University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 12 - Fluid Mechanics - Problems - Exercises - Page 392: 12.43

Answer

$P_1=1.47 \times 10^5~Pa$

Work Step by Step

$P1+\rho gy_1+{1\over 2}\rho v_1^2=P_2+\rho gy_2+{1\over 2}\rho v_2^2$ Since the main has a much larger diameter than the hose, $v_1=0$ and $v_2=0$ because the final velocity at the top of the stream will be 0. There is no pressure at the top of the stream so $P_2=0$. We will consider the main as $y_1=0$. Therefore, we get the following equation: $P_1=\rho g y_2$ $P_1=(1000~kg/m^3)(9.8~m/s^2)(15~m)=1.47 \times 10^5~Pa$
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