A few friends and myself are having some trouble determining how to go about solving this problem for our engineering class.
"An emptying tank in which the water level drops at a constant rate in time can be used as a water clock. Consider a tank where the drainage volumetric flow rate is proportional to h^(1/2), where h is the liquid height. What must be the shape of the tank so that the draining vessel may be used as a water clock?"
The shape of the water clock isn't given, but we all know it's a conical water clock, where the flow rate isn't constant, but dh/dt is, which allows for a linear scale to be read from. The goal is to prove that it the volume is a conical shaped water clock or the flow rate is equivalent to what was stated in the problem. We can assume it's a conical water clock to prove that it equates to the flow rate.
There weren't any equations given, but this is the most generic form used for this engineering type problem.
d/dt(p*V) = p*Qe
The Attempt at a Solution
We've attempted multiple different ways, but it ultimately ended up in the same situation.
I started with
dV(h) = A*dh; A = pi*[r(h)]^2
r(h)=(R*h)/L, where R is the radius of the top of the cone, and L is the height of the entire water clock.
V(h) = (pi*R^2*h^3)/(3L^2)
Took the derivative with respect to time and set it equal to the flow rate
dV/dt = (pi*R^2*h^2/L^2)*dh/dt = k*h^(1/2), where k is a proportionality constant.
This is where we continuously get stuck at, because we see no way to get an h^(1/2) from anywhere.