Bernoulli's equation and flow measurement

Euler and Bernoulli equations, heads and grade lines, the extended energy equation, and flow measurement with venturi, orifice, Pitot tube and tank orifices.

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Why it matters

Bernoulli's equation links pressure, speed and height in a flowing fluid. It is how venturimeters, orifice plates and Pitot tubes turn a pressure reading into a flow rate or a speed, how a tank's drain time is estimated, and — once losses and pump heads are added — the starting point of every pipe-system and pump-sizing calculation.

Key ideas

Euler's equation along a streamline. Applying Newton's second law to a small fluid element moving along a streamline, with only pressure and gravity acting (no viscosity), gives dp/ρ + V·dV + g·dz = 0. Integrating it for constant density gives Bernoulli's equation.

Bernoulli's equation and its assumptions. p/(ρg) + V²/(2g) + z = constant along a streamline, provided that:

  1. the flow is steady;
  2. the fluid is incompressible (liquids; gases below a Mach number of about 0.3);
  3. viscous effects are negligible (no friction losses);
  4. the equation is applied along a single streamline (for irrotational flow the constant is the same everywhere);
  5. no shaft work (pump, turbine) or heat transfer occurs between the two points.

Meaning of each term. Each term is energy per unit weight, measured in metres (a "head"): pressure head p/(ρg), velocity head V²/(2g) and elevation (datum) head z. Their sum is the total head H. Multiplied by ρg, the terms become pressures: static pressure p, dynamic pressure ½ρV² and hydrostatic pressure ρgz. Their sum is the stagnation (total) pressure plus ρgz.

Hydraulic and energy grade lines. The energy grade line (EGL) is drawn at height H; the hydraulic grade line (HGL) at p/(ρg) + z, which is a velocity head below the EGL. In an ideal flow the EGL is horizontal; in a real pipe it slopes down in the flow direction because of losses and jumps up at a pump. Where the HGL falls below the pipe, the pressure is below atmospheric.

Real flows — the extended energy equation. Between sections 1 and 2 of a pipe: H₁ + h_pump = H₂ + h_turbine + h_L, where h_L is the head lost to friction and fittings. A kinetic-energy correction factor α multiplies V²/(2g) when the velocity profile is not uniform (α = 2 for laminar pipe flow, about 1.05 for turbulent).

Flow measurement.

  • Venturimeter: converging cone, throat and gentle diverging cone. Throat velocity rises and pressure falls; the measured difference gives the ideal flow rate, corrected by a discharge coefficient C_d ≈ 0.95–0.99. Low permanent pressure loss.
  • Orifice meter: a sharp-edged plate with a hole. The jet contracts to a vena contracta just downstream, so C_d is much lower (about 0.6–0.65) and the permanent loss is large; it is cheap and compact.
  • Flow nozzle: between the two in both C_d and loss.
  • Pitot tube: a tube facing the flow brings fluid to rest at its tip (a stagnation point), so it senses stagnation pressure. A Pitot-static tube also senses static pressure through side holes; the difference is the dynamic pressure ½ρV², which gives the local speed.
  • Small orifice in a tank (Torricelli): jet speed √(2gH); actual discharge Q = C_d·a·√(2gH), with C_d = C_c × C_v.

Manometer reading as a head. A differential manometer with liquid of relative density S_m under a working liquid of relative density S reads x; the piezometric-head difference is h = x(S_m/S − 1), which is valid whether or not the meter is horizontal, because the manometer measures the change in (p/ρg + z).

Formulas

p₁/(ρg) + V₁²/(2g) + z₁ = p₂/(ρg) + V₂²/(2g) + z₂

  • p = static pressure (Pa), ρ = density (kg/m³), V = velocity (m/s), z = elevation above datum (m), g = 9.81 m/s². Steady, incompressible, inviscid, along a streamline.

p₁/(ρg) + α₁·V₁²/(2g) + z₁ + h_pump = p₂/(ρg) + α₂·V₂²/(2g) + z₂ + h_turbine + h_L

  • h = heads (m); α = kinetic energy correction factor (–). Real pipe flow.

Q = C_d · A₁·A₂ · √(2g·h) / √(A₁² − A₂²)

  • Q = flow rate (m³/s), A₁ = inlet area, A₂ = throat or orifice area (m²), h = piezometric head difference (m of flowing liquid), C_d = discharge coefficient (–). Venturi, orifice and nozzle meters.

h = x · (S_m / S − 1)

  • x = manometer deflection (m), S_m and S = relative densities of manometer and flowing liquids.

V = C_v · √(2·(p₀ − p) / ρ)

  • p₀ = stagnation pressure, p = static pressure (Pa), C_v ≈ 0.98–1 for a Pitot tube.

