combustion air quantities

[Cutter]

Howdy all. I realize that there are a zillion variables when addressing this issue but is there any simple math that lets you know how much combustion air it takes to burn wood properly? I have a Free-Air fireplace that has a four inch fresh air line that comes up just behind the glass doors. Then there are two screw type adjusters just below the doors. The two on front pull air from inside the house and the four inch from outside. The problem is that if I close off the two front controls my fire almost smothers it'self out. I figured that the four inch would supply adequate air for combustion without taking air from inside my heated space. It was installed just as the manufacturer required. The interior grate over the intake stays cold to touch even when the rest of the fireplace is roaring. So I am getting air just not enough. I have tried just about every possible combination of damper and air control and I just wish I could get a better burn without pulling in conditioned air. Even if the front sccrews are opened a half a turn i'm alright. Actually I guess the point is moot because there is no way to change the setup because it is all solid masonry. But Iam curious if the New-Air folks need to address this issue for installers.
I really like the big dude. We heat around 3400 sq" of floorspace on two levels with the aid of the blowers on our furnace. The blowers are piped into seperate ductwork that keeps the majoritty of the house quite comfy.

http://i269.photobucket.com/albums/jj70/brad7254/dogs010.jpg

 

[Wet1]

My guess is your 4" supply is restricted someplace. A 4" OAK is way more than what would be required for proper combustion in that setup. Are you sure the 4" OAK is truly vented? The fact that you can crack the fronts and it burns okay suggests to me the 4" line might not be vented properly (assuming that it's actually for outside air).

 

[CrappieKeith]

My guess is your 4" supply is restricted someplace. A 4" OAK is way more than what would be required for proper combustion in that setup. Are you sure the 4" OAK is truly vented? The fact that you can crack the fronts and it burns okay suggests to me the 4" line might not be vented properly (assuming that it's actually for outside air).

plus 1

 

[Cutter]

Just checked pipeing again. Not even a spiderweb. Still looking for some Math.

 

[LLigetfa]

I'm guessing there is more to the plumbing than what was described, that the OAK supplies a different segment of the fire than the indoor air. While my fireplace for comparison is completely different, the two air supplies serve two separate and distinct needs. It uses a 3" OAK for secondary burn and the glass door air down wash which is considered primary combustion air. Indoor air is funneled to the coal bed via the doghouse.

I suspect in Cutter's case, it's not just about quantity and source but rather where the air goes. Probably the indoor air feeds primary air in a way the OAK does not. I would not be concerned about using some small quantity of conditioned indoor air.

 

[jdinspector]

OK, I'll bite on this one. The math I'm providing is for gas appliances, but I'll bet it's close to what you need for wood...

As a general rule, gas appliances need 1 square inch of free air opening per 1,000 btu/hour of input. So, for example, a 100,000 btu furnace needs 100 square inches (10"x10" vent into the closet or furnace/boiler room)

Here in Chicago, the gas supplier has provided VERY specific requirements for "make up air". If a gas appliance is in a "confined space" then additional air must be provided. The calculation is (Room Volume in cubic feet) divided by 50 cubic feet/btu. This number is then multiplied by 1,000. In a 8'(H) x 33'(W) x 25'(L) room (6,600 cubic feet) the calculation goes like this...6,600 cubic foot room divided by 50 multiplied by 1,000. The result is the maximum total btu/hour that this room will support. The answer is 132,000 btu/hour.

If you can calculate how much air the 4" duct is allowing in, then you could do simple math and figure out how much air is required from inside the house.

As I said, I'm not sure if these calcs are appropriate for wood burning. If they're not... never mind

 

[Corey]

Dug this up from last year: https://www.hearth.com/talk/threads/24880/

Well, lets see…. I’m sure entire books have been dedicated to this subject - but to try and get some quick ballpark…

- If the stove is quoted at 74,000 btu/hr @ 70% efficiency, then that should mean you’re burning 74,000 / .70 or ~106,000 btu of wood per hour.

- Wood generally has about 7,600 btu/lb regardless of species, so that is 106,000 / 7,600 or ~14 lbs of wood per hour

- This article http://mha-net.org/html/p-tieg02.htm says gaseous combustion stops when the air-fuel ratio falls below 15:1, and most stoves are operated with substantial excess air, so 18:1 or even 20:1 is probably closer to reality. So based on those ratios, 14lbs of wood:

@ 15:1 AFR = 14x15 = 210 lbs air
@ 18:1 AFR = 14x18 = 252 lbs air
@ 20:1 AFR = 14x20 = 280 lbs air

So it looks like your initial assumption is at least in the ballpark. As a bonus question - density of air @ standard temp/pressure is 0.075 lb/cu ft - so you’re talking anywhere from 210/0.075 ~ 2,800cu ft per hour to 280/0.075 ~ 3,700 cu ft per hour. Or given a standard 8 foot ceiling, all the air in a room roughly 19’ x 19’ on the small end up to a room 21.5’ x 21.5’ on the large end, or all the air in a 2,000 square foot house roughly every 4-6 hours.

Extra bonus for this year:

2800 cfh through a 4" pipe

Area of pipe = 2^2 x 3.1415 = 12.5 sq. inches / 144 sq inches per sq ft = .087 sq. ft

2800 cu.ft/hr / .087 sq. ft = ~ 32,100 linear feet of air per hour and 32,100/5280 = ~ about a 6 mph breeze through the 4" pipe.

 

[Cutter]

Yeah this is the kind of info I was looking for. Thanks. I am off to dreamland now but I will play with this info in the AM.
Brad