Boundary layer, drag and lift

Why it matters

Drag sets the fuel use of vehicles, the power of fans and conveyors moving air past products, the wind load on structures and the terminal speed of particles in separators. Lift keeps aircraft and drones aloft and drives wind-turbine and propeller blades. Both forces are decided by what happens in the thin boundary layer next to the surface, and by whether that layer stays attached.

Key ideas

The boundary layer. Because of no-slip, fluid at a wall is at rest relative to it, and viscous shear slows a thin layer next to the surface. Outside this boundary layer the flow behaves almost as inviscid, so Bernoulli and potential-flow results can be used there. Prandtl's idea of splitting the flow this way is the basis of external aerodynamics. The layer is thin when Re is large (δ/x ∝ Re_x^(−1/2) for laminar flow).

Thickness measures.

• Boundary-layer thickness δ: distance from the wall where u reaches 99 % of the free-stream speed U.
• Displacement thickness δ*: how far the wall would have to be moved outward to give the same flow rate in an inviscid flow — the "flow-rate deficit".
• Momentum thickness θ: the equivalent momentum deficit; it is directly linked to the drag on the plate.
• For any profile δ > δ* > θ; their ratio H = δ*/θ (the shape factor) is about 2.6 for laminar and about 1.3 for turbulent layers.

Laminar and turbulent layers on a flat plate. Starting at the leading edge the layer is laminar. It becomes turbulent at a critical local Reynolds number Re_x = Ux/ν of about 5 × 10⁵ (lower with roughness or free-stream turbulence). A turbulent layer grows faster (δ ∝ x^0.8 versus x^0.5), has a fuller velocity profile and a much larger wall shear stress, so its skin-friction drag is higher. Underneath a turbulent layer a thin viscous sublayer remains.

Momentum integral (von Kármán). For a flat plate with zero pressure gradient, the wall shear stress equals ρU² dθ/dx. Assuming a velocity profile (linear, parabolic, cubic or 1/7-power) gives approximate δ and drag; the exact laminar (Blasius) solution gives the coefficients in the formulas below.

Pressure gradient and separation. In a favourable pressure gradient (pressure falling downstream, as on the front of a body) the layer is stable. In an adverse gradient (pressure rising, as on the rear of a cylinder or the upper surface of a wing at high angle of attack) the slow near-wall fluid is decelerated further until the wall shear falls to zero and the flow reverses: the layer separates. Behind the separation point a wide, low-pressure wake forms. Turbulent layers carry more momentum near the wall and resist separation longer — which is why golf-ball dimples, or tripping a sphere's boundary layer, cut drag.

Drag. Total drag = skin-friction drag (wall shear) + pressure (form) drag (from the front-to-back pressure difference caused by the wake). Thin plates parallel to the flow have mainly friction drag; bluff bodies such as a flat plate normal to flow, a cylinder or a car have mainly pressure drag. Streamlining reduces pressure drag by delaying separation. The drag coefficient C_D depends on Re and shape; for a sphere it is 24/Re at Re < 1 (Stokes flow), about 0.4–0.5 for 10³ < Re < 2 × 10⁵, and drops sharply (the drag crisis) to about 0.1–0.2 when the layer turns turbulent before separating.

Lift. Lift is the force component normal to the free stream. On an aerofoil, its shape and angle of attack make the flow faster over the upper surface, with lower pressure there, so the net pressure force acts upward. Mathematically, lift per unit span equals ρUΓ (Kutta–Joukowski), where Γ is the circulation around the aerofoil; a spinning cylinder or ball generates lift the same way (Magnus effect). C_L rises roughly linearly with angle of attack until the upper-surface boundary layer separates and the wing stalls.

Formulas

Re_x = U·x / ν

• U = free-stream speed (m/s), x = distance from leading edge (m), ν = kinematic viscosity (m²/s).

δ / x = 5.0 / √Re_x, δ* = 1.72·x / √Re_x, θ = 0.664·x / √Re_x (laminar, Blasius)

• δ, δ*, θ in m. Valid for Re_x < about 5 × 10⁵.

C_f,x = 0.664 / √Re_x, C_D = 1.328 / √Re_L (laminar plate)

• C_f,x = local skin-friction coefficient τ_w/(½ρU²), C_D = average over length L.

δ / x = 0.37 / Re_x^0.2, C_D = 0.074 / Re_L^0.2 (turbulent plate, 1/7-power law, 5 × 10⁵ < Re_L < 10⁷)

F_D = C_D · ½·ρ·U² · A, F_L = C_L · ½·ρ·U² · A

• F = force (N), ρ = fluid density (kg/m³), A = reference area (m²): wetted area for plates, frontal area for bluff bodies, planform area for wings.

