AME 50531: Intermediate Thermodynamics Homework Solutions

AME 50531: Intermediate Thermodynamics Homework Solutions Fall 2010 1 Homework 1 Solutions 1.1 Problem 1: CPIG air enters and isentropic nozzle at 1.30 atm and 25 C with a velocity of 2.5 m/s. The nozzle entrance diameter is 120 mm. The air exits the nozzle at 1.24 atm with a velocity of 90 m/s. Determine the temeprature of the exiting air and the nozzle exit diameter. At the nozzle entrance (State 1): At the nozzle exit (State 2): T 2 and d 2 are unknown. 101.3 kpa P 1 = 1.30 atm = 131.7 kpa 1 atm (1) T 1 = 25 C+273.15 = 298.15 K (2) v 1 = 2.5 m/s (3) A 1 = π (0.120 m) 2 = 0.045 m 2 (4) 01.3 kpa P 2 = 1.24 atm = 125.6 kpa 1 atm (5) v 2 = 90 m/s (6) Because the nozzle is isentropic, we can use the isentropic relation for temperature and pressure to find the exit temperature. For air, k = 1.4. To find the exit diameter, recall that ṁ = ρav. Then, Knowing that ρ 1 = ρ 2, Therefore, d 2 = 0.020 m = 20 mm. T 2 T 1 = ( P2 P 1 ) k 1 k ( ) k 1 ( ) 1.4 1 P2 k 125.6 kpa 1.4 T 2 = T 1 = 298.15 K P 1 131.7 kpa T 2 = 294.14 K = 21 C (9) ṁ = ρ 1 A 1 v 1 = ρ 2 A 2 v 2 (10) A 1 v 1 = A 2 v 2 (11) ( ) ( ) v1 2.5 m/s A 2 = A 1 = 0.045 m 2 = 0.0012 m 2 (12) v 2 90 m/s (7) (8) 1

1.2 Problem 2: 9.47 Consider a steam turbine power plant operation near critical pressure, as shown in Figure 1.2. As a first approximation, it may be assumed that the turbine and pump processes are reversible and adiabatic. Neglecting any changes in kinetic and potential energies, calculate: The specific turbine work output and the turbine exit state. The pump work input and enthalpy at the pump exit state. The thermal efficiency of the cycle. We are given P 1 = P 4 = 20 MPa, T 1 = 800 C, P 2 = P 3 = 10 kpa, and T 3 = 40 C. State 1: Using Table B.1.3, h 1 = 4069.80 /kg, s 1 = 7.0544 kg K State 2: Using Table B.1.2, The specific work output from the turbine, The turbine exit state is a saturated mixture. s 2 = s 1 = 7.0544 kg K (13) s 2 = 0.6492+7.5010 x 2 = 0.8539 (14) h 2 = 191.81+0.8539 2392.82= 2235.04 /kg (15) w t = h 1 h 2 = 4069.80 /kg 2235.04 /kg = 1834.76 /kg (16) State 3: Compressed liquid, using Table B.1.1, h 3 = 167.54 /kg, v 3 = 0.001007 State 4: Using the property relation for constant v, The heat transfer in the boiler is w p = v 3 (P 4 P 3 ) = 0.001007 (20000 10)= 20.13 /kg (17) h 4 = h 3 w p = 167.54 /kg+20.13 /kg = 187.67 /kg (18) q in = h 1 h 4 = 4069.80 /kg 187.67 /kg = 3882.13 /kg (19) w net = w t + w p = 1834.76 /kg 20.13 /kg = 1814.63 /kg (20) η T H = w net = 1814.63 = 0.4674 q in 3882.13 (21) 2

2 Homework 2 Solutions: 2.1 Problem 1: 7.64 Helium has the lowest normal boiling point of any of the elements at 4.2 K. At this temperature the enthalpy of evaporation is 83.3 /kmol. A Carnot refrigeration cycle is analyzed for the production of 1 kmol of liquid helium at 4.2 K from saturated vapor at the same temperature. What is the work input to the refrigerator and the coefficient of performance for the cycle with an ambient temperature at 280 K? For the Carnot cycle the ratio of the heat transfers is the ratio of temperatures: Q L = n h f g = 1 kmol 83.3 /kmol = 83.3 (22) Q H = Q L TH = 83.3 280 = 5553.3 (23) T L 4.2 W IN = Q H Q L = 5553.33 83.3= 5470 (24) β = Q L = 83.3 [ W IN 5470 = 0.0152 = T ] L (25) T H T L 3

2.2 Problem 2: 8.97 A piston/cylinder contains air at 1380 K, 15 MPa, with V 1 = 9 cm 3, A cyl = 5 cm 2. The piston is released, and just before the piston exits the end of the cylinder the pressure inside is 200 kpa. If the cylinder is insulated, what is its length? How much work is done by the air inside? The cylinder is a control volume of air. It is insulated and therfore adiabatic, so Q = 0. By the continuity equation, m 2 = m 1 = m (26) From the energy equation (5.11), From the entropy equation (8.37), m(u 2 u 1 ) = Q 1 2 W 1 2 = W 1 2 (27) m(s 2 s 1 ) = dq T + S 1 2gen = 0+ S 1 2gen (28) Pressure and temperature are known at State 1, but only pressure is known at State 2, so T 2 must be obtained. Assume a reversible process. 1S2gen = 0 s 2 s 1 = 0 (29) State 1: Using Table A.7, u 1 = 1095.2 /kg and s T 1 m = P 1V 1 RT 1 = = 8.5115 kg K 15000 9 10 6 0.287 1380 = 0.000341 kg (30) State 2: From the entropy equation, s 2 = s 1 so from Eq. 8.19, s T 2 = s T 1 + Rln P ( ) 2 200 = 8.5115+0.287ln = 7.2724 P 1 15000 kg K (31) Then, use Table A.7 to interpolate and find T 2 = 447.2 K and u 2 = 320.92 /kg. T 2 P 1 447.2 15000 V 2 = V 1 = 9 T 1 P 2 1380 200 = 218.7 cm3 (32) L 2 = V 2 = 218.7 = 43.74 cm (33) A cyl 5 1w2 = u 1 u 2 = 774.3 /kg (34) W 1 2 = m 1w 2 = 0.264 = 264 J (35) 4

2.3 Problem 3: CPIG air in a cylinder (V 1 = 0.03 m 3, P 1 = 100 kpa, T 1 = 10 C) is compressed reversibly at constant temperature to a pressure of 420 kpa. Determine the entropy change, the heat transferred, and the work done. Also accurately plot this process on T-S and P-V diagrams. State 1: T 1 = 10 C = 283.15 K P 1 = 100 kpa V 1 = 0.03 m 3 State 2: T 2 = T 1 = 283.15 K P 2 = 420 kpa V 2 =? We know that m 1 = m 2, so, m 1 = m 2 = P 1V 1 RT 100 0.03 = = 0.0369 kg (36) 0.287 283.15 V 2 = mrt 0.0369 0.287 283.15 = = 0.00714 m 3 (37) P 2 420 v 1 = V 1 m = 0.03 0.0369 = 0.813 m3 /kg (38) v 2 = V 2 m = 0.00714 ( S 2 S 1 = m c p ln T 2 Rln P 2 T 1 P 1 0.0369 = 0.193 m3 /kg (39) ) ( = 0.0369 1.004ln1 0.287ln 420 ) (40) 100 S 2 S 1 = 0.0152 /K (41) Now, to find the heat transfer and work done, 1Q2 = 1 W 2 = mrt ln v 2 = 0.0369 0.287 283.15ln 0.193 v 1 0.813 (42) 1Q2 = 1 W 2 = 4.312 (43) 300 500 400 T (K) 290 280 P (kpa) 300 200 100 270 0.23 0.24 0.25 0.26 S (/K) (a) T-S Diagram 0 0 0.01 0.02 0.03 0.04 V (m 3 ) (b) P-V Diagram 5

