2300_hw_6_sol

# 2300_hw_6_sol - cycle with three processes. The boundary...

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Homework 6 Solutions. 3-27 Complete the following table for H 2 O : 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 - - - Compressed liquid 350.0 800 3162.2 - - - Superheated vapor 3-28 Complete the following table for Refrigerant-134a : T, ° C P, kPa v , m 3 / kg Phase description -8 320 0.0007569 Compressed liquid 30 770.64 0.015 Saturated mixture -12.73 180 0.11041 Saturated vapor 80 600 0.044710 Superheated vapor 3-88 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, R = 0.4615 kPa·m 3 /kg·K, T cr = 647.1 K, P cr = 22.06 MPa Analysis ( a ) From the ideal gas equation of state, error) (17.6% /kg m 0.03106 3 = = = kPa) (10,000 K) K)(673 /kg m kPa (0.4615 3 P RT ( b ) From the compressibility chart (Fig. A-15), H 2 O 10 MPa 400 ° C

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84 . 0 1.04 K 647.1 K 673 0.453 MPa 22.06 MPa 10 = = = = = = = Z T T T P P P cr R cr R Thus, error) (1.2% /kg m 0.02609 3 = = = /kg) m 3106 (0.84)(0.0 3 ideal v Z ( c ) From the superheated steam table (Table A-6), } /kg m 0.02644 3 = ° = = C 400 MPa 10 T P 4-25 A piston-cylinder device contains air gas at a specified state. The air undergoes a

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Unformatted text preview: cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a). Analysis For the isothermal expansion process: 3 1 1 m 01341 . kPa) (2000 K) 273 50 kJ/kg.K)(3 287 . ( kg) 15 . ( = + = = P mRT V 3 2 2 m 05364 . kPa) (500 K) 273 50 kJ/kg.K)(3 287 . ( kg) 15 . ( = + = = P mRT kJ 37.18 = = =-3 3 3 1 2 1 1 2 1 , m 01341 . m 05364 . ln ) m 41 kPa)(0.013 (2000 ln P W b For the polytropic compression process: 3 3 2 . 1 3 2 . 1 3 3 3 2 2 m 01690 . kPa) (2000 ) m 64 kPa)(0.053 (500 = = = n n P P kJ-34.86 =--=--=-2 . 1 1 ) m 64 kPa)(0.053 (500 ) m 90 kPa)(0.016 (2000 1 3 3 2 2 3 3 3 2 , n P P W b For the constant pressure compression process: kJ-6.97 =-=-=-3 3 1 3 1 3 , 0.01690)m 41 kPa)(0.013 (2000 ) ( P W b The net work for the cycle is the sum of the works for each process kJ-4.65 =-+-+ = + + =---) 97 . 6 ( ) 86 . 34 ( 18 . 37 1 3 , 3 2 , 2 1 , net b b b W W W W Air 2 MPa 350 C...
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## 2300_hw_6_sol - cycle with three processes. The boundary...

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