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Topic: Calculate heat transmission coeficient? (Read 3 times) previous topic - next topic

zoomx


I am not sure what cheap metal you can mix with copper, but if they used copper+aluminum (but aluminum melting point it is very low ~650C vs ~1080C, so I am not sure a mix will be homogeneous), then is 64% copper


I believe that is a copper alloy.

http://en.wikipedia.org/wiki/List_of_copper_alloys

robtillaart

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Is it not enough to heat it up to a certain temperature and measure the cool down temperature versus time?
from my head it would be some negative exponential function like

Tt = Te + c1 * exp(c2 * -t).    // Tt is temp at time t; Te is T environment, the temp it cools down to, c1 and c2 are constants depending on material form etc.

and c1 was the HTC? (I do not recall anymore)

Excel could help you find the equation if you have the sample data.


I want to use a similar formula I found in wikipedia, but then I was wondering the many variables the material form will have, it the shape was just a box, then it is easy, dissipation surfaces are the layers, but having these "pegs" increase the surface. The aluminium ones have another design:


Still if you measure the cooling down curve, the graph would be (reasonably) identical to a box with the same surface area (assuming it can radiate freely)
Rob Tillaart

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mrburnette

I went through reliving my college physics and thermodynamics classes when writing the H,E,A.T. article here: http://www.instructables.com/id/LASER-HEAT/. But when all is said and done, a little bench work can can put those nasty equations into the closet :-)

Use a power-resistor and lab supply to heat the resistor to the same temperature as the Pi CPU when operating... Use an IR thermometer to measure the temp.  Now, put the resistor on the heat sink oriented in the position it will operate and measure the temperature using the same power supply V & I.  Note the stabilized temperature.  Remove thee heatsink, reduce the I &E until the same temperature is reached w/o the heatsink as was recorded with the heatsink. 

Your power supply values, I * E, gives the power in Watts between the no heatsink and heatsinked scenarios.  Of course, our recordings are only approximations since mounting hardware, thermal grease, etc. Will have effects.

Ray


pito

#8
May 31, 2013, 12:54 pm Last Edit: May 31, 2013, 12:57 pm by pito Reason: 1
Farnell lists about 1120 passive heatsinks (natural convection ones), where about 215 are for BGA packages. The typical thermal resistance of an aluminum similarly shaped I see there is 25degC/Watt. So why to waste time with science? :) :)

Jack Christensen


Farnell lists about 1120 passive heatsinks (natural convection ones), where about 215 are for BGA packages. The typical thermal resistance of an aluminum similarly shaped I see there is 25degC/Watt. So why to waste time with science? :) :)


Good observation pito! Never forget the pragmatic approach.
MCP79411/12 RTC ... "One Million Ohms" ATtiny kit ... available at http://www.tindie.com/stores/JChristensen/

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