Incompressible Aerodynamics: Fluid Basics, Viscous Flow, Potential Flow, Aerofoils and Wings

Section 3 of the GATE Aerospace Engineering (AE) paper, Aerodynamics, is several subjects under one heading — basic fluid mechanics, viscous flow and boundary layers, two-dimensional potential flow, aerofoil and finite-wing theory, and compressible flow with shocks and expansions. The section is given two chapters, split where the physics splits: this one treats everything in which density can be held constant, and the second treats compressible flow, where it cannot. This chapter moves from fluid kinematics and the conservation laws, through dimensional analysis and similarity, Couette and Hagen-Poiseuille flow and the boundary layer, to the elementary potential flows and their superposition; then aerofoil nomenclature, the Kutta-Joukowski theorem, the Kutta condition, the starting vortex and thin-aerofoil theory; then the finite wing, induced drag and Prandtl’s lifting-line theory; and it closes with the special topic of pressure measurement by U-tube manometer and Pitot probe.

1. Fluid kinematics and the conservation laws

A flow can be described by following particles (Lagrangian) or by watching fixed points (Eulerian); the link is the material derivative D/Dt = ∂/∂t + V·∇. Three kinds of line are drawn: a streamline is everywhere tangent to the velocity at one instant; a pathline is the track of one particle over time; a streakline joins all particles that have passed through one point, which is what a dye or smoke filament shows. In steady flow all three coincide; in unsteady flow they generally differ. Vorticity ω = ∇ × V measures local rotation, and a flow with ω = 0 everywhere is irrotational.

The conservation laws are written for a control volume (integral form) or at a point (differential form). Mass: ∂/∂t∫ρ dV + ∮ρV·dA = 0, or ∂ρ/∂t + ∇·(ρV) = 0. Momentum: the rate of change of momentum in the volume plus its net outflow equals pressure, body and viscous forces — at a point, ρDV/Dt = −∇p + ρf + viscous terms, which for a Newtonian fluid are the Navier-Stokes equations and without viscosity the Euler equations. Energy: the first law for a flowing fluid, which in steady adiabatic flow with no work reduces to constant stagnation enthalpy h + V²/2. A Newtonian fluid is one whose shear stress is linear in the rate of strain, τ = μ du/dy in simple shear, with μ independent of the strain rate; air and water are Newtonian.

ℹ️ When is a flow incompressible?
Incompressibility is a statement about the flow, not the fluid: the density of each particle does not change, Dρ/Dt = 0, which by continuity is the same as ∇·V = 0. For a gas this holds closely when the Mach number is below about 0.3, where density changes are under about 5% (ρ0/ρ = (1 + 0.2M²)^2.5 = 1.046 at M = 0.3). Air over a car or a landing aeroplane is incompressible in this sense; air in a jet-engine intake at cruise is not.

2. Dimensional analysis and dynamic similarity

Buckingham’s π theorem says a relation among n variables involving k independent dimensions can be written among n − k dimensionless groups. For the force on a body, F = f(ρ, V, l, μ, a) has six variables and three dimensions, so C_F = F/(½ρV²l²) depends on two groups only: the Reynolds number Re = ρVl/μ and the Mach number M = V/a (plus the shape and the angle of attack). Two flows are dynamically similar when they are geometrically similar and every relevant π group is equal; then all the force and pressure coefficients are equal too. This is why a wind-tunnel model result can be scaled to flight — and why it is hard: a 1/10 scale model in air at the same temperature needs ten times the speed to match Re, which destroys the Mach match. Pressurised or cryogenic tunnels raise Re by raising density or lowering viscosity instead.

3. Couette flow, Hagen-Poiseuille flow and the boundary layer

Two exact solutions of the Navier-Stokes equations show viscosity at work. In plane Couette flow a fluid of depth h lies between a fixed plate and one moving at U with no pressure gradient; the velocity is linear, u = Uy/h, and the shear stress is the same everywhere, τ = μU/h. In Hagen-Poiseuille flow a pressure gradient drives fully developed laminar flow along a pipe of radius R: the profile is the paraboloid u = (−dp/dx)(R² − r²)/(4μ), the centre-line speed is twice the mean, the volume flow is Q = πR⁴Δp/(8μL), and the Darcy friction factor is f = 64/Re. Between parallel plates the same analysis gives a parabola whose maximum is 1.5 times the mean.