2.4 Problem 4: 12.97 Consider an ideal air-standard Stirling cycle with an ideal regenerator. The minimum pressure and tmeperature in the cycle are 100 kpa, 25 C, the compression ratio is 11, and the maximum temperature in the cycle is 1000 C. Analyze each of the four processes in this cycle for work and heat transfer, and determine the overall perfomance of the engine. P 1 = 100 kpa, T 1 = T 2 = 25 C, v 1 v 2 = 11, T 3 = T 4 = 1000 C From 1-2 at constant temperature, w 1 2 = q 1 2 = T 1 (s 2 s 1 ) From 2-3 at constant volume, w 2 3 = 0 1w2 = RT 1 ln v 1 = 0.287 298.15ln11 = 205.2 /kg (44) v 2 From 3-4 at constant temperature, w 3 4 = q 3 4 = T 1 (s 4 s 3 ) From 4-1 at constant volume, w 4 1 = 0 Since q 23 is supplied by q 41 (regenerator), NOTE: q 23 = C V0 (T 3 T 2 ) = 0.717(1000 25)= 699 /kg (45) 3w4 = +RT 3 ln v 4 = 0.287 1273.15ln11 = 876.2 /kg (46) v 3 q 41 = C V0 (T 1 T 4 ) = 0.717(25 1000)= 699 /kg (47) w NET = 205.2+0+876.2+0= 671 /kg (48) η T H = q H q L q H = T 3 T 1 q H = q 34 = 876.2 /kg (49) η T H = w NET = 671 = 0.766 (50) q H 876.2 q H = q 34 = RT 3 ln11 (51) q L = q 1 2 = RT 1 ln11 (52) 3 = 975 = 0.766 = Carnot efficiency (53) 1273.15 6

3 Homework 3 Solutions: 3.1 Problem 1: 11:36 Consider an ideal steam reheat cycle where steam enters the high-pressure turbine at 4.0 MPa, 400 C, amd then expands to 0.8 MPa. It is then reheated to 400 C and expands to 10 kpa in the low-pressure turbine. Calculate the cycle thermal efficiency and the moisture content of the steam leaving the low-pressure turbine. State 3: High-pressure turbine entrance. P 3 = 4 MPa, T 3 = 400 C h 3 = 3213.51 /kg, s 3 = 6.7689 State 4: High-pressure turbine exit. s 4 = s 3 h 4 = 2817.77 /kg State 5: Low-pressure turbine entrance. P 5 = 0.8 MPa, T 5 = 400 C h 5 = 3267.07 /kg, s 5 = 7.5715 State 6: Low-pressure turbine exit. Use entropy to find the moisture content x 6 : s 6 = s 5 = 7.5715 kg K kg K (54) (55) two-phase state (56) kg K x 6 = s 6 s f 7.5715 0.6492 = = 0.92285 (57) s f g 7.501 h 6 = h f + x 6 h f g = 191.81+0.92285 2392.82= 2400.02 /kg (58) State 1: Pump entrance. P 1 = 10 kpa, v 1 = 0.00101 m 3 /kg, h 1 = 191.81 /kg State 2: Pump exit. Pump is reversible and adiabatic. Assume incompressible flow. w P = v 1 (P 2 P 1 ) = 0.00101(4000 10)= 3.94 /kg (59) h 2 = h 1 + w P = 191.81+3.94= 195.75 /kg (60) w T,tot = h 3 h 4 + h 5 h 6 = 3213.51 2817.77+3267.07 2400.02= 1262.79 /kg (61) q H1 = h 3 h 2 = 3213.51 195.75= 3017.76 /kg (62) q H = q H1 + h 5 h 4 = 3017.76+3267.07 2817.77= 3467.06 / jg (63) η CYCLE = (w T,tot w P ) 1262.79 3.94 = = 0.3631 (64) q H 3467.06 7

3.2 Problem 2: Consider a 600 MW reheat-rankine cycle steam power plant. The boiler pressure is 1000 psia, the reheater pressure is 200 psia, and the condenser pressure is 1 psia. Both turbine inlet temperatures are 1000 F. The water leaving the condenser is saturated liquid. Determine the thermal efficiency of the power plant and the boiler mass flow rate in lbm/hr. State 3: HP turbine entrance. P 3 = 1000 psia, T 3 = 1000 F, h 3 = 1505.86 Btu/lb m, s 3 = 1.6530 Btu lb m R State 4: HP turbine exit. P 4 = 200 psia, s 4 = s 3 = 1.6530 Btu lb m R. Interpolating, h 4 = 1297.58 Btu/lb m State 5: LP turbine entrance. P 5 = 2000 psia, T 5 = 1000 F, h 5 = 1529.28 Btu/lb m, s 5 = 1.8425 Btu lb m R State 6: LP turbine exit. P 6 = 1 psia, s 6 = s 5 = 1.8425 Btu lb m R x 6 = s 6 s f 1.8425 0.1323 = = 0.9264 (65) s f g 1.8461 h 6 = h f + x 6 h f g = 69.57+0.9264 1036.11= 1029.41 Btu/lb m (66) State 1: Pump entrance, saturated liquid. P 1 = 1 psia, h 1 = 69.57 Btu/lb m, v 1 = 0.016136 ft 3 /lb m State 2: Pump exit. P 2 = 1000 psia w P = h 1 h 2 = v 1 (P 2 P 1 ) h 2 = h 1 + v 1 (P 2 P 1 ) (67) Converting pressures into lb/ ft 2, we get P 1 = 144 lb/ ft 2 and P 2 = 144000 lb/ ft 2. w P = 0.016136(144000 144) 1.285 10 3 Btu 1 lb ft = 2.983 Btu/lb m h 2 = 72.55 Btu/lb m (68) η T = w NET = (h 3 h 4 )+(h 5 h 6 )+(h 1 h 2 ) = 0.4235 (69) Q IN (h 3 h 2 )+(h 5 h 4) Ẇ cycle = 600 MW 1 Btu/s 1.055 10 3 MW = 5.687 106 Btu/s (70) ṁ = Ẇcycle Ẇ cycle = w NET (h 3 h 4 )+(h 5 h 6 )+(h 1 h 2 ) 3600 s = 2.9033 10 7 lb m /hr (71) 1 hr 8

3.3 Problem 3: 11.45 A power plant with one open feedwater heater has a condenser temperature of 45 C, a maximum pressure of 6 MPa, and a boiler exit temperature of 900 C. Extraction steam at 1 MPa to the feedwater heater is mixed with the feedwater line so the exit is saturated liquid into the second pump. Find the fraction of extraction steam flow and the two specific pump work inputs. State 5: Boiler exit/turbine entrance. h 5 = 4375.29 /kg, s 5 = 7.8727 kg K State 6: Turbine is reversible and adiabatic: s 7 = s 6 = s 5 P 6 = 1 MPa, s 6 = 7.8727 kg K h 6 = 3569.39 /kg (72) State 1: Condenser exit/pump 1 entrance. h 1 = 188.42 /kg, v 1 = 0.00101 m 3 /kg, P 1 = 9.593 kpa State 2: Pump 1 exit. P 2 = P 6 w P1 = h 2 h 1 = v 1 (P 2 P 1 ) = 0.00101(1000 9.593)= 1.0 /kg (73) h 2 = h 1 + w P1 = 188.42+1.0= 189.42 /kg (74) State 3: Feedwater exit. P 3 = P 2, h 3 = 762.79, v 3 = 0.001127 m 3 /kg The extraction fraction can be expressed as y = ṁ6 ṁ tot. From the energy equation, (1 y)h 2 + yh 6 = h 3 State 4: Pump 2 exit. P 4 = P 5 y = h 3 h 2 762.79 189.42 = = 0.1696 (75) h 6 h 2 3569.39 189.42 w P2 = h 4 h 3 = v 3 (P 4 P 3 ) = 0.001127(6000 1000)= 5.635 /kg (76) 9

