Thermal conductivity - comprehension

Cathar · 03-01-2003, 10:02 PM

Okay, so thermal conductivity is defined in terms of W/m-K.

Copper has a thermal conductivity of 385W/m-K

What this basically means that if a piece of copper that is 1m long, then if one end is 1C cooler than the other end, then 385W of heat will flow from the warmer end to the cooler end.

I understand this, but where I'm wanting clarification is that this value must surely be related to the cross-sectional area of the substance. ie. the 385W value must be dependent upon the area through which the heat is flowing. Being an SI definition, is it correct to assume that we're talking about a 1m^2 cross-sectional area here?

myv65 · 03-01-2003, 10:19 PM

Q = k / L * A * delta-T

In your example, the "sides" would need to be perfectly insulated and the bar have an area of 1m^2.

Cathar · 03-01-2003, 10:31 PM

Quote:

Originally posted by myv65
Q = k / L * A * delta-T

In your example, the "sides" would need to be perfectly insulated and the bar have an area of 1m^2.

Thanks Dave for the clarification.

This is what I thought and assumed, but just wanted someone who really knew their stuff to confirm it.

msv · 03-08-2003, 10:47 AM

I´m a bit dull for the moment (OK, most of the time), but if I get the mathematix right it means that if the area shrinks and the same amount of heat is to be transferred then there will be a larger remperature difference? Right?
A.k.a. a temp gradient in the heat exchanger?
regards
Mikael S.

myv65 · 03-08-2003, 07:00 PM

It isn't quite that simple, especially when you begin talking about radiators. Nonetheless, if all else remains constant than shrinking area for a given heat transfer means you need a higher differential.

msv · 03-10-2003, 06:29 AM

Right, it´s not that simple.
I tried consulting my old Fundamentals of Classic Thermodynamics friday night. Maybe the combination of coffee, whisky, bag pipe music and thermodynamics wasn´t optimal, but somewhere between Carnot-cycles and sonic walls as a function of preassure and temp I just stopped bother the thermodynamics. I guess the book wasn´t written for water cooled computers.

regards
Mikael S.

gmat · 03-10-2003, 06:38 AM

Yep. And thermodynamics teachers could use PC watercooling as a motivation for their students... At least it would bring some interest, as generally everyone hates thermo. (i did.. until i discovered what it was really about)

Alchemy · 03-10-2003, 06:24 PM

Quote:

Originally posted by msv
Right, it´s not that simple.
I tried consulting my old Fundamentals of Classic Thermodynamics friday night. Maybe the combination of coffee, whisky, bag pipe music and thermodynamics wasn´t optimal, but somewhere between Carnot-cycles and sonic walls as a function of preassure and temp I just stopped bother the thermodynamics. I guess the book wasn´t written for water cooled computers.
regards
Mikael S.

You wouldn't have learned much from it anyway, at least nothing relating to this thread.

Watercooling is entirely based on the science of heat transfer, not thermodynamics. Heat transfer draws many things from thermo, including heat capacity and the concepts of heat and energy, but it is a science unto itself.

Thermodynamics is based on the change in material properties with temperature and pressure, among other things, and as such would help you only in understanding evaporative cooling.

Alchemy

03-01-2003, 10:02 PM	#1
Cathar Thermophile Join Date: Sep 2002 Location: Melbourne, Australia Posts: 2,538	Thermal conductivity - comprehension Okay, so thermal conductivity is defined in terms of W/m-K. Copper has a thermal conductivity of 385W/m-K What this basically means that if a piece of copper that is 1m long, then if one end is 1C cooler than the other end, then 385W of heat will flow from the warmer end to the cooler end. I understand this, but where I'm wanting clarification is that this value must surely be related to the cross-sectional area of the substance. ie. the 385W value must be dependent upon the area through which the heat is flowing. Being an SI definition, is it correct to assume that we're talking about a 1m^2 cross-sectional area here? Last edited by Cathar; 03-01-2003 at 10:11 PM.

03-01-2003, 10:19 PM	#2
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	Q = k / L * A * delta-T In your example, the "sides" would need to be perfectly insulated and the bar have an area of 1m^2.

03-08-2003, 10:47 AM	#4
msv Cooling Savant Join Date: Dec 2002 Location: Sweden Posts: 336	I´m a bit dull for the moment (OK, most of the time), but if I get the mathematix right it means that if the area shrinks and the same amount of heat is to be transferred then there will be a larger remperature difference? Right? A.k.a. a temp gradient in the heat exchanger? regards Mikael S.

03-08-2003, 07:00 PM	#5
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	It isn't quite that simple, especially when you begin talking about radiators. Nonetheless, if all else remains constant than shrinking area for a given heat transfer means you need a higher differential.

03-10-2003, 06:29 AM	#6
msv Cooling Savant Join Date: Dec 2002 Location: Sweden Posts: 336	Right, it´s not that simple. I tried consulting my old Fundamentals of Classic Thermodynamics friday night. Maybe the combination of coffee, whisky, bag pipe music and thermodynamics wasn´t optimal, but somewhere between Carnot-cycles and sonic walls as a function of preassure and temp I just stopped bother the thermodynamics. I guess the book wasn´t written for water cooled computers. regards Mikael S.

03-10-2003, 06:38 AM	#7
gmat Thermophile Join Date: Mar 2001 Location: France Posts: 1,221	Yep. And thermodynamics teachers could use PC watercooling as a motivation for their students... At least it would bring some interest, as generally everyone hates thermo. (i did.. until i discovered what it was really about)

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