Solution Homework set 2-P SP-19-1

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Homework Set 2-P (Chapter 3 and 4)

** Problem 03.021 - Steam tables 1 Refer Table A-5: T, °C P, kPa u, kJ/kg Phase description 143.61 400 1450 Saturated mixture 220 2319.6 2601.3 Saturated vapor 190 2500 805.15 Compressed liquid 466.21 4000 3040 Superheated vapor **Problem 03.022 - Steam tables 2 Refer Table A-4/5: T, °C P, kPa v, m3/kg Phase description 140 361.53 0.035 Saturated mixture 155.46 550 0.001097 Saturated liquid 125 750 0.001065 Compressed liquid 300 1803 0.140 Superheated vapor **Problem 03.023E - Steam tables 3 Refer Table A-4E/5E/6E/7E: T, °F P, psia u, Btu/lbm Phase description 300 67.03 782 Saturated mixture 267.22 40 236.02 Saturated liquid 500 120 1174.4 Superheated vapor 400 400 373.84 Compressed liquid **Problem 03.025 - Steam tables for water Refer Table A-4/5: T, °C P, kPa h, kJ/kg x Phase description 120.21 200 2045.8 0.7 Saturated mixture 140 361.53 1800 0.565 Saturated mixture 177.66 950 752.74 0.0 Saturated liquid 80 500 335.37 No determined value Compressed liquid 350.0 800 3162.2 No determined value Superheated vapor **Problem 03.026E - Steam tables for refrigerant 134a Refer Table Refer Table A-4E/5E/6E/7E: T, °F P, psia h, Btu/lbm x Phase description 65.89 80 78 0.566 Saturated mixture 15 29.759 69.92 0.6 Saturated mixture 10 70 15.36 No determined value Compressed liquid 160 180 129.46 No determined value Superheated vapor 110 161.16 117.25 1.0 Saturated vapor **Problem 03.027 - Steam tables for refrigerant 134a 2 Refer Table A-11, 12, 13: T, °C P, kPa u, kJ/kg Phase description 20 572.07 95 Saturated mixture −12 185.37 35.76 Saturated liquid 86.25 400 300 Superheated vapor 8 600 62.37 Compressed liquid

** Problem 03.029E - Total internal energy and enthalpy of water in container 0.96 lbm of water fills a container whose volume is 1.96 ft3. The pressure in the container is 100 psia. Calculate the total internal energy and enthalpy in the container. Use data from the steam tables. The specific volume is calculated as follows: v = V/m = 1.96 ft 3 /0.96 lbm = 2.0417 ft 3 /lbm At this specific volume and given pressure, the state is saturated mixture. The quality, internal energy, and enthalpy at this state are (Table A-5E) as follows: v f = 0.01774 ft 3 /lbm, v g = 4.4327 ft 3 /lbm u f = 298.19 Btu/lbm, u fg = 807.29 Btu/lbm h f = 298.51 Btu/lbm, and h fg = 888.99 Btu/lbm V avg = v f + x. v fg x = (v – v f )/v fg = (v – v f )/(v g – v f ) = (2.0417 ft 3 /lbm - 0.01774 ft 3 /lbm)/(4.4327 ft 3 /lbm - 0.01774 ft 3/ lbm) = 0.4584 u = u f + xu fg = 298.19 Btu/lbm + (0.4584 × 807.29 Btu/lbm) = 668.2717 Btu/lbm h = h f + xh fg = 298.51 Btu/lbm + (0.4584 × 888.99 Btu/lbm) = 706.045 Btu/lbm The total internal energy and enthalpy are calculated as follows: U = mu = 0.96 lbm × 668.2717 Btu/lbm = 641.5408 Btu H = mh = 0.96 lbm × 706.045 Btu/lbm = 677.8032 Btu

