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 .
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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
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