Q = C_d · a · √(2g·H)

  • a = orifice area (m²), H = head above orifice centre (m). Small orifice in a large tank.

t = 2·A_T·(√H₁ − √H₂) / (C_d · a · √(2g))

  • t = time to fall from H₁ to H₂ (s), A_T = tank cross-section (m²). Uniform tank, quasi-steady draining.

Worked examples

Example 1 — venturimeter (standard). A horizontal venturimeter, 200 mm inlet and 100 mm throat, carries water. A mercury–water differential manometer reads 150 mm. Take C_d = 0.98. Find the flow rate.

  1. Head difference: h = x·(S_m/S − 1) = 0.15 × (13.6 − 1) = 1.89 m of water.
  2. Areas: A₁ = π × 0.2²/4 = 0.03142 m²; A₂ = π × 0.1²/4 = 0.007854 m².
  3. √(2gh) = √(2 × 9.81 × 1.89) = 6.089 m/s.
  4. A₁A₂/√(A₁² − A₂²) = A₂/√(1 − (A₂/A₁)²) = 0.007854/√(1 − 0.0625) = 0.008111 m².
  5. Q = C_d × 0.008111 × 6.089 = 0.98 × 0.04939 = 0.0484 m³/s. Answer: Q ≈ 0.0484 m³/s (48.4 L/s).

Example 2 — Pitot-static tube and tank draining (GATE level). (a) A Pitot-static tube in an air duct (ρ = 1.2 kg/m³) is connected to a water manometer reading 40 mm. Find the air speed (C_v = 1).

  1. Dynamic pressure: Δp = ρ_w·g·x = 1000 × 9.81 × 0.04 = 392.4 Pa.
  2. V = √(2Δp/ρ) = √(2 × 392.4/1.2) = √654 = 25.57 m/s. Answer: about 25.6 m/s.

(b) A vertical cylindrical tank of cross-section 2 m² drains through a 50 mm sharp-edged orifice (C_d = 0.62) in its base. How long does the level take to fall from 4 m to 1 m?

  1. Orifice area a = π × 0.05²/4 = 1.963 × 10⁻³ m².
  2. t = 2·A_T·(√H₁ − √H₂)/(C_d·a·√(2g)) = 2 × 2 × (2 − 1)/(0.62 × 1.963 × 10⁻³ × 4.429).
  3. Denominator = 5.392 × 10⁻³, so t = 4/5.392 × 10⁻³ = 741.8 s. Answer: about 742 s (12.4 min).

Common mistakes

  • Applying Bernoulli across a pump, turbine or a lossy valve without adding the head terms.
  • Using the manometer deflection x directly as the head; convert with h = x(S_m/S − 1).
  • Using (D₂/D₁)² instead of (A₂/A₁)² = (D₂/D₁)⁴ in the venturi formula.
  • Forgetting C_d, or using the venturi value (≈0.98) for an orifice meter (≈0.6).
  • Using air density in Δp = ρgx when the manometer liquid is water.
  • Mixing gauge and absolute pressures between the two points.
  • Saying "faster flow always means lower pressure" — that holds only along a streamline with no work, losses or height change.

For GATE ME

Expect venturi and orifice meter flow rates with inclined meters and manometers, Pitot-tube speeds in air and water, pressure at a point in a siphon or nozzle, time to drain a tank, and energy-equation problems with a pump or turbine and stated losses. Conceptual MCQs test the assumptions of Bernoulli's equation, the meaning of HGL and EGL, and the relative sizes of C_d for different meters. Practise writing the head balance with clear datum and gauge-pressure choices.

Quick check

  1. List the assumptions behind Bernoulli's equation.
  2. What pressure does the tip of a Pitot tube sense?
  3. Why is the discharge coefficient of an orifice meter much lower than that of a venturimeter?
  4. In an ideal pipe flow, how is the EGL drawn?
  5. Water at 3 m/s and 200 kPa in a horizontal pipe speeds up to 5 m/s. What is the new pressure?

Answers: 1. steady, incompressible, inviscid, along a streamline, no shaft work or heat transfer; 2. stagnation (total) pressure; 3. the jet contracts to a vena contracta smaller than the orifice and there are larger losses; 4. horizontal (no losses); 5. 200 − 0.5 × 1000 × (25 − 9)/1000 = 192 kPa.

Interview questions

All Fluid Mechanics and Fluid Power interview questions

Try answering each one aloud before you open it.

  1. 1.What is Bernoulli's equation and what does it describe in fluid mechanics?Concept

    Bernoulli's equation is a principle in fluid mechanics that describes the conservation of energy in a flowing fluid. It states that the sum of the pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline. Mathematically, it is expressed as P + 0.5ρv² + ρgh = constant, where P is the pressure, ρ is the fluid density, v is the fluid velocity, g is the acceleration due to gravity, and h is the height above a reference point.

