After understanding all the pre-processing for race car applications, there are some calculations that the engineer use to perform manually. These will be presented in this article as exercises. However, it is important to review the characteristics of the inflation layers (read more) presented in the previous article.

Inflation layers review

The boundary layer mesh is divided into three sublayers, the viscous sub layer or laminar, the buffer layer and the log-law region. These are measured by the non-dimensional parameter y+ = y/δv and have specific range, which are 1 < y+ < 5, 5 < y+ < 30 and 30 < y+ < 500, respectively. The objective of these calculation is identify the high of the log-law region and set the boundary layer mesh in order to match the log law region. This is one is a turbulent region with high Re, inertia forces and turbulent stress, it requires a high resolution mesh. The laminar layer is characterized by a low Re, laminar forces and turbulent model (SST, read more).

Exercise 1

Compute the boundary layer thicknesses δ_buff and δ_ir⁺, which are the buffer layer and the inner region thicknesses, respectively, according to the following parameters:

• Wheelbase: 2.86 m;
• Reference velocity: V_ref = 50 m/s;
• Density: ρ = 1.225 kg/m³;
• Dynamic viscosity: μ = 1.78∙10^-5 Pa∙s.

The first step is the Reynolds number (Re) calculation:

Re_x = (ρ∙u∙L)/μ = (1.225∙50∙2.86)/(1.78∙10^-5) = 3130.200365

The second step is the calculation of the friction coefficient. Applying the boundary layer theory over a flat plate, the formulae for the friction coefficient (C_f) is given by:

C_f = 0.455/ln²(0.06∙Re_x) = 0.455/ln²(0.06∙3130.200365) = 0.016599873

The third step is the calculation of the shear stress (τ_w):

τ_w = ½∙ρ∙u²∙C_f = ½∙1.225∙50²∙0.016599873 = 25.41855528 Pa

The fourth step is the calculation of the shear velocity (u_τ):

u_τ = √τ_w/ρ = √25.418/1.225 = 4.555199342 m/s

The fifth step is the calculation of the kinematic viscosity (ν):

ν = μ/ρ = 1.78∙10^-5/1.225 = 1.4531∙10^-5 m²/s

The sixth step is the calculation of the convective length (δ_v):

δ_v = ν/u_τ = 1.4531∙10^-5/4.55199342 = 3.192∙10^-6m

The final step is the calculation of the buffer layer and inner region. According to Figure 1 it is possible to notice that these layers are given by a non-dimensional parameter y⁺, which is the ratio between the vertical distance (y) and the convective length (δ_v). Hence, the buffer layer is between 5 and 30 y+, while the entire inner region has 500 y+. Once the convective length is calculated, it possible to calculate these ranges by the following statement:

δ_v∙y_bm⁺ < δ_buff < δ_v∙y_bM⁺

This means that the buffer layer range is between the product of lower limit y+ and the convective length and the product upper limit y+ and the convective length. Hence:

δ_v∙y_bm⁺ < δ_buff < δ_v∙y_bM⁺ → 3.192∙10^-6∙5 < δ_buff < 3.192∙10^-6∙30 → 1.6∙10^-5 < δ_buff < 1.6∙10^-3

Therefore, the inner region thickness δ_ir⁺ can be calculated by the following formulae:

δ_ir⁺ = δ_v∙y⁺ = 3.196∙10^-6∙500 = 1.6∙10^-3

Exercise 2

Compute the boundary layer setup parameter according to a typical near wall treatment and to the following parameters:

• Surface discretization: s0 = 8 mm;
• Wheel base: L = 3000 mm;
• Assymptotic velocity: U_∞ = 50 m/s;
• Density: ρ = 1.225 kg/m³;
• Dynamic viscosity: μ = 1.78∙10^-5 Pa∙s;
• Mean underwing pressure coefficient: C_p = – 2;
• Grow rate: GR = 1.1.

At this exercise it was given the pressure coefficient at the mean underwing, thus the process is different from the previous exercise. The first step is to calculate the velocity at the boundary layer.

