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 December 25th, 2013, 11:53 AM #1 Member   Joined: Sep 2013 Posts: 43 Thanks: 0 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.
December 26th, 2013, 08:32 PM   #2
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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 $\ y^2=2x^2 \$ then the area of flow to the outer tank is $\ y^2-x^2=2x^2-x^2=x^2 \$. So the areas of flow, both out and in, are the same so:

$\frac{dh_1}{dt}+\frac{dh_2}{dt}=0$

which gives:

$h_1+h_2=h_0 \$ where $\ h_0 \$ is the initial height of the water in the inner tank.

Now we have:

$\frac{dV}{dt}=-A\sqrt{2g(h_1-h_2)}=-A\sqrt{2g(h_1-h_0+h1)}=-A\sqrt{2g(2h_1-h_0)}=-A\sqrt{4g\left(h_1-\frac{h_0}{2}\right)}$

$\frac{dV}{dt}=-2A\sqrt{g\left(h_1-\frac{h_0}{2}\right)}$

$h_1=\frac{V}{x^2} \$ 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 $\ h_0x^2 \$ to find the constant of integration (it is not 0).

For V, I get:

$V=\left(x\sqrt{\frac{h_0}{2}}-\frac{A\sqrt{g}}{x} \cdot t \right)^2+\frac{h_0x^2}{2} \text{ for } t \leq \frac{x^2}{A}\sqrt{\frac{h_0}{2g}}$

I have run a numerical simulation with $\ x=10, \ y=10\sqrt{2}, \ h_0=10, \ A=0.1 \text{ Edit: and } A=1.0 \$ 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.

 December 27th, 2013, 05:06 PM #3 Member   Joined: Sep 2013 Posts: 43 Thanks: 0 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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