At high Reynolds number viscosity matters only in a thin boundary layer next to the surface, where the velocity rises from zero (no-slip) to the outer inviscid value. Its thickness δ is conventionally where u = 0.99U. Two integral thicknesses are more useful: the displacement thickness δ* = ∫(1 − u/U)dy, the distance the outer flow is pushed out, and the momentum thickness θ = ∫(u/U)(1 − u/U)dy, a measure of the momentum lost to skin friction. For the laminar flat plate (Blasius) δ ≈ 5.0x/√Re_x, δ* = 1.72x/√Re_x, θ = 0.664x/√Re_x and the local skin-friction coefficient is c_f = 0.664/√Re_x, so δ grows as √x. A turbulent layer grows faster, roughly δ ≈ 0.37x/Re_x^(1/5). Transition on a smooth flat plate is usually taken near Re_x ≈ 5 × 10⁵. An adverse pressure gradient (pressure rising downstream) slows the near-wall fluid until the wall shear falls to zero and the layer separates.

⚠️ δ* and θ are not the boundary-layer thickness
For the Blasius layer δ* ≈ δ/2.9 and θ ≈ δ/7.5, and the shape factor H = δ/θ = 2.59. A question that gives 0.664 or 1.72 is asking for θ or δ, not δ; the coefficient 5.0 belongs to δ alone.

4. Two-dimensional potential flow and Bernoulli’s equation

In irrotational flow the velocity is the gradient of a potential, V = ∇φ, and in incompressible flow continuity makes φ satisfy Laplace’s equation ∇²φ = 0; in two dimensions the stream function ψ (u = ∂ψ/∂y, v = −∂ψ/∂x) satisfies it too, and lines of constant ψ are streamlines. Because Laplace’s equation is linear, elementary solutions can be added. Bernoulli’s equation p + ½ρV² = constant holds along a streamline in steady, inviscid, incompressible flow, and everywhere in the flow field when the flow is also irrotational — which is what turns a potential-flow velocity field into a pressure distribution.

The elementary two-dimensional flows
FlowVelocityφψ
Uniform streamu = V∞V∞xV∞y
Source (Λ > 0) or sink (Λ < 0)V_r = Λ/(2πr)(Λ/2π) ln r(Λ/2π)θ
Doublet of strength κFrom a source-sink pair in the limit(κ/2π) cos θ/r−(κ/2π) sin θ/r
Point vortex of strength Γ (clockwise positive)V_θ = −Γ/(2πr)−(Γ/2π)θ(Γ/2π) ln r
  • Uniform stream + source = Rankine half-body. The stagnation point sits Λ/(2πV∞) upstream of the source, and far downstream the body is Λ/V∞ thick.
  • Uniform stream + source + equal sink = Rankine oval, a closed body.
  • Uniform stream + doublet = non-lifting circular cylinder of radius R = √(κ/(2πV∞)). The surface speed is 2V∞ sin θ, so C_p = 1 − 4 sin²θ: +1 at the stagnation points, −3 at the shoulders, and the symmetric pressure gives zero drag — d’Alembert’s paradox.
  • Add a vortex = lifting cylinder. The lift per unit span is ρ∞V∞Γ. As Γ grows the two stagnation points move together towards the bottom of the cylinder, meet there when Γ = 4πV∞R, and leave the surface above that.

5. Aerofoils — nomenclature, Kutta-Joukowski, the Kutta condition and thin-aerofoil theory

An aerofoil is defined by its chord line (leading to trailing edge, length c), its mean camber line (midway between upper and lower surfaces), the maximum camber and its position, the thickness distribution and maximum thickness, and the leading-edge radius. In the NACA four-digit series, NACA 2412 has a maximum camber of 2% of the chord located at 40% chord and a maximum thickness of 12% of the chord; NACA 0012 is symmetric and 12% thick. The section coefficients are c_l = L′/(q∞c), c_d = D′/(q∞c) and c_m = M′/(q∞c²), the primes denoting per unit span. The centre of pressure is where the resultant acts, and it moves with incidence; the aerodynamic centre is the point about which c_m does not change with α, and it is the more useful reference.