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Problem 03.031 - Heater refrigerant container 10 kg of R-134a fill a 1.115-m 3 rigid container at an initial temperature of – 30°C. The container is then heated until the pressure is 200 kPa. Determine the final temperature and the initial pressure. Use data from the steam tables. This is a constant-volume process. The specific volume is calculated as follows: v 1 = v 2 = V/m = 1.115 m 3 /10 kg = 0.1115 m 3 /kg The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature. From Table A-11, P 1 = P sat @ −30 °C = 84.43 kPa The final state is superheated vapor, and the temperature is determined by interpolation as follows: From Table A-13, P 2 = 200 kPa v 2 = 0.1115 m 3 /kg } T 2 = 14.2°C **Problem 03.033 - Accuracy of specific volume What is the specific volume of water at 5 MPa and 90°C? What would it be if the incompressible liquid approximation were used? Determine the accuracy of this approximation. Use data from the steam tables. The state of water is compressed liquid. From the steam tables, Table A-7, P = 5 Mpa T = 90°C } v = 0.0010339 m 3 /kg Based upon the incompressible liquid approximation, From Table A-4, P = 5 Mpa T = 90°C } v ≅ v f @ 90°C = 0.001036 m 3 /kg The error involved is calculated as follows: Percent Error = (0.001036 m 3 /kg - 0.0010339 m 3 /kg)/0.0010339 m 3 /kg × 100 = 0.2031% which is quite acceptable in most engineering calculations.

Problem 03.037 - Water vapor in left chamber of an evacuated container 0.98 kg of water vapor at 200 kPa fills the 1.1989-m3 left chamber of a partitioned system shown in the figure. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 3°C. Use data from the steam tables. The initial specific volume is calculated as follows: v 1 = V 1 /m = 1.1989 m 3 /0.98 kg = 1.2234 m 3 /kg At the final state, the water occupies three times the initial volume. Then, v 2 = 3v 1 = 3 × 1.2234 m 3 /kg = 3.6701 m 3 /kg Based on this specific volume and the final temperature, the final state is a saturated mixture and the pressure from Table A-4 is determined as follows: P 2 = P sat @ 3°C = 0.768 kPa Problem 03.038E - Pressure cooker absolute pressure The temperature in a pressure cooker during cooking at sea level is measured to be 225°F. Determine the absolute pressure inside the cooker in psia and in atm. Would you modify your answer if the place were at a higher elevation? The saturation pressure of water at 225°F is 18.998 psia. The standard atmospheric pressure at sea level is 1 atm = 14.7 psia. The assumption made here is that the properties of pure water can be used to approximate the properties of juicy water in the cooker. The saturation pressure of water at 225°F is 18.998 psia (Table A-4E). The standard atmospheric pressure at sea level is 1 atm = 14.7 psia. The absolute pressure in the cooker is simply the saturation pressure at the cooking temperature: P abs = P sat @ 225°F = 18.998 psia It is equivalent to P abs = 18.998 psia ×/(14.7 psia) = 1.2924 atm The elevation has no effect on the absolute pressure inside when the temperature is maintained constant at 225°F.

**Problem 03.043 - Refrigerant in piston-cylinder device 100 kg of R-134a at 200 kPa are contained in a piston – cylinder device whose volume is 13.641 m 3 . The piston is now moved until the volume is one-half its original size. This is done such that the pressure of the R-134a does not change. Determine the final temperature and the change in the total internal energy of R-134a. Use data from the steam tables. The initial specific volume is calculated as follows: v 1 = V/m = 13.641 m 3 /100 kg = 0.13641 m 3 /kg The initial state is superheated and the internal energy at this state is determined from the steam tables, Table A-13, as follows: P 1 = 200 kPa v 1 = 0.13641 m3/kg } u 1 = 287.75 kJ/kg The final specific volume is calculated as follows: v 2 = v 1 /2 = 0.13641 m 3 /kg/2 = 0.06821 m 3 /kg This is a constant-pressure process. The final state is determined to be a saturated mixture whose temperature is determined from the steam tables, Table A-12, as follows: T 2 = T sat @ 200kPa = −10.09 °C v f = 0.0007532 m 3 /kg, v fg = vg − vf = (0.099951 − 0.0007532) m 3 /kg u f = 38.26 kJ/kg, and u fg = 186.25 kJ/kg The internal energy at the final state is calculated as follows: x 2 = (v 2 – v f )/v fg = (0.06821 m3/kg - 0.0007532 m 3 /kg/(0.099951 m 3 /kg - 0.0007532 m 3 /kg) = 0.68 u 2 = u f + x 2 u fg = 38.26 kJ/kg + (0.68 × 186.25 kJ/kg) = 164.9049 kJ/kg Hence, the change in the internal energy is calculated as follows: Δu = u 2 - u 1 = 164.9049 kJ/kg - 287.75 kJ/kg = -122.8451 kJ/kg

