Heat Input

A block of aluminium 200 x 200 x 50 is 2 ltrs and will weigh 5400g. Specific heat capacity of aluminium is 0.9 J/g/C.

If you want to raise it by 150C (assuming NO OTHER LOSSES) you will need

One parameter missing from your post is time. If you could insulate the ali block against losses, a very small amount of power would heat it up over a long time. A large amount of power would heat it very quickly.

Assume you wanted to heat the block in 10 minutes (and again, no other losses are catered for here), you would need to expend the 729,000 Joules in 600 seconds. 1J = 1W/s, so this would be 1215W for the 10 minutes. In one minute, 12150W, etc.

That's the theory, the practical side (inc. losses) may be very different depending on the environment the block is in.

As Duncan said, environment is almost everything. The finish on the surface of the block is also important in determining heat loss rate. A painted surface loses heat much faster than a metal one. Shape and proximity to other objects are further factors.

There are plenty of data on this problem. If you want the full calculation, email me directly (I'll need to sent charts and tables).

Tim: Could we modify that statement a little please? Paint is an insulator, and, for example, a white painted ali heatsink would lose heat more slowly than a black anodised one. Temperature differential also plays a big part in heat loss.

Trying not to be too picky.. :-))

Peter

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Bloody hell Duncan - that takes me back 20 years!

Sorry, Peter, but since paint coats are thin, their thermal conductivity is not a significant factor. Moreover, there is little difference between the emissivity of different types of paint. A gloss white surface loses heat almost as quickly as a matt black one at the temperature differences under discussion. Aluminium paint, like the bare metal, loses heat at about one tenth of the rate of normal paints. These rates depend on environment: air cooling and about 30% direct radiative loss is assumed.

I have detailed data and sources if you would like them.

Take your word for it :-))

Goes against all our information on heatsinking.

Peter

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So why aren't computer heat sinks painted?

It's not that I doubt what you say, just seems a remarkable business opportunity going missing. I just need to check just how many heat sinks my

conductivity

First, to answer Kevin's question: for a rectilinear block 200 x 200 x 50 mm with a dull, natural finish suspended in free, still air, about 315 W would be required to maintain a temperature of 170 degC in and ambient temperature of 20 degC. If the block is polished, perhaps 10% less power would be required; if the block was painted (any colour but metallic), about 15% more power would be required. If the block sits on one of its 200 x 200 faces, on a reasonably insulating material such as wood, power required would be down to around 215 W.

Note that this is the surface temperature: the temperature around the heating element will be a few degrees higher, and the temperature of the heating element itself higher still, depending on how good the thermal contact is. Moreover, the temperature is that which would be reached ultimately: the terminal temperature of the system. Since this would be approached asymptotically, it could be hours before the temperature reached, say 165 degC.

A quick (ish) check on this would be to insert a 100W soldering-iron bit in a hole in the block, and insulate the shaft of the iron. Surface temperature rise should approach 60 degC.

Heatsinks: finned heatsinks are a more complex problem. At their normal operating temperature (70 degC or so) around 30% of the heat is lost by radiation. Depending on the complexity of the heatsink, much of this will fall on adjacent fins. Since surface finish primarily affects radiated heat, the actual finish becomes less important. Computer heatsinks are generally fan-cooled. This is yet another ball game.

By radiation, perhaps. But at modest temperatures wouldn't conduction and convection dominate?

Thanks for all the responses, the basic formula is fine for the rough check I want to do. I'll work out the heat for 15 or 30 minutes and depending on losses it will just take a bit longer to get to temperature (on a thermocouple so it won't get too hot).

Just for reference what is the specific heat capacity for steel?

I know I had all these figures, and the equations, but it's over 15 years since I was an apprentice at college!

Regards

Kevin

Theres a table that has a few others at: