Water flows at a rate of 0.035 m3/s in a horizontal pipe | StudySoup

Textbook Solutions for Fluid Mechanics

Chapter 5 Problem 76P

Question

Water flows at a rate of \(0.035 \mathrm{~m}^{3} / \mathrm{s}\)in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pressure at the centerline is measured to be 480 kPa and 445 kPa before and after the reducer, respectively, determine the irreversible head loss in the reducer. Take the kinetic energy correction factors to be 1.05.

Solution

 

Water flows at a rate of \(0.035 \mathrm{~m}^{3} / \mathrm{s}\) in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. The pressure at the centerline is measured to be 480 kPa and 445 kPa before and after the reducer, respectively. The kinetic energy correction factors is 1.05. The irreversible head loss in the reducer can be determined by solving the energy equation for steady incompressible flow through a control volume between points  1 and 2.

 

                                       

Step 1 of 2

Let us take points 1 and 2 along the centerline of the pipe before and after the reducer, respectively \(\left(z_{1}=z_{2}\right)\). The kinetic energy correction factors\(\alpha_{1}=\alpha_{2}=1.05\). The energy equation between the points 1  and 2 reduces to

\(\frac{P_1}{\rho g}+\alpha_1\frac{V_1^2}{2g}+z_1+h_{\text{pump},\ u}=\frac{P_2}{\rho g}+\alpha_2\frac{V_2^2}{2g}+z_2+h_{\text{turbine},\ e}+h_L\)

 

\(h_{L}=\frac{\left(P_{1}-P_{2}\right)}{\rho g}+\alpha \frac{\left(V_{1}^{2}-V_{2}^{2}\right)}{2 g}\)

 

\(\text { Where, } V_{1}=\frac{\frac{d V}{d t}}{A}=\frac{0.035}{\pi D_{1}^{2} / 4}=\frac{0.035}{\pi(0.15)^{2} / 4}=1.98 \mathrm{~m} / \mathrm{s}\)

 

\(V_{2}=\frac{\frac{d V}{d t}}{A}=\frac{0.035}{\pi D_{1}^{2} / 4}=\frac{0.035}{\pi(0.08)^{2} / 4}=6.96 \mathrm{~m} / \mathrm{s}\)

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full solution

Title Fluid Mechanics 2 
Author Yunus A. Cengel, John M. Cimbala
ISBN 9780071284219

Water flows at a rate of 0.035 m3/s in a horizontal pipe

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