[Madeleine P. Vincent]

Hello,

This is a simple question about time averaging.

I was reading in Turbulence Modeling for CFD by Wilcox that the time average of the product of 2 properties has the following identity:

The time average of phi * psi = Phi Psi + time average of phi' psi'

where time average is denoted by an overbar, and the capitalized variables are mean quantities, and the primed values are the fluctuating components.

Then there is the statement:

"where we take advantage of the fact that the product of a mean quantity and a fluctuating quantity has zero mean."

I don't understand this. Why would the time average of Phi * psi' and Psi * phi' be equal to 0? And why wouldn't the time average of phi' * psi' be equal to 0?

Regards,
Madeleine.

[FMDenaro]

Originally Posted by Madeleine P. Vincent

Hello,

This is a simple question about time averaging.

I was reading in Turbulence Modeling for CFD by Wilcox that the time average of the product of 2 properties has the following identity:

The time average of phi * psi = Phi Psi + time average of phi' psi'

where time average is denoted by an overbar, and the capitalized variables are mean quantities, and the primed values are the fluctuating components.

Then there is the statement:

"where we take advantage of the fact that the product of a mean quantity and a fluctuating quantity has zero mean."

I don't understand this. Why would the time average of Phi * psi' and Psi * phi' be equal to 0? And why wouldn't the time average of phi' * psi' be equal to 0?

Regards,
Madeleine.

The issue is general and depends on the definition of time-averaging.
The statistical Reynolds average is defined as


<f>(x) = lim T-> +Inf Int[t0, T] f(x,t) dt (1)


so that a function is decomposed as f(x,t) = <f>(x) + f'(x,t).
You can easily see from (1) that <<f>(x)> =<f>(x).
Therefore a product of functions f*g obeys the rule above

<f*g>(x) = < [<f>(x) + f'(x,t)] [<g>(x) + g'(x,t)] >

Using (1) you see that the cross-product between average and fluctuations vanishes.

It is noteworthy that other types of time-averaging (for example time-filtering used in LES) do not produce the same disregarding of cross-products!

[Madeleine P. Vincent]

Originally Posted by FMDenaro

Using (1) you see that the cross-product between average and fluctuations vanishes.

In fact, this is exactly what I don't understand. I don't see why the product should vanish. Could you try to explain why that is?

Thank you for your help.

[FMDenaro]

Originally Posted by Madeleine P. Vincent

In fact, this is exactly what I don't understand. I don't see why the product should vanish. Could you try to explain why that is?

Thank you for your help.

very simply you see that averaging further:

f'(x,t) =f(x,t) - <f>(x) -> <f'(x,t)> =<f(x,t)> - <<f>(x)> = 0

<f'(x,t) <f>(x) > = <f>(x) <f'(x,t)> = 0

[sbaffini]

You can see this by the definition of average used by Filippo. From the definition it comes out that the average value is not anymore dependent from the time variable, which is saturated by the integration. Roughly speaking, the average of a signal over an infinite amount of time is just a constant value, which can go out of any additional integration operation.

Moreover, if you use the decomposition:

f(x,t) = <f>(x) + f'(x,t)

you can see that applying the average definition to the left side has, by definition, to return only the first member on the right. As the time average returns a constant, its double average is still the same constant. Hence, the second term averaged is zero.

From this you should see why their product has a null average.

For general unsteady flows, you have to use ensemble averages and the probability theory is the most suitable tool to show that the same result holds. Saturation in time is substituted with saturation over an inifinite number of experiments.

[Madeleine P. Vincent]

Originally Posted by FMDenaro

very simply you see that averaging further:

f'(x,t) =f(x,t) - <f>(x) -> <f'(x,t)> =<f(x,t)> - <<f>(x)> = 0

<f'(x,t) <f>(x) > = <f>(x) <f'(x,t)> = 0

So you're saying that

<f'(x,t)> =<f(x,t)> - <<f>(x)> = 0

however, in the Wilcox book he states:

"There is no a priori reason for the time average of the product of two fluctuating quantities to vanish".

This seems to contract what you have written, since isn't the following then true

< <f'(x,t)> <g'(x,t)> > = < 0 * 0 >

??

Why does the average of the product of the 2 fluctuations not go to 0?

Thanks again. I appreciate your help.

[FMDenaro]

Indeed, I stated that only using this type of averaging the fluctuaction field has vanishing average.
In general, You can define a local time-averaging and a different residual fluctuation field for which the averaging is not zero.
The differences lead to either RANS or URANS/LES approaches