Glycerine of density 1.25 × 10³ kg m⁻³ flows through a conical pipe. End areas are 10 cm² and 5 cm², and pressure drop is 3 N m⁻². The volume-flow rate is x × 10⁻⁵ m³ s⁻¹. Find x.

Question

G1 Physics · Accepted Answer

Continuity: A₁v₁ = A₂v₂ With A₁ = 10 cm² and A₂ = 5 cm²: v₁ = v₂/2 Bernoulli for a horizontal pipe: ΔP = (1/2)ρ(v₂² - v₁²) 3 = (1/2)(1.25×10³)[v₂² - (v₂/2)²] = (1/2)(1.25×10³)( 3 4 \frac{3}{4} 4 3 ​ )v₂² Solving gives: v₂ = 8×10⁻² m s⁻¹ Flow rate: Q = A₂v₂ = 5×10⁻⁴ × 8×10⁻² = 4×10⁻⁵ m³ s⁻¹ Final value: 4