Asked by maria

An open-topped tank completely filled with water has a release valve near its bottom. The valve is 1.63 m below the water surface. Water is released from the valve to power a turbine, which generates electricity. The area of the top of the tank, AT, is a = 12.3 times the cross-sectional area, AV, of the valve opening.

Calculate the speed of a drop of water released from rest at h = 1.63 m when it reaches the elevation of the valve. ?
Answered by drwls
It would make more sense to ask for the velocity of water leaving the valve when the column height of water above is 1.63 m. In that case, you could use the Bernoulli equation.

That would give a result of
V = sqrt(2gH)= 5.65 m/s

When water that was originally at the top of the tank reaches the valve, the water colum height will be zero, and the water will just drip through at a slow rate.

The area ratio affects the volume flow rate but not the speed.

This is, in my opinion, either a trick of poorly defined question.
Answered by drwls
Actually a vortex will form and some of the water from top center will reach the valve earlier. The problem as defined is extremely hard to model
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