4 Homework 4 Solutions: 4.1 Problem 1: 12.22 A Brayton cycle produces 14 MW with an inlet state of 17 C, 100 kpa, and a compression ratio of 17 : 1. The heat added in the combustion is 960. What are the highest temperature and mass flow rate of air, using properties from Table A.7.1? The specific heat varies; therefore it is necessary to go through the processes individually to find the net work and highest temperature T 3. State 1: T 1 = 290.15 K, P 1 = 100 kpa. From A.7.1, h 1 = 290.43 /kg, s T 1 State 2: The compression is reversible and adiabatic, so from Eq. 8.19, = 6.83521 kg K s 2 = s 1 s T 2 = s T 1 + Rln P 2 = 6.83521+0.287ln17 = 7.64834 P 1 kg K (77) From the energy equation with compressor work in, State 3: Energy equation for the combustor, T 2 = 642.36 K, h 2 = 652.04 /kg (78) w C = w 1 2 = h 2 h 1 = 652.04 290.43= 361.61 /kg (79) h 3 = h 2 + q H = 652.04+960= 1612.04 /kg (80) T 3 = 1480.33 K, s T 3 = 8.59596 State 4: The expansion is reversible and adiabatic so from Eq. 8.19, kg K s 4 = s 3 s T 4 = s T 3 + Rln P 4 = 8.59596+0.287ln 1 = 7.78283 P 3 17 kg K (81) (82) From the energy equation with turbine work out, The net work is The total power requires a mass flow rate of T 4 = 728.36 K, h 4 = 744.14 /kg (83) w T = h 3 h 4 = 1612.04 744.14= 867.9 /kg (84) w net = w T w C = 867.9 361.61= 506.29 /kg (85) ṁ = Ẇnet = 14000 KW = 27.65 kg/s (86) w net 506.29 /kg 10

4.2 Problem 2: 12.36 A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100 MW to an electric generator. The minimum temperature in the cycle is 290 K, and the maximum temperature is 1600 K. The minimum pressure in the cycle is 100 kpa, and the conmpressor compression ratio is 14 : 1. The compressor has an isentropic efficiency of 85% and the turbine has an isentropic efficiency of 88%. Calculate the power output of the turbine. What fraction of the turbine output is required to drive the compressor? What is the thermal efficiency of the cycle? Solve using constant C P0. For an ideal compressor, s 2 = s 1 Implemented by Eq. 8.23 For the actual compressor, ( ) k 1 P2 k T 2s = T 1 = 290(14) 0.286 = 616.4 K (87) P 1 w Cs = h 2 h 1 = C P0 (T 2 T 1 ) = 1.004(616.4 290)= 327.71 /kg (88) For an ideal turbine, s 4 = s 3 Implemented by Eq. 8.23 w C = w SC η SC = 327.71 0.85 = 385.54 /kg = C P0(T 2 T 1 ) (89) T 2 = T 1 + w C = 290+ 385.54 = 674 K (90) C P0 1.004 ( ) k 1 ( ) P4 k 1 0.286 T 4s = T 3 = 1600 = 752.8 K (91) P 3 14 For the actual turbine, w T s = h 3 h 4 = C P0 (T 3 T 4 ) = 1.004(1600 752.8)= 850.63 /kg (92) w T = η ST w ST = 0.88 850.63 = 748.56 /kg = C P0 (T 3 T 4 ) (93) To calculate the overall net work and the cycle efficiency, Energy input is from the combustor, T 4 = T 3 w t = 1600 748.56 = 854.4 K (94) C P0 1.004 w NET = w T w C = 748.56 385.54= 363.02 /kg (95) ṁ = ẆNET = 100000 kw = 275.47 kg/s (96) w NET 363.02 /kg Ẇ NET = ṁw T = 275.47 kg/s 748.56/kg = 206.203 MW (97) w C = 385.54 = 0.515 (98) w T 748.56 q H = C P0 (T 3 T 2 ) = 1.004(1600 674)= 929.7 /kg (99) η T H = w NET = 363.02 = 0.3905 (100) q H 929.7 11

4.3 Problem 3: 12.41 An air-standard Ericsson cycle has an ideal regenerator. Heat is supplied at 950 C and heat is rejected at 80 C. Pressure at the beginning of the isothermal compression process is 70 kpa. The heat added is 700 /kg. Find the compressor work, the turbine work, and the cycle efficiency. Identify the states: Heat supplied at high temperature: T 3 = T 4 = 950 C = 1223.15 K Heat rejected at low temperature: T 1 = T 2 = 0 C = 353.15 K Beginning of the compression: P 1 = 70 kpa Ideal regenerator: 2q3 = 4 q 1 q H = 3 q 4 = 700 /kg (101) w T = q H = 700 /kg (102) η T H = η CARNOT = 1 353.15 = 0.7113 (103) 1223.15 w NET = η T H q H = 0.7113 700 = 497.895 /kg (104) w C = q L = q H w NET = 700 497.895= 202.105 /kg (105) 12

4.4 Problem 4: 12.53 An afterburner in a jet engine adds fuel after the turbine this raising the pressure and temperature due to the energy of combustion. Assume a standard condition of 800 K, 250 kpa after the turbine into the nozzle that exhausts at 95 kpa. Assume the afterburner adds 425 /kg to that state with a rise in pressure for the same specific volume, and neglect any upstream effects on the turbine. Find the nozzle exit velocity before and after the afterburner is turned on. Without afterburner: T 1 = 800 K, P 1 = 200 kpa, P 2 = 95 kpa With afterburner: v 3 = v 1, P 4 95 kpa Assume reversible adiabatic nozzle flow, then constant s from Eq. 8.23 Energy Equation: ( ) (k 1) ( ) P2 k 95 0.2857 T 2 T 1 = 800 K = 606.8 K (106) P 1 250 1 2 V 2 2 = C P (T 1 T 2 ) V 2 = 2C P (T 1 T 2 ) = 2 1004 Add the q AB at assumed constant volume then energy equation gives Reversible adiabatic expansion, again from Eq. 8.23 J (800 606.8) K = 622.8 m/s (107) kg K T 3 = T 1 + q AB = 800+ 425 = 1392.63 K C v 0.717 (108) ( ) T3 v 3 = v 1 P 3 = P 1 = 25 kpa 1392.63 = 435.2 kpa T 1 800 (109) ( ) (k 1) ( ) P4 k 95 0.2857 T 4 = T 3 = 1392.63 = 901.5 K (110) P 3 435.2 V 2 = J 2C P (T 3 T 4 ) = 2 1004 (1392.63 901.5) K = 993.02 m/s (111) kg K Comment: The real process adds some fuel that burns releasing energy so the temperature goes up and due to the confined space then pressure goes up. As the pressure goes up the exit velocity increases altering the mass flow rate through the nozzle. As in most problems the real device is more complicated than we can describe with our simple analysis. 13

5 Homework 5 Solutions: 5.1 Problem 1: 12.70 A gasoline engine takes air in at 300 K, 90 kpa and then compresses it. The combustion adds 1000 /kg to the air after which the temperature is 2050 K. Use variable heat capacities (Table A.7) and find the compression ratio, the compression specific work and the highest pressure in the cycle. Standard Otto cycle, solve using Table A.7.1. State 3: T 3 = 2050 K, u 3 = 1725.71 /kg State 2: Combustion Process u 2 = u 3 q H = 1725.71 1000= 725.71 /kg (112) T 2 = 960.5 K, s T 2 = 8.0889 kg K State 1: Compression from 1 to 2: u 1 = 214.36 /kg, s 1 = 6.86926 kg K, s 2 = s 1 From Eq. 8.19 ( ) ( ) 0 = s T 2 s P2 T 1 Rln = s T 2 s T2 v 1 T1 Rln P 1 T 1 v 2 ( ) ( ) 960.5 v1 = 8.0089 6.86926 0.287ln 0.287ln 300 Comment: This is much too high for an actual Otto cycle. Highest pressure is after combustion where v 3 = v 2 so we get ( )( ) T3 v1 = 90 v 2 (113) (114) (115) Solving v 1 v 2 = 21.89 (116) w 1 2 = u 2 u 1 = 725.71 214.36= 511.35 /kg (117) P 3 = P 2 T 3 T 2 = P 1 T 1 v 3 ( 2050 300 ) 21.89 = 13.46 MPa (118) 14

