Question

Water is moving with a speed of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$through a pipe with a cross-sectional area of $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}$. The water gradually descends as the pipe cross-sectional area increases to $8.0{\mathrm{cm}}^2$.

(a) What is the speed at the lower level?

(b) If the pressure at the upper level is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{Pa}$, What is the pressure at the lower level?

Short answer

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

b) The pressure of water at the lower level is $2.6\times {10}^5\mathrm{Pa}$.

Step-by-step solution

Given information

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, ${\rho }_u=1.5\times {10}^5\mathrm{Pa}$.

Determining the concept

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:

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

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

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 $\mathrm{\rho }$is density.

(a) Determining the speed of the water at the lower level

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,

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

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

$=2.5\raisebox{1ex}{$\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}$}\right.$.

(b) Determining the pressure of water at the lower level

The water flow obeys Bernoulli’s principle.

So,

$\mathrm{p\gamma }+\frac{1}{2}\mathrm{\rho gh}+{\mathrm{p}}_{\mathrm{\nu }}=\mathrm{\rho \gamma }+\frac{1}{2}\mathrm{\rho gh}$

Simplifying,

${\mathrm{p}}_{\mathrm{i}}\vee {\mathrm{P}}_{\mathrm{u}}\ast \frac{1}{2}\left({\mathrm{oq}}_{\mathrm{i}}^2+7\right)/(\mathrm{u}-1)$

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

Thus, putting the values,

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

$=2.57\times {10}^5\mathrm{Pa}$

$\approx 2.6\times {10}^5\mathrm{Pa}$

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