Practice Question: Time to Discharge Tank from Orifice

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Solution

Answer: C) 150 min

Coefficient of Discharge (given):
\[C_d=0.85\]

 

$$d=dia\ of\ orifice=10\ in=0.833\ ft $$

 

$$a=area\ of\ orifice=\frac{\pi }{4}d^2=0.545\ ft^2 $$

 

$$D=dia\ of\ tank=75\ ft $$

 

$$A=area\ of\ tank=\frac{\pi }{4}D^2=4,418\ ft^2 $$

 

$$H_i=initial\ height\ of\ water=20\ ft $$

 

$$H_f=final\ height\ of\ water=5\ in=0.42\ ft $$

$$Q=C_da\sqrt{2gH} $$

 

The discharge decreases with change in height of water with the passage of time.

 

$$Q=-A\frac{dh}{dt} $$

 

$$-A\frac{dh}{dt}=C_da\sqrt{2gH}$$

 
$$dt=-\frac{A}{C_da\sqrt{2gH}}dh $$

 

Integrate H from \(H_i\) to \(H_f\) and integrate t from 0 to t, then solve for time t, which is the time required for the tank to drain to the orifice level:

 

$$t=\frac{A}{C_d a}\left(\sqrt{H_i}-\sqrt{H_f}\right)\sqrt{\frac{2}{g}}\  $$

 

$$t=\frac{4,418}{0.85\times 0.454}\left(\sqrt{20}-\sqrt{0.42}\right)\sqrt{\frac{2}{32.2}}=9,088\ seconds=\ 151.5\ min$$ 

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