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: 12.45


$P_2=2.03\times 10^4~Pa$

Work Step by Step

$A_2=2A_1$ Use Fluid Continuity equation to find $v_2$: $A_1v_1=A_2v_2$ $v_2=v_1({A_1 \over 2A_1})={v_1 \over 2}={2.5~m/s \over 2}=1.25~m/s$ Since we know the pipe is horizontal there, the terms with vertical component will zero out in Bernoulli's Equation, therefore we get: $P_1+{1 \over 2}\rho v_1^2=P_2+{1 \over 2}\rho v_2^2$ $P_2=P_1+{1 \over 2}\rho v_1^2-{1 \over 2}\rho v_2^2$ $P_2=P_1+{1 \over 2}\rho(v_1^2-v_2^2)$ $P_2=(1.8 \times 10^4~Pa)+{1\over2}(1000~kg/m^3)[(2.5~m/s)^2-(1.25~m/s)^2]=2.03\times10^4~Pa$
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