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
