Outside Air Kit - Oak - My research after investigating

[zrtmatos]

I replaced a pellet stove with 34K output and no OAK with a stove that had slightly more output, 38K but installed an OAK with it. Big difference in cutting down air leaking in through an adjacent room to the stove. This room has newer windows and the leaks of air found their way in without the OAK. I can attest to it working. I probably could have not changed stoves and just installed an OAK. Just do the OAK install in the first place.

 

[F4jock]

I second that F4jock, we're talking about intake air, at ambient, which would be the difference if you were pulling from inside vs. outside. Exhaust will be a much higher volume due to the heat. Everything was somewhat arbitrary in the air turnover calculation I laid out earlier, just food for thought, ball park figures to give a talking point and a better feel for how much of a difference we're talking about.

The argument still stands even in light of a residential structure turning it's air over every other hour without the negative pressure of an appliance on it - if you add that 100CFM pull on all those leaks you're increasing the turn-over rate by that much more.

Mass balance FTW.

 

[Lake Girl]

Trying to find the gas laws that apply to this situation to show increased efficiency of burn using cold air ...

http://www.chemteam.info/GasLaw/GasDensity.html
Air is a mixture of 21% oxygen gas and 79% nitrogen gas (neglect minor components and water vapor). What is the density of air at 30.0 °C and 1.00 atm?

Comment: For both solutions, we need the "molecular weight" of air:

MW(air) = (%O2 x MWO2) + (%N2 x MWN2)
(0.21 x 32) + (0.79 x 28) = 29 g/mol

Solution #1:

1) Use PV = nRT and assume 1.00 L:

(1.00 atm) (1.00 L) = (n) (0.08206) (303 K)
n = 0.0402185 mol (of air at 303 K)

2) Calculate grams of air:

0.0402185 mol times 29 g/mol = 1.17 g
3) Determine density:

1.17 g / 1.00 L = 1.17 g/L

Same equation but using -20C:
1x1=n (.08206) (253.15K)
n=.0478235

.0478235 x 29 g/mol = 1.3868 g
1.3868g/1.00L = 1.3868 g/L

Brain still hurts so someone else can take it from here...

 

[F4jock]

Trying to find the gas laws that apply to this situation to show increased efficiency of burn using cold air ...

http://www.chemteam.info/GasLaw/GasDensity.html
Air is a mixture of 21% oxygen gas and 79% nitrogen gas (neglect minor components and water vapor). What is the density of air at 30.0 °C and 1.00 atm?

Comment: For both solutions, we need the "molecular weight" of air:

MW(air) = (%O2 x MWO2) + (%N2 x MWN2)
(0.21 x 32) + (0.79 x 28) = 29 g/mol

Solution #1:

1) Use PV = nRT and assume 1.00 L:

(1.00 atm) (1.00 L) = (n) (0.08206) (303 K)
n = 0.0402185 mol (of air at 303 K)

2) Calculate grams of air:

0.0402185 mol times 29 g/mol = 1.17 g
3) Determine density:

1.17 g / 1.00 L = 1.17 g/L

Same equation but using -20C:
1x1=n (.08206) (253.15K)
n=.0478235

.0478235 x 29 g/mol = 1.3868 g
1.3868g/1.00L = 1.3868 g/L

Brain still hurts so someone else can take it from here...

And at colder temperature the molecular density also reduces water vapor entrainment thus allowing even more oxygen and other molecules BUT you're unfortunately wasting your time as there is a tendency to deny the physics of the situation and replace it with what is "felt" to be right.

 

[Zebby]

I was told the "gap in the OAK path" you describe was a safety decision in case the OAK became clogged. Many trailer home installs are done with the OAK "drilled & dropped" down thru the trailer floor boards without proper critter guard. Great place to store supplies & make a nest.

My OAK has a screen to prevent critters larger than 1/4 inch from entering. I'll have to cover the opening in warmer weather to prevent bees, spiders, etc. making homes though.

Safety decision?
But if the continuous OAK path were to become clogged, no/reduced air entering would make the owner seek a cause for poor stove performance.
The gap in the OAK would allow the stove to take in room air in violation of mobile home codes, and the stove would continue working.
Which is the more safe condition?

 

[Zebby]

You have to calculate based on intake at ambient because that is the volume the blower is moving. Output = input plus a bit because as well as blower CFM you have stack draft due to heat thus you have a slight negative pressure in the firebox / stack. Volume difference us due to expansion of the same amount of air. Just takes up more space.

