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HEAT TRANSFER CYLINDER WITH HEAT CONVECTION RESULTS FROM THEORICAL SAMPLE

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Message 1 of 3
jorge.lopera
725 Views, 2 Replies

HEAT TRANSFER CYLINDER WITH HEAT CONVECTION RESULTS FROM THEORICAL SAMPLE

Hi, I'm  trying to develop an exercise from Heat Transfer's INCROPERA book.

 

The exercise is to compare a cylinder with a cylinder with flip or wing to disipate.

 

Sample 3.9 ( 4th edition Spanish).

 

I have found big difference results from the book. I have found 561 W and the book says 236 W, any idea?

 

I attached in the dwf model, the .fem model  has not passed by its size.

 

Regards

Technical Specialist
Nexsys de Colombia
2 REPLIES 2
Message 2 of 3
John_Holtz
in reply to: jorge.lopera

Hi Jorge, here is some information about your sample problem.

 

First, I am not sure what material the outer ring is made from. Is it also aluminum? (Why have two cylinders of aluminum?)

 

Second, there is an error in the formula for q. The radius r should be 0.075 m, not 0.025 m because the convection is on the outside of the 150 mm diameter cylinder. So, the theoretical answer (assuming the other terms are correct) should be 708 W. So your answer of 561 W is a lot closer to the theoretical example. You probably just need a better approach to solving the problem. (See below)

 

Third, the .FEM file does not contain the complete model input, so even if you could have attached it, it may not have been helpful. The file we need from a simulation is the .ACH file ("File > Archive > Create"). The Archive Utility compresses the chosen files, which can be the model only or model and results. (I changed the extension from .ACH to .TXT so that it would attach to the forum. You need to change it back to .ACH in order to open it with Autodesk Simulation.)

 

Finally, I suggest that you look at the "Heat Rate Through Face" result instead of the "Heat Flux". The heat rate through face gives a more accurate answer and eliminates the need to calculate the area. (In your formula, you used the inner radius of 25 mm. I assume your heat flux of 23812 is also taken at the same radius, but that is something to check.)

 

So, I created the same model in Autodesk Simulation. Because the model is symmetric (same geometry, loads, and results all around), it could be analyzed using 2D elements, or analyzed using brick elements but modeling a portion of the full cylinder; that is, you could have created a half cylinder, 90 degree segment, or any other smaller portion! I just sliced the cylinder at an arbitrary angle which worked out to about 13.2 degrees. Either approach creates a very small model (easy to move around) and runs fast (in case you need to re-run to get other results). I have attached the archive of my model. I got results of 703 W and 689 W. The only difference between my two design scenarios is in the Element Definition.

 

Let us know if you have any other questions.

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 3 of 3
jorge.lopera
in reply to: John_Holtz

John,

 

thanks for your answer.

 

I have modeled two cylinders because the inside cylinder is the aluminium.

 

And the outer cylinder is air with a temperature (300K) and a convection coefficient (h= 50 w /m2 K) I have put that information  at the DWF file.

 

I have tried with external fluid, but, this external fluid is only a box, not a cylinder, that's why I have modeled as a two cylinder from Inventor.

 

Now that I see your model, I'm wondering that there's no boundary conditions at the inner face that define the polar symmetry.

 

Where do you define that?

 

Really thanks for your advices.

Regards.

 

 

 

Technical Specialist
Nexsys de Colombia

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