m3 = 115 L

5.53 Superheated refrigerant R-134a at 20°C, 0.5 MPa is cooled in a piston/cylinder arrangement at constant temperature to a final two-phase state with quality of 50%. The refrigerant mass is 5 kg, and during this process 500 kJ of heat is removed. Find Continuity: m2 = m1 = m ; Energy: m(u2 -u1) = 1Q2 – 1W2 = -500 – 1W2 State 1: T1 , P1 Table B.5.2, v1 = 0.04226 m3/kg => V1 = mv1 = 0.211 m 3

u1 = h1 -P1v1 = 411.65 – 500 · 0.04226 = 390.52 kJ/kg State 2: T2 , x2 u2 = 227.03 + 0.5 · 162.16 = 308.11 kJ/kg, v = 0.000817 + 0.5 · 0.03524 = 0.018437 m3/kg => V = mv = 0.0922 m3 2 22 1W2 = -500 – m(u2 – u1) = -500 – 5 x (308.11 – 390.52) = -87.9 kJ

5.55 Calculate the heat transfer for the process described in Problem 4.26. A piston cylinder setup similar to Problem 4.24 contains 0.1 kg saturated liquid and vapor water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 500 kPa will float it. The water is heated to 300°C. Find the final pressure, volume and the 12 Solution: Take CV as the water: m2 = m1 = m P 1a 2 Energy: m(u2 -u1) = 1Q2 – 1W2 P lift Process: v = constant until P = Plift To locate state 1: Table B.1.2 P1 1 V v1 = 0.001043 + 0.25· 1.69296 = 0.42428 m3/kg u1 = 417.33 + 0.25· 2088.7 = 939.5 kJ/kg 1a: v1a = v1 = 0.42428 > vg at 500 kPa so state 1a is sup.vapor T1a = 200°C State 2 is 300°C so heating continues after state 1a to 2 at constant P => 2: T2, P2 = Plift => Tbl B.1.3 v2 =0.52256 m3/kg; u2 = 2802.9 kJ/kg 1W2 = Plift m(v2 – v1) = 500 · 0.1 (0.5226 – 0.4243) = 4.91 kJ 1Q2 = m(u2 – u1) + 1W2 = 0.1(2802.9 – 939.5) + 4.91 = 191.25 kJ

5.56 A piston/cylinder, shown in Fig. P5.56, contains R-12 at – 30°C, x = 20%. The volume is 0.2 m3. It is known that Vstop = 0.4 m3, and if the piston sits at the bottom, the spring force balances the other loads on the piston. It is now heated up to 20°C. Find the mass of the fluid and show the P–v diagram. Find the work and heat transfer. Solution: C.V. R-12, this is a control mass. Properties in Table B.3 State 1: v1 = 0.000672 + 0.2*0.1587 = 0.0324 m3/kg P T -5° u1 = 8.79 + 0.2 · 149.4 = 38.67 kJ/kg 2P 1

Continuity Eq.: m2 = m1 = V1/v1 = 6.17 kg, 1 P Energy: E2 – E1 = m(u2 – u1) = 1Q2 – 1W2 1

5.57 Ammonia, NH3, is contained in a sealed rigid tank at 0°C, x = 50% and is then heated to 100°C. Find the final state P2, u2 and the specific work and heat transfer. Cont.: m2 = m1 ; Energy: E2 – E1 = 1Q2 ; (1W2Ê=Ê0/ ) Process: V2 = V1 v2 = v1 = 0.001566 + 0.5 * 0.28783 = 0.14538 Table B.2.2: v2 & T2 between 1000 kPa and 1200 kPa

P 2 P2 = 1187 kPa x2 = undef; h2 = 1658.4 kJ/kg u2 = 1658.4 – 1187 x0.14538 = 1485.83 kJ/kg 1 v u1 = 179.69 + 0.5 · 1138.3 = 748.84 kJ/kg

Ê1w2 = 0 ; 1q2 = u2 – u1 = 1485.83 – 748.84 = 737 kJ/kg 5.58 A house is being designed to use a thick concrete floor mass as thermal storage material for solar energy heating. The concrete is 30 cm thick and the area exposed to the sun during the day time is 4 m · 6 m. It is expected that this mass will undergo an average temperature rise of about 3°C during the day. How much energy will be Concrete V = 4 x 6 x 0.3 = 7.2 m3 ; m = V = 2200x 7.2 = 15 840 kg U = m C T = 15840 x 0.88 x 3 = 41818 kJ = 41.82 MJ 5.59 A car with mass 1275 kg drives at 60 km/h when the brakes are applied quickly to decrease its speed to 20 km/h. Assume the brake pads are 0.5 kg mass with heat capacity of 1.1 kJ/kg K and the brake discs/drums are 4.0 kg steel where both masses are heated uniformly. Find the temperature increase in the brake assembly. 122 m = constant; E2 – E1 = 0 – 0 = mcar (V2 – V1) + mbrake(u2 – u1) 2 The brake system mass is two different kinds so split it, also use Cv since we do not have a u table for steel or brake pad material.

5.60 A copper block of volume 1 L is heat treated at 500°C and now cooled in a 200-L oil bath initially at 20°C, shown in Fig. P5.60. Assuming no heat transfer with the Solution: mmet = V = 0.001×8300 = 8.3kg, moil = V = 0.2×910 = 182kg mmet(u2 – u1)met + moil(u2 – u1)oil = 1Q2 – 1W2 = 0 solid and liquid: u CV T mmetCVmet(T2 – T1,met) + moilCVoil(T2 – T1,oil) = 0 8.3 x 0.42(T2 -500) + 182 x 1.8 (T2 -20) = 0 331.09 T2 – 1743 – 6552 = 0 T2 = 25 °C

5.61 Saturated, x = 1%, water at 25°C is contained in a hollow spherical aluminum vessel with inside diameter of 0.5 m and a 1-cm thick wall. The vessel is heated until the water inside is saturated vapor. Considering the vessel and water together as a control m2 = m1 ; U2 – U1 = 1Q2 -1W2 = 1Q2 State 1: v1 = 0.001003 + 0.01 x 43.359 = 0.4346 m3/kg u1 = 104.88 + 0.01 x 2304.9 = 127.9 kJ/kg State 2: x2 = 1 and constant volume so v2 = v1 = V/m vgÊT2 = v1 = 0.4346 => T2 = 146.1°C; u2 = uG2 = 2555.9 0.06545 VINSIDE = 6 (0.5)3 = 0.06545 m ; mH2O = 0.4346 = 0.1506 kg 3

5.62 An ideal gas is heated from 500 to 1500 K. Find the change in enthalpy using constant specific heat from Table A.5 (room temperature value) and discuss the accuracy of the result if the gas is a. Argon b. Oxygen c. Carbon dioxide Solution: T1 = 500 K, T2 = 1500 K, h = CP0(T2-T1) a) Ar : h = 0.520(1500-500) = 520 kJ/kg b) O2 : h = 0.922(1500-500) = 922 kJ/kg c) CO2: h = 0.842(1500-500) = 842 kJ/kg Polyatomic gas heat capacity changes, see figure 5.11 5.63 A rigid insulated tank is separated into two rooms by a stiff plate. Room A of 0.5 m3 contains air at 250 kPa, 300 K and room B of 1 m3 has air at 150 kPa, 1000 K. The plate is removed and the air comes to a uniform state without any heat transfer. Find the final pressure and temperature.

5.64 An insulated cylinder is divided into two parts of 1 m3 each by an initially locked piston, as shown in Fig. P5.64. Side A has air at 200 kPa, 300 K, and side B has air at 1.0 MPa, 1000 K. The piston is now unlocked so it is free to move, and it conducts heat so the air comes to a uniform temperature TA = TB. Find the mass in both A and C.V. A + B Force balance on piston: PAA = PBA State 1A: Table A.7 uA1 = 214.364 kJ/kg, mA = PA1VA1/RTA1 = 200 x 1/(0.287 x 300) = 2.323 kg State 1B: Table A.7 uB1 = 759.189 kJ/kg, mB = PB1VB1/RTB1 = 1000 x 1/(0.287 x 1000) = 3.484 kg For chosen C.V. 1Q2 = 0 , 1W2 = 0 mA(u2 – u1)A + mB(u2 – u1)B = 0 (mA + mB)u2 = mAuA1 + mBuB1 = 2.323 x 214.364 + 3.484 x 759.189 = 3143 kJ u2 = 3143/(3.484 + 2.323) = 541.24 kJ/kg T2 = 736 K P = (mA + mB)RT2/Vtot = 5.807 x 0.287 x 736 / 2 = 613 kPa

5.65 A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa, 400°C. It is cooled to 40°C, at which point the pressure is 300 kPa. Solution: P 1 Linear spring gives 21 1W2 = ? PdV = (P1 + P2)(V2 – V1) 2 v 1Q2 = m(u2 – u1) + 1W2 Equation of state: PV = mRT State 1: V1 = mRT1/P1 = 2 x 0.18892 x 673.15 /500 = 0.5087 m3 State 2: V2 = mRT2/P2 = 2 x 0.18892 x 313.15 /300 = 0.3944 m3 1 1W2 = (500 + 300)(0.3944 – 0.5087) = -45.72 kJ 2

5.66 A piston/cylinder in a car contains 0.2 L of air at 90 kPa, 20°C, shown in Fig. P5.66. The air is compressed in a quasi-equilibrium polytropic process with polytropic exponent n = 1.25 to a final volume six times smaller. Determine the final pressure, Solution: Continuty: m2 = m1 Energy: m(u2 – u1) = 1Q2 – 1W2 Process: Pvn = const.

P v n = P v n P = P (v /v )n = 90 x 61.25 = 845.15 kPa 11 22 2 112 Substance ideal gas: Pv = RT T2 = T1(P2v2/P1v1) = 293.15(845.15/90 x 6) = 458.8 K

5.67 Water at 20°C, 100 kPa, is brought to 200 kPa, 1500°C. Find the change in the specific internal energy, using the water table and the ideal gas water table in Solution: State 1: Table B.1.1 u1 uf = 83.95 kJ/kg State 2: Highest T in Table B.1.3 is 1300°C Using a u from the ideal gas tables, A.8, we get — h(1500°C) – h(1300°C) = 61367.7 – 51629.5 = 9738.2 kJ/kmol – u1500 – u1300 = h/M – R(1500 – 1300) = 540.56 – 92.3 = 448.26 kJ/kg Since the ideal gas change is at low P we use 1300°C, lowest P available 10 kPa u2 – u1 = (u2 – ux)ID.G. + (ux – u1) = 448.26 + 4683.7 – 83.95 = 5048 kJ/kg 5.68 For an application the change in enthalpy of carbon dioxide from 30 to 1500°C at 100 kPa is needed. Consider the following methods and indicate the most accurate one. b. Constant specific heat, value at average temperature from the equation in Table A.6. Solution: a) h = Cp T = 0.842 (1500 – 30) = 1237.7 kJ/kg b) Tave = 1038.2 K ; = T/100 = 10.382 Table A.6 — Cp = 54.64 Cp = Cp/M = 1.2415 h = Cp,ave T = 1.2415 x 1470 = 1825 kJ/kg c) For the entry to Table A.6: 2 = 17.7315 ; 1 = 3.0315

100 – ? CpdT = M ? pd h= C 100 2 1.5 – 1.5) = 44.01 [-3.7357( 2 – 1) + 3 x 30.529( 2 1 11 – 4.1034 x ( 2 – 2) + 0.024198 x ( 3 – 3)] = 1762.76 kJ/kg 221 321 d) h = (77833 – 189)/44.01 = 1764.3 kJ/kg The result in d) is best, very similar to c). For large T or small T at high Tave a) is very poor.

5.69 Air in a piston/cylinder at 200 kPa, 600 K, is expanded in a constant-pressure process to twice the initial volume (state 2), shown in Fig. P5.69. The piston is then locked with a pin and heat is transferred to a final temperature of 600 K. Find P, T, and h for C.V. Air. Control mass m2 = m3 = m1 1 => 2: u2 – u1 = 1q2 -1w2 ; 1w2 = P dv = P1(v2 -v1) = R(T2 -T1) Ideal gas Pv = RT T2 = T1v2/v1 = 2T1 = 1200 K P2 = P1 = 200 kPa, 1w2 = RT1 = 172.2 kJ/kg Table A.7 h2 = 1277.8 kJ/kg, h3 = h1 = 607.3 kJ/kg 1q2 = u2 – u1 + 1w2 = h2 – h1 = 1277.8 – 607.3 = 670.5 kJ/kg 2? 3: v3 = v2 = 2v1 2w3 = 0, P3 = P2T3/T2 = P1T1/2T1 = P1/2 = 100 kPa 2q3 = u3 – u2 = 435.1 – 933.4 = -498.3 kJ/kg

5.70 An insulated floating piston divides a cylinder into two volumes each of 1 m3, as shown in Fig. P5.70. One contains water at 100°C and the other air at -3°C and both pressures are 200 kPa. A line with a safety valve that opens at 400 kPa is attached to the water side of the cylinder. Assume no heat transfer to the water and that the water is incompressible. Show possible air states in a P–v diagram, and find the air temperature when the safety valve opens. How much heat transfer is needed to bring Solution: C.V. air: CONT: m3 = m2 = m1; ENERGY: mair(u3 – u1) = 1Q3 – 1W3

1: (T,P,V) Ideal gas Table A.5 and A.7 mair = P1V1/RT1 P = 200*1/(0.287*270.15) = 2.578 kg 400 2 2: P2 = 400 kPa, v2 = v1 1 540 P => T2 = T1P2/P1 = 2T1 = 540.3 K 1 v 270 C.V. H2O: It is compressed liquid v1 2v1 2: no H2O out no change in vH O 2

=> no work T2 = T1 no Q 3: to fix it find 2a: T2a = T1P2aV2a/P1V1 = 4T1 = 1080 K < T3 so V3 = V2a 3 1300 K 2a 1080 K

5.71 Two containers are filled with air, one a rigid tank A, and the other a piston/cylinder B that is connected to A by a line and valve, as shown in Fig. P5.71. The initial conditions are: mA = 2 kg, TA = 600 K, PA = 500 kPa and VB = 0.5 m3, TB = 27°C, PB = 200 kPa. The piston in B is loaded with the outside atmosphere and the piston mass in the standard gravitational field. The valve is now opened, and the air comes to a uniform condition in both volumes. Assuming no heat transfer, find the initial mass in B, the volume of tank A, the final pressure and temperature and the work, 1W2. Cont.: m2 = m1 = mA1 + mB1 Energy: m2u2 – mA1uA1 – mB1uB1 = -1W2 ; 1W2 = PB1(V2 – V1) System: PB = const = PB1 = P2 ; Substance: PV = mRT mB1 = PB1VB1/RTB1 = 1.161 kg ; VA = mA1RTA1/PA1 = 0.6888 m3 P2 = PB1 = 200 kPa ; A.7: uA1 = 434.8, uB1 = 214.09 kJ/kg m2u2 + P2V2 = mA1uA1 + mB1uB1 + PB1V1 = m2h2 = 1355.92 kJ h2 = 428.95 kJ/kg T2 = 427.7 K V2 = mtotRT2/P2 = 1.94 m3 1W2 = 200 · (1.94 – 1.1888) = 150.25 kJ

5.73 A piston/cylinder arrangement, shown in Fig. P5.73, contains 10 g of air at 250 kPa, 300°C. The 75-kg piston has a diameter of 0.1 m and initially pushes against the stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools to 20°C as heat is Determine if piston will drop. So a force balance to float the piston gives: mpg 75ÊxÊ9.80665 Pfloat = P0 + = 100 + 2 = 193.6 kPa A ÊxÊ0.1 ÊxÊ0.25ÊxÊ1000 If air is cooled to T2 at constant volume P2 = P1T2/T1 = 250 x 293.15/573.15 = 127.9 kPa < Pfloat State 2: T2, P2 = Pfloat State 1: V1 = mRT1/ P1 = 0.010 x 0.287 x 573.15 / 250 = 0.00658 m3 V1T2P1 0.00658ÊxÊ293.15ÊxÊ250 Ideal gas V2 = = = 0.00434 m3 P2T1 193.65ÊxÊ573.15 1W2 = P dV = Pfloat(V2 - V1) = 193.65(0.00434 – 0.00658) = -0.434 kJ 1Q2 = m(u2 - u1) + 1W2 mCv(T2 - T1) + 1W2 = 0.1 x 0.717 x (20 - 300) - 0.434 = -2.44 kJ

5.74 Oxygen at 300 kPa, 100°C is in a piston/cylinder arrangement with a volume of 0.1 m3. It is now compressed in a polytropic process with exponent, n = 1.2, to a final temperature of 200°C. Calculate the heat transfer for the process.

Continuty: m2 = m1 Energy: m(u2 – u1) = 1Q2 – 1W2 State 1: T1 , P1 & ideal gas, small change in T, so use Table A.5 P1V1 300ÊxÊ0.1Êm 3 m = = = 0.3094 kg RT1 0.25983ÊxÊ373.15 Process: PVn = constant 1 mR 0.3094ÊxÊ0.25983 1W2 = (P2V2 – P1V1) = (T2 – T1) = (200 – 100) 1-n 1-n 1Ê-Ê1.2 = -40.196 kJ 1Q2 = m(u2 – u1) + 1W2 mCv(T2 – T1) + 1W2 = 0.3094 x 0.662 (200 – 100) – 40.196 = -19.72 kJ P P = C v -1.2 T 2 T = C v-0.2 T22 1 T1 v 1

5.75 A piston/cylinder contains 2 kg of air at 27°C, 200 kPa, shown in Fig. P5.75. The piston is loaded with a linear spring, mass and the atmosphere. Stops are mounted so that Vstop = 3 m3, at which point P = 600 kPa is required to balance the piston forces. The air is now heated to a final pressure of 400 kPa. Find the final temperature, volume and the work and heat transfer. Find the work done on the spring.

5.76 A piston/cylinder contains 0.001 m3 air at 300 K, 150 kPa. The air is now compressed in a process in which P V1.25 = C to a final pressure of 600 kPa. Find the work Solution: Continuty: m2 = m1 Energy: m(u2 – u1) = 1Q2 – 1W2 Process : PV1.25 = const.

State 2: V2 = V1 ( P1/P2 )1.25= 0.00033 m 3 600ÊxÊ0.00033 T2 = T1 P2V2/(P1V1) = 300 = 395.85 K 150ÊxÊ0.001 11 1W2 = (P2 V2 – P1V1) = (600 x 0.00033 – 150 x 0.001) = – 0.192 kJ n-1 n-1 P1V1 1Q2 = m(u2 – u1) + 1W2 = Cv (T 2–T1) +1W2 RT1 = 0.001742 x 0.717x 95.85 – 0.192 = – 0.072 kJ 5.77 An air pistol contains compressed air in a small cylinder, shown in Fig. P5.77. Assume that the volume is 1 cm3, pressure is 1 MPa, and the temperature is 27°C when armed. A bullet, m = 15 g, acts as a piston initially held by a pin (trigger); when released, the air expands in an isothermal process (T = constant). If the air pressure is 0.1 MPa in the cylinder as the bullet leaves the gun, find Solution: C.V. Air.

Air ideal gas: mair = P V /RT = 1000 x 10-6/(0.287 x 300) = 1.17×10-5 kg 11 1 Process: PV = const = P1V1 = P2V2 V2 = V1P1/P2 = 10 cm3

P1V1 W = ? PdV = dV = P V ln (V /V ) = 2.32 J 12 ?V 11 21 1W2,ATM = P0(V2 – V1) = 101 x (10-1) x 10- kJ = 0.909 J 6

5.78 A spherical elastic balloon contains nitrogen (N2) at 20oC, 500 kPa. The initial volume is 0.5 m3. The balloon material is such that the pressure inside is proportional to the balloon diameter. Heat is now transferred to the balloon until its volume reaches a) Can the nitrogen be assumed to behave as an ideal gas throughout this process? b) Calculate the heat transferred to the nitrogen.

Solution: Continuty: m2 = m1 Energy: m(u2 – u1) = 1Q2 – 1W2 Process: P D V1/3 => PV-1/3 = constant. Polytropic process n = -1/3. State 1: 20oC = 293.2 K, 500 kPa, Table A.2: TC = 126.2 K, PC = 3.39 MPa V1 = 0.5 m3 => m = P1V1/RT1 = 500 x 0.5/(0.2968 x 293.15) = 2.873 kg Process => P = P [ V / V ]1/3 = 500 [1.0 / 0.5]1/3 = 630 kPa 2121 P2ÊV2 630ÊxÊ1.0 From ideal gas law: T2 = T1 = 293.15 x = 738.7 K P1ÊV1 500ÊxÊ0.5 2 P2ÊV2Ê-ÊP1ÊV1 630ÊxÊ1.0Ê-Ê500ÊxÊ0.5 1W2 = P dV = = = 285 kJ 1 1-Ên 1Ê-Ê(-1/3) 1Q2 = m(u2 – u1) + 1W2 = m CVo (T2 – T1) + 1W2 = 2.873 x 0.745 (738.7 – 293.2) + 285 = 1238.6 kJ

5.79 A 10-m high cylinder, cross-sectional area 0.1 m2, has a massless piston at the bottom with water at 20°C on top of it, shown in Fig. P5.79. Air at 300 K, volume 0.3 m3, under the piston is heated so that the piston moves up, spilling the water out over the side. Find the total heat transfer to the air when all the water has been pushed out. Solution: The water on top is compressed liquid and has volume and mass VH2O = Vtot – Vair = 10 x 0.1 – 0.3 = 0.7 m3 mH2O = VH2O/vf = 0.7 / 0.001002 = 698.6 kg The initial air pressure is then 698.6ÊxÊ9.807 P1 = P0 + mH Og/A = 101.325 + = 169.84 kPa 2 0.1ÊxÊ1000

169.84ÊxÊ0.3 Pair and then mair = PV/RT = = 0.592 kg 0.287ÊxÊ300 1 State 2: No liquid water over the piston so 2 V P2 = P0 + 0/ = 101.325 kPa, V2 = 10×0.1 = 1 m3

T1P2V2 300×101.325×1 State 2: P2, V2 T2 = = = 596.59 K P1V1 169.84×0.3 The process line shows the work as an area 1 1W2 = ? PdV = (P1 + P2)(V2 – V1) 2 1 = (169.84 + 101.325)(1 – 0.3) = 94.91 kJ 2 The energy equation solved for the heat transfer becomes 1Q2 = m(u2 – u1) + 1W2 mCv(T2 – T1) + 1W2 = 0.592 x 0.717 x (596.59 – 300) + 94.91 = 220.7 kJ

5.80 A cylinder fitted with a frictionless piston contains carbon dioxide at 500 kPa, 400 K, at which point the volume is 50 L. The gas is now allowed to expand until the piston reaches a set of fixed stops at 150 L cylinder volume. This process is polytropic, with the polytropic exponent n equal to 1.20. Additional heat is now transferred to the gas, until the final temperature reaches 500 K. Determine Solution: Continuity: m3 = m2 = m1; Energy: m (u3 – u1) = 1Q3 – 1W3 Process 2 – 3: Constant volume V3 = V2 => 2W3= 0 State 1: 400 K, 500 kPa, Ideal gas Table A.5, R = 0.1889 500ÊxÊ0.05 V1 = 50 L => m = P1V1/RT1 = = 0.331 kg 0.1889ÊxÊ400 State 2: Polytropic expansion to stops at V2 = 150 L P = P1 · (V /V )n = 500 · (50 / 150)1.2 = 133.8 kPa 2 12 State 3: Add Q to T3 = 500 K, constant volume V3 = V2 V1 T3 50 500 P3 = P1 · · = 500 · · = 208.3 kPa V3 T1 150 400 1W3 = 1W2 + 2W3= 1W2 + 0 = P dV P2ÊV2Ê-ÊP1ÊV1 133.8ÊxÊ0.15Ê-Ê500ÊxÊ0.05 = = = +24.7 kJ 1-Ên 1Ê-Ê1.2 1Q3 = m(u3 – u1) + 1W3 = m CVo (T3 – T1) + 1W3 = 0.331 x 0.653 (500 – 400) + 24.7 = 21.6 + 24.7 = +46.3 kJ

5.81 A cylinder fitted with a frictionless piston contains R-134a at 40oC, 80% quality, at which point the volume is 10 L. The external force on the piston is now varied in such a manner that the R-134a slowly expands in a polytropic process to 400 kPa, 20oC. Calculate the work and the heat transfer for this process.

Process: PVn = constant => P1V1 = P2V2 nn State 1: (T, x) Table B.5.1 => P1 = Pg = 1017 kPa v1 = 0.000873 + 0.8 · 0.019147 = 0.01619 m3/kg u1 = 255.65 + 0.8 · 143.81 = 370.7 kJ/kg m = V1/v1 = 0.010/0.01619 = 0.618 kg State 2: (P2, T2) Table B.5.2 v2 = 0.05436 m3/kg, h2 = 414.0 kJ/kg u2 = h2 – P2v2 = 414.0 – 400 x 0.05436 = 392.3 kJ/kg V2 = mv2 = 0.618 x 0.05436 = 0.0336 m3 = 33.6 L P1 V2 1017 33.6 0.93315 Process => n = ln / ln = ln / ln = = 0.77 P2 V1 400 10 1.21194 P2ÊV2Ê-ÊP1ÊV1 400ÊxÊ0.0336Ê-Ê1017ÊxÊ0.010 1W2 = P dV = = = +14.2 kJ 1-Ên 1Ê-Ê077 1Q2 = m(u2 – u1) + 1W2 = 0.618 (392.3 – 370.6) + 14.2 = 13.4 + 14.2 = 27.6 kJ

5.83 Water at 150°C, quality 50% is contained in a cylinder/piston arrangement with initial volume 0.05 m3. The loading of the piston is such that the inside pressure is linear with the square root of volume as P = 100 + CV 0.5 kPa. Now heat is transferred to the cylinder to a final pressure of 600 kPa. Find the heat transfer in the process. Continuty: m2 = m1 Energy: m(u2 – u1) = 1Q2 – 1W2 State 1: v1 = 0.1969, u1 = 1595.6 kJ/kg m = V/v1 = 0.254 kg Process equation P1 – 100 = CV11/2 so (V2/V1)1/2 = (P2 – 100)/(P1 – 100) ØP Ê-Ê100 2 Ø 500 2 2Ø Ø V2 = V1 x ? œ = 0.05 x ? œ = 0.0885 ºP1Ê-Ê100ß º475.8Ê-Ê100ß W = ? PdV = ? (100Ê+ÊCV1/2)dV = 100x(V – V ) + 2 C(V 1.5 – V 1.5) 12 21321 = 100(V2 – V1)(1 – 2/3) + (2/3)(P2V2 – P1V1) 1W2 = 100 (0.0885-0.05)/3 + 2 (600 x 0.0885-475.8 x 0.05)/3 = 20.82 kJ State 2: P2, v2 = V2/m = 0.3484 u2 = 2631.9 kJ/kg, T2 196°C 1Q2 = 0.254 x (2631.9 – 1595.6) + 20.82 = 284 kJ

5.84 A piston/cylinder has 1 kg propane gas at 700 kPa, 40°C. The piston cross-sectional area is 0.5 m2, and the total external force restraining the piston is directly proportional to the cylinder volume squared. Heat is transferred to the propane until its temperature reaches 700°C. Determine the final pressure inside the cylinder, the work done by the propane, and the heat transfer during the process.

Process: P = Pext = CV2 PV-2 = const, n = -2 Ideal gas: PV = mRT, and process yields n 700+273.15 2/3 P2 = P1(T2/T1)n-1 = 700 = 1490.7 kPa ?40+273.15 ? ÊÊ2 P2V2Ê-ÊP1V1 mR(T2Ê-ÊT1) 1W2 = ? ÊPdV = = 1Ê-Ên 1Ê-Ên ÊÊ1 1· 0.18855· (700 – 40) = = 41.48 kJ 1- (-2) 1Q2 = m(u2 – u1) + 1W2 = mCv(T2 – T1) + 1W2 = 1 x 1.490 x (700 – 40) + 41.48 = 1024.9 kJ

5.85 A closed cylinder is divided into two rooms by a frictionless piston held in place by a pin, as shown in Fig. P5.85. Room A has 10 L air at 100 kPa, 30°C, and room B has 300 L saturated water vapor at 30°C. The pin is pulled, releasing the piston, and both rooms come to equilibrium at 30°C and as the water is compressed it becomes two- phase. Considering a control mass of the air and water, determine the work done by P2 = PGÊH OÊatÊ30°C = PA2 = PB2 = 4.246 kPa 2

Air, I.G.:PA1VA1 = mARAT = PA2VA2 = PGÊH OÊatÊ30°CVA2 2 100ÊxÊ0.01 ? VA2 = 4.246 m3 = 0.2355 m 3

VB2 = VA1 + VB1 – VA2 = 0.30 + 0.01 – 0.2355 = 0.0745 m3 VB1 0.3 = 9.121×10-3 mB = = kg => vB2 = 8.166 m3/kg vB1 32.89 8.166 = 0.001004 + xB2 x (32.89 – 0.001) xB2 = 0.2483 System A+B: W = 0; UA = 0 ( IG & T = 0 ) uB2 = 125.78 + 0.2483 x 2290.8 = 694.5, uB1 = 2416.6 kJ/kg 1Q2 = 9.121×10-3(694.5 – 2416.6) = -15.7 kJ

5.86 A small elevator is being designed for a construction site. It is expected to carry four 75-kg workers to the top of a 100-m tall building in less than 2 min. The elevator cage will have a counterweight to balance its mass. What is the smallest size (power) m = 4 x 75 = 300 kg ; Z = 100 m ; t = 2 minutes . . Z 300ÊxÊ9.807ÊxÊ100 -W = PE = mg = = 2.45 kW t 1000ÊxÊ2ÊxÊ60

5.88 Consider the 100-L Dewar (a rigid double-walled vessel for storing cryogenic liquids) shown in Fig. P5.88. The Dewar contains nitrogen at 1 atm, 90% liquid and 10% vapor by volume. The insulation holds heat transfer into the Dewar from the ambient to a very low rate, 5 J/s. The vent valve is accidentally closed so that the pressure inside slowly rises. How long time will it take to reach a pressure of 500 kPa?

State 1: T1 = 77.3 K, Vliq1 = 0.9 V, Vvap1 = 0.1 V, Table B.6.1: v1f = 0.00124 m3/kg, v1g = 0.21639 m3/kg, 0.9ÊxÊ0.1 0.1ÊxÊ0.1 mliq1 0.00124 mvap1 0.21639 = = 72.5806 kg; = = 0.0462 kg mtotÊ = mliq1 + mvap1 = 72.6268 kg; x1 = 0.0462 / 72.6268 = 0.000636 u1 = -122.27 + 0.000636×177.04 = -122.16 kJ/kg

v1 = 0.1/ 72.6268 = 0.001377 m3/kg P Process: v2 = v1 vf at T 93.4 K, Table B.6.1 Pg = 483 kPa 483 293.4 K 1v

5.89 A computer in a closed room of volume 200 m3 dissipates energy at a rate of 10 kW. The room has 50 kg wood, 25 kg steel and air, with all material at 300 K, 100 kPa. Assuming all the mass heats up uniformly how long time will it take to increase the temperature 10°C?

· C.V. Air, wood and steel. m2 = m1 ; U2 – U1 = 1Q2 = Q t The total volume is nearly all air, but we can find volume of the solids. V = m/ = 50/510 = 0.098 m3 ; V = 25/7820 = 0.003 m3 wood steel Vair = 200 – 0.098 – 0.003 = 199.899 m3 mair = PV/RT = 101.325 x 199.899/(0.287 x 300) = 235.25 kg We do not have a u table for steel or wood so use heat capacity.

U = [mair Cv + mwood Cv + msteel Cv ] T = (235.25 x 0.717 + 50 x 1.38 + 25 x 0.46) 10 · = 1686.7 + 690 +115 = 2492 kJ = Q x t = 10* t => t = 2492/10 = 249.2 sec = 4.2 minutes 5.90 The heaters in a spacecraft suddenly fail. Heat is lost by radiation at the rate of 100 kJ/h, and the electric instruments generate 75 kJ/h. Initially, the air is at 100 kPa, 25°C with a volume of 10 m3. How long will it take to reach an air temperature of -20°C?

5.93 A spherical balloon initially 150 mm in diameter and containing R-12 at 100 kPa is connected to a 30-L uninsulated, rigid tank containing R-12 at 500 kPa. Everything is at the ambient temperature of 20°C. A valve connecting the tank and balloon is opened slightly and remains so until the pressures equalize. During this process heat is exchanged so the temperature remains constant at 20°C and the pressure inside the balloon is proportional to the diameter at any time. Calculate the final pressure and the C.V.: balloon A + tank B State A1: Table B.3.2; vA1 = 0.19728, u = 203.85 -100*0.19728 = 184.12 kJ/kg VA1 = 6(0.15)3 = 0.001 767 m ; mA1 = 0.001767/0.19728 = 0.009 kg 3

5.94 Calculate the heat transfer for the process described in Problem 4.44. Two springs with same spring constant are installed in a massless piston/cylinder with the outside air at 100 kPa. If the piston is at the bottom, both springs are relaxed and 3 the second spring comes in contact with the piston at V = 2 m . The cylinder (Fig. 3 P4.44) contains ammonia initially at -2°C, x = 0.13, V = 1 m , which is then heated until the pressure finally reaches 1200 kPa. At what pressure will the piston touch the second spring? Find the final temperature and the total work done by the ammonia. Solution : P 3 State 1: P = 399.7 kPa Table B.2.1 v = 0.00156 + 0.13 x 0.3106 = 0.0419 2 u = 170.52 + 0.13 x 1145.78 = 319.47 1 m = V/v = 1/0.0419 = 23.866 kg 0

P 1W2 2W3 V At bottom state 0: 0 m3, 100 kPa 0

0 1 2 V State 2: V = 2 m3 and on line 0-1-2 3

Slope of line 0-1-2: P/ V = (P – P )/ V = (399.7-100)/1 = 299.7 kPa/ m3 10 P2 = P1 + (V2 – V1) P/ V = 399.7 + (2-1) x 299.7 = 699.4 kPa P3 = P2 + (V3 – V2)2 P/ V V3 = V2+ (P3 – P2)/(2 P/ V) V3 = 2 + (1200-699.4)/599.4 = 2.835 m3 v3 = v1V3/V1 = 0.0419 x 2.835/1 = 0.1188 m3/kg T = 51°C u3 = h3 – P3v3 = 1527.92 – 1200 x 0.1188 = 1385 kJ/kg 11 1W3 = 1W2 + 2W3 = (P1 + P2)(V2 – V1) + (P3 + P2)(V3 – V2) 22 = 549.6 + 793.0 = 1342.6 kJ 1Q3 = m(u3-u1) + 1W3 = 23.866 x (1385 -319.47) + 1342.6 = 26773 kJ

5.96 A cylinder fitted with a frictionless piston contains R-134a at 10oC, quality of 50%, and initial volume of 100 L. The external force on the piston now varies in such a manner that the piston moves, increasing the volume. It is noted that the temperature is 25oC when the last drop of liquid R-134a evaporates. The process continues to a final state of 40oC, 600 kPa. Assume the pressure is piecewise linear in volme and determine the final volume in the cylinder and the work and heat transfer for the Solution: State 1: Table B.5.1 (10oC, x1 = 0.50) P = P gÊ10C = 415.8 kPa v1 = 0.000794 + 0.5(0.04866) = 0.02512 m3/kg u1 = 213.25 + 0.5 x 170.42 = 298.46 kJ/kg V1 = 0.1 m3 => m = V1/v1 = 0.1/0.02512 = 3.981 kg State 2: 25oC, x2 = 1.0: P = P gÊ25C = 666.3 kPa v = v = 0.03098 m3/kg => V = mv = 0.1233 m3 2 gÊ25C 2 2 State 3: Table B.5.2 (40oC, 600kPa) v = 0.03796 m3/kg => V = mv = 0.1511 m3 3 33 u3 = h3 – P3v3 = 428.88 – 600*0.03796 = 406.11 kJ/kg T 666 kPa P 600 kPa 40 25 10 2 666 3 600 416 kPa 416 23

5.98 A cylinder fitted with a frictionless piston contains 0.2 kg of saturated (both liquid and vapor present) R-12 at -20oC. The external force on the piston is such that the pressure inside the cylinder is related to the volume by the expression: P = – 47.5 + 4.0 x V1.5 , kPa and L Heat is now transferred to the cylinder until the pressure inside reaches 250 kPa. Calculate the work and heat transfer.