The Kutta-Joukowski theorem says that for any two-dimensional body in steady, inviscid, incompressible flow the lift per unit span is L′ = ρ∞V∞Γ, where Γ is the circulation round the body, and the drag is zero. What fixes Γ for an aerofoil is the Kutta condition: the flow leaves a sharp trailing edge smoothly, with finite velocity there — in vortex-sheet language, γ(TE) = 0. Viscosity is what enforces it in reality. Kelvin’s circulation theorem (circulation round a closed fluid curve is constant in inviscid, barotropic flow) then explains how the circulation appears: as the aerofoil starts moving, a starting vortex of strength −Γ is shed from the trailing edge and left behind, so the total circulation of the fluid stays zero while the aerofoil carries a bound vortex +Γ.

Thin-aerofoil theory replaces the aerofoil by a vortex sheet on its camber line and imposes flow tangency and the Kutta condition. For a symmetric aerofoil it gives c_l = 2πα (α in radians), c_m about the quarter chord equal to zero, the centre of pressure and the aerodynamic centre both at c/4, and c_m,LE = −c_l/4. For a cambered aerofoil c_l = 2π(α − α_L0), with α_L0 negative for positive camber, and c_m,c/4 = constant (negative for positive camber) — so the aerodynamic centre is still at the quarter chord, but the centre of pressure now moves aft as c_l falls. The lift slope 2π per radian (about 0.11 per degree) is the result to remember.

6. Finite wings — induced drag and Prandtl’s lifting-line theory

On a finite wing the higher pressure below spills round the tips, rolling the trailing vortex sheet into two tip vortices. By Helmholtz’s theorems a vortex filament cannot end in the fluid, so the bound vortex turns at the tips and trails downstream — the horseshoe-vortex model. The trailing vortices induce a downwash w at the wing, which tilts the local relative wind down by the induced angle α_i ≈ w/V∞. The wing section therefore sees a smaller effective angle α_eff = α − α_i, and the local lift, perpendicular to the tilted wind, has a rearward component: the induced drag, drag due to lift that exists even in inviscid flow.

Prandtl’s lifting-line theory models the wing as a bound vortex of varying strength Γ(y) along the span, shedding trailing vorticity dΓ/dy; the downwash at each station follows from the Biot-Savart law and the fundamental equation matches it to the section lift. Its central result is the elliptic lift distribution, Γ(y) = Γ0√(1 − (2y/b)²): the downwash is constant along the span, α_i = C_L/(πAR), and C_D,i = C_L²/(πAR), the least induced drag possible for a given lift and span. Any other planform loading gives C_D,i = C_L²/(πeAR) with span efficiency e < 1. The finite wing’s lift slope falls below the section value a0: a = a0/(1 + a0/(πeAR)), so a wing of aspect ratio 6 with a0 = 2π has a = 4.71 per radian. Here AR = b²/S; an untapered, untwisted elliptic planform produces the elliptic loading.

🧠 Induced drag falls with speed, parasite drag rises
In level flight C_L = 2W/(ρV²S), so the induced drag D_i = qS·C_L²/(πeAR) = 2W²/(πeρV²b²) falls as 1/V² and depends on span loading W/b, not on aspect ratio as such. Parasite drag rises as V². That is why induced drag dominates at take-off and climb, why gliders have long spans, and why the minimum-drag speed is where the two are equal.

