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FrozenGate by Avery

Copper vs Aluminum vs Ect.

Diamond heatsinks anyone? :whistle:

However, not so sure about its specific heat capacity, which is more important in this case.

Cheers! :beer:
 





TheDukeAnumber1 said:
Why no one is trying to make a sink out of aluminum oxide.
Click to expand...

What do you mean? Plenty of people use alumina on the surface of their heatmass because of the better heat properties of alumina compared to "bare" aluminum... All of my flashlights and lasers are alumina coated.
 
BShanahan14rulz said:
What do you mean? Plenty of people use alumina on the surface of their heatmass because of the better heat properties of alumina compared to "bare" aluminum... All of my flashlights and lasers are alumina coated.
Click to expand...

Have a look at a thermal conductivity chart of materials.

Alumimum Oxide has poor thermal conductivity and is not a good as a heat sink material
 
I'm not talking about heat conductivity or specific heat capacity. I agree, a heatsink made completely out of alumina would not perform as well as aluminum.
 
BShanahan14rulz said:
I'm not talking about heat conductivity or specific heat capacity. I agree, a heatsink made completely out of alumina would not perform as well as aluminum.
Click to expand...

Yeah we get it your talking about emissivity which depends more on the color and the finish surface than most anything else. If you have some useful data then please present it.
 
All I'm saying is that if you don't think about radiating that heat, or removing it from the system some other way, then you will still see temperature increases at the junction. Not that you wouldn't otherwise, but as most of these threads conclude, it depends on your application. I would rather have an aluminum head coated in either natural oxidation or purposely oxidized (anodized), rather than a tarnished copper head. If this part were simply to hold the heat, or spread it out and pass it on to the radiative part of the heat system, and I didn't mind cleaning the heatsink from time to time, or actually took the care to not let its surface deteriorate, I'd go for copper too.

Emissivity Coefficients of some common Materials < interesting data.
 
BShanahan14rulz said:
All I'm saying is that if you don't think about radiating that heat, or removing it from the system some other way, then you will still see temperature increases at the junction. Not that you wouldn't otherwise, but as most of these threads conclude, it depends on your application. I would rather have an aluminum head coated in either natural oxidation or purposely oxidized (anodized), rather than a tarnished copper head. If this part were simply to hold the heat, or spread it out and pass it on to the radiative part of the heat system, and I didn't mind cleaning the heatsink from time to time, or actually took the care to not let its surface deteriorate, I'd go for copper too.

Emissivity Coefficients of some common Materials < interesting data.
Click to expand...

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

This is from the reference you just used.
Radiation Heat Transfer
q = σ T4 A (1)

where

q = heat transfer per unit time (W)

σ = 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant

T = absolute temperature Kelvin (K)

A = area of the emitting body (m2)



q = ε σ T4 A (2)

where

ε = emissivity of the object (one for a black body)

XXXXXXXXXXXXXXXXXXXXXXX
 
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srs_bsns.jpg
 
You screwed something up. Two things, actually. You're not accounting for the energy the object receives from the surroundings, and you forgot that fourth power.

(5.6703*10^-8)(300-20)^4 = 350 watts of heat

A more typical value of 20C above ambient would be 9mW of heat, so your point is still valid.
 
Cyparagon said:
You screwed something up. Two things, actually. You're not accounting for the energy the object receives from the surroundings, and you forgot that fourth power.

(5.6703*10^-8)(300-20)^4 = 350 watts of heat

A more typical value of 20C above ambient would be 9mW of heat, so your point is still valid.
Click to expand...

Hang on if you use the first equation at 30 deg C or 303 K with ε = 1 and Ac = 1 m^2

q = ε σ T^4 A you get: (5.6703*10^-8)(303)^4 = 477.9 W

Which is basically what Cyparagon said, although he calced it incorrectly. If however, you you use the second formula he found on that page to account for the surroundings and also using ε = 1 and Ac = 1 m^2:

q = ε σ (Th^4 - Tc^4) Ac you get: (5.6703*10^-8)(303^4 - 293^4) = 60W ! Which is relevant!

You must use Kelvin in the calculation and the formula was K2^4 -k1^4 not (K2-K1)^4

:beer:

Edit: All the same it seems a little high, guess you cant really use an emissivity of a black body for the calc...

Using 0.2 - 0.31; which was given in a table for the emissivity co-eff of some common materials ie Al heavily oxidised

We get 0.31 x 60 = 18.6 W or 12 W for 0.2; which would seem much more reasonable and very relevant!

BTW polished Al has a co-eff of only 0.039 - 0.057 suggesting the radiative emission would be considerably worse for a highly polished host, which has already been stated.
 
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Thank both of you for cleaning that up, one more correction though. Area isn't even close to 1m^2.

(5.6703*10^-8)(303^4 - 293^4)(.07) = 4.2W

I was being quite generous at .07m^2
Still higher than what seems like it should be, I still wonder if there is something we are missing.
 
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