Building an Aelph 1.2

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Herm said:

Looks like about .55 degrees C per watt.

Coulomb said:

Thermally it is close, yes. I am using 0.4 to allow for a fudge factor.
Anthony

Keep in mind that that data from Thermalloy was for 3 in. length. And if you're trying to be conservative, your number should get bigger rather than smaller. A 'not-bad' approximation of the equivalency between this sink and yours can be gotten from looking at the in^2/in. numbers of the two sinks.
 
My question is did you get the dimension orientation all mix up or did I misunderstood your post? I personally would always want the fins to point vertically up, not sideways.

Hello Fcel, the answer is yes. :) Between the time I put the first post up and when I put the drawings togeher, I realized the flue's should be vertical as you say and not horizontal. Not only di d this change the orienation, it also changed the heat sink count from 6 to 8 for aesthetics', as I want the finished product to look more conventional in design.

Anthony
 
Anthony,
Sorry, I still don't quite understand how you're aligning the heatsinks. To me, conventional way means having heatsinks fins pointing up which means if you have 4 heatsinks on each side, your amp would be approximately 40" long which would make it the longest amp that I will ever see.
 
your amp would be approximately 44"

Hello Fcel,

They are to be stacked 2 high, if you look at my drawing, imagine you can see the second row of heat sinks on top of the first. The 4 heatsinks becomes 8, the dimensions would be approx 14" wide x 20" deep x 10" high.

I will try to post a photo of the parts in a rough layout tonight.

Anthony
 
Anthony,

I saw your pdf and it remind me my aleph 5 that is under constructions now.
the chassis is almost done, and the electronics just need wiring.

see these two pics to see if it inspires you.
 

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Coulomb;

OK, so the Aleph 1.2 dissipates 500 watts per channel.
One of your heatsink is about .55 degrees C per watt.

Since you have 8 of them, your aggregate rise will be
.55/8=.06875 degrees C per watt.

Lets multiply that times the 500 watts youare trying
to get rid of and we get 500 * .06875 = 34.375 degrees.

Add on your 23 degrees ambient, and your grand total
looks like 57 degrees celsius (or 134 fahrenheit).

-herm

P.S. Your mileage may vary ...
 
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