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Unformatted text preview: u3 we can interpolate in Table A‐17 to find T3 = 1539 K and vr3 = 6.5876. The ideal gas law gives P3 = 3898 kPa. b) An energy balance over the engine gives qin ‐ qout ‐ wnet,out = 0 or wnet,out = qin ‐ qout. (2) The heat input is given and we can find qout from an energy balance on process 1‐4: qout = u4 ‐ u1. (3) To find u4, use v
v r 4 v r 3 4 rv r 3 86.5876 52.70 . v3 Interpolation in Table A‐17 with vr4 gives u4 = 571.74 kJ/kg and T4 = 774.6 K. From (3), qout = 357.67 kJ/kg and from (2) wnet,out = 392.33 kJ/kg. c) The efficiency is th,otto w net,out
qin 392.33 0.5231 or 52.3% 750 d) The mean effective pressure (MEP) is given by MEP w net,out
v1 v 2 w net,out
392.33 494.7 kPa RT1 RT2 300 673.1 0.2870 P1
P2 95 1705 Problem 3 (10 points) A refrigerator uses refrigerant‐134a as the working fluid and operates on an ideal vapor‐
compression refrigeration cycle between 0.12 and 0.7 MPa. The mass flow rate of the refrigerant is 0...
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This homework help was uploaded on 03/02/2014 for the course CHEM 1010 taught by Professor Staff during the Spring '08 term at Utah State University.
- Spring '08