Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 15 - Exercises and Problems: 56

Answer

$1.5\times 10^{-5}\frac{m^3}{s}$

Work Step by Step

According to equation of continuity $A_1V_1=A_2V_2$ $\implies \frac{\pi D_1^2V_1}{4}=\frac{\pi D_2^2V_2}{4}$ This can be rearranged as: $V_1=V_2(\frac{D_2}{D_1})^2$ We plug in the known values to obtain: $V_1=V_2(\frac{0.69}{1.9})^2=0.113V_2$ $\implies V_2^2-(0.113V_2)^2=0.225$ $V_2=\frac{\sqrt{0.225}}{1-0.113^2}=0.477\frac{m}{s}$ Now volume flow rate$=A_2V_2=\frac{\pi(0.64\times 10^{-2})^2}{4}\times 0.477=1.5\times 10^{-5}\frac{m^3}{s}$
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