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: 4-9

Answer

$W = 135.52\times10^3\ kJ$

Work Step by Step

Assuming the process occurs at a negligible pressure change: From the steam tables: The initial point: Saturated, $T_1=T_2=200°C, P_1=P_2 = 1554.9\ kPa, v_{1,L} = 0.001157 m³/kg, v_{1,G} = 0.12721 m³/kg $ Calculating the specific volume at the final state with steam quality $x=0.8$: $v_2 = v_{1,L} + x(v_{1,G} -v_{1,L} )$ $v_2 = 0.102 m³/kg$ The mass, from the initial volume of $V_1 = 1\ m³$ is: $m = V_1/v_{1,L}$ $m = 864.304\ kg$ The work at constant pressure is given by: $W = mP(v_2-v_{1,G})$ $W = 135.52\times10^3\ kJ$
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