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**Problem 03.060 - Temperature at which the water boils Water is being heated in a vertical piston – cylinder device. The piston has a mass of 40 kg and a cross-sectional area of 150 cm 2 . If the local atmospheric pressure is 100 kPa, determine the temperature at which the water starts boiling. Use data from the steam tables. The pressure in the cylinder is determined from a force balance on the piston, PA=P atm A + W or, P = P atm + mg/A = 100 kPa + (40 kg × 9.81 m/s 2 )/0.0150 m 2 × (1 kPa/1000 kg/m ⋅ s 2 ) = 126.16 kPa The boiling temperature is the saturation temperature to this pressure. From Table A-5, T = T sat @ 126.16 kPa = 106.2196°C

**Problem 03.068E - Specific volume of oxygen What is the specific volume of oxygen at 26 psia and 80°F? The gas constant of oxygen is R = 0.3353 psia·ft 3 /lbm·R. The assumption made here is that at specified conditions, oxygen behaves as an ideal gas. Given: The gas constant of oxygen is R = 0.3353 psia·ft 3 /lbm.R According to the ideal-gas equation of state, v = RT/P = 0.3353 psia ⋅ ft 3 /lbm ⋅ R × (80 + 460) R/26 psia = 6.9639 ft 3 /lbm Problem 03.069 - Air pressure A 100-L container is filled with 1 kg of air at a temperature of 25.9°C. What is the pressure in the container? The gas constant of air is R = 0.287 kJ/kg·K. Use data from the steam tables. The assumption made here is that at specified conditions, air behaves as an ideal gas. Given: The gas constant of air is R = 0.287 kJ/kg·K. The definition of the specific volume gives v = V/m = 0.100 m 3 /1 kg = 0.100 m 3 /kg Using the ideal-gas equation of state, the pressure is calculated as follows: P = RT/v = 0.287 kPa ⋅ m 3 /kg ⋅ K × (25.9 + 273 )K/0.100 m 3 /kg = 857.843 kPa **Problem 03.071 - Gage pressure of air A 400-L rigid tank contains 5 kg of air at 25°C. Determine the reading on the pressure gage if the atmospheric pressure is 95.5 kPa. The gas constant of air is R = 0.287 kPa·m 3 /kg·K. The assumption made here is that at specified conditions, air behaves as an ideal gas. Given: The gas constant of air is R = 0.287 kPa·m3/kg·K. Treating air as an ideal gas, the absolute pressure in the tank is determined as follows: P = mRT/V = 5 kg × 0.287 kPa ⋅ m 3 /kg ⋅ K × (25 + 273)K/0.4 m 3 = 1069.075 kPa Thus, the gage pressure is calculated as follows: P g = P − P atm = 1069.1 kPa − 95.5 kPa = 973.575 kPa

** Problem 03.072 - Amount of oxygen The pressure gage on a 2.5-m 3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C and the atmospheric pressure is 97 kPa. The gas constant of air is R = 0.2598 kPa·m 3 /kg·K. The absolute pressure of O 2 is determined as follows: P = P g + P atm = 500 kPa + 97 kPa = 597 kPa Treating O 2 as an ideal gas, the mass of O 2 in the tank is determined as follows: m = PV/RT = 597 kPa × 2.5 m 3 /0.2598 kPa ⋅ m 3 /kg ⋅ K × (28 + 273) K = 19.0857 kg Problem 03.073 - Balloon filled with helium A spherical balloon with a diameter of 9 m is filled with helium at 25.5°C and 200 kPa. Determine the mole number and the mass of the helium in the balloon. The universal gas constant is R u = 8.314 kPa·m 3 /kmol·K. The molar mass of helium is 4.0 kg/kmol. The volume of the sphere is calculated as follows: V = 4 πr 3 /3 = 4 π × 4.5 m 3 /3 = 381.7 m 3 Assuming ideal gas behavior, the mole numbers of He is determined as follows: N = PV/R u T = (200 kPa × 381.7 m 3 )/8.314 kPa ⋅ m 3 /kmol ⋅ K × (25.5 + 273) K= 30.7608 kmol Then the mass of He can be determined as follows: m = NM = 30.7608 kmol × 4.0 kg/kmol = 123.0432 kg

