Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems - Page 287: 50

Answer

$Q=8.8\times10^{-2}\frac{m^3}{s}$

Work Step by Step

$P_1+pgh_1+\frac{1}{2}\rho(v_1)^2=P_2+pgh_2+\frac{1}{2}\rho(v_2)^2$ $P_1+\frac{1}{2}\rho(v_1)^2=P_2+\frac{1}{2}\rho(v_2)^2$ $A_2v_2=A_1v_2$ $v_2=\frac{A_1v_2}{A_2}$ $P_1+\frac{1}{2}\rho(v_1)^2=P_2+\frac{1}{2}\rho(\frac{A_1v_2}{A_2})^2$ $P_1-P_2=\frac{1}{2}\rho\Big((\frac{A_1v_2}{A_2})^2-(v_1)^2\Big)$ $v_1=\sqrt{\frac{2P_1-P_2}{\rho\big(\big(\frac{A_1}{A_2}\big)^2-1\big)}}$ $Q=A_1v_1=((0.030)^2\pi)\sqrt{\frac{2(33500-22600)}{1000\big(\big(\frac{(0.030)^2\pi}{(0.0225)^2\pi}\big)^2-1\big)}}=8.8\times10^{-2}\frac{m^3}{s}$
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