Discharge of Water from a Tank A 4-ft-high, 3-ft-diameter cylindrical water tank whose top is open to the atmosphere is initially filled with water. Now the discharge plug near the bottom of the tank is pulled out, and a water jet whose diameter is 0.5 in streams out (Fig. 5-10).

Question

Discharge of Water from a Tank

A 4-ft-high, 3-ft-diameter cylindrical water tank whose top is open to the atmosphere is initially filled with water. Now the discharge plug near the bottom of the tank is pulled out, and a water jet whose diameter is 0.5 in streams out (Fig. 5-10). The average velocity of the jet is approximated as [latex]V = \sqrt {2 gh}[/latex], where h is the height of water in the tank measured from the center of the hole (a variable) and g is the gravitational acceleration. Determine how long it takes for the water level in the tank to drop to 2 ft from the bottom .

Accepted Answer

The plug near the bottom of a water tank is pulled out. The time it takes for half of the water in the tank to empty is to be determined.

Assumptions 1 Water is an incompressible substance. 2 The distance
between the bottom of the tank and the center of the hole is negligible com pared to the total water height. 3 The gravitational acceleration is 32.2 ft/s[latex]^2[/latex].

Analysis We take the volume occupied by water as the control volume. The size of the control volume decreases in this case as the water level drops, and thus this is a variable control volume. (We could also treat this as a fixed control volume that consists of the interior volume of the tank by disregarding the air that replaces the space vacated by the water.) This is obviously an unsteady-flow problem since the properties (such as the amount of mass) within the control volume change with time.
The conservation of mass relation for a control volume undergoing any process is given in rate form as

[latex]\dot{m}_{ ext {in }}-\dot{m}_{ ext {out }}=\frac{d m_{\mathrm{CV}}}{d t}[/latex] (1)

During this process no mass enters the control volume [latex]\left(\dot{m}_{ ext {in }}=0 ight)[/latex], and the mass flow rate of discharged water can be expressed as

[latex]\dot{m}_{ ext {out }}=( ho V A)_{ ext {out }}= ho \sqrt{2 g h} A_{ ext {jet }}[/latex] (2)

where [latex]A_{ ext {jet }}=\pi D^{2} _ ext{jet}/ 4[/latex] is the cross-sectional area of the jet, which is constant. Noting that the density of water is constant, the mass of water in the tank at any time is

[latex]m_{\mathrm{CV}}= ho= ho A_{ ext {tank }} h[/latex] (3)

where [latex]A_{ ext {tank }}=\pi D^{2}_{ ext {tank }}/ 4[/latex] is the base area of the cylindrical tank. Substituting Eqs. 2 and 3 into the mass balance relation (Eq. 1) gives

[latex]- ho \sqrt{2 g h} A_{\mathrm{jet}}=\frac{d\left( ho A_{\mathrm{tank}} h ight)}{d t} ightarrow- ho \sqrt{2 g h}\left(\pi D_{\mathrm{jec}}^{2} / 4 ight)=\frac{ ho\left(\pi D_{\mathrm{tank}}^{2} / 4 ight) d h}{d t}[/latex]

Canceling the densities and other common terms and separating the variables give

[latex]d t=\frac{D_{ ext {tank }}^{2}}{D_{ ext {jet }}^{2}} \frac{d h}{\sqrt{2 g h}}[/latex]

Integrating from [latex]t=0[/latex] at which [latex]h=h_{0}[/latex] to t=t at which [latex]h=h_{2}[/latex] gives

[latex]\int_{0}^{t} d t=-\frac{D_{ ext {tank }}^{2}}{D_{ ext {jet }}^{2} \sqrt{2 g}} \int_{h_{0}}^{h_{2}} \frac{d h}{\sqrt{h}} ightarrow t=\frac{\sqrt{h_{0}}-\sqrt{h_{2}}}{\sqrt{g / 2}}\left(\frac{D_{ ext {tank }}}{D_{ ext {jet }}} ight)^{2}[/latex]

Substituting, the time of discharge is

[latex]t=\frac{\sqrt{4 \mathrm{ft}}-\sqrt{2 \mathrm{ft}}}{\sqrt{32.2 / 2 \mathrm{ft} / \mathrm{s}^{2}}}\left(\frac{3 imes 12 \mathrm{in}}{0.5 \mathrm{in}} ight)^{2}=757 \mathrm{~s}=12.6 \mathrm{~min}[/latex]

Therefore, half of the tank is emptied in 12.6 min after the discharge hole is unplugged.

Discussion Using the same relation with [latex]h_{2}=0[/latex] gives t=43.1 min for the discharge of the entire amount of water in the tank. Therefore, emptying the bottom half of the tank takes much longer than emptying the top half. This is due to the decrease in the average discharge velocity of water with decreasing h.

Question 5.2: Discharge of Water from a Tank A 4-ft-high, 3-ft-diameter cy...

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