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Old 3rd November 2004, 07:11 AM   #1
Luke is offline Luke  New Zealand
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Default any idea what these will dissipate?

Hi all,

I have 2 of these heatsinks which are 430mm long X 170mm High and fin length is 95mm on a 12 mm slab of aluminium. There are 33 fins in total.
Sorry didnt have the standard measure coke can, but I think beer bottle is universal size too

Question is how do I estimate the C/W on these things?
I am building aleph-x and want both channels in one enclosure.
Tx is 18 0 18 rails 24 volts bias 8 amp

Total power per channel is 384
In thinking one aint gonna quite cut it for near 800 watts?

Your opinions appreciated.
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Old 3rd November 2004, 07:12 AM   #2
Luke is offline Luke  New Zealand
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another one
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Old 3rd November 2004, 07:50 AM   #3
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Hi,

they look like something around 0,2°K/Watt. This would go down if you would cut it into 4 pieces but I donīt think you will get below 0,15°K/W.

So with 2x384 watts this would rise ca. 115°K above ambient wich is a bit too much

William
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Old 3rd November 2004, 07:58 AM   #4
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Default This might be useful

Rod Elliott has made a spreadsheet to estimate thermal resistance which may be useful: http://sound.westhost.com/heatsink.zip

I put some quick figures in... William's estimate is probably about right
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Old 3rd November 2004, 08:00 AM   #5
TroelsM is offline TroelsM  Denmark
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Default Use a fan

I think they need a fan, because of the very small distance between the individual fins. I guess its hard to get any airflow with such a narrow airway without fans.

But on the other hand, -I have no experience with it.

TroelsM
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Old 3rd November 2004, 08:10 AM   #6
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Default Gee, that was fast

3 replies in ten minutes... lol

I had a closer look at the photos, and you say the heatsink is 170mm wide with 33 fins, that means that the space between them is around 5mm. Since air moves slower when in closer contact with the surface (laminar flow), a fan is probably a good idea. Also, the shiny surface ain't the best for radiation...

So maybe the earlier estimates were perhaps a little optimistic
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Old 3rd November 2004, 08:12 AM   #7
AuroraB is offline AuroraB  Norway
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Location: Norway, -north of the moral circle..
Thumb rule calculation C/W= 50 /( Sqr A ) ,- A in sqr cm's suggests somewhere in the vicinity of 0.25....
These will have to used vertically to be most effective, and very long fins loose efficiency somewhat............
Better test it with a power resistor and controlled current, and measure the temp. rise...?? I use some 50 W metal case resitors for heat sink mount.
One solution could be to cut them in two halves mounted sideways, giving a combined 21 cm high 34 cm deep block...??
Do some calcs before you decide...
Here is a link to some heat sink theory and a calculator spreadsheet.
Th calculator valuis far to optimistic, I think.. possibly because of the very long fins
http://sound.westhost.com/heatsinks.htm

And another
http://w1.859.telia.com/~u85920178/begin/heat-0.htm
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Old 3rd November 2004, 08:32 AM   #8
Luke is offline Luke  New Zealand
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Thanks all for your help.
I have never seen one of these calculators, but Rods one tells me that its 0.0896 C/W. Ive jigged my bias etc to get about 6.8 per channel as I have just bought some .22 0hm resistors from Steve at apex and thought I should use them

Any way Im now looking at 650 watts so i get 59 degrees above ambient. This isnt going to work so its long like Im going to have to build monoblocks again. More metalwork and two transformers I have to ship from Aussie.
Thanks all for your help.

BTW any one remember Nelsons rule of thumb on heatsink temp. Was something about if you can touch the sink for a few seconds only its around ......
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Old 3rd November 2004, 08:54 AM   #9
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Luke,

0.08....canīt be right

If you can just touch it for a few seconds itīs around 50-55° , unless your a housewife, then itīs around 75-80°C

william
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Old 3rd November 2004, 08:54 AM   #10
Luke is offline Luke  New Zealand
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Just had another look and its actually 500mm long. Ive just been playing with this calculator and Im not sure if its correct. The change between 250mm to 500mm is 0.0936 to 0.0881? Can this be right or does the calculator assume a point source for the heat and not distributed?
It may make sense to cut one in half and build two monoblocks.


cheers Arthur
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