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16-1 A cylindrical tank 2 m long mounted with its axis horizontal is to separate liquid ammonia from ammonia vapor. The ammonia vapor bubbles through the liquid and 1.2 mvelocity of the vapor is limited to 1 m/s and the vessel is to operate with the liquid level two-thirds of the^3 /s leaves the surface of the liquid. If the diameter from the bottom, what must the diameter of the tank be? Solution: L = 2 m Surface Area = A = (1.2 m^3 /s) / (1 m/s) = 1.2 m^2 Width = W = A/L = (1.2 m^2 ) / (2 m) = 0.6 m
32 D= 21 D+^ x D^2 W^2 2 x =^1 − D^2 W^2 2
D^1
D^1
D D^2 W^2
98 D^2 =W^2 =^ (0.6^ )^2
D = 0.636 m - - - Ans. 16-2. A liquid subcooler as shown in Fig. 16-14 receives liquid ammonia at 30 C and subcools 0.6 kg/s to 5 C.Saturated vapor leaves the subcooler for the high-stage compressor at -1 C. Calculate the flow rate of ammonia that evaporated to cool the liquid. Solution: Refer to Fig. 16-14. Liquid ammonia at 30 C, Table A-3.h Subcooled ammonia at 5 C, Table A-3.^1 = hf^ = 341.769 kJ.kg h 2 = hf = 223.185 kJ/kg Saturated vapor ammonia at -1 C, Table A-3.h 3 = hg = 1460.62 kJ/kg Heat Balance: w 1 (h 1 - h 2 ) = w 2 (h 3 - h 1 )
(0.6)(341.769 - 223.185) = w 2 (1460.62 - 341.769) w 2 = 0.0636 kg/s - - - Ans.
16-3. In a refrigerant 22 refrigeration system the capacity is 180 kw at a temperature of -30 C. The vapor from theevaporator is pumped by one compressor to the condensing pressure of 1500 kPa. Later the system is revised to a two-stage compression operating on the cycle shown in Fig. 16-6 with intercooling but noremoval of flash gas at 600 kPa. (a)(b) Calculate the power required by the single compressor in the original system.Calculate the power required by the two compressor in the revised system.
Solution: (a) Original system
At 1, -30 C, Table A-6.h s^1 = 393.138 kJ/kg 1 = 1.80329 kJ/kg.K At 2, 1500 kPa condensing pressure = 39.095 C condensing temp.Table A-7, constant entropy h 2 = 450.379 kJ/kg h 3 = h 4 = 248.486 kJ/kg w = 180 kw / (h 1 - h 4 ) w = 180 / (393.138 - 248.486)w = 1.2444 kg/s P = w (h 2 - h 1 ) P = 1.2444 (450.379 - 393.138) P = 71.23 kw - - - Ans. (b) Revised system (Fig. 16-6).
w 2 + 1.2444 = w 3 w 2 = w 3 - 1. (w 3 - 1.2444)(248.486) + (1.2444)(424.848) = w 3 (407.446) w 3 = 1.38063 kg/s P = w 1 (h 2 - h 1 ) + w 3 (h 4 - h 3 ) P = (1.2444)(424.848 - 393.138) + (1.38063)(430.094 - 407.446) P = 70.73 kw - - - Ans.
16-4. A refrigerant 22 system has a capacity of 180 kw of an evaporating temperature of -30 C when thecondensing pressure is 1500 kPa. (a)(b) Compute the power requirement for a system with a single compressor.Compute the total power required by the two compressors in the system shown in Fig. 16-7 where there is no intercooling but there is flash-gas removal at 600 kPa? Solution:
(a) Original system
At 1, -30 C, Table A-6.h s^1 = 393.138 kJ/kg 1 = 1.80329 kJ/kg.K At 2, 1500 kPa condensing pressure = 39.095 C condensing temp.Table A-7, constant entropy h 2 = 450.379 kJ/kg h 3 = h 4 = 248.486 kJ/kg w = 180 kw / (h 1 - h 4 ) w = 180 / (393.138 - 248.486)w = 1.2444 kg/s
P = w (h 2 - h 1 ) P = 1.2444 (450.379 - 393.138) P = 71.23 kw - - - Ans.
