- #1

jbarker91

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Hi guys,

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.

2. Homework Equations

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

3. 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.