Blasius solution and the momentum integral equation

The laminar boundary-layer equations, the Blasius similarity solution and its flat-plate results, and von Kármán's momentum integral applied to assumed profiles and turbulent plates, with drag and wall-shear numericals.

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Why it matters

Skin friction is roughly half the cruise drag of a transport aircraft, and its first estimate always comes from flat-plate boundary-layer results. The Blasius solution gives the exact laminar answer; von Kármán's momentum integral gives quick, surprisingly accurate answers for any assumed profile, works with pressure gradients, and is the basis of many fast design methods for wings and ducts.

Key ideas

Boundary-layer equations. For steady, 2-D, incompressible flow with ∂p/∂y ≈ 0: u ∂u/∂x + v ∂u/∂y = U_e dU_e/dx + ν ∂²u/∂y², with ∂u/∂x + ∂v/∂y = 0. Boundary conditions: u = v = 0 at y = 0; u → U_e as y → ∞.

Blasius solution (zero pressure gradient). For a flat plate at zero incidence U_e = U is constant. Blasius showed that velocity profiles at different x are geometrically similar when y is scaled with √(νx/U). With the similarity variable η = y·√(U/(νx)) and a stream function ψ = √(νUx)·f(η), the PDEs reduce to one ordinary differential equation, 2f‴ + f·f″ = 0, with f(0) = f′(0) = 0 and f′(∞) = 1. Here u/U = f′(η). It is solved numerically; the key result is f″(0) = 0.332, which gives the wall shear. u/U reaches 0.99 at η ≈ 5.0, giving δ = 5.0x/√Re_x. Validity: laminar, steady, incompressible, zero pressure gradient, Re_x large enough for boundary-layer theory but below transition (about 5×10⁵ for a quiet stream).

Results. Wall shear decreases along the plate as x^(−1/2) because the layer thickens. The total drag of a plate of length L is the integral of τ_w, which equals ρU²θ(L) per unit width: drag is the momentum removed from the stream.

Von Kármán momentum integral equation. Integrating the boundary-layer momentum equation across the layer gives an ordinary differential equation for the thicknesses: τ_w/(ρU_e²) = dθ/dx + (2 + H)(θ/U_e)(dU_e/dx). For a flat plate (dU_e/dx = 0) it becomes τ_w = ρU²·dθ/dx. It is exact as an integral statement; the approximation enters only through the assumed velocity profile, which must satisfy u = 0 at the wall, u = U at y = δ and ∂u/∂y = 0 at y = δ (and preferably ∂²u/∂y² = 0 at the wall for zero pressure gradient).

Procedure (Pohlhausen-type, flat plate).

  1. Assume u/U = F(y/δ).
  2. Compute θ = c₁δ and τ_w = μU·F′(0)/δ.
  3. Substitute into τ_w = ρU²dθ/dx, giving δ·dδ = (F′(0)/c₁)(ν/U)dx.
  4. Integrate from δ = 0 at x = 0: δ²/2 = (F′(0)/c₁)·νx/U.
  5. Then c_f and C_D follow. Integral results are insensitive to the profile detail: even a straight line gives c_f within about 13 % of Blasius.

Turbulent flat plate. The same integral method with the 1/7-power profile and the empirical wall law τ_w = 0.0225ρU²(ν/(Uδ))^(1/4) gives δ = 0.37x/Re_x^0.2 and C_D = 0.074/Re_L^0.2 (for turbulent flow from the leading edge, roughly 5×10⁵ < Re_L < 10⁷).

Formulas

η = y·√(U/(ν·x)), u/U = f′(η), 2f‴ + f·f″ = 0

  • η dimensionless; U (m/s); ν (m²/s); x, y (m).

Blasius: δ = 5.0·x/√Re_x, δ* = 1.721·x/√Re_x, θ = 0.664·x/√Re_x, τ_w = 0.332·ρ·U²/√Re_x, c_f = 0.664/√Re_x, C_D = 1.328/√Re_L

  • c_f: local skin-friction coefficient; C_D: average over length L (one side); Re_L = UL/ν.

τ_w / (ρ·U_e²) = dθ/dx + (2 + H)·(θ/U_e)·(dU_e/dx)

  • Von Kármán momentum integral; τ_w (Pa); θ (m); H = δ*/θ.

