Question

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 13.9 m below the water level. If the rate of flow from the leak is3.00 ✕ 10⁻³ m³/min, determine the following.

(a) Determine the speed at which the water leaves the hole.
m/s

(b) Determine the diameter of the hole.
mm

Answer 1

a)

Let’s use Bernoulli’s equation.
P1 + ½ * Density * v1^2 + Density * g * y1 = P2 + ½ * Density * v2^2 + Density * g * y2
The left side is at the hole, and the right side is at the surface.
P1 and P2 are the atmospheric pressure. The atmospheric pressure is the same at the surface and at the hole. So, we can neglect the atmospheric pressure.

½ * Density * v1^2 + Density * g * y1 = ½ * Density * v2^2 + Density * g * y2
The density of the water is the same at the surface and at the hole. So we can divide both sides by Density.

½ * v1^2 + g * y1 = ½ * v2^2 + g * y2
Since the cross sectional area of tank is large and the cross sectional area hole is small, the downward velocity of water at the surface is much slower than the velocity of water through the hole. So we can neglect the velocity of water at the surface, v2.

½ * v1^2 + g * y1 = g * y2
Subtract g * y1 from both sides
½ * v1^2 = g * y2 – g * y1
½ * v1^2 = g * (y2 – y1)
Multiply both sides by 2
v1^2 = 2 * g * (y2 – y1)
v1 = √[2 * g * (y2 – y1)]
(y2 – y1) is the distance from the surface to the hole = 13.9 m
v1 = √(2 * 9.8 * 13.9) = 16.50 m/s

b)

Flow rate A₁*V₁ = A₂*V₂

(3*10^-3 m³/min)(1min / 60s) = (d² / 4)*16.50

by solving we get

d = 1.96*10^-3 m = 1.96 mm