In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of radius 0.500 cm at the same elevation as the first tube. The pressure differs by DP between the liquid in one tube and the liquid in the second tube. (a) Find the volume flow rate as a function of DP. Evaluate the volume flow rate for (b) DP 5 6.00 kPa and (c) DP 5 12.0 kPa. 49. The Venturi tube discussed

a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² - 1], b) Q = 3.4 10⁻² m³ / s, c) Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

We replace

ΔP = ½ ρ [ (v₁ A₁ / A₂) ² - v₁²]

ΔP = ½ ρ v₁² [ (A₁ / A₂) ² - 1]

Let's clear the speed

v₁ = √ 2ΔP / ρ[ (A₁ / A₂) ² - 1]

The expression for the flow is

Q = A v

Q = A₁ v₁

Q = A₁ √ 2ΔP / rho [ (A₁ / A₂) ² - 1]

The areas are

A₁ = π r₁

A₂ = π r₂

We replace

Q = π r₁ √ 2ΔP / rho [r₁² / r₂² - 1]

Let's calculate for the different pressures

r₁ = d₁ / 2 = 1.00 / 2

r₁ = 0.500 10⁻² m

r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

Q = π 0.5 10⁻² √ (2 6.00 10³ / (850 (0.5² / 0.25² - 1))

Q = 1.57 10⁻² √ (12 10³/2550)

Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

Q = 1.57 10⁻² √ (2 12 10³ / (850 3)

Q = 4.8 10⁻² m³ / s