91Ó°ÊÓ

i) The speed of the water at the upper level,${\mathrm{v}}_{\mathrm{u}}=5.0\mathrm{m}/\mathrm{s}$ .

ii) The area of the cross-section at the upper level, $A_u=4.0{\mathrm{m}}^2$.

iii) The depth of the lower level, $D=10\mathrm{m}$.

iv) The area of the cross-section at the lower level, $A_1=8.0{\mathrm{m}}^2$.

v) The pressure at the upper level, ${\mathrm{ÒÏ}}_u=1.5×{10}^5\mathrm{Pa}$.

Determine the speed of the water at the lower level using the equation of continuity. Then, using Bernoulli’s equation, find the pressure of water at the lower level. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

$\mathrm{p}∨−\frac{1}{2}{\mathrm{ÒÏg}}^2\mathrm{h}+\mathrm{constant}$

$\mathrm{Av}=\mathrm{constant}$

Where, $\mathrm{p}$is pressure,$\mathrm{v}$is velocity, $\mathrm{h}$is height, $\mathrm{g}$is an acceleration due to gravity, $\mathrm{h}$is height, $\mathrm{A}$is area, and $\mathrm{ÒÏ}$is density.

The water flow through the pipe obeys the equation of continuity,

${\mathrm{A}}_{\mathrm{u}}{\mathrm{V}}_{\mathrm{u}}={\mathrm{A}}_1{\mathrm{V}}_1$

Hence,

${\mathrm{V}}_1=\frac{{\mathrm{A}}_{\mathrm{ii}}{\mathrm{V}}_{\mathrm{u}}}{{\mathrm{A}}_{\mathrm{i}}}$

$=\frac{4.0{\mathrm{m}}^2×5.0\mathrm{m}/\mathrm{s}}{8.0{\mathrm{m}}^2}$

$=2.5\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Hence, the speed of the water at the lower level is $2.5\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

The water flow obeys Bernoulli’s principle.

So,

$\mathrm{\pm èγ}+\frac{1}{2}\mathrm{ÒÏgh}+{\mathrm{p}}_{\mathrm{ν}}=\mathrm{ÒÏγ}+\frac{1}{2}\mathrm{ÒÏgh}$

Simplifying,

${\mathrm{p}}_{\mathrm{i}}∨{\mathrm{P}}_{\mathrm{u}}∗\frac{1}{2}\left({\mathrm{oq}}_{\mathrm{i}}^2+7\right)/(\mathrm{u}−1)$

Where, $h_{is}−h_t=D=10\mathrm{m}$and $\mathrm{ÒÏ}=\text{Density of water}=1000\mathrm{kg}/{\mathrm{m}}^3$.

Thus, putting the values,

role="math" ${\mathrm{p}}_{\mathrm{i}}=\left(1.5×{10}^5\mathrm{Pa}\right)+\frac{1}{2}1000\mathrm{kg}/{\mathrm{m}}^3×\left((5\mathrm{m}/\mathrm{s}{)}^2−(2.5\mathrm{m}/\mathrm{s}{)}^2\right)+1000\mathrm{kg}/{\mathrm{m}}^3×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)×\left(10\mathrm{m}\right)$

$=2.57×{10}^5\mathrm{Pa}$

$≈2.6×{10}^5\mathrm{Pa}$

Hence, the pressure of water at the level is $2.6×{10}^5\mathrm{Pa}$.