5.2 Problem 2: 12.90 A diesel engine has a state before compression of 95 kpa, 300 K, a peak pressure of 6000 kpa, and a maximum temperature of 2400 K. Find the volumetric compression ratio and the thermal efficency. Use the properties from Table A.7. Compression: State 1: T 1 = 300 K, u 1 = 214.36, s 1 = 6.869 kg K State 2: s 2 = s 1 From Eq. 8.19 s T 2 = s T 1 + Rln ( P2 P 1 ) = 6.869+0.287ln ( ) 6000 = 8.0588 95 kg K (119) State 3: h 3 = 2755.78 /kg, s Expansion process v 3 v 4 = v 3 v 1 = s T 4 Rln A.7.1 T 2 = 935.32 K,h 2 = 972.90 /kg (120) T 3 = 9.19586 kg K q H = h 3 h 2 = 2755.78 972.90= 1782.88 /kg (121) CR = v ( )( ) ( )( ) 1 T1 P2 300 6000 = = = 20.26 v 2 935.32 95 (122) T 2 P 1 s T 4 = s T 3 + Rln ( P4 ( v2 v 1 ( v3 v 4 ) ( T3 T 2 ) = P 3 ( T3 ) = s T 3 + Rln ) = s T 3 + Rln ( T4 )( 1 CR T 2 ( T4 T 3 ) = T 3 ( 2400 935.32 ) + Rln Trial and error on T 4 since it appears both in s T 4 and the ln function ( 1300 T 4 = 1300 K, LHS = 8.4405 0.287ln 2400 ( 1250 T 4 = 1250 K, LHS = 8.3940 0.287ln 2400 Now linearly interpolate T 4 = 1280.7 K, u 4 = 1004.81 /kg q H ( v3 v 4 ) )( 1 20.26 (123) ) = 0.1267 (124) ) = 9.1958 + 0.287 ln0.1267 = 8.6029 (125) ) = 8.6160 (126) ) = 8.5812 (127) q L = u 4 u 1 = 1004.81 214.36= 790.45 /kg (128) ( ) ( ) ql 790.45 η = 1 = 1 = 0.5566 1782.88 (129) 15

5.3 Problem 3: 11.122 Consider an ideal dual-loop heat-powered refrigeration cycle using R-134a as the working fluid, as shown in Figure 5.3. Saturated vapor at 90 C leaves the boiler and expands in the turbine to the ocndenser pressure. Saturated vapor at 15 C leaves the evaporator and is compressed to the condenser pressure. The ratio of the flows through the two loops is such that the turbine produces juse enough power to drive the compressor. The two exiting streams mix together and enter the condenser. Saturated liquid leaving the compressor at 45 C is then separated into two streams in the necessary proportions. Determine the ratio of mass flow rate through the power loop the the refrigeration loop. Find also the performance of the cycle, in terms of the ratio Q L Q H. From Table B.5.1, T 6 = 90 C sat. vapor P 5 = P 6 = 3.2445 MPa From Table B.5.1, T 3 = 45 C sat. liquid P 2 = P 3 = P 7 = 1.1602 MPa C.V. Turbine C.V. Compressor CV: turbine + compressor CV: pump CV: evaporator Q L = ṁ 1 (h 1 h 4 ) CV: boiler Q L = ṁ 6 (h 6 h 5 ) T 1 = 15 C, h 1 = 389.20 /kg, h 3 = h 4 = 264.11 /kg, h 6 = 425.70 /kg (130) s 7 = s 6 = 1.6671 kg K = 1.2145+x 7 0.4962, x 7 = 0.912 (131) h 7 = 264.11+0.912 157.85= 408.09 /kg (132) s 2 = s 1 = 0.9258 h 2 = 429.89 /kg (133) Continuity Eq.: ṁ 1 = ṁ 2,ṁ 6 = ṁ 7 (134) Energy Eq.: ṁ 1 h 1 + ṁ 6 h 6 = ṁ 2 h 2 + ṁ 7 h 7 (135) ṁ 6 = (h 2 h 1 ) 429.89 389.20 = = 2.31 (136) ṁ 1 (h 6 h 7 ) 425.20 408.09 w P = v 3 (P 5 P 3 ) = 0.000890(3244.5 1160.2)= 1.855 /kg (137) h 5 = h 3 + w P = 265.97 /kg (138) β = Q L = ṁ1(h 1 h 4 ) Q H ṁ 6 (h 6 h 5 ) = 389.20 264.11 = 0.339 (139) 2.31(425.70 265.97) 16

5.4 Problem 4: 12.112 A small utility gasoline engine of 250 cc runs at 1500 RPM with a compression ratio of 7 : 1. The inlet state is 75 kpa, 17 C and the combustion adds 1500 /kg to the charge. This engine runs a heat pump using R-410a with a high pressure of 4 MPa and an evaporator operating at 0 C. Find the rate of heating the pump can deliver. Overall cycle efficiency is from Eq. 12.12, r v = v 1 v 2 = 7 We also need specific volume to evaluate Eqs. 12.9 to 12.11 η T H = 1 r 1 k = 1 7 0.4 = 0.5408 (140) w NET = η T H q H = 0.5408 1500= 811.27 /kg (141) v 1 = RT 1 P 1 = P me f f = w NET v 1 v 2 = Now we can find the power from Eq. 12.11 (assume 4-stroke engine) Ẇ = P me f f V displ RPM 60 For the refrigeration cycle we have: State 1: h 1 = 279.12 /kg, s 1 = 1.0368 kg K State 2: P 2 = 4 MPa, s 2 = s 1, interpolate h 2 = 323.81 /kg State 3: P 3 = 4 MPa, interpolate h 3 = 171,62 /kg The work out of the heat engine queals the input to the heat pump 0.287 kg K 290 K = 1.1097 m 3 /kg (142) 75 kpa v 2 = v 1 CR = 0.15853 m3 /kg (143) 811.27 kk/kg 1.097 0.15853 m 3 = 852.9 kpa (144) /kg 1 2 = 852.9 2.5 10 4 1500 60 1 = 2.665 kw (145) 2 β HP = q H = h 2 h 3 323.81 171.62 = = 3.405 (146) w C h 2 h 1 323.81 279.12 Q H = β HPẆ = 3.405 2.665 kw = 9.07 kw (147) 17

6 Homework 6 Solutions: 6.1 Problem 1: 13.20 A 100 m 3 storage tank with fuel gases is at 300 K, 100 kpa containing a mixture of acetylene C 2 H 2, propane C 3 H 8 and butane C 4 H 10. A test shows the partial pressure of C 2 H 2 is 15 kpa and that of C 3 H 8 is 65 kpa. How much mass is there of each component? Assume ideal gases, then the ratio of partial to total pressure is the mole fraction, y = P P tot y C2 H 2 = 15 100 = 0.15, y C 3 H 8 = 65 100 = 0.65, y C 4 H 10 = 20 = 0.20 (148) 100 n tot = PV RT = 100 kpa 100 m3 = 4.0091 kmol (149) 8.31451 kmol K 300 K m C2 H 2 = (nm) C2 H 2 = y C2 H 2 n tot M C2 H 2 = 0.15 4.0091 kmol 26.038 kg/kmol = 15.659 kg (150) m C3 H 8 = (nm) C3 H 8 = y C3 H 8 n tot M C3 H 8 = 0.65 4.0091 kmol 44.097 kg/kmol = 114.905 kg (151) m C4 H 10 = (nm) C4 H 10 = y C4 H 10 n tot M C4 H 10 = 0.20 4.0091 kmol 58.124 kg/kmol = 46.604 kg (152) 18