5.99 A certain elastic balloon will support an internal pressure equal to Po = 100 kPa until the balloon becomes spherical at a diameter of Do = 1 m, beyond which P = Po + C(1-x6)x ; x = Do/D because of the offsetting effects of balloon curvature and elasticity. This balloon contains helium gas at 250 K, 100 kPa, with a 0.4 m3 volume. The balloon is heated until the volume reaches 2 m3. During the process the maximum pressure inside the a. What is the temperature inside the balloon when pressure is maximum? b. What are the final pressure and temperature inside the balloon? Balloon becomes spherical at V0 = ( /6) x (1)3 = 0.5236 m3 and the initial mass is P1V1 100ÊxÊ0.4 m = = = 0.077 kg RT1 2.07703ÊxÊ250 dP *Ê-2 *Ê-8 a) = C[-D + 7D ] = 0 at Pmax dD* max max *Ê6 * = 71/6 = 1.38309 or -D + 7 = 0 , Dmax = D max max Vmax = ( /6) Dmax3 = 1.3853 m3, Pmax = 200 kPa Pmax Vmax 200 1.3853 Tmax = T1 x x = 250 x x = 1731.6 K P1 V1 100 0.4 b) 200 = 100 + C(1.38309-1 – 1.38309-7) , => C = 161.36 V2 = 2.0 m3 = ( /6) D23 ? D2 = 1.5632 m P2 = 100 + 161.36(1.5632-1 – 1.5632-7) = 196 kPa P2 V2 196 2.0 T2 = T1 x x = 250 x x = 2450 K P1 V1 100 0.4 ÊÊ2 V2 c) 1W2 =? ÊPdV = P0(V0 – V1) + ? ÊÊ PdV ÊÊ1 Ê V0 V2 = P0(V0 – V ) + P0(V – V0) + ? ÊÊ C(D*Ê-1 – D*Ê-7)dV 12 Ê V0 3 3 D*Ê2dD* V = 6 D3, dV = 6 D dD = 6 Do 23

5.100 A frictionless, thermally conducting piston separates the air and water in the cylinder shown in Fig. P5.100. The initial volumes of A and B are each 500 L, and the initial pressure on each side is 700 kPa. The volume of the liquid in B is 2% of the volume of B at this state. Heat is transferred to both A and B until all the liquid in B evaporates. Notice that PA = PB and TA = TB = Tsat through the process and iterate to find final a) System: Air(A) + H2O(B) 0.02×0.5 0.98×0.5 mB = mLIQB1 + mVAPB1 = + = 10.821 kg 0.001108 0.2729 P1VA1 700ÊxÊ0.5 mA = = = 2.783 kg RATA1 0.287ÊxÊ438.2 ÊTAÊ=ÊTBÊ=ÊTSATÊÊÊÊÊÊÊÊPAÊ=ÊPB At all times: 3 ÊVAÊ+ÊVBÊ=Ê1Êm mARATSAT 2.783ÊxÊ0.287ÊxÊTSAT + mBvG = + 10.821vG = 1.0 P2 P2 Assume P2 = 2.57 MPa TSAT = 225.4°C 2.783ÊxÊ0.287ÊxÊ498.6 + 10.821 x 0.07812 » 1.0 2570 P = 2.57 MPa 2

mVAPÊB1 1.796 x = = = 0.166; u = 696.4 + 0.166 x 1876.1 = 1007.9 B1 m 10.821 B1 B W = 0 for system A+B 12 Q = m (u – u ) + m (u – u ) 1 2 A A2 A1 B B2 B1 = 2.783 x 0.717(225.4 -165) + 10.821(2603.2 -1007.9) = 17383 kJ b) System: Air(A) only At any P between P & P , T = T for H O 1 2 SAT 2 plot or m R T P 2 calculate P(kPa) T(K) V (m3) = A A area A P = PdV A 700 438.2 0.50 900 448.6 0.3981 1 1200 461.2 0.3070 1500 471.5 0.2511 2000 485.6 0.1939 2570 498.6 0.1550 VA

English Unit Problems 5.102E A hydraulic hoist raises a 3650 lbm car 6 ft in an auto repair shop. The hydraulic pump has a constant pressure of 100 lbf/in.2 on its piston. What is the increase in potential energy of the car and how much volume should the pump displace to No change in kinetic or internal energy of the car, neglect hoist mass. 3650 · 32.174 · 6 E2 – E1 = PE2 – PE1 = mg (Z2 – Z1) = = 21900 lbf-ft 32.174 The increase in potential energy is work into car from pump at constant P. W = E2 – E1 = P dV = P V V = (E2 – E1) / P = 21900/(100 x 144) = 1.52 ft3

5.104E Find the missing properties and give the phase of the substance. Solution: a) Table C.8.1: uf < u < ug => 2-phase mixture of liquid and vapor x = (u – uf)/ ufg = (1000 – 238.81)/854.14 = 0.8912 v = vf + x vfg = 0.01717 + 0.8912 x 10.0483 = 8.972 ft3/lbm h = hf + x hfg = 238.95 + 0.8912 x 931.95 = 1069.5 Btu/lbm ( = 1000 + 41.848 x 8.972 x 144/778) b) Table C.8.1: u < uf so compressed liquid B.1.3, x = undefined T = 471.8 F, v = 0.019689 ft3/lbm c) Table B.3.1: P > Psat => x = undef, compr. liquid Approximate as saturated liquid at same T, h hf = 18.61 Btu/lbm d) Table C.11.1: h > hg => x = undef, superheated vapor C.11.2, find it at given T between saturated 243.9 psi and 200 psi to match h: 185-Ê183.63 v 0.1836 + (0.2459 – 0.1836)· = 0.2104 ft3/lbm Ê186.82-183.63 185-Ê183.63 P 243.93 + (200 – 243.93)· = 225 lbf/in2 Ê186.82-183.63 e) Table C.9.1: P < Psat x = undef. superheated vapor C.9.2, v = (6.3456 + 6.5694)/ 2 = 6.457 ft3/lbm u = h-Pv = (1/2)(694.59 + 705.64) – 60 x 6.4575 x (144/778) = 700.115 – 71.71 = 628.405 Btu/lbm PT C.P. C.P.

States shown are placed relative to the two-phase region, not to each other.

5.105E Find the missing properties among (P, T, v, u, h) together with x, if applicable, a. R-22 T = 50 F, u = 85 Btu/lbm b. H2O T = 600 F, h = 1322 Btu/lbm c. R-22 P = 150 lbf/in.2, h = 115.5 Btu/lbm d. R-134aT = 100 F, u = 175 Btu/lbm e. NH3 T = 70 F, v = 2 ft3/lbm Solution: a) Table C.10.1: u < ug => L+V mixture, P = 98.727 lbf/in2 x = (85 – 24.04)/ 74.75 = 0.8155 v = 0.01282 + 0.8155·0.5432 = 0.4558 ft3/lbm h = 24.27 + 0.8155·84.68 = 93.33 Btu/lbm b) Table C.8.1: h > hg => superheated vapor follow 600 F in C.8.2 P 200 lbf/in2 ; v = 3.058 ft3/lbm ; u = 1208.9 Btu/lbm c) Table C.10.1: h > hg => superheated vapor so in C.10.2 T 100 F ; v = 0.3953 ft3/lbm 144 u = h – Pv = 115.5 – 150 x 0.3953 x = 104.5 Btu/lbm 778 d) Table C.11.1: : u > ug => sup. vap., calculate u at some P to end with P » 55 lbf/in2 ; v » 0.999 ft3/lbm; h = 185.2 Btu/lbm e) Table C.9.1: v < vg => L+V mixture, P = 128.8 lbf/in2 x = (2 – 0.02631)/ 2.2835 = 0.864 h = 120.21 + 0.864·507.89 = 559.05 Btu/lbm u = 119.58 + 0.864·453.44 = 511.4 Btu/lbm P C.P.

States shown are placed relative to the two-phase region, not to each other.

5.106E Water in a 6-ft3 closed, rigid tank is at 200 F, 90% quality. The tank is then Solution: C.V.: Water in tank. m2 = m1 ; m(u2 – u1) = 1Q2 – 1W2 Process: V = constant, v2 = v1, 1W2 = 0 State 1: v1 = 0.01663 + 0.9 x 33.6146 = 30.27 ft3/lbm u1 = 168.03 + 0.9 x 906.15 = 983.6 Btu/lbm State 2: T2, v2 = v1 mix of sat. solid + vap. Table C.8.4 v2 = 30.27 = 0.01744 + x2 x 5655 => x2 = 0.00535 u2 = -149.31 + 0.00535 x1166.5 = -143.07 Btu/lbm m = V/v1 = 6 / 30.27 = 0.198 lbm 1Q2 = m(u2 – u1) = 0.198 (-143.07 – 983.6) = -223 Btu

1 T 2 v T 1 2 v 5.107E A cylinder fitted with a frictionless piston contains 4 lbm of superheated refrigerant R-134a vapor at 400 lbf/in.2, 200 F. The cylinder is now cooled so the R-134a remains at constant pressure until it reaches a quality of 75%. Calculate Solution: P C.V.: R-134a

m2 = m1 ; m(u2 – u1) = 1Q2 – 1W2 2 1 Process: P = const. 1W2 = ? PdV 1W2 = P(V2 – V1) = Pm(v2 – v1) 1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) + Pm(v2 – v1) = m(h2 – h1) v

5.108EAmmonia at 30 F, quality 60% is contained in a rigid 8-ft3 tank. The tank and ammonia are now heated to a final pressure of 140 lbf/in.2. Determine the heat Solution: C.V.: NH 3 m = m = m ; m(u – u ) = Q – W P 21 211212 Process: Constant volume v = v & W = 0/ 2112 2 State 1: Table C.9.1 v = 0.02502 + 0.6 · 4.7978 = 2.904 ft3/lbm 1 u = 75.06 + 0.6 x 491.17 = 369.75 Btu/lbm 1 1 m = V/v1 = 8/2.904 = 2.755 lbm V

5.109E A vertical cylinder fitted with a piston contains 10 lbm of R-22 at 50 F, shown in Fig. P5.20. Heat is transferred to the system causing the piston to rise until it reaches a set of stops at which point the volume has doubled. Additional heat is transferred until the temperature inside reaches 120 F, at which point the pressure C.V. R-22. Control mass goes through process: 1 -> 2 -> 3 As piston floats pressure is constant (1 -> 2) and the P volume is constant for the second part (2 -> 3) So we have: v = v = 2 x v 321 3

State 3: Table C.10.2 (P,T) v = 0.2959 ft3/lbm 3 u = h – Pv = 117.0 – 200·0.2959·144/778 1 2 3V = 106.1 Btu/lbm So we can determine state 1 and 2 Table C.10.1: v1 = 0.14795 = 0.01282 + x1(0.5432) => x1 = 0.249 u1 = 24.04 + 0.249·74.75 = 42.6 Btu/lbm State 2: v2 = 0.2959 ft3/lbm, P2 = P1 = 98.7 psia, this is still 2-phase. Ê2 1W3 = 1W2 = ? ÊPdV = P1(V2 – V1) ÊÊ1 = 98.7 · 10(0.295948 – 0.147974) · 144/778 = 27.0 Btu 1Q3 = m(u3 – u1) + 1W3 = 10(106.1 – 42.6) + 27.0 = 662 Btu

5.110E Twenty pound-mass of water in a piston/cylinder with constant pressure is at 1100 F and a volume of 22.6 ft3. It is now cooled to 100 F. Show the P–v diagram and Solution: Constant pressure 1W2 = mP(v2 – v1) P

5.111E A piston/cylinder contains 2 lbm of liquid water at 70 F, and 30 lbf/in.2. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 300 lbf/in.2 with a volume of 4 ft3. Find the final temperature and plot the P-v diagram for the process. Calculate the work and the heat transfer Solution: Take CV as the water.

P m2 = m1 = m ; m(u2 -u1) = 1Q2 – 1W2 2 State 1: Compr. liq., use sat. liq. same T, Table C.8.1 P 2 v = vf = 0.01605, u = uf = 38.09 Btu/lbm 1 P State 2: v = V/m = 4/2 = 2 ft3/lbm and P = 300 psia 1 v => Sup. vapor T = 600 F ; u = 1203.2 Btu/lbm Work is done while piston moves at linearly varying pressure, so we get 144 1W2 = P dV = Pavg(V2 -V1) = 0.5x(30+3000)(4 – 0.0321) = 121.18 Btu 778 Heat transfer is found from energy equation 1Q2 = m(u2 – u1) + 1W2 = 2 x (1203.2 – 38.09) + 121.18 = 2451.4 Btu

5.112E A piston/cylinder arrangement has the piston loaded with outside atmospheric pressure and the piston mass to a pressure of 20 lbf/in.2, shown in Fig P5.28. It contains water at 25 F, which is then heated until the water becomes saturated vapor. Find the final temperature and specific work and heat transfer for the Solution: Continuity: m = m , Energy: u – u = q – w 21 211212 ÊÊ2Ê Process: P = const. = P , => w = ? ÊPÊdv = P (v – v ) 1 12 121 Ê1 State 1: T , P => Table C.8.4 compressed solid, take as saturated solid. 11 v = 0.01746 ft3/lbm, u = -146.84 Btu/lbm 11 State 2: x = 1, P = P = 20 psia due to process => Table C.8.1 21

5.113EA piston/cylinder contains 2 lbm of water at 70 F with a volume of 0.1 ft3, shown in Fig. P5.35. Initially the piston rests on some stops with the top surface open to the atmosphere, Po, so a pressure of 40 lbf/in.2 is required to lift it. To what temperature should the water be heated to lift the piston? If it is heated to saturated vapor find the final temperature, volume, and the heat transfer. Solution: m = m = m ; m(u – u ) = Q – W 21 211212

State 1: 20 C, v1 = V/m = 0.1/2 = 0.05 ft3/lbm x = (0.05 – 0.01605)/867.579 = 0.0003913 u1 = 38.09 + 0.0003913·995.64 = 38.13 Btu/lbm P2 To find state 2 check on state 1a: P 1 P = 40 psia, v = v1 = 0.05 ft3/lbm Table C.8.1: vf < v < vg = 10.501 P

1a 2 1 V State 2 is saturated vapor at 40 psia as state 1a is two-phase. T2 = 267.3 F v = v = 10.501 ft3/lbm , V = m v = 21.0 ft3 u = ug= 1092.27 Btu/lbm 2g 22,2 Pressure is constant as volume increase beyond initial volume.

1W2 = P dV = P (V2-V1) = 40 (21.0 – 0.1) · 144 / 778 = 154.75 Btu lift 1Q2 = m(u2 – u1) + 1W2 = 2 (1092.27 – 38.13) + 154.75 = 2263 Btu

5.115E A water-filled reactor with volume of 50 ft3 is at 2000 lbf/in.2, 560 F and placed inside a containment room, as shown in Fig. P5.41. The room is well insulated and initially evacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room volume so the final pressure does not Mass: m2 = m1 = Vreactor/v1 = 50/0.02172 = 2295.7 lbm Energy m(u2 – u1) = 1Q2 – 1W2 = 0/ u2 = u1 = 552.5 Btu/lbm State 2: 30 lbf/in.2, u2 < ug 2 phase Table C.8.1 u = 552.5 = 218.48 + x2 869.41 x2 = 0.3842 v2 = 0.017 + 0.3842 x 13.808 = 5.322 ft3/lbm V2 = mv2 = 2295.7 x 5.322 = 12218 ft3

5.117E Calculate the heat transfer for the process described in Problem 4.72. A cylinder containing 2 lbm of ammonia has an externally loaded piston. Initially the ammonia is at 280 lbf/in.2, 360 F and is now cooled to saturated vapor at 105 F, and then further cooled to 65 F, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V. Solution: P

280 229 118 State 1: (T, P) Table C.9.2 1 v1 = 1.7672 ft3/lbm 360 F 105 F P2 = 229 psia, v2 = 1.311 ft3/lbm 3 65 F State 3: (T, x) P3 = 118 psia, v v3 = (0.02614+2.52895)/2 = 1.2775 u3 = (113.96 + 572.29)/2 = 343.1 3 P1Ê+ÊP2 P2Ê+ÊP3 1W3 = ? ÊPdV » ( )m(v2 – v1) + ( )m(v3 – v2) 22 1 280Ê+Ê229 229Ê+Ê118 144 = 2[( )(1.311 – 1.7672) + ( )(1.2775 – 1.311)] = -45.1 Btu 2 2 778 144 1: P1,T1 u1 = h1-P1v1 = 794.94 – 280 · 1.767 · = 703.36 778 1Q3 = 2(343.1 – 703.36) – 45.1 = -766 Btu

5.118EAmmonia, NH3, is contained in a sealed rigid tank at 30 F, x = 50% and is then heated to 200 F. Find the final state P2, u2 and the specific work and heat transfer. Solution: Cont.: m2 = m1 ; Energy: E2 – E1 = 1Q2 ; (1W2 = 0) State 1: Table C.9.1, u1 = 75.06 + 0.5 · 491.17 = 320.65 Btu/lbm Process: Const. volume v2 = v1 = 0.02502 + 0.5 · 4.7945 = 2.422 ft3/lbm State 2: v2 , T2 Table C.9.2 P sup. Vap. Between 150 psia and 175 psia 2 P2 = 163 lbf/in2, h2 = 706.6 linear interpolation v u2 = h2-P2v2 = 706.6 – 163· 2.422· 144/778 = 633.5 1

5.119E A car with mass 3250 lbm drives with 60 mi/h when the brakes are applied to quickly decrease its speed to 20 mi/h. Assume the brake pads are 1 lbm mass with heat capacity of 0.2 Btu/lbm R and the brake discs/drums are 8 lbm steel where both masses are heated uniformly. Find the temperature increase in the brake 122 m = constant; E2 – E1 = 0 – 0 = mcar (V2 – V1) + mbrake(u2 – u1) 2 The brake system mass is two different kinds so split it, also use Cv since we do not have a u table for steel or brake pad material.

5.123E A piston/cylinder in a car contains 12 in.3 of air at 13 lbf/in.2, 68 F, shown in Fig. P5.66. The air is compressed in a quasi-equilibrium polytropic process with polytropic exponent n = 1.25 to a final volume six times smaller. Determine the Cont.: m2 = m1 ; Energy: E2 – E1 = m(u2 – u1) = 1Q2 – 1W2 Process: Pvn = const. ; Ideal gas: Pv = RT vn P v n = P v n P = P = 13 · (6) 1.25 = 122.08 lbf/in2 1 1 1 2 2 2 1?v2? T2 = T1(P2v2/P1v1) = 527.67(122.08/13 · 6 ) = 825.9 R

5.124EWater at 70 F, 15 lbf/in.2, is brought to 30 lbf/in.2, 2700 F. Find the change in the specific internal energy, using the water table and the ideal gas water table in State 1: Table C.8.1 u1 uf = 38.09 State 2: Highest T in Table C.8.2 is 1400 F Using a u from the ideal gas table C.7, we get — h2700 – h2000 = 26002 – 11769 = 14233 Btu/lbmol= 790 Btu/lbm u2700 – u1400 = h- R(2700 – 1400) = 790 – 53.34 · Error! Reference source not found. = 700.9 Since ideal gas change is at low P we use 1400 F, lowest P available 1 lbf/in2 u2 – u1 = (u2 – ux)ID.G. + (ux – u1) = 700.9 + 1543.1 – 38.09 = 2206 Btu/lbm 5.125E Air in a piston/cylinder at 30 lbf/in.2, 1080 R, is shown in Fig. P5.69. It is expanded in a constant-pressure process to twice the initial volume (state 2). The piston is then locked with a pin, and heat is transferred to a final temperature of 1080 R. Find P, T, and h for states 2 and 3, and find the work and heat transfer in C.V. Air. Control mass m2 = m3 = m1

1? 2: u2 -u1 = 1q2 -1w2 ; 1w2 = ? Pdv = P1(v2 -v1) = R(T2 -T1)

Ideal gas Pv = RT T2 = T1v2/v1 = 2T1 = 2160 R P2 = P1 = 30 lbf/in2 , h2 = 549.357 1w2 = RT1 = 74.05 Btu/lbm Table C.6 h2 = 549.357 Btu/lbm, h3 = h1 = 261.099 Btu/lbm 1q2 = u2 – u1 + 1w2 = h2 – h1 = 549.357 – 261.099 = 288.26 Btu/lbm 2? 3: v3 = v2 = 2v1 2w3 = 0, P3 = P2T3/T2 = P1/2 = 15 lbf/in2 2q3 = u3 – u2 = 187.058 – 401.276 = -214.2 Btu/lbm

5.127E Oxygen at 50 lbf/in.2, 200 F is in a piston/cylinder arrangement with a volume of 4 ft3. It is now compressed in a polytropic process with exponent, n = 1.2, to a final temperature of 400 F. Calculate the heat transfer for the process. Continuity: m2 = m1 ; Energy: E2 – E1 = m(u2 – u1) = 1Q2 – 1W2 State 1: T, P and ideal gas, small change in T, so use Table C.4 P1V1 50Ê· Ê4Ê· Ê144 m = = = 0.9043 lbm RT1 48.28Ê· Ê659.67 Process: PVn = constant 1 mR 0.9043Ê· Ê48.28 400Ê-Ê200 1W2 = (P2V2 – P1V1) = (T2 – T1) = · 1-n 1-n 1Ê-Ê1.2 778 = – 56.12 Btu 1Q2 = m(u2 – u1) + 1W2 mCv(T2 – T1) + 1W2 = 0.9043 · 0.158 (400 – 200) – 56.12 = – 27.54 Btu P P = C v -1.2 2 T2

5.128EA piston/cylinder contains 4 lbm of air at 100 F, 2 atm, as shown in Fig. P5.75. The piston is loaded with a linear spring, mass, and the atmosphere. Stops are mounted so that Vstop = 100 ft3, at which point P = 6 atm is required to balance the piston forces. The air is now heated to a final pressure of 60 lbf/in.2. Find the final temperature, volume, and the work and heat transfer. Find the work done on the spring.

5.129E An air pistol contains compressed air in a small cylinder, as shown in Fig. P5.77. Assume that the volume is 1 in.3, pressure is 10 atm, and the temperature is 80 F when armed. A bullet, m = 0.04 lbm, acts as a piston initially held by a pin (trigger); when released, the air expands in an isothermal process (T = constant). If the air pressure is 1 atm in the cylinder as the bullet leaves the gun, find C.V. Air. Air ideal gas: mair = P1V1/RT1 = 10Ê· Ê14.7Ê· Ê1 = 4.26· 10-5 lbm 53.34Ê· Ê539.67Ê· Ê12 Process: PV = const = P1V1 = P2V2 V2 = V1P1/P2 = 10 in3

5.131E A cylinder fitted with a frictionless piston contains R-134a at 100 F, 80% quality, at which point the volume is 3 Gal. The external force on the piston is now varied in such a manner that the R-134a slowly expands in a polytropic process to 50 lbf/in.2, 80 F. Calculate the work and the heat transfer for this process. Solution: C.V. The mass of R-134a. Properties in Table C.11.1 v1 = vf + x1 vfg= 0.01387 + 0.8 · 0.3278 = 0.2761 ft3/lbm u1 = 108.51 + 0.8 · 62.77 = 158.73 Btu/lbm; P1 = 138.926 psia m = V/v1 = 3 · 231 · 12 -3 / 0.2761 = 0.401/ 0.2761 = 1.4525 lbm u2 = 181.1 – 50 · 1.0563 · 144/778 = 171.32 P1 V2 138.926 1.0563 Process: n = ln / ln = ln / ln = 0.7616 P2 V1 50 9.2761 P2ÊV2Ê-ÊP1ÊV1 1W2 = P dV = 1-Ên 50Ê· Ê1.0563Ê-Ê138.926Ê· Ê0.2761 144 = · 1.4525· = 16.3 Btu 1Ê-Ê0.7616 778 1Q2 = m(u2 – u1) + 1W2 = 1.4525 (171.32 – 158.73) + 16.3 = 34.6 Btu

5.133E Water at 300 F, quality 50% is contained in a cylinder/piston arrangement with initial volume 2 ft3. The loading of the piston is such that the inside pressure is linear with the square root of volume as P = 14.7 + CV0.5 lbf/in.2. Now heat is transferred to the cylinder to a final pressure of 90 lbf/in.2. Find the heat transfer Continuity: m2 = m1 Energy: m(u2 – u1) = 1Q2 – 1W2 State 1: v1 = 3.245, u1 = 684.76 m = V/v1 = 0.616 lbm Process equation P1 – 14.7 = CV11/2 so (V2/V1)1/2 = (P2 – 14.7)/(P1 – 14.7) ØP Ê-Ê14.7 2 Ø 90Ê-Ê14.7 2 2Ø Ø V2 = V1 · ? œ = 2 · ? œ = 4.149 ft3 ºP1Ê-Ê14.7ß º66.98Ê-Ê14.7ß W = ? PdV = ? (14.7Ê+ÊCV1/2)dV = 14.7· (V – V ) + 2 C(V 1.5 – V 1.5) 12 21321 = (14.7)(V2 – V1)(1-2/3) + (2/3)(P2V2 – P1V1) Ø14.7 2 Ø144 1W2 = ? (4.149-2)+Ê (90· 4.149-66.98· 2)œ = 31.5 Btu º 3 3 ß778 State 2: P2,v2 = V2/m = 6.7354 u2 = 1204.66, T2 573.6 1Q2 = 0.616 · (1204.66 – 684.76) + 31.5 = 351.8 Btu

5.134E A closed cylinder is divided into two rooms by a frictionless piston held in place by a pin, as shown in Fig. P5.85. Room A has 0.3 ft3 air at 14.7 lbf/in.2, 90 F, and room B has 10 ft3 saturated water vapor at 90 F. The pin is pulled, releasing the piston and both rooms come to equilibrium at 90 F. Considering a control mass of the air and water, determine the work done by the system and the heat transfer to the cylinder.

P2 = PGÊH2OÊatÊ90°F = PA2 = PB2 Air, I.G.: PA1VA1 = mARAT = PA2VA2 = PGÊH OÊatÊ90°F VA2 2 14.7Ê· Ê0.3 ? VA2 = 0.6988 = 6.31 ft3 VB2 = VA1 + VB1 – VA2 = 0.30 + 10 – 6.31 = 3.99 ft3 VB1 10 mB = vB1 = 467.7 = 0.02138 lbm ? vB2 = 186.6 ft3/lbm 186.6 = 0.016099 + xB2 · (467.7 – 0.016) => xB2 = 0.39895 System A+B: W12 = 0; UA = 0 ( IG & T = 0 ) uB2 = 58.07 + 0.39895 · 982.2 = 449.9 Btu/lbm; uB1 = 1040.2 1Q2 = 0.02138 (449.9 – 1040.2) = -12.6 Btu

5.136E A computer in a closed room of volume 5000 ft3 dissipates energy at a rate of 10 hp. The room has 100 lbm of wood, 50 lbm of steel and air, with all material at 540 R, 1 atm. Assuming all the mass heats up uniformly how long time will it take to increase the temperature 20 F?

· C.V. Air, wood and steel. m2 = m1 ; U2 – U1 = 1Q2 = Q t The total volume is nearly all air, but we can find volume of the solids. V = m/ = 100/44.9 = 2.23 ft3 ; V = 50/488 = 0.102 ft3 wood steel Vair = 5000 – 2.23 – 0.102 = 4997.7 ft3 mair = PV/RT = 14.7· 4997.7· 144/(53.34· 540) = 367.3 lbm We do not have a u table for steel or wood so use heat capacity.

CHAPTER 6 New Old New Old New Old 1 83 25 112 49 117 2 new 26 113 50 119 3 new 27 new 51 120 4 84 28 new 52 139 5 85 29 92 53 121 6 new 30 96 54 122 7 new 31 new 55 123 8 103 32 115 56 124 9 86 33 88 57 126 10 89 mod 34 new 58 127 11 97 35 93 59 129 12 98 36 87 60 131 13 99 37 101 61 132 14 94 38 102 62 134 15 95 39 104a 63 135 16 new 40 104b 64 136 17 100 41 105 65 137 18 new 42 106 66 new 19 90 43 107 67 new 20 91 44 108 68 new 21 new 45 109 69 118 22 new 46 114 70 125 23 110 47 116 71 128 24 111 48 new 72 130 The advanced problems start with number 6.69.

6.1 Air at 35°C, 105 kPa, flows in a 100 mm · 150 mm rectangular duct in a heating system. The volumetric flow rate is 0.015 m3/s. What is the velocity of the air flowing in the duct?

. . A = 100 x 150×10-6 = 0.015 m V = mv = AV with 2 V 0.015Êm3/s V = = = 1.0 m/s A 0.015Êm2ÊÊ RT 0.287ÊxÊ308.2 3 Êv=Ê Ê=Ê Ê=Ê0.8424Êm /kg P . 105 IdealÊgasÊsoÊnote:Ê . V 0.015 ? mÊ=Ê Ê=Ê Ê=Ê0.0178Êkg/sÊ ? v 0.8424

6.2 A boiler receives a constant flow of 5000 kg/h liquid water at 5 MPa, 20°C and it heats the flow such that the exit state is 450°C with a pressure of 4.5 MPa. Determine the necessary minimum pipe flow area in both the inlet and exit pipe(s) .. 1 Mass flow rate mi = me = (AV/v) i = (AV/v) e = 5000 kg/s 3600 Table B.1.4 v = 0.001 m3/kg, v = 0.07074 m3/kg, both V £ 20 m/s ie . 5000 -5 2 2 Ai ‡ vi m/Vi = 0.001· / 20 = 6.94 · 10 m = 0.69 cm 3600 . 5000 -3 2 2 Ae ‡ ve m/Ve = 0.07074 · / 20 = 4.91 · 10 m = 49 cm 3600

6.3 A natural gas company distributes methane gas in a pipeline flowing at 200 kPa, 275 K. They have carefully measured the average flow velocity to be 5.5 m/s in a 50 cm diameter pipe. If there is a – 2% uncertainty in the velocity measurement how would you qoute the mass flow rate?

6.4 Nitrogen gas flowing in a 50-mm diameter pipe at 15°C, 200 kPa, at the rate of 0.05 kg/s, encounters a partially closed valve. If there is a pressure drop of 30 kPa across the valve and essentially no temperature change, what are the velocities upstream and downstream of the valve?

Same inlet and exit area: A = (0.050)2 = 0.001963 m2 4 RTi 0.2968ÊxÊ288.2 3 Ideal gas: vi = Pi = 200 = 0.4277 m /kg mvi 0.05ÊxÊ0.4277 Vi = A = 0.001963 = 10.9 m/s RTe 0.2968ÊxÊ288.2 3 Ideal gas: ve = = = 0.5032 m /kg Pe 170 mve 0.05ÊxÊ0.5032 Ve = A 0.001963 = 12.8 m/s =

6.5 Saturated vapor R-134a leaves the evaporator in a heat pump system at 10°C, with a steady mass flow rate of 0.1 kg/s. What is the smallest diameter tubing that can be used at this location if the velocity of the refrigerant is not to exceed 7 m/s?

Table B.5.1: vg = 0.04945 m3/kg . 22 AMIN = mvg/VMAX = 0.1×0.04945/7 = 0.000706 m = ( /4) DMIN DMIN = 0.03 m = 30 mm

6.6 Steam at 3 MPa, 400°C enters a turbine with a volume flow rate of 5 m3/s. An extraction of 15% of the inlet mass flow rate exits at 600 kPa, 200°C. The rest exits the turbine at 20 kPa with a quality of 90%, and a velocity of 20 m/s. Determine the volume flow rate of the extraction flow and the diameter of the final exit pipe.

6.7 A pump takes 10°C liquid water in from a river at 95 kPa and pumps it up to an irrigation canal 20 m higher than the river surface. All pipes have diameter of 0.1 m and the flow rate is 15 kg/s. Assume the pump exit pressure is just enough to carry a water column of the 20 m height with 100 kPa at the top. Find the flow work into and out of the pump and the kinetic energy in the flow.

Both states are compressed liquid so Table B.1.1: vi = vf = 0.001 m3/kg Flow rates in and out are the same, pipe size the same so same velocity.

Vi = Ve = m. v/ ( 4 D2 ) = 15 · 0.001/( 4 0.12) = 1.91 m/s e KEi = 1 Vi2 = KEe= 1Ve2 = 1(1.91)2 m2/s2 = 1.824 J/kg 2. i 22 Flow work at the boundary: mPv; the P’s are different Wflow, i = mi Pi vi = 15 · 95 · 0.001 = 1.425 kW Pe = Po + Hg/v = 100 + (20 · 9.807/0.001)/1000 = 100 + 196 = 296 kPa Wflow, e = mPeve = 15 · 296 · 0.001 = 4.44 kW 6.8 A desuperheater mixes superheated water vapor with liquid water in a ratio that produces saturated water vapor as output without any external heat transfer. A flow of 0.5 kg/s superheated vapor at 5 MPa, 400°C and a flow of liquid water at 5 MPa, 40°C enter a desuperheater. If saturated water vapor at 4.5 MPa is LIQ 2 Cont.: m1 + m2 = m3

VAP 1 3 Energy Eq.: m1h1 + m2h2 = m3h3 0.5 x 3195.7 + m2 x 171.97 = (0.5 + m2) 2797.9 .

=> m2 = 0.0757 kg/s 6.9 Carbon dioxide enters a steady-state, steady-flow heater at 300 kPa, 15°C, and exits at 275 kPa, 1200°C, as shown in Fig. P6.9. Changes in kinetic and potential energies are negligible. Calculate the required heat transfer per kilogram of C.V. Heater SSSF single inlet and exit. Energy Eq.: q + hi = he 60145Ê-Ê(-348.3) Table A.8: q = he – hi = = 1374.5 kJ/kg 44.01 (If we use CP0 from A.5 then q 0.842(1200 – 15) = 997.8 kJ/kg) Too large T, Tave to use Cp0 at room temperature.

6.10 Saturated liquid nitrogen at 500 kPa enters a SSSF boiler at a rate of 0.005 kg/s and exits as saturated vapor. It then flows into a superheater also at 500 kPa where it exits at 500 kPa, 275 K. Find the rate of heat transfer in the boiler and the C.V.: boiler SSSF, single inlet and exit, neglict KE, PE energies in flow Continuity Eq.: m1 = m2 = m3 (SSSF)

P T3 1 2 3 Q 500 12312 Q v v

6.12 In a steam generator, compressed liquid water at 10 MPa, 30°C, enters a 30-mm diameter tube at the rate of 3 L/s. Steam at 9 MPa, 400°C exits the tube. Find the rate of heat transfer to the water.

Constant diameter tube: A = A = (0.03)2 = 0.0007068 m2 ie4 Table B.1.4 m = Vi/vi = 0.003/0.0010003 = 3.0 kg/s Vi = Vi/Ai = 0.003/0.0007068 = 4.24 m/s Ve = Vi x ve/vi = 4.24 x 0.02993/0.0010003 = 126.86 m/s .. 22 Q Ø Ê-Êh )Ê+Ê V Ê-ÊV Ê/2Ø = m º(he i ? e i ? ß Ø 126.862Ê-Ê4.242Ø = 3.0 ?3117.8Ê-Ê134.86Ê+Ê œ= 8973 kW º 2ÊxÊ1000 ß

6.13 A heat exchanger, shown in Fig. P6.13, is used to cool an air flow from 800 to 360K, both states at 1 MPa. The coolant is a water flow at 15°C, 0.1 MPa. If the water leaves as saturated vapor, find the ratio of the flow rates mH2O/mair C.V. Heat exchanger, SSSF, 1 inlet and exit for air and water each. The two mairhai + mH2Ohfi = mairhae + mH2Ohge Table A.7: hai = 822.202, hae = 360.863 kJ/kg Table B.1: hfi = 62.99 (at 15°C), hge = 2675.5 (at 100 kPa) mH2O/mair = (hai – hae)/(hge – hfi) = (822.202 – 360.863)/(2675.5 – 62.99) = 0.1766 6.14 A condenser (heat exchanger) brings 1 kg/s water flow at 10 kPa from 300°C to saturated liquid at 10 kPa, as shown in Fig. P6.14. The cooling is done by lake water at 20°C that returns to the lake at 30°C. For an insulated condenser, find the flow rate of cooling water.

300°C sat. liq. C.V. Heat exchanger 1 kg/s . . mcoolh20 + mH2Oh300 = .

6.15 Two kg of water at 500 kPa, 20°C is heated in a constant pressure process (SSSF) Continuity: min = mex = m, Energy: q + hin = hex q = hex – hin steam tables only go up to 1300°C so use an intermediate state at lowest pressure (closest to ideal gas) hx(1300°C, 10 kPa) from Table B.1.3 and table A.8 for the high T change h hex – hin = (hex – hx) + (hx – hin) = (71423 – 51629)/18.015 + 5409.7 – 83.96 = 6424.5 kJ/kg Q = m(hex – hin) = 2 x 6424.5 = 12849 kJ

6.16 A mixing chamber with heat transfer receives 2 kg/s of R-22 at 1 MPa, 40°C in one line and 1 kg/s of R-22 at 30°C, quality 50% in a line with a valve. The outgoing flow is at 1 MPa, 60°C. Find the rate of heat transfer to the mixing chamber. C.V. Mixing chamber. SSSF with 2 flows in and 1 out, heat transfer in.

1 Table B.4: Mixer 3 h1 = 271.04, h3 = 286.97 Heater h2 = 81.25 + 0.5 · 177.87 = 170.18 Q Cont. Eq.: m1 + m2 = m3 ; Energy Eq.: m1h1 + m2h2 + Q = m3h3 m3 = 2 + 1 = 3 kg/s Q = 3 · 286.973 – 2 · 271.04 – 1 · 170.18 = 148.66 kW 6.17 Compressed liquid R-22 at 1.5 MPa, 10°C is mixed in a steady-state, steady-flow process with saturated vapor R-22 at 1.5 MPa. Both flow rates are 0.1 kg/s, and the exiting flow is at 1.2 MPa and a quality of 85%. Find the rate of heat transfer to the mixing chamber.

