A generalized thermal/dynamic wall function: Part 3
In this post i summarize the initial problem and the procedure to determine the wall function value (i.e., the solution) for given
,
,
and
.
We looked for a solution
to the problem:
![\frac{dT^+}{dy^+}=\frac{Pr\left(1+F_T^+y^+\right)}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]} \frac{dT^+}{dy^+}=\frac{Pr\left(1+F_T^+y^+\right)}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}](/Forums/vbLatex/img/6f93b3151b1a95e271960766dd7c8d1d-1.gif)
with:

for given
. This represents, in nondimensional form, a certain set of viscous boundary conditions for velocity/temperature. In practice,
is a wall-function. The solution is given by:
![T^+=Pr \left[f^+\left(1 +F_T^+y^+\right)-g^+\right] T^+=Pr \left[f^+\left(1 +F_T^+y^+\right)-g^+\right]](/Forums/vbLatex/img/f3c170727c6edd66978d053ebcaa9c27-1.gif)
Where,
is given by:
![f^+=\int_0^{y+}{\frac{1}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}dz^+}=\frac{Pr_t}{kPr}\ln\left(\frac{y^++a}{a}\right)+ f^+=\int_0^{y+}{\frac{1}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}dz^+}=\frac{Pr_t}{kPr}\ln\left(\frac{y^++a}{a}\right)+](/Forums/vbLatex/img/c51e02b828e4362aaf78db892b7d6851-1.gif)
![\frac{\alpha}{a+4\alpha}\left\{\left(a-4\alpha\right)\ln\left[\frac{a\left[\left(y^+-\alpha\right)^2+\beta^2\right]}{2\alpha\left(y^++a\right)^2}\right]+\gamma\left[\arctan\left(\frac{y^+-\alpha}{\beta}\right)+\arctan\left(\frac{\alpha}{\beta}\right)\right]\right\} \frac{\alpha}{a+4\alpha}\left\{\left(a-4\alpha\right)\ln\left[\frac{a\left[\left(y^+-\alpha\right)^2+\beta^2\right]}{2\alpha\left(y^++a\right)^2}\right]+\gamma\left[\arctan\left(\frac{y^+-\alpha}{\beta}\right)+\arctan\left(\frac{\alpha}{\beta}\right)\right]\right\}](/Forums/vbLatex/img/555f1e220bd91848e372424c88a2ca07-1.gif)




![\eta = \sqrt[3]{1+\theta\left(\psi+\sqrt{\psi^2+\phi}\right)} \eta = \sqrt[3]{1+\theta\left(\psi+\sqrt{\psi^2+\phi}\right)}](/Forums/vbLatex/img/f2fcd46b8600cca0cf70bb0a2e3bf6fb-1.gif)



and
is given by a (typically) straightforward integration:

In practice, given
above, the integral in
is easily computable for
in the form of a polynomial of arbitrary degree. For the sake of conciseness we only considered
as constant, the resulting integral being
times the following function:
https://www.wolframalpha.com/input/?...Batan(b%2Fc)))
The whole procedure is implemented in the attached MATLAB/Octave script, where the analytical solution above (ta) is compared with the numerical one (tn), obtained by numerically integrating
.
The usage should be straightforward:
1) Pick up values for
,
,
and
(lines 8-11).
2) Choose plotting options (lines 5-6)
3) Choose how many points you want to use to integrate numerically
(line 7).
4) Run the script.
Note that the constant
has been calibrated using as reference the mixing length model with a constant
. The relative turbulent mixing length viscosity has been left commented in the script (line 60), so that you can use it to calibrate the model for different values of the constant (line 13). Note that the calibration is always done once using
and
. After that, any value of those parameters will be taken into account automatically by the formula.
In general, the value
is such that
at
. Hence, it could be also used as a mean to introduce roughness effects in the formulation.
In conclusion, it is worth mentioning what are the limitations of the present wall function with respect to the full solution of the turbulent equations with, say, the SA model:
1) The equations are not in their full form, but in the incompressible TBLE form, with constant RHS.
2) The underlying turbulent viscosity model is not formally sensitized to the pressure gradient.
However, note that the same limitations also apply to, say:
K. Suga et al. / Int. J. Heat and Fluid Flow 27 (2006) 852–866
which provides a relation similar to the present one but with a higher approximation on the turbulent viscosity.




We looked for a solution

![\frac{dT^+}{dy^+}=\frac{Pr\left(1+F_T^+y^+\right)}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]} \frac{dT^+}{dy^+}=\frac{Pr\left(1+F_T^+y^+\right)}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}](/Forums/vbLatex/img/6f93b3151b1a95e271960766dd7c8d1d-1.gif)
with:

for given


![T^+=Pr \left[f^+\left(1 +F_T^+y^+\right)-g^+\right] T^+=Pr \left[f^+\left(1 +F_T^+y^+\right)-g^+\right]](/Forums/vbLatex/img/f3c170727c6edd66978d053ebcaa9c27-1.gif)
Where,

![f^+=\int_0^{y+}{\frac{1}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}dz^+}=\frac{Pr_t}{kPr}\ln\left(\frac{y^++a}{a}\right)+ f^+=\int_0^{y+}{\frac{1}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}dz^+}=\frac{Pr_t}{kPr}\ln\left(\frac{y^++a}{a}\right)+](/Forums/vbLatex/img/c51e02b828e4362aaf78db892b7d6851-1.gif)
![\frac{\alpha}{a+4\alpha}\left\{\left(a-4\alpha\right)\ln\left[\frac{a\left[\left(y^+-\alpha\right)^2+\beta^2\right]}{2\alpha\left(y^++a\right)^2}\right]+\gamma\left[\arctan\left(\frac{y^+-\alpha}{\beta}\right)+\arctan\left(\frac{\alpha}{\beta}\right)\right]\right\} \frac{\alpha}{a+4\alpha}\left\{\left(a-4\alpha\right)\ln\left[\frac{a\left[\left(y^+-\alpha\right)^2+\beta^2\right]}{2\alpha\left(y^++a\right)^2}\right]+\gamma\left[\arctan\left(\frac{y^+-\alpha}{\beta}\right)+\arctan\left(\frac{\alpha}{\beta}\right)\right]\right\}](/Forums/vbLatex/img/555f1e220bd91848e372424c88a2ca07-1.gif)




![\eta = \sqrt[3]{1+\theta\left(\psi+\sqrt{\psi^2+\phi}\right)} \eta = \sqrt[3]{1+\theta\left(\psi+\sqrt{\psi^2+\phi}\right)}](/Forums/vbLatex/img/f2fcd46b8600cca0cf70bb0a2e3bf6fb-1.gif)



and


In practice, given





https://www.wolframalpha.com/input/?...Batan(b%2Fc)))
The whole procedure is implemented in the attached MATLAB/Octave script, where the analytical solution above (ta) is compared with the numerical one (tn), obtained by numerically integrating

The usage should be straightforward:
1) Pick up values for




2) Choose plotting options (lines 5-6)
3) Choose how many points you want to use to integrate numerically

4) Run the script.
Note that the constant




In general, the value



In conclusion, it is worth mentioning what are the limitations of the present wall function with respect to the full solution of the turbulent equations with, say, the SA model:
1) The equations are not in their full form, but in the incompressible TBLE form, with constant RHS.
2) The underlying turbulent viscosity model is not formally sensitized to the pressure gradient.
However, note that the same limitations also apply to, say:
K. Suga et al. / Int. J. Heat and Fluid Flow 27 (2006) 852–866
which provides a relation similar to the present one but with a higher approximation on the turbulent viscosity.
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