6.2 Problem 2: 13.29 A flow of 1 kg/s argon at 300 K and another flow of 1 kg/s CO 2 at 1600 K both at 200 kpa are mixed without any heat transfer. Find the exit T,P using variable specific heats. No work implies no pressure change for a simple flow. The energy equation becomes Trial and error on T e using Table A.8 for h eco2 P e = 200 kpa (153) ṁh i = ṁh e = (ṁh i ) Ar +(ṁh i ) CO2 = (ṁh e ) Ar +(ṁh e ) CO2 (154) ṁ CO2 (h e h i ) CO2 + ṁ Ar C PAr (T e T i ) Ar = 0 (155) 1 kg/s (h e 1748.12 /kg)+(1 0.52) kw/k (T e 300 K) = 0 (156) h eco2 + 0.52T e = 1748.12+0.52 300= 1904.12 /kg (157) T e = 1200 K : LHS = 1223.34+0.52 1200= 1847.34 (158) Interpolating, T e = 1300 K : LHS = 1352.28+0.52 1300= 2028.28 (159) 1904.12 1847.34 T e = 1200+100 = 1231.4 K (160) 2028.28 1847.34 19

6.3 Problem 3: 13.46 A mixture of 2 kg oxygen and 2 kg argon is in an insulated piston cylinder arrangement at 150 kpa, 300 K. The piston now compresses the mixture to half its initial volume. Find the final pressure, temperature, and the piston work. C.V. Mixture. Control mass, boundary work, adiabatic and assume reversible. Process: cosntant s Pv k = constant, v 2 = v 1 2, Assume ideal gases(t 1 T c ) and use k mix and C vmix for properties. Energy Eq. 5.11: u 2 u 1 = q 1 2 w 1 2 = w 1 2 (161) Entropy Eq. 8.37: s 2 s 1 = 0+0 = 0 (162) Eq. 13.15: R mix = c i R i = 0.5 0.25983+0.5 0.20813= 0.234 Eq. 13.23: C Pmix = c i C Pi = 0.5 0.9216+0.5 0.5203= 0.721 C vmix = C Pmix R mix = 0.487 kg K kg K kg K (163) (164) (165) Ratio of specific heats: k mix = C Pmix C vmix = 1.4805 The relations for the polytropic process ( ) k v1 Eq. 8.25: P 2 = P 1 = P 1 (2) k = 150(2) 1.4805 = 418.5 kpa (166) v 2 Work from the energy equation ( ) k 1 v1 Eq. 8.24: T 2 = T 1 = T 1 (2) k 1 = 300(2) 0.4805 = 418.6 K (167) v 2 1W2 = m tot(u 1 u 2 ) = m tot C v (T 1 T 2 ) = 4 kg 0.487 (300 418.6) K = 231 (168) kg K 20

6.4 Problem 4: 13.54 Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in an insulated mixing chamber. Both flows are at 100 kpa and the mass ratio of carbon dioxide ot nitrogen is 2 : 1. Find the exit temprature and total entropy generation per kg of the exit mixture. CV mixing chamber. The inlet ratio is so ṁ CO2 = 2ṁ N2 and assume no external heat transfer, no work involved. Take 300 K as reference and write h = h 300 +C Pmix (T mixex 300) Continuity Eq. 6.9: ṁ N2 + 2ṁ N2 = ṁ ex = 3ṁ N2 (169) Energy Eq. 6.10: ṁ N2 (h N2 2h CO2 ) = 3ṁ N2 h mixex (170) C PN2 (T in2 300)+2C PCO2 (T ico2 300) = 3C Pmix (T mixex 300) (171) C Pmix = c i C Pi = 2 3 0.842+ 1 1.042 = 0.9087 3 kg K (172) 3C Pmix T mixex = C PN2 T in2 + 2C PCO2 T ico2 = 830.64 /kg (173) T mixex = 304.7 K (174) To find the entropies we need the partial pressures, which assuming ideal gas are equal to the mole fractions times the total pressure: y N2 = y i = [ ] ci M i c j [ 0.3333 28.013 0.3333 28.013 + 0.6666 44.01 (175) M j ] = 0.44 (176) y CO2 = 1 y N2 = 0.56 (177) Ṡ gen = ṁ ex s ex (ṁs) ico2 (ṁs) in2 = ṁ N2 (s e s i ) N2 + 2ṁ N2 (s e s i ) CO2 (178) Ṡ gen = 1 [ C PN2 ln T ] ex R N2 lny N2 + 2 [ C PCO2 ln T ] ex R CO2 lny CO2 (179) 3ṁ N2 3 T in2 3 T ico2 = 1 [ ( ) ] 304.7 1.042 ln 0.2968 ln0.44 + 2 [ ( ) ] 304.7 0.842 ln 0.1889 ln0.56 (180) 3 280 3 320 Ṡ gen = 0.1561 kg mix K (181) 21

7 Homework 7 Solutions: 7.1 Problem 1: 13.73 Ambient moist air enters a steady-flow air-conditioning unit at 105 kpa, 30 C, with a 60% relative humidity. The volume flow rate entering the unit is 100 L/s. The moist air leaves the unit at 95 kpa, 15 C, with a relative humidity of 100%. Liquid condensate also leaves the unit at 15 C. Determine the rate of heat transfer for this process. State 1: State 2: Energy Eq. 6.10: P V1 = φ 1 P G1 = 0.60 4.246 kpa = 2.5476 kpa (182) 2.5476 w 1 = 0.622 = 0.0154668 105 2.5476 (183) ṁ A = P A1V 1 102.45 0.1 = = 0.1178 kg/s R A T 1 0.287 303.15 (184) P V2 = P G2 = 1.705 kpa (185) 1.705 w 2 = 0.622 = 0.1137 (186) 95 1.705 Q CV + ṁ A h A1 + ṁ V1 h V1 = ṁ A h A2 + ṁ V2 h V2 + ṁ 3 h L3 (187) Q CV ṁ A = C P0A (T 2 T 1 )+w 2 h V2 w 1 h V1 +(w 1 w 2 )h L3 (188) = 1.004(15 30)+0.1137 2528.91 0.0154668 2556.25+0.00456 63.0= 26.732 /kg air (189) Q CV = 0.1178( 26.732)= 3.088 kw (190) 22

7.2 Problem 2: 13.103 Use a psychrometric chart to find the missing property of: φ, ω, T wet, T dry a. T dry = 25 C, φ = 80% b. T dry = 15 C, φ = 100% c. T dry = 20 C, ω = 0.008 d. T dry = 25 C, T wet = 23 C a. 25 C, φ = 80% ω = 0.016, T wet = 22.3 C (191) b. 15 C, φ = 100% ω = 0.0106, T wet = 15 C (192) c. 20 C, ω = 0.008 φ = 57%, T wet = 14.4 C (193) d. 25 C, T wet = 23 C ω = 0.017, φ = 86% (194) 23

7.3 Problem 3: 14.35 The Joule-Thomson coefficient µ J is a measure of the direction and magnitude of the temperature change with pressure in a throttling process. For any three properties x,y,z use the mathematical relation ( ) ( ) ( ) x y z = 1 (195) y z x z x y to show the following relations for the Joule-Thomson coefficient: ( ) ( ) T T v T v P µ J = = = RT 2 ( ) Z P h C P PC p T P (196) Let x = T, y = P and z = h and substitute into the relations as ( ) ( ) ( ) T P h P h T Then we have the definition of specific heat as C P = ( h T h ( ) T µ J = P h = The last derivative is substituted with Eq. 14.25 so we get ) T P 1 ( C P ) P h T ( ) T µ J = = P h If we use the compressibility factor then we get ( ) v Pv = ZRT = ZR T P P + RT P so then and we have shown the last expression also. T ( ) v v = v+ RT 2 T P P ( ) T µ J = = P h P = 1 (197) so solve for the first term = 1 C P ( ) h P T (198) ( ) T v T v P (199) C P ( ) Z = v T P T + RT P ( ) Z v = RT 2 T P P ( ) T v T v P = RT 2 C P PC p ( ) Z T P ( ) Z T P ( ) Z T P (200) (201) (202) 24

