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 0.700 m. A worker accidently pokes a circular hole with diameter 0.0210 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 times 10^3 Pa at the surface of the water. Ignore any effects of viscosity. 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

Part a)

For this problem we use the expression Bernoulli

P + ½ v2 + g y = cte

We write this expression for two points, the water surface and exit through the hole

Point 1 surface water

point 0 hole level (tank bottom)

P1 + g y1 = Po + ½ vo2 + g yo

We assume that as the gap is small speed water height is zero

P1-Po + g (y1- yo ) = ½ v02

= 1000 Kg/m³

Given the pressure manometrical is Pm = P - Patm accordingly

P = Pm + Patm

P1 -Po = Pm + Patm - Patm = Pm

vo2= (P1-Po + g (y1- yo )) 2/

vo2= (5 103 + 1000 9.8 ( 0.7-0) ) 2 / 1000

vo2= (11860) 2 /1000

vo2= 23.72

vo = 4.87 m/s

Part b)

If the tank is opened

pressure on the water surface is equal to atmospheric pressure

P1 = Patm

P1- Po= Patm – Patm = 0

v22= g (y1- yo) 2 = 2g y1

v22= 2 9.8 0.7 =13.72

v2= 3.70 m/s

Vclosed/Vopen = Vo/V2 = 4.87/3.70 = 1.316

Part c)

for this is an approximation since by reducing the height of water surface velocity changes

Let's use the expression for flow

Q = A v

Q = r2 v

Q = 0.0212 4.87

Q = 6.747 10-3 m³/s

This is the outflow, which we assume constant

the tank capacity is

V = R2 y1

V = 2.22 0.7

V = 10.64 m³

We use a rule of proportions to find the time

6.747 10-3 m³---- 1 s

10.64 m³ -------t

t = 10.64 / 6.747 10-3

t = 1.578 103s

60s = 1 min

t = 26.29 min

Part d)

We calculate the new flow

Qopen= A v2

Q = r2 v2

Q = 0.0212 3.70

Qopen = 5.126 10-3 m³/s

5.126 10-3 m³---- 1 s

10.64 m³ -------t

t = 10.64 / 6.747 10-3

t = 2.0756 103s

60s = 1 min

topen= 34.59 min

topen/t= 34.59/26.29 = 1.316

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