  2. 2.Explain the assumptions made in deriving Bernoulli's equation.Concept

    Bernoulli's equation comes from integrating Euler's equation along a streamline, so it assumes steady flow, incompressible fluid (constant density), negligible viscosity (no friction losses) and that the two points lie on the same streamline. It also assumes no shaft work (pumps, turbines) and no heat transfer between the points. If the flow is also irrotational, the Bernoulli constant is the same on all streamlines, so any two points can be linked. For real pipes we keep the form but add pump, turbine and loss heads.

  3. 3.How is Bernoulli's equation applied in measuring fluid flow using a Venturi meter?Application

    A Venturi meter uses the principle of Bernoulli's equation to measure fluid flow. It consists of a converging section, a throat, and a diverging section. As fluid flows through the converging section into the throat, its velocity increases, causing a decrease in pressure. By measuring the pressure difference between the inlet and the throat, the flow rate can be calculated using Bernoulli's equation and the continuity equation.

  4. 4.Why is a Pitot tube used in aircraft to measure airspeed?Application

    A Pitot tube measures airspeed by utilizing Bernoulli's principle. It captures the dynamic pressure of the airflow, which is the difference between the total pressure (measured by the Pitot tube) and the static pressure (measured by a separate static port). This pressure difference is proportional to the square of the airspeed, allowing the calculation of the aircraft's speed relative to the air.

  5. 5.What happens to the pressure in a fluid as its velocity increases, according to Bernoulli's equation?Application

    Along a streamline at constant elevation, with no losses and no pump or turbine in between, the sum p + ½ρV² stays constant, so a rise in velocity must be paid for by a fall in static pressure. This is why pressure drops at a venturi throat or nozzle exit. The trade is in energy per unit volume: Δp = ½ρ(V₂² − V₁²). The statement does not hold blindly when elevation changes, when there are friction losses, or across a pump, which raises both pressure and velocity.

  6. 6.Explain how Bernoulli's equation is used in the design of carburetors in internal combustion engines.Application

    In carburetors, Bernoulli's equation is used to mix air and fuel efficiently. As air flows through a narrow section of the carburetor, its velocity increases, causing a drop in pressure. This pressure drop creates a suction effect that draws fuel into the airflow from a jet, mixing it with the air. The resulting air-fuel mixture is then delivered to the engine for combustion.

  7. 7.What is the role of the continuity equation in conjunction with Bernoulli's equation in fluid flow analysis?Concept

    In steady flow the mass flow rate is the same at every section; for an incompressible fluid this means the volume flow rate Q = A₁V₁ = A₂V₂ is constant. Bernoulli's equation alone has two unknown velocities when areas change, so continuity supplies the second equation linking them. That pairing gives the venturi and orifice meter formula, Q = C_d·A₁A₂√(2gh)/√(A₁² − A₂²), and lets you find pressure changes in nozzles and reducers.

  8. 8.Calculate the flow rate of water through a horizontal pipe with a diameter of 0.1 m at the inlet and 0.05 m at the throat, given that the pressure difference between these points is 5000 Pa. Assume water density is 1000 kg/m³ and no losses.Numerical

    From continuity, v₂ = (A₁/A₂)v₁ = (0.1/0.05)²v₁ = 4v₁. Bernoulli for a horizontal pipe gives Δp = ½ρ(v₂² − v₁²) = ½ × 1000 × 15v₁², so 5000 = 7500v₁² and v₁ = 0.8165 m/s. With A₁ = π × 0.1²/4 = 0.007854 m², Q = A₁v₁ ≈ 0.00641 m³/s (6.41 L/s). A real venturi would deliver slightly less, by its discharge coefficient.

  9. 9.A fluid flows through a pipe with a velocity of 3 m/s and a pressure of 200 kPa. If the pipe narrows and the velocity increases to 5 m/s, what is the new pressure? Assume the fluid density is 850 kg/m³.Numerical
    1. Apply Bernoulli's equation: P₁ + 0.5ρv₁² = P₂ + 0.5ρv₂².
    2. Substitute known values: 200,000 + 0.5 * 850 * 3² = P₂ + 0.5 * 850 * 5².
    3. Calculate: 200,000 + 3825 = P₂ + 10625.
    4. Solve for P₂: P₂ = 200,000 + 3825 - 10625 = 193,200 Pa.
  10. 10.Why might Bernoulli's equation not be applicable in real-world fluid flow scenarios?Application

    Bernoulli's equation may not be applicable in real-world scenarios due to its assumptions. Real fluids often have viscosity, leading to energy losses due to friction, which Bernoulli's equation does not account for. Additionally, if the flow is turbulent or if there are energy inputs or outputs (like pumps or turbines), the equation's assumptions are violated. Compressibility effects in gases at high speeds can also invalidate the equation.

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