91Ó°ÊÓ

The given data can be listed below as,

• The vertical cylinder tank diameter and pokes diameter are, d₁= 2.00 m and d₂= 0.0200m respectively.
• The depth of water is, h = 0.800m .
• The gauge pressure at the surface of water is, $p_g=5×{10}^3\text{ P²¹}$.

The total energy per unit mass of a fluid flowing i.e. the sum of the fluid's kinetic energy, potential energy, and pressure energy at any point in its interior is equal to a constant value.

Part (a)

The relation of Bernoulli’s equation is expressed as,

$p_g+\mathrm{ÒÏ}gh+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2=\frac{1}{2}\mathrm{ÒÏ}{v}_2^2$

Here, $p_g$is the gauge pressure, $\mathrm{ÒÏ}$ is the density of water and g is the gravitational acceleration, v₁ and v₂ are the velocity at top and bottom level of water.

The relation of continuity equation is expressed as,

$A_1{v}_1=A_2{v}_2$

Here A₁ and A₂ are the area of vertical cylinder tank and pokes. (Here $A=\frac{\mathrm{Ï€}}{4}{d}^2$).

Then continuity equation can be expressed as,

$\begin{array}{c}\frac{\mathrm{Ï€}}{4}{d}_1^2{v}_1=\frac{\mathrm{Ï€}}{4}{d}_2^2{v}_2\\ d_1^2{v}_1=d_2^2{v}_2\\ v_1=\frac{d_2^2}{d_1^2}{v}_{2.......\left(i\right)}\end{array}$

Substitute $v_1=\frac{{d_2}^2}{{d_1}^2}{v}_2$in Bernoulli’s equation then,

$\begin{array}{c}{p}_g+\mathrm{ÒÏ}gh+\frac{1}{2}\mathrm{ÒÏ}{\left(\frac{d_2^2}{d_1^2}{v}_2\right)}^2=\frac{1}{2}\mathrm{ÒÏ}{v}_2^2\\ \frac{1}{2}\mathrm{ÒÏ}{v}_2^2−\frac{1}{2}\mathrm{ÒÏ}{\left(\frac{d_2^2}{d_1^2}{v}_2\right)}^2=p_g+\mathrm{ÒÏ}gh\\ \frac{1}{2}\mathrm{ÒÏ}{v}_2^2\left(1−{\left(\frac{d_2}{d_1}\right)}^4\right)=p_g+\mathrm{ÒÏ}gh\\ v_2=\sqrt{\frac{2\left(p_g+\mathrm{ÒÏ}gh\right)}{\mathrm{ÒÏ}\left(1−{\left(\frac{d_2}{d_1}\right)}^4\right)}}\end{array}$.......(ii)

Substitute 5000Pa for p_g , 1000kg/m³ for $\mathrm{ÒÏ}$ , 9.81 m/s²for g, 0.8m for h, 0.02 m for d₂, and 2m for d₁ in equation (ii).

$\begin{array}{c}{v}_2=\sqrt{\frac{2\left(5000\text{ P²¹}+1000\text{ k²µ}/{\text{m}}^{\text{3}}×9.81\text{ }{\text{m}/\text{s}}^2×0.8\text{″¾}\right)}{1000\text{ k²µ}/{\text{m}}^{\text{3}}\left(1−{\left(\frac{0.02\text{″¾}}{2\text{″¾}}\right)}^4\right)}}\\ =5.07\text{″¾}/\text{s}\end{array}$

This is the value of speed of the water as it emerges.

The value of the speed v₁, when the top of the tank is opened to air,

$v_1=\sqrt{2gh}$

Substitute 9.81 m/s²for g , 0.8m for h in above equation.

$\begin{array}{c}{v}_1=\sqrt{2×9.81\text{″¾}/{\text{s}}^2×0.8\text{″¾}}\\ v_1=3.96\text{″¾}/\text{s}\end{array}$

The ratio speeds will be,

$\begin{array}{c}\frac{v_2}{v_1}=\frac{5.07\text{″¾}/\text{s}}{3.96\text{″¾}/\text{s}}\\ =1.28\end{array}$

Hence the speed of the water as it emerges from the hole is 5.07 m/s, and the ratio of this speed to the efflux speed if the top of the tank is open to the air is 1.28.

Part (b)

The initial velocity of dissension of water is expressed as,

$v_1^2=\frac{2\left(p_g+\mathrm{ÒÏ}gh\right)}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}$

Substitute 5000Pa for p_g , 1000kg/m³ for $\mathrm{ÒÏ}$ , 9.81 m/s²for g, 0.8m for h, 0.02 m for d₂, and 2m for d₁ in above equation.

