problem calculation thermal resistance

[ansyolitico]

Hi.
I'm trying to make a thermal simulation
I have a resistance of 40 w which has to heat a block of metals...as loads I used generation of internal heat calculated with power/volume resistance i.e. 40/770=0.05 w/mm3, the other load I used the natural convention air 22 w/mm2 applied on the 6 faces of the block...the problem that in the simulation as a result the block does not exceed 50 degrees while in reality comes over 260... what is wrong?



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[MassiVonWeizen]

Why don't you put decent titles instead of puerile battutines, which help you understand the subject and facilitate any searches by other users?

 

[meccanicamg]

- Are you sure about the units?
- Did you fix the start-up temperature?
Are you sure you're studying the real behavior of the object?
Do you have the experimental data of reality with all the values to simulate it?
- Did you consider internal conductivity?

I don't know.

 

[ansyolitico]

I'm sorry if I only answer now, but I've had a little to do in the last few months, and I've been out of my mind all...
mammals You're right. If I can change it, I just don't know how to do it.

when it comes to the rest

40 watt resistance has a diameter of 6 mm long 28 m, then with a volume of 770 mm3
using an internal heat generation load I suppose I should have split 40/770=0.05 w/mm3
as a boundary condition I have a set a natural confectory of 22 w/mm2 on the six face of the cube

the start temperature I fixed it at 20 degrees and the internal conductivity does not impose it by dissolving the material?

I have a real model to refer to...but it does not reflect this simulation
You don't want me, but it's the first time I do a thermal analysis, I've always done the structural ones.

Thank you.

 

[MassiVonWeizen]

If I can change it, I just don't know how to do it.

interpella the moderators

 

[exxon]

I think you're not setting the problem properly. I don't see what sense it has to divide the power by the volume of resistance. In addition to this, the heat exchange between two bodies only happens if the same have different temperature, therefore those 22 w/mm2 are a nonsensical. the heat exchange coefficient is expressed in power per surface for temperature difference.

 

[ansyolitico]

is that I do not know how to treat an electrical resistance

I've been intuited....also because I hypothesized an internal generation load that has a w/mm3 measuring unit, so I'm not sure I'm treating that load

while the 22 w/mm2 are the boundary conditions ....in the end the block is subject to convention. I'm trying to get this...

 

[exxon]

the generation is given to you with that unit of measurement because then it must be multiplied by the volume of the heating element. All you need is to know the power that is, as you say, 40 w.

The other value I don't know where you get it out. a flow of 22 w/mm2 is a stratospheric quantity that by convection you would get with a temperature difference of one million degrees (although convection laws would no longer be valid at such temperatures).

and repeat, w/mm2 is a flow, which you only get by setting a certain temperature difference. a metal surface dissipates for natural convection from 1 to 20 w/m2k (watt per square meter for kelvin). and well understood is square meter, not square millimeter.

 

[ansyolitico]

My mom somaro, the units of measure!!! !
that given I had taken it but as a fool I did not look at the unit of measure


so I imposed only the internal generation ... and in fact doing the result more reflects the reality....only that with that only given tells me that the matrix is bad place and to be careful to interpret the results....if I carry the convention of 20 w/m2k in mm2?

 

[exxon]

I can't understand what you mean. However, to switch from m2 to mm2 to denominator. . . divide the value for a million. would be 20 uw/mm2k.

 

[ansyolitico]

I corrected the value of the convention, but the temperatures rose to the stars....7000 degrees

I mean, if I only put the internal generation of 40 w, the result is almost equal to reality.
If I enter the convention, it's all wrong.
and in theory I should also insert irradiation

 

[exxon]

I don't know what you did with that simulator, but before using it, you should know it sufficiently, and I think it's not your case... Entering data at home, hoping to get a likely result, is not the correct method of proceeding.

If you place the size of the object, I can make you analytical calculation in more or less realistic conditions.

 

[ansyolitico]

allego step step
load 40 w/mm3
radiation convention on block faces
2,2x 10^-5 w/mm2
emmigrant coef 0,4
tempen environment 22 cc

 

[exxon]

from the calculations (very approximate) of an object of that kind it turns out that the thermal balance is 360 °c, where you have about 10 w for irradiation and 30 w for convection. the temperature will probably be lower because at those temperatures the air motion will be strongly turbulent around the object resulting in increased dissipation.

Ansyolithic... but look at what you write?? ?

1) the "load" (which is actually the input power) is 40 w, not 40 w/mm3!! !
2) we are talking about convection, not convention !!!
3) thermal transmittance is measured in w/m2k, not in w/m2 !!!

Note, the calculations are performed with the data in #13, therefore considering also the fins (only for convection) and the surfaces exposed outside for radiation.

 

[ansyolitico]

anxiety,
I'm writing from cel. I'm sorry to fuck with the concealer.

for the load I have as unit of measure w/mm3....according to me is there the casino

because I am tempted max 6000 degrees. .

 

[exxon]

in #1 you wrote that you have a 40w resistance from which you calculated 440/770=0.05 w/mm3, now write that you have 40 w/mm3.

Maybe it's better if you clarify your ideas first.

 

[ansyolitico]

in #4 I seem to have well described my reasoning

I have a resistance of 40 w, I decided to treat it as an internal generation, ansys has as a unit of measurement for the internal generation w/mm3, then I decided to take the power of 40 w and diveder it for the volume of the resistance that generates 770mm3 that from as a result 0.05 w/mm3 and insert this data as load

I have well specified that I do not know how to treat this resistance

after you arrived you and told me that it does not need to divide the strength of the resistance for its volume only that this type of simulator has as a unit of measure w/mm3 and I inserted 40 as load... but so I arrive at 6000 degrees

Now my question is, since your account is right even if I wonder how you did it analytically when from the memories of technical physics it was already challenging to do it in xy... what number do I have to enter as a load? Is my reasoning right?

Why?

 

[exxon]

I have a resistance of 40 w, I decided to treat it as an internal generation, ansys has as a unit of measurement for the internal generation w/mm3, then I decided to take the power of 40 w and diveder it for the volume of the resistance that generates 770mm3 that from as a result 0.05 w/mm3 and insert this data as load

If in your simulator, the volume of the solid you call "load" is 770 mm3, then it may also be right.

after you arrived you and told me that it does not need to divide the strength of the resistance for its volume only that this type of simulator has as a unit of measure w/mm3 and I inserted 40 as load... but so I arrive at 6000 degrees

what I told you is that you do not need the specific power (power/volume), but only the total power that I seem to be always 40 w...

Now my question is, since your account is right, even if I wonder how you did it.

is not a difficult calculation: just match the power that enters (40 w) with the one coming out (always 40 w) and you have the heat balance. Since the power that comes out does so in two ways, irrational and convection, just take the two equations that manage these energy flows (each reported to its reference surface) and find the temperature to which you have the balance. for irradiation use the equation of stefan-boltzmann, for convection it is sufficient that of thermal transmittance.

What number do I need to insert as a load? Is my reasoning right?

I think you're leaving from wrong assumptions: You're a student and I think this is an exercise you've been asked to solve. exercises are used to connect theory to the reality of the phenomenal world around us, not to put the magic number in the simulator that makes you get the right result.

I have already told you (and in the last post you confirm it explicitly) that you cannot use that simulator. Now you have two roads: I have studies of manuals to learn how to use it (long road...) or do some tests to understand empirically that numbers slip into it.

What would I do if I didn't have the manuals but I just knew how to set the operation? I would take a radius ball 1 m, I would put a source of 1 w and I would calculate analytically the result, comparing it with that of the simulator... It also seems to me an interesting thing for a student, as long as you get into what you're studying.