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^2.
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
\dot{m}_{\text {in }}-\dot{m}_{\text {out }}=\frac{d m_{\mathrm{CV}}}{d t} (1)
During this process no mass enters the control volume \left(\dot{m}_{\text {in }}=0\right), and the mass flow rate of discharged water can be expressed as
\dot{m}_{\text {out }}=(\rho V A)_{\text {out }}=\rho \sqrt{2 g h} A_{\text {jet }} (2)
where A_{\text {jet }}=\pi D^{2} _\text{jet}/ 4 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
m_{\mathrm{CV}}=\rho=\rho A_{\text {tank }} h (3)
where A_{\text {tank }}=\pi D^{2}_{\text {tank }}/ 4 is the base area of the cylindrical tank. Substituting Eqs. 2 and 3 into the mass balance relation (Eq. 1) gives
-\rho \sqrt{2 g h} A_{\mathrm{jet}}=\frac{d\left(\rho A_{\mathrm{tank}} h\right)}{d t} \rightarrow-\rho \sqrt{2 g h}\left(\pi D_{\mathrm{jec}}^{2} / 4\right)=\frac{\rho\left(\pi D_{\mathrm{tank}}^{2} / 4\right) d h}{d t}
Canceling the densities and other common terms and separating the variables give
d t=\frac{D_{\text {tank }}^{2}}{D_{\text {jet }}^{2}} \frac{d h}{\sqrt{2 g h}}
Integrating from t=0 at which h=h_{0} to t=t at which h=h_{2} gives
\int_{0}^{t} d t=-\frac{D_{\text {tank }}^{2}}{D_{\text {jet }}^{2} \sqrt{2 g}} \int_{h_{0}}^{h_{2}} \frac{d h}{\sqrt{h}} \rightarrow t=\frac{\sqrt{h_{0}}-\sqrt{h_{2}}}{\sqrt{g / 2}}\left(\frac{D_{\text {tank }}}{D_{\text {jet }}}\right)^{2}
Substituting, the time of discharge is
t=\frac{\sqrt{4 \mathrm{ft}}-\sqrt{2 \mathrm{ft}}}{\sqrt{32.2 / 2 \mathrm{ft} / \mathrm{s}^{2}}}\left(\frac{3 \times 12 \mathrm{in}}{0.5 \mathrm{in}}\right)^{2}=757 \mathrm{~s}=12.6 \mathrm{~min}
Therefore, half of the tank is emptied in 12.6 min after the discharge hole is unplugged.
Discussion Using the same relation with h_{2}=0 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.
