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

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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