Question

A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidentally pokes a circular hole with diameter 0.0200 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. (a) Just after the hole is made, what is the speed of the water as it emerges from the hole? What is the ratio of this speed to the efflux speed if the top of the tank is open to the air? (b) How much time does it take for all the water to drain from the tank? 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?

Answer 1

Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

Explanation:

Using the Bernoulli's laws to use the conserved energy

a. Solve the speed and the radio of this speed of the tank is open to the air

p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 25.68  m²/s²

v1 = 5.06 m/s

Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 15.68  m²/s²

v2 = 3.96 m/s

The ratio on the air is solve using both velocities so:

R = v1/v2 = 5.06 m/s / 3.96 m/s

R = 1.27

b. Now to find the time it takes for the tank to drain if the tank is open to the air

dh/dt = -u

dh/dt = -v * A/A'

dh/dt = v*(.02m)²/(2.0m)² = -v / 10000

and we can further substitute for v:

dh/dt = -(1/1e4)*√[(p+9800h)/500]

Solve replacing

-(1000/49)*√(49000h) = t + C

-(1000/49)*√(49000*0.8) = 0 + C

C = - 4040.6

Then when h = 0,

t = 4286 s

t = 1 hr, 11 min, 26 sec