Question

A horizontal tube has different cross-sectional areas at point a and
b.The diameter of a is 4 cm and that of bis 2 cm. Two manometer limbs are attached at a and
b. When a liquid of density 8.0 g cm−3cm-3 flows through the tube, the pressure difference between the limbs of the manometer is 8 cm. Calculate the rate of flow of the liquid in the tube .

Answer 1

The rate of flow of the liquid in the tube is 406 cm³ / sec

Given -

• Diameter of A = 4 cm
• Diameter of B = 2 cm
• Density of liquid = 8.0 g/cm³
• The pressure difference between the limbs of the manometer is = 8 cm

Area (A₁) = π r² = π (2)² = 4 π cm²

Area (A₂) = π (1)² = π cm²

V₂ > V₁

P₁ + 1/2 ρ V₁² = P₂ + 1/2 ρ V₂²

P₁ - P₂ = 1/2 ρ (V₂² - V₁²)

2 ΔP /ρ = V₂² - V₁²

We can write V₁ in terms of area.

By equation of continuity

A₁ V₁ = A₂ V₂

V₁ = A₂ V₂/ A₁

By substituting the value of V₁ we get

2 ΔP /ρ = V₂² - A₂² V₂²/A₁²

2 ΔP /ρ = V₂² ( A₁² - A₂²/A₁)

V₂ = ( A₁²2Δ P / A₁² - A₂² ρ)^1/2

V₂ = A₁ ( 2ΔP/(A₁² - A₂²) ρ)^1/2

Q = A₂V₂ = A₁V₁ where Q is the volume flow rate

Q = A₁ A₂ ( 2ΔP/(A₁² - A₂²) ρ) ^1/2

By substituting the values

Q = 4 π × π ( 2 × 8 × 980 / 15 π²)

Q = 406 cm³ / sec