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I was told that I have a cylindrical water tank $10$ ft tall that can store $5000 $ ft$^3$ of water, and that the water drains from the bottom of the tank at a rate proportional to the instantaneous water level. After $6$ hours, half of the tank has drained out from the bottom. If the drain is opened and water is added to the tank at a constant rate of $100$ $\frac{ft^3}{hr}$, at what height above the drain will the elevation remain constant?

My attempt:

We are given that $$\frac{dV}{dt}=kh$$ where $\frac{dV}{dt}$ is the rate at which water leaves the tank, $V$ is the total change in water that has left the tank, and $h$ is the water's height above the drain.

If the elevation must remain constant in the tank, then I am merely finding the height in which the water that enters the tank equals to the water that leaves the tank, and therefore $\frac{dV}{dt}=100\frac{ft^3}{hr}$.

To find $k$, I must do separation of variables and integrate both sides, so: $$\int{dV}=\int{khdt}$$ $$V=kht+C$$ I know that $C=0$ because at $t=0$ hr, no water has left the tank ($V=0$), and at $t=6$ hr, $V=2500$ ft$^3$ and $h=5$ ft, since half the tank has drained out.

Solving for $k$: $$2500=k(5)(6)+0$$ $$k=\frac{250}{3}$$ And solving for $h$ when $\frac{dV}{dt}=100\frac{ft^3}{hr}$:$$100=\frac{250}{3}h$$ $$h=1.2ft$$ However, the answer is actually about $1.73ft$. Where is my mistake?

Niwde Aup
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2 Answers2

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Your mistake is in the solution of differential equation and in the initial boundary condition.

see the following solution: $$\frac{dV}{dt}=kh$$ $$A\frac{dh}{dt}=kh$$ $$500\frac{dh}{h}=kdt$$ integrate $$500\log h=kt+C$$ at $t=0, h=10ft$

so $$C=1151.3$$ at $t=6 hr, h=5 ft$ $$k=-57.76$$ now use that $$\frac{dV}{dt}=-57.76h$$ $$-100=-57.76h$$ so $$h=1.73 ft$$

E.H.E
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We have a differential equation that defines the rate that water leaves the drain.

$\frac{dx}{dt} = -ax\\ x = C e^{-at}$

We know the initial conditions and after 6 hours to find our constants.

$x(0) = 10$

This gives us $C$

$x(6) = 5\\ 5 = 10e^{-6a}\\ 6a = \ln 2\\ a = \frac{\ln2}{6}$

What do we know about the dimensions of the tank?

$v = \pi r^2 x\\ \pi r^2 = 500$

The flow out the drain equals the flow into the tank.

$\frac{dv}{dt} = \frac {dv}{dx}\frac{dx}{dt}\\ 500 \frac {dx}{dt} = 100\\ 500(\frac{\ln 2}{6})x = 100\\ x = \frac{6}{5\ln 2}$

user317176
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