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In Reply to: You're killin me... posted by jneutron on August 25, 2003 at 15:56:18:
You said Cu at 400w/mk. there's gotta be a length in there somewhere..Cu is 10.2 watts/inch-degree C. (sorry about the units, they say memory is the second thing to go)I found a couple of references to this number:
4 Watts per cm per degree C, which agrees with your 10.2 watts per inch per degree. Converting to MKS should give 400W per meter per Kelvin, unless I got the conversion the wrong way round and it's 0.04 W per meter per K....
Let me think this through....
The equation is actually something like:
dQ/dt (watts) = -k dT/dx (degrees/meter)
(the whole assumes unit area, BTW...)since dT/dx is degrees/distance, I think k has to be watts per degree per distance to be dimensionally correct....
Example: a 1 degree per centimeter gradient = a 100 degree per meter gradient, so....
I GOT IT THE WRONG WAY ROUND.... the thermal conductivity if Cu is 0.04W/m/K. Damn! So what effect does that have.....
The right calc is:
1.5x10-7 Watts = 0.04W/mK x 10-6m2 x (temp gradient)
(using 0.04W/mK for the approximate thermal conducivity of CU and Ag at 300K)So, temp gradient is 1.5/4 x 10 = 3.75K/m.
Ok, now I need to go think about your Seebeck effect theory here...
Maybe tomorrow!
Peter
Follow Ups:
Again, havin fun with the subject line....When you said 400 w/mK, I thought you meant millikelvin...sorry..the guys here use millikelvin numbers in the dewars..ya had me major confused..that and a martini..
Ya gotta admit...it is interesting, isn't it??
Tomorrow, I gotta figure out how an optical encoder at 216,000 counts per revolution is losing 7.5 counts per turn..the magnetic analysis guy is roasting me cause of some harmonics on a serpentine quad I made using a computer thingy.....
I'm running out of things to test to find the drift error..pissin me off..big time..It must be the mechanical engineer's fault..it's always their fault...:-)
Peter, it's been a pleasure..later..
Cheers, John
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