Solve Volume of Water Clock Homework

AI Thread Summary
The discussion revolves around solving a homework problem related to the design of a conical water clock, where the drainage flow rate is proportional to the square root of the liquid height. Participants agree that the tank must be conical to achieve a constant rate of height change over time, allowing for a linear measurement scale. They derive the volume of the cone and attempt to relate the flow rate to the height using calculus, but encounter difficulties in connecting the flow rate to the required proportionality of h^(1/2). The conversation suggests the need for further mathematical exploration or possibly moving the topic to a more specialized forum for calculus assistance. Ultimately, the goal is to confirm the conical shape of the tank aligns with the specified flow rate characteristics.
jbarker91
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Homework Statement


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.

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

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.
 
Physics news on Phys.org
Perhaps ask a moderator to move this to the calculus sub-forum?
 
volume remaining after time t
= V₀ - ∫k.√h.dt
= ...

height of cone having this volume = ...
 
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