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

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

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

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

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

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:

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

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

The water flow should obey Bernoulli’s principle,

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

Simplifying,

$$\begin{gathered} p_{l}v+\frac{1}{2}p{g}^{2}h+p_l=p_{0}v+\frac{1}{2}p{g^2}_{0}h+{}_0 \\ {p}_{l}v{p}_{0}v\frac{1}{2}\left(p{g}^2-h{l}^2\right)h\left(o-l\right) \\ \Delta pv\frac{1}{2}v\left({{{}^{}}^2}_0-p{g}^2\right)hh\left({}_0-l\right) \end{gathered}$$

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

Thus, putting the values,

$$\begin{gathered} \Delta p=\frac{1}{2}1000kg/m^3\times \left({\left(0.4m/s\right)}^2-{\left(9.5m/s\right)}^2+1000kg/m^3\times \left(9.8m/s^2\right)\times \left(180m\right)\right) \\ =1.7\times {10}^{6}pa \end{gathered}$$

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