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 400: 7-49

Answer

$\dot{W}_s=2285\ kW$

Work Step by Step

From table A-6: Outlet 2 ($P_3=50\ kPa, T_3=100°C$): $s_3=7.6953\ kJ/kg.K,\ h_3=2682.4\ kJ/kg$ Inlet ($P_1=3\ MPa, s_1=s_3$): $h_1=3851.2\ kJ/kg$ Outlet 1 ($P_2=0.5\ MPa, s_2=s_3$): $h_2=3206.5\ kJ/kg$ From the data given: $\dot{m}_1=2\ kg/s,\ \dot{m}_2=0.05\dot{m}_1,\ \dot{m}_3=0.95\dot{m}_1$ From the energy balance: $\dot{m}_1h_1=\dot{m}_2h_2+\dot{m}_3h_3+\dot{W}_s$ $\dot{W}_s=\dot{m}_1(h_1-0.05h_2-0.95h_3)$ $\dot{W}_s=2285\ kW$
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