A math or a simple fluid problem?

[hanson]

Hi all!
I am solving for the height of a tank as a function of time.
The tank has a constant inflow of a and is subjected to small fluctuation of bsin(wt).
So the inflow is simply a+bsin(wt).
The outflow should be proportional to the square root of the height, H.
So outflow = c*sqrt(H)
Therefore the below differential equation is obtained with k=the area of the tank:
$$\frac{a+bsinwt-c\sqrt{H}}{k}=\frac{dH}{dt}$$


but the problem is that I don't know how to solve this ODE...
Can anyone solve the problem?

 

[Clausius2]

Have you been given this equation or have you set it up by yourself?

 

[hanson]

I set it up myself. And I have show my deduction:
The tank has a constant inflow of a and is subjected to small fluctuation of bsin(wt).
So the inflow is simply a+bsin(wt).
The outflow should be proportional to the square root of the height, H.
So outflow = c*sqrt(H)
Therefore the above differential appears.
Am i correct?

 

[Clausius2]

I set it up myself. And I have show my deduction:
The tank has a constant inflow of a and is subjected to small fluctuation of bsin(wt).
So the inflow is simply a+bsin(wt).
The outflow should be proportional to the square root of the height, H.
So outflow = c*sqrt(H)
Therefore the above differential appears.
Am i correct?

I am not sure about your assumption of the outflow. I would revise that.

 

[batman394]

this problem really has nothing to do with fluids..its just a math problem..

inflow is a+bsin(wt) correct.
outflow you said is proportional to square root of the height, h..
is then right too.. c*sqrt(h)...

now if you divide volume by cross sectional area you get height.. I am thinking you need to find the radius in terms of height also by rearranged the volume equation... that way your equation will be composed only of H terms.