7.4 Problem 4: 14.77 Develop general expressions for changes in internal energy, enthalpy, and entropy for a gas obeying the Redlich- Kwong equation of state. From Eq. 14.30 2 2 u 2 u 1 = C v (T,v)dT + 1 1 Redlich-Kwong equation of state:p = RT v b a v(v+b)t 1/2 (203) ( ) P = R T v v b + a 2v(v+b)T 3/2 (204) 3a 2 2v(v+b)T 1/2 = 1 C v ( ˆT,v)d ˆT 3a ln 2bT 1/2 [( v2 + b v 2 )( )] v1 v 1 + b We find the change in h from change in u, so we do not do the derivative in Eq. 14.27. This is due to the form of the EOS. 2 h 2 h 1 = C P ( ˆT,v)d ˆT + P 2 v 2 P 1 v 1 3a [( )( )] v2 + b v1 ln (206) 1 2bT 1/2 v 1 + b Entropy follows from Eq. 14.35 2 s 2 s 1 = C v (T,v) dt 2 [ ] R 1 T + 1 v b + a/2 v(v+b)t 3/2 dv (207) 2 s 2 s 1 = C v ( ˆT,v) d ( ) ˆT 1 ˆT + Rln v2 b a ln v 1 b 2bT 3/2 v 2 [( v2 + b v 2 )( )] v1 v 1 + b (205) (208) 25

7.5 Problem 5: 14.81 A flow of oxygen at 235 K, 5 MPa is throttled to 100 kpa in a steady flow process. Find the exit temperature and the specific entropy generation using the Redlich-Kwong equation of state and ideal gas heat capacity. Notice this becomes iterative due to the non-linearity coupling h, P v and T. C.V. Throttle. Steady single flow, no heat transfer and no work. Find the change in h from Eq. 14.26 assuming C P is constant. From Eq. 14.31 Energy Eq.: h 1 + 0 = h 2 + 2 so constant h (209) Entropy Eq.: s 1 + s gen = s 2 so entropy generation (210) Redlich-Kwong equation of state:p = RT v b a v(v+b)t 1/2 (211) ( ) P = R T v v b + a 2v(v+b)T 3/2 (212) 2 (u 2 u 1 ) T = 1 3a 2v(v+b)T 3a = ln 1/2 2bT 1/2 [( v2 + b v 2 )( )] v1 v 1 + b We find change in h from change in u, so we do not do the derivative in Eq. 14.27. This is due to the form of the EOS. (h 2 h 1 ) T = P 2 v 2 P 1 v 1 3a [( )( )] v2 + b v1 ln (214) 2bT 1/2 v 1 + b Entropy follows from Eq. 14.35 (s 2 s 1 ) T = 2 1 [ R v b + a/2 v(v+b)t 3/2 ] dv = Rln ( ) v2 b v 1 b v 2 a ln 2bT 3/2 P c = 5040 kpa, T c = 154.6 K, R = 0.2598 kg K [( v2 + b v 2 )( )] v1 v 1 + b (213) (215) (216) b = 0.08664 RT c 0.2598 154.6 = 0.08664 = 0.0006905 m 3 /kg (217) P c 5040 a = 0.42748 R2 Tc 5/2 = 0.42748 (0.2598)2 (154.6) 5/2 = 1.7013 (218) P c 5040 We need to find T 2 so the energy equation is satisfied h 2 h 1 = h 2 h x + h x h 1 = C p (T 2 T 1 )+(h 2 h 1 ) T = 0 (219) and we will evaluate it similar to figure 13.4, where the first term is done from state x to 2 and the second term is done from state 1 to state x (at T 1 = 235 K). We do this as we assume state 2 is close to ideal gas, but we do not know T 2. We first need to find v 1 from the EOS, so guess v and find P v 1 = 0.011 m 3 /kg P = 5059 too high (220) v 1 = 0.01114 m 3 /kg P = 5000.6 OK (221) Now evaluate the change in h along the 235 K from state 1 to state x, that requires a value for v x. Guess ideal gas at T x = 235 K, v x = RT x = 0.2598 235 P 2 100 = 0.6105 m3 /kg (222) 26

From the EOS: P 2 = 99.821 kpa (close) A few more guesses and adjustments gives v x = 0.6094 m 3 /kg, P 2 = 100.0006 kpa OK (223) (h x h 1 ) T = P x v x P 1 v 1 3a [( )( )] vx + b v1 ln = 19.466 /kg 2bT 1/2 v 1 + b (224) v x From energy eq.: T 2 = T 1 = (h x h 1 ) T = 235 19.466 = 213.51 K (225) C p 0.9055 Now the change in s is done in a similar fashion, s gen = s 2 s 1 = (s x s 1 ) T + s 2 s x = Rln ( ) vx b a ln v 1 b 2bT 3/2 [( vx + b v x )( )] v1 +C p ln T 2 (226) v 1 + b T x = 1.0576+0.0202 0.0868= 0.9910 kg K (227) 27

7.6 Problem 6: Consider a thermodynamic system in which there are two reversible work modes: compression and electrical. So take the version of the text s Eq. 4.16 giving dw to be δw = PdV E dz, (228) where E is the electrical potential difference and dz is the amount of charge that flows into the system. Extend the Gibbs equation to account for electrical work. Find the Legendre transformation which renders the independent variables to be P, E and T and show how the other variables can be determined as functions of these independent variables. Find all Maxwell equations associated with this Legendre transformation. Gibbs equation: Legendre transformation: du = δq δw, δq = Tds (229) du = TdS PdV +E dz (230) ψ 1 = T, ψ 2 = P, ψ 3 = E, x 1 = S, x 2 = V, x 3 = Z (231) F 1 = U ψ 1 x 1 = U TS (232) F 1 = U ψ 2 x 2 = U PV (233) F 1 = U ψ 3 x 3 = U E Z (234) F 1,2,3 = U TS+PV E Z (235) F(P,E,T) = SdT +VdP ZdE (236) Maxwell equations: S = ( ) U, V = T P,E ) ( S P T,E ( ) U, Z = P T,E ) = ( ) S = PE T,P ( V T P,E ( ) Z T P,E ( ) ( ) V Z = E T,P P T,E ( ) U E T,P (237) (238) (239) (240) 28

7.7 Problem 7: 14.113 A 2 kg mixture of 50% argon and 50% nitrogen by mass is in a tank at 2 MPa, 180 K. How large is the volume using a model of (a) ideal gas and (b) Redlich-Kwong equation of state with a, b for a mixture. a) Ideal gas mixture: Eq. 13.15: R mix = c i R i = 0.5 0.2081+0.5 0.2968= 0.25245 kg K (241) V = mr mixt 2 0.25245 180 = = 0.0454 m 3 (242) P 2000 b) Redlich-Kwong equation of state: Before we can do the parameters a, b for the mixture we need the individual component parameters, Eq. 14.54, 13.55. Now the mixture parameters are from Eq. 14.84 Using now Eq. 14.53: a Ar = 0.42748 R2 T 5/2 C = 0.42748 (0.2081)2 (150.8) 2.5 = 1.06154 (243) P c 4870 a N2 = 0.42748 R2 T 5/2 2.5 C (0.2081)2 (126.2) = 0.42748 = 1.98743 (244) P c 3390 a mix = b Ar = 0.08664 RT C 0.2081 150.8 = 0.08664 = 0.000558 (245) P c 4870 b N2 = 0.08664 RT C 0.2081 126.2 = 0.08664 = 0.000957 (246) P c 3390 ( ) c i a 1/2 2 ( i = 0.5 1.06154+ 0.5 1.98743) 2 = 1.4885 (247) b mix = c i b i = 0.5 0.000558+0.5 0.000957= 0.000758 (248) 2000 = By trial and error we find the specific volume, v = 0.02102 m 3 /kg P = RT v b a v(v+b)t 1/2 (249) 0.25245 180 v = 0.000758 1.4885 v(v+0.000758)180 1/2 (250) V = mv = 0.04204 m 3 (251) 29

