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 7 - Entropy - Problems - Page 398: 7-31

Answer

$\Delta S=-0.132\ kJ/K$

Work Step by Step

From table A-6: Initial ($P_1=0.2\ MPa, T_1=150°C$): $v_1=0.95986\ m³/kg,\ s_1=7.2810\ kJ/kg.K$ Final ($v_2=v_1, T_2=40°C$): $x_2=0.04914,\ s_2=0.9499\ kJ/kg.K$ $m=V/v,\ V=0.02\ m^3$ $m=0.02084\ kg$ $\Delta S=m(s_2-s_1)=-0.132\ kJ/K$
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