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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The intake in Figure has cross-sectional area of ${\mathbf{0}}{\mathbf{.}}{\mathbf{74}}{{\mathbit{m}}}^{{\mathbf{2}}}$ and water flow at ${\mathbf{0}}{\mathbf{.}}{\mathbf{40}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$. At the outlet, distance ${\mathbit{D}}{\mathbf{=}}{\mathbf{180}}{\mathbit{m}}$ below the intake, the cross-sectional area is smaller than at the intake, and the water flows out at ${\mathbf{9}}{\mathbf{.}}{\mathbf{5}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$ into the equipment. What is the pressure difference between inlet and outlet?

The pressure difference between inlet and outlet of the pipe is $1.7×{10}^{6}Pa$.

See the step by step solution

## Step 1: Given data

1. The cross-sectional area of the inlet, ${\mathbit{A}}_{\mathbf{t}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{74}{\mathbit{m}}^{\mathbf{2}}$.

2. The speed of the water at the inlet, ${\mathbit{v}}_{\mathbf{t}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbit{m}\mathbf{/}\mathbit{s}$.

3. The depth of the outlet,$\mathbit{D}\mathbf{=}\mathbf{180}\mathbit{m}$.

4. The speed of the water at the outlet, ${\mathbit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{5}\mathbit{m}\mathbf{/}\mathbit{s}$.

## Step 2: Determining the concept

By applying Bernoulli’s principle, determine the pressure difference between inlet and outlet. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

The equation is as follows:

$\mathbit{p}\mathbit{v}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbit{p}{\mathbit{g}}^{\mathbf{2}}\mathbit{h}\mathbf{+}\mathbf{}\mathbit{c}\mathbit{o}\mathbit{n}\mathbit{s}\mathbit{t}\mathbit{a}\mathbit{n}\mathbit{t}$

Where, p is pressure, v is velocity, h is height, g is the acceleration due to gravity, h is height and $p$ is density.

## Step 3: Determining the pressure difference between the inlet and outlet of the pipe

The water flow should obey Bernoulli’s principle,

${\mathbit{p}}_{\mathbf{t}}\mathbit{v}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbit{p}{\mathbit{g}}^{\mathbf{2}}\mathbit{h}\mathbf{+}\mathbf{}\mathbf{}\mathbit{p}\mathbf{}{\mathbf{1}\mathbf{}\mathbf{=}\mathbf{}{{\mathbf{p}}_{\mathbf{0}}}_{}}_{}\mathbit{v}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbit{p}{{\mathbit{g}}^{\mathbf{2}}}_{\mathbf{0}}\mathbit{h}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{2}}_{}\mathbf{}$

Simplifying,

${p}_{l}v+\frac{1}{2}p{g}^{2}h+{p}_{l}={p}_{0}v+\frac{1}{2}p{{g}^{2}}_{0}h+{}_{0}\phantom{\rule{0ex}{0ex}}{p}_{l}v{p}_{0}v\frac{1}{2}\left(p{g}^{2}-h{l}^{2}\right)h\left(o-l\right)\phantom{\rule{0ex}{0ex}}\Delta pv\frac{1}{2}v\left({{{}^{}}^{2}}_{0}-p{g}^{2}\right)hh\left({}_{0}-l\right)$

Where, ${\mathbit{h}}_{\mathbf{0}}\mathbf{-}{\mathbit{h}}_{\mathbf{l}\mathbf{}}\mathbf{=}\mathbit{D}\mathbf{=}\mathbf{180}\mathbit{m}$ and density of water $\mathbit{p}\mathbf{=}\mathbf{1000}\mathbit{k}\mathbit{g}\mathbf{/}{\mathbit{m}}^{\mathbf{3}}$

Thus, putting the values,

$\Delta p=\frac{1}{2}1000kg/{m}^{3}×\left({\left(0.4m/s\right)}^{2}-{\left(9.5m/s\right)}^{2}+1000kg/{m}^{3}×\left(9.8m/{s}^{2}\right)×\left(180m\right)\right)\phantom{\rule{0ex}{0ex}}=1.7×{10}^{6}pa$

Hence, the pressure difference between inlet and outlet of the pipe is$\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{6}\mathbf{}}\mathbit{p}\mathbit{a}$ .