Water with a density of 62.4 lbm/ft3 enters a 36” diameter cylindrical container

Question

Water with a density of 62.4 lbm/ft3 enters a 36” diameter cylindrical container at 85 gpm. Water is allowed to exit the container through a short tube in the bottom that is 3/4” in diameter. The velocity of water exiting the tank is proportional to the height of the water in the tank according to the relationship
V= (2gh)^(1/2)
a) How long will it take for the tank to fill with 150 gallons of water if the tank was initially empty? b) How much water will exit the tank during this filling process?
c) If allowed to run indefinitely what will be the final height of the water in the container?

Explanation / Answer

Assume the water height is h, the the mass m in the container is

m = V = R2h    =>    dm/dt = R2 dh/dt

Qin = 85 gpm =85*( 0.002228) = 0.18938 ft/s3

Qout = AV =[ r2](2gh)1/2

Therefore, from conservation of mass, dm/dt = Qin - Qout , we have
R2 dh/dt =0.18938- r2(2gh)1/2

R2 dh/dt =0.18938- r2(2gh)1/2

This is the differential equation to solve for h as a function of t. Unfortunately, I could not solve it analytically, even using Mathematica. You can solve for it numerically, using Matlab.

Ater solve for the differential equation, you can answer all the question.

Hope that helps.

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