Question

A 5.0mm diameter hole is 1.0m m below the surface of a 2.0m diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?

answer in mm/min

Answer 1

Let us consider the followings:

ρ = density of the water = 999.97 kg/m³
h = depth of the water = 1.0 m
g = gravitational acceleration = 9.81 m/s²
P = static pressure of the water at h depth = to be determined
d = diameter of the hole = 5.0 mm = 0.005 m
D = diameter of the water tank = 2.0 m
A = cross-sectional area of the hole to be determined
As = area of the surface of the water to be determined
v = velocity of the water going through the hole to be determined
R = volumetric flow rate to be determined

Now, first we will calculate the volume flow rate through the hole in L/min.
A = πd²/4
A = (3.14159)(0.005 m)²/4
A = 19.6 x 10^-6 m²

v = √(2gh)
v = √[2(9.81 m/s²)(1.0 m)]
v = 4.43 m/s

R = A∙v
R = (19.6 x 10^-6 m²)(4.43 m/s)
R = 0.000086828 m³/s = 8.68 x 10^-5 m³/s
R = 0.0868 L/s
R = 60 s/min x 0.037877 L/s
R = 5.208 L/min Volumetric Flow Rate

Again calculate the rate in mm/min at which the water level in the tank will drop if the
water is not replenished.

As = πD²/4
As = 3.14159(2.0 m)²/4
As = 3.14159 m² = 3141590 mm²

R = 5.208 L/min = 5208000 mm³/min

h' = rate of change in water level to be determined
h' = R/As
h' = (5208000 mm³/min)/(3141590 mm²)
h' = 1.658 mm/min Water Level Change Rate (Answer)