warm bodies

From: glen mccready <tyra@dak.pbz>
Forwarded-by: Keith Bostic <obfgvp@obfgvp.pbz>
From: sevrqzna@fcybqr.pbz (Noah Friedman)

Someone at MIT wrote this.  I can't believe I didn't save the headers,
but I can't find the original message (although I did find one reply).

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Subject: The CP/Donut Heat Engine

Feel free to forward this to any 2.40 types you feel might be interested.
I'm such a nerd and love it so....

Thursday, there was some conference for campus police officers in 10-250.
I made the mistake of walking by this ill-fated room and discovered (quite
to my surprise, of course) the largest array of donuts I have ever seen
in my life.  They had six full-sized folding tables absolutely FILLED with
donuts.

If we consider 3m^2 of space per table and six tables, that's 18m^2 of
space for donuts.  A donut on its side is approximately 3cm x 15cm or
.0045m^2.  This makes 4000 donuts!  10-250 seats a maximum of 300 people,
which gives us an incredible 13.3 cream-filled chocolate-glazed
confectionaries per police officer!

At a conservative 350 calories/donut, that means that each CP consumed
4600  calories at the conference yesterday, which happens to be just about
double the entire reccomended caloric intake of a sedentary middle-aged
male.

Let's model 10-250 as a closed system.  Consider it a triangular prism
formed by cutting a 5m X 30m x 35m rectangular solid across its diagonal,
resulting in an enclosed volume of 2625m^2.  Through PV=mRT we find that
the mass of air enclosed in this room is m=PV/RT (Rair=287, T=300K, P=10e5
Pa), or 305 kg of air.

If 25% of the donuts' energy is converted to heat by the body (the
remainder going to the production of fat and the recombination of chemical
bonds after digestion), we see that (.25 x 4000 donuts x 350 kcal/donut
x 4.16 kJ/kcal) 364000 kJ of energy is released into the room.

Now, if we use U=mc(T2-T1), we can find the final temperature of the room.
U=364,000 kJ, m= 305 kJ, T1 = 300 K, c(air)= .716 kJ/kg-K

The final temperature in the room would end up being 1395 K or 1122
degrees C.  This is just about the melting point of copper....

This suggests that 10-250 is NOT an open system or that less than 25% of
the donuts' energy actually gets converted to heat.

Now, what exactly is the implication of 364 MJ?  It may seem like a lot
of energy (and it is) but what exactly is it in terms of power?  As the
egalitarian's credo tells us, power is more important than work, and the
demands of this problem also state that instantaneous output is more
important than the integrated function.

I think the conference was eight hours long.  Instantaneous power output
is measured in Kilowatts, which is a J/sec.  Eight hours is (8 hours x 60
min/hr x 60 sec/min) 28800 sec, giving us a power output of 12.64 kW
total.  We previously assumed 300 people in the room, or 42 Watts/cop.

Thus, each cop is putting out about 2/3 as much heat as a standard
incandescent light bulb.  This is completely reasonable.

I feel as if I have just hit upon some great truth of humanity here, but
I'm not sure what it is.

In Nerd Hell

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From: obo@tenpxyr.fgbpxoevqtr.zn.hf (Robert J. Chassell)
To: sevrqzna@tah.nv.zvg.rqh
Subject: Re: nerd hell
Date: Fri, 28 Jan 94 07:44:17 EST

    ..., each cop is putting out about 2/3 as much heat as a standard 
    incandescent light bulb.  This is completely reasonable.

An average person puts out about 100 watts:

    2000 kcal/day in food approx.
    4.16 kj/kcal
    60 sec/min
    60 min/hr
    24 hr/day

    (/ (* 4160 2000) (* 24 60 60))
         ==> 96 j/sec 

This is a useful rule of thumb:

When you host a group of twenty people, you should expect that your room
will warm up as if it contained two 1kw electric heaters, and plan to cool
the room appropriately.

Energy thoughput is the bread and butter of engineering and technology.
I cannot imagine how anyone at MIT can avoid viewing other people as 100
watt lightbulbs.

    Robert J. Chassell               obo@tah.nv.zvg.rqh
    Rattlesnake Mountain Road        obo@tenpxyr.fgbpxoevqtr.zn.hf
    Stockbridge, MA 01262-0693 USA   (413) 298-4725