Paul350Z

iTrader: (1)

I pulled this out of a recent thread so that it wouldn't get lost. Many people are putting amplifiers into their closed cubby's. I had to take a wag on the actually thickness of the cubby's plastic. Next time I take it out I'll put a micrometer on it. Anyone see anything wrong with my math?

Let's see - one watt-hour of electricty produces 3.412 BTUH.

The thermal conductivity of plastic is about 0.190 (and that's not good). The heat fins of an amp are made of aluminum which has a thermal conductivity of 250 watts/meter-C.

Fourier's law of thermal conduction is q=-(k * A * Delta T)/S

q is the heat transfer in watts
k is the thermal conductivity of the material
A is the heat transfer area in square meters
Delta T is the demperature difference across the material is degrees Celsius
s is the material thickness in meters

SO

q is the number of watts that we can expect to transfer into the second thermal trap of the inside of the rear of the car. I am not going to even think about how long you'd have to have the amp up to over heat this much larger compartment.

Using some WASG (wild *** sciencitific guesses)

The k of plastic is 0.190

The A of the cubby hole is about 840 square inches or 0.54 square meters

Figure the temperture difference from a comfortable passenger cabin of 78 degrees F to the manufactures maximum temperture of 110 degrees is 32 degrees F or 25.5 degrees C to 43.3 degrees C or 17.8 degrees C.

The thickness of the cubby is about 0.0075 meters thick.

Plugging this number's in the the good Mr. Fourier's formula gives me:

q = -((0.190)*(0.54)*(17.8)/0.0075 = 243.5 watts worth of amplier power. Most amplifiers are about 50-80% efficient so figure that the actual speaker power of 121 to 195 watts worth of amplifier.

So if you're going to put a larger amplifier than 120-200 watts or so you're going to run the temperture of the cubby hole beyond the maximum recommended operating temperture of an amplifier. Putting a CD, DVD, or even a playstation in the cubby would be fine.

So venting that heat to the outside is critical. Open the bottom of the cubby up with a series (three on a side to start with) of large 1 to 1.5 inch holes on either side of your amplifier. Cut as many matching ones above the amplifier. If you can use forced air cooling with a small 80 cm computer fan mounted above the cubby to pull about 20-30 cubic feet of air per minute across the amp.

gsazabi

iTrader: (51)

Thanks for your reply on my other post.
Well, I don't plan on drilling holes on my cubby, so
would it be the best idea to move the amp elsewhere?

Deceit

Interesting read. I've been finding my JL 300/2 getting a little bit hot in the cubby lately. I've seen a post from a member on here that ran some tubing from one of his ac ducts to the glove box and supposedly worked pretty well.

JimRHIT

iTrader: (6)

I run a class D 1180w RMS JBL amp, as well as an old school PG M44i and have only once had heat issues in the cubby. That was when I was driving in 112 degree temps in Vegas w/o AC

buzzardmountain

iTrader: (17)

JimRHIT,
Sent you a pm.

Paul350Z

iTrader: (1)

Starting with a hot car is going to make things worse. Forced air is the way to go to get more cooling when conditions warrant.

You could get away with a few holes on each side of the amp and a small computer fan on the top.

If your amplifier is a class D amp it will actually draw less power when cranked up than when idleing which is @ss backwards than other classes of audio amplifiers.

I plan on mounting my amplifiers vertically on the front of my e-bay Zenclosure's under strut woofer box. I'll some vibration isolation but not a whole lot - mounting an amplifier to a woofer box isn't the best but depending on how well I screw the thing into place those vibrations will be damped by the car body.

I've priced getting a large copper plate and mounting in under the amplifiers and having it extend out a bit acting as a large heat sink/dump. I do not listen to music at crazy loud levels but look at the heat sink as more of a fashion statement. I've used brushed aluminum before and it's looked nice too.

karmstro

I have a little problem with this I guess

q = -((0.190)*(0.54)*(17.8)/0.0075 = 243.5 watts worth of amplier power. Most amplifiers are about 50-80% efficient so figure that the actual speaker power of 121 to 195 watts worth of amplifier.

the "q" you came up with is the amount of joules (energy) your cubby hole can dissipate per second. However since your amp is powering speakers, most of the power is going outside the cubby hole (you hope!!)

Thus the max amp you could have using your numbers would be

(243.5 watts) / (1 - the efficiency of the amp) = 243.5 / (1-0.80)
for a class D amps
your new calculation would be 1217.5 watts

On a side note, there seems to be too many uncertainties built into you problem, thus accuracy in this answer shouldn’t be relied on too heavily.

If you want to make sure the amp stays cool, I would maybe install a computer fan blowing out the stock bose sub cover. See if that helps!!

Good luck

karmstro

That shouldn't be true. The class D should be more efficient when it’s near full power. But, I would highly doubt the parasitic losses (ie idle losses) from the amp would be higher then the heat generated with the amp was running.

My 2 cents

Paul350Z

iTrader: (1)

Good point on the power being delivered to the load ...

I went to school when we only had A, B, AB, and C amplifiers but from what I've read of D's they draw more power at idle - like a television transmitter.

karmstro

I think you just read something a little wrong. Its the efficiency you are thinking of not the heat generated. the efficency goes up as the load does. this acually happends in many power applications

karmstro

Acually thinking about it. This would be the correct way to solve it:

total amp power - heat generated = power to speakers
heat generated is the maximum amount you think the cubby can cool = 243.5 watts
total amp power - 243.5 watts = power to speakers
factor out total amp power:
total amp power * (1 - (243.5 watts / total amp power)) = power to speakers
divide by total amp power:
(1 - (243.5 watts / total amp power)) = (power to speakers / total amp power)
and (since power to speakers / total amp power) = efficiency
and efficiency = 0.8
we get:
(1 - (243.5 watts / total amp power)) = 0.8
(243.5 watts / total amp power) = 0.2
thus
total amp power = 1217.5
and since
total amp power * efficiency = power to speakers
1217.5 * 0.8 = 974 watts

So, you want an amp around a 1000 watts or less to get proper cooling!!
thanks for making me think lol!!

unstable

This thread made me dizzy... lol