Water flowing from one vertical cylinder to another

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SUMMARY

The discussion centers on calculating the time it takes for water levels to equalize between two vertical cylinders with diameters D1=3 m and D2=4 m, and a hole diameter of d=10 cm. The initial height of water in the first cylinder is H=5 m, while the second cylinder is empty. The continuity equation and Bernoulli's equation are applied to derive the relationship between the water heights and flow rates. The final expression for equilibrium height is debated, with a need to clarify the variables involved to simplify the integration process.

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1. The problem statement, all variables, and given/known data
Two vertical cylinders rest on a flat surface, next to each other. Both have a hole on the side at the bottom so that the water can flow from one cylinder to the other without leakage. The diameter of the first cylinder is D1=3 m, the second D2 = 4 m, and the hole diameter is d=10 cm.
If the first cylinder is filled with water to the height of H=5m, and the other one is empty, find the time τ it takes for water levels to equalize.

Homework Equations


Continuity equation: Sv = const
Bernoulli equation:


The Attempt at a Solution


First, I determined the "equalized" height as follows:
(total volume) = (volume in the first cylinder after equalization) x (volume in the second cylinder after equalization)
Which gives:
D12H = D12he + D22he
So from here we get
he = 9H/25
This seemed like a simple problem but ended up being quite challenging (at least to me). First, let v1, v3 and v2 be speeds of water (at any moment) in the first cylinder, through the opening, and in the second cylinder respectively.
upload_2017-5-18_11-47-26.png

(sorry for writing like this, but I was already comfortable with latex, so it seemed a faster approach )

I was very tempted to integrate here, but the problem is - h1 and h2 do not change in the same manner.
I cannot seem to be able to get past here.
 
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Probably a typo, but you give two different expressions for the equilibrium height. One has 26 in the denominator and the other 25.

More importantly, your final expression has three variables, h, h1 and h2. There should be only one, say h representing the height in the cylinder that was empty originally. Since the volume of the liquid is fixed, you can express the height of the liquid in the other cylinder as a function of h. Then get an expression for ##\frac{dt}{dh}##.
 
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