Question

The circular stream of water from a faucet is observed to taper from a diameter of 21 mm to 10 mm in a distance of 48 cm. Determine the flowrate

Answer 1

Answer:

2.47x10⁻⁴ m³/s

Explanation:

The water goes from point (1), with a diameter of 21 mm, to the point (2), with a diameter of 10 mm. By the continuity equation, the flow rate must be constant in all faucet, and the flow rate in the area (A) multiplied by the velocity (V), so:

A₁V₁ = A₂V₂

The area of a circuference cross section is (π/4)*d², where d is the diameter (in meters) so:

(π/4)*d₁² *V₁ = (π/4)*d₂² *V₂

(π/4)*(0.021)²*V₁ = (π/4)*(0.01)²*V₂

V₁ = (0.01)²*V₂/(0.021)²

V₁ = 0.2268V₂

V₂ = 4.41V₁

By the Bernouli's equation:

p₁ + ρV₁²/2 + γz₁ = p₂ + ρV₂²/2 + γz₂

Where p is the pressure, ρ is the density of the liquid, γ is the specific weight of the liquid (ρ*g, where g is the gravity acceleration), and z is the high of the point. Because both points are subjected to the same surrounding pressure, p₁ = p₂, and the terms are canceled.

ρV₁²/2 + γ*(z₁ - z₂) = ρV₂²/2

z₁ - z₂ = h, which is the distance of the points (48 cm = 0.48 m), so:

(ρ/2)*(V₂² - V₁²) = γh

(V₂² - V₁²) = 2γh/ρ

(V₂² - V₁²) = 2ρgh/ρ

(V₂² - V₁²) = 2gh

With V₂ = 4.41V₁, and g = 9.8 m/s²:

(4.41V₁)² - V₁² = 2*9.8*0.48

18.45V₁² = 9.408

V₁² = 0.51

V₁ = 0.714 m/s

The flow rate (ΔV) is A₁*V₁, so:

ΔV = (π/4)*(0.021)²*0.714

ΔV = 2.47x10⁻⁴ m³/s