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 269: 5-175

Answer

$\dot{W}_e=18.8\ kW$ $\dot{W}_e^*=10.8\ kW$ $C_s=\$0.15$

Work Step by Step

$\dot{m}=\rho \dot{V},\ \rho=1\ kg/L,\ \dot{V}=10\ L/min$ $\dot{m}=10\ kg/min$ $\dot{W}_e+\dot{m}h_1=\dot{m}h_2$ $\dot{W}_e=\dot{m}c_p(T_2-T_1)$ $c_p=4.18\ kJ/kg.°C,\ T_2=43°C,\ T_1=16°C$ $\dot{W}_e=18.8\ kW$ $\dot{Q}=\epsilon \dot{m}c_p(T_2^*-T_1)$ $\epsilon=0.5,\ T_2^*=39°C$ $\dot{Q}=8.0\ kW$ $\dot{W}_e^*=10.8\ kW$ $C_s=\dot{Q}\times10\ min\times\$0.11/kWh$ $C_s=\$0.15$
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