7. Pressure measurement — U-tube manometers and the Pitot probe

A U-tube manometer measures a pressure difference by the height of a liquid column: with manometric liquid of density ρ_m under a fluid of density ρ, p1 − p2 = (ρ_m − ρ)gh, and for air over water or mercury the ρ term is usually negligible. Inclining the tube at angle θ to the horizontal spreads the same vertical height over a length L = h/sin θ, raising the sensitivity. A Pitot tube facing the flow brings the fluid to rest at its mouth and senses the stagnation (total) pressure p0; a static port on the side senses the static pressure p; a Pitot-static tube combines both. In incompressible flow Bernoulli gives V = √(2(p0 − p)/ρ). In subsonic compressible flow the isentropic relation p0/p = (1 + (γ − 1)M²/2)^(γ/(γ−1)) must be used instead; in supersonic flow a bow shock stands ahead of the probe, and the Rayleigh Pitot formula relates the measured total pressure behind the shock to the free-stream Mach number.

⚠️ A Pitot tube alone does not give velocity
The Pitot mouth reads p0 only. Velocity needs p0 − p, so the static pressure must be measured as well — by a separate static port or the static holes of a Pitot-static tube. In a manometer calculation, the density in V = √(2Δp/ρ) is the density of the flowing air, while the density in Δp = ρ_m g h is the manometer liquid’s; mixing the two is the commonest arithmetic error.

Key takeaways

  • Streamlines, pathlines and streaklines coincide in steady flow; incompressible means Dρ/Dt = 0 ⇔ ∇·V = 0, a good model below M ≈ 0.3; a Newtonian fluid has τ = μ du/dy.
  • Couette: linear profile, τ = μU/h. Hagen-Poiseuille: parabolic, u_max = 2ū, Q = πR⁴Δp/(8μL), f = 64/Re. Blasius: δ = 5.0x/√Re_x, δ* = 1.72x/√Re_x, θ = 0.664x/√Re_x.
  • Source, sink, doublet and vortex superpose: uniform + doublet = cylinder with C_p = 1 − 4 sin²θ; adding a vortex gives L′ = ρ∞V∞Γ (Kutta-Joukowski).
  • Thin aerofoil: c_l = 2π(α − α_L0), aerodynamic centre at c/4; symmetric: c_m,c/4 = 0, c_m,LE = −c_l/4. The Kutta condition fixes Γ; the starting vortex balances it (Kelvin).
  • Lifting line: elliptic loading ⇒ constant downwash, C_D,i = C_L²/(πAR), minimum induced drag; in general C_D,i = C_L²/(πeAR) and a = a0/(1 + a0/(πeAR)).
  • Manometer: Δp = (ρ_m − ρ)gh; Pitot-static: V = √(2(p0 − p)/ρ) in incompressible flow, isentropic relation when compressible, Rayleigh Pitot formula when supersonic.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following statements are correct?

    1. In steady flow, streamlines, pathlines and streaklines coincide
    2. A dye filament released continuously from one point marks a streakline
    3. For an incompressible flow, the divergence of the velocity field is zero
    4. An incompressible flow must also be irrotational
    Show answer

    Answer: A — In steady flow, streamlines, pathlines and streaklines coincide; B — A dye filament released continuously from one point marks a streakline; C — For an incompressible flow, the divergence of the velocity field is zero

    With a velocity field that does not change in time, particles follow the instantaneous streamlines, so all three lines coincide; a dye stream from one point is by definition a streakline; and Dρ/Dt = 0 with continuity gives ∇·V = 0. Incompressibility says nothing about vorticity: Couette and Poiseuille flows are incompressible and rotational.
  2. A 1/10 scale model of a wing is tested in air at the same temperature and pressure as the full-size flight condition. To match the Reynolds number, the tunnel speed must be:

    1. The same as the flight speed
    2. Ten times the flight speed, which makes the Mach numbers unequal
    3. One tenth of the flight speed
    4. √10 times the flight speed
    Show answer

    Answer: B — Ten times the flight speed, which makes the Mach numbers unequal

    Re = ρVl/μ with ρ and μ unchanged, so V must rise in proportion as l falls: ten times. The speed of sound is unchanged, so the model Mach number becomes ten times the flight value — Re and M cannot both be matched this way, which is why pressurised and cryogenic tunnels exist.
  3. Oil of viscosity 0.1 Pa·s fills a 5 mm gap between a fixed plate and a plate moving at 2 m/s, with no pressure gradient. What is the shear stress on the plates, in Pa?

