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Interpretation of Spalding's viscosity damping coefficient

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Old   March 6, 2019, 10:19
Default Interpretation of Spalding's viscosity damping coefficient
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According to Spalding's law of the wall we have

y^+ = u^+ + exp(-kB)(exp(ku^+) - 1 - ku^+ - \frac{(ku^+)^2}{2} - \frac{(ku^+)^3}{3})

leading to

\mu_t = \mu *  exp(-kB)(exp(ku^+) - 1 - ku^+ - \frac{(ku^+)^2}{2}) , which can be interpreted as a damping function applied to the viscosity.

How does this damping function behave (or should behave) as y^+ becomes large. Also up till what values of u^+ does the expression for y^+ hold? It seems that in order for the latter to be equal to, say, 100, a negative value for the former is needed. How is this possible?
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Old   March 7, 2019, 05:39
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Quote:
Originally Posted by Lmath View Post
According to Spalding's law of the wall we have

y^+ = u^+ + exp(-kB)(exp(ku^+) - 1 - ku^+ - \frac{(ku^+)^2}{2} - \frac{(ku^+)^3}{3})

leading to

\mu_t = \mu *  exp(-kB)(exp(ku^+) - 1 - ku^+ - \frac{(ku^+)^2}{2}) , which can be interpreted as a damping function applied to the viscosity.

How does this damping function behave (or should behave) as y^+ becomes large. Also up till what values of u^+ does the expression for y^+ hold? It seems that in order for the latter to be equal to, say, 100, a negative value for the former is needed. How is this possible?
Let first generalize a little bit. Spalding Law of the wall is of the form:

y^+ = u^+ + e^{-kB}\left[e^{ku^{+}}-\sum_{i=0}^N \frac{\left(ku^+\right)^i}{i!}\right]

while the turbulent viscosity ratio implied by equilibrium wall model is:

\frac{\mu_t}{\mu}=\frac{1}{\frac{du^+}{dy^+}}-1

From the first one you can obtain by straightforward differentiation:

\frac{dy^+}{du^+}=1+ke^{-kB}\left[e^{ku^+}-\sum_{i=1}^N \frac{\left(ku^+\right)^{i-1}}{\left(i-1\right)!}\right] = 1+k\left[\left(y^+-u^+\right)+e^{-kB}\frac{\left(ku^+\right)^N}{N!}\right]

where the original law has been used again in the last passage.

It follows that:

\frac{\mu_t}{\mu}=k\left[\left(y^+-u^+\right)+e^{-kB}\frac{\left(ku^+\right)^N}{N!}\right]

Note that u^+\le y^+, and this is a damping just approaching the viscous sublayer (where u^+ \approx y^+). But even in that case, talking of damping is misleading, it is just that the turbulent viscosity is nondimensionalized by the molecular viscosity.
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Old   March 10, 2019, 10:29
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Originally Posted by sbaffini View Post
Let first generalize a little bit. Spalding Law of the wall is of the form:

y^+ = u^+ + e^{-kB}\left[e^{ku^{+}}-\sum_{i=0}^N \frac{\left(ku^+\right)^i}{i!}\right]

while the turbulent viscosity ratio implied by equilibrium wall model is:

\frac{\mu_t}{\mu}=\frac{1}{\frac{du^+}{dy^+}}-1

From the first one you can obtain by straightforward differentiation:

\frac{dy^+}{du^+}=1+ke^{-kB}\left[e^{ku^+}-\sum_{i=1}^N \frac{\left(ku^+\right)^{i-1}}{\left(i-1\right)!}\right] = 1+k\left[\left(y^+-u^+\right)+e^{-kB}\frac{\left(ku^+\right)^N}{N!}\right]

where the original law has been used again in the last passage.

It follows that:

\frac{\mu_t}{\mu}=k\left[\left(y^+-u^+\right)+e^{-kB}\frac{\left(ku^+\right)^N}{N!}\right]

Note that u^+\le y^+, and this is a damping just approaching the viscous sublayer (where u^+ \approx y^+). But even in that case, talking of damping is misleading, it is just that the turbulent viscosity is nondimensionalized by the molecular viscosity.

