Flow measurement: venturi, orifice and Pitot-static tube

Why it matters

Every aircraft shows its airspeed from a Pitot-static system, every wind tunnel sets its test speed from a pressure difference, and fuel, hydraulic and coolant lines are metered with Venturis and orifice plates. All of these devices are Bernoulli plus continuity, corrected with a coefficient for real-fluid losses. Getting the coefficient, the area ratio and the manometer reading right is what separates a correct flow rate from one that is 30 % off.

Key ideas

Differential-pressure meters. A restriction speeds the flow up; continuity gives the speed ratio from the area ratio, and Bernoulli turns the speed change into a pressure drop. Measuring the pressure drop therefore gives the flow rate, with Q proportional to √Δp.

Venturi meter. A smooth converging cone (about 20°), a throat, and a long diverging diffuser (5–15°) that recovers most of the pressure. Flow stays attached, so losses are small and the discharge coefficient C_d is high (about 0.95–0.99). Permanent pressure loss is typically 10–20 % of the measured difference. Drawback: long and expensive.

Orifice meter. A thin plate with a sharp-edged hole. The jet keeps contracting downstream to a vena contracta (area ratio C_c ≈ 0.61–0.65), then expands abruptly with large losses. C_d is about 0.6–0.65 and depends on the diameter ratio β = d/D, the tapping positions and Reynolds number; take it from the standard or calibration. It is cheap and compact but wastes much more pressure. A flow nozzle sits between the two.

Coefficients. C_d = actual Q / ideal Q = C_c × C_v, where C_v (velocity coefficient, about 0.97–0.99) accounts for friction and C_c for jet contraction. The "velocity of approach" factor 1/√(1 − β⁴) corrects for the upstream velocity not being zero.

Pitot tube and Pitot-static tube. A tube facing the flow brings fluid to rest at its mouth and reads stagnation (total) pressure p₀. Side holes parallel to the flow read static pressure p. The difference is the dynamic pressure, so V = √(2(p₀ − p)/ρ) in incompressible flow. A Pitot-static probe measures velocity at a point; integrating across a duct gives flow rate. Errors come from misalignment (yaw over about 10–15°), probe blockage, turbulence and, at high speed, compressibility.

Airspeeds in aviation. The airspeed indicator is calibrated with sea-level density ρ₀ = 1.225 kg/m³, so it shows (after instrument and position corrections) equivalent airspeed V_E = √(2q/ρ₀). True airspeed is V = V_E·√(ρ₀/ρ), higher at altitude. Above about Mach 0.3 the compressible Pitot relation must be used, and in supersonic flow a bow shock forms in front of the probe (Rayleigh Pitot formula; see compressible flow).

Manometer readings. For a U-tube with manometric liquid ρ_m under a flowing fluid ρ, the pressure difference is (ρ_m − ρ)·g·h, independent of the meter's inclination when both limbs are filled with the flowing fluid.

Formulas

Q = C_d · A₁·A₂ / √(A₁² − A₂²) · √(2Δp/ρ)

• Venturi or nozzle; A₁: inlet area (m²); A₂: throat area (m²); Δp = p₁ − p₂ (Pa); ρ (kg/m³); C_d dimensionless.

Q = C_d · A₀ · √(2Δp / (ρ·(1 − β⁴)))

• Orifice; A₀: orifice area (m²); β = d/D (dimensionless).

Δp = (ρ_m − ρ)·g·h or as head h_f = h·(ρ_m/ρ − 1)

• h: manometer deflection (m); h_f: head of flowing fluid (m).

V = √(2(p₀ − p)/ρ)

• Pitot-static, incompressible; p₀: stagnation pressure (Pa); p: static pressure (Pa).

V_E = √(2q/ρ₀) and V_TAS = V_E·√(ρ₀/ρ)

• q = p₀ − p (Pa); ρ₀ = 1.225 kg/m³.

Worked examples

Example 1 (standard): Venturi with a mercury manometer. Given: water, D₁ = 200 mm, d₂ = 100 mm, mercury deflection h = 250 mm, C_d = 0.98.

1. Areas: A₁ = 0.031416 m², A₂ = 0.0078540 m².
2. Head difference in water: h_f = h(ρ_m/ρ − 1) = 0.25 × 12.6 = 3.15 m.
3. Ideal: Q = A₁A₂/√(A₁² − A₂²) · √(2g·h_f) = 0.0081118 × √(61.803) = 0.06377 m³/s.
4. Actual: 0.98 × 0.06377 = 0.06249 m³/s. Answer: Q ≈ 0.0625 m³/s (62.5 L/s).

Example 2 (standard): orifice plate. Given: water, D = 200 mm, orifice d = 100 mm (β = 0.5), C_d = 0.62 (from the standard), Δp = 20 kPa.

1. A₀ = 0.0078540 m², 1 − β⁴ = 0.9375.
2. Q = C_d·A₀·√(2Δp/(ρ(1 − β⁴))) = 0.62 × 0.0078540 × √(40 000/937.5) = 0.62 × 0.0078540 × 6.532. Answer: Q ≈ 0.0318 m³/s.

Example 3 (GATE level): airspeed. (a) A Pitot-static tube in a wind tunnel (air ρ = 1.2 kg/m³) is connected to a water U-tube reading 50 mm.

1. Δp = (ρ_w − ρ_air)gh ≈ 1000 × 9.81 × 0.05 = 490.5 Pa (air column negligible).
2. V = √(2 × 490.5/1.2) = 28.6 m/s. (b) An aircraft's Pitot-static system measures q = 2000 Pa at an altitude where ρ = 0.9093 kg/m³.
3. Equivalent airspeed V_E = √(2q/ρ₀) = √(4000/1.225) = 57.1 m/s.
4. True airspeed V = √(2q/ρ) = √(4000/0.9093) = 66.3 m/s. Answer: V_E ≈ 57.1 m/s, V_TAS ≈ 66.3 m/s.

Common mistakes

• Forgetting the area-ratio factor A₁/√(A₁² − A₂²) (the velocity of approach) and using Q = A₂√(2Δp/ρ).
• Using h of mercury directly as a head of water. Convert with (ρ_m/ρ − 1), not ρ_m/ρ.
• Using C_d of a Venturi (≈ 0.98) for an orifice (≈ 0.6), or omitting C_d altogether.
• Using sea-level density at altitude when true airspeed is wanted.
• Using incompressible Pitot relations at high subsonic or supersonic speed.
• Unit slips between mm and m and between diameters and areas.

For GATE AE

Expect Pitot-static airspeed with a manometer reading, equivalent versus true airspeed at a given altitude, Venturi and orifice discharge with C_d, manometer conversion to head, and conceptual MCQs on why Venturis have lower permanent loss and where the vena contracta forms. Practise the full Venturi formula until it is automatic, and keep the ISA density table in mind.

Quick check

1. Air (ρ = 1.2 kg/m³), p₀ − p = 200 Pa. What is V?
2. A mercury–water manometer reads 100 mm. What is the head of water?
3. Which meter has the larger permanent pressure loss for the same flow and area ratio, Venturi or orifice?
4. q = 1500 Pa at sea level. What is V?

Answers: 1. 18.3 m/s 2. 1.26 m 3. Orifice 4. 49.5 m/s