Engineering

Determine the minimimum depth of a water reservoir that will provide a flow rate of 1200 gpm in a 4 inch diameter horizontal pipe that opens to the atmosphere


You ought to consider Bernoulli's equation for this, it is in your text.

You can model it as this:

A drop of water at the top has potential energy m*g*h
This has to equal the kinetic enery of the same drop going out of the horizontal tank
1/2 m v^2

Now veloicty v can be manipulated equal
area*velocity = volume/time, and you know the volume/time as 1200g/min. Change that to m^3/sec.
set the two energies equal.
1/2 mv^2=mgh
h= 1/2 v^2/g
put in for v the velocity (convert 1200gal/min to inch^3/sec, divide by the area of a 4 inch pipe. Then solve for h.



im sorry but i am completly lost.
i have h= 1/2 v^2/g
i know h is height v is velocity and g is gravity but how do i find g?
another thing what is the easiest way to convert gal/min to inch/sec?


gal convert to inches cubed.
min convert to sec

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