large cylindrical water tank is mounted on a platform with its central axis vertical. The water level is 3.75 m above the base of the tank, and base is 6.50 m above the ground. A small hole 2.22 mm in diameter has formed in the base of the tank. Both the hole and the top of the tank are open to the air. We can ignore air resistance and treat water as an ideal fluid with a density of 1000 kg/m3. How many cubic meters of water per second is this tank losing

0.0000333 m³/s

Explanation:

using Bernoulli's equation

P1 + 1/2ρv1² + ρh1g = P2 + 1/2ρv2² + ρh2g

P1 = P2 both are atmospheric pressure; they cancel out

1/2ρv1² + ρh1g = 1/2ρv2² + ρh2g divide both side with ρ (rho)

1/2v1² + h1g = 1/2v2² + h2g

collect the like terms

h1g - h2g = 1/2v2² + 1/2v1²

multiply both side by 2 and take g out from the left equation

2g (h1 - h2) = v2² + v1² where v1 is approximately 0 assuming the surface of the tank is large

v2² = 2g (h1 - h2)

take square root of both sides

v2 = √2g (h1 - h2)

h1 = 3.75 + 6.5 = 10.25

h2 = 6.5

v2 = √2g (h1 - h2) = √ (2*9.81 * (10.25 - 6.5)) = 8.6 m/s

Volume in m³/s = A * v where A is the area of the hole and the radius of the hole = 2.22 mm / 2 (where 2.22 mm is the diameter of the hole) = 1.11 mm and 1.11mm / 1000 to convert it to meters

Area of the hole (assuming it is a circle) = πr² = 3.142 * (1.11/1000) ²

Area = 0.00000387 m²

Volume in M³/s = 0.00000387 * 8.6 = 0.0000333 m³/s