Time taken for water to drain from a tank through a pipe (fluid mech./Bernoulli)

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SUMMARY

The discussion focuses on calculating the time taken for a tank of water to drain through a pipe using Bernoulli's principle. The initial height of the water is denoted as H, and the vertical distance the pipe extends below the tank is h. The derived equation for the velocity of water exiting the pipe is v = √(2gh), leading to the differential equation dt = dh/√(2gh). The integration of this equation is necessary to find the total time, with the constant of integration C needing to be determined based on the initial conditions of the system.

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Homework Statement


A) A tank of water of cross-sectional area A empties through a pipe
of cross-sectional area a, where a A so that the speed of the water flow in the tank can be neglected compared to that in the pipe. The initial height of the water in the tank is H and the pipe extends a vertical distance h below the bottom of the tank. Find an expression for the time taken for the tank to empty. You may assume an inviscid flow. [Take the height of the water in the tank to be z. The pressure at the water surface and at the bottom of the pipe are both atmospheric.]

B) Explain what would happen if a small hole was made:
a) halfway up the side of the tank;
b) halfway along the pipe.

Only a brief calculation/explanation is needed for this part.

Homework Equations


\frac{v^{2}}{2}+gz+\frac{p}{ρ}=const.

The Attempt at a Solution


Ok, from Bernoulli we can see that at the exit of the pipe:

gz=0

and at the entrance to the pipe at the bottom of the tank

v=0

As the speed is said to be negligible in the tank.
Since p and ρ are constant, final Bernoulli equation is:

\frac{v^{2}}{2}=gh

Rearranging we get:

v=\sqrt{2gh}

\frac{∂h}{∂t}=\sqrt{2gh}

∂t=(2gh)^{-1/2}∂h

And I'm not really sure where to go from here, or if it's correct up until this point. I assume you integrate both sides but I'm not entirely sure what the boundaries would be on the right hand side, and also I'm not sure where the areas come into it? And part b I don't know where to start...

Thanks in advance!
 
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∂t=(2gh)−1/2∂h
OR
dt=dh/Sqrt(2gh)

t=2*sqrt(h)/(2g)+C
I don't know, How to find the value of 'C' (A Constant value of integration).
 
Last edited:
You cannot ignore the pressure created by the water still in the tank.
 

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