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**Problem 03.075 - Tanks connected through a valve A 1-m 3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m 3 /kg·K. The assumption made here is that at specified conditions, air behaves as an ideal gas. Given: The gas constant of air is R = 0.287 kPa·m 3 /kg·K (Table A-1). Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined as follows: V B = (m 1 RT 1 /P 1 ) B = 3 kg × 0.287 kPa ⋅ m 3 /kg ⋅ K × 308 K 1 /150 kPa = 1.768 m 3 m A = (P 1 V/RT 1 ) A = 350 kPa × 1.0 m 3 /(0.287 kPa ⋅ m 3 /kg ⋅ K × 283 K) = 4.309 kg Thus, V = V A + V B = 1.0 m 3 + 1.768 m 3 = 2.768 m 3 m = m A + m B = 4.309 kg + 3 kg = 7.309 kg Then the final equilibrium pressure is calculated as follows: P 2 = mRT 2 /V = 7.309 kg × 0.287 kPa ⋅ m 3 /kg ⋅ K × (19 + 273)K/2.768 m 3 = 221.2874 kPa

Problem 03.077 - Temperature change of helium in rigid vessel A mass of 0.1 kg of helium fills a 0.2 m 3 rigid vessel at 350 kPa. The vessel is heated until the pressure is 650 kPa. Calculate the temperature change of helium (in K) as a result of this heating. The gas constant of helium is R = 2.0769 kPa·m 3 /kg·K. The assumption made here is that at specified conditions, helium behaves as an ideal gas. Given: The gas constant of helium is R = 2.0769 kPa·m 3 /kg·K. (Table A-1). According to the ideal-gas equation of state, the initial temperature is calculated as follows: T 1 = P 1 V/mR = 350 kPa × 0.2 m 3 /(0.1 kg × 2.0769 kPa ⋅ m 3 /kg ⋅ K ) = 337 K Since the specific volume remains constant, the ideal-gas equation gives v 1 = RT 1 /P 1 = v 2 = RT 2 /P 2 T 2 = T 1 P 2 /P 1 = 337 K × 650 kPa/350 kPa = 625.8571 K The temperature change is calculated as follows: Δ T = T 2 − T 1 = 625.8571 K − 337 K = 288.8571 K

** Problem 03.088 - Heated ethylene under constant pressure Ethylene is heated at constant pressure of 5 MPa from 20°C to 200°C. Using the compressibility chart, determine the change in ethylene’s specific volume as a result of this heating. The gas constant, the critical pressure, and the critical temperature of ethane are R = 0.2964 kPa·m 3 /kg·K, T cr = 282.4 K, and P cr = 5.12 MPa. Use data from the steam tables. Given: The gas constant, the critical pressure, and the critical temperature of ethane, from Table A-1, are R = 0.2964 kPa·m 3 /kg·K, T cr = 282.2 K, and P cr = 5.12 MPa. P R1 = P 1 /P cr = 5 Mpa/5.12 MPa = 0.977} T R1 = T 1 /T cr = 293 K/282.2 K = 1.038 } Z 1 = 0.56 P R2 = P R1 = 0.977 T R2 = T 2 /T cr = 473 K/282.4 K = 1.675 } Z 2 = 0.961 The specific volume change is calculated as follows: Δ v = R/P(Z 2 T 2 − Z 1 T 1 ) = 0.2964 kPa ⋅ m 3 /kg ⋅ K/5000 kPa × ((0.961 × 473 K) − (0.56 × 293 K)) = 0.0172 m 3 /kg

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Problem 03.093 - Error without using specific volume values What is the percentage of error involved in treating carbon dioxide at 5 MPa and 25°C as an ideal gas? The critical pressure and the critical temperature of CO 2 are T cr = 304.2 K and P cr = 7.39 MPa. Use data from the steam tables. The critical pressure and the critical temperature of CO 2 , from Table A- 1, are T cr = 304.2 K and P cr = 7.39 MPa. From the compressibility chart (Fig. A-15), P R = P/P cr = 5 Mpa/7.39 MPa = 0.677 T R = T/T cr = 298 K/304.2 K = 0.98 } Z = 0.69 Then, the error involved in treating CO 2 as an ideal gas is calculated as follows: Percentage error = (v − vactual)/v = 1 − 1/Z = 1 − 1/0.69 = −0.449 = 44.9% Problem 03.117 - Heated refrigerant in rigid container One kilogram of R-134a fills a 0.090-m 3 rigid container at an initial temperature of -36°C. The container is then heated until the pressure is 280 kPa. Determine the initial pressure and final temperature. Use data from the steam tables. From Table A-11, T 1 = -36 ° C v 1 = 0.090 m3/kg } P 1 = P sat @ -36 ° C = 62.95 kPa This is a constant-volume cooling process ( v = V / m = constant). The final state is superheated vapor, and the final temperature is determined as follows: From Table A-13, P 2 = 280 k v 2 = v 1 = 0.090 m3/kg } T 2 = 50 ° C