(b) For Fig. 16-7.
At 1, -30 C, Table A-6h s^1 = 393.138 kJ/kg 1 = 1.80329 kJ/kg.K At 2, 1500 kPa, Sat. Temp. = 39.095 C (Table A-7)Constant Entropy h 2 = 450.379 kJ/kg At 3, 600 kPa, Sat. Temp. = 5.877 C (Table A-6)h s^3 = 407.446 kJ/kg 3 = 1.74341 kJ/kg.K At 4, 1500 kPa, Sat. Temp. = 39.095 C (Table A-7)h 4 = 430.094 kJ/kg At 5, 1500 kPa, Sat. Temp. = 39.095 C (Table A-6)h 5 = 248.486 kJ/kg At 7, 600 kPa, Say. Temp. = 5.877 C (Table A-6)h 7 = 206.943 kJ/kg h 6 = h 5 = 248.486 kJ/kg h 8 = h 7 = 206.943 kJ/kg
At 1, -40 C, Table A-3.h s^1 = 1407.26 kJ/kg 1 = 6.2410 kJ/kg.K At 2, 0 C, Fig. A-1, Constant Entropyh 2 = 1666 kJ/kg At 3, 0 C, Table A-3h 3 = 1461.70 kJ/kg At 4, 35 C, Fig. A-1h 4 = 1622 kJ/kg At 5, 35 C, Table A-3.h 5 = 366.072 kJ/kg At 6, h 6 = h 5 = 366.072 kJ/kg At 7, 0 C, Table A-3h 7 = 200 kJ/kg At 8, h 8 = h 7 = 200 kJ/kg. w 1 = entering low-stage compressor w 1 = 250 / (h 1 - h 8 ) w 1 = 250 / (1407.26 - 200) w 1 = 0.2071 kg/s w 2 = entering high-stage compressor leaving intercooler and flashtank Heat balance through intercooler and flashtank. w 2 (h 3 - h 6 ) = w 1 (h 2 - h 7 ) w 2 (1461.70 - 366.072) = (0.2071)(1666 - 200) w 2 = 0.2771 kg/s w 3 = entering intermediate temperature evaporator w 3 = 150 kw / (h 3 - h 6 ) = 150 / (1461.70 - 366.072) w 3 = 0.1369 kg/s Total refrigerant compressed by high=pressure compressor= w = 0.4140 kg/s - - - Ans^2 + w^3 = 0.2771 + 0.1369. 16-6. A two-stage refrigerant 22 system that uses flash-gas removal and intercooling serves a single low-temperature evaporator, as in Fig. 16-10a. The evaporator temperature is -40 C, and the condensing temperature is 30 C. The pumping capacity of the high- and low-stage compressors is shown in Fig. 16-18.What is (a) the refrigerating capacity of the system and (b) the intermediate pressure?
Solution: Refer to Fig. 16-18 and Fig. 16-10a.
At 1, -40 C, Table A-6h s^1 = 388.609 kJ/kg At 5, 30 C, Table A-6^1 = 1.82504 kJ/kg.K h 5 = 236.664 kJ/kg Trial 1
pi = ( 105 )( 1192 ) = 354 kPa
At 354 kPa, Sat. Temp. = -10 C At 2, -10 C, Constant Entropy, Table A-7h 2 = 417.46 kJ/kg At 3, -10 C, Table A-6h s^3 = 401.555 kJ/kg 3 = 1.76713 kJ/kg.K At 4, 30 C, Constant Entropy, table A-7h 4 = 431.787 kJ/kg At 6, h6 = h5 = 236.664 kJ/kgAt 7, -10 C, Table A- h 7 = 188.426 kJ/kg At 8, h 8 = h 7 - 188.426 kJ/kg w 1 = low-stage compressor w 2 = high-stage compressor
