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Lift and drag evaluation in external CFD

Short answer

Lift and drag are computed by integrating surface pressure and shear stress and resolving along and perpendicular to the free stream. With planform area 1 m², dynamic pressure 960 Pa, C_L=0.9, C_D=0.012: L=864 N, D=11.5 N, L/D=75.

Explain how lift and drag on an airfoil are evaluated from a CFD solution. A wing section of chord 0.5 m and span 2 m moves through air (ρ=1.2\rho = 1.2 kg/m³) at 40 m/s. A CFD simulation gives CL=0.9C_L = 0.9 and CD=0.012C_D = 0.012. Find the lift, drag and lift-to-drag ratio.

Evaluating forces from CFD

The aerodynamic force on a body is the integral of the surface pressure and wall shear stress over its surface SS:

F⃗=∮S(−p n^+τ⃗w)dS\vec F=\oint_S\left(-p\,\hat n+\vec\tau_w\right)dS

where n^\hat n is the outward unit normal. The force is resolved relative to the free-stream direction:

  • Drag DD: component parallel to the free stream, made up of pressure (form) drag and skin friction drag.
  • Lift LL: component perpendicular to the free stream.

Non-dimensional coefficients:

CL=L12ρU2S,CD=D12ρU2SC_L=\frac{L}{\tfrac12\rho U^2S},\qquad C_D=\frac{D}{\tfrac12\rho U^2S}

with SS = planform area (chord ×\times span). In practice, the CFD code sums the force on each wall face and reports CLC_L and CDC_D through "force monitors"; the angle of attack is set by the flow direction. Accuracy requires a fine boundary-layer mesh (y+≈1y^+\approx1 for k-ω\omega SST), correct transition and separation modelling, and a mesh-independence check. A far-field distance of 10 to 20 chords is used.

Numerical

Planform area S=0.5×2=1.0S=0.5\times2=1.0 m². Dynamic pressure:

q=12ρU2=0.5×1.2×402=960 Paq=\tfrac12\rho U^2=0.5\times1.2\times40^2=960\ \text{Pa} L=CLqS=0.9×960×1.0=864 NL=C_LqS=0.9\times960\times1.0=864\ \text{N} D=CDqS=0.012×960×1.0=11.52 ND=C_DqS=0.012\times960\times1.0=11.52\ \text{N} LD=CLCD=0.90.012=75\frac LD=\frac{C_L}{C_D}=\frac{0.9}{0.012}=75

Answer: L=864L=864 N, D=11.5D=11.5 N, L/D=75L/D=75.

✓ Answer checked by Claude Opus 5.5 (high effort)

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