A Venturi meter has upstream diameter 0.15 m and throat diameter 0.10 m. If Δp between upstream and throat is 20 kPa, compute the discharge Q (ρ = 1000 kg/m^3, neglect losses).

• A. Q ≈ 0.050 m^3/s
• B. Q ≈ 0.056 m^3/s
• C. Q ≈ 0.060 m^3/s
• D. Q ≈ 0.070 m^3/s

Answer: (B)

In a Venturi meter, the flow is steady and incompressible, so the discharge is set by continuity and the Bernoulli energy balance. The smaller throat accelerates the fluid, causing a pressure drop Δp from upstream to throat. Using A1 v1 = A2 v2 and p1 + ½ρv1^2 = p2 + ½ρv2^2, we relate the pressure difference to the velocities and areas, then find the discharge Q = A2 v2.

Compute areas: A1 = π(0.15)^2/4 ≈ 0.01767 m^2, A2 = π(0.10)^2/4 ≈ 0.00785 m^2. Continuity gives v1 = (A2/A1) v2 = (4/9) v2.

Bernoulli gives Δp = p1 − p2 = ½ρ(v2^2 − v1^2) = ½ρ(1 − (A2/A1)^2) v2^2. Here (A2/A1)^2 = (4/9)^2 = 16/81, so 1 − (A2/A1)^2 = 65/81 ≈ 0.8025.

Thus v2^2 = 2Δp / [ρ(1 − (A2/A1)^2)] = 2×20000 / (1000×0.8025) ≈ 49.9, so v2 ≈ 7.07 m/s.

Discharge is Q = A2 v2 ≈ 0.00785 × 7.07 ≈ 0.0555 m^3/s, about 0.056 m^3/s.

So the best answer is approximately 0.056 m^3/s.