6.18 Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated 12 Energy Eq.: h1 + ? = h2 + V2 2 2 V2 = 2 ( h1 – h2 ) 2 CPN (T1 – T2 ) = 2 x1.042 (400 – 330) 2

= 145.88 kJ/kg = 145 880 J/kg V2 = 381.94 m/s 6.19 Superheated vapor ammonia enters an insulated nozzle at 20°C, 800 kPa, shown in Fig. P6.19, with a low velocity and at the steady rate of 0.01 kg/s. The ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature 22 Energy Eq.: q + hi + V /2 = he + V /2, q = 0, Vi = 0 ie Table B.2.2: hi = 1464.9 = he + 4502/(2×1000) he = 1363.6 kJ/kg Pe = 300 kPa Sat. state at -9.2°C : 1363.6 = 138.0 + xe x 1293.8, => xe = 0.947, ve = 0.001536 + xe x 0.4064 = 0.3864 Ae = meve/Ve = 0.01 x 0.3864 / 450 = 8.56 x 10-6 m2

6.23 Helium is throttled from 1.2 MPa, 20°C, to a pressure of 100 kPa. The diameter of the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities are equal. Find the exit temperature of the helium and the ratio of the pipe Energy Eq.: hi = he, Ideal gas => Ti = Te = 20°C m = But m, V, T are constant => PiAi = PeAe RT/P De Pi 1/2 1.2 1/2 = = = 3.464 Di Pe ?0.1? ??

6.24 Water flowing in a line at 400 kPa, saturated vapor, is taken out through a valve to 100 kPa. What is the temperature as it leaves the valve assuming no changes in kinetic energy and no heat transfer?

C.V. Valve (SSSF) 1 2 Cont.: m1 = m2 ; Energy: m1h1 + Q = m2h2 + W Small surface area: Q = 0; No shaft: W = 0 Table B.1.2: h2 = h1 = 2738.6 kJ/kg T2 = 131.1°C

6.26 Water at 1.5 MPa, 150°C, is throttled adiabatically through a valve to 200 kPa. The inlet velocity is 5 m/s, and the inlet and exit pipe diameters are the same. Determine the state and the velocity of the water at the exit.

CV: valve. m = const, A = const Ve = Vi(ve/vi) 1 2 1 2 1 2 Øve 2 Ø hi + Vi = Ve + he or (he – hi) + Vi ? -Ê1œ= 0 2 2 2 º?vi ? ß

(5)2 Ø ve 2 Ø h – 632.87 + ? -1œ= 0 e 2ÊxÊ1000 º?0.00109? ß Table B.1.2: he = 504.7 + xe x 2201.9, ve = 0.001061 + xe x 0.8846 Substituting and solving, xe = 0.04885 Ve = 5 (0.04427 / 0.00109) = 203 m/s

6.27 An insulated mixing chamber receives 2 kg/s R-134a at 1 MPa, 100°C in a line with low velocity. Another line with R-134a as saturated liquid 60°C flows through a valve to the mixing chamber at 1MPa after the valve. The exit flow is saturated vapor at 1 MPa flowing at 20 m/s. Find the flow rate for the second line. … …12 Cont.: m1 + m2 = m3 ; Energy Eq.: m1h1 + m2h2 = m3( h3 + V3 ) 2 . 12 . 12 m2 (h2 – h3 – V3 ) = m1 ( h3 + V3 – h1 ) 22 1: Table B.5.2: 1MPa, 100°C, h1 = 483.36 kJ/kg 2: Table B.5.1: x = ? , 60°C, h2 = 287.79 kJ/kg 3: Table B.5.1: x = 1, 1 MPa, 20 m/s, h3 = 419.54 kJ/kg . 1 1 1 202 m2 = 2 · [419.54 + 202 · – 483.36] / [287.79 – 419.54 – ] 2 1000 2 1000 = 2 · ( -63.82 + 0.2) / ( -131.75 – 0.2) = 0.964 kg/s Notice how kinetic energy was insignificant.

6.28 A mixing chamber receives 2 kg/s R-134a at 1 MPa, 100°C in a line with low velocity and 1 kg/s from a line with R-134a as saturated liquid 60°C flows through a valve to the mixing chamber at 1 MPa after the valve. There is heat transfer so the exit flow is saturated vapor at 1 MPa flowing at 20 m/s. Find the … ….12 Cont.: m1 + m2 = m3 ; Energy Eq.: m1h1 + m2h2 + Q = m3( h3 + V3 ) 2

1 Mixer 3 Table B.5.2: 1: 1MPa, 100°C, Heater h1 = 483.36 kJ/kg 2 . 2: Table B.5.1: x = ? , 60°C, Q h2 = 287.79 kJ/kg 3: Table B.5.1 x = 1, 1 MPa, 20 m/s, h3 = 419.54; v3 = 0.02038 m3 = 1 + 2 = 3 kg/s Q. = 3(419.54 + 1 202/1000) – 2 · 483.36 – 1 · 287.79 = 4.71 kW 2 Ae = m3v3/V3 = 3 · 0.02038/20 = 0.003057 m2, 4 De = [ Ae ] 1/2 = 0.062 m

6.29 A steam turbine receives water at 15 MPa, 600°C at a rate of 100 kg/s, shown in Fig. P6.29. In the middle section 20 kg/s is withdrawn at 2 MPa, 350°C, and the rest exits the turbine at 75 kPa, and 95% quality. Assuming no heat transfer and no changes in kinetic energy, find the total turbine power output.

C.V. Turbine SSSF, 1 inlet and 2 exit flows. 1 Table B.1.3 h1 = 3582.3 kJ/kg, h2 = 3137 kJ/kg Table B.1.2 : h3 = hf + x3hfg = 384.3 + 0.95· 2278.6 = 2549.1 kJ/kg 3 2

6.30 A small, high-speed turbine operating on compressed air produces a power output of 100 W. The inlet state is 400 kPa, 50°C, and the exit state is 150 kPa, -30°C. Assuming the velocities to be low and the process to be adiabatic, find the C.V. Turbine, no heat transfer, no KE, no PE hin = hex + wT wT = hin – hex Cp(Tin – Tex) = 1.004(50 – (-30)) = 80.3 kJ/kg W = mwT m = W/wT = 0.1/80.3 = 0.00125 kg/s 6.31 A steam turbine receives steam from two boilers. One flow is 5 kg/s at 3 MPa, 700°C and the other flow is 15 kg/s at 800 kPa, 500°C. The exit state is 10 kPa, with a quality of 96%. Find the total power out of the adiabatic turbine. C.V. whole turbine SSSF, 2 inlets, 1 exit, no heat transfer Q = 0 Continuity Eq.: m1+ m2 = m3 = 5 + 15 = 20 kg/s Energy Eq.: m1h1 + m2h2 = m3h3 + WT 1 2 Table B.1.3: h1 = 3911.7 kJ/kg, h2 = 3480.6 kJ/kg Table B.1.2: h3 = 191.8 + 0.96 · 2392.8 WT = 2488.9 kJ/kg 3 WT = 5 · 3911.7 + 15 · 3480.6 – 20 · 2488.9 = 21990 kW = 21.99 MW 6.32 A small turbine, shown in Fig. P6.32, is operated at part load by throttling a 0.25 kg/s steam supply at 1.4 MPa, 250°C down to 1.1 MPa before it enters the turbine and the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhaust 110 C.V. Turbine, SSSF, no heat transfer, specific work: w = = 440 kJ/kg 0.25 Energy Eq.: h1 = h2 = h3 + w = 2927.2 (B.1.3) h3 = 2927.2 – 440 = 2487.2 kJ/kg

Table B.1.2: 2487.2 = 191.83 + x3 x 2392.8 T = 45.8°C , x3 = 0.959 T 1 2

6.33 Hoover Dam across the Colorado River dams up Lake Mead 200 m higher than the river downstream. The electric generators driven by water-powered turbines deliver 1300 MW of power. If the water is 17.5°C, find the minimum amount of water running through the turbines.

C.V.: H2O pipe + turbines, min(h+ V2/2 + gz)in = mex(h+ V2/2 + gz)ex + WT Lake Water states: hin hex ; vin vex so Mead wT = gzin – gzex = 9.807 x 200/1000 = 1.961 kJ/kg . . 1300×103ÊkW m = WT/wT = = 6.63×105 kg/s 1.961ÊkJ/kg . . v = 6.63×105 x 0.001001 = 664 m V = m 3/s DAM H T

6.34 A large SSSF expansion engine has two low velocity flows of water entering. High pressure steam enters at point 1 with 2.0 kg/s at 2 MPa, 500°C and 0.5 kg/s cooling water at 120 kPa, 30°C enters at point 2. A single flow exits at point 3 with 150 kPa, 80% quality, through a 0.15 m diameter exhaust pipe. There is a heat loss of 300 kW. Find the exhaust velocity and the power output of the engine.

6.35 A small water pump is used in an irrigation system. The pump takes water in from a river at 10°C, 100 kPa at a rate of 5 kg/s. The exit line enters a pipe that goes up to an elevation 20 m above the pump and river, where the water runs into an open channel. Assume the process is adiabatic and that the water stays at 10°C. Find C.V. pump + pipe. SSSF, 1 inlet, 1 exit. Cont.: min = mex = m

Energy Eq.: m(h + (1/2)V 2 + gz ) = m(h + (1/2) V 2 + gz ) + W in in in ex ex ex .. i W = m(gzin – gzex) = 5 x 9.807 x (0 – 20)/1000 = -0.98 kW I.E. 0.98 kW required input e

6.37 Two steady flows of air enters a control volume, shown in Fig. P6.37. One is 0.025 kg/s flow at 350 kPa, 150°C, state 1, and the other enters at 350 kPa, 15°C, both flows with low velocity. A single flow of air exits at 100 kPa, -40°C through a 25-mm diameter pipe, state 3. The control volume rejects 1.2 kW heat to the surroundings and produces 4.5 kW of power. Determine the flow rate of air at the inlet at state 2.

A3 = D3 = (0.025) = 4.909×10-4 m 222 44 v = RT3/P3 = 0.287ÊxÊ233.2 = 0.6693 m3/kg 3 100 V3 = = = 1363.5(0.025 + m2) A3 4.909×10-4 Energy Eq.: QCV + m1h1 + m2h2 = m3(h3 + V /2) + WCV 3 -1.2 + 0.025 x 1.004 x 423.2 + m2 x 1.004 x 288.2 Ø . 2Ø . (1363.5(0.025Ê+Êm2)) = (0.025 + m )?1.004ÊxÊ233.2Ê+Ê œ+ 4.5 2 º 2ÊxÊ1000 ß Solving, m2 = 0.01815 kg/s

6.38 An air compressor takes in air at 100 kPa, 17°C and delivers it at 1 MPa, 600 K to a constant-pressure cooler, which it exits at 300 K. Find the specific compressor work and the specific heat transfer.

1 C.V. air co.mpressor q = 0 W Energy: h1 + wc = h2 c 23

6.39 The following data are for a simple steam power plant as shown in Fig. P6.39. State 1 2 3 4 5 6 7 P MPa 6.2 6.1 5.9 5.7 5.5 0.01 0.009 T °C 45 175 500 490 40 State 6 has x6 = 0.92, and velocity of 200 m/s. The rate of steam flow is 25 kg/s, with 300 kW power input to the pump. Piping diameters are 200 mm from steam generator to the turbine and 75 mm from the condenser to the steam generator. Determine the power output of the turbine and the heat transfer rate in the Turbine A = ( /4)(0.2)2 = 0.03142 m2 5 V5 = mv5/A5 = 25 x 0.06163/0.03142 = 49 m/s h6 = 191.83 + 0.92 x 2392.8 = 2393.2 w = 3404.2 – 2393.2 – (2002 – 492)/(2 x 1000) = 992.2 WT = mwT = 25 x 992.2 = 24805 kW

6.40 For the same steam power plant as shown in Fig. P6.39 and Problem 6.39, determine the rate of heat transfer in the economizer which is a low temperature heat exchanger and the steam generator. Determine also the flow rate of cooling water through the condenser, if the cooling water increases from 15° to 25°C in the condenser.

6.41 Cogeneration is often used where a steam supply is needed for industrial process energy. Assume a supply of 5 kg/s steam at 0.5 MPa is needed. Rather than generating this from a pump and boiler, the setup in Fig. P6.41 is used so the supply is extracted from the high-pressure turbine. Find the power the turbine C.V. Turbine, SSSF, 1 inlet and 2 exit flows, assume adiabatic, QCV = 0

Supply state 1: 20 kg/s at 10 MPa, 500C 1 Process steam 2: 5 kg/s, 0.5 MPa, 155 C, Exit state 3: 20 kPa, x = 0.9 Table B.1: h1 = 3373.7, h2 = 2755.9, h3 = 251.4 + 0.9 x 2358.3 = 2373.9 2

WCV = 20 x 3373.7 – 5 x 2755.9 – 15 x 2373.9 = 18.084 MW 3 WT

6.42 A somewhat simplified flow diagram for a nuclear power plant shown in Fig. 1.4 is given in Fig. P6.42. Mass flow rates and the various states in the cycle are shown in the accompanying table. The cycle includes a number of heaters in which heat is transferred from steam, taken out of the turbine at some intermediate pressure, to liquid water pumpedfrom the condenser on its way to the steam drum. The heat exchanger in the reactor supplies 157 MW, and it may be a. Assume the moisture separator has no heat transfer between the 44 d. Find the ratio of the total power output of the two turbines to the total power delivered by the reactor.

a) Moisture Separator, SSSF, no heat transfer, no work 62.874 x 2517 = 58.212 x h4 + 4.662 x 558 h4 = 2673.9 = 566.18 + x4 x 2160.6 => x4 = 0.9755 b) Low Pressure Turbine, SSSF no heat transfer m4h4 = m5h5 + m8h8+ WCV(LP) 58.212 x 2673.9 = 55.44 x 2279 + 2.772 x 2459 + WCV(LP) WCV(LP) = 22489 kW = 22.489 MW c) High Pressure Turbine, SSSF no heat transfer m2h2 = m3h3 + m12h12 + m17h17 + WCV(HP) 75.6×2765=62.874×2517+8.064×2517+4.662×2593+WCV(HP) WCV(HP) = 18 394 kW = 18.394 MW d) (WHP + WLP)/QREACT = 40.883/157 = 0.26 b.What is the power to the pump that feeds water to the reactor?

a. Determine the temperature of the water leaving the intermediate pressure 13 b. Determine the pump work, between states 13 and 16.

a) Intermediate Pressure Heater Energy Eq.: m11h11 + m12h12 + m15h15 = m13h13 + m14h14 75.6×284.6 + 8.064×2517 + 4.662×584 = 75.6xh13 + 12.726×349 h13 = 530.35 ? T13 = 126.3°C b) The high pressure pump Energy Eq.: m13h13 = m16h16 + WCv,P WCv,P = m13(h13 – h16) = 75.6(530.35 – 565) = -2620 kW a. Find the power removed in the condenser by the cooling water (not shown). c. Do the energy terms balance for the low pressure heater or is there a a) Condenser: QCV + m5h5 + m10h10 = m6h6 QCV + 55.44 x 2279 + 20.16 x 142.51 = 75.6 x 138.3 QCV = -118765 kW = -118.77 MW b) The condensate pump WCv,P = m6(h6 – h7) = 75.6(138.31 – 140) = -127.8 kW c) Low pressure heater Assume no heat transfer m14h14 + m8h8 + m7h7 + m9h9 = m10h10 + m11h11 LHS = 12.726×349 + 2.772×2459 + 75.6×140 + 4.662×558 = 24443 kW RHS = (12.726 + 2.772 + 4.662) x 142.51 + 75.6 x 284.87 = 24409 kW A slight imbalance, but OK.

6.46 A proposal is made to use a geothermal supply of hot water to operate a steam turbine, as shown in Fig. P6.46. The high-pressure water at 1.5 MPa, 180°C, is throttled into a flash evaporator chamber, which forms liquid and vapor at a lower pressure of 400 kPa. The liquid is discarded while the saturated vapor feeds the turbine and exits at 10 kPa, 90% quality. If the turbine should produce 1 MW, find the required mass flow rate of hot geothermal water in kilograms per hour. h1 = 763.5 = 604.74 + x x 2133.8 x = 0.07439 = m2/m1 Table B.1.2: h2 = 2738.6; h3 = 191.83 + 0.9 x 2392.8 = 2345.4 . . . 1000 W = m2(h2 – h3) m2 = = 2.543 2738.6Ê-Ê2345.4 m1 = 34.19 kg/s = 123075 kg/h

6.47 A R-12 heat pump cycle shown in Fig. P6.47 has a R-12 flow rate of 0.05 kg/s with 4 kW into the compressor. The following data are given State 1 2 3 4 5 6 P kPa 1250 1230 1200 320 300 290 T °C 120 110 45 0 5 Calculate the heat transfer from the compressor, the heat transfer from the R-12 in the condenser and the heat transfer to the R-12 in the evaporator.

6.48 A rigid 100-L tank contains air at 1 MPa, 200°C. A valve on the tank is now opened and air flows out until the pressure drops to 100 kPa. During this process, heat is transferred from a heat source at 200°C, such that when the valve is closed, the temperature inside the tank is 50°C. What is the heat transfer?

1 : 1 MPa, 200°C, m1 = P1V1/RT1 = 1000 · 0.1/(0.287 · 473.1) = 0.736 kg 2 : 100 kPa, 50°C, m2 = P2V2/RT2 = 100 · 0.1/(0.287 · 323.1) = 0.1078 kg mex = m1 – m2 = 0.628 kg, m2u2 – m1u1 = – mex hex + 1Q2 Table A.7: u1 = 340.0 kJ/kg, u2 = 231.0 kJ/kg, heÊave = (h1 + h2)/2 = (475.8 + 323.75)/2 = 399.8 kJ/kg 1Q2 = 0.1078 · 231.0 – 0.736 · 340.0 + 0.628 · 399.8 = +25.7 kJ

6.49 A 25-L tank, shown in Fig. P6.49, that is initially evacuated is connected by a valve to an air supply line flowing air at 20°C, 800 kPa. The valve is opened, and air flows into the tank until the pressure reaches 600 kPa.Determine the final temperature and mass inside the tank, assuming the process is adiabatic. Develop an expression for the relation between the line temperature and the final temperature using constant specific heats.

6.52 A nitrogen line, 300 K and 0.5 MPa, shown in Fig. P6.52, is connected to a turbine that exhausts to a closed initially empty tank of 50 m3. The turbine operates to a tank pressure of 0.5 MPa, at which point the temperature is 250 K. C.V. turbine & tank USUF Conservation of mass: mi = m2 m Energy Eq: mihi = m2u2 + WCV ; WCV = m(hi – u2) Table B.6 : i : Pi = 0.5 MPa, Ti = 300K, Nitrogen; hi = 310.276 kJ/kg 2: P2 = 0.5 MPa, T2 = 250 K, u2 = h2 – P2v2 u2 = 257.799 – 500(0.14782) = 180.89 kJ/kg m2 = V/v2 = 50/0.14782 = 338.25 kg WCV = 338.25(310.276 – 180.89) = 43764.8 kJ = 43.765 MJ 6.53 An evacuated 150-L tank is connected to a line flowing air at room temperature, 25°C, and 8 MPa pressure. The valve is opened allowing air to flow into the tank until the pressure inside is 6 MPa. At this point the valve is closed. This filling process occurs rapidly and is essentially adiabatic. The tank is then placed in storage where it eventually returns to room temperature. What is the final pressure? C.V. Tank: mi = m2 Energy Eq.: mihi = m2u2 constant CPo: T2 = (CP/CV) Ti = kTi= 1.4 x 298.2 = 417.5 K Process: constant volume cooling to T3: P3 = P2 x T3/T2 = 6.0 x 298.15/ 417.5 = 4.29 MPa

P2V2Ê-ÊP1V1 300×0.221Ê-Ê200×0.0654 1W2 =? PdV = = = 39.92 kJ 1Ê-Ên 1Ê+Ê1/3 m2 = P2V2/RT2 = (300 x 0.221)/(0.287 x 350) = 0.66 kg m1 = P1V1/RT1 = (200 x 0.0654)/(0.287 x 300) = 0.152 kg 1Q2 = 0.66×250.32 – 0.152×214.364 – 0.508×401.299 + 39.92 = -31.31 kJ

6.55 A 500-L insulated tank contains air at 40°C, 2 MPa. A valve on the tank is opened, and air escapes until half the original mass is gone, at which point the me = m1 – m2, m2 = m1/2 me = m2 = 5.5625 kg 1st law: 0 = m2u2 – m1u1 + meheÊAV 0 = 5.5625×0.717 T2 – 11.125×0.717×313.2 + 5.5625×1.004 (313.2 + T2)/2 Solving, T2 = 239.4 K m2RT2 5.5625ÊxÊ0.287ÊxÊ239.4 P2 = = = 764 kPa V 0.5

6.57 A 2-m3 insulated vessel, shown in Fig. P6.57, contains saturated vapor steam at 4 MPa. A valve on the top of the tank is opened, and steam is allowed to escape. During the process any liquid formed collects at the bottom of the vessel, so that only saturated vapor exits. Calculate the total mass that has escaped when the C.V. Vessel: Mass flows out. me = m1 – m2 0 = m2u2 – m1u1 + (m1-m2)he or m2(he-u2) = m1(he-u1) But he » (hG1+hG2)/2 = (2801.4+2778.1)/2 = 2789.8 m1 = V/v1 = 40.177 kg, m2 = V/v2 2 (2789.8-u2) = 40.177(2789.8-2602.3) = 7533.19 v2 But v2 = .001 127 + .193 313 x2 and u2 = 761.7 + 1822 x2 Substituting and solving, x2 = 0.7936 m2 = V/v2 = 12.94 kg, me = 27.24 kg

6.58 A 1-m3 insulated, 40-kg rigid steel tank contains air at 500 kPa, and both tank and air are at 20°C. The tank is connected to a line flowing air at 2 MPa, 20°C. The valve is opened, allowing air to flow into the tank until the pressure reaches 1.5 MPa and is then closed. Assume the air and tank are always at the same temperature and find the final temperature.

6.59 A 750-L rigid tank, shown in Fig. P6.59, initially contains water at 250°C, 50% liquid and 50% vapor, by volume. A valve at the bottom of the tank is opened, and liquid is slowly withdrawn. Heat transfer takes place such that the temperature remains constant. Find the amount of heat transfer required to the CV: vessel 0.375 0.375 mLIQ1 = = 299.76 kg; mVAP1 = = 7.48 kg 0.001251 0.05013 m1 = 307.24 kg; me = m2 = 153.62 kg 0.75 v2 = = 0.004882 = 0.001251 + x2 x 0.04888 153.62 x2 = 0.07428 ; u2 = 1080.39 + 0.07428 x 1522 = 1193.45 m1u1 = 299.76 x 1080.39 + 7.48 x 2602.4 = 343324 kJ QCV = m2u2 – m1u1 + mehe = 153.62 x 1193.45 – 343324 + 153.62 x 1085.36 = 6744 kJ 6.60 An initially empty bottle, V = 0.25 m3, is filled with water from a line at 0.8 MPa, 350°C. Assume no heat transfer and that the bottle is closed when the pressure reaches line pressure. Find the final temperature and mass in the bottle. C.V. bottle + valve, 1Q2 = 0, 1W2 = 0, USUF Continuity Eq.: m2 – m1 = mi ; m1 = 0 ; Energy Eq.: m2u2 = mihi State 2: P2 = Pline , u2 = hi = 3161.7 kJ/kg T2 520°C, v2 = 0.4554 m2 = V/v2 = 0.25/0.4554 = 0.549 kg

Assume saturated mixture: m2 = V/v2 = V/(uf2 + x2vfg); u2 = uf + x2ufg 134.063 + x2 x 1175.26 – 180.36 = 42.157(0.001534 + x2 x 0.41684)/0.05 x2 = 0.05777 => Therefore, state 2 is saturated v2 = 0.02561 m3/kg T2 = -10°C m2 = V/v2 = 1.952 kg

6.63 A mass-loaded piston/cylinder, shown in Fig. P6.63, containing air is at 300 kPa, 17°C with a volume of 0.25 m3, while at the stops V = 1 m3. An air line, 500 kPa, 600 K, is connected by a valve that is then opened until a final inside pressure of 400 kPa is reached, at which point T = 350 K. Find the air mass that enters, the work, and heat transfer.

Open to P2 = 400 kPa, T2 = 350 K 300ÊxÊ0.25 m1 = = 0.90 kg 0.287ÊxÊ290.2 P1 ? const P to stops, then const V to P2 400ÊxÊ1 m2 = = 3.982 kg 0.287ÊxÊ350 mi = 3.982 – 0.90 = 3.082 kg CV: inside of cylinder 1W2 = P1(V2 – V1) = 300(1 – 0.25) = 225 kJ QCV + mihi = m2u2 – m1u1 + 1W2 QCV = 3.982 x 0.717 x 350 – 0.90 x 0.717 x 290.2 + 225 – 3.082 x 1.004 x 600 = -819.2 kJ

AIR 6.64 An elastic balloon behaves such that pressure is proportional to diameter and the balloon contains 0.5 kg air at 200 kPa, 30°C. The balloon is momentarily connected to an air line at 400 kPa, 100°C. Air is let in until the volume doubles, during which process there is a heat transfer of 50 kJ out of the balloon. Find the m2 – m1 = mi , m2u2 – m1u1 = mihi + 1Q2 – 1W2

1Q 2 P Process: P ~ D ~ V1/3 2 so P = P1(V/V1)1/3 1 v

m2u2 – m2hi = m1u1 – m1hi + 1Q2 – 1W2 hi = ui + RTi m2(u2 – ui – RTi) = m1(u1 – ui – RTi) + 1Q2 – 1W2 (P2V2/RT2)(Cv(T2 – Ti) – RTi) = m1(Cv(T1 – Ti) – RTi) + 1Q2 – 1W2 = 0.5(0.7165(30 – 100) – 0.287 x 373.15) – 50 – 49.583 = -178.2 T2 = 316.5 K = 43.4°C m2 = P2V2/RT2 = 1.207 kg mi = m2 – m1 = 0.707 kg

6.65 A 2-m3 storage tank contains 95% liquid and 5% vapor by volume of liquified natural gas (LNG) at 160 K, as shown in Fig. P6.65. It may be assumed that LNG has the same properties as pure methane. Heat is transferred to the tank and saturated vapor at 160 K flows into the a steady flow heater which it leaves at 300 K. The process continues until all the liquid in the storage tank is gone. Calculate the total amount of heat transfer to the tank and the total amount of heat transferred to the heater.

6.67 A rigid tank initally contains 100 L of saturated-liquid R-12 and 100 L of saturated-vapor R-12 at 0°C. A valve on the bottom of the tank is connected to a line flowing R-12 at 10°C, 900 kPa. A pressure-relief valve on the top of the tank is set at 745 kPa (when tank pressure reaches that value, mass escapes such that the tank pressure cannot exceed 745 kPa). The line valve is now opened, allowing 10 kg of R-12 to flow in from the line, and then this valve is closed. Heat is transferred slowly to the tank, until the final mass inside is 100 kg, at which point a) How much mass exits the pressure-relief valve during the overall process? b) How much heat is transferred to the tank?

6.70 Air is contained in the insulated cylinder shown in Fig. P6.70. At this point the air is at 140 kPa, 25°C, and the cylinder volume is 15 L. The piston cross-sectional area is 0.045 m2, and the spring is linear with spring constant 35 kN/m. The valve is opened, and air from the line at 700 kPa, 25°C, flows into the cylinder until the pressure reaches 700 kPa, and then the valve is closed. Find the final temperature. m2 = m1 + mi 1st law: mihi = m2u2 – m1u1 + WCV Ideal gas, const. specific heat: (m2 – m1)CPoTi = m2CVoT2 – m1CVoT1 + WCV Also P2V2 = m2RT2 Linear spring relation: P2 = P1 + (K/A2)(V2-V1) or 700 = 140 + 35 (V – 0.015); V = 0.0474 m3 22 2 (0.045) 700 x 0.0474 = m2 x 0.287 x T2; m2 = 115.61/T2 P1V1 140ÊxÊ0.015 Also m1 = = = 0.02454 kg RT1 0.287ÊxÊ298.2 W = ? ÊPdV = ØP Ê+Ê(K/A2)Ê(VÊ-ÊV )ØdV CV ? º 1 1 ß

6.71 An inflatable bag, initially flat and empty, is connected to a supply line of saturated vapor R-22 at ambient temperature of 10°C. The valve is opened, and the bag slowly inflates at constant temperature to a final diameter of 2 m. The bag is inflated at constant pressure, Po = 100 kPa, until it becomes spherical at Do = 1 m. After this the pressure and diameter are related according to A maximum pressure of 500 kPa is recorded for the whole process. Find the heat transfer to the R-22 10°C = T0 x = 1.0 P0 = 100 kPa Balloon spherical at D0 = 1 m For D > D0 , P = P0 + C(D*Ê-1 – D*Ê-7), D* = D/D0 slowly inflates ( T = const ) D2 = 2 m , PMAX = 500 kPa dPMAX *Ê-2 *Ê-8 * = 71/6 = 1.38309 = C(-D + 7D ) = 0 D dD*ÊÊ MAX MAX MAX 500 = 100 + C(0.72302 – 0.10329), C = 645.44 P2 = 100 + 645.44(2-1 – 2-7) = 417.7 kPa V = ( /6)D3, dV = ( /2)D2 dD = ( /2)D3 D*Ê2 dD* 0 ÊÊD*=2 3 3 D*Ê2 dD* W = ? PdV = P0 ( /6)D0 + ? ÊÊÊP D0 2 ÊÊD*=1 Ê 3 3 + C(D*Ê-1 – D*Ê-7)] D*Ê2 dD* = P0( /6)D0 +? Ê D0[P0 2 Ê

ENGLISH UNIT PROBLEMS 6.73E Air at 95 F, 16 lbf/in.2, flows in a 4 in. · 6 in. rectangular duct in a heating system. The volumetric flow rate is 30 cfm (ft3/min). What is the velocity of the air flowing in the duct?

A= 4 · 6 · 1 = 0.167 ft2 144 . . V 30 V = mv = AV V = = = 3.0 ft/s A 60Ê· Ê0.167

RT 53.34Ê· Ê554.7 3 ÊnoteÊidealÊgas:ÊÊÊÊvÊ=Ê Ê=Ê Ê=Ê12.842Êft /lbm P 16Ê· Ê144 . V 30 ? ÊÊÊÊÊÊÊÊÊmÊ=Ê Ê=Ê Ê=Ê0.0389Êlbm/sÊ ? v 60Ê· Ê12.842 6.74E Saturated vapor R-134a leaves the evaporator in a heat pump at 50 F, with a steady mass flow rate of 0.2 lbm/s. What is the smallest diameter tubing that can be used at this location if the velocity of the refrigerant is not to exceed 20 ft/s?

Table C.11.1: vg = 0.792 ft3/lbm m = AV/v A = mv/V = 0.2 · 0.792/20 = 0.00792 ft2

6.76E Carbon dioxide gas enters a steady-state, steady-flow heater at 45 lbf/in.2 60 F, and exits at 40 lbf/in.2, 1800 F. It is shown in Fig. P6.9, where changes in kinetic and potential energies are negligible. Calculate the required heat transfer per lbm C.V. heater: q + hi = he 20470.8Ê-Ê(-143.4) Table C.7: q = he – hi = = 468.4 Btu/lbm 44.01 (Use CP0 then q 0.203(1800 – 60) = 353.2 Btu/lbm) Too large T, Tave to use Cp0 at room temperature.

6.77E In a steam generator, compressed liquid water at 1500 lbf/in.2, 100 F, enters a 1-in. diameter tube at the rate of 5 ft3/min. Steam at 1250 lbf/in.2, 750 F exits the tube. Find the rate of heat transfer to the water.

6.79E A condenser, as the heat exchanger shown in Fig. P6.14, brings 1 lbm/s water flow at 1 lbf/in.2 from 500 F to saturated liquid at 1 lbf/in.2. The cooling is done by lake water at 70 F that returns to the lake at 90 F. For an insulated condenser, find the flow rate of cooling water.

500 F sat. liq. C.V. Heat exchanger 1 lbm/s . . . mcoolh70 + mH2Oh500 = m cool

6.82E A diffuser shown in Fig. P6.20 has air entering at 14.7 lbf/in.2, 540 R, with a velocity of 600 ft/s. The inlet cross-sectional area of the diffuser is 0.2 in.2. At the exit, the area is 1.75 in.2, and the exit velocity is 60 ft/s. Determine the exit .. 22 Cont: mi = AiVi/vi = me = AeVe/ve , Energy: hi + (1/2)Vi = he + (1/2)Ve h – h = (1/2)x(6002 – 602)/(32.174×778) = 7.119 Btu/lbm ei Te = Ti + 7.119/0.24 = 569.7 R ve = vi(AeVe/AiVi) = (RTi/Pi)(AeVe/AiVi) = RTe/Pe Pe = Pi(Te/Ti)(AiVi/AeVe) = 14.7(569.7/540)[0.2 ·600/1.75·60] = 17.72 lbf/in.2

6.83E Helium is throttled from 175 lbf/in.2, 70 F, to a pressure of 15 lbf/in.2. The. diameter of the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities are equal. Find the exit temperature of the helium and the ratio of the pipe diameters. Energy Eq.: he = hi, Ideal gas ? Te = Ti = 75 F , m = AV/(RT/P) . 1/2 1/2 But m, V & T constant D2/D1 = (P1/P2) = (175/15) = 3.416

6.84E Water flowing in a line at 60 lbf/in.2, saturated vapor, is taken out through a valve to 14.7 lbf/in.2. What is the temperature as it leaves the valve assuming no changes in kinetic energy and no heat transfer?

6.85E An insulated mixing chamber receives 4 lbm/s R-134a at 150 lbf/in.2, 220 F in a line with low velocity. Another line with R-134a as saturated liquid 130 F flows through a valve to the mixing chamber at 150 lbf/in.2 after the valve. The exit flow is saturated vapor at 150 lbf/in.2 flowing at 60 ft/s. Find the mass flow rate . . . . . . 12 Cont.: m1 + m2 = m3 ; Energy Eq.: m1h1 + m2h2 = m3( h3 + V3 ) 2 . 12 . 12 m2 (h2 – h3 – V3 ) = m1 ( h3 + V3 – h1 ) 22 1 3 1: Table C.11.1: 150 psia, 220 F, Mixer Heater h1 = 209.63 Btu/lbm 2 . 2: Table C.11.1: x = ? , 130 F, Q h2 = 119.88 Btu/lbm

State 3: x = 1, 150 psia, h3 = 180.61 121 2 V3 = 2 · 602 /(32.174 · 778) = 0.072 Btu/lbm . . 12 12 m2 = m1 (h3 2 V3 – h1)/ (h2 – h3- 2 V3 ) + = 4 (180.61 + 0.072 – 209.63)/ (119.88 – 180.61- 0.072) = 1.904 lbm/s 6.86E A steam turbine receives water at 2000 lbf/in.2, 1200 F at a rate of 200 lbm/s as shown in Fig. P6.29. In the middle section 40 lbm/s is withdrawn at 300 lbf/in.2, 650 F and the rest exits the turbine at 10 lbf/in.2, 95% quality. Assuming no heat transfer and no changes in kinetic energy, find the total turbine work.

C.V. Turbine SSSF, 1 inlet and 2 exit flows. 1 Table C.8.2 h1 = 1598.6 , h2 = 1341.6 Btu/lbm Table C.8.1 : h3 = hf + x3hfg = 161.2 + 0.95 · 982.1 = 1094.2 Btu/lbm Cont.: m1 = m2 + m3 => m3 = 160 lbm/s Energy: m1h1 = WT + m2h2 + m3h3 WT = m1h1-m2h2 -m3h3 = 9.1· 104 Btu/s 2 3

6.87E A small, high-speed turbine operating on compressed air produces a power output of 0.1 hp. The inlet state is 60 lbf/in.2, 120 F, and the exit state is 14.7 lbf/in.2, -20 F. Assuming the velocities to be low and the process to be adiabatic, find the mhi = mhe + W hi – he = Cp(Tin – Tex) = 0.24(120 – (-20)) = 33.6 Btu/lbm m = 0.1 · 550/(778 · 33.6) = 0.0021 lbm/s = 7.57 lbm/h 6.88E Hoover Dam across the Colorado River dams up Lake Mead 600 ft higher than the river downstream. The electric generators driven by water-powered turbines deliver 1.2 · 106 Btu/s. If the water is 65 F, find the minimum amount of water min(h+ V2/2 + gz)in = mex(h+ V2/2 + gz)ex + WT Lake DAM H Water states: hin hex ; vin vex so Mead T

6.90E An air compressor takes in air at 14 lbf/in.2, 60 F and delivers it at 140 lbf/in.2, 1080 R to a constant-pressure cooler, which it exits at 560 R. Find the specific compressor work and the specific heat transfer.