Profile results (flat plate, laminar): linear δ/x = 3.46/√Re_x, c_f = 0.577/√Re_x; parabolic u/U = 2(y/δ) − (y/δ)²: δ/x = 5.48/√Re_x, c_f = 0.730/√Re_x; cubic u/U = 1.5(y/δ) − 0.5(y/δ)³: δ/x = 4.64/√Re_x, c_f = 0.646/√Re_x.

Turbulent: δ = 0.37·x/Re_x^0.2, c_f = 0.0576/Re_x^0.2, C_D = 0.074/Re_L^0.2.

D = C_D·½·ρ·U²·b·L = ρ·U²·θ(L)·b

  • D: drag on one side of a plate of width b (N).

Worked examples

Example 1 (standard): momentum integral with a parabolic profile. Given: u/U = 2(y/δ) − (y/δ)², flat plate.

  1. θ = δ∫₀¹ (2s − s²)(1 − 2s + s²) ds = 2δ/15.
  2. τ_w = μ(∂u/∂y)_wall = 2μU/δ.
  3. τ_w = ρU²·dθ/dx: 2μU/δ = ρU²(2/15)dδ/dx, so δ dδ = 15(ν/U)dx.
  4. Integrate: δ² = 30νx/U, so δ/x = √30/√Re_x = 5.48/√Re_x.
  5. c_f = τ_w/(½ρU²) = 4ν/(Uδ) = 4/(5.48√Re_x) = 0.730/√Re_x. Answer: δ/x = 5.48/√Re_x, c_f = 0.730/√Re_x (Blasius: 5.0 and 0.664; within 10 %).

Example 2 (GATE level): laminar plate drag and local shear. Given: air, ρ = 1.225 kg/m³, ν = 1.5×10⁻⁵ m²/s, U = 5 m/s; plate L = 1 m, width b = 0.5 m, one side wetted.

  1. Re_L = U·L/ν = 5/1.5×10⁻⁵ = 3.33×10⁵, laminar throughout. √Re_L = 577.4.
  2. C_D = 1.328/√Re_L = 0.002300.
  3. ½ρU² = 0.5 × 1.225 × 25 = 15.31 Pa.
  4. D = 0.002300 × 15.31 × 0.5 × 1 = 0.0176 N.
  5. Check: θ(L) = 0.664 × 1/577.4 = 1.150 mm; ρU²θb = 1.225 × 25 × 0.00115 × 0.5 = 0.0176 N.
  6. Local wall shear at the trailing edge: τ_w = 0.332ρU²/√Re_L = 0.332 × 30.625/577.4 = 0.0176 Pa. Answer: D ≈ 0.0176 N; τ_w(L) ≈ 0.0176 Pa.

Example 3: turbulent plate. U = 50 m/s, L = 3 m, same air: Re_L = 10⁷, C_D = 0.074/Re_L^0.2 = 0.00295, drag per metre width = 0.00295 × ½ × 1.225 × 50² × 3 = 13.5 N.

Common mistakes

  • Mixing the local c_f = 0.664/√Re_x with the average C_D = 1.328/√Re_L (a factor of 2).
  • Using Blasius when Re_x exceeds the transition value, or for flows with pressure gradients.
  • Writing η = y·√(Ux/ν) (wrong placement of x).
  • In the momentum integral, using δ* instead of θ in τ_w = ρU²dθ/dx.
  • Forgetting that drag on both sides of a plate is twice the one-side value.
  • Choosing an assumed profile that does not satisfy u = U and ∂u/∂y = 0 at y = δ.

For GATE AE

Expect: δ, τ_w and drag of laminar plates from Blasius coefficients; applying the momentum integral to linear, parabolic, cubic or sine profiles to obtain δ/x and c_f; ratios such as drag of the front half versus the whole plate (laminar drag ∝ √L, so the front half carries 1/√2 ≈ 71 %); and turbulent plate drag with the 1/5-power laws. Practise the four-step integral method until it takes two minutes.

Quick check

  1. What is f″(0) in the Blasius solution?
  2. Laminar plate: what fraction of total drag acts on the front half?
  3. Momentum integral for a flat plate in one line?
  4. Average laminar C_D at Re_L = 10⁶?

Answers: 1. 0.332 2. 1/√2 ≈ 0.707 3. τ_w = ρU²·dθ/dx 4. 1.328/1000 = 0.00133

Interview questions

All Fluid Mechanics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Blasius solution in fluid mechanics?Concept

    The Blasius solution is a mathematical solution to the boundary layer equations for a steady, incompressible flow over a flat plate. It provides a similarity solution for the velocity profile in the boundary layer, assuming a laminar flow and no pressure gradient along the plate. The solution is derived using a similarity transformation that reduces the partial differential equations to an ordinary differential equation.