Sorry if I'm being a PITA on this.....but my combustion blower is on the hot side of the firebox, so why would it's cfm (cubic feet per minute) rating not then apply to the volume of heated exhaust gas?
If it was mass airflow, then I'd see that (intake air mass + mass of carbon from the burning wood = exhaust mass).

I admit that I wasn't so good at compressible flow and I know you must be, especially having jet engine theory experience.

Thanks

 

[acammer]

Sorry if I'm being a PITA on this.....but my combustion blower is on the hot side of the firebox, so why would it's cfm (cubic feet per minute) rating not then apply to the volume of heated exhaust gas?
If it was mass airflow, then I'd see that (intake air mass + mass of carbon from the burning wood = exhaust mass).

I admit that I wasn't so good at compressible flow and I know you must be, especially having jet engine theory experience.

Thanks

The intake volume doesn't change as the intake air temperature doesn't change - the temperature and volume change is happening at the combustion event inside the stove, after the intake. That's why the exhaust volume is higher than the intake volume.

 

[F4jock]

The intake volume doesn't change as the intake air temperature doesn't change - the temperature and volume change is happening at the combustion event inside the stove, after the intake. That's why the exhaust volume is higher than the intake volume.

Thank you!

 

[bags]

Also IF your OAK was completely clogged meaning zero air I'd say you would be experiencing a poor burn scenario. They should be sealed as to not leak any cold air between the exterior intake and delivery to the burn. Not always the case and I am sure some installs would allow air to be sucked in regardless. Mine is sealed with very secure hose clamps and high temp silicone.

PITA if I need to pull it apart? No. A quick simple roll of a razor knife cuts silicone really quick and easy in a few seconds. I am all about NOT having leaky air jacking my program up. I found an adjustable door sill that had a nice long gap this fall. Couldn't see it at all from anywhere until I laid on the floor doing some caulking and was pissed when I spotted it. I was wondering why I felt a cold draft and some itsy bitsy bugs seemed to be found inside by this door. DUH! High use door that's 4 years old so I assume non stop stepping on the threshold or sill readjusted it over time.

 

[F4jock]

Sorry if I'm being a PITA on this.....but my combustion blower is on the hot side of the firebox, so why would it's cfm (cubic feet per minute) rating not then apply to the volume of heated exhaust gas?
If it was mass airflow, then I'd see that (intake air mass + mass of carbon from the burning wood = exhaust mass).

I admit that I wasn't so good at compressible flow and I know you must be, especially having jet engine theory experience.

Thanks

Please see acammer's reply above.

 

[F4jock]

My OAK has a screen to prevent critters larger than 1/4 inch from entering. I'll have to cover the opening in warmer weather to prevent bees, spiders, etc. making homes though.

Safety decision?
But if the continuous OAK path were to become clogged, no/reduced air entering would make the owner seek a cause for poor stove performance.
The gap in the OAK would allow the stove to take in room air in violation of mobile home codes, and the stove would continue working.
Which is the more safe condition?

If you want to screen your intake more finely you can upsize the intake (Easy calculation if you know combustion blower capacity.) to be sure that you can pass enough air with a given percent of the size mesh you choose clogged, reduce it to the OAK diameter and inspect the screen daily. Best of all worlds.

 

[woodmakesheat]

Curious about posters using the combustion blower cfm rating as the standard to calculate air exchange in a heated space.
Assuming the stove is running, the exhaust consists of products of combustion, excess air and a small amount of unburnables (ash).
All of this except the ash is expanded to a larger volume by being at a much higher temperature than the ~70 degree F indoor air used without an OAK.
I assume the air actually taken from the source to be of a smaller volume than that which the combustion blower sends out, therefore a calculation of air exchange via a hot stove should be based on intake air volume.

I know it is still a substantial amount being drawn into the firebox, and therefore taken out of the heated space, just not quite as much as the combustion fan cfm would suggest.
Anybody got thoughts on that? Is the difference between firebox intake volume and exhaust volume considered not significant? Or are people just using combustion blower cfm to simplify the discussion?

Using 20 degree cold air and 400 degree hot air, the latter seems to be greater in volume by 86% by Charles' law. V2/V1 = T2/T1 in degrees K at constant pressure. A pellet stove is not as simple as expansion of plain air, admittedly.

I think this is a good question, so I decided to measure my stove's actual input CFM. I did a more accurate measurement of my CPM-10 running full tilt (9/9) which is about 36k BTU/hr input. This time I removed the OAK tube from the back of the stove so I could measure the airflow entering the stove @ a standard temperature (70F) and through an easy to measure rigid pipe.

