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Old 02-24-2015, 06:20 PM   #1
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Brilliant idea gone wrong...


I had a brilliant idea last year in converting a coat closet into an electronic room and put all of my electrical equipment there. The surprise I ran into was the amount of heat generated by all the equipment. I had a second brilliant idea of cutting a large hole in the ceiling and putting a fan to vent the hot air into the attic. And the idea worked perfectly...

But... Now learning more about making the home more efficient, I went up to the attic and sealed all light fixtures, ac cables and any holes obvious to air leaks; but am stuck with my brilliant idea of cutting a 8" hole and putting a fan to suck all of the air into the attic. Let me remind you that this is a coat closet with no insulation. There's a 1" gap on the bottom of the door and no insulation. How can I not waste the heated/cooled air to escape through this closet yet maintain the equipment from heating up?

I was thinking, I have a washer room next door with a vent. Was thinking of cutting that vent and looping it to this closet. That way instead of sucking the air into the attic, I would concentrate it to the washer room instead. At lease it won't be a constant waste into the attic. Not sure if this is also a brilliant idea or a bad idea.

Thanks in advance.
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Old 02-25-2015, 07:30 AM   #2
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Depends on how hot the room gets.

Perhaps some sort of damper system that would allow you to recirculate the warmed air into the rest of the house during the heating season and out of the house (and ideally out of the attic) in the cooling season.

Is there any duct work in the attic?
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Old 02-25-2015, 09:24 AM   #3
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Use a louvered door, no vent needed.
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Old 02-25-2015, 09:26 AM   #4
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Quote:
Originally Posted by joecaption View Post
Use a louvered door, no vent needed.
Exactly, I'd seal back up the attic and either put in a louvered door or cut vents in to the existing door. If it still gets really warm in stall a fan to blow the air out of the vents.

or just leave the door off.
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Old 02-25-2015, 12:32 PM   #5
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Quote:
I was thinking, I have a washer room next door with a vent.
Ayuh,.... While I agree that a vented or louvered door is the easiest answer,....

If ya wanta vent into the next room, through a common wall, that ain't that bigga deal either,...
1st step is findin' the framin', 'n wirin' lay-out, so's ya put the "Hole" in an open space,...
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Old 03-14-2015, 09:01 PM   #6
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Have louvers both at the top and the bottom of the door, if not running the full height.
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Old 03-14-2015, 10:19 PM   #7
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The louvers will actually work very well.

And don't feel bad. What you did is a very common mistake.

There is actually an easy way to figure out how much 'potential' heat your stuff is going to generate.

Look at the wattage listing on the label of the device. If they don't show wattage, but Volts/amps....then calculate watts from that.

For example, if you have a wireless router that says it's need 12Vdc @ 500ma, that works out to 12x0.5=6w.

If you have a DVR for a security camera system in there and it shows the power as being 120Vac @ 2A, then 120x2=240w.

Now, with all that said. Just because something 'can' use up to 6w of energy, that does not mean it will. But, for heat load calculations, you have to assume the worse.

Next.....add up all those watts. That is your heat load.

To give you an idea of how much heat that is, think of it this way. Say your total heat load was 100w. Now, think of how hot a 100w incandescent light bulb is? Oh yea, start to see the picture? If you have 10 60w incandescent light bulbs in your house, that is 600W of heat energy. If it's summer time, that is heat your having to 'remove' from the house.

Anyway, getting back on topic. If we know the wattage, it's easy to calculate the BTU's (heat). As a 'rough' estimate, 100w of energy is equal to 341 BTU's.

There are plenty internet references.....but here is a quicky

--------------------------------------
CFM * 1.08 * tRise = Btu/hr
--------------------------------------

or....CFM = BTU / (1.08 x tRise)

General information follows.

That factor is based on standard conditions which are at sea level and A standard humidity level which i do not remember off hand. As altitude increases the factor decreases. At 2100 feet it is roughly 1.0.
The humidity also affects this number. Generally the greater the humidity the greater the factor.

So....if you look at the above, plug in your total heat load and divide it by 1.08 and the number of degrees you will allow the temp to rise. So, if your have 1000 BTU of heat and you don't want a temp rise of more than 10 deg...then....1000/(1.08*10) = 92.59 CFM

As a general rule, you want to try and keep the ambient temp for your electronics below 100 deg. Most stuff is good for around 110....(most stuff)....but heat is the worst enemy to electronics.

Also as a general rule....keep sensitive stuff closer to the ground.....heat rises.
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