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 201: 4-63

Answer

$w_e = 3.27\ kW$

Work Step by Step

The energy balance for this system reduces to: $-W_e=\Delta U=mc_v\Delta T$ For an ideal gas: $PV=mRT$ Plugging in the values $P_1=100kPa, T_1=5°C=278.15K,V=120m³, R=0.2870\ kJ/kg.K$ We get $m=150.32\ kg$ With the values of $c_v=0.718\ kJ/kg.K, \Delta T=20 K$ We derive $-W_e=2158.61\ kJ=-w_e\Delta t$ Since $\Delta t = 11min = 660 s $ $w_e = 3.27\ kW$
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