Problem 03.118 - Cooled refrigerant in rigid container A rigid tank with a volume of 0.12355 m 3 contains 1 kg of refrigerant-134a vapor at 240 kPa. The refrigerant is now allowed to cool. Determine the pressure when the refrigerant first starts condensing. Use data from the steam tables. This is a constant-volume process ( v = V / m = constant), and the specific volume is determined as follows: v = V/m = 0.12355 m 3 /1 kg = 0.12355 m 3 /kg When the refrigerant starts condensing, the tank will contain saturated vapor only. Thus, v 2 = v g = 0.12355 m 3 /kg The pressure at this point is the pressure that corresponds to this v g value: P 2 = Psat @ v g = 0.12355 m 3 /kg = 160 kPa **Problem 03.122 - Vapor and liquid phases A 4-L rigid tank contains 1.95 kg of saturated liquid – vapor mixture of water at 50°C. The water is now slowly heated until it exists in a single phase. At the final state, will the water be in the liquid phase or the vapor phase? What would your answer be if the volume of the tank were 400 L instead of 4 L? Use data from the steam tables. v 2 = v 1 The critical specific volume of water is 0.003106 m 3 /kg. Thus, if the final specific volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor. V = 4L → v = V/m = 0.004 m 3 /1.95 kg = 0.0021 m 3 /kg Since, 0.0021 m 3 /kg < v cr , it is in liquid phase. V = 400L → v = V/m = 0.4 m 3 /1.95 kg = 0.2051 m3/kg Since, 0.2051 m 3 /kg > v cr , it is in vapor phase.

**Problem 03.123 - Volume change for steam in piston-cylinder device A piston – cylinder device initially contains 0.22 kg of steam at 200 kPa and 300°C. Now, the steam is cooled at constant pressure until it is at 150°C. Determine the volume change of the cylinder during this process using the compressibility factor, and compare the result to the actual value. The gas constant, the critical pressure, and the critical temperature of steam are R = 0.4615kPa·m 3 /kg·K, T cr = 647.1 K, and P cr = 22.06 MPa. Use data from the tables. Given: The gas constant, the critical pressure, and the critical temperature of steam are R = 0.4615kPa·m 3 /kg·K, T cr = 647.1 K, and P cr = 22.06 MPa (Table A-1). The exact solution is given by the following: From Table A-6, P = 200 kPa T 1 = 300 ° C } v 1 = 1.31623 m 3 /kg P = 200 kPa T 2 = 150 ° C } v 2 = 0.95986 m 3 /kg Δ V exact = m(v 1 − v 2 ) = 0.22 kg × (1.31623 m 3 /kg − 0.95986 m 3 /kg) = 0.0784 m 3 Using compressibility chart (an appropriate software function for compressibility factor is used ), P R = P 1 / P cr = 0.2 Mpa/22.06 MPa = 0.0091 T R,1 = T 1 / T cr = ( 300 + 273 K)/647.1 K = 0.886 } Z 1 = 0.9956 P R = P 2 / P cr = 0.2 Mpa/22.06 MPa = 0.0091 T R,2 = T 2 / T cr = ( 150 + 273 K)/647.1 K = 0.65 } Z 2 = 0.9897 V 1 = Z 1 mRT 1 /P 1 = 0.9956 × 0.22 kg × 0.4615 kPa·m 3 /kg·K × 300 + 273) K/200 kPa = 0.2896 m3 V 2 = Z 2 mRT 2 /P 2 = 0.9897 × 0.22 kg × 0.4615 kPa·m 3 /kg·K × (150 + 273) K/200 kPa = 0.21252 m3 Δ V chart = V 1 − V 2 = 0.2896 m 3 − 0.21252 m 3 = 0.07708 m 3 The percentage error is calculated as follows: Percentage error = ( 0.07708 m 3 − 0.0784 m 3 )/ 0.0784 m 3 × 100 = -1.6862%

