Question

1. A cylindrical tube that is 2.20 m long and has a radius of 150 mm is filled with water. It is oriented with its long central axis vertical, and it is open to the air at the upper end. A hole 80.0 mm in radius is drilled in the bottom, and the water is allowed to drain out. When the water level is half the height of the tube, (a) what is the speed of the water exiting through the hole, (b) at what rate (in meters cubed per second) is the water leaving the tube, and (c) how fast is the top of the water column falling?

Answer 1

The flow rate has to be constant at both ends so

A₁v₁ = A₂v₂

=> (150mm)² v₁ = (80mm)² v₂

=> (3.515625 ) v₁ = v₂

Now Using bernoulli's theorem,

P + gh + v²/2 = constant

Now,

The pressure at both ends is same as P_atm .

And that if we consider the hole to be the base so the surface will be at a height of h

So,

P_atm + gh + v₁²/2 = P_atm + g0 + v₂²/2

=> g x 1.10 m = v₂²/2 - v₁²/2

Using The relation between velocities;

g x 1.10m = (3.515625 v₁)²/2 - v₁²/2

g x 1.10m = 5.68 x v₁²

=> v₁ = 1.3776 m/s

Now v₂ = 4.843 m/s

So the speed of water exiting through hole = v₂ = 4.843 m/s

b) flow rate = A₁v₁ = (150mm)² v₁ = 0.0973768 m³/s

c) The speed at which height is falling = v₁ = 1.3776 m/s