A closed and elevated vertical cylindrical tank with diameter 2.20 m contains wa

Question

A closed and elevated vertical cylindrical tank with diameter 2.20 m contains water to a depth of 0.800 m . A worker accidently pokes a circular hole with diameter 0.0220 m in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of 5.00×103Pa at the surface of the water. Ignore any effects of viscosity.

Part A

Just after the hole is made, what is the speed of the water as it emerges from the hole?

5.07

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Part B

What is the ratio of this speed to the efflux speed if the top of the tank is open to the air?

Part C

How much time does it take for all the water to drain from the tank?

Part D

What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

5.07

Explanation / Answer

part A)

v1 = sqrt(2*( (dp/rho)+gH))

v1 = sqrt(2*(5+(9.81*0.8))) = 5.07 m/s

part B)

efflux speed v2 = sqrt(2*g*H) = sqrt(2*9.81*0.8) = 3.96 m/s

part C)

volume flow rate = A*v1 = pi*r^2*v1 m^3/s

volume of the water in the tank V = pi*R^2*H = pi*1.1^2*0.8 = 3.04 m^3

time taken to drain t1 = 3.04/pi*0.011^2*5.07 = 1575.13 s

partD)

time taken t2 = 3.04/A*v2 = 3.04/(pi*r^2*v2)

t2 = 3.04/(pi*0.011^2*3.96) = 2019.5 s

ratio = 1575.13 /2019.5 = 0.78

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