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A math or a simple fluid problem?

by hanson
Tags: fluid, math, simple
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hanson
#1
Jan29-06, 11:47 AM
P: 320
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:
[tex]\frac{a+bsinwt-c\sqrt{H}}{k}=\frac{dH}{dt}[/tex]


but the problem is that I don't know how to solve this ODE...
Can anyone solve the problem?
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Clausius2
#2
Jan29-06, 01:17 PM
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Have you been given this equation or have you set it up by yourself?
hanson
#3
Jan29-06, 11:13 PM
P: 320
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
#4
Jan30-06, 08:45 PM
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A math or a simple fluid problem?

Quote Quote by 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?
I am not sure about your assumption of the outflow. I would revise that.
batman394
#5
Feb21-06, 01:09 AM
P: 37
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.. im 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.


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