1 C.V. air. compressor q = 0 W Energy: h1 + wc = h2 c 23

Q wc = h2 – h1 = 261.1 – 124.3 = 136.8 Btu/lbm C.V. cooler w = 0/ Cont.: m3 = m1 Energy: h2 = q + h3 q = h2 – h3 = 261.1 – 133.98 = 127.12 Btu/lbm

6.92E For the same steam power plant as shown in Fig. P6.39 and Problem 6.91 determine the rate of heat transfer in the economizer which is a low temperature heat exchanger and the steam generator. Determine also the flow rate of cooling water through the condenser, if the cooling water increases from 55 to 75 F in the 200000Ê· Ê0.01617 Condenser: V7 = = 18 ft/s 3600Ê· Ê0.0491 0.182Ê-Ê62 q = 78.02 – 1028.7 + = -957.9 Btu/lbm 5

Q. COND = 200000(-957.9) = -1.916· 108 Btu/h Economizer V3 » V2 , Liquid v ~ const q = 323.0 – 85.3 = 237.7 Btu/lbm Q. ECON = 200000(237.7) = 4.75· 107 Btu/h Generator: 0.9505 V3 » 20 ft.s , V4 = 153 · = 151 ft/s 0.964 1.512Ê-Ê0.22 q = 1467.8 – 323.0 + = 1145.2 Btu/lbm 5

6.94E A 1-ft3 tank, shown in Fig. P6.49, that is initially evacuated is connected by a valve to an air supply line flowing air at 70 F, 120 lbf/in.2. The valve is opened, and air flows into the tank until the pressure reaches 90 lbf/in.2. Determine the final temperature and mass inside the tank, assuming the process is adiabatic. Develop an expression for the relation between the line temperature and the final temperature using constant specific heats.

a) C.V. Tank, USUF: Continuity Eq.: mi = m2 Energy Eq.: mihi = m2u2 u2 = hi = 293.64 ( Table C.6 ) T2 = 740 R P2V 90Ê· Ê144Ê· Ê1 m2 = = = 0.3283 lbm RT2 53.34Ê· Ê740 Assuming constant specific heat, hi = ui + RTi = u2 , RTi = u2 – ui = CVo(T2 – Ti) CVoT2 = (CVo + R)Ti = CPoTi , T2 = (CPo/CVo) Ti = kTi For Ti = 529.7 R & constant CPo, T2 = 1.40 · 529.7 = 741.6 R

6.95E A 20-ft3 tank contains ammonia at 20 lbf/in.2, 80 F. The tank is attached to a line flowing ammonia at 180 lbf/in.2, 140 F. The valve is opened, and mass flows in until the tank is half full of liquid, by volume at 80 F. Calculate the heat transferred from the tank during this process.

6.96E A 18-ft3 insulated tank contains air at 100 F, 300 lbf/in.2. A valve on the tank is opened, and air escapes until half the original mass is gone, at which point the me = m1 – m2, m2 = m1/2 me = m2 = 13.025 lbm 1st law: 0 = m2u2 – m1u1 + meheÊAV 0 = 13.025 · 0.171 T2 – 26.05 · 0.171 · 559.67 + 13.025 · 0 .24 (559.67 + T2)/2 Solving, T2 = 428 R m2RT2 P2 = = P1T2/2T1 = 300 · 428/2 · 559.67 = 114.7 lbf/in2 V

6.98E A 35-ft3 insulated, 90-lbm rigid steel tank contains air at 75 lbf/in.2, and both tank and air are at 70 F. The tank is connected to a line flowing air at 300 lbf/in.2, 70 F. The valve is opened, allowing air to flow into the tank until the pressure reaches 250 lbf/in.2 and is then closed. Assume the air and tank are always at the same 1st law: mihi = (m2u2 – m1u1)AIR + mST(u2 – u1)ST P1V 75Ê· Ê144Ê· Ê35 m1ÊAIR = = = 13.37 lbm RT1 53.34Ê· Ê530 P2V 250Ê· Ê144Ê· Ê35 23622 m2ÊAIR = = = RT2 53.34Ê· ÊT2 T2 23622 mi = (m2 – m1)AIR = – 13.37 T2 23Ê622 23622 Ê-Ê13.37 · 0.24 · 530 = · 0.171 · T2 ? T2 ? T2 – 13.37 · 0.171 · 530 + 90 · 0.107(T2 – 530) Solving, T2 = 589.3 R

6.99E A cylinder fitted with a piston restrained by a linear spring contains 2 lbm of R-22 at 220 F, 125 lbf/in.2. The system is shown in Fig. P6.72 where the spring constant is 285 lbf/in., and the piston cross-sectional area is 75 in.2. A valve on the cylinder is opened and R-22 flows out until half the initial mass is left. Heat is transferred so the final temperature of the R-22 is 30 F. Find the final state of the P – P = (k /A2)(V – V ) = (k /A2) (m v – m v ) 2 1 S 2 1 S 22 11 v1 = 0.636, h1 = 138.96, u1 = 138.96 – 125 · 0.636(144/778) = 124.25 P2 – 125 = (285 · 144 · 12 / 752 ) (1 · v2 – 2 · 0.636) If state 2 is 2-phase, P2= P sat(30F) = 69.591 lbf/in2 v2 = 0.63913 < vg ? 2-phase OK 0.63913 = 0.01243 + x2 · 0.7697 x2 = 0.8142 u2 = 18.45 + 0.8142 · 78.76 = 82.58 1 WCV = ? ÊPdV = (P1 + P2)(V2 - V1) 2 1 = (125 + 69.591) (1· 0.63913 - 2· 0.636)(144/778) = -11.4 Btu 2

he AVG = (h1 + h2) /2 = (138.96 + 82.58 + 8.23) /2 = 114.9 QCV = m2u2 – m1u1 + meheÊAVE + WCV = 1· 82.58 – 2 · 124.25 + 1 · 114.9 · 11.4 = – 62.42 Btu 6.100E An initially empty bottle, V = 10 ft3, is filled with water from a line at 120 lbf/in.2, 500 F. Assume no heat transfer and that the bottle is closed when the pressure reaches line pressure. Find the final temperature and mass in the bottle. C.V. bottle + valve, 1Q2 = 0, 1W2 = 0, USUF m-m=m;mu=mh 2 _1 i 2 2 i i State 2: P2 = Pline , u2 = hi = 1277.1 Btu/lbm T2 764 F, v2 = 6.0105 m2 = V/v2 = 10/6.0105 = 1.664 lbm

6.101E A mass-loaded piston/cylinder containing air is at 45 lbf/in.2, 60 F with a volume of 9 ft3, while at the stops V = 36 ft3. An air line, 75 lbf/in.2, 1100 R, is connected by a valve, as shown in Fig. P6.63. The valve is then opened until a final inside pressure of 60 lbf/in.2 is reached, at which point T=630R. Find the air mass that enters, the work, and heat transfer.

Open to: P2 = 60 lbf/in2 hi = 366.13 P1V1 45Ê· Ê9Ê· Ê144 m1 = = RT1 53.34Ê· Ê519.7 · = 2.104 lbm AIR 2 P = 45 lbf/in A1 T = 60°F 1 V = 9 ft3 1 3 V = 36 ft stop

7.1 Calculate the thermal efficiency of the steam power plant cycle described in Solution: From solution to problem 6.39, W = 24805 – 300 = 24505 kW NET Total Q = 13755 + 67072 = 80827 kW H . . 24505 = W /Q = = 0.303 TH NET H 80827

7.2 Calculate the coefficient of performance of the R-12 heat pump cycle described in Solution: From solution to problem 6.47, -W = 4.0 kW; -Q = 8.654 kW IN COND . . 8.654 Heat pump: ¢ = Q /W = = 2.164 H IN 4.0 7.3 Prove that a cyclic device that violates the Kelvin–Planck statement of the second law also violates the Clausius statement of the second law.

H.E. violating Kelvin receives Q from TH H HH LL H.P. discharges Q + Q to T . Net Q to HLH HHHLL H.E. + H.P. together transfers Q from T LL to T with no W thus violates Clausius. TL H QH

7.5 Discuss the factors that would make the heat pump described in Problem 6.47 an Solution: 2. Pressure loss (back flow leak) in compressor 5. Throttling process 3 => 4.

7.6 Calculate the thermal efficiency of a Carnot-cycle heat engine operating between reservoirs at 500°C and 40°C. Compare the result with that of Problem 7.1. Solution: T = 500°C = 773.2 K; T = 40°C = 313.2 K HL T Ê-ÊT 773.2Ê-Ê313.2 HL Carnot: = = = 0.595 (7.1 has : 0.3) TH T 773.2 H

7.9 In a steam power plant 1 MW is added at 700°C in the boiler, 0.58 MW is taken out at 40°C in the condenser and the pump work is 0.02 MW. Find the plant thermal efficiency. Assuming the same pump work and heat transfer to the boiler is given, how much turbine power could be produced if the plant were running in a Carnot cycle?

Solution: Q CV. Total: Q + WP,in = WT + Q H HL .

WT = 1 + 0.002 -0.58 = 0.44 MW WP, in WT TH = (WT – WP,in)/QH = 0.42 Carnot = Wnet/Q = 1 – T /T H LH .

L = 1 – 313.15 = 0.678 Q 973.15 WT – WP,in = CarnotQH = 0.678 MW WT = 0.698 MW

7.10 At certain locations geothermal energy in undergound water is available and used as the energy source for a power plant. Consider a supply of saturated liquid water at 150°C. What is the maximum possible thermal efficiency of a cyclic heat engine using this source of energy with the ambient at 20°C? Would it be better to locate a source of saturated vapor at 150°C than use the saturated liquid at 150°C?

Solution: MAX = 150°C = 423.2 K = TH TMin L T ; = 20°C = 293.2 K = T T Ê-ÊT 130 HL = = = 0.307 THÊMAX T 423.2 H Yes. Saturated vapor source at 150°C would remain at 150°C as it condenses to liquid, providing a large energy supply at that temperature.

7.11 Find the maximum coefficient of performance for the refrigerator in your kitchen, Solution:

The refrigerator coefficient of performance is = Q /W = Q /(Q – Q ) = T /(T – T ) LLHLLHL Assuming T ~ 0°C, T ~ 35°C, LH £ 273.15/(35 – 0) = 7.8 Actual working fluid temperatures must be such that T < Trefrigerator and T > Troom LH

7.12 An air-conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric air at 35°C. Estimate the amount of power needed to operate the air-conditioner. Solution: Qair = m h m Cp T = 1 · 1.004 · 20 = 20 kW Assume Carnot cycle refrigerator QL . . . TL 273Ê+Ê15 = . = QL / (QH – QL ) = = 14.4 W THÊ-ÊTL 35Ê-Ê15 W = QL / = 20.07 / 14.4 = 1.39 kW This estimate is the theoretical maximum performance. To do the required heat transfer TL 5°C and TH = 45°C are more likely; secondly < carnot 7.13 A sales person selling refrigerators and deep freezers will guarantee a minimum coefficient of performance of 4.5 year round. How would you evaluate that? Are Solution: Assume a high temperature of 35°C. If a freezer compartment is included T ~ -20°C (deep freezer) and fluid temperature is then T ~ -30°C LL deepÊfreezer £ T /(T - T ) = (273.15 - 30)/[35 - (-30)] = 3.74 LHL A hot summer day may require a higher T to push Q out into the room, so HH Claim is possible for a refrigerator, but not for a deep freezer.

7.14 A car engine operates with a thermal efficiency of 35%. Assume the air- conditioner has a coefficient of performance that is one third of the theoretical maximum and it is mechanically pulled by the engine. How much fuel energy should you spend extra to remove 1 kJ at 15°C when the ambient is at 35°C? Solution: Maximum for air-conditioner is for a Carnot cycle carnot = QL / W = T / (T – T ) = 288 / 20 = 14.4 LHL actual = 14.4 / 3 = 4.8 W = Q / = 1 / 4.8 = 0.2083 L

QH,engine = W/ eng = 0.2083 / 0.35 = 0.595 kJ 7.15 We propose to heat a house in the winter with a heat pump. The house is to be maintained at 20°C at all times. When the ambient temperature outside drops to -10°C, the rate at which heat is lost from the house is estimated to be 25 kW. What is the minimum electrical power required to drive the heat pump?

Solution: Minimum power if we QL QH assume a Carnot cycle . . HP QH = Qleak = 25 kW .

Q T 293.2 . 25 .H H ¢ = = = = 9.773 W = = 2.56 kW W T -T 30 IN 9.773 IN H L W Q leak

7.16 Electric solar cells can produce power with 15% efficiency. Assume a heat engine with a low temperature heat rejection at 30°C driving an electric generator with 80% efficiency. What should the effective high temperature in the heat engine be to have the same overall efficiency as the solar cells.

Solution: Wel = Q cell = gen Weng = gen eng QHeng => cell = gen eng H

7.17 A cyclic machine, shown in Fig. P7.17, receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200 kJ of work as output. Is this cycle reversible, irreversible, or impossible? Solution: T L Carnot = 1 – = 1 – 400/1000 = 0.6 T H

eng = W/Q = 200/325 = 0.615 > Carnot H 7.18 A household freezer operates in a room at 20°C. Heat must be transferred from the cold space at a rate of 2 kW to maintain its temperature at -30°C. What is the Solution: Assume a Carnot cycle between T = -30°C and T = 20°C: Q LHH = QL/Win = TL/(TH – TL) W = (273.15 – 30)/[20 – (-30)] = 4.86 REF Win = Q / = 2/4.86 = 0.41 kW L 2 kW QL

This is the theoretical minimum power input. Any actual machine requires a larger input.

7.19 A heat pump has a coefficient of performance that is 50% of the theoretical maximum. It maintains a house at 20°C, which leaks energy of 0.6 kW per degree temperature difference to the ambient. For a maximum of 1.0 kW power input find the minimum outside temperature for which the heat pump is a sufficient heat source.

7.20 A heat pump cools a house at 20°C with a maximum of 1.2 kW power input. The house gains 0.6 kW per degree temperature difference to the ambient and the heat pump coefficient of performance is 60% of the theoretical maximum. Find the maximum outside temperature for which the heat pump provides sufficient cooling.

Solution: W = 1.2 kW Here: Q TL = Thouse QH QL leak HP TH = Tamb T L

In this setup the low temperature space is the house and the high temperature space is the ambient. The heat pump must remove the gain or Qleak = 0.6 (Tamb – Thouse) = Q which must be removed by the heat pump. L ‘ = Q / W = 1 + Q / W = 0.6 ‘carnot = 0.6 Tamb / (Tamb – Thouse ) HL Substitute in for Q and multiply with (Tamb – Thouse): L (Tamb – Thouse ) + 0.6 (Tamb – Thouse )2 / W = 0.6 Tamb Since Thouse = 293.15 K and W = 1.2 kW it follows 2 Tamb – 585.5 Tamb + 85350.6 = 0 Solving = > Tamb = 311.51 K = 38.36 °C

7.21 Differences in surface water and deep water temperature can be utilized for power generation. It is proposed to construct a cyclic heat engine that will operate near Hawaii, where the ocean temperature is 20°C near the surface and 5°C at some depth. What is the possible thermal efficiency of such a heat engine?

7.22 A thermal storage is made with a rock (granite) bed of 2 m3 which is heated to 400 K using solar energy. A heat engine receives a QH from the bed and rejects heat to the ambient at 290 K. The rock bed therefore cools down and as it reaches 290 K the process stops. Find the energy the rock bed can give out. What is the heat engine efficiency at the beginning of the process and what is it at the end of the process?

Solution: Assume the whole setup is reversible and that the heat engine operates in a Carnot cycle. The total change in the energy of the rock bed is u2 – u1 = q = C T = 0.89 (400 – 290) = 97.9 kJ/kg m = V = 2750 · 2 = 5500 kg , Q = mq = 5500 · 97.9 = 538 450 kJ To get the efficiency use the CARNOT as = 1 – To/TH = 1 – 290/400 = 0.275 at the beginning of process = 1 – To/TH = 1 – 290/290 = 0.0 at the end of process

7.23 An inventor has developed a refrigeration unit that maintains the cold space at – 10°C, while operating in a 25°C room. A coefficient of performance of 8.5 is claimed. How do you evaluate this?

7.24 A steel bottle V = 0.1 m3 contains R-134a at 20°C, 200 kPa. It is placed in a deep freezer where it is cooled to -20°C. The deep freezer sits in a room with ambient temperature of 20°C and has an inside temperature of -20°C. Find the amount of energy the freezer must remove from the R-134a and the extra amount of work input to the freezer to do the process.

Solution: Energy equation: m(u2 – u1) = 1Q2 – 1W2 Process : V = Const => v2 = v1 => 1W2 = 0 Table B.5.2: v1 = 0.11436, u1 = 418.145 – 200 · 0.11436 = 395.273 m = V/ v1 = 0.87443 kg State 2: v2 = v1 < vg = 0.14649 Table B.5.1 => 2 phase => x2 = (0.11436 – 0.000738)/0.14576 = 0.77957 u2 = 173.65 + 0.77957*192.85 = 323.99 kJ/kg 1Q2 = m(u2 – u1) = – 62.334 kJ Assume Carnot cycle = QL / Win = T /(T -T ) = 253.15 / [ 20 – (-20)] = 6.33 LHL Win = QL / = 62.334 / 6.33 = 9.85 kJ 7.25 A certain solar-energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it, if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?

7.26 Liquid sodium leaves a nuclear reactor at 800°C and is used as the energy souce in a steam power plant. The condenser cooling water comes from a cooling tower at 15°C. Determine the maximum thermal efficiency of the power plant. Is it misleading to use the temperatures given to calculate this value?

Solution: o 800 C ENERGY TO H2O ENERGY FROM COOLING TOWERSTEAM POWER PLANT LIQ Na

T = 800°C = 1073.2 K, T = 15°C = 288.2 K HL T Ê-ÊT 1073.2Ê-Ê288.2 HL = = = 0.731 THÊMAX T 1073.2 H It might be misleading to use 800°C as the value for TH, since there is not a supply of energy available at a constant temperature of 800°C (liquid Na is Similarly, the H2O leaves the cooling tower and enters the condenser at 15°C, The water does not provide for condensing steam at a constant temperature of 15°C.

7.27 A 4L jug of milk at 25°C is placed in your refrigerator where it is cooled down to 5°C. The high temperature in the Carnot refrigeration cycle is 45°C and the properties of milk are the same as for liquid water. Find the amount of energy that must be removed from the milk and the additional work needed to drive the refrigerator.

C.V milk + out to the 5 °C refrigerator space Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Process : P = constant = 1 atm => 1W2 = Pm (v2 – v1) State 1: Table B.1.1, v1 vf = 0.001003 m3/kg, h1 hf = 104.87 kJ/kg m2 = m1 = V1/v1 = 0.004 / 0.001003 = 3.988 kg State 2: Table B.1.1, h2 hf = 20.98 kJ/kg 1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) + Pm (v2 – v1) = m(h2 – h1) 1Q2 = 3.998 (20.98 – 104.87) = -3.988 · 83.89 = – 334.55 kJ C.V. Refrigeration cycle TL = 5 °C ; TH = 45 °C, assume Carnot Ideal : = QL / W = QL / (QH – QL ) = TL/ (TH – TL) = 278.15 / 40 = 6.954 W = QL / = 334.55 / 6.954 = 48.1 kJ

7.28 A house is heated by a heat pump driven by an electric motor using the outside as the low-temperature reservoir. The house loses energy directly proportional to the temperature difference as Qloss = K(TH – TL). Determine the minimum electric power to drive the heat pump as a function of the two temperatures.

Solution: Coefficient of performance QL QH less than or equal to Carnot heat pump.

7.29 A house is heated by an electric heat pump using the outside as the low- temperature reservoir. For several different winter outdoor temperatures, estimate the percent savings in electricity if the house is kept at 20°C instead of 24°C. Assume that the house is losing energy to the outside as described in the previous problem.

Solution: Heat Pump Qloss (T – T ) HL . )2 Max Q T K(T Ê-ÊT ) . K(T Ê-ÊT . H = H. L H HL = ,W= Perf. W T Ê-ÊT W IN T IN H L IN H

A: T = 24°C = 297.2 K B: T = 20°C = 293.2 K HH AB T ,°C W /K W /K % saving L IN IN AB -20 6.514 5.457 16.2 % -10 3.890 3.070 21.1 % 0 1.938 1.364 29.6 % 10 0.659 0.341 48.3 % 7.30 An air-conditioner with a power input of 1.2 kW is working as a refrigerator ( = 3) or as a heat pump ( ‘ = 4). It maintains an office at 20°C year round which exchanges 0.5 kW per degree temperature difference with the atmosphere. Find the maximum and minimum outside temperature for which this unit is sufficient.

Solution: Analyse the unit in heat pump mode Replacement heat transfer equals the loss: H amb Q = 0.5 (T – T ) W = Q / ‘ = 0.5 (T – Tamb) / 4 HH T – Tamb = 4 W / 0.5 = 9.6 H

7.31 A house is cooled by an electric heat pump using the outside as the high- temperature reservoir. For several different summer outdoor temperatures, estimate the percent savings in electricity if the house is kept at 25°C instead of 20°C. Assume that the house is gaining energy from the outside directly proportional to the temperature difference.

Solution: Air-conditioner (Refrigerator) Q (T – T ) LEAK H L .

Max Q T K(T Ê-ÊT ) . K(T Ê-ÊT )2 . L L H. L H L = = ,W= Perf. W T Ê-ÊT W IN T IN H L IN L

A: T = 20°C = 293.2 K B: T = 25°C = 298.2 K LL AB T ,°C W /K W /K % saving H IN IN AB 45 2.132 1.341 37.1 % 40 1.364 0.755 44.6 % 35 0.767 0.335 56.3 % 7.32 Helium has the lowest normal boiling point of any of the elements at 4.2 K. At this temperature the enthalpy of evaporation is 83.3 kJ/kmol. A Carnot refrigeration cycle is analyzed for the production of 1 kmol of liquid helium at 4.2 K from saturated vapor at the same temperature. What is the work input to the refrigerator and the coefficient of performance for the cycle with an ambient at 300 K?

7.33 We wish to produce refrigeration at -30°C. A reservoir, shown in Fig. P7.33, is available at 200°C and the ambient temperature is 30°C. Thus, work can be done by a cyclic heat engine operating between the 200°C reservoir and the ambient. This work is used to drive the refrigerator. Determine the ratio of the heat transferred from the 200°C reservoir to the heat transferred from the -30°C reservoir, assuming all processes are reversible.

TH To = 30 C THÊ-ÊT0 = 200 C W=Q H1 T Q Q ?H? H1 H2 also W HE REF T0Ê-ÊTL W=Q L2 T ?L? Q L1 Q L2 To= 30 C TL =- 30 C

7.34 A combination of a heat engine driving a heat pump (similar to Fig. P7.33) takes waste energy at 50°C as a source Qw1 to the heat engine rejecting heat at 30°C. The remainder Qw2 goes into the heat pump that delivers a QH at 150°C. If the total waste energy is 5 MW find the rate of energy delivered at the high Solution: Waste supply: QW1 + QW2 = 5 MW Heat Engine: W = QW1 = ( 1 – TL1 / TH1 ) QW1 Heat pump: W = QH / HP = QW2 / ‘ = QW2 / [TH1 / (TH – TH1 )] Waste source Qw1 W HE QL HEAT 150 C QH

7.35 A temperature of about 0.01 K can be achieved by magnetic cooling, (magnetic work was discussed in Problems 4.41 and 4.42). In this process a strong magnetic field is imposed on a paramagnetic salt, maintained at 1 K by transfer of energy to liquid helium boiling at low pressure. The salt is then thermally isolated from the helium, the magnetic field is removed, and the salt temperature drops. Assume that 1 mJ is removed at an average temperature of 0.1 K to the helium by a Carnot-cycle heat pump. Find the work input to the heat pump and the coefficient of performance with an ambient at 300 K.

Solution: . . TL 0.1 = Q /W = = = 0.00033 L IN T Ê-ÊT 299.9 HL . 1· 10-3 W = =3J IN 0.00033 7.36 The lowest temperature that has been achieved is about 1 · 10-6 K. To achieve this an additional stage of cooling is required beyond that described in the previous problem, namely nuclear cooling. This process is similar to magnetic cooling, but it involves the magnetic moment associated with the nucleus rather than that associated with certain ions in the paramagnetic salt. Suppose that 10 µJ is to be removed from a specimen at an average temperature of 10-5 K (ten microjoules is about the potential energy loss of a pin dropping 3 mm). Find the work input to a Carnot heat pump and its coefficient of performance to do this assuming the ambient is at 300 K.

Solution: Q = 10 µJ = 10· 10-6 J at T = 10-5 K LL Q = Q · TH = 10· 10-6 · 300 = 300 J H L T 10-5 L

7.37 A heat pump heats a house in the winter and then reverses to cool it in the summer. The interior temperature should be 20°C in the winter and 25°C in the summer. Heat transfer through the walls and ceilings is estimated to be 2400 kJ per hour per degree temperature difference between the inside and outside. a. If the winter outside temperature is 0°C, what is the minimum power required b.For the same power as in part (a), what is the maximum outside summer tem- perature for which the house can be maintained at 25°C?

Solution: a) Winter: QL QH House is T and ambient H is at T L W Q leak HP

T = 20°C = 293.2 K , T = 0°C = 273.2 K and HL Q = 2400(20 -0) kJ/h H

. . 2400(20Ê-Ê0) TH 293.2 ¢ = Q /W = . = = H IN W T Ê-ÊT 20 IN H L W = 3275 kJ/h = 0.91 kW (For Carnot cycle) IN

b) W Summer: Q TL = Thouse QH QL leak HP TH = Tamb T L

T = 25°C = 298.2 K, W = 3275 kJ/h and L IN Q = 2400(T – 298.2) kJ/h LH .

7.38 It is proposed to build a 1000-MW electric power plant with steam as the working fluid. The condensers are to be cooled with river water (see Fig. P7.38). The maximum steam temperature is 550°C, and the pressure in the condensers will be 10 kPa. Estimate the temperature rise of the river downstream from the power Solution:

W = 106 kW, T = 550°C = 823.3 K NET H P = 10 kPa ? T = T (P = 10 kPa) = 45.8°C = 319 K COND L G TH L 823.2Ê-Ê319 Ê-ÊT = = = 0.6125 THÊCARNOT T 823.2 H . 1Ê-Ê0.6125 Q = 106 = 0.6327· 106 kW LÊMIN ? 0.6125 ? . 60Ê· Ê8Ê· Ê10/60 But m = = 80000 kg/s H O 0.001 2

. . 0.6327· 106 T = Q /m C = = 1.9°C H OÊMIN MIN H O PÊLIQÊH O 80000Ê· Ê4.184 222

7.39 Two different fuels can be used in a heat engine, operating between the fuel- burning temperature and a low temperature of 350 K. Fuel A burns at 2500 K delivering 52000 kJ/kg and costs $1.75 per kilogram. Fuel B burns at 1700 K, delivering 40000 kJ/kg and costs $1.50 per kilogram. Which fuel would you buy Solution:

350 Fuel A: = 1 – T /T = 1 – = 0.86 TH,A L H 2500 WA = · Q = 0.86 · 52000 = 44720 kJ/kg TH,A A W /$ = 44720/1.75 = 25554 kJ/$ AA 350 Fuel B: = 1 – T /T = 1 – = 0.794 TH,B L H 1700 W = · Q = 0.794 · 40000 = 31760 kJ/kg B TH,B B W /$ = 31760/1.5 = 21173 kJ/$ BB Select fuel A for more work per dollar.

7.40 A refrigerator uses a power input of 2.5 kW to cool a 5°C space with the high temperature in the cycle as 50°C. The QH is pushed to the ambient air at 35°C in a heat exchanger where the transfer coefficient is 50 W/m2K. Find the required Solution: W = 2.5 kW = QH / HP QH = W · HP = 2.5 · [323 / (50 – 5)] = 17.95 kW = h A T A = Q / h T = 17.95 / 50 · 10-3 · 15 = 23.9 m2 H

7.41 Refrigerant-12 at 95°C, x = 0.1 flowing at 2 kg/s is brought to saturated vapor in a constant-pressure heat exchanger. The energy is supplied by a heat pump with a low temperature of 10°C. Find the required power input to the heat pump. Solution:

1 · Q H 2 · W · Q L TL Assume Carnot heat pump ‘ = Q /W = T /(T – T ) HHHL T = 368.2, T = 283.2, => ‘ = 4.332 HL Table B.3.1: h = 147.23, h = 211.73 12 Q = m (h – h ) = 129.0 kW H R-12 2 1 W = Q / = 129.0/4.332 = 29.8 kW H

7.42 A furnace, shown in Fig. P7.42, can deliver heat, QH1 at TH1 and it is proposed to use this to drive a heat engine with a rejection at Tatm instead of direct room heating. The heat engine drives a heat pump that delivers QH2 at Troom using the atmosphere as the cold reservoir. Find the ratio QH2/QH1 as a function of the temperatures. Is this a better set-up than direct room heating from the furnace? Solution: C.V.: Heat Eng.: W = Q where = 1 – Tatm/T HE H1 H1 C.V.: Heat Pump: W = Q / ¢ where ¢ = Trm/(Trm – Tatm) HP H2 Work from heat engine goes into heat pump so we have Q = ¢W = ¢ Q H2 HP H1 and we may substitute T’s for ¢, . If furnace is used directly Q = Q , H2 H1 so if ¢ > 1 this proposed setup is better. Is it? For T > Tatm formula shows H1 that it is good for Carnot cycles. In actual devices it depends wether ¢ > 1 is obtained.

7.43 A heat engine has a solar collector receiving 0.2 kW per square meter inside which a transfer media is heated to 450 K. The collected energy powers a heat engine which rejects heat at 40 C. If the heat engine should deliver 2.5 kW what is the minimum size (area) solar collector?

Solution: TH = 450 K TL = 40 oC = 313.15 K HE = 1 – TL / TH = 1 – 313.15 / 450 = 0.304 W = QH = > QH = W / = 2.5 / 0.304 = 8.224 kW QH = 0.2 A = > A = QH / 0.2 = 41 m2

7.44 In a cryogenic experiment you need to keep a container at -125°C although it gains 100 W due to heat transfer. What is the smallest motor you would need for a heat pump absorbing heat from the container and rejecting heat to the room at 20°C?

Solution: ¢ = Q / W = = = 2.022 = 1 + Q / W HP H T Ê-ÊT 20Ê-Ê(-125) L HL => W = Q /( ‘ – 1) = 100/1.022 = 97.8 W L

7.45 Sixty kilograms per hour of water runs through a heat exchanger, entering as saturated liquid at 200 kPa and leaving as saturated vapor. The heat is supplied by a Carnot heat pump operating from a low-temperature reservoir at 16°C. Find the Solution: C.V. Heat exchanger …..1 m1 = m2 ; m1h1 + Q = m1h2 HH Table B.1.2: h1 = 504.7, h2 = 2706.7 TH = Tsat(P) = 120.93 +273.15 = 394.08 · Q L

. 1 TL Q = = (2706.7 – 504.7) = 36.7 kW H 60 2 · Q

7.46 Air in a rigid 1 m3 box is at 300 K, 200 kPa. It is heated to 600 K by heat transfer from a reversible heat pump that receives energy from the ambient at 300 K besides the work input. Use constant specific heat at 300 K. Since the coefficient of performance changes write dQ = mair Cv dT and find dW. Integrate dW with Solution:

Solution: TH W Ê-ÊT L W = Q = · K(T – T ) ; maximize W(T ) = 0/ TH res T TH H H H H W = K(T – T )T T -2- K(1 – T /T ) = 0/ T res H L H L H H

T=TT H res L 7.49 A 10-m3 tank of air at 500 kPa, 600 K acts as the high-temperature reservoir for a Carnot heat engine that rejects heat at 300 K. A temperature difference of 25°C between the air tank and the Carnot cycle high temperature is needed to transfer the heat. The heat engine runs until the air temperature has dropped to 400 K and then stops. Assume constant specific heat capacities for air and find how much work is given out by the heat engine.

Solution: AIR TH = Tair – 25°C TL = 300 K Q P1V 500Ê· Ê10 H W mair = RT1 = 0.287Ê· Ê600 = 29.04 kg HE TL dW = dQ = 1Ê-Ê dQ Q H ? TairÊ-Ê25? H L

300 K dQH = -mairdu = -mairCvdTair Ø TL Ø Ø Ta2-25Ø W = ? dW = -mairCv ?1Ê-Ê œdTa = -mairCv?T -T -T ÊlnÊ œ ? º Ta-25ß º a2 Ta1-25ß a1 L

7.50 Consider a Carnot cycle heat engine operating in outer space. Heat can be rejected from this engine only by thermal radiation, which is proportional to the radiator area and the fourth power of absolute temperature, Qrad ~ KAT4. Show that for a given engine work output and given TH, the radiator area will be minimum when Solution: T Ê-ÊT T Ê-ÊT HL HL 4 W = Q = Q ; also Q = KATL NET H T L T L ?H??L? 4 W AT T ØT 3 T 4Ø NET L H L L = Ê-Ê1 = A ? Ê-Ê œ= const 4 4T T T KT T ? L ? º? H? ? H? ß HH Differentiating,

ØTL 3 TL 4Ø Ø TL 2 TL 3Ø TL dA ? Ê-Ê œ+ A ?3 Ê-Ê4 œd = 0 TTTTT º? H? ? H? ß º ? H? ? H? ß ? H? dA Ø TL 2 TL 3Ø TL 3 TL 4 = – A ?3 Ê-Ê4 œ/ [ – ] = 0 d(T /T ) T T T T L H º ? H? ? H? ß ? H? ? H?

7.51 Air in a piston/cylinder goes through a Carnot cycle with the P-v diagram shown in Fig. 7.24. The high and low temperatures are 600 K and 300 K respectively. The heat added at the high temperature is 250 kJ/kg and the lowest pressure in the cycle is 75 kPa. Find the specific volume and pressure at all 4 states in the cycle assuming constant specific heats at 300 K..

7.52 Hydrogen gas is used in a Carnot cycle having an efficiency of 60% with a low temperature of 300 K. During the heat rejection the pressure changes from 90 kPa to 120 kPa. Find the high and low temperature heat transfer and the net cycle Solution:

= 0.6 = 1 – TL / TH = > TH = TL /(1 – 0.6) = 750 K v3 / v4 = ( RT3 / P3 ) / ( RT4 / P4 ) = P4 / P3 = 120 / 90 = 1.333 qL = RTL ln (v3/v4 ) = 355.95 kJ/kg ; R = 4.1243 qH = qL / (1 – 0.6) = 889.9 kJ/kg ; w = qH – qL = 533.9 kJ/kg

7.53 Obtain information from manufacturers of heat pumps for domestic use. Make a listing of the coefficient of performance and compare those to corresponding Carnot cycle devices operating between the same temperature reservoirs.

7.54E Calculate the thermal efficiency of the steam power plant cycle described in Solution:

From solution to problem 6.91, W. NET = 33000 – 400 = 32600 hp = 8.3·107 Btu/h .

W = . = 0.30 Q H 7.55E Calculate the thermal efficiency of a Carnot-cycle heat engine operating between reservoirs at 920 F and 110 F. Compare the result with that of Problem 7.54.

Solution: T = 920 F , T = 110 F HL TL 110Ê+Ê459.67 Carnot = 1 – = 1 – = 0.587 (about twice 7.54: 0.3) T 920Ê+Ê459.67 H

7.56E A car engine burns 10 lbm of fuel (equivalent to addition of QH) at 2600 R and rejects energy to the radiator and the exhaust at an average temperature of 1300 R. If the fuel provides 17 200 Btu/lbm what is the maximum amount of work the engine can provide?

7.57E In a steam power plant 1000 Btu/s is added at 1200 F in the boiler, 580 Btu/s is taken out at 100 F in the condenser and the pump work is 20 Btu/s. Find the plant thermal efficiency. Assume the same pump work and heat transfer to the boiler as given, how much turbine power could be produced if the plant were running in a Carnot cycle?

Solution: Q C.V. Total: QH + WP,in = WT + QL H W = 1000 + 20 – 580 = 440 Btu/s T WP, in WT = (W – WP,in)/Q = 420/1000 = 0.42 TH T H carnot = W / Q = 1 – TL/TH net H Q L 100Ê+Ê459.67 = 1 – = 0.663 1200Ê+Ê459.67

. . . . Btu – WP,in = carnotQ WT W = 663 Btu/s => = 683 THs

7.58E An air-conditioner provides 1 lbm/s of air at 60 F cooled from outside atmospheric air at 95 F. Estimate the amount of power needed to operate the air- conditioner. Clearly state all assumptions made.

7.59E A car engine operates with a thermal efficiency of 35%. Assume the air- conditioner has a coefficient of performance that is one third of the theoretical maximum and it is mechanically pulled by the engine. How much fuel energy should you spend extra to remove 1 Btu at 60 F when the ambient is at 95 F? Solution:

= TL / (TH – TL ) = (60 + 459.67) / (95 – 60) = 14.8 actual = / 3 = 4.93 W = QL / = 1 / 4.93 = 0.203 Btu QHengine = W / = 0.203 / 0.35 = 0.58 Btu 7.60E We propose to heat a house in the winter with a heat pump. The house is to be maintained at 68 F at all times. When the ambient temperature outside drops to 15 F, the rate at which heat is lost from the house is estimated to be 80000 Btu/h. What is the minimum electrical power required to drive the heat pump?

Solution: Minimum power if we QL QH assume a Carnot cycle QH = Qleak = 80 000 Btu/h W Q leak HP

7.61E A heat pump cools a house at 70 F with a maximum of 4000 Btu/h power input. The house gains 2000 Btu/h per degree temperature difference to the ambient and the heat pump coefficient of performance is 60% of the theoretical maximum. Find the maximum outside temperature for which the heat pump provides Solution: W = 4000 Btu/h Here: TL = Thouse QH QL Qleak TH = Tamb HP T L

In this setup the low temperature space is the house and the high temperature space is the ambient. The heat pump must remove the gain or Qleak = 2000 (Tamb – Thouse) = Q which must be removed by the heat pump. L ‘ = Q / W = 1 + Q / W = 0.6 ‘carnot = 0.6 Tamb / (Tamb – Thouse ) HL Substitute in for Q and multiply with (Tamb – Thouse): L (Tamb – Thouse ) + 2000 (Tamb – Thouse )2 / W = 0.6 Tamb Since Thouse = 529.7 R and W = 4000 Btu/h it follows 2 Tamb – 1058.6 Tamb + 279522.7 = 0 Solving => Tamb = 554.5 R = 94.8 F

7.62E A thermal storage is made with a rock (granite) bed of 70 ft3 which is heated to 720 R using solar energy. A heat engine receives a QH from the bed and rejects heat to the ambient at 520 R. The rock bed therefore cools down and as it reaches 520 R the process stops. Find the energy the rock bed can give out. What is the heat engine efficiency at the beginning of the process and what is it at the end of Solution:

7.63E An inventor has developed a refrigeration unit that maintains the cold space at 14 F, while operating in a 77 F room. A coefficient of performance of 8.5 is claimed. How do you evaluate this?

