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 196: 4-6


a)$W = -44 kJ$ $W = -55kJ$ c)$T = 151.83°C$

Work Step by Step

a) At the first state, the steam at $1\ MPa$ and $400°C$ has a specific volume of $v_1 = 0.30661\ m³/kg$, therefore, given its mass of $0.6\ kg$: $V_1 = 0.184 m³$ Similarly at the final state, $1\ MPa$ and $250°C$ $v_2 = 0.23275\ m³/kg$: $V_2 = 0.140 m³$ The work from isobaric cooling is $W = P(V_2-V_1)$ $W = -44 kJ$ b) With the final volume being 40% of the original volume: $V_2 = 0.074 m³$ $W = P(V_2-V_1)$ $W = -55kJ$ c)The specific volume in this final state is $v_2 = 0.1227\ m³/kg$ By the steam tables, since this value is lower than saturated steam and bigger than saturated water the final state is saturated with part of the steam already condensed: $T = 151.83°C$
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