Why did you install an outdoor air kit?

FirepotPete said:

I think that's just a air circulation and insulation problem. If your oil burner was just in one room, a space heater, you would see the same results. The OAK doesn't cause circulation problems it cures them.

Click to expand...

Should have included I only had the zone on where the pellet stove is, not the entire house.

and thats no OAK

and thats no OAK

biomass burner said:

and thats no OAK

Click to expand...

And that's no JOAK.

Do I see signs of "cabin fever" in this thread?

Tstark,
I'm having trouble with your medieval units : but from what I can translate they don't ring right.

I looked up the standard renewal rates for air in different rooms and found non less than once per hour.
Usually much more.
So for your 10K cf house I would suggest the MINIMUM air changes should be 240K cf a day.

A well regulated pellet stove running at 3 KW/hr ( this should be enough to heat that house)
will use about 1.65 m3/hr of air in the OAK .
If we multiply this by 24 for the hours and by 35 to use your barbaric units , I get 1400 cf / day.
This is running at stoichiometric plus 20%.
Even in a badly insulated house the figures are not in the same ballpark.

If someone could confirm ( or correct ) my arithmetic , I would be very grateful.

Monica in France said:

Tstark,
I'm having trouble with your medieval units : but from what I can translate they don't ring right.

If someone could confirm ( or correct ) my arithmetic , I would be very grateful.

Click to expand...

I might be able to but I would give you angst by using medieval and barbaric units, so won't bother.

Monica in France said:

I would be very grateful.

Click to expand...

Doubt it

Monica in France said:

Tstark,
I'm having trouble with your medieval units : but from what I can translate they don't ring right.

I looked up the standard renewal rates for air in different rooms and found non less than once per hour.
Usually much more.
So for your 10K cf house I would suggest the MINIMUM air changes should be 240K cf a day.

A well regulated pellet stove running at 3 KW/hr ( this should be enough to heat that house)
will use about 1.65 m3/hr of air in the OAK .
If we multiply this by 24 for the hours and by 35 to use your barbaric units , I get 1400 cf / day.
This is running at stoichiometric plus 20%.
Even in a badly insulated house the figures are not in the same ballpark.

If someone could confirm ( or correct ) my arithmetic , I would be very grateful.

Click to expand...

Monica,

Check your calculation for how much air required to burn the fuel. That's all screwed up. Check google for conversion of English to metric units. 1.65 m3/hr is waaaay too low. That's only 0.97 CFM in medieval real world units. Where did you come up with the 1.65 cu. m/hr number? Do you mean 1.65 m3/min? That would be closer. BTW, how many days in a metric month?

You're making my head hurt.

Oh yah, if God had meant for man to use the metric system, he/she would have given us 10 fingers........ah....never mind.

Don't remember anyone saying not to use the metric system. I'm just too much of a troglodyte to do the conversion

bogieb said:

Don't remember anyone saying not to use the metric system

Click to expand...

I think the Americans in the 70's said not to use it, here,
after a badly failed attempt to get us to use it...

Pete Zahria said:

I think the Americans in the 70's said not to use it, here,
after a badly failed attempt to get us to use it...

Click to expand...

Okay, I worded that badly - I meant in this thread

Thanks skibumm100 ,
I too felt that the figure of 1.65 m3/hr was too low ,
but thats what came out of a simulator.
On the other hand the factor of 35 between SI units and Cubic feet I think is right.( 39.25/12 ^^3 )
Don't forget we are talking of the OAK air flow , in the chimney the flows are much higher because of the temperature.

So just trying an off-the-cuff estimate :
Taking pellets at 4.7 KW/Kg ( net , after burning/evaporating off the humidity ).
Ideally for 3 KW/hr you are burning 3 / 4.7 = 0.65 Kg /Hr. ( ~ one of my sacks of 15Kg a day ).
Adding for losses we have : 0.80 Kg/hr.
The stoichiometric ratio for wood is not easy to find : but I found two sources giving the value of 6.4 ( but this needs to be re-checked)
This means we are using 0.80 x 6.4 = 5.12 kgs of air / hr.
Adding extra air we have : ~ 6 kgs/Hr.
This in volume (at sea level and 0°c) is 6 / 1.3 = 4.7 m3.
This would be about 3 cu.ft a minute.
A figure which is still intuitively on the low side.
So please poke holes in my logic so we can be sure of the right answer.


