With this entry I’d like to tackle the issue of thermal conductivity vs. thermal impedence in adhesive bondlines. Although some adhesive manufacturers may claim to have bulk thermal conductivity values higher than 30 W/m°K, device manufacturers need to appreciate that the bulk value is only an indicator of potential for heat transfer through the adhesive bondline. The material with the lowest thermal impedance should actually be the goal, as it is with this that the best heat transfer can actually be obtained.
The material with the lowest thermal impedance should actually be the goal
The factors that affect thermal impedance include: adhesion at the interface, surface wetting, thin adhesive bondlines and void-free bondlines. If the interfacial adhesion is weak or impacted by filler alignment, then the thermal resistance across the interface will be so great that the bulk resistivity becomes meaningless. If the bondline can be kept thin, for example by using a solventless adhesive, then heat transfer will be very efficient regardless of the difference in the bulk thermal conductivity.
Consider for a moment that air has a thermal conductivity of 0.0261 W/m°K (@25°C). Even a moderately-filled silica epoxy has a thermal conductivity of 0.3 W/m°K which is already more than 10 times more conductive than air. By the same token, consider a highly solvent-filled adhesive. Should the least amount of air be present, whether through a void or delamination, then the bulk thermal conductivity is immediately no longer the same.
With this in mind, I believe that the highest solvent-free, organic-filled pastes we will see will never be higher than 10 W/m°K, but that these pastes are still good candidates for applications requiring higher thermal conductivity - even those applications requesting values greater than 10 W/m°K. I remain open to the possibility that higher thermal-conductivity products may appear on the market, but I would look first to the test method used to determine the thermal conductivity and then to the void-free bondlines that these adhesives can achieve before being convinced that they make superior candidates for high thermally conductive applications.
12 responses so far ↓
1 ajay vasudev // Sep 21, 2007 at 3:13 pm
dear sir,
i am more concerned about how to increase the thermal conductivity so that i can get less timing for cooling the reactor & therby increasing my production.
product- styropore
preavious k value-490 w/mk
resent k value- 465 w/mk
k value decreased to 25 units
am gettin more time to cool the reaction mixture, which affect my production.
am more concerned about how to increase the k value, there is no diposition on jacket side.
2 Linqblog // Sep 28, 2007 at 3:24 pm
Sorry, our specialty is working with thermally conductive adhesives, or other specialty chemicals. Since the styropore falls outside this category, our comments may not be of much use to you.
3 vahik // Oct 13, 2007 at 6:46 am
I need your catalogues
4 rahul // Oct 25, 2007 at 11:52 am
Where the Styropore available? We require the list of suppliers. Better, if get the Indian/ Asian Supplier.
5 Linqblog // Oct 25, 2007 at 9:10 pm
CAPLINQ can supply the Styropore or equivalent material. What we require with these inquiries are the size of product required and the quantity requested. Once this is received, CAPLINQ can provide a quotation for you.
6 Lee // Nov 16, 2007 at 9:16 pm
i have a question about heat transfer
let say: one face of aluminum plat 1.5cm thick is maintained 700 degree celsius and the other face is maintained at 20 degree celsius (air outside).
therma conductivity for aluminum is 202w/mxk
area is 0.0491 m square
so the answer is approximate 449.63Kw/m square
so… it take how long to balance the both temperature????
thank you
7 Linqblog // Nov 21, 2007 at 9:42 pm
Though this isn’t the right forum to discuss thermodynamics equations, this is indeed a simple formula that follows Newton’s law of cooling which states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. The law is:
Q = h · A(To - Ta)
where
Q = Heat transfer in Watts
h = Heat transfer coefficient
A = Surface area of the heat being transferred
To = Temperature of the object’s surface
Ta = Temperature of the surroundings
Grab a paper and pencil and you’ll get your answer in no time!
8 Eric Nana Owusu // Dec 19, 2007 at 1:37 pm
please i need table for as many thermal conductivity of substances as possible.
9 Linqblog // Dec 19, 2007 at 2:28 pm
The WebElements website is the best source to obtain the thermal conductivity value of the entire periodic table. http://www.webelements.com
10 Doodee // Feb 1, 2008 at 9:45 am
Thanks for sharing
11 BlogBacker » Bulk Thermal Conductivity vs. Thermal Impedence by Linqblog // Feb 5, 2008 at 2:51 am
[...] backed up on 02:04:2008Originally Published: Sat, 01 Sep 2007 20:04:44 +0000http://caplinq.com/blog/bulk-thermal-conductivity-vs-therm… With this entry I’d like to tackle the issue of thermal conductivity vs. thermal impedence in [...]
12 spouspeli // Feb 8, 2008 at 5:56 pm
I’d prefer reading in my native language, because my knowledge of your languange is no so well. But it was interesting!
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