Solution: Assume Carnot cycle then Q T 14Ê+Ê459.67 LL = = = = 7.5 Win TH-TL 77Ê-Ê14 Claim is impossible 7.64E Liquid sodium leaves a nuclear reactor at 1500 F and is used as the energy source in a steam power plant. The condenser cooling water comes from a cooling tower at 60 F. Determine the maximum thermal efficiency of the power plant. Is it misleading to use the temperatures given to calculate this value?

Solution: 1500 F ENERGY TO H2O ENERGY FROM REACTOR STEAM POWER PLANT LIQ Na 60 F

COOLING TOWER LIQ H O 2 T = 1500 F = 1960 R, T = 60 F = 520 R HL T Ê-ÊT 1960Ê-Ê520 HL = = = 0.735 THÊMAX T 19860 H It might be misleading to use 1500 F as the value for T , since there is not a H supply of energy available at a constant temperature of 1500 F (liquid Na is cooled to a lower temperature in the heat exchanger).

7.65E A house is heated by an electric heat pump using the outside as the low- temperature reservoir. For several different winter outdoor temperatures, estimate the percent savings in electricity if the house is kept at 68 F instead of 75 F. Assume that the house is losing energy to the outside directly proportional to the temperature difference as Q. loss = K(TH – TL).

Solution: Heat Pump Q (T – T ) LOSS H L .

Max QH TH K(THÊ-ÊTL) . K(THÊ-ÊTL)2 .==.,W= Perf. Win THÊ-ÊTL Win in TH A: T = 75 F = 534.7 R B: T = 68 F = 527.7 R HA HB T , F WINA/K WINB/K % saving L -10 13.512 11.529 14.7 % 10 7.902 6.375 19.3 % 30 3.787 2.736 27.8 % 50 1.169 0.614 47.5 % 7.66E A house is cooled by an electric heat pump using the outside as the high- temperature reservoir. For several different summer outdoor temperatures estimate the percent savings in electricity if the house is kept at 77 F instead of 68 F. Assume that the house is gaining energy from the outside directly proportional to the temperature difference.

Solution: Air-conditioner (Refrigerator) Q (T – T ) LEAK H L .

7.67E We wish to produce refrigeration at -20 F. A reservoir is available at 400 F and the ambient temperature is 80 F, as shown in Fig. P7.33. Thus, work can be done by a cyclic heat engine operating between the 400 F reservoir and the ambient. This work is used to drive the refrigerator. Determine the ratio of the heat transferred from the 400 F reservoir to the heat transferred from the -20 F reservoir, assuming all processes are reversible.

TH= 860 R To = 540 R T Ê-ÊT HO Q H1 QH2 W = QH1 ? TH ? W also HETOÊ-ÊTL Q L1 Q L2 W = QL2 ? TL ? REF

To= 540 R TL = 440 R QH TOÊ-ÊTL TH 100 860 = = · = 0.611 QL TL ??THÊ-ÊTO? 440 320 ?

7.68E Refrigerant-22 at 180 F, x = 0.1 flowing at 4 lbm/s is brought to saturated vapor in a constant-pressure heat exchanger. The energy is supplied by a heat pump with a low temperature of 50 F. Find the required power input to the heat pump.

7.69E A heat engine has a solar collector receiving 600 Btu/h per square foot inside which a transfer media is heated to 800 R. The collected energy powers a heat engine which rejects heat at 100 F. If the heat engine should deliver 8500 Btu/h what is the minimum size (area) solar collector?

Solution: TH = 800 R TL = 100 + 459.67 = 560 R = 1 – TL/ TH = 1 – 560/800 = 0.3 W = QH => QH = W / = 8500/0.3 = 28333 Btu/s QH = 600 A => A = QH / 600 = 47.2 ft2 7.70E Six-hundred pound-mass per hour of water runs through a heat exchanger, entering as saturated liquid at 30 lbf/in.2 and leaving as saturated vapor. The heat is supplied by a Carnot heat pump operating from a low-temperature reservoir at 60 F. Find the rate of work into the heat pump.

Solution: C.V. Heat exchanger ….. 1 m1 = m2 ; m1h1 + QH 1h2 =m Table C.8.1: h1 = 218.92 h2 = 1164.3 T = Tsat(P) = 250.34 F = 710 R H .1 Q = (1164.3- 218.92) = 157.6 Btu/s H6 Assume a Carnot heat pump.

. . TH 710 = Q /W = = = 3.73 H T Ê-ÊT 190.34 HL W = Q / = 157.6/3.73 = 42.25 Btu/s H 2 · Q H

7.71E Air in a rigid 40 ft3 box is at 540 R, 30 lbf/in.2. It is heated to 1100 R by heat transfer from a reversible heat pump that receives energy from the ambient at 540 R besides the work input. Use constant specific heat at 540 R. Since the coefficient of performance changes write dQ = mair Cv dT and find dW. Integrate dW with temperature to find the required heat pump work.

Solution: = QH / W = QH / (QH – QL) TH / (TH – TL) mair = P1V1 / RT1 = (30 · 40 · 144) / (540 · 53.34) = 6.0 lbm dQ = mair Cv dTH = dW = [TH / (T – T )] dW H HL = > dW = mair Cv [TH / (T – T )] dTH HL 1W2 = mair Cv ( 1- T / T ) dT = mair Cv ( 1 – T / T ) dT LL = mair Cv [T2 – T1 – TL ln (T2 / T1)] 1W2 = 6.0 · 0.171 [1100 – 540 – ln (1100/540)] = 180.4 Btu

7.72E A 350-ft3 tank of air at 80 lbf/in.2, 1080 R acts as the high-temperature reservoir for a Carnot heat engine that rejects heat at 540 R. A temperature difference of 45 F between the air tank and the Carnot cycle high temperature is needed to transfer the heat. The heat engine runs until the air temperature has dropped to 700 R and then stops. Assume constant specific heat capacities for air and find how much Solution:

AIR TH = Tair – 45 , TL = 540 R P1V 80Ê· Ê350Ê· Ê144 Q H mair = = = 69.991 lbm W RT1 53.34Ê· Ê1080 HE TL dW = dQH = 1Ê-Ê dQH Q TairÊ-Ê45 L ?? 300 K dQH = -mairdu = -mairCvdTair

7.73E Air in a piston/cylinder goes through a Carnot cycle with the P-v diagram shown in Fig. 7.24. The high and low temperatures are 1200 R and 600 R respectively. The heat added at the high temperature is 100 Btu/lbm and the lowest pressure in the cycle is 10 lbf/in.2. Find the specific volume and pressure at all 4 states in the cycle assuming constant specific heats at 80 F.

CHAPTER 8 The correspondence between the new problem set and the previous 4th edition chapter 7 problem set.

New Old New Old New Old 1 new 26 new 51 61 2 1 27 29 52 new 3 2 28 28 53 42 4 3 29 27 54 43 5 new 30 new 55 44 6 4 31 new 56 45 7 5 32 new 57 46 8 new 33 20 58 new 9 7 34 21 59 48 10 8 35 22 60 new 11 9 36 30 61 49 12 10 37 31 62 51 13 11 38 new 63 53 14 new 39 new 64 54 15 13 40 33 65 new 16 new 41 34 66 56 17 15 42 new 67 57 18 6 43 35 68 new 19 16 44 36 69 58 20 12 45 37 70 55 21 17 46 38 71 60 22 new 47 39 72 59 23 new 48 new 73 14 24 new 49 40 74 52 25 25 50 41 75 new

8.1 Consider the steam power plant in Problem 7.9 and the heat engine in Problem Solution: Show Clausius: dQ/T £ 0 For problem 7.9 we have : QH / TH – QL / TL = 1000/973.15 – 580/313.15 = 1.0276 – 1.852 = -0.825 < 0 OK For problem 7.17 we have: QH / TH - QL / TL = 325/1000 - 125/400 = 0.325 - 0.3125 = 0.0125 > 0 This is impossible 8.2 Find the missing properties and give the phase of the substance a) Table B.1.1 T = Tsat(P) = 64.97°C 7.70Ê-Ê0.893 x = (s – sf)/sfg = = 0.981 Ê6.9383 h = 271.9 + 0.981 · 2346.3 = 2573.8 kJ/kg b) Table B.1.2 u > ug => Sup.vap Table B.1.3, x = undefined T 682°C , s 7.1223 kJ/kg K c) Table B.3.2, sup. vap., x = undefined, s = 0.7139 kJ/kg K d) Table B.5.1 v = vf + xvfg = 0.000755 + 0.45 · 0.098454 = 0.04506 m3/kg s = sf + xsfg = 0.9507 + 0.45 · 0.7812 = 1.3022 kJ/kg K e) Table B.2.1, s > sg => Sup.vap. Table B.2.2, x = undefined u = h-Pv = 1492.8 – 439.18 · 0.3100 = 1356.7 kJ/kg

8.3 Consider a Carnot-cycle heat engine with water as the working fluid. The heat transfer to the water occurs at 300°C, during which process the water changes from saturated liquid to saturated vapor. The heat is rejected from the water at 40°C. Show the cycle on a T–s diagram and find the quality of the water at the beginning and end of the heat rejection process. Determine the net work output per kilogram of water and the cycle thermal efficiency.

From the definition of the Carnot cycle, two constant s and two constant T processes.

T s3 = s2 = 5.70438 300 1 2 = 0.57243 + x3(7.68449) x3 = 0.6678 40 s4 = s1 = 3.25327 43 = 0.57243 + x4(7.68449) s x4 = 0.3489 c) TH = wNET/qH = (TH – TL)/TH = 260/573.2 = 0.4536 qH = TH(s2 – s1) = 573.2(5.70438 – 3.25327) = 1405.0 kJ/kg wNET = TH · qH = 637.3 kJ/kg

8.4 In a Carnot engine with water as the working fluid, the high temperature is 250°C and as QH is received, the water changes from saturated liquid to saturated vapor. The water pressure at the low temperature is 100 kPa. Find TL, the cycle thermal efficiency, the heat added per kilogram, and the entropy, s, at the beginning of the heat rejection process.

8.5 Water is used as the working fluid in a Carnot cycle heat engine, where it changes from saturated liquid to saturated vapor at 200°C as heat is added. Heat is rejected in a constant pressure process (also constant T) at 20 kPa. The heat engine powers a Carnot cycle refrigerator that operates between -15°C and +20°C. Find the heat added to the water per kg water. How much heat should be added to the water in the heat engine so the refrigerator can remove 1 kJ from the cold space? Solution: Carnot cycle: qH = TH (s2 – s1 ) = hfg = 473.15 (4.1014) = 1940 kJ/kg TL = Tsat (20 kPa) = 60.06 oC ref = QL / W = TL / (TH – TL ) = (273 – 15) / (20 – (-15)) = 258 / 35 = 7.37 W = QL / = 1 / 7.37 = 0.136 kJ W = HE QHÊH2O HE = 1 – 333/473 = 0.29 QHÊH2O = 0.136 / 0.296 = 0.46 kJ 8.6 Consider a Carnot-cycle heat pump with R-22 as the working fluid. Heat is rejected from the R-22 at 40°C, during which process the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 0°C. b. Find the quality of the R-22 at the beginning and end of the isothermal heat c. Determine the coefficient of performance for the cycle.

a)T 40 0 c) b) s4 = s3 = 1.1909 kJ/kg K 3 2 = 1.00 + x4(0.7262) => x4 = 0.2629 s1 = s2 = 1.7123 kJ/kg K 41 = 1.00 + x1(0.7262) s => x1 = 0.9809 ¢ = qH/wIN = TH/(TH – TL) = 313.2/40 = 7.83 8.8 Water at 200 kPa, x = 1.0 is compressed in a piston/cylinder to 1 MPa, 250°C in a reversible process. Find the sign for the work and the sign for the heat transfer. Solution: State 1: Table B.1.1: v1 = 0.8857 ; u1 = 2529.5 kJ/kg; s1 = 7.1271 kJ/kg K State 2: Table B.1.3: v2 = 0.23268; u2 = 2709.9 kJ/kg; s2 = 6.9246 kJ/kg K v2 < v1 => 1w2 = P dv < 0 s2 < s1 => 1q2 = T ds < 0 PT 2 2 1 1

v s 8.9 One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isothermal process to 100 kPa. Find the work and heat transfer for this C.V.: NH3 m2 = m1 ; m(u2 – u1) = 1Q2 – 1W2 P 1 2 Rev.: 1W2 = ? ÊPdV 1Q2 = ? ÊTds = T(s2 – s1) State 1: u1 = 1391.3 ; s1 = 5.265 T

8.10 One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isobaric process to 140°C. Find the work and heat transfer for this Control mass. P T 2 m(u2 – u1) = 1Q2 – 1W2 1 2 1 Process: P = constant T 1W2 = mP(v2 – v1) vs

State 1: v1 = 0.145 m3/kg, Table B.2.2 u1 = h1 – P1v1 = 1536.3 – 1000 · 0.145 = 1391.3 kJ/kg State 2: v2 = 0.1955 m3/kg, Table B.2.2 u2 = h2 – P2v2 = 1762.2 – 1000 · 0.1955 = 1566.7 kJ/kg 1W2 = 1 · 1000(0.1955 – 0.145) = 50.5 kJ 1Q2 = m(u2 – u1) + 1W2 = 1 · (1566.7 – 1391.3) + 50.5 = 225.9 kJ

8.11 One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible adiabatic process to 100 kPa. Find the work and heat transfer for this Control mass: Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen Process: 1Q2 = 0/ ; 1S2,gen = 0/ s2 = s1 = 5.2654 kJ/kg K State 1: (P, T) Table B.2.2, u1 = h1 – P1v1 = 1536.3 – 1000· 0.14499 = 1391.3 State 2: P2 , s2 2 phase Table B.2.1 Interpolate: sg2 = 5.8404 kJ/kg K, sf = 0.1192 kJ/kg K x2 = (5.2654-0.1192)/5.7212 = 0.90, u2 = 27.66 + 0.9· 1257.0 = 1158.9 1W2 = 1 · (1391.3 – 1158.9) = 232.4 kJ

8.12 A cylinder fitted with a piston contains ammonia at 50°C, 20% quality with a volume of 1 L. The ammonia expands slowly, and during this process heat is transferred to maintain a constant temperature. The process continues until all the liquid is gone. Determine the work and heat transfer for this process.

Table B.2.1: T1 = 50°C, x1 = 0.20, V1 = 1 L v1 = 0.001777 + 0.2 · 0.06159 = 0.014095 T s1 = 1.5121 + 0.2 · 3.2493 = 2.1620 m = V1/v1 = 0.001/0.014095 = 0.071 kg 50 C 1 2 S v2 = vG = 0.06336, s2 = sG = 4.7613 Process: T = constant to x2 = 1.0, P = constant = 2.033 MPa 1W2 = ? PdV = Pm(v2 – v1) = 2033 · 0.071 · (0.06336 – 0.014095) = 7.11 kJ 1Q2 = ? TdS = Tm(s2 – s1) = 323.2 · 0.071(4.7613 – 2.1620) = 59.65 kJ or 1Q2 = m(u2 – u1) + 1W2 = m(h2 – h1) h1 = 421.48 + 0.2 · 1050.01 = 631.48, h2 = 1471.49 1Q2 = 0.071(1471.49 – 631.48) = 59.65 kJ 8.13 An insulated cylinder fitted with a piston contains 0.1 kg of water at 100°C, 90% quality. The piston is moved, compressing the water until it reaches a pressure of Energy Eq.: 1Q2 = 0 = m(u2 – u1) + 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 0 + 0 (assume reversible) NH3

P 2 State 1: 100°C, x1 = 0.90: Table B.1.1, s1 = 1.3068 + 0.90· 6.048 1 = 6.7500 kJ/kg K u1 = 418.91 + 0.9 · 2087.58 = 2297.7 kJ/kg s2Ê=Ês1Ê=Ê6.7500 ÊT2Ê=Ê232.3°C P2Ê=Ê1.2ÊMPa Êu2Ê=Ê2672.9 1W2 = -0.1(2672.9 – 2297.7) = -37.5 kJ v T 2

8.14 A cylinder fitted with a frictionless piston contains water. A constant hydraulic pressure on the back face of the piston maintains a cylinder pressure of 10 MPa. Initially, the water is at 700°C, and the volume is 100 L. The water is now cooled and condensed to saturated liquid. The heat released during this process is the Q supply to a cyclic heat engine that in turn rejects heat to the ambient at 30°C. If the overall process is reversible, what is the net work output of the heat engine?

C.V.: H2O, 1‡3, this is a control mass: Continuity Eq.: m1 = m3 = m Process: P = C => 1W3 = P dV = Pm(v3-v1) State 1: 700oC, 10 MPa, V1 = 100 L Table B.1.4 v1 = 0.04358 m3/kg => m = m1 = V1/v1 = 2.295 kg h1 = 3870.5 kJ/kg, s1 = 7.1687 kJ/kg K State 3: P3 = P1 = 10 MPa, x3 = 0 Table B.1.2 h3 =hf = 1407.5 kJ/Kg, s3 = sf = 3.3595 kJ/Kg K P 321

v T1 2 3 s 1Q3 = m(u3-u1) + Pm(v3 – v1) = m(h3 – h1) = -5652.6 kJ Heat transfer to the heat engine: QH = -1Q3 = 5652.6 kJ Take control volume as total water and heat engine. H.E.

8.15 One kilogram of water at 300°C expands against a piston in a cylinder until it reaches ambient pressure, 100 kPa, at which point the water has a quality of 90%. It may be assumed that the expansion is reversible and adiabatic. What was the initial pressure in the cylinder and how much work is done by the water?

T a) b) P C.V. Water. Process: Rev., Q = 0 1

1 T1 m(u2 – u1) = 1Q2 – 1W2 = – 1W2 P 2 m(s2 – s1) = dQ/T + 1S2Êgen = 0 + 0 => s2 = s1 2 P2 = 100 kPa, x2 = 0.90 => s s2 = 1.3026 + 0.9 · 6.0568 = 6.7537 u2 = 417.36 + 0.9 · 2088.7 = 2297.2 At T1 = 300°C, s1 = 6.7537 P1 = 2.048 MPa, u1 = 2771.5 kJ/kg 1W2 = m(u1 – u2) = 1(2771.5 – 2297.2) = 474.3 kJ

8.16 A piston/cylinder has 2 kg ammonia at 50°C, 100 kPa which is compressed to 1000 kPa. The process happens so slowly that the temperature is constant. Find the heat transfer and work for the process assuming it to be reversible.

CV : NH3 Control Mass Energy: m(u2 – u1) = 1Q2 – 1W2 ; Entropy: m(s2 – s1) = dQ/T + 1S2Êgen Process: T = constant and assume reversible process 1S2Êgen = 0 1: (T,P), v = 1.5658, u1 = 1581.2 – 100· 1.5668 = 1424.62, s1 = 6.4943 2: (T,P), v = 0.1450, u2 = 1536.3 – 1000· 0.145 = 1391.3, s2 = 5.2654

8.17 A heavily insulated cylinder/piston contains ammonia at 1200 kPa, 60°C. The piston is moved, expanding the ammonia in a reversible process until the temperature is -20°C. During the process 600 kJ of work is given out by the State 1: Table B.2.2 v1 = 0.1238, s1 = 5.2357 kJ/kg K u1 = h – Pv = 1553.3 – 1200· 0.1238 = 1404.9 kJ/kg Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen Process: reversible (1S2Êgen = 0) and adiabatic (dQ = 0) => s2 = s1

8.18 A closed tank, V = 10 L, containing 5 kg of water initially at 25°C, is heated to 175°C by a heat pump that is receiving heat from the surroundings at 25°C. Assume that this process is reversible. Find the heat transfer to the water and the Process: constant volume (reversible isometric) 1: v1 = V/m = 0.002 x1 = (0.002 – 0.001003)/43.358 = 0.000023 u1 = 104.86 + 0.000023· 2304.9 = 104.93 kJ/kg s1 = 0.3673 + 0.000023· 8.1905 = 0.36759 kJ/kg K Continuity eq. (same mass) and V = C fixes v2 2: T2, v2 = v1 x2 = (0.002-0.001121)/0.21568 = 0.004075 u2 = 740.16 + 0.004075· 1840.03 = 747.67 kJ/kg s2 = 2.0909 + 0.004075· 4.5347 = 2.1094 kJ/kg K Energy eq. has W = 0, thus provides heat transfer as 1Q2 = m(u2 – u1) = 3213.7 kJ P

8.19 A rigid, insulated vessel contains superheated vapor steam at 3 MPa, 400°C. A valve on the vessel is opened, allowing steam to escape. The overall process is irreversible, but the steam remaining inside the vessel goes through a reversible adiabatic expansion. Determine the fraction of steam that has escaped, when the C.V.: steam remaining inside tank. Rev. & Adiabatic (inside only) Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen T 1 P 1

2 2 C.V. m 2 v s Rev ( 1S2Êgen = 0) Adiabatic ( Q = 0) => s2 = s1 = 6.9212 = sG at T2 T2 = 141°C, v2 = vGÊATÊT2 = 0.4972 me m1-m2 m2 v1 0.09936 = = 1 – = 1 – = 1 – = 0.80 m1 m1 m1 v2 0.4972 8.20 A cylinder containing R-134a at 10°C, 150 kPa, has an initial volume of 20 L. A piston compresses the R-134a in a reversible, isothermal process until it reaches the saturated vapor state. Calculate the required work and heat transfer to Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen State 1: (T, P) Table B.5.2 u1 = 410.6 – 0.14828· 150 = 388.36, s1 = 1.822 m = V/v1 = 0.02/0.148283 = 0.1349 kg State 2: (10°C, sat. vapor) u2 = 383.67, s2 = 1.7218 Process: T = constant, reversible 1S2Êgen = 0 => P 2 1 T

8.21 An insulated cylinder fitted with a piston contains 0.1 kg of superheated vapor steam. The steam expands to ambient pressure, 100 kPa, at which point the steam inside the cylinder is at 150°C. The steam does 50 kJ of work against the piston during the expansion. Verify that the initial pressure is 1.19 MPa and find the initial temperature.

C.V. Water in cylinder. Control mass insulated so no heat transfer Q = 0. Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen Process: Q = 0 and assume reversible 1S2Êgen = 0 => s = constant.

8.22 A heavily-insulated cylinder fitted with a frictionless piston contains ammonia at 6°C, 90% quality, at which point the volume is 200 L. The external force on the piston is now increased slowly, compressing the ammonia until its temperature reaches 50°C. How much work is done on the ammonia during this process? Solution: Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen State 1: T1 = 6oC, x1 = 0.9, V1 = 200 L = 0.2 m3 Table B.2.1 saturated vapor, P1 = Pg = 534 kPa v1 = vf + x1vfg = 0.21166 m3/kg, u1 = uf + x1ufg = 207.414 + 0.9· 1115.3 = 1211.2 kJ/kg s1 = sf + x1sfg = 0.81166 + 0.9· 4.4425 = 4.810 kJ/kg-K, m1 = V1/v1 = 0.2 / 0.21166 = 0.945 kg Process: 1‡2 Adiabatic 1Q2 = 0 & Reversible 1S2Êgen = 0 => s1 = s2 State 2: T2 = 50oC, s2 = s1 = 4.810 kJ/kg-K superheated vapor, interpolate in Table B.2.2 => P2 = 1919 kPa, v2 = 0.0684 m3/kg, h2 = 1479.5 kJ/kg u2 = h2 – P2v2 = 1479.5 – 1919· 0.0684 = 1348.2 kJ/kg

8.23 A piston/cylinder with constant loading of piston contains 1L water at 400 kPa, quality 15%. It has some stops mounted so the maximum possible volume is 11L. A reversible heat pump extracting heat from the ambient at 300 K, 100 kPa heats the water to 300°C. Find the total work and heat transfer for the water and the Solution: Take CV around the water and check possible P-V combinations. State 1: v1 = 0.001084 + 0.15·0.46138 = 0.07029 u1 = 604.29 + 0.15 · 1949.26 = 896.68 kJ/kg s1 = 1.7766 + 0.15 · 5.1193 = 2.5445 kJ/kg K m1 = V1/v1 = 0.001/0.07029 = 0.0142 kg

8.24 A piston/cylinder contains 2 kg water at 200°C, 10 MPa. The piston is slowly moved to expand the water in an isothermal process to a pressure of 200 kPa. Any heat transfer takes place with an ambient at 200°C and the whole process may be assumed reversible. Sketch the process in a P-V diagram and calculate both the Solution: Energy: m(u2 – u1) = 1Q2 – 1W2 Entropy m(s2 – s1) = 1Q2 / T State 1: Table B.1.4 : v1 = 0.001148, u1 = 844.49, s1 = 2.3178, V1 = mv1 = 0.0023 m3 State 2: Table B.1.3 : v2 = 1.08034, s2 = 7.5066, u2 = 2654.4 V2 = mv2 = 2.1607 m3, P 1 2 T

v T 12 s 1Q2 = mT(s2 – s1) = 2 · 473.15 (7.5066 – 2.3178) = 4910 kJ 1W2 = 1Q2 – m(u2 – u1) = 1290.3 kJ

8.25 An insulated cylinder/piston has an initial volume of 0.15 m3 and contains steam at 400 kPa, 200°C. The steam is expanded adiabatically, and the work output is measured very carefully to be 30 kJ. It is claimed that the final state of the water is in the two-phase (liquid and vapor) region. What is your evaluation of the claim?

T P1 Energy Eq.: 1 1Q2 = 0 = m(u2 – u1) + 1W2 V1 0.15 130 C u = 2540 v1 = 0.5342 o m = = 0.2808 kg 30 u2 = 2646.8 – = 2540.0 kJ/kg 0.2808 7.0259 s Entropy Eq.: s2 = s1 = 7.1706 kJ/kg K State 2 given by (u, s) check Table B.1.1: sG (at uG = 2540) = 7.0259 < s1 State 2 must be in superheated vapor region.

8.26 An amount of energy, say 1000 kJ, comes from a furnace at 800°C going into water vapor at 400°C, from which it goes to a solid metal at 200°C and then into some air at 70°C. For each location calculate the flux of s through a surface as Solution:

T1 => T2 => T3 => T4 AIR FLOW furnace vapor metal air FURNACE Flux of s: Fs = Q/T Fs1 = 1000/1073.15 = 0.932 kJ/K, Fs2 = 1000/673.15 = 1.486 kJ/K Fs3 = 1000/473.15 = 2.11 kJ/K, Fs4 = 1000/343.15 = 2.91 kJ/K Q over T is irreversible processes 8.27 An insulated cylinder/piston contains R-134a at 1 MPa, 50°C, with a volume of 100 L. The R-134a expands, moving the piston until the pressure in the cylinder has dropped to 100 kPa. It is claimed that the R-134a does 190 kJ of work against State 1: Table B.5.2, v1 = 0.02185, u1 = 431.24 – 1000· 0.02185 = 409.4, s1 = 1.7494, m = V1/v1 = 0.1/0.02185 = 4.577 kg Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 = 0/ – 190 u2 = 367.89 kJ/kg State 2: P2 , u2 Table B.5.2: T2 = -19.25°C ; s2 = 1.7689 kJ/kg K Entropy Eq.: m(s2 – s1) = ? dQ/T + 1S2,gen = 1S2,gen = 0.0893 kJ/K This is possible since 1S2,gen > 0/

8.28 A piece of hot metal should be cooled rapidly (quenched) to 25°C, which requires removal of 1000 kJ from the metal. The cold space that absorbs the energy could be one of three possibilities: (1) Submerge the metal into a bath of liquid water and ice, thus melting the ice. (2) Let saturated liquid R-22 at -20°C absorb the energy so that it becomes saturated vapor. (3) Absorb the energy by vaporizing a. Calculate the change of entropy of the cooling media for each of the three cases. a) Melting or boiling at const P & T 1Q2 = m(u2 – u1) + Pm(v2 – v1) = m(h2 – h1) 1000 1) Ice melting at 0°C , Table B.1.5: m = 1Q2 /hig = = 2.9993 kg 333.41 SH O = 2.9993(1.221) = 3.662 kJ/K 2

1000 2) R-22 boiling at -20°C, Table B.4.1: m = 1Q2 /hfg = = 4.539 kg 220.327 SR-22 = 4.539(0.8703) = 3.950 kJ/K 1000 3) N2 boiling at 101.3 kPa, Table B.6.1: m = 1Q2 /hfg = = 5.029 kg 198.842 SN = 5.029(2.5708) = 12.929 kJ/K 2

8.30 A cylinder fitted with a movable piston contains water at 3 MPa, 50% quality, at which point the volume is 20 L. The water now expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300°C. It is claimed that the water Solution: C.V.: H2O in Cylinder State 1: 3 MPa, x1 = 0.5, V1 = 20L = 0.02 m3, Table B.1.2: T1 = 233.9oC , v1 = vf + x1vfg = 0.001216 + 0.5· 0.06546 = 0.033948 m3/kg u1 = uf + x1ufg = 1804.5 kJ/kg, s1 = sf + x1sfg = 4.4162 kJ/kg-K m1 = V1/v1 = 0.589 kg solve for u2 = 1804.5 + (600 – 124)/0.589 = 2612.6 kJ/kg P State 2: P2 = 1.2 MPa, u2 = 2612.6 kJ/kg T2 200oC, s2 = 6.5898 kJ/kgK 2 1 T1

8.32 A piston/cylinder contains 1 kg water at 150 kPa, 20°C. The piston is loaded so pressure is linear in volume. Heat is added from a 600°C source until the water is at 1 MPa, 500°C. Find the heat transfer and the total change in entropy. Solution: CV H2O 1 => 2 1Q2 & 1W2 m(u2 – u1) = 1Q2 – 1W2 ; 1W2 = P dV = _ (P1 + P2 ) (V2 – V1) State 1: B.1.1 Compressed liq. use sat. liq. at same T: v1 = 0.001002 u1 = 83.94 ; s1 = 0.2966 P 2 v2 = 0.35411 u2 = 3124.3 ; s2 = 7.7621 1 T1

8.33 Water in a piston/cylinder is at 1 MPa, 500°C. There are two stops, a lower one at which Vmin = 1 m3 and an upper one at Vmax = 3 m3. The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 500 kPa. This setup is now cooled to 100°C by rejecting heat to the surroundings at 20°C. Find the total entropy generated in the process.

Initial state: Table B.1.3: v1 = 0.35411, u1 = 3124.3, s1 = 7.7621 m =V/v1 = 3/0.35411 = 8.472 kg P 1000 500 1

8.34 Two tanks contain steam, and they are both connected to a piston/cylinder as shown in Fig. P8.34. Initially the piston is at the bottom and the mass of the piston is such that a pressure of 1.4 MPa below it will be able to lift it. Steam in A is 4 kg at 7 MPa, 700°C and B has 2 kg at 3 MPa, 350°C. The two valves are opened, and the water comes to a uniform state. Find the final temperature and the total Control mass: All water mA + mB.

B.1.3: vA1 = 0.06283, uA1 = 3448.5, sA1 = 7.3476 , VA = 0.2513 m3 B.1.3: vB1 = 0.09053, uB1 = 2843.7, sB1 = 6.7428, VB = 0.1811 m3 Continuity Eq.: m2 = mA + mB = 6 kg Energy Eq.: m2u2 – mAuA1 – mBuB1 = 1Q2 – 1W2 = – 1W2 Entropy Eq.: m2s2 – mAsA1 – mBsB1 = 1S2Êgen P 1400 2

8.35 A cylinder/piston contains 3 kg of water at 500 kPa, 600°C. The piston has a cross-sectional area of 0.1 m2 and is restrained by a linear spring with spring constant 10 kN/m. The setup is allowed to cool down to room temperature due to heat transfer to the room at 20°C. Calculate the total (water and surroundings) State 1: Table B.1.3, v1 = 0.8041, u1 = 3299.6 , s1 = 7.3522

State 2: T2 & on line in P-V diagram. P P = P1 + (ks/A2 )(V – V1) 1 cyl Assume state 2 is two-phase, P2 = Psat(T2) = 2.339 kPa v2 = v1 + (P2 – P1)A2cyl/mks 2 v v2 = 0.8041 + (2.339 – 500)0.01/(3 · 10) = 0.6382 = vf + x2vfg x2 = (0.6382 – 0.001002)/57.7887 = 0.011, u2 = 109.46, s2 = 0.3887 1 1W2 = (P1 + P2)m · (v2 – v1) 2 1 = (500 + 2.339) · 3 · (0.6382 – 0.8041) = -125 kJ 2 1Q2 = m(u2 – u1) + 1W2 = 3(109.46 – 3299.6) – 125 = -9695.4 kJ Stot = Sgen,tot = m(s2 – s1) – 1Q2/Troom = 3(0.3887 – 7.3522) + 9695.4/293.15 = 12.18 kJ/K 8.36 A cylinder/piston contains water at 200 kPa, 200°C with a volume of 20 L. The piston is moved slowly, compressing the water to a pressure of 800 kPa. The loading on the piston is such that the product PV is a constant. Assuming that the room temperature is 20°C, show that this process does not violate the second law. C.V.: Water + cylinder out to room at 20°C Process: PV = constant = Pmv v2 = P1v1/P2 1w2 = ? Pdv = P1v1 ln(v2/v1) State 1: Table B.1.3, v1 = 1.0803, u1 = 2654.4 , s1 = 7.5066 State 2: P2 , v2 = P1v1/P2 = 200 · 1.0803/800 = 0.2701 Table B.1.3: u2 = 2655.0 kJ/kg , s2 = 6.8822 kJ/kg K 1w2 = 200 · 1.0803 ln(0.2701/1.0803) = -299.5 kJ/kg 1q2 = u2 – u1 + 1w2 = 2655.0 – 2654.4 – 299.5 = -298.9 1ss,gen = s2 – s1 – 1q2/Troom = 6.8822 – 7.5066 + 298.9/293.15 = 0.395 kJ/kg K > 0/ satisfy 2nd law.

8.37 One kilogram of ammonia (NH3) is contained in a spring-loaded piston/cylinder as saturated liquid at -20°C. Heat is added from a reservoir at 100°C until a final condition of 800 kPa, 70°C is reached. Find the work, heat transfer, and entropy C.V. = NH3 Cont. m2 = m1 = m Energy: E2 – E1 = m(u2 – u1) = 1Q2 – 1W2 Entropy: S2 – S1 = ? dQ/T + 1S2,gen 11 Process: 1W2 = ? PdV = (P1 + P2)(V2 – V1) = (P1 + P2)m(v2 – v1) 22 State 1: Table B.2.1 P 2

1 P1 = 190.08, v1 = 0.001504 u1 = 88.76, s1 = 0.3657 v2 = 0.199, s2 = 5.5513 u2 = 1597.5 – 800· 0.199 = 1438.3 v T 2 P2 1

8.39 An insulated cylinder fitted with a frictionless piston contains saturated vapor R- 12 at ambient temperature, 20°C. The initial volume is 10 L. The R-12 is now expanded to a temperature of -30°C. The insulation is then removed from the cylinder, allowing it to warm at constant pressure to ambient temperature. Calculate the net work and the net entropy change for the overall process. C.V.: R-12 State 1: T = 20oC, V = 10 L = 0.01 m3, Sat. Vapor ‡ x = 1.0 11 1 P1 = Pg = 567 kPa, v1 = vg = 0.03078 m3/kg, m1 =V1/v1 = 0.325 kg u1 = ug = 178.32 kJ/kg, s1 = sg = 0.68841 kJ/kg-K State 2: T2 = -30oC Assume 1‡2 Adiabatic & Reversible: s2 = s1 = 0.68841 kJ/kg-K s2 = sf + x2sfg; => x2 = 0.95789, P2 = Pg = 100.4 kPa v2 = vf + x2vfg = 0.15269 m3/kg, h2 = hf + x2hfg = 167.23 u2 = h2 – P2v2 = 151.96 kJ/kg State 3: T3 = 20oC, P3 = P2 = 100.41 kPa v3 = 0.19728 m3/kg, h3 = 203.86 kJ/kg, s3 = 0.82812 kJ/kg-K

P 1 3 2 v T 13 2 s 1st Law: 1‡2, 1Q2 = m(u2 – u1) + 1W2 ; 1Q2 = 0 1W2 = m(u1 – u2) = 8.57 kJ 2‡3: Process: P = constant => 2W3 = Pm dv = Pm(v3 – v2) = 1.45 kJ WTOT = 1W2 + 2W3 = 8.57 + 1.45 = 10.02 kJ b) 2nd Law: 1‡3, Snet = m(s3 – s1) – QCV/To; To = 20 C o

8.40 A foundry form box with 25 kg of 200°C hot sand is dumped into a bucket with 50 L water at 15°C. Assuming no heat transfer with the surroundings and no boiling away of liquid water, calculate the net entropy change for the process. C.V. Sand and water, constant pressure process msand(u2 – u1)sand + mH O(u2 – u1)H O = -P(V2 – V1) 22

msand hsand + mH O hH O = 0 22 For this problem we could also have said that the work is nearly zero as the solid sand and the liquid water will not change volume to any measurable extent. Now we get changes in u’s instead of h’s. For these phases CV =CP = C 25 · 0.8· (T2 – 200) + (50· 10-3/0.001001) · 4.184 · (T2 – 15) = 0 T2 = 31.2°C 304.3 304.3 S = 25 · 0.8 ln + 49.95 · 4.184 ln = 2.57 kJ/K ?473.15? ?288.15?

