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Markat October 15, 2012 12:41

Cooling- Heat transfer model
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Hello everybody,

i'm glad to have joined the CFD Online community. I am a new ANSYS Fluent and generally CFD user (though I have relatively much experience in structural models). I'm trying to build the following CFD Model as part of my bachelor thesis: A free elliptical solid rigid body (2D Model) with T=700K is being cooled through convection and radiation. The fluid (environment) is air and the boundary conditions are velocity inlet (x: 2m/s, y=0, T=300K) and pressure outlet (gauge pressure=0, T=300K). The flow is turbulent due to Reynolds number for L= string of ellipse= 1m (I have used a k-e model). As radiation model I have used the DO model. Through this model I would like to calculate:

1. The cooling rate of the elliptical body in K/s
2.The temperature gradient in the body

First I have built a flow model with energy equation off, which has converged.Then I tried to calculate the convection model. Unfortunately I couldnīt find a way to define the temperature of the body so that it wouldn't be constant (I tried to define the body's wall temperature, which apparently remains constant during the calculation of the solution).
What are your suggestions?
What strategies should I follow to build my model?
Do you have further suggestions and recommendation for a new user like me?

Thank you in advance.

P.S. I have attached a .jpg of the model

Markat October 16, 2012 14:31

  1. Could you please help me defining the boundary conditions in the conjugate heat transfer problem (without taking the radiation into account). I have red older threads about wall and wall-shadow but in my model both the wall and the wall shadow are aluminium. I wasn't able to turn one of them to fluid. I have already turned the 2 parts into one part with 2 bodies so that I do not have to use contacts.
  2. I would like also to mention that when I try to solve k-e Modell (a this point energy equation off only flow model) the solution converged and the flow was laminar.When I try to solve a laminar model the solution does not corverge. Do you have any idea why?

Markat October 17, 2012 13:09

Does anybody have any suggestion? Should I preferably mesh the solid or not?

Markat October 18, 2012 17:58

Has anyone any idea?

eng_s_sadeghi October 20, 2012 13:23

Boundary conditions are OK. Why can't get answer?

achal February 8, 2013 01:45

hi , i am also working on a similar kind of problem. I was wondering if you could help me.

Problem Description : An iron brick is to be cooled by the use of
water jet, this would include convection from the top surface of brick
and conduction within the brick.
Is that what we call a conjugate heat transfer ?
Can it modelled with out the use of UDF ?
How do I model this, because there is no temperature change observed ?
Any sort of help would highly be appreciated.

msaeedsadeghi February 8, 2013 09:55

There is no need of udf. Fluent can model it by default settings. Think of boundary conditions. There would be fundamental problems.

achal February 8, 2013 11:08

I tried the problem a couple of hours earlier, there is no need of udf.
Also, to give the initial value to the hot brick, we will have to use Patch option ?

and in the post processing, can I get a temperature profile at a particular time of a particular cross section ?



Markat February 8, 2013 19:45

What temperature does the brick have? In high temperatures you should also consider and model radiation. Patching is goof way to define an initial temperature, but only if you mesh your brick. Otherwise you have to define manually an appropriate region to define the initial temperature.
Did you use shell conduction or did you mesh your solid?

I would also like to ask, if does have the same effect giving an initial temperature (by solution initialization) and patching a temperature?

achal February 9, 2013 01:22

temperature is not high to consider radiation. no, i am not using shell conduction as i have brick a solid entity.

patching and solution initialization are 2 different thins.... it does not have the same effect



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