calculating surface temperature 1

[Bencycle]

Hi there,

I am a new user, and a relatively new mechanical engineer. I have been given a heat transfer problem at work, and could use a little help. I am looking to find the surface temperature on a 1 dimensional simple wall. I have the temperature on one side (Ts,1=800F) ,and am trying to find the temperature on the other side (Ts,2=?). The material is 304SS with a convective heat transfer coefficient of 21.4W/mK @500C (i found this number online, please correct me if its not accurate). The material dimensions are as follows 0.75 x 1 x 6 inches long, with the heat running along the 6in length. I have been trying to use fourier's law for 1d steady state conduction, but i don't know q and i don't know Ts,2, so one equation two unknowns. It seems like there should be a pretty simple solution to this, but i have yet to be able to find it. I am ignoring convective and radiation effects currently, but if i should not be please let me know.

 

[IRstuff]

"convective heat transfer coefficient of 21.4W/mK @500C"

That ought not be convective, but conductive, given the units.

However, you cannot ignore convection from the plate, since that's what determines the surface temperature, i.e., the heat flow limit is the convection of the air.

The end result is a equilibrium wherein the conduction T.inside-T.surface across the plate results in the same heat flux that results from the convective T.surface-T.ambient

TTFN
faq731-376

 

[IRstuff]

Oh, and if the surface temp is sufficiently high, you'd need to include radiation

TTFN
faq731-376

 

[MintJulep]

That's a tiny "wall". 1 dimensional assumptions probably don't hold.

Gut answer is that the temperature of the whole thing is "pretty close to 800F".

 

[Bencycle]

Ok wow thanks for the help.
Yes IRstuff you are correct that is supposed to be "conductive"
So i can calculate the flux of the convective heat transfer, including radiation effects, and then equate that to the conductive heat transfer flux to solve for the unknown temperature?

Mintjulep; as for the "wall" yes it is sufficiently small ,and that was my hunch as well.

 

[IRstuff]

No, it's a simultaneous solution, but you might be able to iterate on it.

So, heat is boing through the 0.75 thickness?

TTFN
faq731-376

 

[Bencycle]

so solve convective with conductive and radiation?
The heat would be boring through the 6 in length, with a cross sectional area of 0.75in^2

 

[IRstuff]

780F

I can't see if the screen cap got loaded

TTFN
faq731-376

 

[25362]

Assuming a steady state heat transfer situation, what about heat loss through the prism lateral 21 in[sup]2[/sup] surface?

 

[Bencycle]

Thanks for the help guys i think i got a good answer