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PS9_sol_S08(2)

# 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 T C = (given conditions are in bold characters) State ( ) P Pa ( ) o T C ( / ) h kJ 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 T C = 1 2 3 4 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) 34.39 o env T C = 1 2 3 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 4 4 1 2 2 3 3 4 3 4 3 4 3 4 3 4 3 2 3 2 3 4 3 4 1 3 5 5 R R R R R c act evap act o env o freez P P P h h P P s s W m h h h h W m h h h h Q m h h h h COP W m h h h h T T C T T C η = = = = = = = = = = = = = − = + = & & & & & & & & env T 15 258.15 o freez T C K = − =

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