rokicki@polya.Stanford.EDU (Tomas G. Rokicki) (07/14/88)
You think a 2000 heats up the place? You should try a laser printer;
the Canon CX engine in mine is a bona fide heater. (I think the
average power dissipation is somewhere around 500W, but I could easily
be wrong.) Any engineers out there want to calculate the temperature
change per watt, in a 15 by 15 foot room (say 1800 cubic feet) with
perhaps 50 cubic feet of air per minute being pumped through?
Hmm, a steady-state system, with air coming in and leaving, and the
temperature increase we can say is half the temperature difference
between the air coming in and the air leaving.
Let's see, the specific heat of air at constant pressure we'll
estimate at 5/2 R, or 20.786 joules per degree kelvin per mole.
If a joule is a watt second, and a degree kelvin is 1.8 degrees
F, and 60 seconds are a minute, and a mole is 22.414
liters, and a liter is 1000 cc, and a square foot is (2.54*12)^3
cc, we have
Temperature change =
(Power dissipated in watts) / (Air flow in cubic feet per minute)
times
(2 * 22.414 * 1.8 * 60) / (5 * 8.3144 * (2.54 * 1.2)^3 * 2) = 2.1
So, a 300W computer system (including the monitor and Amiga) with
a fan going at 50 cubic feet per minute would heat up a room by
12.6 degrees Fahrenheit. Ouch, something's got to be wrong with that.
Anyone want to correct my figures? Maybe fans move more air than
what I've indicated. After all, a person consumes 90W, right?
I'll measure air flow when I get home today . . .
-tom
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