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: 4-34

Answer

$V = 223.8\ V$

Work Step by Step

At the initial state (175 kPa, saturated water): $v_1=0.001057\ m³/kg,\ h_1=487.01\ kJ/kg$ At the final state (175 kPa, saturated mixture, with 50% quality): $h_2=1593.56\ kJ/kg$ For this electrically heated, insulated, and agitated constant-pressure stationary closed system, the energy balance reduces to: $-W_e-W_{sh}=\Delta H$ $W_e=-VI\Delta t$ $W_p=-400kJ$ $\Delta H = m(h_2-h_1)$ $m = \frac{V_1}{v_1} = 4.73\ kg$ $W_e = -4834.39\ kJ$ $V = 223.8\ V$ On the P-v diagram, the process starts at the saturation curve and proceeds in a straight horizontal line to a point below the curve.
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