Boundary layer theory, drag and lift

Why it matters

Viscous effects in most external flows are confined to a thin layer next to the surface. Prandtl's boundary-layer idea explains skin-friction drag on plates and tubes, the separation that causes form drag behind cylinders and spheres, and — through the analogy between momentum, heat and mass transfer — the film coefficients used for heat exchangers, dryers and catalytic surfaces. It also explains why dimples on a golf ball or roughness on a tube bank can reduce drag.

Key ideas

The boundary layer. At high Reynolds number, flow past a body can be split into an outer inviscid region (where Bernoulli applies) and a thin boundary layer at the wall in which velocity rises from zero (no slip) to about the free-stream value U. Its thickness δ is conventionally where u = 0.99U. Pressure is impressed on the boundary layer by the outer flow: ∂p/∂y ≈ 0 across it.

Growth on a flat plate (zero pressure gradient).

• Local Reynolds number Re_x = Ux/ν, with x from the leading edge.
• Laminar boundary layer (Blasius solution): δ ≈ 5x/√Re_x (4.91 in the exact solution), so δ grows as √x.
• Transition at a critical Re_x of about 5 × 10⁵ for a smooth plate in a quiet stream (it can range from about 3 × 10⁵ to 3 × 10⁶ depending on roughness and free-stream turbulence).
• Turbulent boundary layer (1/7 power-law): δ ≈ 0.37x/Re_x^0.2, so δ grows as x^0.8 — much faster.
• Other thicknesses: displacement thickness δ* (how far the outer flow is pushed away; δ* ≈ δ/3 for laminar) and momentum thickness θ (momentum deficit; enters the von Kármán momentum-integral equation τ_w = ρU²·dθ/dx).

Skin friction. The wall shear stress is high near the leading edge and falls as the boundary layer thickens. For a turbulent boundary layer it is much higher than for a laminar one at the same Re_x because eddies bring fast fluid close to the wall.

Separation. In an adverse pressure gradient (pressure rising in the flow direction, as on the rear of a cylinder or in a diffuser), the slow fluid near the wall is decelerated until the wall shear becomes zero — the separation point — after which reversed flow and a wake form. Separation causes large form (pressure) drag and loss of lift. A turbulent boundary layer has more momentum near the wall and resists separation better, so it separates later.

Drag and lift.

• Total drag = skin-friction drag (shear) + form or pressure drag (separation). Streamlined bodies are dominated by friction; bluff bodies by form drag.
• Drag coefficient C_D = F_D/(½ρU²A); lift coefficient C_L = F_L/(½ρU²A), with A the frontal (projected) area for bluff bodies or the plan area for plates and wings.
• Drag crisis: for a smooth sphere or cylinder at Re ≈ 2–3 × 10⁵, the boundary layer becomes turbulent before separating, the wake narrows and C_D drops sharply (from about 0.47 to about 0.1 for a sphere).
• Lift arises from a pressure difference between the lower and upper surfaces of an airfoil, set up by its shape and angle of attack (circulation); stall occurs when the boundary layer on the upper surface separates at high angle of attack.

Formulas

• Re_x = U·x/ν — local Reynolds number; U free-stream velocity (m/s), x (m), ν (m²/s).
• δ = 5·x/√Re_x — laminar thickness (m); Re_x < ≈5 × 10⁵.
• δ = 0.37·x/Re_x^0.2 — turbulent thickness (m), assumed turbulent from the leading edge.
• C_f,x = 0.664/√Re_x, C_D = 1.328/√Re_L — laminar local and average skin-friction coefficients.
• C_D = 0.074/Re_L^0.2 — turbulent average, 5 × 10⁵ < Re_L < 10⁷.
• C_D = 0.074/Re_L^0.2 − 1742/Re_L — mixed laminar–turbulent plate, transition at 5 × 10⁵.
• F_D = C_D·(½·ρ·U²)·A — drag force (N); A wetted area of one side for plate friction (m²).
• F_L = C_L·(½·ρ·U²)·A — lift force (N).

Worked examples

Example 1 (standard). Air (ρ = 1.2 kg/m³, ν = 1.5 × 10⁻⁵ m²/s) flows at 5 m/s along both sides of a smooth plate 0.5 m long (in the flow direction) and 1 m wide. Find the boundary-layer thickness at the trailing edge and the total friction drag.

1. Re_L = UL/ν = 5 × 0.5 / 1.5 × 10⁻⁵ = 1.67 × 10⁵ < 5 × 10⁵ → laminar throughout.
2. δ = 5L/√Re_L = 2.5/408.2 = 6.1 × 10⁻³ m (6.1 mm).
3. C_D = 1.328/√Re_L = 1.328/408.2 = 3.25 × 10⁻³.
4. Dynamic pressure ½ρU² = 0.5 × 1.2 × 25 = 15 Pa.
5. Drag on one side = 3.25 × 10⁻³ × 15 × 0.5 = 0.0244 N; both sides = 0.0488 N. δ ≈ 6.1 mm; F_D ≈ 0.049 N.

Example 2 (GATE level). Water (ρ = 1000 kg/m³, ν = 1 × 10⁻⁶ m²/s) flows at 2 m/s along one side of a plate 3 m long and 1 m wide. Take transition at Re = 5 × 10⁵. Find the transition point, the friction drag, and the boundary-layer thickness at the trailing edge.

1. Transition: x_cr = Re_cr·ν/U = 5 × 10⁵ × 10⁻⁶ / 2 = 0.25 m.
2. Re_L = 2 × 3/10⁻⁶ = 6 × 10⁶ → mixed boundary layer.
3. C_D = 0.074/Re_L^0.2 − 1742/Re_L = 0.074/22.68 − 1742/(6 × 10⁶) = 3.263 × 10⁻³ − 0.290 × 10⁻³ = 2.97 × 10⁻³.
4. F_D = C_D × ½ρU² × A = 2.97 × 10⁻³ × 2000 × 3 = 17.8 N.
5. δ at x = 3 m (turbulent): 0.37 × 3 / (6 × 10⁶)^0.2 = 1.11/22.68 = 0.049 m. x_cr = 0.25 m; F_D ≈ 17.8 N; δ ≈ 49 mm. Treating the whole plate as turbulent would overestimate the drag (19.6 N) by about 10 %.

Common mistakes

• Using the pipe transition value (about 2100–2300) for a flat-plate boundary layer; the plate value is about 5 × 10⁵ based on x.
• Using diameter or plate width as the length in Re_x; it is distance from the leading edge.
• Forgetting to double the drag when both sides of a plate are wetted.
• Saying turbulence always increases drag: it raises skin friction but can cut form drag on bluff bodies by delaying separation.
• Mixing up frontal area (bluff bodies) and wetted plan area (plates).

For GATE CH

Expect numericals on boundary-layer thickness and skin-friction drag on plates (laminar, turbulent and mixed), the location of transition, ratio questions (how δ or drag changes with U or x), and conceptual questions on separation, adverse pressure gradient and the drag crisis. The momentum-integral method with an assumed velocity profile is also a common derivation-style question.

Quick check

1. How does laminar δ change if the free-stream velocity is quadrupled?
2. What is the condition at the separation point?
3. Which separates earlier on a cylinder, a laminar or a turbulent boundary layer?
4. Find x_cr for air (ν = 1.5 × 10⁻⁵ m²/s) at 10 m/s with Re_cr = 5 × 10⁵.

Answers: 1. It halves (δ ∝ 1/√U). 2. Wall shear stress is zero, (∂u/∂y) at the wall = 0. 3. Laminar. 4. 0.75 m.