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 253: 5-14

Answer

a) $\dot{V}_1=0.3079\ m³/s$ b) $\dot{m}=2.696\ kg/s$ c) $\dot{V}_2=0.3705\ m³/s$, $\nu_2=6.02\ m/s$

Work Step by Step

From table A-13: Inlet ($P_1=200\ kPa, T_1=20°C$): $v_1=0.1142\ m³/kg$ $\dot{V}_1=\frac{\pi}{4}D^2\nu_1$ With $D=0.28m,\ \nu_1=5\ m/s$: $\dot{V}_1=0.3079\ m³/s$ $\dot{m}=\dot{V}_1/v_1=2.696\ kg/s$ Outet ($P_2=180\ kPa, T_2=40°C$): $v_2=0.1374\ m³/kg$ $\dot{V}_2=\dot{m}v_2=0.3705\ m³/s$ $\dot{V}_2=\frac{\pi}{4}D^2\nu_2$ $\nu_2=6.02\ m/s$
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