8 Homewrok 8 Solutions: 8.1 Problem 1: 15.20 Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion of ethanol, C 2 H 5 OH. Reaction Eq.: C 2 H 5 OH + ν O2 (O 2 + 3.76N 2 ) aco 2 + bh 2 O+cN 2 (252) Do the atom balance Balance C: 2 = a (253) Balance H: 6 = 2b b = 3 (254) Balance O: 1+2ν O2 2a+b = 4+3 = 7 ν O2 = 3 (255) ( ) A = ν O 2 (1+3.76) = 3 4.76 = 14.28 (256) F mol 1 ( ) A = ν O 2 M O2 + ν N2 M N2 3 31.999+11.28 28.013 = = 8.943 (257) F mass M Fuel 46.069 30

8.2 Problem 2: 16.20 A container has liquid water at 15 C, 100 kpa in equilibrium with a mixture of water vapor and dry air also at 15 C, 100 kpa. How much is the vater vapor pressure and what is the saturated water vapor pressure? From the steam tables we have for a saturated liquid: P g = 1.705 kpa, v f = 0.001001 m 3 /kg (258) The liquid is at 100 kpa so it is compressed liquid still at 15 Cso from Eq. 14.15 at constant T g liq g f = vdp = v f (P P g ) (259) The vapor in the moist air is at the partial pressure P v also at 20 C so we assume ideal gas for the vapor g vap g g = vdp = RT ln P v P g (260) We have two saturated phases so g f g g (q = h f g = T s f g ) and now for equilibrium the two Gibbs functions must be the same as g vap = g liq = RT ln P v P+ g + g g = v f (P P g )+g f (261) leaving us with ln P v = v f(p P g ) 0.001001(100 1.705) = = 0.0007466 (262) P g RT 0.4615 288.15 P v = P g e 0.0007466 = 1.7063 kpa (263) This is only a minute amount above the saturation pressure. For moist air applications such differences were neglected and assumed the partial water vapor pressure at equilibrium (100% relative humidity) is P g. The pressure has to be much higher for this to be asignificant difference. 31

8.3 Problem 3: 16.24: Calculate the equilibrium constant for the reaction H 2 2H at a temperature of 2000 K, using properties from Table A.9. Compare the result with the value listed in Table A.11. From Table A.9 at 2000 K we find: Table A.11: lnk = 12.841 OK h H2 = 52942 /kmol, s H2 = 188.419 h H2 = 35375 /kmol, s H2 = 154.279 kmol K, h f = 0 (264) kmol K, h f = 217999 (265) G = H T S = H RHS H LHS T(S RHS S LHS) (266) = 2 (35375 217999) 52943 2000(2 154.279 182.419)= 213528 /kmol (267) lnk = G RT = 213528 = 12.8407 (268) 8.3145 2000 K = 2.6507 10 6 (269) 32

8.4 Problem 4: 16.34 Pure oxygen is heated from 25 C to 3200 K in a steady flow process at a constant pressure of 300 kpa. Find the exit composition and the heat transfer. The only reaction will be the dissociation of oxygen Look at initially 1 mol Oxygen and shift reaction with x O 2 2O, From A.11: K(3200) = e 3.069 = 0.046467 (270) n O2 = 1 x, n O = 2x, n tot = 1+x, y i = n i n tot (271) ( ) K = y2 O P 2 1 y 2 = 4x2 1+x O 2 P 0 (1+x) 2 1 x 3 = 12x2 1 x 2 (272) x 2 = K/12 x = 0.06211 (273) 1+K/12 y O2 = 1 x 1+x = 0.883, y O = 1 y O2 = 0.117 (274) q = n O2 ex h O2 ex + n Oex h Oex h O2 in = (1+x)(y O2 h O2 + y O h O ) 0 (275) h O2 = 106022 /kmol, h O = 249170+60767= 309937 /kmol q = 137947 /kmol O 2 (276) q = q = 4311 /kg (277) 32 33

9 Homework 9 Solutions: 9.1 Problem 1: 15:70 An isobaric combustion process receives gaseous benzene C 6 H 6 and air in a stoichiometric ratio at P 0, T 0. To limit the product temperature to 2200 k, liquid water is sprayed in after the combustion. Find the kmol of liquid water adder per kmol of fuel and the dew point of the combined products. The reaction for stoichiometric mixture with C and H balance done is: With x kmol of water added per kmol fuel the products are C 6 H 6 + ν O2 (O 2 + 3.76N 2 ) 3H 2 O+6CO 2 + cn 2 (278) O balance: 2ν O2 = 3+6 2 = 15 v O2 = 7.5 (279) N balance: c = 3.76ν O2 = 3.76 7.5 = 28.2 (280) Products: (3+x)H 2 O+6CO 2 + 28.2N 2 (281) Energy Eq.: H R = HR + xh f H 2 O liq = H P + H P = HP + x H 2 O vap +(3+x) h H2 O + 6 h CO2 + 28.2 h N2 (282) Where the extra water is shown explicitly. Rearrange to get h f HR HP 6 h CO2 28.2 h N2 3 h H2 O = x( h f H 2 O vap h f H 2 O liq + h H2 O) (283) 40756 78.114 6 103562 28.2 63362 3 83153= x[ 241826 ( 285830)+83153] (284) 525975 = x(127157) x = 4.1364 kmol/kmol f uel (285) Dew point: y V = 3+x = 0.1862 (286) 6+28.2+x P V = y V P = 0.1862 101.325= 18.86 kpa (287) B.1.2 : T dew = 58.7 C (288) 34

9.2 Problem 2: 15.81 A stoichimetric mixture of benzene C 6 H 6 and air is mixed from the reactants flowing at 25 C, 100 kpa. Find the adiabatic flame temperature. What is the error if constant specific heat at T 0 for the products from Table A.5 are used? Linear interpolation T AD = 2529 K C 6 H 6 + ν O2 O 2 + 3.76ν O2 N 2 3H 2 O+6CO 2 + 3.76ν O2 N 2 (289) ν O2 = 6+ 3 2 = 7.5 ν N 2 = 28.2 (290) H P = H P + H P = H R = H R (291) H P = H RP = 40576 78.114= 3169554 /kmol (292) H P = 6 h CO2 + 3 h H2 O + 28.2 h N2 (293) H P 2600K = 6(128074)+3(104520)+28.2(77963)= 3280600 (294) H P 2400K = 6(115779)+3(93741)+28.2(70640)= 2968000 (295) ν i C Pi = 6 0.842 44.01+3 1.872 18.015+28.2 1.042 28.013= 1146.66 kmol K (296) T = H P = 3169554 = 2764 (297) ν i C Pi 1146.66 T AD = 3062 K, 21% high (298) 35

9.3 Problem 3: 16.49 Water from the combustion of hydrogen and pure oxygen is at 3800 K and 50 kpa. Assuming we only have H 2 O, O 2 and H 2 as gases, find the equilibrium composition. With only the given components we have the reaction which at 3800 K has an equilibrium constant from A.11 of ln K = 1.906. Assume we start with 2 kmol water and let it dissociate x to the left then 2H 2 O 2H 2 + O 2 (299) Species H 2 O H 2 O 2 Initial 2 0 0 Change 2x 2x x Final 2 2x 2x x Total: 2+x Then we have which reduces to K = e 1.906 = y2 H 2 y O2 y 2 H 2 O 0.148674 = 1 4x 3 (1 x) 2 2+x ( ) P 2+1 2 = Trial and error to solve for x = 0.54, then the concentrations are P 0 ( 2x ) 2 x 2+x 2+x ( 2 2x 2+x 50 ) 2 100 (300) 1 1 4 2 or x3 = 0.297348(1 x) 2 (2+x) (301) y H2 O = 2 2x 2+x = 0.362, y O 2 = x 2+x = 0.213, y H 2 = 2x = 0.425 (302) 2+x 36

