91Ó°ÊÓ

i) The depth of the reservoir dam, $D=15\mathrm{m}$.

ii) Depth of horizontal pipe, $d=6.0\mathrm{m}$.

iii) Diameter of the horizontal pipe, $4.0\mathrm{cm}=0.04\mathrm{m}$.

iv) $t=3\mathrm{hr}=10800\mathrm{s}$.

By calculating the difference in the pressure $\left(P_d-P_a\right)$at the surface $\mathit{s}$of the water and at the depth $\mathit{d}$, find the magnitude of the frictional force between the plug and pipe wall. By applying Bernoulli's equation and calculating the volume rate of flow of water, find how much water volume exits the pipe in $\mathbf{3}\mathbf{.}\mathbf{0}$$\mathit{h}\mathit{r}$if the plug is removed. According to Bernoulli's equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

i) The magnitude of the frictional force between plug and pipe wall, $f=\left(P_d-P_{\mathrm{a}}\right)A$

ii) Bernoulli's equation,

PV $\frac{1}{2}\mathrm{ÒÏ}{g}^{2}y+=$constant

iii) Rate of flow of water $Q,Q-VA$

iv) The volume of water that flows out from the pipe in time $t$ is $R=Qt$

Where, $P$ is pressure, $V,V$ are velocities, $y$ is distance, $g$ is an acceleration due to gravity, $h$ is height, $A$ is the area, $R$ is the rate of flow of water, $f$ is the frictional force, and $\mathrm{ÒÏ}$ is density.

The force acting on the plug due to the depth of the water is,

$F_d=P_{d}A$

The force acting on the plug due to the air pressure on the surface of the water is,

$F_{\mathrm{θ}}=P_{\mathrm{θ}}A$

The force of friction acting between plug and pipe wall is,

$f=F_d-F_a$

$=P_{d}A-P_{\mathrm{Î\pm }}A$

$=\left(P_{\mathrm{σ}}-P_{\mathrm{σ}}\right)A$

Now,

$\left(\mathrm{β}gh{P}_z\right)=$

$\mathrm{ÒÏ}$is the density of fresh water $\left(\mathrm{ÒÏ}=1000\mathrm{kg}/{\mathrm{m}}^2\right)$and $h=d$.

$\left(P_d-P_{\mathrm{a}}\right)=1000\mathrm{kg}/{\mathrm{m}}^3×9.8\mathrm{m}/{\mathrm{s}}^2×6\mathrm{m}$

$=58800Pa$

Now, the area of the pipe having diameter $=0.04\mathrm{m}$that passes through the dam is,

$A=\frac{\mathrm{π}×{\text{Diameter}}^2}{4}$

$=\frac{3.14×(0.04\mathrm{m}{)}^2}{4}$

$=0.001256{\mathrm{m}}^2$

Putting these values in the equation,

$f=\left(P_d-P_a\right)A$

$=58800\mathrm{Pa}×0.001256{\mathrm{m}}^2$

$=73.853\mathrm{N}$

$≈74\mathrm{N}$

Hence, the magnitude of the frictional force between plug and pipe wall is role="math" localid="1657634728141" $74\mathrm{N}$.

Applying Bernoulli's equation, the total energy flow for the flow of water just outside the pipe is,

$pl4\frac{1}{2}\mathrm{ÒÏ}{g}^{2}y+=$constant

The total energy flow for the flow of water just inside the pipe is,

$\mathrm{ÒÏ}$и $\frac{1}{2}\mathrm{ÒÏ}{g}^{2}y12=$constant

The energy flow rate of the water in the pipe is,

$\mathrm{β}V+\frac{1}{2}\mathrm{ÒÏ}{g}^{2}y+p_1=\mathrm{ÒÏ}V+\frac{1}{2}\mathrm{ÒÏ}{g}^{2}y+{}_2=\text{constant}$

But the flow of fluid is on the same level as the ground. Thus,

$y_1=y_2$

And the speed of the water just outside the pipe is equal to zero. Thus, $V=0$It gives,

$\mathrm{β}v=p+\frac{1}{2}2$

$v=\sqrt{\frac{2(P-p)}{\mathrm{ÒÏ}}}$

The pressure difference between the outside and inside of the pipe is equal to the pressure due to depth $h=d$.

Thus,

$\left(Pgh{P}_{\mathrm{θ}}\right)=(P-p)=$

$v=\sqrt{2gd}$

Putting this value,

$v=\sqrt{\frac{2\mathrm{ÒÏ}gd}{\mathrm{ÒÏ}}}$

$=\sqrt{2×9.8\mathrm{m}/{\mathrm{s}}^2×6\mathrm{m}}$

$=10.8444\mathrm{m}/\mathrm{s}$

Now, the volume rate of flow of water from the horizontal pipe is,

$Q=AV$

$=0.001256{\mathrm{m}}^2×10.8444\mathrm{m}/\mathrm{s}$

$=0.013621{\mathrm{m}}^3/\mathrm{s}$

The volume of water that flows out from the pipe in time $t=3\mathrm{hr}$,

$R=Qt$

$=0.013621{\mathrm{m}}^3/\mathrm{s}×10800\mathrm{s}$

$=147.1021{\mathrm{m}}^3$

$≈150{\mathrm{m}}^3$

Hence, the water volume exiting the pipe in $3.0\mathrm{hr}$, if the plug is removed, is $150m^3$.