Heat transfer in vacuum 2

[knowlittle]

2 metallic objects separated by vacuum.
Vacuum level 10e-6 torr.
HOT 120 C, COOL 20 C.
Distance 8 mm (1/3").
Surface areas same (25 x 250 mm, 1"x10")
HOT 25 mm thick (1"), COOL 6 mm (1/4")
HOT electrical heating strips on 2 sides.
COOL water cooled copper line brazed on the back.
Temperature: measured with thermister from the back
Temperature control: 1 C OR BETTER.

When the back plate reads 20 C, how realistic is it for the front (facing 120 C object) not to exceed 21 C? Can someone do rough calculation for me? Both objects are aluminum with electroless nickel plating. Ignore edge effect. Thanks.

 

[knowlittle]

I just did some rough estimation. The radiation energy is pretty miniscule. I am pretty sure when the back side reads 20 C, the front side facing the HOT object is about the same. If anyone finds a different result, please let me know.

 

[MintJulep]

Temperature is not heat.

Radiation calculations need to use K, not C.

 

[IRstuff]

Worst case heat transfer appears to be about 5.8W.

Hypothetically, if the coolant flow is sufficient, there should be no more than about 0.9K drop between the front and the back, assuming a worst-case heat flow through the copper.

TTFN
faq731-376

 

[EdStainless]

I actually came up with a value just under 5W. If you heat input is any greater than this you will not be able to cool it in this system.
The hot temp is just too low the generate much radiation.
With special surface treatments to enhance radiation (emission and absorption) in the required wavelengths would help.

= = = = = = = = = = = = = = = = = = = =
Plymouth Tube

 

[Maui]

Is this a homework question?

Maui

http://www.EngineeringMetallurgy.com

 

[Metalhead97]

Sounds like something right out of Poirier-Geiger!

MH

http://www.linkedin.com/pub/luke-autry/1b/510/566

 

[knowlittle]

This is my rough calculation.
Heat = emissivity x sigma x area x (T1^4 - T2^4), assuming T2 is much larger is T1 in size (conservative estimate)

Using emissivity = 0.5
sigma = 5.67 x 10^-8
area = 6.25 x 10^-3
T1= 393
T2=293,
I got heat = 3 watts.

Conduction heat transfer
heat = area x conductivity x (T2-T1)/distance.
Using
Heat=3 W
area=6.25 x 10^-3
conductivity=160
thickness=6x10^-3,
I got (T2-T1)=0.02 deg C.

How did you get 0.9 deg C? Where did I miss? Thanks.

 

[IRstuff]

I generally don't retain calculations for postings like this. However, my recollection was that I attempted to account for the thermal spreading resistance, since your coolant lines are unlikely to fully cover the back surface of the plate.

TTFN
faq731-376

 

[knowlittle]

Thanks for the note. I am designing this new setup and all your thoughts are well taken into consideration.

 

[racookpe1978]

OK, so the part 1 and the part 2 are in a vacuum. So all parts inside the tank are radiating into all other parts of the tank from time t - 0 through time t = finish.

Heat loss is from Part 1 to part 2 at Kelvin temperatures T time = 0 (assuming part 1 is the hottest), which will be a function of total area of part 1 (but only its front side facing part 2); assuming realistically that emissivity of part 1 for the entire exposed area of part 1 towards part2 is the same on both sides
PLUS (or minus - depending on your attitude)
part 2 (at temperature 2 (time 0) and emissivity 2 and area 2 (exposed towards part 1 at temperature 1 (time 0).

But part 1 is also radiating into the vacuum to be absorbed by the tank walls (at temperature time = 0 of the tank and emissivity tank and area of the tank exposed to part 1)
BUT
part 2 is also re-radiating from its back wall into the back wall of the tank, so
emissivity of part 2 at temperature part 2 into the tank walls at temperature tank time 0 and emissivity tank wall and area of the tank walls towards part 2 need to be also included in heat loss and heat gain of the system.

Best if the vacuum walls are insulated/non-radiating, right?

 

[davefitz]

normal aluminum has an IR emmissivity of aobut 0.1, but this can be increased to 0.9 by anodizing the material.

As EdStainless said, NASA has used special surface treatments to control absorptance and emmmission for each wavelength range, and such an approach can provide better control of surface temperatures.