The intake in Figure has cross-sectional area of $\mathbf{0}\mathbf{.}\mathbf{74}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$and water flow at $\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. At the outlet, distance $\mathbf{D}\mathbf{=}\mathbf{180}\mathbf{}\mathbf{m}$below the intake, the cross-sectional area is smaller than at the intake, and the water flows out at $\mathbf{9}\mathbf{.}\mathbf{5}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$ into the equipment. What is the pressure difference between inlet and outlet?

i) The cross-sectional area of the inlet, ${\mathrm{A}}_{\mathrm{i}}-0.74{\mathrm{m}}^2$.

ii) The speed of the water at the inlet, $v_1=0.40\mathrm{m}/\mathrm{s}$.

iii) The depth of the outlet, $\mathrm{D}=180\mathrm{m}$.

iv) The speed of the water at the outlet, $v_0=9.5\mathrm{m}/\mathrm{s}$.

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:

$\mathrm{p}\leftrightarrow \frac{1}{2}{\mathrm{\rho g}}^2\mathrm{h}+\mathrm{constant}$

Where, $\mathrm{p}$is pressure, $\mathrm{v}$is velocity, $\mathrm{h}$is height, $\mathrm{g}$is the acceleration due to gravity, $\mathrm{h}$is height and $\mathrm{\rho }$is density.

The water flow should obey Bernoulli’s principle,

${\mathrm{p}}_{\mathrm{i}}\mathrm{v}+\frac{1}{2}{\mathrm{\rho g}}_{\mathrm{h}}\mathrm{h}+{\mathrm{p}}_1={\mathrm{\rho }}_0\mathrm{v}+\frac{1}{2}{\mathrm{\rho g}}_{\mathrm{b}}\mathrm{h}+2$

Simplifying,

$\mathrm{pv}{+}_2^1\mathrm{\rho gh}+{\mathrm{p}}_{\mathrm{i}}={\mathrm{\rho }}_0\mathrm{v}+\frac{1}{2}\mathrm{\rho gh}+0$

${\mathrm{\rho }}_{\mathrm{i}}\mathrm{\forall }{\mathrm{p}}_0\mathrm{\forall }\frac{1}{2}\left({\mathrm{\rho gh}}_2^2\right)\mathrm{A}\left({\mathrm{n}}_0^2{-}_{\mathrm{i}}\right)$

role="math" localid="1657632752895" $\mathrm{\Delta p}\ne \frac{1}{2}\left({}_0^2-\mathrm{\rho g}\right)\#\#\left({\mathrm{\rho }}_0-\mathrm{j}\right)$

Where, ${\mathrm{h}}_0-{\mathrm{h}}_1=\mathrm{D}=180\mathrm{m}$and density of water$\mathrm{\rho }=1000\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right.$

Thus, putting the values,

$\mathrm{\Delta p}=\frac{1}{2}1000\mathrm{kg}/{\mathrm{m}}^3\times \left((0.4\mathrm{m}/\mathrm{s}{)}^2-(9.5\mathrm{m}/\mathrm{s}{)}^2\right)+1000\mathrm{kg}/{\mathrm{m}}^3\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \left(180\mathrm{m}\right)$

role="math" localid="1657632844989" $=1.7\times {10}^6\mathrm{Pa}$

Hence, the pressure difference between inlet and outlet of the pipe is role="math" localid="1657632849774" $1.7\times {10}^6\mathrm{Pa}$.