Euler's equation and Bernoulli's equation

Why it matters

Bernoulli's equation is the most used, and most misused, equation in aerodynamics. It converts a Pitot reading into airspeed, relates the pressure distribution on a wing to the local velocity, and sizes nozzles, siphons and tank outlets. Euler's equation, from which it comes, shows exactly which assumptions are being made, so you know when Bernoulli can be trusted and when it cannot.

Key ideas

Euler's equation is Newton's second law for an inviscid fluid element: mass times acceleration equals the pressure force plus the body force. In vector form ρ DV/Dt = −∇p + ρg. The acceleration includes both local and convective parts. Euler's equation is valid for compressible or incompressible, steady or unsteady flow, as long as viscous stresses are negligible, which is true outside boundary layers and wakes.

Along a streamline (s-direction). For steady flow, Euler's equation becomes dp/ρ + V dV + g dz = 0. Integrating with ρ constant gives Bernoulli's equation: p + ½ρV² + ρgz is the same at every point on one streamline.

Normal to a streamline (n-direction). ∂p/∂n = ρV²/R (ignoring gravity), with n pointing away from the centre of curvature. Pressure rises outward across curved streamlines; this is what holds a curved flow on its curved path and what causes suction over a curved upper wing surface.

Assumptions of Bernoulli's equation.

1. Steady flow.
2. Incompressible flow (ρ constant; for air, M below about 0.3).
3. Inviscid flow (no friction losses, no shaft work, no heat transfer).
4. Applied along one streamline. If the flow is also irrotational, the constant is the same throughout the whole flow field, so Bernoulli can be applied between any two points. That is the usual situation for flow approaching a wing from a uniform free stream.

Pressures in aerodynamics.

• Static pressure p: the pressure felt by a probe moving with the flow.
• Dynamic pressure q = ½ρV².
• Total (stagnation) pressure p₀ = p + ½ρV², the pressure at a point where the flow is brought to rest without losses.
• Pressure coefficient Cp = (p − p∞)/q∞ = 1 − (V/V∞)² for incompressible flow. Cp = 1 at a stagnation point and is negative where flow is faster than the free stream.

Energy form. Dividing by ρg gives heads: p/(ρg) + V²/(2g) + z = H. The energy grade line (EGL) is at height H, the hydraulic grade line (HGL) is V²/(2g) below it. With real losses, add a head loss term h_L between sections (covered in pipe flow).

Unsteady Bernoulli. For unsteady irrotational flow, ∂φ/∂t + p/ρ + V²/2 + gz = C(t). It is needed for oscillating columns and starting flows.

Compressible form. For isentropic flow of a perfect gas the integral of dp/ρ gives γ/(γ − 1)·p/ρ + V²/2 = constant. Using incompressible Bernoulli above M ≈ 0.3 underestimates the stagnation pressure rise.

Lift, correctly. Faster flow over the upper surface goes with lower pressure (Bernoulli), but the speed difference is set by the circulation around the aerofoil (Kutta condition), not by any "equal transit time". Air over the top actually reaches the trailing edge first.

Formulas

ρ·DV/Dt = −∇p + ρ·g

• Euler's equation; ρ in kg/m³, p in Pa, V in m/s, g in m/s².

∂p/∂n = ρ·V² / R

• R: radius of curvature of the streamline (m); n points away from the centre of curvature.

p₁ + ½ρV₁² + ρgz₁ = p₂ + ½ρV₂² + ρgz₂

• Steady, incompressible, inviscid, along a streamline (anywhere if irrotational). z: elevation (m).

p/(ρg) + V²/(2g) + z = H

• Head form; H: total head (m).

p₀ = p + ½ρV², q = ½ρV², Cp = (p − p∞)/q∞ = 1 − (V/V∞)²

• p₀: stagnation pressure (Pa); q: dynamic pressure (Pa); Cp dimensionless.

V = √(2gh)

• Torricelli: ideal efflux speed from a large open tank with head h (m) above the outlet.

Worked examples

Example 1 (standard): tank outlet. Given: an open tank with water 5 m above a 50 mm sharp outlet (ideal flow).

1. Bernoulli from the free surface (p = p_a, V ≈ 0) to the jet (p = p_a): V = √(2gh) = √(2 × 9.81 × 5) = 9.905 m/s.
2. Q = A·V = (π/4)(0.05²) × 9.905 = 0.01945 m³/s. Answer: V ≈ 9.90 m/s, Q ≈ 19.4 L/s (a real orifice gives less; see flow measurement).

Example 2 (standard): rising, contracting pipe. Given: water; section 1: d = 0.2 m, V₁ = 2 m/s, p₁ = 150 kPa; section 2: d = 0.1 m, 3 m higher. Find p₂, neglecting losses.

1. Continuity: V₂ = V₁(d₁/d₂)² = 2 × 4 = 8 m/s.
2. p₂ = p₁ + ½ρ(V₁² − V₂²) − ρg(z₂ − z₁).
3. = 150 000 + 500 × (4 − 64) − 1000 × 9.81 × 3 = 150 000 − 30 000 − 29 430. Answer: p₂ = 90.57 kPa.

Example 3 (GATE level): airspeed and wing pressure. Given: aircraft at an altitude where ρ = 0.909 kg/m³; Pitot-static reading p₀ − p = 2.0 kPa. At a point on the upper surface the local speed is 1.3 V∞.

1. V∞ = √(2(p₀ − p)/ρ) = √(2 × 2000/0.909) = 66.3 m/s.
2. q∞ = p₀ − p = 2000 Pa.
3. Cp = 1 − (V/V∞)² = 1 − 1.69 = −0.69.
4. p − p∞ = Cp·q∞ = −0.69 × 2000 = −1380 Pa. Answer: V∞ ≈ 66.3 m/s; local pressure 1.38 kPa below ambient.

Common mistakes

• Applying Bernoulli across a pump, fan, turbine or a region with large losses without adding work or loss terms.
• Applying it between points on different streamlines in rotational flow (e.g. across a boundary layer).
• Using it for air at high Mach number. Above M ≈ 0.3 use the compressible form.
• Forgetting continuity: V₂ must come from A₁V₁ = A₂V₂ before Bernoulli is used.
• Mixing gauge and absolute pressures in the same equation.
• Explaining lift with "equal transit time". It is false.
• Using the static pressure where the total pressure is meant, or vice versa.

For GATE AE

Expect airspeed from Pitot-static readings at a given altitude, Cp and local velocity on aerofoils and cylinders, pressure change in contractions and ducts, efflux from tanks, and conceptual MCQs on the assumptions of Bernoulli and the pressure gradient across curved streamlines. Practise combining continuity with Bernoulli and switching between Cp, velocity ratio and pressure difference.

Quick check

1. What is Cp at a stagnation point in incompressible flow?
2. Find the dynamic pressure of air (ρ = 1.225 kg/m³) at 30 m/s.
3. In which direction does pressure increase across curved streamlines?
4. Can Bernoulli be applied across a fan?

Answers: 1. 1 2. 551.25 Pa 3. Away from the centre of curvature 4. No, not without adding the fan's work as a head or pressure rise