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 - Page 280: 54



Work Step by Step

We know that according to Bernoulli's principle $P_1+\frac{1}{2}\rho V_1^2+\rho gh_1=P_2+\frac{1}{2}\rho V_2^2$...............eq(1) We also know that $P_2=P_{atm}+\rho gh_2$ According to equation of continuity $A_1V_1=A_2V_2$ $\implies AV_1=\frac{A}{2}V_2$ so $V_2=2V_1$ We plug in the known values in equation(1) to obtain: $P_{atm}+\frac{1}{2}\rho V_1^2+\rho gh_1=P_{atm}+\rho gh_2+\frac{1}{2}\rho (2V_1)^2$ This simplifies to: $\rho g(h_1-h_2)=\frac{1}{2}\rho (4V_1^2-V_1^2)$ $\implies \frac{2g(h_1-h_2)}{3}=V_1^2$ Thus $V=A_1V_1=A\sqrt{\frac{2g(h_1-h_2)}{3}}$
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