A closed and elevated vertical cylindrical tank with diameter 1.60 m contains wa

Question

A closed and elevated vertical cylindrical tank with diameter 1.60 m contains water to a depth of 0.500 m . A worker accidently pokes a circular hole with diameter 0.0190 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?

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?

Explanation / Answer

A)

speed of efflux, v = sqrt(2*delta_P/rho + 2*g*h)

= sqrt(2*5*10^3/1000 + 2*9.8*0.5)

= 4.45 m/s

B) v_open = sqrt(2*g*h)

= sqrt(2*9.8*0.5)

= 3.13 m/s

v/v_open = 4.45/3.13

= 1.42

C) volume flo rate = A*v

= (pi*d^2/4)*v

= (pi*0.019^2/4)*4.45

= 0.00126 m^3/s

time taken to drain out the tank, t = Volume of the water/volue flow rate

= pi*r^2*h/volume flow rate

= pi*0.8^2*0.5/0.00126

= 798 s

= 798/60

= 13.3 min

D)

t/t_open = v_open/v

= 3.13/4.45

= 0.703

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