8.41 A large slab of concrete, 5 · 8 · 0.3 m, is used as a thermal storage mass in a solar-heated house. If the slab cools overnight from 23°C to 18°C in an 18°C C.V.: Control mass concrete. V = 5 · 8 · 0.3 = 12 m3 m = V = 2300 · 12 = 27600 kg 1Q2 = mC T = 27600 · 0.65(-5) = -89700 kJ T2 291.2 SSYST = mC ln = 27600 · 0.65 ln = -305.4 kJ/K T1 296.2 SSURR = -1Q2/T0 = +89700/291.2 = +308.0 kJ/K SNET = -305.4 + 308.0 = +2.6 kJ/K

8.45 A mass of 1 kg of air contained in a cylinder at 1.5 MPa, 1000 K, expands in a reversible isothermal process to a volume 10 times larger. Calculate the heat m(u2 – u1) = 1Q2 – 1W2 = 0/ (T = constant so ideal gas => u2 = u1 )

8.47 Consider a Carnot-cycle heat pump having 1 kg of nitrogen gas in a cylinder/piston arrangement. This heat pump operates between reservoirs at 300 K and 400 K. At the beginning of the low-temperature heat addition, the pressure is 1 MPa. During this processs the volume triples. Analyze each of the four processes in the cycle and determine a. The pressure, volume, and temperature at each point b. The work and heat transfer for each process T

4 3 T1 = T2 = 300 K, T3 = T4 = 400 K, P1 = 1 MPa, V2 = 3 · V1 a) P2V2 = P1V1 => P2 = P1/3 = 0.3333 MPa 2 mRT1 1Ê· Ê0.2968Ê· Ê300 1 V1 = = = 0.08904 m3 P1 1000 s V2 = 0.26712 m3 P = P (T /T ) k = 0.3333 400 3.5 = 0.9123 MPa 3 2 3 2 k-1 ?300? P2 T3 0.3333 400 V3 = V2 · · = 0.26712 · · = 0.1302 m3 P3 T2 0.9123 300 N 2

8.48 A rigid tank contains 2 kg of air at 200 kPa and ambient temperature, 20°C. An electric current now passes through a resistor inside the tank. After a total of 100 kJ of electrical work has crossed the boundary, the air temperature inside is 80°C. Solution: C.V.: Air in Tank; Ideal gas, R = 0.287 kJ/kg-K, Cv = 0.717 kJ/kg-K 1st Law: 1‡2, 1Q2 = m(u2 – u1) + 1W2 , 1W2 = -100 kJ State 1: T1 = 20oC, P1 = 200 kPa, m1 = 2 kg State 2: T2 = 80oC Assume Constant Specific Heat 1Q2 = mCv(T2 – T1) + 1W2 = -14.0 kJ 2nd Law: 1‡2, Snet = m(s2 – s1) – Qcv/To, QCV = 1Q2 v2 v2 s2 – s1 = Cv ln (T2/T1) + Rln ; v2 = v1, ln = 0 v1 v1 s2 – s1 = Cvln (T2/T1) = 0.1335 kJ/kg-K Snet = 0.3156 kJ/kg-K > 0, Process is Possible T2 T2 P2 Note: P2 = P1 , s2 – s1 = Cp ln – R ln , Results in the same answer T1 T1 P1

8.49 A handheld pump for a bicycle has a volume of 25 cm3 when fully extended. You now press the plunger (piston) in while holding your thumb over the exit hole so that an air pressure of 300 kPa is obtained. The outside atmosphere is at P0, T0. Consider two cases: (1) it is done quickly ( 1 s), and (2) it is done very slowly ( 1 h). C.V. Air in pump. Assume that both cases result in a reversible process. Case I) Quickly means no time for heat transfer u2 – u1 = -1w2 s2 = s1 = 0/ Pr2 = Pr1 · P2/P1 = 1.0907 · (300/100) = 3.2721 T2 = 407.5 K V2 = P1V1T2/T1P2 = 11.39 cm3 The process is then a reversible isothermal compression.

8.50 An insulated cylinder/piston contains carbon dioxide gas at 120 kPa, 400 K. The gas is compressed to 2.5 MPa in a reversible adiabatic process. Calculate the final temperature and the work per unit mass, assuming a. Variable specific heat, Table A.8 b. Constant specific heat, value from Table A.5 c. Constant specific heat, value at an intermediate temperature from Table A.6 a) Table A.8 for CO2 s- – s- = s- o – s- o – 2 1 T2 T1 – R ln(P2/P1) s-o = 225.314 + 8.3145 ln(2.5/0.12) = 250.561 T2 T2 = 697.3 K — 1w2 = -(u2 – u1) = -( (h2-h1)-R(T2-T1)) /M

8.52 A rigid storage tank of 1.5 m3 contains 1 kg argon at 30°C. Heat is then transferred to the argon from a furnace operating at 1300°C until the specific entropy of the argon has increased by 0.343 kJ/kg K. Find the total heat transfer Solution: C.V. Argon. Control mass. R = 0.20813, m = 1 kg Energy Eq.: m (u2 – u1 ) = m Cv (T2 – T1) = 1Q2 Process: V = constant => v2 = v1 State 1: P1 = mRT/V = 42.063 kPa State 2: s2 = s1 + 0.343, s2 – s1 = Cp ln (T2 / T1 ) – R ln (T2 / T1 ) = Cv ln (T2 / T1 ) ln (T2 / T1 ) = (s2 – s1)/ Cv = 0.343/0.312 = 1.0986 Pv = RT => (P2 / P1) (v2 / v1) = T2 / T1 = P2 / P1 T2 = 2.7 · T1 = 818.3, P2 = 2.7 · P1 = 113.57

8.54 Two rigid tanks each contain 10 kg N2 gas at 1000 K, 500 kPa. They are now thermally connected to a reversible heat pump, which heats one and cools the other with no heat transfer to the surroundings. When one tank is heated to 1500 K the process stops. Find the final (P, T ) in both tanks and the work input to the Control volume of hot tank B, process = constant volume & mass U2 – U1 mCv(T2 – T1) = 1Q2 = 10 · 0.7448 · 500 = 3724 kJ P2 = P1T2/T1 = 1.5(P1) = 750 kPa

A 1 -> 3 W HE Q 13 Q 12 B 1 -> 2 State: 1 = initial, 2 = final hot 3 = final cold

8.55 Repeat the previous problem, but with variable heat capacities. C.V. Hot tank B. Constant volume and mass

A 1 -> 3 W HE Q 13 Q 12 B 1 -> 2 State: 1 = initial, 2 = final hot 3 = final cold

8.56 We wish to obtain a supply of cold helium gas by applying the following technique. Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is b. Calculate the final temperature and the net work per kilogram of helium. c. If a diatomic gas, such as nitrogen or oxygen, is used instead, would the final temperature be higher, lower, or the same?

a) T T1= T2 c) Diatomic gas: P 2 P 1= P3 2 k < 1.67 (probably 1.40) 1 T3 > 143.2 K 3 Higher s =s s 23 b) 1w2 = -RT1 ln(P2/P1) = -2.0771 · 293.15 · ln(600/100) = -1091.0 kJ/kg k-1 T3 = T2(P3/P2) k = 293.15 (100/600)0.4 = 143.15 K 2w3 = CVo(T2-T3) = 3.116 (293.15 – 143.15) = +467.4 kJ/kg wNET = -1091.0 + 467.4 = -623.6 kJ/kg 8.57 A 1-m3 insulated, rigid tank contains air at 800 kPa, 25°C. A valve on the tank is opened, and the pressure inside quickly drops to 150 kPa, at which point the valve is closed. Assuming that the air remaining inside has undergone a reversible Cont.Eq.: m2 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 0 + 0 P 1

8.60 Nitrogen at 600 kPa, 127°C is in a 0.5 m3 insulated tank connected to a pipe with a valve to a second insulated initially empty tank of volume 0.5 m3. The valve is opened and the nitrogen fills both tanks. Find the final pressure and temperature and the entropy generation this process causes. Why is the process irreversible?

CV Both tanks + pipe + valve Insulated : Q = 0 Rigid: W = 0 m(u2 – u1) = 0 – 0 => u2 = u1 = ua1 m(s2 – s1) = dQ/T + 1S2Êgen = 1S2Êgen (dQ = 0) 1: P1 , T1 , Va => m = PV/RT = (600 · 0.5)/ (0.2968 · 400) = 2.527 2: V2 = Va + Vb ; uniform state v2 = V2 / m ; u2 = ua1

P 1 2 v T P1 1 2 P2 s Ideal gas u (T) => u2 = ua1 => T2 = Ta1 = 400 K P2 = mR T2 / V2 = (V1 / V2 ) P1 = _ · 600 = 300 kPa Sgen = m(s2 – s1) = m[sT2 – sT1 – R ln(P2 / P1)] = m [0 – R ln (P2 / P1 )] = -2.527 · 0.2968 ln _ = 0.52 kJ/K Irreversible due to unrestrained expansion in valve P ? but no work out.

4 Eqs and 4 unknowns : P2 , Ta2 , Tb2 , x = ma2 / ma1 V2 / ma1 = x va2 + (1 – x) vb2 = x · (R Ta2 /P2)+ (1 – x) (R Tb2 / P2) ma2 ( ua2 – ua1 ) + mb2 (ub2 – ua1) = 0 x Cv ( Ta2 – Ta1 ) + (1 – x) (Tb2 – Ta1) Cv = 0 x Ta2 + (1 – x)Tb2 = Ta1 P2 V2 / ma1 R = x Ta2 + (1 – x) Tb2 = Ta1) P2 = ma1 R Ta1 / V2 = ma1 R Ta1 / 2Va1 = _ Pa1 = 300 kPa sa2 = sa1 => Ta2 = Ta1 (P2 / Pa1)k-1 / k = 400 · (1/2)0.2857 = 328.1 K Now we have final state in A va2 = R Ta2 / P2 = 0.3246 ; ma2 = Va / va2 = 1.54 kg x = ma2 / ma1 = 0.60942 mb2 = ma1 – ma2 = 0.987 kg Substitute into energy equation Tb2 = ( Ta1 – x Ta2 ) / (1 – x) = 512.2 K Sgen = mb2 ( sb2 – sa1) = mb2 [ Cp ln (Tb2 / Ta1 ) – R ln (P2 / Pa1 )] = 0.987 [ 1.0416 ln (512.2/400) – 0.2968 ln (1/2) ] = 0.4572 kJ/K 8.61 Neon at 400 kPa, 20°C is brought to 100°C in a polytropic process with n = 1.4. Give the sign for the heat transfer and work terms and explain.

P T Neon Table A.5 2 2 k = = 1.667 so n < k 1 Cv = 0.618, R = 0.412 T=c 1 s v

8.62 A cylinder/piston contains carbon dioxide at 1 MPa, 300°C with a volume of 200 L. The total external force acting on the piston is proportional to V3. This system is allowed to cool to room temperature, 20°C. What is the total entropy generation m = P1V1/RT1 = (1000 · 0.2)/(0.18892 · 573.2) = 1.847 kg Process: P = CVÊ3 or PV-3 = constant, polytropic with n = -3 n P2 = P1(T2/T1)n-1 = 1000(293.2/573.2)3/4 = 604.8 kPa 1 V2 = V1(T1/T2)n-1 = 0.16914 m3 1W2 =? ÊPdV = (P2V2-P1V1)/(1-n) = [604.8 · 0.16914 – 1000 · 0.2]/[1-(-3)] = -24.4 kJ 1Q2 = 1.847 · 0.6529(20 – 300) – 24.4 = -362.1 kJ = 1.847[0.8418 ln 293.2 604.8] SYST 573.2 – 0.18892 ln 1000 S = 1.847[-0.4694] = -0.87 kJ/K SSURR = +362.1/293.2 = +1.235 kJ/K SNETÊ = -0.87 + 1.235 = +0.365 kJ/K

8.64 The power stroke in an internal combustion engine can be approximated with a polytropic expansion. Consider air in a cylinder volume of 0.2 L at 7 MPa, 1800 K. It now expands in a reversible polytropic process with exponent, n = 1.5, through a volume ratio of 8;1. Show this process on P–v and T–s diagrams, and calculate the work and heat transfer for the process.

P T P1 = 7 MPa, T1 = 1800 K, 1 1 V1 = 0.2 L 2 Rev. PV1.50 = const, V 2 S V2/V1 = 8

P V 7000Ê· Ê0.2Ê· Ê10-3 m1 = R1T1 = 0.287Ê· Ê1800 = 2.71· 10-3 kg T2 = T1(V1/V2)n-1 = 1800(1/8)0.5 = 636.4 K 1W2 = ? ÊPdV = mR(T2-T1)/(1-n) 2.71· 10-3Ê· Ê0.287(636.4Ê-Ê1800) = = 1.81 kJ 1Ê-Ê1.5 1Q2 = m(u2 – u1) + 1W2 = 2.71· 10-3 · (463.05 – 1486.331) + 1.81 = -0.963 kJ

8.66 A cylinder/piston contains air at ambient conditions, 100 kPa and 20°C with a volume of 0.3 m3. The air is compressed to 800 kPa in a reversible polytropic process with exponent, n = 1.2, after which it is expanded back to 100 kPa in a c. What is the potential refrigeration capacity (in kilojoules) of the air at the final state?

a) P T 2 2 1 3 b) T2 = T1(P2/P1) n 800 0.167 V s = 293.2 = 414.9 K ?100? 2 P2v2-P1v1 R(T2-T1) 0.287(414.9-293.2) 1w2 = ? ÊPdv = = = = -174.6 kJ/kg 1-n 1-n 1-1.20 Ê1 3 m = P1V1/RT1 100Ê· Ê0.3 = = 0.3565 kg 1 0.287Ê· Ê293.2 n-1

k-1 100 0.286 T3 = T2(P3/P2) k = 414.9?800? = 228.9 K 2w3 = CV0(T2-T3) = 0.717(414.9 – 228.9) = +133.3 kJ/kg wNET = 0.3565(-174.6 + 133.3) = -14.7 kJ c) Refrigeration: warm to T0 at const P 3Q1 = mCP0(T1 – T3) = 0.3565 · 1.004 (293.2 – 228.9) = 23.0 kJ

8.67 An ideal gas having a constant specific heat undergoes a reversible polytropic expansion with exponent, n = 1.4. If the gas is carbon dioxide will the heat Solution:

T n=k n>k 2 1 CO2: k = 1.289 < n Since n > k and P=const n

8.69 A cylinder/piston contains 100 L of air at 110 kPa, 25°C. The air is compressed in a reversible polytropic process to a final state of 800 kPa, 200°C. Assume the heat transfer is with the ambient at 25°C and determine the polytropic exponent n and the final volume of the air. Find the work done by the air, the heat transfer and m = (P1V1)/(RT1) = (110 · 0.1)/(0.287 · 298.15) = 0.1286 kg n-1 n-1 473.15 800 n n-1 T2/T1 = (P2/P1) n = = 0.2328 298.15 ?110? n 1 110 0.7672 3 n = 1.3034, V2 = V1(P1/P2)n = 0.1 = 0.02182 m ?800? P2V2Ê-ÊP1V1 800Ê· Ê0.02182Ê-Ê110Ê· Ê0.1 1W2 = ? PdV = = = -21.28 kJ 1Ê-Ên 1Ê-Ê1.3034 1Q2 = mCv(T2 – T1) + 1W2 = 0.1286 · 0.7165 · (200 – 25) – 21.28 = -5.155 kJ s2 – s1 = CP0ln(T2/T1) – R ln(P2/P1) 473.15 800 kJ = 1.004 ln – 0.287 ln = -0.106 ?298.15? ?110? kgÊK 1S2,gen = m(s2 – s1) – 1Q2/T0 = 0.1286 · (-0.106) + (5.155/298.15) = 0.00366 kJ/K 8.70 A mass of 2 kg ethane gas at 500 kPa, 100°C, undergoes a reversible polytropic expansion with exponent, n = 1.3, to a final temperature of the ambient, 20°C. Calculate the total entropy generation for the process if the heat is exchanged with the ambient.

8.73 A spring-loaded piston/cylinder, shown in Fig. P8.73, contains water at 100 kPa with v = 0.07237 m3/kg. The water is now heated to a pressure of 3 MPa by a reversible heat pump extracting Q from a reservoir at 300 K. It is known that the water will pass through saturated vapor at 1.5 MPa and that pressure varies linearly with volume. Find the final temperature, the heat transfer to the water and the work input to the heat pump.

C.V.: H2O, 1? 2 m1 = m2 = m, m(u2 – u1) = 1Q2 – 1W2 Ê2 1 1 2=? 2 1 2 2 1 W ÊPdV = (P + P )(V – V ) q Ê1 12 w 1: 100 kPa, 0.07237 m3/kg hp q x1 = 0.04213 u1 = 505.36 s1 = 1.5578 res 2¢: 1500 kPa, x = 1 v = 0.13177 slope = 23569 2: P2 = P1 + C(v2 – v1) slope & v2 = 0.1954 2 P T2 u2 2¢ = 1000°C, = 4045.4 2 s2 = 8.4009 2¢ P

1 P2¢ – P 1

1w2 = 2 (P1 + P2)(v2 – v1) P1 1 slope v – v = 2¢ 1 = 190.734 kJ/kg 1q2 = u2 – u1 +1w2 = 3730.7 1 vvv 2¢ 2

C.V.: H2O + heat pump everything else reversible mH O(u2 – u1) = Qres – 1W2 + Wh.p. wh.p. = u2 – u1 + 1w2 – qres 2

Advanced Problems 8.76 An insulated cylinder with a frictionless piston, shown in Fig. P8.76, contains water at ambient pressure, 100 kPa, a quality of 0.8 and the volume is 8 L. A force is now applied, slowly compressing the water until it reaches a set of stops, at which point the cylinder volume is 1 L. The insulation is then removed from the cylinder walls, and the water cools to ambient temperature, 20°C. Calculate the work and the heat transfer for the overall process.

H2O P1 = 100 kPa x1 = 0.80 V1 = 8 L Slowly compress to stops, V2 = 1 L Insulation removed, cool to T3 = 20°C u1 = uf + x1 · ufg = 417.36 + 0.8 · 2088.7 = 2088.3 kJ/kg Process 1? 2: Rev., Q = 0 s2 = s1 = 1.3026 + 0.8 · 6.0568 = 6.1480 v1 = 0.001043 + 0.8 · 1.6930 = 1.3554 v2 = (1/8)v1 = 0.16943 s2 & v2 fix state 2, trial and error on P2: find x2: s2 = sf + x2 · sfg = 6.1480 find u2: u2 = uf + x2 · ufg check v2: v2 = vf + x2 · (vg – vf) = 0.16943

8.77 Consider the process shown in Fig. P8.77. The insulated tank A has a volume of 600 L, and contains steam at 1.4 MPa, 300°C. The uninsulated tank B has a volume of 300 L and contains steam at 200 kPa, 200°C. A valve connecting the two tanks is opened, and steam flows from A to B until the temperature in A reaches 250°C. The valve is closed. During the process heat is transferred from B to the surroundings at 25°C, such that the temperature in B remains at 200°C. It may be assumed that the steam remaining in A has undergone a reversible adiabatic expansion. Determine the final pressure in tank A, the final pressure and mass in tank B, and the net entropy change, system plus surroundings, for the a) mA1 = 0.6/0.18228 = 3.292; mB1 = 0.3/1.0803 = 0.278 kg sA2 = sA1 = 6.9534, TA2 = 250°C => PA2 = 949.5 kPa b) mA2 = 0.6/0.2479 = 2.42 kg mAe = mBi = 3.292 – 2.42 = 0.872 kg => mB2 = 0.278 + 0.872 = 1.15 kg vB2 = 0.3/1.15 = 0.2609, TB2 = 200°C => PB2 = 799.8 kPa c) 1Q2 = (mA2uA2+ mB2uB2) – (mA1uA1+ mB1uB1) = (2.42 · 2711.3 + 1.15 · 2630.6) – (3.292 · 2785.2 + 0.278 · 2654.4) = -320.3 kJ SSYST = (mA2sA2+ mB2sB2) – (mA1sA1+ mB1sB1) = (2.42 · 6.9534 + 1.15 · 6.8159) – (3.292 · 6.9534 + 0.278 · 7.5066) = -0.3119 kJ/K SSURR = – 1Q2/T0 = +320.3/298.2 = +1.0743 kJ/K SNETÊ = -0.3119 + 1.0743 = +0.7624 kJ/K

8.78 A vertical cylinder/piston contains R–22 at -20°C, 70% quality, and the volume is 50 L, shown in Fig. P8.78. This cylinder is brought into a 20°C room, and an electric current of 10 A is passed through a resistor inside the cylinder. The voltage drop across the resistor is 12 V. It is claimed that after 30 min the P1 = P2 = 0.245 MPa, m = V1/v1 = 0.05/0.06521 = 0.767 kg WELEC = -Ei t = -12 · 10 · 30 · 60/1000 = -216 kJ 1Q2 = m(u2 – u1) + WBDRY + WELEC = m(h2 – h1) + WELEC = 0.767(282.2 – 176.0) – 216 = -134.5 kJ SSYST = 0.767(1.1014 – 0.6982) = 0.3093 kJ/K SSURR = -1Q2/T0 = +134.5/293.15 = 0.4587 kJ/K SNET = +0.3093 + 0.4587 = +0.768 kJ/K Claim is OK.

8.79 Redo Problem 8.57, but calculate the mass withdrawn by a first-law, control- volume analysis. Compare the result to that obtained in Problem 8.57. Show from a differential step of mass out that the first law leads to the same result. (Find the relation between dP and dT) i) CV: Tank : 0 = m2u2 – m1u1 + (m1 – m2)heÊAVG or 0 = m2CVoT2 – m1CVoT1 + (m1-m2) CPo(T1+T2)/2 Also, m2T2 = P2V/R = (150 · 1)/0.287 150 0 = 150/0.287 – 9.35 · 298.2 + 9.35Ê-Ê · 1.4 · (298.2+T2)/2 ? 0.287ÊT2? T2 = 191.1 K, m2 = 2.74 kg, me = m1-m2 = 6.61 kg Approximate answer because of heÊAVE value used. Answer will be closer to ii) solve as in i), except in 2 steps Solving from 1-2: T2 = 245.2 K & m2 = 5.684 Now using 2 as the initial state and 3 as the final Note that final T and m are closer to those in 8.57. To generalize this solution, substitute the equation of state for m1 & m2 into the 1st law of i). Then, dividing by P1, get P2 T2-(P2/P1)T1 k 0 = – 1 + (T1+T2) P1 T1T2 2 ?? Let P = P2-P1 & P = P1 & T = T2-T1 & T = T1 The above equation becomes k-1 P T In the limit, ¢s ? d¢s and integrate, = ? k ?P T get same answer as in i).

8.80 A vertical cylinder is fitted with a frictionless piston that is initially resting on stops. The cylinder contains carbon dioxide gas at 200 kPa, 300 K, and at this point the volume is 50 L. A cylinder pressure of 400 kPa is required to make the piston rise from the stops. Heat is now transferred to the gas from an aluminum a) What is the temperature of the aluminum block when the piston first begins to b) The process continues until the gas and block reach a common final Solution: Energy Eq.: m(u2 – u1)AL + m(u2 – u1)CO2 = 1Q2 – 1W2 = – 1W2 Process: V = constant => 1W2 = 0 Properties Table A.3 and A.5 CO2 ideal gas CO2: R = 0.1889 kJ/kg K, Cp = 0.842 kJ/kg K, Cv = 0.653 kJ/kg K Al: = 2700 kg/m3, C = 0.9 kJ/kg K, VAl = 0.001 m3, mAl = V = 2.7 kg State 1: CO2, V1 = 50 L = 0.05 m3, m1 = P1V1/(RT1) = 0.1764 kg State 2: CO2, P2 = PEXT = 400 kPa, V2 = V1 = 0.05m3 T2 = T1 P2/P1 = 600 K a) Assume Constant Specific Heat mCO2Cv(T2 – T1) + mALC(T2 – T1) = 0 = 34.55 + 2.7· 0.9(T2 – 700) Solve for T2 of the Aluminum: T2,AL = 685.8 K b) C.V.: Aluminum & CO2, where 2Q3 = 0 1st Law: 2Q3 = mAL(u3 – u2)AL + mCO2(u3 – u2)CO2 + 2W3;

8.81 A piston/cylinder contains 2 kg water at 5 MPa, 800°C. The piston is loaded so pressure is proportional to volume, P = CV. It is now cooled by an external reservoir at 0°C to a final state of saturated vapor. Find the final pressure, work, C.V. Water. Control mass with external irreversibility due to heat transfer. Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 1Q2/To + 1S2Êgen (if CV to To) State 1: Table B.1.3: s1 = 7.744, v1 = 0.09811, u1 = 3646.6

8.85E In a Carnot engine with water as the working fluid, the high temperature is 450 F and as QL is received, the water changes from saturated liquid to saturated vapor. The water pressure at the low temperature is 14.7 lbf/in.2. Find TL, cycle thermal efficiency, heat added per pound-mass, and entropy, s, at the beginning of the heat rejection process.

T 1 Constant T constant P from 1 to 2 2 qH = h2 – h1 = hfg = 775.4 Btu/lbm States 3 & 4 are two-phase 4 3 s TL = T3 = T4 = 212 F 212Ê+Ê459.67 cycle = 1 – TL/TH = 1 – = 0.262 450Ê+Ê459.67 s3 = s2 = sg(TH) = 1.4806 Btu/lbm R

8.86E Consider a Carnot-cycle heat pump with R-22 as the working fluid. Heat is rejected from the R-22 at 100 F, during which process the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 30 F. b. Find the quality of the R-22 at the beginning and end of the isothermal heat c. Determine the coefficient of performance for the cycle.

a) T 100 30 Table C.10.1 b) s4 = s3 = 0.0794 = 0.0407 + x4(0.1811) 3 2 x4 = 0.214 s1 = s2 = 0.2096 = 0.0407 + x1(0.1811) x1 = 0.9326 41 c) ¢ = qH/wIN = TH/(TH – TL) s = 559.67/(100 – 30) = 7.995

8.88E Water at 30 lbf/in.2, x = 1.0 is compressed in a piston/cylinder to 140 lbf/in.2, 600 F in a reversible process. Find the sign for the work and the sign for the heat transfer. Solution: Table C.8.1: s1 = 1.70 Btu/lbm R v1 = 13.76 ft3/lbm Table C.8.2: s2 = 1.719 Btu/lbm R v2 = 4.41 ft3/lbm => ds >0 : dq = Tds > 0 => q is positive dv < 0 : dw = Pdv < 0 => w is negative

P 2 1 v T 2 1 s 8.89E Two pound-mass of ammonia in a piston/cylinder at 120 F, 150 lbf/in.2 is expanded in a reversible adiabatic process to 15 lbf/in. 2. Find the work and heat Control mass: m(u2 – u1) = 1Q2 – 1W2 Ê2 m(s2 – s1) = ? Ê ÊdQ/T + 1S2,gen 1 Process: 1Q2 = 0/ , 1S2,gen = 0/ s2 = s1 = 1.2504 Btu/lbm R State 2: P2 , s2 2 phase Table C.9.1 (sat. vap. C.9.2 also) Interpolate: sg2 = 1.3921 Btu/lbm R, sf = 0.0315 Btu/lbm R x2 = (1.2504-0.0315)/1.3606 = 0.896 , u2 = 13.36 + 0.896 · 539.35= 496.6 1W2 = 2 · (596.6 – 496.6) = 100 Btu PT 1 1

8.90E A cylinder fitted with a piston contains ammonia at 120 F, 20% quality with a volume of 60 in.3. The ammonia expands slowly, and during this process heat is transferred to maintain a constant temperature. The process continues until all the liquid is gone. Determine the work and heat transfer for this process.

120 NH 3 T1 = 120 F, x1 = 0.20, V1 = 60 in3 T = const to x2 = 1, Table C.9.1: P = 286.5 lbf/in2 T v1 = 0.02836 + 0.2 · 1.0171 = 0.2318 60 v2 = 1.045, m = V/v = = 0.15 lbm 1728· 0.2318 12S 286.5· 144 1 2 = 778 W · 0.15 · (1.045 – 0.2318) = 6.47 Btu s1 = 0.3571 + 0.2 · 0 .7829 = 0.5137 Btu/lbm R, s2 = 1.140 Btu/lbm R 1Q2 = 579.7 · 0.15(1.1400 – 0.5137) = 54.46 Btu – or – h1 = 178.79 + 0.2 · 453.84 = 269.56 Btu/lbm; h2 = 632.63 Btu/lbm 1Q2 = m(h2 – h1) = 0.15(632.63 – 269.56) = 54.46 Btu

8.91E One pound-mass of water at 600 F expands against a piston in a cylinder until it reaches ambient pressure, 14.7 lbf/in.2, at which point the water has a quality of b. How much work is done by the water?

T C.V. Water m = 1 lbm, T1 = 600 F P 1 1 Process: Q = 0, 1S2Êgen = 0 => s2 = s1 T1 Table C.9.1: P2 = 14.7 lbf/in2, x2 = 0.90 P = 14.7 2 a) s1 = s2 = 0.31212 + 0.9 · 1.4446 = 1.6123 Table C.8.2: at T1 = 600 F, s1 2

8.92E A closed tank, V = 0.35 ft3, containing 10 lbm of water initially at 77 F is heated to 350 F by a heat pump that is receiving heat from the surroundings at 77 F. Assume that this process is reversible. Find the heat transfer to the water and the Process: constant volume (reversible isometric) 1: v = V/m = 0.35/10 = 0.035 ft3/lbm x = 2.692· 10-5 11 u1 = 45.11 Btu/lbm, s1 = 0.08779 Btu/lbm R Continuity eq. (same mass) and constant volume fixes v2 State 2: T2, v2 = v1 x2 = (0.035 – 0.01799) / 3.3279 = 0.00511 u2 = 321.35 + 0.00511· 788.45 = 325.38 Btu/lbm s2 = 0.5033 + 0.00511· 1.076 = 0.5088 Btu/lbm R Energy eq. has zero work, thus provides heat transfer as 1Q2 = m(u2 – u1) = 10(325.38 – 45.11) = 2802.7 Btu

Q 12 Entropy equation for the total control volume gives for a reversible process: m(s2 – s1) = QL/T0 QL = mT0(s2 – s1) = 10(77 + 459.67)(0.5088 – 0.08779) = 2259.4 Btu H.P.

8.93E A cylinder containing R-134a at 50 F, 20 lbf/in.2, has an initial volume of 1 ft3. A piston compresses the R-134a in a reversible, isothermal process until it reaches the saturated vapor state. Calculate the required work and heat transfer to Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 1Q2 /T + 0 State 1: Table C.11.2 at 50F: Since v is nonlinear in P interpolate in Pv v = [(2/3)· 3.4859· 15 + (1/3)· 1.6963· 30] / 20 = 2.591 ft3/lbm m = V/v1 = 1/2.591 = 0.3859 lbm u1 = [(2/3)· 176.96 + (1/3)· 175.99] – 20· 2.591· 144/778 = 167.05 Btu/lbm, s1 = [(2/3)· 0.443 + (1/3)· 0.42805] = 0.438 Btu/lbm R,

8.94E A rigid, insulated vessel contains superheated vapor steam at 450 lbf/in.2, 700 F. A valve on the vessel is opened, allowing steam to escape. It may be assumed that the steam remaining inside the vessel goes through a reversible adiabatic expansion. Determine the fraction of steam that has escaped, when the final state Cont.Eq.: m2 = m1 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 0 + 0 .

P 1 2 v 2 s State 1: v1 = 1.458 ft3/lbm, s1 = 1.6248 Btu/lbm R State 2: Table C.8.1 s2 = s1 = 1.6248 = sg at P2 P2 = 76.67 lbf/in2, v2 = vg = 5.703 me m1Ê-Êm2 m2 v1 1.458 = = 1 – = 1 – = 1 – = 0.744 m1 m1 m1 v2 5.703 T 1

8.95E A cylinder/piston contains 5 lbm of water at 80 lbf/in.2, 1000 F. The piston has cross-sectional area of 1 ft 2 and is restrained by a linear spring with spring constant 60 lbf/in. The setup is allowed to cool down to room temperature due to heat transfer to the room at 70 F. Calculate the total (water and surroundings) State 1: Table B.8.2 v1 = 10.831 , u1 = 1372.3 , s1 = 1.9453

8.96E An insulated cylinder/piston contains R-134a at 150 lbf/in.2, 120 F, with a volume of 3.5 ft3. The R-134a expands, moving the piston until the pressure in the cylinder has dropped to 15 lbf/in.2. It is claimed that the R-134a does 180 Btu of State 1: v1 = 0.33316, u1 = 175.33, s1 = 0.41586 m = V/v1 = 3.5/0.33316 = 10.505 lbm Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 = 0/ – 180 u2 = 158.196 Btu/lbm State 2: P2, u2 T2 = -2 F s2 = 0.4220 m(s2 – s1) = ? dQ/T + 1S2Êgen = 1S2Êgen = 0.0645 Btu/R This is possible since 1S2Êgen > 0/

8.97E A mass and atmosphere loaded piston/cylinder contains 4 lbm of water at 500 lbf/in.2, 200 F. Heat is added from a reservoir at 1200 F to the water until it reaches 1200 F. Find the work, heat transfer, and total entropy production for the system and surroundings.

Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 = 1Q2 – Pm(v2 – v1) Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen P T State 1: Table C.8.3, v1 = 0.01661 2 1 2 h1 = 169.18, s1 = 0.2934 1 Process: P = constant = 500 lbf/in2 vv

8.99E Water in a piston/cylinder is at 150 lbf/in.2, 900 F, as shown in Fig. P8.33. There are two stops, a lower one at which Vmin = 35 ft3 and an upper one at Vmax = 105 ft3. The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 75 lbf/in.2. This setup is now cooled to 210 F by rejecting heat to the surroundings at 70 F. Find the total entropy generated in the process.

State 1: Table C.8.2 v1 = 5.353, u1 = 1330.2, s1 = 1.8381 m = V/v1 = 105/5.353 = 19.615 lbm

P T 1000 500 1 v=C 1 2 2 v s

8.100EA cylinder/piston contains water at 30 lbf/in.2, 400 F with a volume of 1 ft3. The piston is moved slowly, compressing the water to a pressure of 120 lbf/in.2. The loading on the piston is such that the product PV is a constant. Assuming that the room temperature is 70 F, show that this process does not violate the second law. C.V.: Water + cylinder out to room at 70 F Process: PV = constant = Pmv v2 = P1v1/P2 1w2 = P dv = P1v1 ln(v2/v1) State 1: v1 = 16.891 , u1 = 1144.0 , s1 = 1.7936 State 2: P2, v2 = P1v1/P2 = 30· 16.891/120 = 4.223 ft3/lbm => T2 = 425.4 F, u2 = 1144.4, s2 = 1.6445 1w2 = 30 · 16.891(144/778) ln(4.223/16.891) = -130.0 Btu 1q2 = u2 – u1 + 1w2 = 1144.4 – 1144 – 130 = -129.6 Btu 1ss,gen = (s2 – s1) – 1q2/Troom = 1.6445 – 1.7936 + 129.6/529.67 = 0.0956 Btu/lbm R > 0/ satisfy 2nd law.

8.101EOne pound mass of ammonia (NH3) is contained in a linear spring-loaded piston/cylinder as saturated liquid at 0 F. Heat is added from a reservoir at 225 F until a final condition of 125 lbf/in.2, 160 F is reached. Find the work, heat transfer, and entropy generation, assuming the process is internally reversible. C.V. = NH3 Cont. m2 = m1 = m Energy: E2 – E1 = 1Q2 – 1W2 = m(u2 – u1) Entropy: S2 – S1 = dQ/T + 1S2,gen 11 Process: 1W2 = P dV = (P1 + P2)(V2 – V1) = (P1 + P2)m(v2 – v1) 22

8.102EA foundry form box with 50 lbm of 400 F hot sand is dumped into a bucket with 2 ft3 water at 60 F. Assuming no heat transfer with the surroundings and no boiling away of liquid water, calculate the net entropy change for the process. msand(u2 – u1)sand + mH O(u2 – u1)H O = -P(V2 – V1) 22

2 msand hsand + mH O hH O =0/ , mH O = = 124.73 lbm 2 2 2 0.016035 50 · 0.19(T2 – 400) + 124.73 · 1.0(T2 – 60) = 0/ , T2 = 84 F 544 544 S = 50 · 0.19 · ln + 124.73 · 1.0 · ln = 1.293 Btu/R ?860? ?520?

8.104EA handheld pump for a bicycle has a volume of 2 in.2 when fully extended. You now press the plunger (piston) in while holding your thumb over the exit hole so an air pressure of 45 lbf/in.2 is obtained. The outside atmosphere is at Po, To. Consider two cases: (1) it is done quickly (~1 s), and (2) it is done slowly (~1 h). C.V. Air in pump. Assume that both cases result in a reversible process. Case I) Quickly means no time for heat transfer u2 – u1 = -1w2 , s2 = s1 = 0/ Table C.6 Pr2 = Pr1 · P2/P1 = 1.0925(45/14.7) = 3.344 T2 = 737.7 R , V2 = P1V1T2/T1P2 = 0.898 in3 Case II) Slowly, time for heat transfer so T = T0. The process is then a reversible isothermal compression.

