Re: Uh oh, cooling problem (with Dell computers)
- From: "Tom Scales" <tjscales@xxxxxxxxx>
- Date: Sun, 25 Mar 2007 07:42:50 -0400
"w_tom" <w_tom1@xxxxxxx> wrote in message
news:1174821617.192721.138770@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 24, 9:16 pm, "Tom Scales" <tjsca...@xxxxxxxxx> wrote:
Yeah, my reading is telling me that. It's becoming a project that may
require a professional. We'll see
No reason to avoid working out this problem first yourself. Missing
in every response is the most important information. No one - not
even a professional - can help without energy numbers. How many watts
are created by servers? That is the very first number that you must
provide. Using that and air temperature of incoming air, then
calculate necessary CFMs and how much warmer the closet can be
compared to incoming air. For example, if incoming air from room is
70 degrees, then acceptable closet temperature might be 75 F. CFMs are
then a simple multiplication.
Next select a fan of sufficient CFMs and dB noise - two most
important parameters provided for every fan.
Problem with dumping air into an attic is that inside air may be
moist - cause roof rotting. Advantage is that hot attic spaces should
be cooled anyway by moving air through attic. An attic should have a
fan just to blow out excessively hot air anyway. That vent is where a
closet fan would be channeled into.
Using household air conditioning is not a good idea since that
closet requires a constant airflow - constant CFM. Furnace and air
conditioner cycle - and would not cycle sufficiently when outside
temperatures are too ideal.
Very first task - how many watts of heat are created? Without that
number, then no good solution can be engineered.
Getting down to the component detail may be a challenge, but possible. Is
it reasonable to use the maximum wattage of the power supply, on theory that
I wouldn't be exceeding that number?
Say, for example, I have 5 machines with roughly 400watt power supplies, I
would be generating 2000 watts. Two monitors for another 200 watts.
Say 2500 watts total.
The room itself is roughly 7x10x8, so 560 cubic feet.
Now I'm lost. How do I factor watts into CFM? For example, this fan looks
interesting:
http://tinyurl.com/2rq922
200cfm and very quiet. The entire line looks to be quiet and is available
from 100cfm to 2000cfm (very pricey at the high end).
I chose this one a bit at random, and it is $155., but 200cfm would exchange
all the air every three minutes, which from most of the charts I've read
would be adequate -- most suggest 2 to 15. If not, I could go up to the 400
for $269. The tradeoff is higher CFM are louder, so I would need to balance
the two objectives. I just noticed the 250 is only $4 more, so perhaps it
is the best choice.
With these fans, I could put one in the ceiling, duct through the attic and
exhaust outside. I looked into the attic and could come out easier than I
expected.
Power is simple too. I'm beginning to think this is possible to be a DIY
project, but need to size it properly.
More suggestions welcome :)
Tom
.
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