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I'm trying to understand the derivation of the turbulent kinetic energy equation, as described in this link: Evaluation of RANS turbulence models for flow problems with significant impact of boundary layers.

I'm able to follow up to equation 2.26 on slide 11 (i.e. page 9), which states

$$\rho \overline{\frac{\partial u_i'}{\partial t}u_i'} + \rho\left( \overline{u_j\frac{\partial u_i}{\partial x_j}u_i} -\bar{u}_j\frac{\partial u_i'}{\partial x_j}\bar{u}_i\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i,$$

where the overbars denote the ensemble averages of the quantities under the bars. Note that here, the average of a product is not necessarily equal to the product of the average of each term.

According to the link, using only averaging rules, the term

$$\left( \overline{u_j\frac{\partial u_i}{\partial x_j}u_i} -\bar{u}_j\frac{\partial u_i'}{\partial x_j}\bar{u}_i\right)$$

can be simplified to become

$$\overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \overline{\frac{\partial u_i'}{\partial x_j}u_j'}\bar{u}_i + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}.$$

If I substitute this directly into the previous equation, I get

$$\rho \left( \overline{\frac{\partial u_i'}{\partial t}u_i'} + \overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \underbrace{\overline{\frac{\partial u_i'}{\partial x_j}u_j'}\bar{u}_i}_{\text{my 4th term}} + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i.$$

However, their derivation yields

$$\rho \left( \overline{\frac{\partial u_i'}{\partial t}u_i'} + \overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \underbrace{\overline{\frac{\partial u_i'u_j'}{\partial x_j}}\bar{u}_i}_{\text{their 4th term}} + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i.$$

The only difference between my derivation and theirs is the 4th term on the left hand side of the equation. I'm sure that their 4th term is correct, since it is supposed to cancel out with a term on the right hand side of the equation. However, I can't seem to figure out how they obtained their 4th term on the LHS of the equation. The link suggests that the chain rule and the incompressiblity assumption are involved, but I'm not sure how.

In particular, the overbar term $\overline{u_i'u_j'}$ is treated as a single term. Thus, how can I apply the chain rule to $\frac{\partial (\overline{u_i'u_j'})}{\partial x_j}$?

Any hints to complete the missing steps would be greatly appreciated.

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The two terms are equal due to the product rule and continuity equation. The use of the continuity equation might not be obvious, however.

In their notation, the Reynolds decomposition is $u_i = \overline{u}_i + u^\prime_i$. Taking the continuity equation and averaging leads to:

$$\frac{\partial \overline{u}_j}{\partial x_j} = 0$$

Subtract the equation above from the non-averaged continuity equation to see that the fluctuations also are divergence free:

$$\frac{\partial u^\prime_j}{\partial x_j} = 0$$

Now, apply the product rule to the term of interest (before averaging):

$$\frac{\partial u^\prime_i u^\prime_j}{\partial x_j} = u^\prime_j \frac{\partial u^\prime_i}{\partial x_j} + u^\prime_i \frac{\partial u^\prime_j}{\partial x_j}$$

The second term is zero by continuity as seen above, and this completes the derivation. The averages commute, so the average of a derivative is the same as the derivative of the average.

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    $\begingroup$ The term "chain rule" here is misleading... The product rule is what is really being applied, not the chain rule. $\endgroup$ – Paul Apr 5 '16 at 1:36
  • $\begingroup$ Yes, you are correct. I was thinking what you suggested but ended up doing something different. I've corrected the answer. Thanks. $\endgroup$ – Ben Trettel Apr 5 '16 at 1:38

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