8.106EA 25-ft3 insulated, rigid tank contains air at 110 lbf/in.2, 75 F. A valve on the tank is opened, and the pressure inside quickly drops to 15 lbf/in.2, at which point the valve is closed. Assuming that the air remaining inside has undergone a reversible adiabatic expansion, calculate the mass withdrawn during the process.

Cont.Eq.: m2 = m ; Energy Eq.: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 0 + 0

8.108E Nitrogen at 90 lbf/in.2, 260 F is in a 20 ft3 insulated tank connected to a pipe with a valve to a second insulated initially empty tank of volume 20 ft3. The valve is opened and the nitrogen fills both tanks. Find the final pressure and temperature and the entropy generation this process causes. Why is the process irreversible? C.V. Both tanks + pipe + valve. Insulated : Q = 0, Rigid: W = 0 m(u2 – u1) = 0 – 0 => u2 = u1 = ua1 Entropy Eq.: m(s2 – s1) = dQ/T + 1S2Êgen = 1S2Êgen State 1: P1 , T1 , Va => Ideal gas m = PV/RT = (90 · 20 · 144)/ (55.15 · 720) = 6.528 lbm 2: V2 = Va + Vb ; uniform final state v2 = V2 / m ; u2 = ua1

P 1 2 v T P1 1 2 P2 s Ideal gas u (T) => u2 = ua1 => T2 = Ta1 = 720 R P2 = mR T2 / V2 = (V1 / V2 ) P1 = _ · 90 = 45 lbf/in.2 Sgen = m(s2 – s1) = m (sT2o – s – R ln (P2 / P1 ) o T1 = m (0 – R ln (P2 / P1 ) = -6.528 · 55.15 · (1/778)ln _ = 0.32 Btu/R Irreversible due to unrestrained expansion in valve P ? but no work out.

P2 = ma1 R Ta1 / V2 = ma1 R Ta1 / 2Va1 = _ Pa1 = 45 lbf/in.2 sa2 = sa1 => Ta2 = Ta1 (P2 / Pa1)k-1 / k = 720 · (1/2)0.2857 = 590.6 R Now we have final state in A va2 = R Ta2 / P2 = 5.0265 ; ma2 = Va / va2 = 3.979 lbm x = ma2 / ma1 = 0.6095 mb2 = ma1 – ma2 = 2.549 lbm Substitute into energy equation Tb2 = ( Ta1 – x Ta2 ) / (1 – x) = 922 R 1S2Êgen = mb2 ( sb2 – sa1) = mb2 ( Cp ln (Tb2 / Ta1 ) – R ln (P2 / Pa1 ) = 2.549 [ 0.249 ln (922/720) – (55.15/778) ln (1/2) ] = 0.2822 Btu/R

8.109EA cylinder/piston contains carbon dioxide at 150 lbf/in.2, 600 F with a volume of 7 ft3. The total external force acting on the piston is proportional to V3. This system is allowed to cool to room temperature, 70 F. What is the total entropy State 1: P = 150 lbf/in2, T = 600 F = 1060 R, V = 7 ft3 Ideal gas 111 P1V1 150Ê· Ê144Ê· Ê7 m = = = 4.064 lbm RT1 35.10Ê· Ê1060 Process: P = CV3 or PVÊ-3 = const. polytropic with n= -3.

8.111E A cylinder/piston contains air at ambient conditions, 14.7 lbf/in.2 and 70 F with a volume of 10 ft3. The air is compressed to 100 lbf/in.2 in a reversible polytropic process with exponent, n = 1.2, after which it is expanded back to 14.7 lbf/in.2 in c. What is the potential refrigeration capacity (in British thermal units) of the air at the final state?

a) P T 2 2 1 1

8.112E A cylinder/piston contains 4 ft3 of air at 16 lbf/in.2, 77 F. The air is compressed in a reversible polytropic process to a final state of 120 lbf/in.2, 400 F. Assume the heat transfer is with the ambient at 77 F and determine the polytropic exponent n and the final volume of the air. Find the work done by the air, the heat transfer Solution: m = (P1V1)/(RT1) = (16 · 4 · 144)/(53.34 · 537) = 0.322 lbm n-1 n-1 T2/T1 = (P2/P1) n = ln(T2 / T1) / ln(P2 / P1) = 0.2337 n

CHAPTER 9 The correspondence between the new problem set and the previous 4th edition second half of chapter 7 problem set.

New Old New Old New Old 1 63 28 87 55 96 2 64 29 new 56 117 3 66 30 88 57 97 4 new 31 90 58 99 5 104 mod 32 92 59 101 6 105 mod 33 new 60 102 7 675 34 new 61 new 8 new 35 new 62 104 9 new 36 new 63 105 10 new 37 new 64 108 11 new 38 77 65 110 12 65 39 110 mod 66 111 13 69 40 78 67 new 14 70 41 new 68 112 15 new 42 new 69 113 16 72 43 new 70 115 17 new 44 new 71 120 18 82 45 new 72 118 19 new 46 68 73 121 20 74 47 80 74 new 21 75 48 84 75 new 22 79 49 new 76 new 23 new 50 71 77 new 24 109 mod 51 103 78 new 25 85 52 94 79 new 26 91 53 106 80 new 27 86 54 95 81 109 82 98

9.1 Steam enters a turbine at 3 MPa, 450°C, expands in a reversible adiabatic process and exhausts at 10 kPa. Changes in kinetic and potential energies between the inlet and the exit of the turbine are small. The power output of the turbine is 800 Mass: mi = me = m, Energy Eq.: mhi = mhe + WT, Entropy Eq.: msi + 0/ = mse ( Reversible Sgen = 0 )

P Explanation for the work term is in 9.3 Eq. (9.19) 1 2 v T 1

2 s Inlet state: Table B.1.3 hi = 3344 kJ/kg, si = 7.0833 kJ/kg K Exit state: Pe , se = si Table B.1.2 sat. as se < sg xe = (7.0833 - 0.6492)/7.501 = 0.8578, he = 191.81 + 0.8578· 2392.82 = 2244.4 kJ/kg m = WT/wT = WT/(hi - he) = 800/(3344 - 2244.4) = 0.728 kg/s

9.2 In a heat pump that uses R-134a as the working fluid, the R-134a enters the compressor at 150 kPa, -10°C at a rate of 0.1 kg/s. In the compressor the R-134a is compressed in an adiabatic process to 1 MPa. Calculate the power input C.V.: Compressor (SSSF reversible: Sgen = 0 & adiabatic: Q = 0.) Inlet state: Table B.5.2 h1 = 393.84 kJ/kg, s1 = 1.7606 kJ/kg K Exit state: P2 = 1 MPa & s2 = s1 h2 = 434.9 kJ/kg Wc = mwc = m(h1 – h2) = 0.1 · (393.84 – 434.9) = -4.1 kW

P Explanation for the work term is in 9.3 Eq. (9.19) 1 2 T 2

9.3 Consider the design of a nozzle in which nitrogen gas flowing in a pipe at 500 kPa, 200°C, and at a velocity of 10 m/s, is to be expanded to produce a velocity of 300 m/s. Determine the exit pressure and cross-sectional area of the nozzle if the mass flow rate is 0.15 kg/s, and the expansion is reversible and adiabatic. 22 Energy Eq.: hi + Vi /2 = he + Ve/2 Entropy: si + 0 = se C (T – T ) = 1.0416(T – 473.2) = (102 – 3002)/(2· 1000) Po e i e k 430 3.5 Te = 430 K, Pe = Pi(Te/Ti)k-1 = 500 = 357.6 kPa ?473.2? ve = RTe/Pe = (0.2968 · 430)/357.6 = 0.35689 m3/kg A = . /V = ((0.15 · 0.35689)/300)· 10+6 = 178 mm2 e mve e 9.4 A compressor is surrounded by cold R-134a so it works as an isothermal compressor. The inlet state is 0°C, 100 kPa and the exit state is saturated vapor. Energy Eq.: hi + q = w + he; Entropy Eq.: si + q/T = se Inlet state: Table B.5.2, hi = 403.4 kJ/kg, si = 1.8281 kJ/kg K Exit state: Table B.5.1, he = 398.36 kJ/kg, se = 1.7262 kJ/kg K q = T(se – si) = 273.15(1.7262 – 1.8281) = – 27.83 kJ/kg w= 403.4 + (-27.83) – 398.36 = -22.8 kJ/kg

P e Explanation for the work term is in 9.3 Eq. (9.19) i v T ei

9.5 Air at 100 kPa, 17°C is compressed to 400 kPa after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle are both reversible and adiabatic and kinetic energy in/out of the compressor can be neglected. Find the compressor work and its exit temperature and find the nozzle Solution:

1 T C 2 3 SSSF separate control volumes around P compressor and nozzle. For ideal 2 2 compressor we have inlet : 1 and exit : 2 P1 Reversible: sgen = 0 Entropy Eq.: s1 + 0/T + 0 = s2

– wC = h2 – h1 , s2 = s1 State 1: Air Table A.7: h1 = 290.43 kJ/kg Pr2 = Pr1· P2/P1 = 0.9899 · 400 / 100 = 3.98 State 2: Pr2 = 3.98 in Table A.7 gives T2 = 430.5 K, h2 = 432.3 kJ/kg -wC = 432.3 – 290.43 = 141.86 kJ/kg The ideal nozzle then expands back down to state 1 (constant s) so energy equation gives: 1V2 = h – h = -w =141860 J/kg (remember conversion to J) 221C

9.6 A small turbine delivers 150 kW and is supplied with steam at 700°C, 2 MPa. The exhaust passes through a heat exchanger where the pressure is 10 kPa and exits as saturated liquid. The turbine is reversible and adiabatic. Find the specific turbine work, and the heat transfer in the heat exchanger.

Continuity Eq.: (SSSF) 1 2 m1 = m2 = m3 = m Turbine: Energy Eq.: WT wT = h1 – h2 Entropy Eq.: s2 = s1 + sTÊgen Heat exch: Energy Eq.: q = h3 – h2 , Entropy Eq.: s3 = s2 + ? Êdq/T + Inlet state: Table B.1.3 h1 = 3917.45 kJ/kg, s1 = 7.9487 kJ/kg K Ideal turbine sTÊgen = 0, s2 = s1 = 7.9487 = sf2 + x sfg2 State 3: P = 10 kPa, s2 < sg => saturated 2-phase in Table B.1.2 x2,s = (s1 – sf2)/sfg2 = (7.9487 – 0.6492)/7.501 = 0.9731 h2,s = hf2 + x· hfg2 = 191.8 + 0.9731· 2392.8 = 2520.35 kJ/kg wT,s = h1 – h2,s = 1397.05 kJ/kg q = h3 – h2,s = 191.83 – 2520.35 = -2328.5 kJ/kg

P Explanation for the work term is in 9.3 Eq. (9.19) 1 32 T 1

9.7 A counterflowing heat exchanger, shown in Fig. P9.7, is used to cool air at 540 K, 400 kPa to 360 K by using a 0.05 kg/s supply of water at 20°C, 200 kPa. The air flow is 0.5 kg/s in a 10-cm diameter pipe. Find the air inlet velocity, the water exit Air in 1: Table A.7, h1 = 544.686 kJ/kg, s°T1 = 7.46642 kJ/kg K, v1 = RT/P = 0.287· 540/400 = 0.3824 m3/kg Air out 2: h2 = 360.863, s°T2 = 7.05276 H2O in 3: Table B.1.1 h3 = 83.96, s3 = 0.2966 H2O out: 4 A = D2/4 = 0.007854 m2 . v/A = 24.34 m/s => V = m As the lines exchange energy select a control volume that includes both with no external heat transfer. Energy and entropy equations for the heat exchanger give Energy Eq.: mair(h1 – h2) = mH2O(h4 – h3) h4 = 83.96 + (0.5/0.05)(544.69-360.86) = 1922.2 kJ/kg h4 < hg Table B.1.2 x4 = (1922.2-504.68)/2202 = 0.64375 T4 = Tsat(P) = 120.23°C , s4 = 1.530 + 0.64375· 5.597 = 5.1331 kJ/kg K Entropy Eq.: 0 = mair(s1 - s2) + mH2O(s3 - s4) + Sgen Sgen = mH2O(4.8365) - mair(0.41366) = 0.02017 kW/K 9.8 Analyse the steam turbine described in Problem 6.29. Is it possible?

C.V. Turbine. SSSF and adiabatic. 1 Energy: m1h1 = m2h2 + m3h3 + W Entropy: m1s1 + Sgen = m2s2 + m3s3 2 3

9.9 A coflowing heat exchanger has one line with 2 kg/s saturated water vapor at 100 kPa entering. The other line is 1 kg/s air at 200 kPa, 1200 K. The heat exchanger is very long so the two flows exit at the same temperature. Find the exit temperature by trial and error. Calculate the rate of entropy generation. Solution: C.V. Heat exchanger. No W, no external Q 1 2 H2O Flows: m1 = m2 = mH2O; m3 = m4 = mair . . 3 4 Air Energy: mH2O (h2 – h1) = mair (h3 – h4) State 1: Table B.1.2 h1 = 2675.5 kJ/kg State 2: 100 kPa, T2 State 3: Table A.7 h3 = 1277.8 kJ/kg, State 4: 200 kPa, T2 Only one unknown T2 and one equation the energy equation: 2( h2 – 2675.5) = 1(1277.8 – h4) => 2h2 + h4 = 6628.8 kW At 500 K: h2 = 2902.0, h4 = 503.36 => LHS = 6307 too small At 700 K: h2 = 3334.8, h4 = 713.56 => LHS = 7383 too large Linear interpolation T2 = 560 K, h2 = 3048.3, h4 = 565.47 => LHS = 6662 Final states are with T2 = 554.4 K = 281 °C H2O: Table B.1.3, h2 = 3036.8 kJ/kg, s2 = 8.1473, s1 = 7.3593 kJ/kg K AIR: Table A.7, h4 = 559.65 kJ/kg, sT4 = 7.4936, sT3 = 8.3460 kJ/kg K The entropy balance equation is solved for the generation term: Sgen = mH2O (s2 – s1) + mair (s4- s3) = 2(8.1473 – 7.3593) +1 (7.4936 – 8.3460) = 0.724 kW/K No pressure correction is needed as the air pressure for 4 and 3 is the same.

9.10 Atmospheric air at -45°C, 60 kPa enters the front diffuser of a jet engine with a velocity of 900 km/h and frontal area of 1 m2. After the adiabatic diffuser the velocity is 20 m/s. Find the diffuser exit temperature and the maximum pressure C.V. Diffuser, SSSF single inlet and exit flow, no work or heat transfer. 22 Energy Eq.: hi + Vi /2 = he + Ve/2, and he – hi = Cp(Te – Ti) Entropy Eq.: si + dq/T + sgen = si + 0 + 0 = se (Reversible, adiabatic) Heat capacity and ratio of specific heats from Table A.5 in the energy equation then gives: 1.004[ T – (-45)] = 0.5[(900· 1000/3600)2 – 202 ]/1000 = 31.05 kJ/kg e

=> Te = -14.05 °C = 259.1 K k Constant s: Pe = Pi (Te/Ti)k-1 = 60 (259.1/228.1)3.5 = 93.6 kPa

9.11 A Hilch tube has an air inlet flow at 20°C, 200 kPa and two exit flows of 100 kPa, one at 0°C and the other at 40°C. The tube has no external heat transfer and no work and all the flows are SSSF and have negligible kinetic energy. Find the fraction of the inlet flow that comes out at 0°C. Is this setup possible? C.V. The Hilch tube. SSSF, single inlet and two exit flows. No q or w. Continuity Eq.: m1 = m2 + m3 ; Energy: m1h1 = m2h2 + m3h3 Entropy: m1s1 + Sgen = m2s2 + m3s3 States all given by temperature and pressure. Use constant heat capacity to evaluate changes in h and s. Solve for x = m2/m1 from the energy equation m3/m1 = 1 – x; h1 = x h2 + (1-x) h3 => x = (h1 – h3)/(h2 – h3) = (T1 – T3)/(T2 – T3) = (20-40)/(0-40) = 0.5 Evaluate the entropy generation Sgen/m1 = x s2 + (1-x)s3 – s1 = 0.5(s2 – s1 ) + 0.5(s3 – s1 ) = 0.5 [Cp ln(T2 / T1) – R ln(P2 / P1)] + 0.5[Cp ln(T3 / T1) – R ln(P3/ P1)] 273.15 100 = 0.5 [1.004 ln( ) – 0.287 ln( )] 293.15 200 313.15 100 + 0.5 [1.004 ln( ) – 0.287 ln( )] 293.15 200 = 0.1966 kJ/kg K > 0 So this is possible.

9.12 Two flowstreams of water, one at 0.6 MPa, saturated vapor, and the other at 0.6 MPa, 600°C, mix adiabatically in a SSSF process to produce a single flow out at 0.6 MPa, 400°C. Find the total entropy generation for this process.

1: B.1.2 h1 = 2756.8 kJ/kg, s1 = 6.760 kJ/kg K 1 3 2: B.1.3 h2 = 3700.9 kJ/kg, s2 = 8.2674 kJ/kg K 3: B.1.3 h3 = 3270.3 kJ/kg, s3 = 7.7078 kJ/kg K 2 Cont.: m3 = m1 + m2, Energy Eq.: m3h3 = m1h1 + m2h2 => m1/m3 = (h3 – h2) / (h1 – h2) = 0.456 Entropy Eq.: m3s3 = m1s1 + m2s2 + Sgen => Sgen/m3 = s3 – (m1/m3) s1 – (m2/m3) s2 = 7.7078 – 0.456· 6.760 – 0.544· 8.2674 = 0.128 kJ/kg K 9.13 In a heat-driven refrigerator with ammonia as the working fluid, a turbine with inlet conditions of 2.0 MPa, 70°C is used to drive a compressor with inlet saturated vapor at -20°C. The exhausts, both at 1.2 MPa, are then mixed together. The ratio of the mass flow rate to the turbine to the total exit flow was measured to be 0.62. Can this be true?

9.14 A diffuser is a steady-state, steady-flow device in which a fluid flowing at high velocity is decelerated such that the pressure increases in the process. Air at 120 kPa, 30°C enters a diffuser with velocity 200 m/s and exits with a velocity of 20 m/s. Assuming the process is reversible and adiabatic what are the exit pressure C.V. Diffuser, SSSF single inlet and exit flow, no work or heat transfer. 22 Energy Eq.: hi + Vi /2 = he + Ve/2, => he – hi = CPo(Te – Ti) Entropy Eq.: si + dq/T + sgen = si + 0 + 0 = se (Reversible, adiabatic) Energy equation then gives: C (T – T ) = 1.004(T – 303.2) = (2002 – 202)/(2·1 000) => T = 322.9 K Po e i e e k Pe = Pi(Te/Ti)k-1 = 120(322.9/303.2)3.5 = 149.6 kPa

9.16 One technique for operating a steam turbine in part-load power output is to throttle the steam to a lower pressure before it enters the turbine, as shown in Fig. P9.16. The steamline conditions are 2 MPa, 400°C, and the turbine exhaust pressure is fixed at 10 kPa. Assuming the expansion inside the turbine to be reversible and adiabatic, determine a. The full-load specific work output of the turbine b. The pressure the steam must be throttled to for 80% of full-load output c. Show both processes in a T–s diagram.

T a) C.V Turbine. Full load reversible 1= 2a s3 = s1 = 7.1271 = 0.6493 + x3a · 7.5009 2b h = C => x3a = 0.8636 h3a = 191.83 + 0.8636 · 2392.8 = 2258.3 kJ/kg 3a 3b 1w3a = h1 – h3a s = 3247.6 – 2258.3 = 989.3 kJ/kg

9.17 Carbon dioxide at 300 K, 200 kPa is brought through a SSSF device where it is heated to 500 K by a 600 K reservoir in a constant pressure process. Find the C.V. Heater and walls out to the source. SSSF single inlet and exit flows. Since the pressure is constant and there are no changes in kinetic or potential energy between the inlet and exit flows the work is zero. w = 0 Continuity: mi = me = m; Energy Eq.: hi + q = he; Entropy Eq.: si + dq/T + sgen = se = si + q/Tsource + sgen Properties are from Table A.8 (mole basis so divide by M = 44.01) q = he – hi = (8305 – 69)/44.01 = 187.1 kJ/kg sgen = se – si – q/Tsource = (234.9 – 214.02)/44.01 – 187.1/600 = 0.4744 – 0.312 = 0.1626 kJ/kg K

9.21 Air enters a turbine at 800 kPa, 1200 K, and expands in a reversible adiabatic process to 100 kPa. Calculate the exit temperature and the work output per kilogram of air, using a. The ideal gas tables, Table A.7 b. Constant specific heat, value at 300 K from table A.5 c. Constant specific heat, value at an intermediate temperature from Fig. 5.10 Discuss why the method of part (b) gives a poor value for the exit temperature air C.V. Air turbine. Adiabatic; q = 0, reversible: sgen = 0 W Energy: wT = hi – he , Entropy Eq.: se = si Turbine a) Table A.7: hi = 1277.8, Pri = 191.174 Pre = Pri · Pe/Pi = 191.174 · 100/800 = 23.897 e Te = 706 K, he = 719.9 kJ/kg w = hi – he = 557.9 kJ/kg k-1 100 0.286 b) Te = Ti (Pe/Pi) k = 1200 ?800? = 662.1 K w = CPo(Ti – Te) = 1.004(1200 – 662.1) = 539.8 kJ/kg —- c) Fig. 5.10 at ~ 1000 K: CPo ~ 32.5, CVo = CPo-R ~ 24.2 – – 0.255 k = CPo/CVo ~ 1.343, Te = 1200 (100/800) = 706.1 K w = (32.5/28.97)(1200 – 706.1) = 554.1 kJ/kg 9.22 Consider a steam turbine power plant operating at supercritical pressure, as shown in Fig. P9.22. As a first approximation, it may be assumed that the turbine and the pump processes are reversible and adiabatic. Neglecting any changes in kinetic and potential energies, calculate a. The specific turbine work output and the turbine exit state b. The pump work input and enthalpy at the pump exit state c. The thermal efficiency of the cycle 25 MPa a) 1: h1 = 3777.51, s1 = 6.67074

9.23 A supply of 5 kg/s ammonia at 500 kPa, 20°C is needed. Two sources are available one is saturated liquid at 20°C and the other is at 500 kPa, 140°C. Flows from the two sources are fed through valves to an insulated SSSFmixing chamber, which then produces the desired output state. Find the two source mass flow rates Continuity Eq.: m1 + m2 = m3; Energy Eq.: m1 h1 + m2h2 = m3h3 Entropy Eq.: m1 s1 + m2s2 + Sgen = m3s3 T 1 2

2 MIXING CHAMBER 3 3 1 s

9.24 A turbo charger boosts the inlet air pressure to an automobile engine. It consists of an exhaust gas driven turbine directly connected to an air compressor, as shown in Fig. P9.24. For a certain engine load the conditions are given in the figure. Assume that both the turbine and the compressor are reversible and adiabatic having also the same mass flow rate. Calculate the turbine exit temperature and CV: Turbine, SSSF, 1 inlet and 1 exit, adibatic: q = 0, reversible: sgen = 0 Energy: wT = h3 – h4 , Entropy Eq.: s4 = s3 k-1 100 0.286 s4 = s3 ? T4 = T3(P4/P3) k = 923.2 = 793.2 K ?170? wT = h3 – h4 = CP0(T3 – T4) = 1.004(923.2 – 793.2) = 130.5 kJ/kg WT = mwT = 13.05 kW Energy: -wC = h2 – h1 , Entropy Eq.: s2 = s1 -wC = wT = 130.5 = CP0(T2 – T1) = 1.004(T2 – 303.2) T2 = 433.2 K k 433.2 3.5 s2 = s1 ? P2 = P1(T2/T1)k-1 = 100 = 348.7 kPa ?303.2?

9.25 A stream of ammonia enters a steady flow device at 100 kPa, 50°C, at the rate of 1 kg/s. Two streams exit the device at equal mass flow rates; one is at 200 kPa, 50°C, and the other as saturated liquid at 10°C. It is claimed that the device operates in a room at 25°C on an electrical power input of 250 kW. Is this Control volume: SSSF device out to ambient 25°C.

9.26 An initially empty 0.1 m3 cannister is filled with R-12 from a line flowing saturated liquid at -5°C. This is done quickly such that the process is adiabatic. Find the final mass, liquid and vapor volumes, if any, in the cannister. Is the C.V. cannister USUF where: 1Q2 = 0/ ; 1W2 = 0/ ; m1 = 0/ Mass: m2 – 0/ = min ; Energy: m2u2 – 0/ = minhline + 0/ + 0/ u2 = hline 2: P2 = PL ; u2 = hL 2 phase u2 > uf ; u2 = uf + x2ufg Table B.3.1: uf = 31.26 ; ufg = 137.16 ; hf = 31.45 all kJ/kg x2 = (31.45 -31.26)/137.16 = 0.001385 v2 = vf + x2vfg = 0.000708 + 0.001385· 0.06426 = 0.000797 m3/kg m2 = V/v2 = 125.47 kg ; mf = 125.296 kg; mg = 0.174 kg V = m v = 0.0887 m3; V = m v = 0.0113 m3 f ff g gg Process is irreversible (throttling) s2 > sf

9.27 Air from a line at 12 MPa, 15°C, flows into a 500-L rigid tank that initially contained air at ambient conditions, 100 kPa, 15°C. The process occurs rapidly and is essentially adiabatic. The valve is closed when the pressure inside reaches some value, P2. The tank eventually cools to room temperature, at which time the pressure inside is 5 MPa. What is the pressure P2? What is the net entropy change CV: Tank. Mass flows in, so this is USUF. Find the mass first

m1 = P1V/RT1 = 100· 0.5/(0.287· 288.2) = 0.604 kg T Fill to P2, then cool to T3 = 15°C, P3 = 5 MPa 12 MPa m3 = m2 = P3V/RT3 line 5 MPa = (5000 · 0.5)/(0.287 · 288.2) = 30.225 kg Mass: mi = m2 – m1 = 30.225 – 0.604 = 29.621 kg In the process 1-2 heat transfer = 0 1st law: mihi = m2u2 – m1u1 ; miCP0Ti = m2CV0T2 – m1CV0T1 (29.621· 1.004Ê+Ê0.604· 0.717)· 288.2 T2 = 30.225Ê· Ê0.717 = 401.2 K P2 = m2RT2/V = (30.225 · 0.287 · 401.2)/0.5 = 6.960 MPa Consider now the total process from the start to the finish at state 3.

9.28 An initially empty spring-loaded piston/cylinder requires 100 kPa to float the piston. A compressor with a line and valve now charges the cylinder with water to a final pressure of 1.4 MPa at which point the volume is 0.6 m3, state 2. The inlet condition to the reversible adiabatic compressor is saturated vapor at 100 kPa. After charging the valve is closed and the water eventually cools to room temperature, 20°C, state 3. Find the final mass of water, the piston work from 1 to 2, the required compressor work, and the final pressure, P3.

in Process 1? 2: USUF, adiabatic. · for C.V. compressor + cylinder -W c Assume process is reversible Continuity: m2 – 0 = min , Energy: m2u2 – 0/ = (minhin) – Wc – 1W2 Entropy Eq.: m2s2 – 0/ = minsin + 0 s2 = sin Inlet state: Table B.1.2, hin = 2675.5 kJ/kg, sin = 7.3594 kJ/kg K 11 1W2 = ? PdV = (Pfloat+ P2)(V2 – 0/ ) = (100+1400)0.6 = 450 kJ 22 State 2: P2 , s2 = sin Table B.1.3 v2 = 0.2243, u2 = 2984.4 kJ/kg m2 = V2/v2 = 0.6/0.2243 = 2.675 kg Wc = minhin – m2u2 – 1W2 = 2.675 · (2675.5 – 2984.4) – 450 = -1276.3 kJ

9.29 An initially empty cannister of volume 0.2 m3 is filled with carbon dioxide from a line at 1000 kPa, 500 K. Assume the process is adiabatic and the flow continues until it stops by itself. Find the final mass and temperature of the carbon dioxide C.V. Cannister + valve out to line. No boundary/shaft work, m1 = 0; Q = 0. Continuity Eq.: m2 – 0 = mi Energy: m2 u2 – 0 = mi hi Entropy Eq.: m2s2 – 0 = misi + 1S2Êgen State 2: P2 = Pi and u2 = hi = hline = h2 – RT2 (ideal gas) To reduce or eliminate guess use: h2 – hline = CPo(T2 – Tline) Energy Eq. becomes: CPo(T2 – Tline) – RT2 = 0 T2 = Tline CPo/(CPo – R) = Tline CPo/CVo = k Tline Use A.5: CP = 0.842, k = 1.289 => T2 = 1.289· 500 = 644 K m2 = P2V/RT2 = 1000· 0.2/(0.1889· 644) = 1.644 kg 1S2Êgen = m2 (s2 – si) = m2[ CP ln(T2 / Tline) – R ln(P2 / Pline)] = 1.644[0.842· ln(1.289) – 0] = 0.351 kJ/K If we use A.8 at 550 K: CP = 1.045, k = 1.22 => T2 = 610 K, m2 = 1.735 kg

9.30 A 1-m3 rigid tank contains 100 kg R-22 at ambient temperature, 15°C. A valve on top of the tank is opened, and saturated vapor is throttled to ambient pressure, 100 kPa, and flows to a collector system. During the process the temperature inside the tank remains at 15°C. The valve is closed when no more liquid remains inside. Calculate the heat transfer to the tank and total entropy generation in the process. C.V. Tank out to surroundings. This is USUF. Rigid tank so no work term. Energy Eq.: m2u2 – m1u1 = QCV – mehe Entropy Eq.: m2s2 – m1s1 = QCV/TSUR – mese + Sgen State 1: Table B.3.1, v1 = V1/m1 = 1/100 = 0.000812 + x1 0.02918 x1 = 0.3149, u1 = 61.88 + 0.3149 · 169.47 = 115.25 s1 = 0.2382 + 0.3149 · 0.668 = 0.44855; he = hg = 255.0 State 2: v2 = vg = 0.02999, u2 = ug = 231.35, s2 = 0.9062 Exit state: he = 255.0, Pe = 100 kPa ? Te = -4.7°C, se = 1.0917 m2 = 1/0.02999 = 33.34 kg; me = 100 – 33.34 = 66.66 kg QCV = m2u2 – m1u1 + mehe = 33.34· 231.35 – 100· 115.25 + 66.66· 255 = 13 186 kJ SCVÊÊ = m2s2 – m1s1 = 33.34(0.9062) – 100(0.44855) = -14.642 SSUR = – QCV/TSUR + mese = -13186/288.2 + 66.66(1.0917) = +27.012 Sgen = SNETÊ = -14.642 + 27.012 = +12.37 kJ/K

9.31 An old abandoned saltmine, 100000 m3 in volume, contains air at 290 K, 100 kPa. The mine is used for energy storage so the local power plant pumps it up to 2.1 MPa using outside air at 290 K, 100 kPa. Assume the pump is ideal and the process is adiabatic. Find the final mass and temperature of the air and the required pump work. Overnight, the air in the mine cools down to 400 K. Find the (USUF) C.V. = Air in mine + pump Continuity Eq.: m2 – m1 = min Energy: m2u2 – m1u1 = 1Q2 – 1W2 + minhin Entropy: m2s2 – m1s1 = ? dQ/T + 1S2Êgen + minsin Process: Adiabatic 1Q2 = 0 , Process ideal 1S2Êgen = 0 , s1 = sin m2s2 = m1s1 + minsin = (m1 + min)s1 = m2s1 s2 = s1 Const. s Pr2 = Pr1P2/P1 = 0.9899(21) = 20.788 T2 = 680 K , u2 = 496.97 kJ/kg m1 = P1V1/RT1 = 100· 105/(0.287 · 290) = 1.20149· 105 kg m2 = P2V2/RT2 = 100 · 21· 105/(0.287 · 680) = 10.760· 105 kg min = 9.5585· 105 kg 1W2 = minhin + m1u1 – m2u2 = min(290.43) + m1(207.19) – m2(496.97) = -2.322· 108 kJ 2? 3: Process : V3 = V2 = V1 => 2W3 = 0 P3 = P2T3/T2 = 1235 kPa 2Q3 = m2(u3 – u2) = 10.760· 105(286.49 – 496.97) = -2.265· 108 kJ

9.33 An insulated 2 m3 tank is to be charged with R-134a from a line flowing the refrigerant at 3 MPa. The tank is initially evacuated, and the valve is closed when the pressure inside the tank reaches 3 MPa. The line is supplied by an insulated compressor that takes in R-134a at 5°C, quality of 96.5 %, and compresses it to 3 MPa in a reversible process. Calculate the total work input to the compressor to C.V.: Compressor, R-134a. SSSF, 1 inlet and 1 exit, no heat transfer.

9.36 A frictionless piston/cylinder is loaded with a linear spring, spring constant 100 kN/m and the piston cross-sectional area is 0.1 m2. The cylinder initial volume of 20 L contains air at 200 kPa and ambient temperature, 10°C. The cylinder has a set of stops that prevent its volume from exceeding 50 L. A valve connects to a line flowing air at 800 kPa, 50°C. The valve is now opened, allowing air to flow in until the cylinder pressure reaches 800 kPa, at which point the temperature inside the cylinder is 80°C. The valve is then closed and the process ends. b) Taking the inside of the cylinder as a control volume, calculate the heat c) Calculate the net entropy change for this process.

800 P line 500 Air from Table A.5: x 200 V R = 0.287 kJ/kg-K, Cp = 1.004 kJ/kg-K 20 50 Cv = 0.717 kJ/kg-K, k = 1.4 Ap = 0.1 m2, Vstop = 50 L State 1: T1 = 10 C, P1 = 200 kPa, V1 = 20 L = 0.02 m3, o

9.37 An insulated piston/cylinder contains R-22 at 20°C, 85% quality, at a cylinder volume of 50 L. A valve at the closed end of the cylinder is connected to a line flowing R-22 at 2 MPa, 60°C. The valve is now opened, allowing R-22 to flow in, and at the same time the external force on the piston is decreased, and the piston moves. When the valve is closed, the cylinder contents are at 800 kPa, 20°C, and a positive work of 50 kJ has been done against the external force. What is the final volume of the cylinder? Does this process violate the second law of thermodynamics?

State 1: T1 = 20oC, x1 = 0.85, V1 = 50 L = 0.05 m3 P1 = Pg = 909.9 kPa, u1 = uf + x1ufg = 208.1 kJ/kg v1 = vf + x1vfg = 0.000824 + 0.85· 0.02518 = 0.022226 m3/kg, s1 = sf + x1sfg = 0.259 + 0.85· 0.6407 = 0.8036 kJ/kg K m1 = V1/v1 = 2.25 kg State 2: T2 = 20oC, P2 = 800 kPa, superheated, v2 = .030336 m3/kg, h2 = 258.7 kJ/kg, u2 = h2 – P2v2 = 234.4 kJ/kg, s2 = 0.91787 kJ/kg K Inlet: Ti = 60oC, Pi = 2 MPa, hi = 271.6 kJ/kg, si = 0.8873 kJ/kg K Energy: 1Q2 + mihi = m2u2 – m1u1 + 1W2; 1Q2 = 0, me = 0, 1W2 = 50 kJ mi = m2 – m1; (m2 – m1)hi = m2u2 – m1u1 solve for m2 = 5.185 kg, V2 = m2v2 = 0.157 m3 Qcv 2nd Law: Snet = m2s2 – m1s1 – misi – , QCV = 0, To = 20oC To Snet = m2s2 – m1si – (m2 – m1)si = 0.3469 kJ/K > 0, Satisfies 2nd Law 9.38 Liquid water at ambient conditions, 100 kPa, 25°C, enters a pump at the rate of 0.5 kg/s. Power input to the pump is 3 kW. Assuming the pump process to be reversible, determine the pump exit pressure and temperature.

9.39 A firefighter on a ladder 25 m above ground should be able to spray water an additional 10 m up with the hose nozzle of exit diameter 2.5 cm. Assume a water pump on the ground and a reversible flow (hose, nozzle included) and find the C.V.: pump + hose + water column, total height difference 35 m. Here V is Continuity Eq.: min = mex = ( AV)nozzle ..2.2 Energy Eq.: mwp + m(h + V /2 + gz)in = m(h + V /2 + gz)ex hin hex , Vin Vex = 0 , zex – zin = 35 m , = 1/v 1/vf wp = g(zex – zin) = 9.81· (35-0) = 343.2 J/kg The velocity in the exit nozzle is such that it can rise 10 m, so make that column C.V.

gz + 1V2 = gz + 0 noz 2 noz ex10 m 35 m Vnoz = 2g(zexÊ-Êznoz) = 2Ê· Ê9.81Ê· Ê10 = 14 m/s . D2 vf?2 ? Wp = mwp = 2.36 kW 9.40 A large storage tank contains liquefied natural gas (LNG), which may be assumed to be pure methane. The tank contains saturated liquid at ambient pressure, 100 kPa; it is to be pumped to 500 kPa and fed to a pipeline at the rate of 0.5 kg/s. How much power input is required for the pump, assuming it to be reversible? C.V. Pump, liquid is assumed to be incompressible.

