What is the water level above the orifice as time goes to infinity?

[process91]

Homework Statement

A tank with vertical sides has a square cross-section of area 4 ft². Water is leaving the tank through an orifice of area 5/3 in². Water also flows into the tank at the rate of 100 in³/s. Show that the water level approaches the value (25/24)² ft above the orifice.

Homework Equations

Rate of discharge of volume through the orifice is 4.8A_0 \sqrt{y} cubic feet per second, where A_0 = size of orifice in square feet

The Attempt at a Solution

\dfrac{dV}{dt}=100/12^3-4.8A_0 \sqrt{y}

Also, V(y)=\int_0^y 4 du so
\dfrac{dV}{dy}=4

By the chain rule, \dfrac{dV}{dt}=\dfrac{dV}{dy} \dfrac{dy}{dt} so
100/12^3-4.8A_0 \sqrt{y} = 4 \dfrac{dy}{dt}

The problem right before this was the same except water was not being added at all, and that was an easily solvable differential equation. I am stuck on this, and when I got the answer from WolframAlpha it did not look encouraging that I was on the right track.

 

[process91]

I think, perhaps, the solution is not to actually solve the differential equation, but rather to do something like this:

Instead of trying to solve my differential equation, I just examine it at y=(25/24)^2, and see that it is zero. Examining the second derivative, we find that it is always negative so this is a maximum of y. For y less than this, the derivative is positive and above this value the derivative is negative. Hence no matter what value of y the problem starts with, as t increases it will approach (25/24)^2.

Does that seem valid?

 

[LawrenceC]

You can demonstrate the discharge rate at y = 25/24 equals the mass influx so the level won't rise anymore.

Also you could write a DE for the rate of volume of water increase in the tank.

Atank * (dV/dt) = 100/12**3 - 4.8 * A0 * (y**2)** .5

Set dV/dt to zero and solve for y. You should get 25/24.