  2. 2.Explain the momentum integral equation in the context of boundary layer theory.Concept

    The momentum integral equation is a simplified form of the Navier-Stokes equations applied to the boundary layer. It relates the rate of change of momentum in the boundary layer to the shear stress at the wall and the pressure gradient. This equation is derived by integrating the momentum equation across the boundary layer thickness and is used to estimate boundary layer thickness, shear stress, and drag force on the surface.

  3. 3.How does the Blasius solution help in understanding boundary layer behavior?Concept

    The Blasius solution provides a detailed description of the velocity profile within the boundary layer for a laminar flow over a flat plate. It helps in understanding how the velocity changes from zero at the wall to the free stream velocity outside the boundary layer. This solution is crucial for predicting the boundary layer thickness, shear stress, and drag force, which are important for designing aerodynamic surfaces.

  4. 4.Why is the Blasius solution limited to laminar flow over a flat plate?Application

    The Blasius solution is derived under the assumption of a steady, incompressible, and laminar flow with no pressure gradient along the plate. These assumptions limit its applicability to situations where the flow remains laminar and the surface is flat. In cases of turbulent flow or curved surfaces, the assumptions break down, and the Blasius solution is no longer valid.

  5. 5.What would happen if the pressure gradient is not zero in the Blasius solution?Application

    If the pressure gradient is not zero, the assumptions used to derive the Blasius solution are violated. The presence of a pressure gradient would affect the velocity profile and the boundary layer thickness, leading to a different flow behavior. In such cases, more complex solutions or numerical methods are required to accurately describe the boundary layer.

  6. 6.How is the momentum integral equation used to estimate drag force on a flat plate?Application

    The momentum integral equation can be used to estimate the drag force by relating the shear stress at the wall to the change in momentum within the boundary layer. By integrating the shear stress over the surface of the plate, the total drag force can be calculated. This approach provides an approximate method to estimate drag without solving the full Navier-Stokes equations.

  7. 7.What are the limitations of using the momentum integral equation for boundary layer analysis?Application

    The momentum integral equation provides an approximate solution and is based on assumptions such as a thin boundary layer and a known velocity profile. It may not accurately capture complex flow phenomena such as separation, transition to turbulence, or three-dimensional effects. Additionally, it requires empirical correlations or assumptions about the velocity profile, which can introduce errors.

  8. 8.Calculate the laminar boundary layer thickness at a distance of 1 meter from the leading edge of a flat plate, given a free stream velocity of 10 m/s and kinematic viscosity of 1.5 x 10^-5 m²/s.Numerical

    Re_x = Ux/ν = 10 × 1/1.5×10⁻⁵ = 6.67×10⁵. Using Blasius, δ = 5.0x/√Re_x = 5.0/816.5 = 6.1×10⁻³ m, about 6.1 mm. Note that Re_x is slightly above the usual flat-plate transition value of about 5×10⁵, so this assumes a quiet free stream that keeps the layer laminar; in a disturbed stream the layer here would already be turbulent and much thicker.

  9. 9.Determine the shear stress at the wall for the same conditions as the previous question.Numerical

    The shear stress (τ_w) at the wall can be calculated using τ_w = μ * (du/dy)|_wall. For a Blasius boundary layer, τ_w = 0.332 * (ρ * U^2 / Re_x^0.5). Given ρ = 1.225 kg/m³ (density of air at sea level), U = 10 m/s, and Re_x = 666,667, τ_w = 0.332 * (1.225 * 10^2 / 666,667^0.5) ≈ 0.332 * (122.5 / 816.5) ≈ 0.0498 N/m².

  10. 10.Explain how the Blasius solution and momentum integral equation complement each other in boundary layer analysis.Concept

    The Blasius solution provides a detailed velocity profile for a laminar boundary layer over a flat plate, which is useful for understanding the flow characteristics and calculating parameters like shear stress and boundary layer thickness. The momentum integral equation, on the other hand, offers a more general approach to estimate these parameters by integrating the momentum equation across the boundary layer. Together, they provide both detailed and approximate methods for analyzing boundary layers, allowing engineers to choose the appropriate level of complexity for their analysis.

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