9.41 A small dam has a pipe carrying liquid water at 150 kPa, 20°C with a flow rate of 2000 kg/s in a 0.5 m diameter pipe. The pipe runs to the bottom of the dam 15 m lower into a turbine with pipe diameter 0.35 m. Assume no friction or heat transfer in the pipe and find the pressure of the turbine inlet. If the turbine exhausts to 100 kPa with negligible kinetic energy what is the rate of work? C.V. Pipe. No work, no heat transfer, v » const. Bernoulli so find kin energy at 1, 2 1 v2 » v1 » vf = 0.001002 .2 Êm = AV = AV/vTurbine3

V1 . 1Ê/A1 Ê= 2000 · 0.001002 / ( 4 0.5 2) = 10.2 m s -1 Ê = Êmv . 2 -1 V2Ê = Êmv2Ê/AÊ2 = 2000 · 0.001002 / ( 0.35 ) = 20.83 m s 4 122 v(Pe-Pi) + (Ve-Vi ) + g (Ze – ZiÊ ) = ? 2

9.43 Saturated R-134a at -10°C is pumped/compressed to a pressure of 1.0 MPa at the rate of 0.5 kg/s in a reversible adiabatic SSSF process. Calculate the power required and the exit temperature for the two cases of inlet state of the R-134a: C.V.: Pump/Compressor, m = 0.5 kg/s, R-134a a) State 1: Table B.5.1, T1 = -10oC, x1 = 1.0 Saturated vapor P1 = Pg = 202 kPa, h1 = hg = 392.3 kJ/kg, s1 = sg = 1.7319 kJ/kg K Assume Compressor is Isentropic, s2 = s1 = 1.7319 kJ/kg-K h2 = 425.7 kJ/kg, T2 = 45oC 1st Law: qc + h1 = h2 + wc; qc = 0 wcs = h1 – h2 = -33.4 kJ/kg; => WC = mwC = -16.7 kW P1 = 202 kPa, h1 = hf = 186.72 kJ/kg, v1 = vf = 0.000755 m3/kg 1st Law: qp + h1 = h2 + wp; qp = 0 Assume Pump is isentropic and the liquid is incompressible: wps = – v dP = -v1(P2 – P1) = -0.6 kJ/kg h2 = h1 – wp = 186.72 – ( – 0.6) = 187.3 kJ/kg, P2 = 1MPa Assume State 2 is a saturated liquid => T2 -9.6oC WP = mwP = -0.3 kW

9.44 A small water pump on ground level has an inlet pipe down into a well at a depth H with the water at 100 kPa, 15°C. The pump delivers water at 400 kPa to a building. The absolute pressure of the water must be at least twice the saturation pressure to avoid cavitation. What is the maximum depth this setup will allow?

C.V. Pipe in well, no work, no heat transfer P inlet pump ‡ 2 Psat, 15C = 2· 1.705 = 3.41 v P + g H = 0 => 1000 · 0.001001 ( 3.41 – 100) + 9.80665 · H = ? i H = 9.86 m e

H Since flow has some kinetic energy and there are losses in the pipe the height is overestimated. Also the start transient would generate a very low inlet pressure (it moves flow by suction) 9.45 Atmospheric air at 100 kPa, 17°C blows at 60 km/h towards the side of a building. Assume the air is nearly incompressible find the pressure and the C.V. A stream line of flow from the freestream to the wall.

122 V v(Pe-Pi) + (Ve-Vi ) + g(Ze – Zi) = ? 2

1 2 1 60Ê· Ê1000 P= Vi = /(0.287 · 290.15/100) 2v 2 3600 = 1 16.6672 / (0.8323 · 1000) = 0.17 kPa 2

Pe = Pi + P = 100.17 kPa Te = Ti (Pe/Pi)0.286 = 290.15 · 1.0005 = 290.3 K Very small efffect due to low velocity and air is light (large specific volume) 9.46 A small pump takes in water at 20°C, 100 kPa and pumps it to 2.5 MPa at a flow C.V. Pump. Assume reversible pump and incompressible flow.

9.47 Helium gas enters a steady-flow expander at 800 kPa, 300°C, and exits at 120 kPa. The mass flow rate is 0.2 kg/s, and the expansion process can be considered as a reversible polytropic process with exponent, n = 1.3. Calculate the power output of the expander.

i exp Q n-1 0.3 Pe n 120 1.3 Te = Ti = 573.2 = 370 K Pi ?800? ?? Work evaluated from Eq.9.20 nR w = – ? vdP = – (Te – Ti) e n-1 -1.3Ê· Ê2.07703 = (370 – 573.2) = 1828.9 kJ/kg 0.3 W = 0.2 · 1828.9 = 365.8 kW PT i n = k = 1.667 n=1i n=1 e n = 1.3 e n = 1.3 v s .

9.48 A pump/compressor pumps a substance from 100 kPa, 10°C to 1 MPa in a reversible adiabatic SSSF process. The exit pipe has a small crack, so that a small amount leaks to the atmosphere at 100 kPa. If the substance is (a) water, (b) R-12, find the temperature after compression and the temperature of the leak flow as it enters the atmosphere neglecting kinetic energies.

2 C.V.: Compressor, reversible adiabatic 3 h1 – wc = h2 ; s1 = s2 -W. State 2: P2, s2 = s1 c

1 C.V.: Crack (SSSF throttling process) h3 = h2 ; s3 = s2 + sgen State 3: P3, h3 = h2 a) Water 1: compressed liquid, Table B.1.1 -wc = +? vdP = vf1(P2 – P1) = 0.001 · (1000 – 100) = 0.9 kJ/kg h2 = h1 – wc = 41.99 + 0.9 = 42.89 kJ/kg => T2 = 10.2°C P3 , h3 compressed liquid at ~10.2°C

P 2 1, 3 T 2 3 1 v States 1 and 3 are at the same 100 kPa, and same h. You cannot separate them in the P-v fig.

s b) R-12 1: superheated vapor, Table B.3.2, s1 = 0.8070 kJ/kg K s2 = s1 & P2 T2 = 98.5°C , h2 = 246.51 kJ/kg -wc = h2 – h1 = 246.51 – 197.77 = 48.74 kJ/kg P3 , h3 T3 = 86.8°C

9.50 A mixing chamber receives 5 kg/min ammonia as saturated liquid at -20°C from one line and ammonia at 40°C, 250 kPa from another line through a valve. The chamber also receives 325 kJ/min energy as heat transferred from a 40°C reservoir. This should produce saturated ammonia vapor at -20°C in the exit line. What is the mass flow rate in the second line and what is the total entropy generation in the process?

1 2 CV: Mixing chamber out to reservoir m1 + m2 = m3 . m1h1 + m2h2 + Q = m3h3 Q Tres m1s1 + m2s2 + Q/Tres + Sgen = m3s3

From the energy equation: m2 = [(m1(h1 – h3) + Q]/(h3 – h2) = [5 · (89.05 – 1418.05) + 325]/(1418.05 – 1551.7) = 47.288 kg/min m3 = 52.288 kg/min Sgen = m3s3 – m1s1 – m2s2 – Q/Tres = 52.288 · 5.6158 – 5 · 0.3657 – 47.288 · 5.9599 – 325/313.15 = 8.94 kJ/K min 9.51 A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the isentropic compressor efficiency and the entropy generation.

9.52 Liquid water enters a pump at 15°C, 100 kPa, and exits at a pressure of 5 MPa. If the isentropic efficiency of the pump is 75%, determine the enthalpy (steam table CV: pump QCV » 0, KE » 0, PE » 0 2nd law, reversible (ideal) process Êes ses = si ws = – ? vdP » -vi(Pe-Pi) = -0.001001(5000 – 100) = -4.905 kJ/kg Êi Real process: w = ws/ s = -4.905/0.75 = -6.54 kJ/kg and he = hi – w = 62.99 + 6.54 = 69.53 kJ/kg

9.53 A centrifugal compressor takes in ambient air at 100 kPa, 15°C, and discharges it at 450 kPa. The compressor has an isentropic efficiency of 80%. What is your State 1: Table A.7: Pr1 = 1.2055 Pr2s = 1.2055 · (450/100) = 5.4248 ? T2s = 442.1 K ws = h1 – h2s = 288.36 – 443.75 = -155.39 wac = -155.39/0.8 = -194.23 h2 = 194.23 + 288.36 = 482.59, T2 = 480.1 K

9.55 A small air turbine with an isentropic efficiency of 80% should produce 270 kJ/kg of work. The inlet temperature is 1000 K and it exhausts to the atmosphere. Find C.V. Turbine actual: w = hi – he,ac = 270 he,ac = 776.22 , Te = 757.9 K C.V. Ideal turbine: ws = w/ s = 270/0.8 = 337.5 = hi – he,s he,s = 708.72 Te,s = 695.5 si = se,sÊÊ Pe/Pi = Pre/Pri [ = (Te/Ti)k/(k-1) for constant Cp] Pi = PePri/Pre = 101.3 · 91.651 / 22.607 = 410.8 kPa [ = 101.3 (1000/695.5)3.5 = 361 kPa for constant Cp] PT i i Pi

e, ac e, s s =C Pe e, s e, ac v s 9.56 Carbon dioxide, CO2, enters an adiabatic compressor at 100 kPa, 300 K, and exits at 1000 kPa, 520 K. Find the compressor efficiency and the entropy generation for C.V. Ideal compressor P2 k-1 1000 0.2242 wc = h1 – h2, s2 = s1 : T2s = T1 k = 300 = 502.7 K P1 ?100 ? ?? wcs = Cp(T1 – T2s) = 0.8418(300-502.7) = -170.63 kJ/kg C.V. Actual compressor wcac = Cp(T1 – T2ac) = 0.8418(300 – 520) = -185.2 kJ/kg c = wcs/wcac = -170.63/(-185.2) = 0.92 sgen = s2ac – s1 = Cp ln (T2ac/T1) – R ln (P2/P1) = 0.8418 ln(520 / 300) – 0.18892 ln(1000 / 100) = 0.028 kJ/kg K Constant heat capacity is a poor approximation.

9.57 Repeat Problem 9.22 assuming the turbine and the pump each have an isentropic P = T = 85% wT,AC = TwT,s = 1343.25 = h1 – h2,AC x2,AC = (2434.3 – 251.4)/2358.3 = 0.926 , T2,AC=60.06°C wP,AC = wP,s/ = 29.5 = h4,AC – h3 h4,AC = 197.0 T4,AC 42°C wT,ACÊ-ÊwP,AC 1313.78 TH = = = 0.367 h1Ê-Êh4,AC 3580.49

9.59 Steam enters a turbine at 300°C and exhausts at 20 kPa. It is estimated that the isentropic efficiency of the turbine is 70%. What is the maximum turbine inlet pressure if the exhaust is not to be in the two-phase region?

P i es e 300 v Tw Turbine: s = = 0.70, i Pi ws w = hi – he 20 es e ws = hi – hes, ses = si s Increasing Pi shifts process left.

For state e to stay out of 2 phase but with max Pi, xe = 1.0 he = 2609.7 Assume Pi = 0.66 MPa Then, at Ti = 300°C; hi = 3060.1, si = 7.3305 ses = si = 7.3305 = 0.8320 + xes · 7.0766 => xes = 0.9183 hes = 251.4 + 0.9183 · 2358.3 = 2417.0 w = 3060.1 – 2609.7 = 450.4 ws = 3060.1 – 2417.0 = 643.1 s = (450.4/643.1) = 0.700 OK Pi = 0.66 MPa

9.62 Assume both the compressor and the nozzle in Problem 9.5 have an isentropic efficiency of 90% the rest being unchanged. Find the actual compressor work and its exit temperature and find the actual nozzle exit velocity.

Solution: 1 C 3 5 T C.V. Ideal compressor, inlet: 1 exit: 2. 3 2 P2 = P3 wC = h2 – h1 , s2 = s1 Pr2 = Pr1P2/P1 = 3.98 State 2: A.7 T2 = 430.5 K, P 1 4 5 1 h2 = 432.3 s wCs = 141.86 kJ/kg

9.63 The small turbine in Problem 9.6 was ideal. Assume instead the isentropic turbine efficiency is 88%. Find the actual specific turbine work, the entropy generated in the turbine and the heat transfer in the heat exchanger.

Continuity Eq.: (SSSF) 1 2 m1 = m2 = m3 = m Turbine: Energy Eq.: WT wT = h1 – h2 Entropy Eq.: s2 = s1 + sTÊgen Heat exch: Energy Eq.: q = h3 – h2 , Entropy Eq.: s3 = s2 + ? Êdq/T + Inlet state: Table B.1.3 h1 = 3917.45 kJ/kg, s1 = 7.9487 kJ/kg K Ideal turbine sTÊgen = 0, s2 = s1 = 7.9487 = sf2 + x sfg2 State 3: P = 10 kPa, s2 < sg => saturated 2-phase in Table B.1.2 x2,s = (s1 – sf2)/sfg2 = (7.9487 – 0.6492)/7.501 = 0.9731 h2,s = hf2 + x· hfg2 = 191.8 + 0.9731· 2392.8 = 2520.35 kJ/kg wT,s = h1 – h2,s = 1397.05 kJ/kg

P Explanation for the work term is in 9.3 Eq. (9.19) 1 32 T 1

9.64 A geothermal supply of hot water at 500 kPa, 150°C is fed to an insulated flash evaporator at the rate of 1.5 kg/s. A stream of saturated liquid at 200 kPa is drained from the bottom of the chamber, and a stream of saturated vapor at 200 kPa is drawn from the top and fed to a turbine. The turbine has an isentropic efficiency of 70% and an exit pressure of 15 kPa. Evaluate the second law for a control volume that includes the flash evaporator and the turbine.

1 HO 2 FLASH OUT Turb 3 h1 = 632.2 = 504.7 + x · 2201.9 => x = 0.0579 = mVAP/m1 mVAP = 0.0579 · 1.5 = 0.08686 kg/s mLIQ = 1.413 kg/s w Turbine: s3s = s2 = sg at 200 kPa s3s = 7.1271 = 0.7594 + x3s · 7.2536 => x3s = 0.8785 h3s = 225.94 + 0.8785· 2373.1 = 2310.7

ws = h2 – h3s = 2706.7 – 2310.7 = 396 kJ/kg w = sws = 0.7 · 396 = 277.2 kJ/kg h3 = h2 – w = 2706.7 – 277.2 = 2429.5 kJ/kg = 225.94 + x3 · 2373.1; => x3 = 0.9286 s3 = 0.7594 + 0.9286 · 7.2536 = 7.4903 Sgen = SNET = SSURR = m4s4 + m3s3 – m1s1 = 1.413· 1.5301 + 0.08686· 7.4903 – 1.5 · 1.8418 = +0.05 kW/K > 0

9.65 Redo Problem 9.39 if the water pump has an isentropic efficiency of 85% (hose, C.V.: pump + hose + water column, height difference 35 m. V is velocity. ..2.2 Energy Eq.: m(-wp) + m(h + V /2 + gz)in = m(h + V /2 + gz)ex

Process: hin hex , Vin Vex = 0 , 35 m zex – zin = 35 m , = 1/v 1/vf 10 m

-wp = g(zex – zin) = 9.80665(35 – 0) = 343.2 J/kg gz + 1V2 = gz + 0 noz 2 noz ex

9.66 A flow of 20 kg/s steam at 10 MPa, 550°C enters a two-stage turbine. The exit of the first stage is at 2 MPa where 4 kg/s is taken out for process steam and the rest continues through the second stage, which has an exit at 50 kPa. Assume both stages have an isentropic efficiency of 85% find the total actual turbine work and the entropy generation.

1 T1 2 T2 C.V.: T1 Ideal . State 1: Table B.1.2, h1 = 3500.9 kJ/kg, s1 = 6.7561 W h1 = h2s + wT1,s ; s1 + 0/ = s2s T

9.67 Air flows into an insulated nozzle at 1MPa, 1200 K with 15 m/s and mass flow rate of 2 kg/s. It expands to 650 kPa and exit temperature is 1100 K. Find the exit 22 Energy: hi + (1/2)Vi = he + (1/2)Ve Entropy: si + sgen = se Ideal nozzle sgen = 0 and assume same exit pressure as actual Pe / Pi = Pre / Pri Pre = Pri Pe/ Pi = 191.174 · 650 / 1000 = 124.26 TeÊs = 1078.2 K, heÊs = 1136 kJ/kg 1212 12 VeÊs = Vi + hiÊ – heÊs = · 15 + (1277.8 – 1136) · 1000 22 2 = 112.5 + 141800 = 141913 J/kg VeÊs =533 m/s Actual nozzle with given exit temperature 12 12 VeÊac = Vi + hi – heÊac = 112.5 + (1277.8 – 1161.2) · 1000 = 116712.5 22 VeÊac = 483 m/s 12121212 noz = ( VeÊac – Vi )/ ( VeÊs – Vi ) = (hiÊ – he, AC)/(hi – he, s) 2222 1277.8Ê-Ê1161.2 = = 0.8 1277.8Ê-Ê1136 9.68 A nozzle is required to produce a steady stream of R–134a at 240 m/s at ambient conditions, 100 kPa, 20°C. The isentropic efficiency may be assumed to be 90%. Find by trial and error or verify that the inlet pressure is 375 kPa. What is the rquired inlet temperature in the line upstream of the nozzle?

9.70 A two-stage compressor having an interstage cooler takes in air, 300 K, 100 kPa, and compresses it to 2 MPa, as shown in Fig. P9.70. The cooler then cools the air to 340 K, after which it enters the second stage, which has an exit pressure of 15.74 MPa. The isentropic efficiency of stage one is 90% and the air exits the second stage at 630 K. Both stages are adiabatic, and the cooler dumps Q to reservoir at T0. Find Q in the cooler, the efficiency of the second stage, and the total entropy generated in this process.

9.71 A two-stage turbine receives air at 1160 K, 5.0 MPa. The first stage exit at 1 MPa then enters stage 2, which has an exit pressure of 200 kPa. Each stage has an isentropic efficiency of 85%. Find the specific work in each stage, the overall isentropic efficiency, and the total entropy generation.

T C.V. around each turbine for first the ideal and then 1 P2 the actual produces for stage 1: 2s 2ac P Ideal T1: s2s=s1 => Pr2 = Pr1P2/P1 = 33.297 3 h2s = 789.93; wt1s = h2s – h1 = 441.04 3ss 3s 3ac s

9.72 A paper mill, shown in Fig. P9.72, has two steam generators, one at 4.5 MPa, 300°C and one at 8 MPa, 500°C. Each generator feeds a turbine, both of which have an exhaust pressure of 1.2 MPa and isentropic efficiency of 87%, such that their combined power output is 20 MW. The two exhaust flows are mixed adiabatically to produce saturated vapor at 1.2 MPa. Find the two mass flow rates and the entropy produced in each turbine and in the mixing chamber.

H2O P1=4.5MPa H2O P2=8MPa h1 = 2943.1 T.1=300°C T2=500°C 1 m1 2 . s1 = 6.2828 m2 Power20oMuWtput h2 = 3398.3 T1 T2 to electric s2 = 6.7240 generator h5 = 2784.8 3 P3=1.2MPa 4 P4=1.2MPa s5 = 6.5233 CV T1: wT1,s = h1 – h3s 5 P5=1.2MPa sat. vapor s3s = s1 Insulated Mixer

9.73 A heat-powered portable air compressor consists of three components: (a) an adiabatic compressor; (b) a constant pressure heater (heat supplied from an outside source); and (c) an adiabatic turbine. The compressor and the turbine each have an isentropic efficiency of 85%. Ambient air enters the compressor at 100 kPa, 300 K, and is compressed to 600 kPa. All of the power from the turbine goes into the compressor, and the turbine exhaust is the supply of compressed air. If this pressure is required to be 200 kPa, what must the temperature be at the exit of the heater? Heater 2 q H

C 1 3 T 4 P1 = 100 kPa T1 = 300 K P2 = 600 kPa P4 = 200 kPa sc = 0.85 st = 0.85

P k-1 T2s = T1 2 k = 300(6)0.286 = 500.8 K ?P1? -wsc = CP0(T2s – T1) = 1.004(500.8 – 300) = 201.5 -wc = (-wsc/ sc) = (201.5/0.85) = 237.1 = wT wsT = (wT/ sT) = (237.1/0.85) = 278.9 = CP0(T3 – T4s) k-1 T4s = T3(P4/P3) k = T3(200/600)0.286 = 0.7304 T3 278.9 = 1.004 T3(1 – 0.7304) => T3 = 1030.9 K

C.V.: Compressor s = 0.8 , R-134a inlet: T1 = 5oC, x1 = 0.965 s1 = sf + x1sfg = 1.0243 + 0.965*0.6995 = 1.6993 kJ/kg K, h1 = hf + x1hfg = 206.8 + 0.965*194.6 = 394.6 kJ/kg exit: P2 = 3 MPa For ideal process s2s = s1 = 1.6993 kJ/kg K; T2s = 90oC, h2s = 436.2 kJ/kg 1st Law: qc + h1 = h2 + wc; qc = 0 wcs = h1 – h2s = -41.6 kJ/kg => w = ws/ s = 52.0 kJ/kg h2 = h1 – w = 446.6 kJ/kg C.V.: Tank; VT = 2 m3, PT = 3 MPa QCV = 0, WCV = 0, me = 0 m1=0, m2=mi; u2 = hi = 446.6 kJ/kg Pf = 3 MPa, u2 = 446.6 kJkg ‡ Tf = 110oC, vf = 0.007339m3/kg mT = VT/vf = 272.5 kg; Wc = mTwc = 14197 kJ

9.77 Saturated vapor R-22 enters an insulated compressor with an isentropic efficiency of 75% and the R-22 exits at 3.5 MPa, 120°C. Find the compressor inlet temperature by trial and error.

9.78 Air enters an insulated turbine at 50°C, and exits the turbine at – 30°C, 100 kPa. The isentropic turbine efficiency is 70% and the inlet volumetric flow rate is 20 L/s. What is the turbine inlet pressure and the turbine power output?

C.V.: Turbine, s = 0.7, Insulated Air: Cp = 1.004 kJ/kg-K, R = 0.287 kJ/kg-K, k = 1.4 .

Inlet: Ti = 50oC, V = 20 L/s = 0.02 m3/s i Exit: Te = -30oC, Pe = 100 kPa a) 1st Law SSSF: qT + hi = he + wT; qT = 0 Assume Constant Specific Heat wT = hi – he = Cp(Ti – Te) = 80.3 kJ/kg wTs = w/ = 114.7 kJ/kg, wTs = Cp(Ti – Tes) Solve for Tes = 208.9 K k Isentropic Process: P = P (T / T )k-1 => P = 461 kPa eiei i b) WT = mwT; m = PV/RT = 0.099 kg/s => WT = 7.98 kW

9.83 An air turbine with inlet conditions 1200 K, 1 MPa and exhaust pressure of 100 kPa pulls a sledge over a leveled plane surface, T = 20°C. The turbine work overcomes the friction between the sledge and the surface. Find the total entropy Assume an adiabatic reversible turbine Energy Eq.: wT = hi – he , Entropy Eq.: si = se Exit state: Pe , se = si Table A.7 Pre = PriPe/Pi = 191.174 · 100/1000 = 19.117 Te = 665 K, he = 676, wT = 1277.8 – 676 = 602 kJ/kg The work is dissipated at the surface as frictional heat sgen = wT/Tsurf = 602 / 293.15 = 2.05 kJ/kg K

9.86 A certain industrial process requires a steady stream of saturated vapor water at 200 kPa at a rate of 2 kg/s. There are two alternatives for supplying this steam from ambient liquid water at 20°C, 100 kPa. Assume pump efficiency of 80%. 1. Pump the water to 200 kPa and feed it to a steam generator (heater). 2. Pump the water to 5 MPa, feed it to a steam generator and heat to 450°C, then expand it through a turbine from which the steam exhausts at the desired a. Compare these two alternatives in terms of heat transfer and work terms. 1) PUMP LIQ 123 BOILER H O 200 kPa 2 200 kPa . 20 oC . . m = 2 kg/s -W p Q B P1 = 100 kPa, T1 = 20°C, v1 = 0.001002 -ws » v1(P2-P1) = 0.001002 (200 – 100) = 0.1 kJ -wP = = 0.125 kJ, -WP = m(-w) = 0.25 kW 0.8 qB = h3-h2, h2 = h1- wP = 83.96 + 0.13 = 84.1 qB = 2706.7 – 84.1 = 2622.6, QB = 5245.2 kW 2)

1 LIQ HO 2 o 20 C PUMP 32 5 MPa -W p BOILER 5 MPa w T QBTURB

9.88 A cylinder fitted with a spring-loaded piston serves as the supply of steam for a steam turbine. Initially, the cylinder pressure is 2 MPa and the volume is 1.0 m3. The force exerted by the spring is zero at zero cylinder volume, and the top of the piston is open to the ambient. The cylinder temperature is maintained at a constant 300°C by heat transfer from a source at that temperature. A pressure regulator between the cylinder and turbine maintains a steady 500 kPa, 300°C at the turbine inlet, such that when the cylinder pressure drops to 500 kPa, the process stops. The turbine process is reversible and adiabatic, and the exhaust is to a condenser c) What is the total heat transfer to the cylinder during the process?

T W 2000 500 100 P T = Constant, Po = 100 kPa V When Vo = 0, Fspr = 0 1 C.V. Cylinder: State 1: v1 = 0.12547 m3/kg, u1 = 2772.6 kJ/kg, h1 = 3023.5 kJ/kg, m1 = V1/v1 = 7.97 kg Linear P-V relation: V2 – V0 = (V1 – V0) · (P2 – P0) / (P1 – P0) => V2 = 0.2105 m3 State 2: T2 = 300oC, P2 = 500 kPa, v2 = 0.52256 m /kg, u2 = 2802.9 kJ/kg, 3

9.89 Supercharging of an engine is used to increase the inlet air density so that more fuel can be added, the result of which is an increased power output. Assume that ambient air, 100 kPa and 27°C, enters the supercharger at a rate of 250 L/s. The supercharger (compressor) has an isentropic efficiency of 75%, and uses 20 kW of power input. Assume that the ideal and actual compressor have the same exit pressure. Find the ideal specific work and verify that the exit pressure is 175 kPa. Find the percent increase in air density entering the engine due to the supercharger and the entropy generation.

English Unit Problems 9.91E Steam enters a turbine at 450 lbf/in.2, 900 F, expands in a reversible adiabatic process and exhausts at 2 lbf/in.2. Changes in kinetic and potential energies between the inlet and the exit of the turbine are small. The power output of the turbine is 800 Btu/s. What is the mass flow rate of steam through the turbine? Mass: mi = me = m, Energy Eq.: mhi = mhe + WT, Entropy Eq.: msi + 0/ = mse ( Reversible Sgen = 0 )

H O Inlet state: Table C.8 hi = 1468.3, si = 1.7113 2

9.93E Air at 1 atm, 60 F is compressed to 4 atm, after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle are both reversible and adiabatic and kinetic energy in/out of the compressor can be neglected. Find the compressor work and its exit temperature and find the nozzle exit velocity.

9.95E Two flowstreams of water, one at 100 lbf/in.2, saturated vapor, and the other at 100 lbf/in.2, 1000 F, mix adiabatically in a SSSF process to produce a single flow out at 100 lbf/in.2, 600 F. Find the total entropy generation for this process.

9.97E One technique for operating a steam turbine in part-load power output is to throttle the steam to a lower pressure before it enters the turbine, as shown in Fig. P9.16. The steamline conditions are 200 lbf/in.2, 600 F, and the turbine exhaust pressure is fixed at 1 lbf/in.2. Assuming the expansion inside the turbine to be reversible and adiabatic, determine a. The full-load specific work output of the turbine b. The pressure the steam must be throttled to for 80% of full-load output c. Show both processes in a T–s diagram.

T 1 2a P2b s3a = s1 = 1.6767 = 0.132 66 + x3a · 1.8453 x3a = 0.8367 2b

h3a = 69.74 + 0.8367 · 1036.0 = 936.6 w = h1 – h3a 3a 3b = 1322.1 – 936.6 = 385.5 Btu/lbm s

9.101EAn old abandoned saltmine, 3.5 · 106 ft3 in volume, contains air at 520 R, 14.7 lbf/in.2. The mine is used for energy storage so the local power plant pumps it up to 310 lbf/in.2 using outside air at 520 R, 14.7 lbf/in.2. Assume the pump is ideal and the process is adiabatic. Find the final mass and temperature of the air and the required pump work. Overnight, the air in the mine cools down to 720 R. Find the C.V. = Air in mine + pump (USUF) Cont: m2 – m1 = min Energy: m2u2 – m1u1 =1Q2 – 1W2 + minhin

Entropy: m2s2 – m1s1 = ? dQ/T + 1S2Êgen + minsin Process: 1Q2 = 0, 1S2Êgen = 0, s1 = sin m2s2 = m1s1 + minsin = (m1 + min)s1 = m2s1 s2 = s1 Const. s Pr2 = Pr1P2/P1 = 0.9767 · 310/14.7 = 20.597 T2 = 1221 R , u2 = 213.09 14.7Ê·Ê3.5·106Ê·Ê144 m1 = P1V1/RT1 = = 2.671· 105 lbm 53.34Ê· Ê520 310Ê·Ê3.5·106Ê·Ê144 m2 = P2V2/RT2 = = 2.4· 106 kg 53.34Ê· Ê1221 min = m2 – m1 = 2.1319· 106 lbm W2 = minhin + m1u1 – m2u2 = 2.1319· 106 · 124.38 + 2.671· 10 5 1

9.102EA rigid 35 ft3 tank contains water initially at 250 F, with 50 % liquid and 50% vapor, by volume. A pressure-relief valve on the top of the tank is set to 150 lbf/in.2 (the tank pressure cannot exceed 150 lbf/in.2 – water will be discharged instead). Heat is now transferred to the tank from a 400 F heat source until the tank contains saturated vapor at 150 lbf/in.2. Calculate the heat transfer to the tank C.V. Tank. vf1 = 0.017 vg1= 13.8247 m LIQ =V LIQ / vf1 = 0.5 · 35/0.017 = 1029.4 lbm m VAP=V VAP / vg1 = 0.5 · 35/13.8247 = 1.266 lbm mÊ = 1030. 67 xÊ= m VAP / (m LIQ + m VAP) = 0.001228 uÊ= uf+ xÊ ufg = 218.48 + 0.001228 · 869.41 = 219.55 sÊ = sf+ x Êsfg = 0.3677 + 0.001228 · 1.3324 = 0.36934 state 2: v2Ê = vg= 3.2214 u2Ê = 1110.31 h2Ê = 1193.77 s2Ê = 1.576 m2Ê = V/v2Ê = 10.865 lbm Q = m2Êu2Ê – mÊ1u1 + meh e+ W = 10.865 · 1110.31 – 1030.67· 219.55 + 1019.8· 1193.77 = 1003187 Btu Sgen = m2Ês2Ê – mÊ1s1 – mese – Ê1Q2Ê / Tsource = 10.865 · 1.576 – 1030.67 · 0.36934 + 1019.8 · 1.57 – 1003187/860 = 77.2 Btu/s R 9.103E Liquid water at ambient conditions, 14.7 lbf/in.2, 75 F, enters a pump at the rate of 1 lbm/s. Power input to the pump is 3 Btu/s. Assuming the pump process to be reversible, determine the pump exit pressure and temperature.

9.104EA fireman on a ladder 80 ft above ground should be able to spray water an additional 30 ft up with the hose nozzle of exit diameter 1 in. Assume a water pump on the ground and a reversible flow (hose, nozzle included) and find the gÊ Z 32.2Ê· Ê130 -wp = PE13 = = = 0.167 Btu/lbm gC 32.2Ê· Ê778 32.2Ê· Ê30 V2 Nozzle: KE = – PE23 = = 32.2Ê· Ê778 2Ê· Ê32.2Ê· Ê778 V2 = 2 · 32.2 · 30 = 1932, V = 43.95 ft/s A = ( /4) · (12/144) = 0.00545 ft2 Assume: v = vF,70F = 0.01605 ft3/lbm m = AV/v = 0.00545 · 43.95 / 0.01605 = 14.935 lbm/s Wpump = mwp = 14.935 · 0.167 · (3600/2544) = 3.53 hp 9.105E Saturated R-134a at 10 F is pumped/compressed to a pressure of 150 lbf/in.2 at the rate of 1.0 lbm/s in a reversible adiabatic SSSF process. Calculate the power required and the exit temperature for the two cases of inlet state of the R-134a: b) quality of 0 %.

9.107E Helium gas enters a steady-flow expander at 120 lbf/in.2, 500 F, and exits at 18 lbf/in.2. The mass flow rate is 0.4 lbm/s, and the expansion process can be considered as a reversible polytropic process with exponent, n = 1.3. Calculate the i Pi = 120 lbf/in2 Ti = 500 F . Pe = 18 lbf/in2 m. = 0.4 lbm/s W n-1 0.3 Pe n 18 1.3 Te = Ti = 960 120 e = 619.6 R ?Pi? ? ? nR 1.3Ê· Ê386 w = – nÊ-Ê1 (Te – Ti) = – 0.3Ê· Ê778 (619.6 – 960) = +731.8 Btu/lbm ? vdP = – . . 3600 W = mw = 0.4 · 731.8 · = 414.2 hp 2544 9.108E A mixing chamber receives 10 lbm/min ammonia as saturated liquid at 0 F from one line and ammonia at 100 F, 40 lbf/in.2 from another line through a valve. The chamber also receives 340 Btu/min energy as heat transferred from a 100-F reservoir. This should produce saturated ammonia vapor at 0 F in the exit line. What is the mass flow rate at state 2 and what is the total entropy generation in the process?

1 2 Q CV: Mixing chamber out to reservoir m1 + m2 = m3 m1h1 = m2h2 + Q = m3h3 m1s1 + m2s2 + Q/Tres + Sgen = m3s3 Tres

9.109E A compressor is used to bring saturated water vapor at 150 lbf/in.2 up to 2500 lbf/in.2, where the actual exit temperature is 1200 F. Find the isentropic Inlet: hi = 1194.9 si = 1.5704 IDEAL EXIT: Pe, se,s = si he,s = 1523.8 ws = hi – he,s = 1194.9 – 1523.8 = -328.9 Btu/lbm ACTUAL EXIT: he,AC = 1587.7, se,AC = 1.6101 wAC = hi – he,ac = 1194.9 – 1587.7 = -392.8 Btu/lbm c = ws/wAC = 328.9/392.8 = 0.837 sGEN = se,AC – si = 0.0397 Btu/lbm R

9.112E Air at 1 atm, 60 F is compressed to 4 atm, after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle both have efficiency of 90% and kinetic energy in/out of the compressor can be neglected. Find the actual compressor work and its exit temperature and find the actual nozzle exit velocity.

9.113E A geothermal supply of hot water at 80 lbf/in.2, 300 F is fed to an insulated flash evaporator at the rate of 10,000 lbm/h. A stream of saturated liquid at 30 lbf/in.2 is drained from the bottom of the chamber and a stream of saturated vapor at 30 lbf/in.2 is drawn from the top and fed to a turbine. The turbine has an isentropic efficiency of 70% and an exit pressure of 2 lbf/in.2. Evaluate the second law for a control volume that includes the flash evaporator and the turbine.

9.115E A nozzle is required to produce a steady stream of R-134a at 790 ft/s at ambient conditions, 14.7 lbf/in.2, 70 F. The isentropic efficiency may be assumed to be 90%. What pressure and temperature are required in the line upstream of the nozzle? T KE2 = 7902/2 · 32.174 · 778 = 12.466 Btu/lbm KE2s = KE2/ = 13.852 1 h1 = h2 + KE2 = 180.981 + 12.466 2s = 193.447 Btu/lbm 2 h1 h2s = h1 – KE2s = 193.447 – 13.852 s1 s = 179.595 Btu/lbm 2s: P2, h2s T2s = 63.12, s2s = 0.4485 1: h1, s1 = s2s T1 = 137.25 F P1 = 55 lbf/in2 9.116E A two-stage turbine receives air at 2100 R, 750 lbf/in.2. The first stage exit at 150 lbf in.2 then enters stage 2, which has an exit pressure of 30 lbf/in.2. Each stage has an isentropic efficiency of 85%. Find the specific work in each stage, the overall isentropic efficiency, and the total entropy generation.

9.117E A watercooled air compressor takes air in at 70 F, 14 lbf/in.2 and compresses it to 80 lbf/in.2. The isothermal efficiency is 80% and the actual compressor has the same heat transfer as the ideal one. Find the specific compressor work and the exit temperature.

9.119E A paper mill has two steam generators, one at 600 lbf/in.2, 550 F and one at 1250 lbf/in.2, 900 F. The setup is shown in Fig. P9.72. Each generator feeds a turbine, both of which have an exhaust pressure of 160 lbf/in.2 and isentropic efficiency of 87%, such that their combined power output is 20000 Btu/s. The two exhaust flows are mixed adiabatically to produce saturated vapor at 160 lbf/in.2. Find the two mass flow rates and the entropy produced in each turbine and in the mixing chamber.

H O P1=600 psi H O P2=1250 psi Inlet States: 22 1 T.1=550 F 2 T2=900 F h1 = 1255.4 .

m1 m2 20000 Btu/s s1 = 1.499 Power output h2 = 1438.4 T1 T2 to electric generator s2 = 1.582 3 P3=160 psi 4 P4=160 psi Ideal & Actual T1: s3s = s1 5 x3s = 0.9367 P5=160 psi sat. vapor h3s = 1141.55 Insulated Mixer