91Ó°ÊÓ

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:

$\mathrm{p}↔\frac{1}{2}{\mathrm{Òϲ\mu }}^2\mathrm{h}+\mathrm{constant}$

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 $\mathrm{ÒÏ}$is density.

The water flow should obey Bernoulli’s principle,

${\mathrm{p}}_{\mathrm{i}}\mathrm{v}+\frac{1}{2}{\mathrm{Òϲ\mu }}_{\mathrm{h}}\mathrm{h}+{\mathrm{p}}_1={\mathrm{ÒÏ}}_0\mathrm{v}+\frac{1}{2}{\mathrm{Òϲ\mu }}_{\mathrm{b}}\mathrm{h}+2$

Simplifying,

$\mathrm{pv}{+}_2^1\mathrm{Òϲ\mu h}+{\mathrm{p}}_{\mathrm{i}}={\mathrm{ÒÏ}}_0\mathrm{v}+\frac{1}{2}\mathrm{Òϲ\mu h}+0$

${\mathrm{ÒÏ}}_{\mathrm{i}}\mathrm{âˆ\P Ä}{\mathrm{p}}_0\mathrm{âˆ\P Ä}\frac{1}{2}\left({\mathrm{Òϲ\mu h}}_2^2\right)\mathrm{A}\left({\mathrm{n}}_0^2{−}_{\mathrm{i}}\right)$

role="math" localid="1657632752895" $\mathrm{Δ\pm è}≠\frac{1}{2}\left({}_0^2−\mathrm{Òϲ\mu }\right)\#\#\left({\mathrm{ÒÏ}}_0−\mathrm{j}\right)$

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

Thus, putting the values,

$\mathrm{Δ\pm è}=\frac{1}{2}1000\mathrm{kg}/{\mathrm{m}}^3×\left((0.4\mathrm{m}/\mathrm{s}{)}^2−(9.5\mathrm{m}/\mathrm{s}{)}^2\right)+1000\mathrm{kg}/{\mathrm{m}}^3×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)×\left(180\mathrm{m}\right)$

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

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