However, how would this function behave on the interval 0\leq y^+ \leq 100? I mean, we would expect the damping ( or ratio for that matter) to become large as we approach y^+ = 100. But in the expression u^+ is found. The only way to get rid of that is to substitute either u^+ \approx y^+ or u^+ \approx \frac{1}{k} * ln(y^+) + B. But then the idea of one expression for the whole region is lost, right?
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Old   March 10, 2019, 10:46
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Quote:
Originally Posted by Lmath View Post
However, how would this function behave on the interval 0\leq y^+ \leq 100? I mean, we would expect the damping ( or ratio for that matter) to become large as we approach y^+ = 100. But in the expression u^+ is found. The only way to get rid of that is to substitute either u^+ \approx y^+ or u^+ \approx \frac{1}{k} * ln(y^+) + B. But then the idea of one expression for the whole region is lost, right?
I don't think I understand your question. If you plot that function, it is cristal clear how that behaves (you just need to treat u as independent variable instead of y).
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Old   March 10, 2019, 10:50
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Let me state it more clearly: Spalding's law of the wall is not a relation valid for any law of the wall (i.e., any form of u+), so that you just put something in u+ and expect it to be valid.

It is a law of the wall, just expressed in reverse form (I don't have the original paper so I can't say where this comes from), as u+ in function of y+. That's it.
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Old   March 10, 2019, 11:10
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Originally Posted by sbaffini View Post
I don't think I understand your question. If you plot that function, it is cristal clear how that behaves (you just need to treat u as independent variable instead of y).
What I mean is best explained using another example. It is known that for the mixing length Van Driest damping can be used to correct for near wall situations, right? Van Driest's damping function is a function of the nondimensionalized wall distance only (D(y^+)), whereas the 'damping function' we are considering is now a function of the nondimensionalized wall distance and velocity(D(y^+, u^+)). But I want to get insight in how this function behaves with y^+. Since u^+ is a function of y^+, I looked for a proper substitution such that I can see the behavior.
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Old   March 10, 2019, 11:22
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Quote:
Originally Posted by Lmath View Post
What I mean is best explained using another example. It is known that for the mixing length Van Driest damping can be used to correct for near wall situations, right? Van Driest's damping function is a function of the nondimensionalized wall distance only (D(y^+)), whereas the 'damping function' we are considering is now a function of the nondimensionalized wall distance and velocity(D(y^+, u^+)). But I want to get insight in how this function behaves with y^+. Since u^+ is a function of y^+, I looked for a proper substitution such that I can see the behavior.
Then you need to collect y+ out of the square parenthesis in my formula for mut/mu, what remains in the square parenthesis is the analogue of the vanDriest damping in the Spalding formulation.

Here it is a simple MATLAB script to have all such quantities for N=4:

%U is the independent variable in Spalding Law of the Wall
u=(0:0.01:30);

%Original Spalding constants
k=0.4;
B=5.5;
c=exp(-k*B);

%Original Spalding Formula
y=u+c*(exp(k*u)-1-k*u-(k*u).^2/2-(k*u).^3/6-(k*u).^4/24);
%Resulting viscosity ratio
mutmu=k*(y-u+c*(k*u).^4/24);
%Resulting wall damping
D=1-u./y+((c/24)*(k*u).^4)./y;

%Velocity plot
figure()
semilogx(y,u)

%Turbulent viscosity ratio plot
loglog(y,mutmu)

%Damping plot
semilogx(y,D)
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Old   March 10, 2019, 11:22
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Quote:
Originally Posted by Lmath View Post
What I mean is best explained using another example. It is known that for the mixing length Van Driest damping can be used to correct for near wall situations, right? Van Driest's damping function is a function of the nondimensionalized wall distance only (D(y^+)), whereas the 'damping function' we are considering is now a function of the nondimensionalized wall distance and velocity(D(y^+, u^+)). But I want to get insight in how this function behaves with y^+. Since u^+ is a function of y^+, I looked for a proper substitution such that I can see the behavior.