If you think you are joking for the metric month - read your history of post revolutionary France.
They even had clocks with 10 hours on them.

I'll take:
"Things way beyond my comprehensive ability" for $200, Alex


Dan

Growing up with imperial measurements, I still struggle with metric. The only time I don't have to guess on temps is when it hits -40 ... thats the point where celsius and fahrenheit are equal.

Lake Girl said:

Growing up with imperial measurements, I still struggle with metric. The only time I don't have to guess on temps is when it hits -40 ... thats the point where celsius and fahrenheit are equal.

Click to expand...

That's why you live where you do!!

Monica in France said:

Thanks skibumm100 ,
I too felt that the figure of 1.65 m3/hr was too low ,
but thats what came out of a simulator.
On the other hand the factor of 35 between SI units and Cubic feet I think is right.( 39.25/12 ^^3 )
Don't forget we are talking of the OAK air flow , in the chimney the flows are much higher because of the temperature.

So just trying an off-the-cuff estimate :
Taking pellets at 4.7 KW/Kg ( net , after burning/evaporating off the humidity ).
Ideally for 3 KW/hr you are burning 3 / 4.7 = 0.65 Kg /Hr. ( ~ one of my sacks of 15Kg a day ).
Adding for losses we have : 0.80 Kg/hr.
The stoichiometric ratio for wood is not easy to find : but I found two sources giving the value of 6.4 ( but this needs to be re-checked)
This means we are using 0.80 x 6.4 = 5.12 kgs of air / hr.
Adding extra air we have : ~ 6 kgs/Hr.
This in volume (at sea level and 0°c) is 6 / 1.3 = 4.7 m3.
This would be about 3 cu.ft a minute.
A figure which is still intuitively on the low side.
So please poke holes in my logic so we can be sure of the right answer.


If you think you are joking for the metric month - read your history of post revolutionary France.
They even had clocks with 10 hours on them.

Click to expand...

OK Monica. Let me work on this. The make-up air flow still sounds low but I'll do some digging to see where it takes me. You are right that CFM in exhaust does not equal CFM in the OAK. This ratio will vary significantly because the heat transfer efficiency varies a lot between stoves and firing rates. I suspect that there is a pretty high fouling factor in most stoves under actual running conditions. The conversion factor of m^3/hr to CFM at 35 is correct. I cheated as I use both metric and imperial systems at work so I have ConvertPad on my phone. When I get to a conversion that's not on ConvertPad, I do it the old fashioned way....by hand.

Your stoich ratio number for wood is pretty close to what I have found on the net. It ranged from 4.58 for generic biomass to 5.75-6.3 for wood. My combustion book says 6.4 so I think that number is good. That's for air, not O2.

According to my Fossil Power Combustion book, "theoretical air" (A), or what's referred to as "stoich", for dry wood ranges from about 710-720 pounds of air per MMBTU of wood fuel. That does not include any excess air.

HHV of various species of wood, dry (in BTU/lb) of 8200-9800 BTU/lb.

Using pellets at 9000 BTU/lb you would need about 6.4 lb of air to combust 1 lb of pellets. At 30 deg F that would be about 79 ft^3 of air. Now add the excess air (15% excess O2) and you get 91 ft^3 per pound of pellets.

Just for comparison, the amount of air needed at 70 deg F (heated room air, not OAK) is 85 ft^3 w/o excess air and 98 ft^3 with 15% excess O2.

So that's just for 1 lb of pellets. Now multiply by the amount of pellets you burn per hour.

If we convert this to your example:

0.8 kg/hr = 1.76 lb/hr

1.76 X 91 ft^3 OAK air = 160 ft^3/hr

160 ft^3/hr = 2.7 CFM

Your math seems to check out. Surprising isn't it?

BTW, if it's cold, I burn twice that much. I also wouldn't be surprised if the efficiency isn't quite a bit lower than 80% in real life.

Regarding flue gas volume vs OAK air volume:

at 30 deg F (-1 deg C) air density is 0.081 lb/ft^3
at 70 deg F (21 deg C) air density is 0.075 lb/ft^3
at 300 deg F (149 deg C) air density is 0.052 lb/ft^3

......so on a mass flow basis this would make 2.7 ft^3/min equal to 0.22 lb/min at 30 deg F
and 0.22 lb/min at 300 deg F would be 4.23 Ft^3/min (CFM), so that's an increase of 56% in volume as the air heats up.

