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1 KWH = 3.6 Million joules (i.e. 1 watt = 1 joule/sec)

Specific heat of air at ~room temp @ 1 ATM ~ 720 J/Kg/oK

Density of air at ~room temp @ 1ATM ~ 1.2 Kg/m3

1 foot = 0.3048 m

So how big is the room ?  Assuming say 20x20x12 ft = 136 m3 = 165 Kg

Let's say the room is drawing a total of 10 amps, so 1200 watts or 4.32
M joules per hour.  If there were no ventilation or other dissipation of
heat, in one hour the room would be expected to go up 36oC (.01oC/sec) !

Ordinary industrial ventilation should have a minimum of 10 air
exchanges per hour.  However, it's not as though the air is really
replaced 10x though, rather the rate of exchange is 10/3600 volume per
second, and the amount of unreplaced air remaining is (1-exchange
fraction)^t.  On this basis about 1/2 the air is removed every 5 minutes
so, instead of going up 3oC in 5 minutes, it would go up more like 1.5oC
in 5 minutes, or  a guesstimate would be 4-18oC/Hr.

dT/dt=(c-1/18)- T(t)/360 (or something like that)

~ln(360*.045555*t) ~11oC (~20oF) in an hour

The other way to look at it is you need ventilation+cooling to equal the
energy going into the room. For 4.32 million joules you need about 4100
BTU of cooling.

We have freezer & equipment rooms that need up to 18000 BTU & special
ventilation ducts for individual pieces of equipment in the lab.

David Looney

Tracy R Reed wrote:
> At work last month we had half of our servers go down because a power
> strip which was supposed to have blown its breaks at 15 amps blew it at
> half that. At home my computer room is already getting quite warm and it
> is barely summer. The computer room has always been the hottest part of
> the house yet I need it to be the most comfortable because that is where
> I spend most of my time and need to be able to think. So this has me
> thinking about a few things...
> 
> Can anyone recommend a simple device which can measure the amperage
> used? Preferably without having to place it inline (I think some can
> measure the EMF) but inline is doable if that is all there is.
> 
> Let's say a computer draws 1 amp. What is the rule on how much heat this
> adds to a room? Any way to calculate how many degrees it might raise the
> temperature of the room? I think we can assume that pretty much all of
> that 2 amps turns into heat. I know power for electricity is voltage
> times current so that would be 110 volts times 1 amp which is 110 watts.
> A watt is 1 joule of energy per second. So at 110 watts I am dumping 110
> joules of energy per second into the room. If I run the computer for an
> hour I have added 396000 joules of energy to the room. If the room is
> 10x10x8 and we assume no outside circulation how fast will the room heat
> up? The goal being to see if that it is reasonable that the computer is
> making this room noticeably warmer (say, 5 degrees or more warmer) than
> it would be without the computer.
> 
> Does anyone think it would accomplish much to somehow vent the hot air
> coming out of my computer to the outdoors instead of into my room? It
> just struck me that it must be more efficient to vent the hot air
> outdoors and cool the ambient air than to try to cool the hot air from
> the computer. I am surprised more datacenters are not trying to do this.
> And if it does make sense to do this (I strongly suspect that it does)
> has anyone seen kits or cases for this? I think I might go down to the
> hardware store tomorrow and buy some of that flexible plastic dryer vent
> hose to attach to the computer and then hang out the window. But how to
> attach it to the back of the case? Hmm...just had an idea: I could get
> one of those metal fittings that go on the inside of the wall that the
> dryer hose attaches to, drill four holes in it, and put the bolts that
> hold the fan in place through it and then through the case and the fan.
> Yep, a run to the hardware store is definitely in order. I'll let you
> know how it goes.
> 

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