$\begin{array}{c}{v}_1^2=\frac{2\left(5000\text{ P²¹}+1000\text{ k²µ}/{\text{m}}^{\text{3}}×9.81\text{″¾}/{\text{s}}^{\text{2}}×0.8\text{″¾}\right)}{1000\text{ k²µ}/{\text{m}}^{\text{3}}\left({\left(\frac{2\text{″¾}}{0.02\text{″¾}}\right)}^4−1\right)}\\ v_1=5.069×{10}^{−4}\text{″¾}/\text{s}\end{array}$

The relation of moment when the water level drops, then it is expressed as,

$\begin{array}{c}v{\text{'}}^2=\frac{2\left(\mathrm{ÒÏ}g\left(h−x\right)+p_g\right)}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}\\ =\frac{2\mathrm{ÒÏ}g\left(h−x\right)}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}+\frac{2p_g}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}\end{array}$

Here, $v{{\text{'}}^2}^{}$is the velocity of water and x is height of water drop.

When , x = h then the final speed of dissension is expressed as,

$v_{1f}^2=\frac{2p_g}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}$

Here, v_1fis the final speed of dissension.

Substitute 5000Pa for p_g , 1000kg/m³ for $\mathrm{ÒÏ}$ , 9.81 m/s²for g, 0.8m for h, 0.02 m for d₂, and 2m for d₁ in above equation.

$\begin{array}{c}{v}_{1f}^2=\frac{2×5000\text{ P²¹}}{1000\text{ k²µ}/{\text{m}}^{\text{3}}\left({\left(\frac{2\text{″¾}}{0.02\text{″¾}}\right)}^4−1\right)}\\ v_{1f}=3.16×{10}^{−4}\text{″¾}/\text{s}\end{array}$

When the water level behaves like uniform decelerating motion then,

$\begin{array}{c}{t}_{desc}=\frac{h}{v_{avg}}\\ =\frac{h}{\frac{v_1+v_{1f}}{2}}\\ =\frac{2h}{v_1+v_{1f}}\end{array}$

Here, t_desc is the time of deceleration.

Substitute 5000Pa for p_g , 1000kg/m³ for $\mathrm{ÒÏ}$ , 9.81 m/s²for g, 0.8m for h, 0.02 m for d₂, and 2m for d₁ in above equation.

$\begin{array}{c}{t}_{desc}=\frac{2×0.8\text{″¾}}{5.069×{10}^{−4}\text{″¾}/\text{s}+3.16×{10}^{−4}\text{″¾}/\text{s}}\\ =1944.3\text{ s}×\frac{1\text{ }\min }{60\text{ s}}\\ t_{desc}=32.4\text{″¾in}\end{array}$

Because air will not be compressed above the water surface then p_g= 0 similarly,

The initial velocity of dissension water is expressed as

$v_1^{\text{'}2}=\frac{2\left(p_g+\mathrm{ÒÏ}gh\right)}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}$

Substitute p_g = 0in the above equation.

$\begin{array}{c}{v}_1^{\text{'}2}=\frac{2\left(0+\mathrm{ÒÏ}gh\right)}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}\\ =\frac{2\mathrm{ÒÏ}gh}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}\\ =\frac{2gh}{\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}\end{array}$

Here v₁^' is the velocity of water at the top level of water.

Substitute 5000Pa for p_g , 1000kg/m³ for $\mathrm{ÒÏ}$ , 9.81 m/s²for g, 0.8m for h, 0.02 m for d₂, and 2m for d₁ in above equation.

$\begin{array}{c}{v}_1^{\text{'}2}=\frac{2×9.81\text{″¾}/{\text{s}}^{\text{2}}×0.8\text{″¾}}{\left({\left(\frac{2\text{″¾}}{0.02\text{″¾}}\right)}^4−1\right)}\\ v_1^{\text{'}}=3.96×{10}^{−4}\text{″¾}/\text{s}\end{array}$

When, x=h then the final speed of dissension is expressed as,

The initial velocity of dissension of water is expressed as,

$v_{1f}^{\text{'}2}=\frac{2p_g}{\mathrm{ÒÏ}\left({\left(\frac{d_1}{d_2}\right)}^4−1\right)}$

Here v^'_1f is the final speed of dissension. Substitute p_g = 0 in the above equation.

v^'_1f = 0

When the water level behaves like uniform decelerating motion then,

$\begin{array}{c}{t}_{desc}^{\text{'}}=\frac{h}{v_{avg}}\\ =\frac{h}{\frac{v_1^{\text{'}}+v_{1f}^{\text{'}}}{2}}\\ =\frac{2h}{v_1^{\text{'}}+v_{1f}^{\text{'}}}\end{array}$

Here, is the time of deceleration.

Substitute 0.8 m for h , 3.94 x 10^-4 for v₁^' , and 0 for v_1f in the above equation.

$\begin{array}{c}{t}_{desc}^{\text{'}}=\frac{2×0.8\text{″¾}}{3.96×{10}^{−4}\text{″¾}/\text{s}+0}\\ =4040.4\text{ s}×\frac{1\text{ }\min }{60\text{ s}}\\ t_{desc}^{\text{'}}=67.34\text{″¾in}\end{array}$

So, the ratio is expressed as,

$\begin{array}{c}\frac{{t^{\text{'}}}_{desc}}{t_{desc}}=\frac{67.3\text{″¾in}}{32.4\text{″¾in}}\\ =2.08\end{array}$

Hence, the time taken for all the water to drain from the tank is 32.4 min and the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air is 2.08.