--- title: "First layer height from y+ estimate" subject: "Applied Computational Fluid Dynamics" chapter: 4 key: "semester_7/applied_computational_fluid_dynamics:first-cell-height-yplus-numerical" source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/first-cell-height-yplus-numerical site: BCT Engineering --- # First layer height from y+ estimate Applied Computational Fluid Dynamics · Chapter 4: Mesh Generation and Solver Setup - Asked in: year not given, 8 marks - Video: [Calculate y+ and First-Cell Height for ANSYS Meshing](https://www.youtube.com/watch?v=oMczkq1b07c) — Vyadh Aerospace - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/first-cell-height-yplus-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-4 ## Question Air ($\rho = 1.2$ kg/m³, $\mu = 1.8\times10^{-5}$ Pa·s) flows at 30 m/s over a flat plate 1.5 m long. A wall-resolved turbulence model requires $y^+ = 1$ at the first cell centre. Estimate (a) the plate Reynolds number, (b) the skin friction coefficient using $C_f=0.026\,Re_L^{-1/7}$, (c) the wall shear stress and friction velocity, (d) the first cell height, and (e) the number of prism layers with growth ratio 1.2 needed to cover the boundary layer thickness estimated from $\delta = 0.37\,L\,Re_L^{-1/5}$. ## Short answer Re_L = 3×10^6; Cf = 0.00309; wall shear stress = 1.67 Pa; friction velocity = 1.18 m/s. First cell height = 2.55×10^-5 m (0.025 mm). With growth ratio 1.2, about 30 prism layers are needed to cover the boundary layer thickness of 0.028 m. ## Answer Definition: $y^+=\dfrac{\rho u_\tau y}{\mu}$ with $u_\tau=\sqrt{\tau_w/\rho}$. The estimate is done before meshing using flat-plate correlations. #### (a) Reynolds number $$ Re_L=\frac{\rho UL}{\mu}=\frac{1.2\times30\times1.5}{1.8\times10^{-5}}=3.0\times10^{6} $$ (turbulent over most of the plate). #### (b) Skin friction coefficient $$ C_f=0.026\,(3\times10^6)^{-1/7}=0.026\times0.1188=3.09\times10^{-3} $$ #### (c) Wall shear stress and friction velocity $$ \tau_w=\tfrac12C_f\rho U^2=0.5\times3.09\times10^{-3}\times1.2\times30^2=1.67\ \text{Pa} $$ $$ u_\tau=\sqrt{\frac{\tau_w}{\rho}}=\sqrt{\frac{1.667}{1.2}}=1.179\ \text{m/s} $$ #### (d) First cell height Kinematic viscosity $\nu=\mu/\rho=1.5\times10^{-5}$ m²/s. For $y^+=1$ the cell centre is at $$ y=\frac{y^+\nu}{u_\tau}=\frac{1\times1.5\times10^{-5}}{1.179}=1.27\times10^{-5}\ \text{m} $$ The first cell has its centre at half its height, so the first cell height is $$ \Delta y_1=2y=2.55\times10^{-5}\ \text{m}\approx 0.025\ \text{mm} $$ #### (e) Number of layers Boundary layer thickness: $$ \delta=0.37\,L\,Re_L^{-1/5}=0.37\times1.5\times(3\times10^6)^{-0.2}=0.0281\ \text{m} $$ For $n$ layers with growth ratio $g=1.2$, total thickness is $\Delta y_1\dfrac{g^n-1}{g-1}\ge\delta$: $$ g^n\ge1+\frac{\delta(g-1)}{\Delta y_1}=1+\frac{0.0281\times0.2}{2.55\times10^{-5}}=221.4 $$ $$ n\ge\frac{\ln221.4}{\ln1.2}=29.6\ \Rightarrow\ n=30\ \text{layers} $$ Check: with 30 layers the total is $2.55\times10^{-5}\times\dfrac{1.2^{30}-1}{0.2}=0.0301$ m, which is above $\delta$. **Answer:** $Re_L=3\times10^6$; $C_f=3.09\times10^{-3}$; $\tau_w=1.67$ Pa; $u_\tau=1.18$ m/s; first cell height $\approx2.5\times10^{-5}$ m; about 30 prism layers. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For air at 30 m/s over a 1.5 m plate, what are Re_L and the skin friction coefficient? A. Re_L = 3 x 10^6; Cf = 3.09 x 10^-3 B. Re_L = 3 x 10^5; Cf = 3.09 x 10^-3 C. Re_L = 3 x 10^6; Cf = 0.0119 D. Re_L = 3 x 10^6; Cf = 0.026 Answer: A. Re_L = 1.2 x 30 x 1.5/1.8e-5 = 3e6, and Cf = 0.026 x Re^(-1/7) = 0.026 x 0.1188 = 3.09e-3. (Key point: Reynolds number and Cf) 2. How are the wall shear stress and friction velocity found? A. tau_w = Cf rho U^2 = 3.3 Pa; u_tau = tau_w/rho B. tau_w = 0.5 Cf rho U^2 = 1.67 Pa; u_tau = sqrt(tau_w/rho) = 1.18 m/s C. tau_w = 0.5 Cf rho U^2 = 1.67 Pa; u_tau = tau_w/rho = 1.39 m/s D. tau_w = Cf U = 0.09 Pa; u_tau = 0.3 m/s Answer: B. tau_w = 0.5 x 3.09e-3 x 1.2 x 900 = 1.67 Pa, and u_tau = sqrt(1.667/1.2) = 1.18 m/s. (Key point: Wall shear and friction velocity) 3. For y+ = 1 at the cell centre, what is the first cell height? A. 1.27 x 10^-5 m, because y = nu/u_tau is the cell height B. 2.55 x 10^-5 m, because the centre is at half the cell height C. 2.55 x 10^-4 m D. 5.1 x 10^-5 m Answer: B. y = y+ nu/u_tau = 1.5e-5/1.179 = 1.27e-5 m is the centre distance, so the cell height is twice that, 2.55e-5 m. (Key point: First cell height) 4. What condition determines the number of prism layers with growth ratio g = 1.2? A. n x first height = delta B. First height x (g^n - 1)/(g - 1) >= delta, giving about 30 layers C. First height x g^n = delta, giving 10 layers D. Layers = delta/first height, about 1100 layers Answer: B. The geometric sum of layer heights must cover delta = 0.0281 m, giving g^n >= 221.4 and n of about 29.6, so 30 layers. 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