Draining irrigation tank: separable DE, validity limit and scaling
A vertical cylindrical irrigation tank has internal cross-section $2.0\ \text{m}^2$ and depth $4.0\ \text{m}$, with a small valve at the base. The tank is full and there is no inflow when the caretaker opens the valve at $t = 0$. Torricelli's law gives $\dfrac{dh}{dt} = -k\sqrt{h}$, where $h$ metres is the depth at time $t$ seconds and $k = 0.0025\ \text{m}^{1/2}\text{s}^{-1}$, a value measured on this tank. The caretaker must switch off the intake pump before the tank empties, and wants the time at which the depth reaches 1.0 m as well as the predicted emptying time. A second tank on the property, with the same valve and the same starting depth, has cross-section $3.0\ \text{m}^2$. She also notices the outflow slows to a trickle near the end and asks whether the formula should be trusted right down to $h = 0$.
1. Solve the differential equation with $h(0) = 4$ and find the time at which the depth reaches 1.0 m. Also give the depth at $t = 600\ \text{s}$.
Read the worked answer
Separate and integrate.
$$h^{-1/2},dh = -k,dt \ \Rightarrow\ \int h^{-1/2},dh = -k\int dt \ \Rightarrow\ 2\sqrt{h} = -kt + C .$$
Apply the condition. At $t = 0$, $h = 4$, so $C = 2\sqrt{4} = 4$, giving
$$2\sqrt{h} = 4 - 0.0025t \ \Rightarrow\ \sqrt{h} = 2 - 0.00125t \ \Rightarrow\ h = (2 - 0.00125t)^2 .$$
Depth 1.0 m. $\sqrt{h} = 1$, so $1 = 2 - 0.00125t \Rightarrow t = \dfrac{1}{0.00125} = 800\ \text{s}$, i.e. 13 min 20 s.
Depth at 600 s. $\sqrt{h} = 2 - 0.00125(600) = 2 - 0.75 = 1.25$, so $h = 1.5625 \approx 1.56\ \text{m}$.
Check. Differentiating $h = (2 - 0.00125t)^2$ gives $\dfrac{dh}{dt} = -2(0.00125)(2 - 0.00125t) = -0.0025\sqrt{h}$, which is exactly the original equation - the implicit constant and the negative sign are both correct.
2. Find the time at which the model says the tank empties, and explain, using the algebra of your solution, why the formula must not be applied for times beyond that value.
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Emptying time. $h = 0$ requires $\sqrt{h} = 0$, so $2 - 0.00125t = 0$ and $t = 1600\ \text{s}$, i.e. 26 min 40 s.
Why the formula stops there. For $t > 1600$, the quantity $2 - 0.00125t$ is negative. Squaring it in $h = (2 - 0.00125t)^2$ makes $h$ positive again and increasing without limit - the algebra would have the tank refilling itself at an accelerating rate, which is physically absurd. The correct interpretation is that the solution of the initial value problem is
$$h(t) = (2 - 0.00125t)^2, \qquad 0 \le t \le 1600,$$
and $h(t) = 0$ for $t \ge 1600$ if the tank is simply left open.
Additional modelling caveat. As $t \to 1600$, $\dfrac{dh}{dt} = -0.0025\sqrt{h} \to 0$: the model has the outflow dying away completely as the depth vanishes. In reality the flow through the valve stays finite until air reaches the outlet, so 1600 s should be treated as a lower bound on the true emptying time. The measured 'trickle' the caretaker notices is consistent with the model being optimistic at small $h$.
3. For the second tank (cross-section $3.0\ \text{m}^2$, same valve, same starting depth), state the new value of $k$, find the new emptying time, and explain the proportional reasoning behind it.
Read the worked answer
How $k$ depends on the tank. Torricelli's law in the form $\dfrac{dh}{dt} = -\dfrac{A_{\text{out}}}{A_{\text{tank}}}\sqrt{2gh}$ shows $k \propto \dfrac{A_{\text{out}}}{A_{\text{tank}}}$. The outlet is unchanged, so
$$k_2 = 0.0025 \times \frac{2.0}{3.0} = 0.001667\ \text{m}^{1/2}\text{s}^{-1} \ (=\tfrac{2}{3} \times 0.0025).$$
Emptying time. From $\sqrt{h} = \sqrt{h_0} - \dfrac{k_2 t}{2}$, emptying occurs at
$$t = \frac{2\sqrt{h_0}}{k_2} = \frac{2 \times 2}{0.001667} = 2400\ \text{s} = 40\ \text{min}.$$
Proportional reasoning. $t \propto \dfrac{1}{k} \propto A_{\text{tank}}$: a tank 1.5 times the cross-section holds 1.5 times the water for the same depth, and the same valve discharges the same volume per second at a given depth, so the level falls 1.5 times more slowly. The new time is $1.5 \times 1600 = 2400\ \text{s}$, and the time to reach 1.0 m becomes $1.5 \times 800 = 1200\ \text{s}$. Scaling arguments like this are a fast sanity check on any re-solved differential equation.