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 252: 5-6E

Answer

$\nu_1=13.8\ ft/s$ $\dot{m}=35.1\ lbm/min$

Work Step by Step

$\nu_1=\frac{\dot{V}_1}{A_1}=\frac{4\dot{V}_1}{\pi D^2}$ With $\dot{V}_1=450\ ft³/min,\ D=10\ in$: $\nu_1=13.8\ ft/s$ $\dot{m}=\rho_1\dot{V}_1$ With $\rho_1=0.078\ lbm/ft³$ $\dot{m}=35.1\ lbm/min$
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