9.4 Problem 4: 16.54 A tank contains 0.1 kmol hydrogen and 0.1 kmol of argon gas at 25 C, 200 kpa and the tank keeps constant volume. To what T should it be heated to have a mole fraction of atomic hydrogen, H, of 8%? ( ) For the reaction H 2 2H, K = y2 H P 2 1 (303) y H2 P 0 Assume the dissociation shifts right with an amount x, then we get Assume we start with 2 kmol water and let it dissociate x to the left then Reaction H 2 2H also, Ar Initial 0.1 0 0.1 Change x 2x 0 Equilibrium 0.1 x 2x 0.1 y H = Total: 0.2+x 2x = 0.08 x = 0.008333 (304) 0.2+x We need to find T so K will take on the proper value; since K depends on P we need to evaluate P first. where we have n 1 = 0.2 and n 2 = 0.2+x = 0.2083 P 1 V = n 1 RT 1, P 2 V = n 2 RT 2 P 2 = P 1 n 2 T 2 n 1 T 1 (305) ( ) K = y2 H P 2 1 = (2x)2 200 y H2 P 0 (0.1 x)n 2 100 Now it is trial and error to get T 2 so the above equation is satisfied with K from A.11 at T 2. n 2 T 2 0.2 298.15 = 0.0001016T 2 (306) 3400 K : ln K = 1.519, K = 0.2189, RHS = 0.34544, error = 0.1266 (307) 3600 K : lnk = 0.611, K = 0.5428, RHS = 0.36576, error = 0.1769 (308) Linear interpolation between the two to make zero error 0.1769 T = 3400+200 = 3483.43 K (309) 0.1769+0.1266 37

9.5 Problem 5: Consider the reaction of heptane and air: ν 1 C 7H 16 + ν 2 (O 2 + 3.76N 2 ) ν 3 CO 2 + ν 4 H 2O+ν 5 CO+ν 6 NO+ν 7 NO 2 + ν 8 N 2 (310) Find the most general set of stoichiometric coefficients for the reaction. Row echelon form: C H O N 7 0 1 0 1 0 0 0 16 0 0 2 0 0 0 0 0 2 2 1 1 1 2 0 0 7.52 0 0 0 1 1 2 ν 1 ν 2 ν 3 ν 4 ν 5 ν 6 ν 7 ν 8 1 0 0 0 1/22 34/1019 16/203 14/579 0 1 0 0 0 25/188 25/188 25/94 0 0 1 0 15/22 103/441 1141/2068 175/1034 0 0 0 1 4/11 67/251 326/517 53/274 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ν 1 ν 2 ν 3 ν 4 From there the four stoichiometric equations can be written out 22 1 ν 34 5 1019 ν 16 6 203 ν 7 579 14 ν 8 = 0 188 25 ν 6 188 25 ν 7 94 25ν 8 15 22 ν 5 103 441 ν 6 1141 2068 ν 175 7 1034 ν 8 4 11 ν 5 251 67 ν 6 517 326ν 53 7 274 ν 8 = /0 (311) { ν } = /0 (312) (313) ν 1 = 1 22 ν 5 + 34 1019 ν 6 + 16 203 ν 7 14 579 ν 8 (314) ν 2 = 25 188 ν 6 25 188 ν 7 25 94 ν 8 (315) ν 3 = 15 22 ν 5 103 441 ν 6 1141 2068 ν 7 + 175 1034 ν 8 (316) ν 4 = 4 11 ν 5 67 251 ν 6 326 517 ν 7 + 53 274 ν 8 (317) 38

10 Homework 10 Solutions: 10.1 Problem 1: Plot ū R versus T [300 K,5000 K] for He, Ar, H, O, H 2, O 2, H 2 O, and H 2 O 2. Use the chemkin software package to generate values of ū(t). 12000 Helium 12000 Argon 10000 10000 8000 8000 u He /R [K] 6000 u Ar /R [K] 6000 4000 4000 2000 2000 0 0 1000 2000 3000 4000 5000 Temperature [K] 0 0 1000 2000 3000 4000 5000 Temperature [K] 3.8 x 104 Hydrogen, Monatomic 4.2 x 104 Oxygen, Monatomic 3.6 4 3.4 3.8 u H /R [K] 3.2 u O /R [K] 3.6 3.4 3 3.2 2.8 3 2.6 0 1000 2000 3000 4000 5000 Temperature [K] 2.8 0 1000 2000 3000 4000 5000 Temperature [K] 39

2.5 x 104 Hydrogen, Diatomic 2.5 x 104 Oxygen, Diatomic 2 2 1.5 1.5 u H2 /R [K] 1 u O2 /R [K] 1 0.5 0.5 0 0 1000 2000 3000 4000 5000 Temperature [K] 0 0 1000 2000 3000 4000 5000 Temperature [K] 0.5 x 104 Water 3 x 104 Hydrogen Peroxide 0 2.5 2 0.5 1.5 /R [K] u H2 O 1 1.5 u H2 O 2 /R [K] 1 0.5 0 2 0.5 2.5 1 1.5 3 0 1000 2000 3000 4000 5000 Temperature [K] 2 0 1000 2000 3000 4000 5000 Temperature [K] 40

10.2 Problem 2: Consider a Hydrogen dissociation reaction, For this reaction H 2 + H 2 2H + H 2 (318) a = 2.23 10 12 cm3 K 1 2 mole s (319) β = 0.5 (320) Ē = 92600 cal (321) mole Consider an isochoric, isothermal reaction in which n H2 = 1 kmole, n H = 0 kmole at t = 0, and for which T = 5200 K and P = 1500 kpa. a) Formulate the reaction kinetics in the form b) Find all equilibria. c) Ascertain the stability of each equilibrium point. d ρ H2 dt = f( ρ H2 ) (322) d) Use any appropriate numerical method such as Mathematica s NDSolve, MATLAB, or a straightforward Euler method to integrate the governing differential equation from the initial state to the equilibrium state. e) Repeat this usingchemkin to generate the reaction rates anddlsode to integrate the differential equations. f) Plot the Gibbs free energy, G = n H2 ḡ H2 + n H ḡ H, as a function of time. Formulation of the reaction kinetics: d ρ H2 dt = v 1 at β e Ē RT ( N ρ v k k k=1 )( 1 1 k c N k=1 ρ v k k ) (323) a = 2.23 10 9 m3 K 2 1, Ē = 387,623.6 /kmol kmol s (324) ( ρ H2 ) t=0 = P 0 RT = 1500 kpa 8.314 kmol K 5200 K = 0.034696 kmol/m3 (325) V = (n H 2 ) ( t=0 rho = 28.82 m H2 )t=0 (326) dn H2 = 1 2 n H ( ) 1 ( ) n H2 (n H2 ) t=0 = nh2 (n H2 ) t=0 (327) d ρ H2 dt n H = 2(1 kmol n H2 ), rho H = 2 2 ( 1 kmol ) 28.82 m 3 rho H2 ( = v H2 at β e Ē RT ( ρ H2 ) v H 2 ( ρ H ) v H 1 1 ( ρ H2 ) v H 2 ( ρh ) v H k c ) (328) (329) v H 2 = 2, v H = 0, v H 2 = 1, v H = 2 (330) d ρ H2 = at β e Ē RT ( ρ H2 ) (1 2 1 ρ 2 ) H dt k c ρ H2 (331) 41

Find all equilibria: k(t) = at β e Ē RT = 2.0532 10 7 (332) ( ) N P0 i=1 v i k c = e RT RT ), G = 179564.2 /kmol k c = 0.14723 (333) d ρ H2 = 2.0532 10 7 ρ 2 1 H dt 2 (1 0.14723 [2(0.034696 ρ ) H 2 )] 2 (334) Equilibria are located at the points where d ρ H 2 dt = 0. Ascertain stability: For stability, d 2 ρ H 2 dt 2 ρ = 0 is unstable ρ = 0.0129 is stable d ρ H2 dt must be negative. Therefore, ρ = 0.0933 is unstable and non-physical ρ H2 = 0 when ρ H2 = {0,0.0129,0.0933} (335) 42