Help needed to resolve a CFD differential equation

Tibo99 · June 8, 2021, 14:45

Hi everyone,

I wondering if it's possible to resolve the following differential equation for 'k'? If so, could someone guide me through the process of resolving it?

Thank you very much!

Best Regards

sbaffini · June 9, 2021, 05:55

Do you mean analytically? I doubt it, as it is non-linear, but to be sure you need someone who is into ODE and math more than me (i.e., there might be a very simple solution method I don't know). Nonetheless, I would probably put it down more simply as follows:

$x^2 \frac{d^2y}{dx^2} +x\frac{dy}{dx} = ay^2 - b$

where

,

, $a = \sigma_k \left(\frac{u^*}{\kappa}\right)^2$ and $b = C_{\mu} \frac{\sigma_k}{\left(u^* \kappa\right)^2}$

From here on you can maybe seek for some transofrmation that will change it in a linear equation, like, say,

. Still, you probably need to first let b, somehow, disappear, in order for such trick to work (maybe including it in the transformation itself).

sbaffini · June 9, 2021, 06:16

Maybe try with the following substitution:

$y = \frac{\sqrt{b}+\frac{1}{f}}{\sqrt{a}}$

so that:

$ay^2 - b = \frac{1}{f^2} + 2\frac{\sqrt{b}}{f}$

then apply the chain rule on the left hand side and see what happens. If you are lucky you will be left with something linear and integrable.

EDIT: nope, b is a bitch and won't let you have a linear form in this case

Tibo99 · June 9, 2021, 08:50

Thank you very much for answering!

I was trying to make an attempt on trying to find for 'k' on the turbulent kinetic energy equation (k-epsilon) by using the turbulent viscosity equation (linear) instead of using the assumption that the productivity (Gk) is equal to the dissipation (epsilon). That why the last two term doesn't vanish and the equation become in this form.

There is maybe an alternative by using the dissipation equation instead, but I think the equation become non-linear also.

I ended up with the following equation, but I wasn't sure if it was correct. Even though, it's clearly non-linear and because of $dx^{2}$ I wasn't sure how to integrate the equation.

Again, thank you for your time and your help!

sbaffini · June 9, 2021, 12:51

Quote:

Originally Posted by Tibo99

Thank you very much for answering!

I was trying to make an attempt on trying to find for 'k' on the turbulent kinetic energy equation (k-epsilon) by using the turbulent viscosity equation (linear) instead of using the assumption that the productivity (Gk) is equal to the dissipation (epsilon). That why the last two term doesn't vanish and the equation become in this form.

There is maybe an alternative by using the dissipation equation instead, but I think the equation become non-linear also.

I ended up with the following equation, but I wasn't sure if it was correct. Even though, it's clearly non-linear and because of $dx^{2}$ I wasn't sure how to integrate the equation.

Again, thank you for your time and your help!

If you are looking for the rans models behavior near the wall, then you can look at the following paper https://citeseerx.ist.psu.edu/viewdo...=rep1&type=pdf

Tibo99 · June 10, 2021, 13:01

Thank you very much for sharing this information! It obviously a good paper and I'll take look in depth what the author did.

What I was trying to attempt is to solve for 'k' in order to have another analytic profile. Not at the wall, but for the domain. I've seen in the literature that other 'k' profile has been obtain by assuming that the production is equal to the dissipation (https://www.sciencedirect.com/scienc...67610508001815). I found out by doing so that the turbulent viscosity is no longer linear.

So, I want to keep the turbulent viscosity linear, and trying to find another 'k' profile by resolving the TKE equation and then, I end up with the differential equation in my first post.

Thank for the discussion.

June 9, 2021, 05:55		#2
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,151 Blog Entries: 29 Rep Power: 39	Do you mean analytically? I doubt it, as it is non-linear, but to be sure you need someone who is into ODE and math more than me (i.e., there might be a very simple solution method I don't know). Nonetheless, I would probably put it down more simply as follows: $x^2 \frac{d^2y}{dx^2} +x\frac{dy}{dx} = ay^2 - b$ where , , $a = \sigma_k \left(\frac{u^}{\kappa}\right)^2$ and $b = C_{\mu} \frac{\sigma_k}{\left(u^ \kappa\right)^2}$ From here on you can maybe seek for some transofrmation that will change it in a linear equation, like, say, . Still, you probably need to first let b, somehow, disappear, in order for such trick to work (maybe including it in the transformation itself). aero_head likes this. Last edited by sbaffini; June 9, 2021 at 07:05.

June 9, 2021, 06:16		#3
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,151 Blog Entries: 29 Rep Power: 39	Maybe try with the following substitution: $y = \frac{\sqrt{b}+\frac{1}{f}}{\sqrt{a}}$ so that: $ay^2 - b = \frac{1}{f^2} + 2\frac{\sqrt{b}}{f}$ then apply the chain rule on the left hand side and see what happens. If you are lucky you will be left with something linear and integrable. EDIT: nope, b is a bitch and won't let you have a linear form in this case aero_head likes this.

June 10, 2021, 13:01		#6
Tibo99 Senior Member René Thibault Join Date: Dec 2019 Location: Canada Posts: 114 Rep Power: 6	Thank you very much for sharing this information! It obviously a good paper and I'll take look in depth what the author did. What I was trying to attempt is to solve for 'k' in order to have another analytic profile. Not at the wall, but for the domain. I've seen in the literature that other 'k' profile has been obtain by assuming that the production is equal to the dissipation (https://www.sciencedirect.com/scienc...67610508001815). I found out by doing so that the turbulent viscosity is no longer linear. So, I want to keep the turbulent viscosity linear, and trying to find another 'k' profile by resolving the TKE equation and then, I end up with the differential equation in my first post. Thank for the discussion. sbaffini likes this.

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