# Use of formulas to apply surface heat flux

Hello everyone.
I need to simulate the heating from laser beam almost to the center of my structure. I’m trying to use formulas to simulate a circular surface heat flux of 20 mW/m^2 at the center of my structure. Using bigger values of surface heat flux I’m able to get the temperature gradient on my structure, using 20 mW/m^2 I don’t get anything from post-processing of the simulation. I don’t know if I’m applying exactly 20 mW in this case. I am reporting here the link of my project.

hi @mrosalia,
Look like the same question you were asked a few days back.

than what your problem right now?
and how this problem was different than the previous one?
can you elaborate?

can you share the amount of that surface heat flux?

best,
Rohit.

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@rszoeke solved my problems adjusting the boundary conditions. He applied a a value of 20 W/m^2 of surface heat flux and the simulation worked properly. When I tried to drop the value of surface heat flux to 20 mW /m^2 I’m able as well to launch the simulation. When I check the post-process simulation it’s not possible to visualize anymore the temperature gradient. I see just one value of temperature spread all over my sample. I am trying to raise the radius of circumference on surface heat flux applied but it doesn’t seem to work anyway. I don’t understand why I’m not able to apply 20 mW on the sample.
I hope I was enough clear to expose my problem.

@mrosalia I think you should have to apply the realistic surface heat flux value to visualize the temprature gradient clearly. In your case the outer convective heat flux is around 50 W/(K·m²) means your outer surfaces are capable of loosing heat more rapidly than you are providing by input surface heat flux of 20 mW/m^2. Just put the realistic values and then try simulating.
Regards

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Thanks for the advice @HamzaBaig. I’ll try to do that.
Best regards

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I was having the same problem. I used solidworks previously for some thermal simulations and I could simply put 50 W of power to a specific surface. Here it seems like it is different and would like to hear others’ opinion.

My interpretation was that when we enter 50 (w/m^2) heat flux the power input is very little for that 1 m^2 area…

Therefore I thought I have to enter a value of 50 (w/m^2) / a (m^2) in order to apply 50 W of power to my “a m^2” surface area, which is less than 1 m^2. I just applied linear ratio pretty much.
Once I did that, it did work and comply with Q=m.c.Dt and temperature change for that mass was similar.

What do you guys think?