Question

The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphere. An orifice of diameter D with a smooth entrance (i.e., negligible losses) is open at the bottom. Develop a relation for the time required for the tank (a) to empty halfway (5-point) and (b) to empty completely (5-point).

Answer 1

Answer:

a. The time required for the tank to empty halfway is presented as follows;

$t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$

b. The time it takes for the tank to empty the remaining half is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time 't', is presented as follows;

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

$\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}$

$\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}$

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

$dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh$

The time for the tank to drop halfway is given as follows;

$\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh$

$t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)$

$t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)$

$t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$The time required for the tank to empty halfway, t₁, is given as follows;

$t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

$\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh$

$t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)$

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time, t, to empty the tank is given as follows;

$t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}$

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$