Combined conduction & convection 1

[swguru]

I have a device that is cooled by both conduction and forced confection heat transfer. I am having trouble determiing how much heat is transfered by condcution and convection. I can easily calculate power disspated by conduction only, as well as for force convection only. How do I determine how much heat is transfered in the combined condition?

As an example, I have a component that is mounted to a PWB. Path for conductive heat transfer is through its leads. The device dissipates .25W. Total area and length of leads is 9.1e-7m^2 and .0029m, respectively. The PWB is at temperature of 71C. Thus the calculated temperature rise from device base to PWB is 4.7C so device case temperaure of 75.7C.

With the influence of forced convection the temperaure rise through the leads will be decreased due to heat being transfered into the fluid film. With an ambient air temperature of 40C and flow of 1.3m/s, the air boundary layer temp rise is 24C. So does this now mean that the component case temperature is reduced to 75.7-24=51.7C?

How much dissipated power is now conducted through the leads?

 

[IRstuff]

It's a simultaneous solution, not superpositionable

TTFN

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[prex]

With your data the air will cool not only the device but also the PWB, as both are hotter than air. Also one would expect a device temperature of 40+24=64 °C, not what you state, supposing the air flow rate is much higher than what is required by the power to be removed.
So it seems that your primary method for heat removal is convection. However that's not so simple to calculate: you could need to account for local velocity distribution, local temperature rise of air flow.
The only sensible way seems to me the following one:
-consider the air is cooling the whole PWB as a single device and determine its temperature as a function of total power to be removed, and of air conditions
-now examine single devices on the PWB and determine their temperature rise with respect to the PWB considering conduction only as in the second paragraph of your post.

prex

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[ko99]

swguru,

Consider it a problem of parallel resistances.

The case-to-PWB is about 19 C/W using L/KA through the leads.

The case-to-air is ??? C/W in natural convection. Component suppliers often have test data but be aware it includes die-air plus die-pwb. The pwb will likely be a JEDEC std and may differ from yours. You can calculate this but it's not trivial

The case-to-air is ??? C/W in forced convection. Again, suppliers often provide this test data or you can try to calculate this.

I don't undertsand your statement "air boundary layer temp rise is 24C". Where exactly is this measurement taken?

ko (

http://www.ecooling.biz)

 

[eaboujaoudeh]

all heat transfer problems that i know of, can be solved as resistance pbs, u know you have either parallel or series resistance

Elie Abou Jaoudeh

 

[swguru]

Thanks all. I ended up solving by looking at resistances. Calculations aggreed well with some CFD modeling that I ran.

Thanks again for the tips!

Ron