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Lets say I exhaust 800cfm when it is 0* outside and 70* inside. How many BTUs will it take per hour to maintain the building temperature. (above what it would take if I were not exhausting anything of course)

Presumably there is a simple formula for this; I just don't know what it is. Thanks.
 

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The formula is 1.08 x delta-t x cfm for sensible heat.

If you're running a humidifier you would have to take that into account - latent heat.
 

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Latent Load
BtuH = 4500 x (air exchanges x (volume) /60) x
(W Final – W Initial)
(W Final – W Initial) = Difference Ratio Pounds of Moisture per dry air

from: http://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0ahUKEwjx3afansbJAhVEMz4KHeVlDyMQFgggMAE&url=http%3A%2F%2Fclasses.sdc.wsu.edu%2Fclasses%2Farch432%2Flectures%2FHeating%2520Loads%2520Infiltration.ppt&usg=AFQjCNEXYob_uOkP895kWEofttaBxOj9Kg&bvm=bv.108538919,d.eWE&cad=rja

Best practice would be to have some make-up air supplied to the same room so you don't end it up with a high heating bill. Very wasteful to be exhausting and reheating 800 cfm - it's the equivalent of running a furnace constantly.




 

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If you were actually creating 800 cfm of new infiltration the calculations above would apply. But, at times when the house is experiencing some natural ventilation, a portion of that air flow is offset by the fan flow. This is known as the "Half Fan Rule". With a 100 cfm bath fan it would often mean 100 cfm would only generate 50 cfm of fresh (cold) air. At 800 cfm only part of that would apply and this on;y applies when the house is experiencing a natural air flow.

Confusing, but for you it just means the actual btu's lost will be something less than the calculated.
Judy Roberson did a good job of summarizing the work that came to this conclusion.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.61.4432&rep=rep1&type=pdf

As an added note, is this an exhaust fan for a house? If so, does the house have any naturally drafted combustion appliances, furnace, boiler, fireplace, or water heater? A fan that big would typically backdraft them all.

Bud
 

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Lets say I exhaust 800cfm when it is 0* outside and 70* inside. How many BTUs will it take per hour to maintain the building temperature. (above what it would take if I were not exhausting anything of course)

Presumably there is a simple formula for this; I just don't know what it is. Thanks.
60,480 BTUs.

Of course that would vary with where the fresh air enters the house. If its pulled in through a duct to the furnace return, then 60,48 is correct. If it comes in through a fresh air duct near the exhaust fan(if its a range hood), then it would be a much lower amount.

As said earlier, temp difference times 1.08 times CFM will give you the required btus.

The half fan rule doesn't apply to your question.
 

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Discussion Starter · #6 ·
Okay, 60,000 btus. I am paying about $1/therm, so exhausting 800cfm would cost about a dollar an hour. Is that approximately correct?
It is for a woodworking dust collector.

My furnace intakes and exhaust outside so this can't affect the furnace, but the water heater uses inside air. Is that likely to be an issue?

I googled on half fan rule, but didn't find anything. I am guessing it means that a 100cfm fan will only draw 50cfm because of resistance on getting fresh air in?
 

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Beenthere is correct that the half fan rule doesn't need to be considered here as your fan flow is so much more than the natural ventilation.

Your estimate of $1 per hour is correct.

A naturally drafted hot water heater, no fan for the exhaust, is the most susceptible appliance in a home. Your 800 cfm could depressurize that area by -25 Pascals, almost 10 times what a water heater can tolerate. One option would be to replace the water heater with electric or sealed combustion, like your furnace. Or a heat pump water heater, basically electric.

The other approach would be a damper to allow 800 cfm return air controlled by the fan. That way the damper remains closed until the fan is turned on. But some calculations might say, the damper would also need a fan.

The basic problem is the replacement air has to come from somewhere, either you provide it or the house will through all of the leaks.

Bud
 

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Okay, 60,000 btus. I am paying about $1/therm, so exhausting 800cfm would cost about a dollar an hour. Is that approximately correct?
It is for a woodworking dust collector.

My furnace intakes and exhaust outside so this can't affect the furnace, but the water heater uses inside air. Is that likely to be an issue?

I googled on half fan rule, but didn't find anything. I am guessing it means that a 100cfm fan will only draw 50cfm because of resistance on getting fresh air in?

Half fan rule assumes that the other 50 CFM will come from the already existing infiltration. So the 100 CFM fan would be removing 100 CFM. But only 50 CFM from a fresh air intake.

Your current furnace may not deal with that much cold fresh air too well. Most modern furnaces have a min return air temp of 50°F. So 800 CFM of 0°F air, and 800 CFM of 70°F room air mixed together, is 1600 CFM of 35°F air. Which would cause condensation in the primary heat exchanger of a condensing furnace. And rot it out.

Yes, unless your make up air is powered, you will end up backdrafting your water heater.
 

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Is this hypothetical or for commercial use, because I don't see why anyone would need to exhaust 800 cfm in a residential application.

200 is good for a range, 50-100 for a bathroom.
 

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Why can't a dust collector be used that vents indoors with a good hepa filter?

You would pretty much have to have a passive or powered 10-12" pipe coming in for make-up air and the workshop would get cold. You can't pipe more than maybe 5-10% fresh air into the return duct without reducing the return below 60f.

Would also have to put in a sealed combustion water heater
 

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Discussion Starter · #12 ·
Why can't a dust collector be used that vents indoors with a good hepa filter?

You would pretty much have to have a passive or powered 10-12" pipe coming in for make-up air and the workshop would get cold. You can't pipe more than maybe 5-10% fresh air into the return duct without reducing the return below 60f.

Would also have to put in a sealed combustion water heater
I have it on a filter now, but the "new thing" among woodworkers is venting it out; so I was looking into what was involved.

Since most days are much warmer than 0* (and assuming A/C cost very roughly the same as heat) then using the dust collector 30 minutes a day would cost about $30 a year. Presumably that could be further reduced by opening a window in my shop and closing the door to let the shop go towards the outside temperature.

I have no heating ducts in my shop, it stays warm/cool because it is surrounded and dirt.

My water heater is power vented, but uses inside air for combustion. Presumably that is problem. Since the water heater is not in my shop, would an open window in the shop make it okay?
 

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I don't know how those water heaters perform against negative pressure.

Cooling uses far less energy than heat, moving heat instead of producing it but electricity often costs far more than gas. Lower temperature delta too.

The hot humid air may create condensation in the basement during the summer though.

You would want make-up air for sure; that 30 minutes per day would really dry out the house.
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I'm sure exhaust to the outside is a bit better than iaq, but you're not dealing with anything toxic, right?

I would want exhaust to the outside for cutting plastics or composite materials with formaldehyde glue.
 

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Since it's exhausting particulate laden air, an HRV is probably out of the question. (air to air heat recovery) It wouldn't help with moisture but it would help to offset the temp difference. Although at an estimated cost of $30/year it's not even worth the effort.

If the water tank can be sealed from the workshop by a door, and you open the window, you probably won't have any issues. I'd check for correct drafting with all exhaust fans on and the tank running just to be sure.

Cheers!
 
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