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December 25th, 2013, 11:53 AM   #1
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Water remaining in inner cubic tank bounded by outer tank

A cubic tank initially consist of water at height h, width x and length x as well is bounded by another cubic tank of width y and length y, the outer tank is initially empty.
given y^2=2x^2
The water is allowed to flow to the outer tank from the inner tank to the outer tank through a hole at the bottom with surface area A.
I want to compute the water remaining in the inner tank after time t.

since, pressure P=pgh
and pressure=0.5pv^2 as well
velocity of water flow through velocity, v=(2gh)^0.5
dv/dt=velocity*area=A(2gh)^0.5

the velocity of water should be depend on the difference in pressure between both tank so
Pinner-Pouter=pg(h1-h2)
so dv/dt=(2g(h1-h2))^0.5
now i got h1 and h2.
Not sure how to proceed to the integration.

Please give me some idea on this question.Thank you.
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December 26th, 2013, 08:32 PM   #2
jks
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Re: Water remaining in inner cubic tank bounded by outer tan

Hi chuackl,

First, be careful with your variables. I believe that you use v for both velocity and volume which may contribute to the variable A being left out of the equation for dV/dt in your second set of equations:

Quote:
Originally Posted by chuackl
velocity of water flow through velocity, v=(2gh)^0.5
dV/dt=velocity*area=-A(2gh)^0.5

the velocity of water should be depend on the difference in pressure between both tank so
Pinner-Pouter=pg(h1-h2)
so dV/dt=-A(2g(h1-h2))^0.5
Where appropriate, I changed to capital V, added A, and added minus signs to dV/dt in bold in the quote above.

Quote:
now i got h1 and h2.
Not sure how to proceed to the integration.
First, since then the area of flow to the outer tank is . So the areas of flow, both out and in, are the same so:



which gives:

where is the initial height of the water in the inner tank.

Now we have:





so substitute that into the equation. You should be able to separate the variables V, dV from dt and integrate. Use the initial condition that at t=0, V is to find the constant of integration (it is not 0).

For V, I get:



I have run a numerical simulation with and the results match the equation well.

Note: This analysis assumes the tanks are sitting on top of one another to be consistent with the equations given. The height dimension of the area A is assumed to be small and the difference in pressure across the height is neglected.
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December 27th, 2013, 05:06 PM   #3
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Re: Water remaining in inner cubic tank bounded by outer tan

Yes! that does make sense.
Irrespect on how h1 and h2 change, the sum of h1+h2 is always equal to h0.
Thank you!
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