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A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.



Answer :

MathPhys

Answer:

4.68×10⁻⁴ m³/s

Explanation:

Using Bernoulli's equation, we can relate the initial and final pressures, velocities, and elevations of the water stream. Volumetric flow rate equals velocity times cross sectional area.

(a) Bernoulli's equation is:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

where P is pressure, ρ is density, v is speed, g is gravity, and h is height.

Take position 1 to be the top of the tank, and position 2 to be the exit at the hole.

The top of the tank is open to air, and the water exits at atmospheric pressure, so P₁ = P₂ = 0.

The speed of the water at the top of the tank is negligible, so v₁ = 0.

Using the height of the hole as reference, h₂ = 0.

The equation simplifies to:

ρgh₁ = ½ ρ v₂²

v₂² = 2gh₁

Plugging in values:

v₂² = 2 (9.8) (14.0)

v₂ = 16.6 m/s

(b) Volumetric flow rate is equal to the velocity times cross sectional area.

Q = vA

Q = (16.6 m/s) (¼ π (0.006 m)²)

Q = 0.000468 mÂł/s

Q = 4.68×10⁻⁴ m³/s

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