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 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 267: 5-161

Answer

$\dot{W}_{flow}=0\ kW$

Work Step by Step

From table A-12: Inlet ($P_1=320\ kPa=P_2=P_3, x_1=0.55$): $v_1=0.03537\ m³/kg,\ v_G=v_2=0.06368\ m³/kg,\ v_L=v_3=0.0007771\ m³/kg$ $\dot{m}=\dot{V}/v,\ \dot{V}_1=0.006\ m³/s$ $\dot{m}_1=0.1696\ kg/s$ $\dot{m}_2=x_1\dot{m}_1=0.09329\ kg/s$ $\dot{m}_3=(1-x_1)\dot{m}_1=0.07633\ kg/s$ $\dot{W}_{flow}=\dot{m}_2P_2v_2+\dot{m}_3P_3v_3-\dot{m}_1P_1v_1$ $\dot{W}_{flow}=0\ kW$
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