|
Propeller Head Plaza Technical and scientific discussion of amps, cables and other topics. |
|
|
|
Propeller Head Plaza Technical and scientific discussion of amps, cables and other topics. |
|
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
This post is made possible by the generous support of people like you and our sponsors:
Follow Ups
- I think we're using the same numbers.. - Commuteman 17:28:12 08/25/03 (1)
- AHA!!!!! So that's where the error is... - jneutron 17:46:53 08/25/03 (0)
Top of Page ]
[ Contact Us ]
[ Support/Wish List ]
[ Copyright © 2024, Audio Asylum, all rights reserved ]