    Numerical answer — type the value.

    Show answer

    Answer: 40

    Plane Couette flow has a linear profile, so du/dy = U/h everywhere and τ = μU/h = 0.1 × 2/0.005 = 40 Pa. Leaving the gap in millimetres gives 0.04 Pa, a thousand times too small.
  4. In fully developed laminar (Hagen-Poiseuille) flow in a circular pipe, the ratio of the centre-line velocity to the mean velocity is:

    1. 1.5
    2. 2
    3. 1.22
    4. 4/3
    Show answer

    Answer: B — 2

    The profile u = u_max(1 − r²/R²) averages to u_max/2 over the circular area, so u_max/ū = 2. The value 1.5 belongs to laminar flow between parallel plates, where the parabola is averaged over a line rather than a circle.
  5. Air (kinematic viscosity 1.5 × 10⁻⁵ m²/s) flows at 10 m/s over a flat plate. Using the Blasius result δ = 5.0x/√Re_x, what is the boundary-layer thickness 0.5 m from the leading edge, in mm (to two decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 4.33

    Re_x = Vx/ν = 10 × 0.5/1.5 × 10⁻⁵ = 3.33 × 10⁵ (laminar, below 5 × 10⁵), √Re_x = 577.4, and δ = 5.0 × 0.5/577.4 = 4.33 × 10⁻³ m = 4.33 mm. Using 0.664 would give the momentum thickness, about 0.58 mm, not δ.
  6. In inviscid flow past a non-lifting circular cylinder, what is the pressure coefficient at the top of the cylinder (θ = 90° from the front stagnation point)?

    Numerical answer — type the value.

    Show answer

    Answer: -3

    The surface speed is 2V∞ sin θ, so C_p = 1 − (V/V∞)² = 1 − 4 sin²θ = 1 − 4 = −3 at θ = 90°. Forgetting to square the factor 2 gives −1. The pressure is symmetric fore and aft, so the drag is zero — d’Alembert’s paradox.
  7. A uniform stream is superposed on a doublet. The resulting flow represents:

    1. A Rankine half-body
    2. Flow past a non-lifting circular cylinder
    3. Flow past a lifting circular cylinder
    4. A Rankine oval
    Show answer

    Answer: B — Flow past a non-lifting circular cylinder

    Uniform stream plus doublet gives the closed streamline r = R = √(κ/(2πV∞)), a circle, with no circulation and hence no lift. A half-body needs a source, an oval needs a separated source and sink, and a lifting cylinder needs a vortex added as well.
  8. An aerofoil section in a stream of density 1.2 kg/m³ and speed 50 m/s carries a circulation of 20 m²/s. By the Kutta-Joukowski theorem, what is its lift per unit span, in N/m?

    Numerical answer — type the value.

    Show answer

    Answer: 1200

    L′ = ρ∞V∞Γ = 1.2 × 50 × 20 = 1200 N/m. There is no factor ½: that belongs to the dynamic pressure in the definition of c_l, and inserting it here gives 600 N/m.
  9. By thin-aerofoil theory, what is the lift coefficient of a symmetric aerofoil at 4° angle of attack (to two decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 0.44

    c_l = 2πα with α in radians: 4° = 0.0698 rad, so c_l = 2π × 0.0698 = 0.4386 ≈ 0.44. Using 2π × 4 treats degrees as radians and gives 25.1, an impossible lift coefficient.
  10. For a thin symmetric aerofoil in inviscid incompressible flow, which statements are correct?

