Solving a DE for Water Leaking from a Tank

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SUMMARY

The discussion focuses on deriving a differential equation for the height of water, h, in a cubical tank with a circular hole at the bottom, where water leaks at a rate defined by the equation c.Ah.(2gh)^(0.5). The area of the hole, Ah, is constant and does not depend on the height of the water. The relationship between the volume of water, V, and the height, h, is established as V = Aw * h, leading to the differential equation dh/dt = (1/Aw) * dV/dt. The empirical constant c is crucial for determining the flow rate.

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Homework Statement



Suppose water is leaking from a tank through a circular hole of area Ah at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to

c.Ah.(2gh)^.5

, where c (0 < c < 1) is a empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown. The radius of the hole is 2 in., and g = 32ft/s2

Homework Equations





The Attempt at a Solution



I am always confused in setting up a DE. Over here, the answer says that V= Aw(area of water). h . The next step written was that dh/dt = 1/Aw * dV/dt ... How does this make sense.. I am confused here. I understand that the "c.Ah.(2gh)^.5" given back in the question is dv/dt . But when you multiply it with 1/Aw, how do you get dh/dt ? Thanks for any help!
 
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ehabmozart said:

Homework Statement



Suppose water is leaking from a tank through a circular hole of area Ah at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to

c.Ah.(2gh)^.5

, where c (0 < c < 1) is a empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown. The radius of the hole is 2 in., and g = 32ft/s2

Homework Equations





The Attempt at a Solution



I am always confused in setting up a DE. Over here, the answer says that V= Aw(area of water). h . The next step written was that dh/dt = 1/Aw * dV/dt ... How does this make sense.. I am confused here. I understand that the "c.Ah.(2gh)^.5" given back in the question is dv/dt . But when you multiply it with 1/Aw, how do you get dh/dt ? Thanks for any help!

The area of the hole is Ah? How can the hole's area depend on the height of water in the tank?
 
Ray Vickson said:
The area of the hole is Ah? How can the hole's area depend on the height of water in the tank?
I think the OP means this as Ah, not A * h. Also, I think Aw means Aw.

ehabmozart,
You can make what you write clearer by using the features available on this site. For example, to write exponents and subscripts, click the Go Advanced button below the input area, which causes the advanced menu to open across the top. One button is X2, which you can use to write exponents. Another button is X2, which you can use to write subscripts.
 
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ehabmozart said:
I am always confused in setting up a DE. Over here, the answer says that V= Aw(area of water). h . The next step written was that dh/dt = 1/Aw * dV/dt ... How does this make sense.. I am confused here. I understand that the "c.Ah.(2gh)^.5" given back in the question is dv/dt . But when you multiply it with 1/Aw, how do you get dh/dt ? Thanks for any help!

If you take the equation V = Aw * h and differentiate with respect to 't', what do you get?
 

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