If I remember correctly, the Spalding's law is a proposal to express the law of wall u+(y+) in a unique expression (but without taking into account the pressure gradient) in all the regions. I don't think that is directly related to a dumping viscosity concept.
Have a look here https://ntrs.nasa.gov/archive/nasa/c...9990081113.pdf
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Old   March 10, 2019, 11:25
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Quote:
Originally Posted by Lmath View Post
What I mean is best explained using another example. It is known that for the mixing length Van Driest damping can be used to correct for near wall situations, right? Van Driest's damping function is a function of the nondimensionalized wall distance only (D(y^+)), whereas the 'damping function' we are considering is now a function of the nondimensionalized wall distance and velocity(D(y^+, u^+)). But I want to get insight in how this function behaves with y^+. Since u^+ is a function of y^+, I looked for a proper substitution such that I can see the behavior.
Obviously, you can't obtain u+ in terms of y+ directly (so that you have an explicit relation for D in terms of y+ only), because that's not how the Spalding law is built. I mean, no such relation is known to be available... maybe you can try to find one by yourself.
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Old   January 26, 2020, 15:45
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Thank you for this matlab code, the only thing I do not really get is how to go from the equation of eddy viscosity to its damping factor equation. Do you also know in addition how to derive the damping factor for the mixing length as it would follow out of Spaldings law?

Quote:
Originally Posted by sbaffini View Post
Then you need to collect y+ out of the square parenthesis in my formula for mut/mu, what remains in the square parenthesis is the analogue of the vanDriest damping in the Spalding formulation.

Here it is a simple MATLAB script to have all such quantities for N=4:

%U is the independent variable in Spalding Law of the Wall
u=(0:0.01:30);

%Original Spalding constants
k=0.4;
B=5.5;
c=exp(-k*B);

%Original Spalding Formula
y=u+c*(exp(k*u)-1-k*u-(k*u).^2/2-(k*u).^3/6-(k*u).^4/24);
%Resulting viscosity ratio
mutmu=k*(y-u+c*(k*u).^4/24);
%Resulting wall damping
D=1-u./y+((c/24)*(k*u).^4)./y;

%Velocity plot
figure()
semilogx(y,u)

%Turbulent viscosity ratio plot
loglog(y,mutmu)

%Damping plot
semilogx(y,D)
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Old   January 27, 2020, 03:55
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Originally Posted by velkon View Post
Thank you for this matlab code, the only thing I do not really get is how to go from the equation of eddy viscosity to its damping factor equation. Do you also know in addition how to derive the damping factor for the mixing length as it would follow out of Spaldings law?
I'm not sure if I understood correctly what you are asking (two questions which really are one) but, if you followed my derivation in the first post, until I get:

\frac{\mu_t}{\mu}=k\left[\left(y^+-u^+\right)+e^{-kB}\frac{\left(ku^+\right)^N}{N!}\right]

then the last step to obtain D, the damping factor for the mixing length in the Spalding formulation, is to simply collect y+ out of the square parenthesis in the formula above, resulting in:

\frac{\mu_t}{\mu}=k y^+\left[\left(1-\frac{u^+}{y^+}\right)+\frac{e^{-kB}}{y^+}\frac{\left(ku^+\right)^N}{N!}\right] = k y^+ D\left(y^+,u^+\right) = l^+ D\left(y^+,u^+\right)

So, D (the square parenthesis), as derived above, is the damping factor for the mixing length underlying the Spalding formulation. I can say so because I first derived mut/mu from the Spalding formulation, and then expressed it as a mixing length (l^+) times a damping term (D).

So, the general approach is: you have a mut/mu formulation, you collect ky^+, what remains is some factor with respect to the mixing length. If this factor is really damping just depends from the original mut/mu model. Often this damping factor is just introduced artificially (that is the mut/mu formulation is built from a mixing length and a purposedly designed damping factor).

Note that I bravely divided by y+ wherever I needed; I knew I could do this because I have u+ at numerator in those terms, and the ratio is finite for y+ going to 0 (at least it is within the formulation adopted).
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Old   January 28, 2020, 02:29
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Aha thank you, so if I understand well the damping factor for the eddy viscosity is the same as for the mixing length? Or is there a different damping factor for the eddy viscosity following from Spaldings Law?

