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: 4-5

Answer

$W = 2.56 kJ$

Work Step by Step

Assuming that the nitrogen is an ideal gas: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $ With: $P_1 = 130 kPa, P_2 = 100kPa, V_1 = 0.07 m³$ $T_1 = 120°C + 273.15 = 393.15 K, T_2 = 100°C + 273.15 = 373.15 K$ $V_2 = 0.0864 m³$ Therefore: $P_1V_1^n = P_2V_2^n \rightarrow \ln(\frac{P_1}{P_2}) = n\ln(\frac{V_2}{V_1})$ $n = 1.25$ The work for a polytropic process is given by: $W = \frac{P_2V_2-P_1V_1}{1-n}$ $W = 2.56 kJ$
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