Answer:
1999.46 mm
45.59 s
Explanation:
given that
cylindrical water tank with diameter, D = 3 m
Height of the tank above the ground, h = 2 m
Depth of the water in the tank, d = 2 m
Diameter of hole, d = 0.420 cm
We start by calculating the volume of water in the tank, which is given as
Volume = πr²h
V = (πD²)/4 * h
V = (3.142 * 3²)/4 * 2
V = 28.278/4 * 2
V = 7.07 * 2
V = 14.14 m³
If 1.0 gal of water is equal to 0.0038m³, then
1 gal is 0.0038 = A * h
the area of the tank is 7.07 m²
therefore, 0.0038 = 7.07 * h
h₁ =0.00054 m = 0.54 mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m
b)
Like we stated earlier, 1.0 gal of water is 0.0038m³
to solve this we use the formula
Q = Cd * A * √2gH
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole
A = (πD²)/4
A = (π * 0.0042²)/4
A = 5.54*10^-5 / 4
A = 1.39*10^-5 m²
H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m
g = 9.8 m/s²
Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958
Q = 1.251*10^-5 * 6.25
Q = 7.82*10^-5 m³/s
Q = V/t
t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s
t = 45.59 s