Now, something is rotten in Denmark. If most combustion fans are running at something like 20-30 CFM, per the experts, what does that tell us? Maybe there's a lot of tramp air coming in through the air wash for the glass? Maybe sneaking in somewhere else? It's hard to believe you could lose that much. I suspect most pellet stoves run at a much higher excess air number. For a well regulated industrial burner you would need 10-20% excess air to run efficiently. Most pellet stoves are pretty crude.

My apologies to the Danes for bringing them into this as this is pretty much a Franco-American thing.......

Anybody see any mistakes in my analysis?

Monica in France said:

Thanks skibumm100 ,
I too felt that the figure of 1.65 m3/hr was too low ,
but thats what came out of a simulator.
On the other hand the factor of 35 between SI units and Cubic feet I think is right.( 39.25/12 ^^3 )
Don't forget we are talking of the OAK air flow , in the chimney the flows are much higher because of the temperature.

So just trying an off-the-cuff estimate :
Taking pellets at 4.7 KW/Kg ( net , after burning/evaporating off the humidity ).
Ideally for 3 KW/hr you are burning 3 / 4.7 = 0.65 Kg /Hr. ( ~ one of my sacks of 15Kg a day ).
.

Click to expand...

3KW is a pretty small heat input. That's only 10,200 BTU/hr. Is this for our theoretical 10,000 ft^3 house?

My scientific calculations are a little more simple.
I put my hand over the end of my stove pipe, and feel very lucky
that I am not sucking in that much cold air into the house.
I figure it to be about a lot x 1.736/12.


Dan

kinsmanstoves said:

I have been known to stir the burn pot a little. Why did you install an OAK?

1.) Is it to cut down the clearance to a window or door? If there is an air wash in the door (Breckwell, Napoleon, US Stove) that allows room air into the unit or any other air inlet to the stove this will not work such as an unsealed ash pan or not a direct sealed connection to the combustion blower.

2.) Is it to let moisture into an appliance that has electrical connections and made of metal so it rusts faster?

3.) Building code. Cause the government knows what is best and I don't break the law (55 MPH means 55 MPH)?

4.) The manufacturer said so?

Please let me know if I forgot to list a response.


Eric

Click to expand...

I didn't...

all these scientific calculations are making my head spin

Consider that It fundamentally doesn't matter if the stove has a flame, is lit, or not.
There is stil a fan, pointed outside, pulling 1/2" H20 draft on the stove whether the stove is stone cold or at temp.
Stoichiometry has virtually no relevance.

Its basically having a bathroom exhaust fan running all day and night.

If one wants to nitpick, a 90-100cfm freerun combustion fan, in place, pulling a 0.45-0.55" H20 draft, installed, may net 50-55 cfm actual.

Still, a day's running is still many times the volume of a typical residence, any why people feel real-world improvements in comfort and efficiency.

TStark said:

Consider that It fundamentally doesn't matter if the stove has a flame, is lit, or not.
There is stil a fan, pointed outside, pulling 1/2" H20 draft on the stove whether the stove is stone cold or at temp.
Stoichiometry has virtually no relevance.

Its basically having a bathroom exhaust fan running all day and night.

If one wants to nitpick, a 90-100cfm freerun combustion fan, in place, pulling a 0.45-0.55" H20 draft, installed, may net 50-55 cfm actual.

Still, a day's running is still many times the volume of a typical residence, any why people feel real-world improvements in comfort and efficiency.

Click to expand...

That may be true for your stove but not everybody's. The combustion fan does not generally run flat out. My stove uses an air flow sensor and varies the speed of the fan so it is not a fixed speed fan. There is also substantial natural draft in some installations and not in others. So, yeah, stoichiometry does have something to do with it. This is more an exercise in trying to quantify the amount of make-up air required for combustion. Based on what the calculations say, it may not be as many CFM as you would think. It's not the same as a bathroom fan although it's affects are similar. The other question is how much of that air going up the flue is used in combustion vs. tramp air. The OAK will have virtually no affect on that.