Water flows through a horizontal pipe that reduces from 200 mm to 100 mm diameter. The velocity in the larger section is 2 m/s. Neglecting losses, find the flow rate, the velocity in the smaller section, and the pressure drop between the two sections. Take ρ = 1000 kg/m³.
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Use continuity for velocity, then Bernoulli with z₁ = z₂.
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Worked answer.
Step 1: Flow rate.
A₁ = (π/4) × 0.2² = 0.031416 m².
Q = A₁V₁ = 0.031416 × 2 ≈ 0.0628 m³/s (about 62.8 litres per second).
Step 2: Velocity in the smaller pipe (continuity). For an incompressible fluid, A₁V₁ = A₂V₂, so
V₂ = V₁ × (d₁/d₂)² = 2 × (200/100)² = 2 × 4 = 8 m/s.
Step 3: Pressure drop (Bernoulli). Along a streamline, with no losses and z₁ = z₂:
p₁ + ½ρV₁² = p₂ + ½ρV₂².
p₁ − p₂ = ½ρ(V₂² − V₁²) = 0.5 × 1000 × (64 − 4) = 30,000 Pa = 30 kPa.
In terms of head: 30,000 / (1000 × 9.81) ≈ 3.06 m of water.
Check. A₂ = (π/4) × 0.1² = 0.007854 m², and A₂V₂ = 0.007854 × 8 = 0.0628 m³/s, matching Q. The pressure falls as the velocity rises, which is what Bernoulli's equation predicts: the kinetic energy increase comes from the pressure energy.
Where this is used. This is the principle of the venturimeter. Measuring p₁ − p₂ with a manometer and knowing the area ratio gives the flow rate. In a real meter, a discharge coefficient (typically around 0.95–0.99 for a venturi) accounts for losses.
Common mistake. Assuming velocity scales with the diameter ratio (giving 4 m/s) instead of the area ratio, which is the diameter ratio squared.
Likely follow-up. "What if the pipe is vertical with the small end 1 m higher?" Add the elevation term: p₁ − p₂ = 30 kPa + ρg × 1 = 30 + 9.81 ≈ 39.8 kPa.
Original UniCircle question. Last reviewed 28 Oct 2026.