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 393: 12.47



Work Step by Step

Convert units: ${dV\over dt}=(7200~cm^3/s)({1~m\over 100~cm})^3=0.0072~m^3/s$ $r_1=4~cm=0.04~m$ $r_2=2~cm=0.02~m$ $P_1=2.4\times10^5~Pa$ Find $v_1$: $v_1={dV/dt \over A}={0.0072 ~m^3/s\over \pi (0.04~m)^2}=1.43~m/s $ Find $v_s$ using Fluid Continuity Equation: $A_1v_1=A_2v_2$ $v_2={\pi(0.04~m)^2(1.43~m/s) \over \pi(0.02~m)^2}=5.73~m/s$ Since we know that the pipe is horizontal, the terms with vertical components will be zero in Bernoulli's Equation: $P_2=P_1+{1\over2}\rho (v_1^2-v_2^2)$ $P_2=(2.4\times10^5~Pa)+{1\over2}(1000~kg/m^3)[(1.43~m/s)^2-(5.73~m/s)^2]=2.25\times10^5~Pa$
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