F_D = 3·π·μ·V·D, C_D = 24 / Re (Stokes, Re < 1)

V_t = (ρ_s − ρ_f)·g·D² / (18·μ)

• V_t = terminal velocity of a small sphere (m/s), ρ_s, ρ_f = sphere and fluid density (kg/m³), D = diameter (m), μ = fluid viscosity (Pa·s). Re < 1.

L′ = ρ·U·Γ

• L′ = lift per unit span (N/m), Γ = circulation (m²/s).

Worked examples

Example 1 — laminar flat plate (standard). Air (ρ = 1.2 kg/m³, ν = 1.5 × 10⁻⁵ m²/s) flows at 5 m/s along a smooth plate 1 m long and 0.5 m wide. Find the boundary-layer thickness at the trailing edge and the drag on one side.

1. Re_L = U·L/ν = 5 × 1/1.5 × 10⁻⁵ = 3.33 × 10⁵ < 5 × 10⁵, so the layer is laminar throughout.
2. δ = 5.0·L/√Re_L = 5.0 × 1/577.4 = 8.66 × 10⁻³ m.
3. C_D = 1.328/√Re_L = 1.328/577.4 = 0.00230.
4. Dynamic pressure ½ρU² = 0.5 × 1.2 × 25 = 15 Pa; area 0.5 m².
5. F_D = 0.00230 × 15 × 0.5 = 0.0173 N. Answer: δ ≈ 8.7 mm; drag ≈ 0.017 N per side. Transition would start at x = 5 × 10⁵ × 1.5 × 10⁻⁵/5 = 1.5 m, beyond the plate.

Example 2 — falling-sphere viscometer and car drag (GATE level). (a) A 2 mm steel ball (ρ_s = 7800 kg/m³) falls through oil (ρ_f = 900 kg/m³, μ = 0.5 Pa·s). Find its terminal velocity and check the Stokes assumption.

1. V_t = (ρ_s − ρ_f)·g·D²/(18μ) = 6900 × 9.81 × (0.002)²/(18 × 0.5) = 0.2708/9 = 0.0301 m/s.
2. Re = ρ_f·V·D/μ = 900 × 0.0301 × 0.002/0.5 = 0.108 < 1, so Stokes' law applies. Answer: V_t ≈ 30.1 mm/s. Timing such a ball is how a falling-sphere viscometer measures μ.

(b) A car with C_D = 0.30 and frontal area 2.2 m² travels at 30 m/s (108 km/h) in still air (ρ = 1.2 kg/m³). Find the drag and the power to overcome it.

1. F_D = C_D·½ρU²·A = 0.30 × 0.5 × 1.2 × 900 × 2.2 = 356.4 N.
2. P = F_D·U = 356.4 × 30 = 10 692 W. Answer: F_D ≈ 356 N; P ≈ 10.7 kW. Because P ∝ U³, driving 10 % faster needs about 33 % more aerodynamic power.

Common mistakes

• Using a turbulent formula when Re_x < 5 × 10⁵, or the laminar one far beyond it.
• Using x in Re_x but L in the average drag coefficient inconsistently — local values use x, average values use L.
• Using frontal area for flat-plate skin friction or wetted area for a bluff body; the reference area must match the C_D source.
• Forgetting that a plate in a stream has two wetted sides when asked for total drag.
• Thinking a turbulent boundary layer always increases total drag; it raises friction but can cut pressure drag sharply on bluff bodies.
• Applying Stokes' law without checking that Re < 1.
• Calling separation a result of the boundary layer "getting thick"; it is caused by an adverse pressure gradient.

For GATE ME

Expect calculations of δ, δ* and θ for given velocity profiles (linear, parabolic, sinusoidal) using the momentum integral, drag on flat plates (laminar and turbulent parts), ratios of thickness or drag at two stations or speeds (δ ∝ √x, drag ∝ U^1.5 for laminar), terminal velocity of small spheres, and drag and lift from C_D and C_L. Conceptual MCQs test separation, adverse pressure gradient, the drag crisis and streamlining. Practise integrating simple profiles for δ* and θ.

Quick check

1. At what local Reynolds number does a flat-plate boundary layer usually become turbulent?
2. How does laminar boundary-layer thickness vary with distance x from the leading edge?
3. What causes boundary-layer separation?
4. Which is larger for any profile: δ, δ* or θ?
5. Why do golf-ball dimples reduce drag?

Answers: 1. about 5 × 10⁵; 2. δ ∝ √x; 3. an adverse pressure gradient that decelerates near-wall fluid until the wall shear becomes zero; 4. δ is the largest, then δ*, then θ; 5. they trip the layer to turbulence, which stays attached longer and shrinks the low-pressure wake.