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Incorrect lift co-efficients with buoyantBoussinesqPimpleFoam

Hello,

I have been trying to validate flow over a heated square cylinder using buoyantBoussinesqPimpleFoam. However, the lift coefficient is seen to be positive instead of negative values. Also, the wake is not captured properly. I am using OpenFOAM 5 as the newer versions consider Boussinesq approximation as an equation of state which am afraid is not a proper way.

Please see below the link to my case file and share your suggestions if you have had any similar experiences.

https://drive.google.com/file/d/1ktKLMg27_hI1Q8Dk86jGKkYpsNSPTYpE/view?usp=sharing

Thank you

Damu


OpenFOAM 28-12-20, 12:44 p.m. damu1414
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Hi,
Can you attach the paper you are trying to validate?
Regards,
Divyesh Variya

29-12-20, 7:15 p.m. divyesh


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Hi Divyesh,

Thanks for the response. Please see below the url for literature.

https://drive.google.com/file/d/1GlBfJgXhLQbyC0NfPUZJn107PRGhrI0A/view?usp=sharing
29-12-20, 10:09 p.m. damu1414

Hi,

Can you cross-check values of velocity, viscosity & Temperature difference?
To use the Boussinesq approximation method, there is a constraint of Temperature difference. In your case, it is 31.5 Which is very large. It seems due to these reasons you are not getting the correct results.

Thanks,
Divyesh Variya
 


30-12-20, 10:35 a.m. divyesh

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Hi Divyesh, I calculated the parameters as follows. For a particular case, Gr=10^4 and Re=200(as per the literature previously attached), I assumed temperature diff=1. My square cylinder side is of 1 unit. beta=3.33e-03. Hence, V=0.36. nu=1.8e-03. All dimensions as per SI. I ran the case and attached is the Cl-time plot. The value is way too high. Please advise.
30-12-20, 1:07 p.m. damu1414

the case file has velocity value 1 m/s
Temperature difference is 31.5. I think you need to correct and run again.


31-12-20, 10:37 a.m. divyesh

Hi Divyesh,

This output belongs to the new case with parameters as mentioned in my previous comment. 

Gr=10^4, Re=200(as per the literature previously attached)

Temperature diff.(assumed)=1. Square cylinder side is of 1 unit. beta=3.33e-03. V=0.36. nu=1.8e-03. All dimensions as per SI


31-12-20, 11:09 a.m. damu1414

Looks like velocity-viscosity values are not working. Could you try some other combinations by changing temp. diff. or you could get more specific values if you dig some more literature. There are a couple of people who did the same simulations for validation. You could look at their papers.


31-12-20, 12:54 p.m. divyesh

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Hi Divyesh, I tried another case as suggested by you. For Gr=10e4 and Re=200 T(air)=300, T(hot)=301, g=9.81, nu=1.58e-05, beta=3.33e-03. I then calculated characteristic length=0.042 and velocity=0.075. Please see attached the cl-time plot.
31-12-20, 10:14 p.m. damu1414

The characteristic length will be the height of the cylinder and based on the above values grashoff number will give you a temperature difference.
I hope you are using the same domain inlet length. And calculating velocity based on that length with given Reynolds no.
 


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