Typically when I shower, I am stricken with fear. I like to do my showering at a leisurely pace, but it’s inevitable that if I spend more than a few minutes on my ablutions, the water level begins to rise around my feet. Then the panic sets in. I am sure that if I let the water run any longer, it will overflow the threshold and run out onto the bathroom floor, soaking all my clean clothes and pissing off the other residents of my dorm.
But the other day, all of that changed. I realized that I was being stupid. My fears had no basis in reality. Unless the drain is clogged, there’s no reason to expect that a shower will ever overflow, for the simple reason that the rate at which water flows through the drain is proportional to the water level. As more water fills the shower, its own weight pushes it out of the drain faster and faster. If the drain is clear and reasonably large, the water should stop rising long before it overflows, because it will be flowing out at the same rate that it is coming in from the showerhead. Since it’s IAP and I had nothing better to do, I thought I’d work out the math and measure the rate of outflow through my shower drain as a function of the water level.
Suppose V is the total volume of water in the shower, and A is the area of the shower floor. Then the height of the water is V/A, and the rate of outflow through the drain will be kV/A, where k is some constant of proportionality with units of area divided by time. If r is the rate of inflow through the showerhead, then we have:
dV/dt = r – kV/A.
This is an extremely elementary equation, and we can see by inspection that the general solution should be:
V = C * exp(-kt/A) + Ar/k,
where C is some constant that depends on our initial conditions. We don’t actually care what C is, because it is part of the transient solution. That bit with the declining exponential will die out – that’s why the shower doesn’t actually run over – and the water volume will eventually stay fixed at Ar/k. That’s the steady-state solution.
Our goal is to determine k! In order to do that, we need to measure r and the steady-state value of V/A (we don’t need to know both V and A as long as we know how they’re related). To measure r, the output of the showerhead, I needed to find some container of known volume. I rummaged around for a while and the only thing I could find was an empty can that had once held cononut milk. It would have to do.
The can held 754 ml, and the shower filled it in about 2 seconds, for a flow rate of 0.377 l/s. Then I let the shower run until it was about as full as I thought it would get, and stuck a ruler into the water, which came up to about the 1.5 cm mark. That’s all the data we need! (1.5 cm) * k = 0.377 l/s gives k = 0.251333… cm^2 / s, which when we account for sigfigs and wild inaccuracies is more like 0.3 cm^2 / s.
Well. That was fun.