    1. The aerodynamic centre is at the quarter-chord point
    2. The centre of pressure is at the quarter-chord point at every angle of attack
    3. The moment coefficient about the leading edge is −c_l/4
    4. The zero-lift angle of attack is negative
    Show answer

    Answer: A — The aerodynamic centre is at the quarter-chord point; B — The centre of pressure is at the quarter-chord point at every angle of attack; C — The moment coefficient about the leading edge is −c_l/4

    With no camber, c_m,c/4 = 0, so the quarter chord is both the aerodynamic centre and the centre of pressure, and transferring the lift L′ acting at c/4 to the leading edge gives c_m,LE = −c_l/4. A symmetric section has zero lift at zero incidence; only positive camber makes α_L0 negative.
  11. In the NACA four-digit designation 2412, the digits "4" and "12" indicate respectively:

    1. Maximum camber of 4% and thickness of 12% at 20% chord
    2. Position of maximum camber at 40% chord and maximum thickness of 12% chord
    3. Design lift coefficient 0.4 and thickness 12%
    4. Position of maximum thickness at 40% chord and camber of 12%
    Show answer

    Answer: B — Position of maximum camber at 40% chord and maximum thickness of 12% chord

    First digit: maximum camber in per cent of chord (2%); second: its position in tenths of chord (0.4c); last two: maximum thickness in per cent of chord (12%). A design lift coefficient is encoded only in the five- and six-digit series.
  12. When an aerofoil starts from rest and acquires a bound circulation Γ, the starting vortex shed from its trailing edge has strength:

    1. Zero, because the flow is inviscid
    2. Γ in the same sense as the bound vortex
    3. Γ in the opposite sense to the bound vortex
    4. 2Γ in the opposite sense
    Show answer

    Answer: C — Γ in the opposite sense to the bound vortex

    Kelvin’s theorem keeps the circulation round a large fluid contour enclosing both aerofoil and wake at its initial value, zero. So the shed starting vortex must carry −Γ, equal and opposite to the bound vortex. Viscosity at the sharp trailing edge is what creates the pair, but the balance itself is inviscid bookkeeping.
  13. A wing with an elliptic lift distribution and aspect ratio 8 operates at C_L = 0.8. What is its induced drag coefficient (to four decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 0.0255

    For elliptic loading e = 1 and C_D,i = C_L²/(πAR) = 0.64/(π × 8) = 0.64/25.13 = 0.0255. Omitting π gives 0.08, and using C_L instead of C_L² gives 0.0318.
  14. An untwisted wing of aspect ratio 6 with an elliptic lift distribution uses aerofoil sections whose lift slope is 2π per radian. By lifting-line theory, what is the wing’s lift-curve slope, in per radian (to two decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 4.71

    a = a0/(1 + a0/(πAR)) = 2π/(1 + 2π/(6π)) = 2π/(1 + 1/3) = 0.75 × 2π = 4.71 per rad. The induced angle C_L/(πAR) reduces the effective incidence, so the slope falls below 2π; forgetting it leaves 6.28.
  15. For a finite wing with an elliptic spanwise lift distribution, which statements are correct according to Prandtl’s lifting-line theory?

    1. The downwash is constant along the span
    2. The induced drag is the minimum possible for the given lift and span
    3. The span efficiency factor e equals 1
    4. The induced drag is independent of the lift coefficient
    Show answer

    Answer: A — The downwash is constant along the span; B — The induced drag is the minimum possible for the given lift and span; C — The span efficiency factor e equals 1

    Elliptic Γ(y) gives w = Γ0/(2b) at every station, so α_i = C_L/(πAR) is uniform and C_D,i = C_L²/(πAR), the minimum, with e = 1. Induced drag is drag due to lift and grows as C_L², so the fourth statement is false.
  16. A Pitot-static tube in an air stream (density 1.225 kg/m³) is connected to a U-tube water manometer (water density 1000 kg/m³) that shows a deflection of 100 mm. Taking g = 9.81 m/s², what is the air speed, in m/s (nearest integer)?

    Numerical answer — type the value.

    Show answer

    Answer: 40

    Δp = (ρ_w − ρ_air)gh = (1000 − 1.225) × 9.81 × 0.1 = 979.8 Pa, and V = √(2Δp/ρ_air) = √(2 × 979.8/1.225) = √1599.7 = 40.0 m/s (neglecting the air column gives 40.02, the same to the nearest integer). Putting water density into the square root gives about 1.4 m/s.