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I am a structural engineering student and have limited experience in fluid dynamics. I'm not quite sure if I'm going about the problem in the correct way.

The question is:

An open cylinder of height 5ft and cross sectional area of $1~ft^2$ is initially empty. There is a small hole at the bottom of the cylinder with an area of $0.005~ft^2$. Water is drawn into the tank at a rate of $4.8~ft^3/min$. At the same time water is discharged out of the cylinder through a small hole at the bottom of the tank. The outlet velocity may be estimated from Torricelli’s theorem. Estimate the time required for the water level in the tank to reach one foot depth.

For Torricelli's Theorem, I used the equation $$\frac{dv}{dt}=-\pi a^2 \sqrt{2 g h}$$ to get $-15.123~ft^3/min$.

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closed as off-topic by user10851, HDE 226868, user36790, John Rennie, Kyle Kanos Dec 16 '15 at 11:06

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I think we have the following equation that describes your problem:

$$Q_{in} = A_{tank}\frac{dH}{dT} + A_{orifice}\sqrt{2g}H^{1/2}$$

Rewritten;

$$\Big(\frac{Q_{in} - A_{orifice}\sqrt{2g}H^{1/2}}{A_{tank}}\Big){dT} = {dH} $$

Rewritten;

$${dT} = \Big(\frac{A_{tank}}{Q_{in} - A_{orifice}\sqrt{2g}H^{1/2}} \Big){dH} $$

Therefore;

$${T} = \int_{H=0}^{H=1}\Big(\frac{A_{tank}}{Q_{in} - A_{orifice}\sqrt{2g}H^{1/2}} \Big){dH} $$

I would have to type quite a bit more to complete the integral. I'll lazily leave that to you. It definitely is solvable though. At first glance, I think you need to do a U-substitution on $H^{1/2}$. Then, go to any set of generally solved integral lists you might find in a college calculus textbook and find the proper formula. In fact, I think this is the formula, but you'll need to do the U-substitution to verify:

$$\int\Big(\frac{u}{a + bu} \Big){du} = \frac{1}{b^2}(a+bu - a\ln|a+bu|) + C $$

Double check on that to make sure I've selected the proper scenario. If you don't want to integrate the right side of the time formula by hand, you can go to Wolfram instead. WATCH YOUR UNITS!!!!

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