Integral of Flow Rate

Question

Please Help me to find anti-derivative of system. water that fills the tank, with the flow rate is known, and flow rate out

there is a flowing water with flow rate Qin = 1.5x10^-3 m^3/s that fills the tank. the height of the initial tank is 1 meter

At the bottom of the tank, there is a hole where the water comes out with Qout = 10^-3 x H^1/2 m^ 3 / s

calculate the equation H with respect of time, and determine at what height Q in = Q out

The integral equation that i've found, 1/ (10^-3 H^1/2 - 1.5x10^-3) dH = - dt

but i have a problem inegrate the (10^-3 H^1/2 - 1.5x10^-3) dH

Answer 1

First of all Heidi T. is correct: you need the relationship between H and V in order to solve this problem as stated.

If your differential equation eventually becomes something like this:

dH/dt = A - BH^1/2 where A and B are constants,

THEN

dH/(A-BH^1/2) = dt

now make the substitution I suggested: x²=H and 2x dx = dH so that

-2x dx/(Bx-a) = dt

(-2/B) dx/(x- (A/B))=dt = (-2/B)[1 - (A/B)/(x- (A/B))

and the last term on the right can be integrated as a ln which solves your problem!

I am sorry that this editor makes writing out solutions complicated...but I hope you will get the idea.