hw12_s13_sol

# Hw12_s13_sol

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online