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Problem 03.133.a - Hot air balloon burning propane Consider an 18-m-diameter hot-air balloon that, together with its cage, has a mass of 116 kg when empty. The air in the balloon, which is now carrying two 85-kg people, is heated by propane burners at a location where the atmospheric pressure and temperature are 93 kPa and 12°C, respectively. Determine the average temperature of the air in the balloon when the balloon first starts rising. The gas constant of air is R = 0.287 kPa·m 3 /kg·K. Use data from the steam tables. The assumption made here is that air is an ideal gas. Given: The gas constant of air is R = 0.287 kPa·m 3 /kg·K. The buoyancy force acting on the balloon is calculated as follows: V balloon = 4 πr 3 / 3 = 4 π (9 m) 3 / 3 = 3053.6281 m 3 ρ cool air = P/RT = 93 kPa/(0.287 kPa ⋅ m 3 /kg ⋅ K × 285 K) = 1.137 kg/m 3 F B = ρ cool air gV balloon = 1.137 kg/m 3 × 9.81 m/s 2 × 3053.6281 m 3 × ( 1 N/1 kg ⋅ m/s 2 ) = 34060.0762 N The vertical force balance on the balloon gives F B = W hotair + W cage + W people = (m hotair + m cage + m people )g Substituting, 34060.0762 N = (m hotair + 116 kg + 170 kg) × 9.81 m/s 2 × ( 1 N/1 kg ⋅ m/s 2 ) which gives m hotair = 3185.9751 kg Therefore, the average temperature of the air in the balloon is T = PV/mR = 93 kPa × 3053.6281 m 3 /( 3185.9751 kg × 0.287 kPa ⋅ m 3 /kg ⋅ K) = 310.5809 K

**Problem 04.006 - Work for an isothermal process, nitrogen Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m 3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process. The following assumptions have been made here: 1. The process is quasi-equilibrium. 2. Nitrogen is an ideal gas. Wb, out=∫ 2 1 PdV=P 1 V 1 lnV 2 /V 1 =P 1 V 1 lnP 1 /P 2 Wb, out=(150 kPa)(0.2 m 3 )(ln150 kPa/850 kPa)(1 kJ/1 kPa ⋅ m 3 )=-50.2192 kJ ** Problem 04.009 - Work for an isobaric process, water A mass of 2 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate the work done by the steam during this process. Use steam tables. The assumption made here is that the process is quasi-equilibrium. Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4 through A-6) P 1 =100 kPa Sat. Vapor } v 1 =vg @100 kPa =1.6941 m 3 /kg P 2 =100 kPa T 2 =200 ° C } v 2 =2.1724 m 3 /kg The boundary work is determined from its definition to be W b, out = ∫ 2 1 Pdν = P(v 2 − v 1 ) = mP(v 2 − v 1 ) W b, out = (2 kg)(100 kPa)(2.1724−1.6941) m 3 /kg(1 kJ/1 kPa ⋅ m 3 )=95.66 kJ

Problem 04.010E - Work for water, piston-cylinder device A frictionless piston – cylinder device contains 15 lbm of superheated water vapor at 40 psia and 600°F. Steam is now cooled at constant pressure until 70 percent of it, by mass, condenses. Determine the work done during this process. Use steam tables. The assumption made here is that the process is quasi-equilibrium . Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4E through A-6E) P 1 =40 Psia T 1 =600 ° F } v 1 =15.686 ft 3 /lbm P 2 =40 psia x 2 =0.3 } v 2 =v 1 +x 2 v fg =0.01715+0.3(10.501−0.01715)=3.162 ft 3 /lbm The boundary work is determined from its definition to be W b, out = ∫ 2 1 Pdν =P(v 2 − v 1 ) = mP(v 2 − v 1 ) W b, out =(15 lbm)(40 psia)(3.1623−15.686) ft 3 /lbm(1 Btu/5.4093 psia ⋅ ft 3 )=-1390.5179 Btu ** Problem 04.012 - Polytropic work, argon Argon is compressed in a polytropic process with n = 1.2 from 150 kPa and 30°C to 1200 kPa in a piston – cylinder device. Determine the final temperature of argon. The assumption made here is that the process is quasi-equilibrium. For a polytropic expansion or compression process, PV n =Constant For an ideal gas, PV=RT Combining these equations, we get T 2 =T 1 (P 2 /P 1 ) (n-1)/n =303 K(1200 kPa/150 kPa) (1.2-1)/1.2 =428.51 K

