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 261: 5-104

Answer

$\dot{m}=0.4005\ kg/s$ $\dot{Q}=45.1\ kW$

Work Step by Step

From table A-6: Inlet ($P_1=2\ MPa, T_1=300°C$): $v_1=0.12551\ m³/kg,\ h_1=3024.2\ kJ/kg$ Outlet ($P_2=1.8\ MPa, T_2=250°C$): $h_2=2911.7\ kJ/kg$ The mass flowrate is given by: $\dot{m}=\frac{A_1\mathcal{V}_1}{v_1}$ Given $A_1=\frac{\pi}{4}D_1^2,\ D_1=0.08\ m,\ \mathcal{V}_1=2.5\ m/s$: $\dot{m}=0.4005\ kg/s$ From the energy balance: $\dot{m}h_1=\dot{Q}+\dot{m}h_2$ Hence $\dot{Q}=45.1\ kW$
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