Question

A large storage tank open at the top and filled with water develops a small hole in its side at a point 16.0 m below the water level. The rate of flow from the leak is found to be 2.50 × 10^(-3) m^3/min.
Determine; (a) the speed (m/s) at which the water leaves the hole and (b) the diameter of the hole (mm).

Answer 1

Answer:

(A) v2 = 17.7 m/s

(b) diameter = 1.73 mm

Explanation:

height of hole below the water level (h2) = 16 m

rate of flow (V) = 2.5 x 10^{-3} m^{3} / min = 4.17 x 10^{-5} m^{3}/s

(A) applying Bernoulli's theorem we can determine the speed of the water at the hole

$\frac{p1}{ρg1} + \frac{V1^{2}}{2g} + h1 = \frac{p2}{ρg1} + \frac{V2^{2}}{2g} + h2$

where

p1 = pressure at the top surface

ρ = density of water

g = acceleration due to gravity = 9.8 m/s

v1 = velocity at the top

h1 = height at the top

p1 = pressure at the top surface

p1 = pressure at the hole

v2 = velocity at the hole

h2 = height of the hole from the top

• p1 and p2 are under atmospheric pressures and are hence both equal in magnitude to the atmospheric pressure (patm)

• the area of the top surface is very large compared to the area of the hole, hence velocity at the top v1 is very small compared to the velocity at the hole v2 hence the velocity at the top is negligible
• let the height of the top be L and the height of the hole from the top be L - 16
• applying this considerations to the equation we have

$\frac{patm}{ρg1} + \frac{(0)^{2}}{2g} + L = \frac{patm}{ρg1} + \frac{V2^{2}}{2g} + L-16$

$0 = \frac{V2^{2}}{2g} - 16$

v2 = $\sqrt{2.g.16}$

v2 = $\sqrt{2x9.8 x 16}$

v2 = 17.7 m/s

(B)  area = rate of flow / velocity

     area = $\frac{4.17 x 10^{-5}}{17.7}$

area = $2.35 x 10^{-6} m^{2}$

area = π($\frac{diameter}{2} ^{2}$)

diameter =2($\sqrt{\frac{area}{π} }$

diameter =2($\sqrt{\frac{2.35 x 10^{-6}}{π} }$

diameter = 1.73 x 10^{-3} m = 1.73 mm