U_bl = U – U_∞

C_p = 1 – (U/U_∞)² = – 2 → – (U/U_∞)² = – 3 → (U/U_∞)² = 3 → U² = 3∙50² → U = 50∙√3 = 86.60254038 m/s

U_bl = 86.6 – 50 = 36.60254038 m/s

Now the procedure is the same as the previous exercise, the second step is to calculate Re:

Re = ρ∙U∙L/μ = (1.225∙36.6∙3000∙10^-3)/(1.78∙10^-5) = 7556985.162

The third step is the calculation of the friction coefficient at the mean under wing. In this case, it is assumed that this component is a flat plate, thus the boundary layer theory for flat plates results in the following formulae:

C_f = 0.455/ln²(0.06∙Re_x) = 0.455/ln²(0.06∙7556985.162) = 0.002682159

The fourth step is the calculation of the shear stress over the flat plate:

τ_w = ½∙ρ∙U²∙C_f = ½∙1.225∙36.60²∙0.0027 = 2.200964379 Pa

The fifth step is the calculation of the shear velocity:

U_τ = √τ_w/ρ = √2.20/1.225 = 1.340412480 m/s

The sixth step is the calculation of the convective length:

ν = μ/ρ = 1.78∙10^-5/1.225 = 1.4531∙10^-5 m²/s

δ_v = ν/Uτ = 1.45∙10^-5/1.34 = 0.000010841 m

Hence, it is possible to apply the same procedure of the previous exercise to calculate the the boundary layer thicknesses, δ_buff and δ_ir.

δ_v∙y_bm⁺ < δ_buff < δ_v∙y_BM⁺ → 1.0841∙10^-5∙30 < δ_buff < 1.0841∙10^-5∙80

3.25221∙10^-4 < δ_buff < 8.67255∙10^-4

δ_ir⁺ = δ_v∙y⁺ = 1.0841∙10^-5∙500 = 5.420346∙10^-3 = 5.42 mm

Calculating the boundary layer size

With the information obtained from the last steps it is possible to size the boundary layer. The first step is to define the centroid cell height y_cfc. Considering the y+ of the first cell y_fc⁺ = 50:

y_cfc = y_fc⁺∙δ_v = 50∙1.0841∙10^-5 = 0.000542050 = 5.4205∙10^-4 m = 0.54205 mm

Understanding that the height of the element is approximately two times the centroid height (read more), it is possible to calculate h₀:

h₀ = 2∙y_cfc = 1.0841∙10-3 m = 1.0841 mm

The first aspect ratio method is applied:

FAR = s₀/h₀ = 8 mm/1.0841 mm = 7.379393045 mm

The height of the boundary layer inner region was previously calculated:

H₀ = δ_ir⁺ = δ_v∙y⁺ = 1.0841∙10^-5∙500 = 5.4205 mm

Hence, the last part of the exercise is to calculate the amount of layer for the boundary layer since the boundary layer thickness is already known, 5.42 mm. The formulae for the layers height is given by:

h_n = h₀∙GRⁿ

h1 = h₀∙GR = 1.0841∙1.1 = 1.19251 mm

h₂ = h₀∙GR² = 1.0841∙1.1² = 1.311761 mm

h₃ = h₀∙GR³ = 1.0841∙1.1³ = 1.4429371 mm

h₄ = h₀∙GR⁴ = 1.0841∙1.1⁴ = 1.587230810 mm

Hence, if all the layer were summed, it is possible to obtain the following value:

h₀ + h₁ + h₂ + h₃ + h₄ = 1.0841 + 1.19251 + 1.311761 + 1.4429371 + 1.587230810 = 6.61853891 mm

However, this result is incorrect since the boundary layer thickness is 5.42 mm. If the boundary layer mesh is set in this way, it will account the separation layer, thus reducing the accuracy of the mesh. To solve this problem the amount of layer is reduced.

h₀ + h₁ + h₂ + h₃ + h₄ = 1.0841 + 1.19251 + 1.311761 + 1.4429371 = 5.0313081 mm

Now the boundary layer mesh is totally inside the physical boundary layer, the result can be considered satisfactory. However, it is possible to adjust the grow rate in order to match the boundary layer mesh with the physical boundary layer more perfectly.