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Originally Posted by sbaffini View Post
I'm not sure if I understood correctly what you are asking (two questions which really are one) but, if you followed my derivation in the first post, until I get:

\frac{\mu_t}{\mu}=k\left[\left(y^+-u^+\right)+e^{-kB}\frac{\left(ku^+\right)^N}{N!}\right]

then the last step to obtain D, the damping factor for the mixing length in the Spalding formulation, is to simply collect y+ out of the square parenthesis in the formula above, resulting in:

\frac{\mu_t}{\mu}=k y^+\left[\left(1-\frac{u^+}{y^+}\right)+\frac{e^{-kB}}{y^+}\frac{\left(ku^+\right)^N}{N!}\right] = k y^+ D\left(y^+,u^+\right) = l^+ D\left(y^+,u^+\right)

So, D (the square parenthesis), as derived above, is the damping factor for the mixing length underlying the Spalding formulation. I can say so because I first derived mut/mu from the Spalding formulation, and then expressed it as a mixing length (l^+) times a damping term (D).

So, the general approach is: you have a mut/mu formulation, you collect ky^+, what remains is some factor with respect to the mixing length. If this factor is really damping just depends from the original mut/mu model. Often this damping factor is just introduced artificially (that is the mut/mu formulation is built from a mixing length and a purposedly designed damping factor).

Note that I bravely divided by y+ wherever I needed; I knew I could do this because I have u+ at numerator in those terms, and the ratio is finite for y+ going to 0 (at least it is within the formulation adopted).
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Old   January 28, 2020, 03:59
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Quote:
Originally Posted by velkon View Post
Aha thank you, so if I understand well the damping factor for the eddy viscosity is the same as for the mixing length? Or is there a different damping factor for the eddy viscosity following from Spaldings Law?
No. Formally, a mixing length model has the following form:

\frac{\mu_t}{\mu}=k y^+ = l^+

and, as you see, it has no damping factor at all. While it has the correct form for the logarithmic layer, it is known to be wrong in the viscous sublayer, where the turbulent viscosity is known to scale as {y^+}^3. So people have added to it empirical damping factors to make it scale correctly. For example, the Van Driest one has the general following form:


D\left(y^+,A^+,n,m\right) = \left[1-e^{-\left(\frac{y^+}{A^+}\right)^n}\right]^m

where m=2 gives the correct near wall scaling. Then the resulting model is built, empirically (as opposed to derived) as:


\frac{\mu_t}{\mu}=k y^+ D\left(y^+,A^+,n,m\right)

Now, in contrast to this line of thinking, there are models that don't formally start from a mixing length hypothesis. The one of Spalding is such a model. Another example is the Musker one (which is very similar to the one underlying the Spalart-Allmaras model near the wall):


\frac{\mu_t}{\mu}= \frac{\left(ky^+\right)^3}{\left(ky^+\right)^2+\left(ka\right)^3-\left(ka\right)^2}

In these cases you don't formally have a mixing length and a damping factor, but you can always collect ky^+, which is the mixing length, and whatever remains is, as a matter of fact, the underlying damping factor. We already did this exercise for the Spalding model; it is much simpler for the Musker one:

\frac{\mu_t}{\mu}= \frac{\left(ky^+\right)^3}{\left(ky^+\right)^2+\left(ka\right)^3-\left(ka\right)^2} = k y^+ \left[\frac{\left(ky^+\right)^2}{\left(ky^+\right)^2+\left(ka\right)^3-\left(ka\right)^2}\right] = k y^+ D\left(y^+,k,a\right)

As for the Spalding model, it turns out that the damping factor we derived from the Musker model is actually a damping.

So, to answer your question, mixing length actually means just a ky^+ scaling for the turbulent viscosity and, by definition, it has no damping factor. People have added damping factor to the mixing length, empirically, to actually make it well behaved in the viscous layer. This is one way of formulating the near wall viscosity model (i.e., starting from a mixing length times a damping factor.) There are, however, other means of doing it; in this case you don't explicitly start from the mixing length and the damping factor but, usually, they are easily obtained by just collecting the correct terms as I have shown you for two such models (Spalding and Musker).
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