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 5 - Mass and Energy Analysis of Control Volumes - Problems - Page 268: 5-166

Answer

$\dot{Q}=2.23\ kW$

Work Step by Step

$\dot{m}=\frac{P\dot{V}}{RT}$ $P=95\ kPa,\ R=0.287\ kJ/kg.°C,\ T=-5°C,\ \dot{V}=0.060\ m³/s$ $\dot{m}=0.0741\ kg/s$ $\dot{Q}+\dot{m}h_1=\dot{m}h_2$ $\dot{Q}=\dot{m}c_p(T_2-T_1)$ $c_p=1.005\ kJ/kg.K,\ T_2=25°C,\ T_1=-5°C$ $\dot{Q}=2.23\ kW$
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