I know there is quite a simple answer to this, but I can't quite remember
it. I want to know how much (in wattage terms) heat I need to put into a
block of metal to heat it up. For example If I have a block of aluminium
200 x 200 x 50 and I fit two electric cartridge heaters into it, and want to
raise the temperature from room temperature (about 20 degrees C IIRC) to 170
degrees C (a rise of 150 degrees) how do I calculate the minimum required
wattage of the heaters (too much power will just heat it up faster as there
will be a thermocouple fitted). I know this has something to do with the
specific heat capacity of the material, but it was a few years ago since I
calculated anything like this.
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
0.9 x 5400 * 150 = 729,000 Joules of energy.
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).
On Wed, 7 Jan 2004 08:22:38 -0000, "Tim Christian"
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 & Rita Forbes
Engine pages for preservation info:
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.
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
6 year old niece can paint a day while off school with chickenpox :)
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.
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!
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