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**Problem 04.017 - Polytropic work, nitrogen A frictionless piston – cylinder device contains 5 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly according to the relation PV 1.4 = constant until it reaches a final temperature of 450 K. Calculate the work input during this process. The gas constant for nitrogen is R = 0.2968 kJ/kg·K. The following assumptions have been made here: 1. The process is quasi-equilibrium. 2. Nitrogen is an ideal gas. The boundary work for this polytropic process can be determined from W b = ∫ 2 1 PdV = (P 1 V 1 − P 2 V 2 )/(1− n) = mR(T 1 − T 2 )/(1− n ) W b = (5 kg)(0.2968 kJ/kg ⋅ K )(250−450) K/(1−1.4) = 742 kJ ** Problem 04.028 - Energy balance for a rigid container A rigid container equipped with a stirring device contains 4 kg of motor oil. Determine the rate of specific energy increase when heat is transferred to the oil at a rate of 1 W, and 1.5 W of power is applied to the stirring device. This is a closed system since no mass enters or leaves. The energy balance for closed system can be expressed as E in − E out = D E system Net energy transfer by heat, work, and mass Change in internal, kinetic, potential etc. energies Q˙ in + W˙ sh, in = ΔE˙ Then, ΔE˙=Q˙ in + W˙ sh, in = 1 W + 1.5 W=2.5 W Dividing this by the mass in the system gives Δe˙=ΔE˙/m=2.5 J/s/4 kg = 0.625 J/ kg ⋅ s

**Problem 04.030 - Energy balance for an insulated rigid container with a stirring device A substance is contained in a well-insulated rigid container that is equipped with a stirring device, as shown in the figure. Determine the change in the internal energy of this substance when 15 kJ of work is applied to the stirring device. The following assumptions have been made here: 1. The tank is stationary, and thus the kinetic and potential energy changes are zero. 2. The tank is insulated, and thus heat transfer is negligible. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out = D e system Net energy transfer by heat, work, and mass = Change in internal, kinetic, potential etc. energies W sh, in = ΔU (Since KE=PE=0) Then, ΔU=15 kJ

**Problem 04.033 - Energy balance for water in a rigid tank A rigid 11-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is then heated until its temperature is 180°C. Calculate the heat transfer required for this process. Use data from the steam tables. The following assumptions have been made here: 1. The system is stationary, and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved. 3. The thermal energy stored in the vessel itself is negligible. We take water as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out = D e system Net energy transfer by heat, work, and mass = Change in internal, kinetic, potential etc. energies Q in = Δ U = m(u 2 −u 1 ) (since KE=PE=0) The properties at the initial and final states are (Tables A-4 and A-6) T 1 = 100°C x 1 =0.123 } v 1 = v f + x v fg =0.001043 m 3 /kg+(0.123)(0.16720−0.001043) m 3 /kg = 0.2066 m 3 /kg, u 1 = u f + x u fg = 419.06 kJ/kg + (0.123)(2087 kJ/kg)=675.76 kJ/kg T 2 =180°C v 2 = v 1 = 0.2066 m 3 /kg } u 2 = 2586.3 kJ/kg The mass in the system is m = V 1 /v 1 = 0.011 m 3 /0.2066 m 3 /kg = 0.053242982 kg Substituting, Q in = m(u 2 −u 1 ) = (0.053242982 kg)(2586.3−675.76) kJ/kg = 101.7228461 kJ

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**Problem 04.035 - Energy balance for R134a in a piston-cylinder device A piston – cylinder device contains 5.5 kg of refrigerant-134a at 800 kPa and 70°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 15°C. Determine the amount of heat loss. The following assumptions have been made here: 1. The cylinder is stationary, and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. E in − E out = D e system Net energy transfer by heat, work, and mass = Change in internal, kinetic, potential etc. energies − Q out − W b,out = Δ U = m(u 2 −u 1 ) (since KE=PE=0) − Q out = m(h 2 −h 1 ) since U + Wb = Δ H during a constant pressure quasi-equilibrium process. The properties of R-134a are (Tables A-11 through A-13) P 1 = 800 kPa T 1 = 70°C } h 1 =306.90 kJ/kg P 2 = 800 kPa T 2 =1 5°C } h 2 ≅ hf @15°C =72.33 kJ/kg Substituting, Q out = −(5.5 kg)(72.33−306.90) kJ/kg = 1290.14 kJ

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**Problem 04.051E - Change of internal energy for air What is the change in the internal energy, in Btu/lbm, of air as its temperature changes from 100 to 200°F? Is there any difference if the temperature were to change from 0 to 100°F?The constant-volume specific heat of air at room temperature is c v = 0.171 Btu/lbm·R. Consider the variation of specific heat with temperature. c v = 0.1725 Btu/lbm·R at 150°F and c v = 0.1712 Btu/lbm·R at 50°F. The assumption made here is that at specified conditions, air behaves as an ideal gas. Using the specific heat at constant volume, Δ u = c v Δ T = (0.171 Btu/lbm ⋅ R )(200−100)R = 17.12 Btu/lbm If we use the same room temperature specific heat value, the internal energy change will be the same for the second case. However, if we consider the variation of specific heat with temperature, Δ u 1 = c v Δ T 1 = (0.1725 Btu/lbm ⋅ R )(200−100)R = 17.25 Btu/lbm Δ u 2 = c v Δ T 2 = (0.1712 Btu/lbm ⋅ R )(100−0)R=17.12 Btu/lbm The two results differ from each other by about 0.759%.

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Problem 04.053 - Change of enthalpy for oxygen What is the change in the enthalpy, in kJ/kg, of oxygen as its temperature changes from 150 to 250°C? The constant-pressure specific heat of oxygen at room temperature is c p = 0.918 kJ/kg·K. Is there any difference if the temperature were to change from 0 to 100°C? Consider the variation of specific heat with temperature. c p = 0.964 kJ/kg·K at 200°C and c p = 0.934 kJ/kg·K at 100°C. The assumption made here is that at specified conditions, oxygen behaves as an ideal gas. Using the specific heat at constant pressure, Δh = cpΔT =(0.918 kJ/kg ⋅ K )(250−150) K=91.8 kJ/kg If we use the same room temperature specific heat value, the internal energy change will be the same for the second case. However, if we consider the variation of specific heat with temperature, Δh 1 = c p ΔT 1 = (0.964 kJ/kg ⋅ K )(250−150) K = 96.4 kJ/kg Δh 2 = c p ΔT 2 =(0.934 kJ/kg ⋅ K )(100−0) K=93.4 kJ/kg The two results differ from each other by about 3.21%.

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**Problem 04.064 - Energy balance for a room A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10,000 kJ/h, and a 100-W fan is used to distribute the warm air in the room. The rate of heat loss from the room is estimated to be about 5000 kJ/h. If the initial temperature of the room air is 10°C, determine how long it will take for the air temperature to rise to 21°C. Assume constant specific heats at room temperature. The gas constant of air is R = 0.287 kPa·m 3 /kg·K (Table A-1). Also, c v = 0.718 kJ/kg·K for air at room temperature (Table A-2). The following assumptions are made here: 1. Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of – 141ºC and 3.77 MPa. 2. The kinetic and potential energy changes are negligible, Δke ≅ Δpe ≅ 0. 3. Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 4. The local atmospheric pressure is 100 kPa. 5. The room is airtight so that no air leaks in and out during the process. We take the air in the room to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary constant-volume closed system can be expressed as E in − E out = D e system Net energy transfer by heat, work, and mass = Change in internal, kinetic, potential etc. energies Q in + W fan, in − Q out = Δ U ≅ mcv, avg(T 2 −T 1 ) (Since KE=PE=0) or, (Q˙ in + W˙ fan, in −Q˙ out ) Δ t=m cv, avg (T 2 −T 1 ) The mass of air is: V=(4×5×7) m=140 m 3 m=P 1 V/RT 1 =(100 kPa)(140 m 3 )(0.287 kPa ⋅ m 3 /kg ⋅ K)(283 K)=172.4 kg Using the c v value at room temperature, ((10000−5000)/3600 kJ/s+0.1 kJ/s)Δ t = (172.4 kg)(0.718 kJ/kg ⋅ ° C)(21−10) °C It yields, Δ t=914.52 s

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Solution Homework set 2-P SP-19-1

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