PS9_sol_S08(2) - Problem Set 9 (2.006, Spring 08) Problem 1...

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Problem Set 9 (2.006, Spring 08) Problem 1 a) and h) b) R-134a, 21.71 o env TC = (given conditions are in bold characters) State () P Pa o (/) hk J kg (/ ) s kJ kg K x Note 1 7×10 5 26.71 71.183 0.259 0 Saturated liquid 2 1.3273×10 5 (1) 71.183 (4) 0.2976 (9) 2 phase 3 1.3273×10 5 -20 220.74 0.87227 1 Saturated vapor 4 7×10 5 (2) 57.38 (8) 278.167 (7) Superheated vapor 4R 7×10 5 (3) 255.196 (6) 0.87227 (5) Reversible state 4 Start from state 1 and 3. Since you know both temperature and quality, you can determine all the properties at state 1 and 3 from pure substance tables. From the information, use the listed condition in the figure to find the properties at the other states. The red numbers in parenthesis show an example of the order of your filling in the table. (Sure, there are so many other possible orders.) Remember that you should have two independent properties at a certain state to determine other properties. 0.0 0.2 0.4 0.6 0.8 1.0 -30 0 30 60 90 120 R-134a Temperature ( o C) Specific entropy (kJ/kgK) 21.71 o env = 1 2 3 4 0.0 0.2 0.4 0.6 0.8 1.0 -30 0 30 60 90 120 Specific entropy (kJ/kgK) 34.39 o env = 1 2 3 4 144 12 23 34 32 1 3 5 5 R R RR R c act evap act o env o freez PPP hh PP ss W m Qm h h COP Wm h h h h TT C C η = = = = = =− == −= =+ & & & & & & & & env T 15 258.15 o freez K =
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(1): 23 P P = , (2,3): 144 R P PP == , (4): 12 hh = , (5): 34 R s s = , (6): interpolation between h=250.79kJ/kg, s=0.8579kJ/kgK at T=30 o C and h=255.88kJ/kg, s=0.8745kJ/kgK at T=35 o C from the P=7×10 5 table. (7): () ( ) /0 . 6 Rc η −− = = (8): interpolation between h=275.81kJ/kgK at T=55 o C and h=280.76kJ/kgK at T=60 o C The temperature of the fluid exiting the compressor = T 4 = 57.4 o C c) State 2 is in 2 phase state. (9): 22 , 2 2, 71.183 7.8255 / 0.2976 212.91 / f fg kJ kg x hk J k g = d) ( ) 220.74 / 278.167 / 57.43 / 11 comp act W W h h kJ kg kJ kg kJ kg kg kg = = e) ( ) 32 220.74 71.183 / 149.56 / 1 evap Q h h kJ kg kJ kg kg =−= = f) 149.56 / 2.60 57.43 / evap comp Q kJ kg COP Wk J k g = 258.15 7.03 21.71 15 freez C Carnot H C env freez T T K COP TT T T K = = g) Heat transfer due to the temperature difference between the freezer compartment and the environment is ( ) 0.54 / 21.71 15 19.823 oo env freez QU AT T WK C C W =− = = & This is the amount of heat flux that we should get rid of from the freezer compartment. Since the compressor works only 1/6 of the time, the capacity of heat removal should be 6 times as much as Q & .
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This note was uploaded on 10/06/2011 for the course MECHANICAL 2.006 taught by Professor Blah during the Spring '08 term at MIT.

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PS9_sol_S08(2) - Problem Set 9 (2.006, Spring 08) Problem 1...

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