# Fundamentals of Thermodynamics Solutions 01-09

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.