Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 4 - Energy Analysis of Closed Systems - Problems - Page 198: 4-30E

Answer

a) $T_2=40.23°F$ b) $70\%$ c) $Q = -4166.33\ Btu$

Work Step by Step

The initial state, saturated R-134a steam at 160 psia, from R-134a tables: $v_1=0.29339\ ft³/lbm,\ u_1 = 108.51\ Btu/lbm$ Therefore, knowing the volume to be 20 ft³, the mass is: $m = 68.169\ lbm$ At the final state: $P_2=50\ psia,\ v_2=v_1$, saturated mixture $v_{2,L} =0.01252\ ft³/lbm,\ v_{2,G} = 0.94909\ ft³/lbm$ $u_{2,L} =24.824\ Btu/lbm,\ u_{2,G} = 100.05\ Btu/lbm$ a) The final temperature is the saturation temperature: $T_2=40.23°F$ b)From $v_2 = v_{2,L} +x(v_{2,G} -v_{2,L} )$ We have the quality: $x= 0.30$ The ammount of refrigerant that has condensed is: $(1-x)\times 100\% = 70\%$ c) For a closed system stationary at constant volume: $\Delta U = Q$ $u_2 = u_{2,L} +x(u_{2,G} -u_{2,L} )$ $u_2 = 47.39\ Btu/lbm$ Therefore: $Q = -4166.33\ Btu$ On a P-v diagram, the process would start at the saturation curve and proceed on a straight vertical line downwards to below the saturation curve.
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