# Melting and solidification with free surface problem?

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 July 22, 2013, 02:45 Melting and solidification with free surface problem? #1 New Member   wendy Join Date: Jul 2013 Posts: 8 Rep Power: 5 I am model the welding process. The steel (Q235) undergoes melting and solidification under the heating arc and, at the same time, molten steel drop into the molten pool. I want to know the shape of welding bead and it is regarding two phase (air and liquid), free surface, and heating problem. First, I modeled the steel as a liquid. There are steel and air in the domain. After adding a volumetric heating source onto the steel, there can get a right temperature field. Then, I add General Momentum Source as following to force the steel under temperature 1753K to a velocity of 0. I have added a Continue source of Q235 to model molten steel drop into the molten pool. It is named Rongdi. But this time, the temperature is not right. Sx=C*step(1753-Q235.T/1[K])(Q235.velocity u-0) Sy=C*step(1753-Q235.T/1[K])(Q235.velocity v-0) Sz=C*step(1753-Q235.T/1[K])(Q235.velocity w-0) C=-10^7 I have enclosed the .def documents, and please give some suggestion for me. How to add a general momentum source and continum source in my case? Does my boundary wrong? .def file download In my domain, the steel is 3 mm in thickness (from z=-0.006mm to z=-0.003mm), others is air. Last edited by cqlwj123; July 22, 2013 at 09:16.

 July 22, 2013, 05:44 #2 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 12,832 Rep Power: 100 What do you mean the temperature is not right? Does it stop the fluid or not? Have you included a source term linearisation factor? You are not including sub cooling, latent heat and many other important processes with this approach. I would not be surprised if it does not prove veyr accurate.

July 22, 2013, 09:27
#3
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wendy
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Quote:
 Originally Posted by ghorrocks Have you included a source term linearisation factor?
I have included a source coefficient C=-10e^7. Does linearisation factor mean this?

Quote:
 Originally Posted by ghorrocks You are not including sub cooling, latent heat and many other important processes with this approach. I would not be surprised if it does not prove veyr accurate.
I set a large thermal capacity on the melting point to present the latent heat; add General Momentum Source to force the steel under temperature melting point (1753K) to a velocity of 0 as sub cooling. Does this settings right?
Sorry for my poor English. Many thanks.

July 22, 2013, 09:30
#4
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wendy
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Quote:
 Originally Posted by ghorrocks What do you mean the temperature is not right? Does it stop the fluid or not?
the temperature distribution seems not reasonable.

 July 22, 2013, 18:44 #5 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 12,832 Rep Power: 100 No, the source term linearisation is a different thing. Have a look in the documentation for it. You can put a lump in the thermal capacity to represent latent heat, but be aware that it is still an approximation. I think you will find this is because the simplifications you have used are not realistic. Read the literature to find out how other people are doing it. You will need to model the thermal conditions more accurately.

July 24, 2013, 23:39
#6
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wendy
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Quote:
 Originally Posted by ghorrocks No, the source term linearisation is a different thing. Have a look in the documentation for it.
I search 'linearisation' within the documentation, only one hit. As my understand, Momentum Source Coefficient is the same with linearisation factor in this case.

 July 25, 2013, 02:46 #7 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 12,832 Rep Power: 100 Don't forget I am Australian so I spell linearisation with an "s". ANSYS is American so they spell it as linearization. The coefficient is the same concept for all the equations, I recall. But some applications of source terms have the linearisation built in so the user does not need to specify it.

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