Input Airflow: 820 LFM
Pipe diameter: 1.75", Cross-sectional Area = 0.0733 sq ft.

Inlet Airflow = 820 ft/min * 0.0733 ft^2 = 60.1 CFM

This is for a stove that is on the low end of BTU output.

Assuming proportional airflow, a large stove like the P61A input probably flows about 100 CFM running full tilt.

You could assume that a typical stove on a cold day is drawing about 60-80 CFM.

Agree/Disagree?

 

[709GADE]

If you have an HRV in your home, which removes stale warm air and returns fresh partially heated air, do you think an OAK is needed? I have a rather new home that has an HRV and this is my first season with a pellet stove. I had the company that I bought the stove from install it and the technician said an OAK is not needed in my house. Do you agree?

 

[woodmakesheat]

If you have an HRV in your home, which removes stale warm air and returns fresh partially heated air, do you think an OAK is needed? I have a rather new home that has an HRV and this is my first season with a pellet stove. I had the company that I bought the stove from install it and the technician said an OAK is not needed in my house. Do you agree?

In my opinion, an OAK is never required unless the manufacturer says it is. Oh the other hand, an OAK is always preferable.

 

[acammer]

In my opinion, an OAK is never required unless the manufacturer says it is. Oh the other hand, an OAK is always preferable.

Agreed. I also think your CFM calculation looks right.

 

[Zebby]

I think this is a good question, so I decided to measure my stove's actual input CFM. I did a more accurate measurement of my CPM-10 running full tilt (9/9) which is about 36k BTU/hr input. This time I removed the OAK tube from the back of the stove so I could measure the airflow entering the stove @ a standard temperature (70F) and through an easy to measure rigid pipe.
Input Airflow: 820 LFM
Pipe diameter: 1.75", Cross-sectional Area = 0.0733 sq ft.
Inlet Airflow = 820 ft/min * 0.0733 ft^2 = 60.1 CFM
This is for a stove that is on the low end of BTU output.
Assuming proportional airflow, a large stove like the P61A input probably flows about 100 CFM running full tilt.
You could assume that a typical stove on a cold day is drawing about 60-80 CFM.
Agree/Disagree?

Sounds good.
My initial assertion was that air volume being drawn out of the house with no OAK (and thus replaced by drafty inflow) should not be 1 to 1 with combustion blower rated cfm, as some comments have suggested.
It's a trivial issue, I admit, but the devil's in the details. I'm one of those kind of guys.
If you'd be so inclined, and when you have a little time, could you take volume measurements of both inlet air @ 70 and vent gasses?

Thanks.

p.s. I just found , , and in with the smilies.

 

[F4jock]

Sounds good.
My initial assertion was that air volume being drawn out of the house with no OAK (and thus replaced by drafty inflow) should not be 1 to 1 with combustion blower rated cfm, as some comments have suggested.
It's a trivial issue, I admit, but the devil's in the details. I'm one of those kind of guys.
If you'd be so inclined, and when you have a little time, could you take volume measurements of both inlet air @ 70 and vent gasses?

Thanks.

p.s. I just found , , and in with the smilies.

You're assertion is essentially incorrect. Now it's up to you to find out why. I suggest your local high school physics dept.

 

[Zebby]

You're assertion is essentially incorrect. Now it's up to you to find out why. I suggest your local high school physics dept.

Ok, I'll look into it some more.
Thanks

 

[F4jock]

Ok, I'll look into it some more.
Thanks

Try asking about this term for starters: Mass Balance

 

[woodmakesheat]

Sounds good.
My initial assertion was that air volume being drawn out of the house with no OAK (and thus replaced by drafty inflow) should not be 1 to 1 with combustion blower rated cfm, as some comments have suggested.
It's a trivial issue, I admit, but the devil's in the details. I'm one of those kind of guys.
If you'd be so inclined, and when you have a little time, could you take volume measurements of both inlet air @ 70 and vent gasses?

Thanks.

p.s. I just found , , and in with the smilies.

I am not sure my anemometer can accurately measure the flow of the hot gasses since I don't have a way to cleanly insert the probe into the exhaust. The temperature of the probe in the exhaust stream makes a difference in getting an accurate number. Also not sure the probe would be happy in those temperatures and contamination levels. I would agree that the nameplate rating of blower is not an accurate method of determining the actual inlet flow of a stove - it's a ballpark method one uses if no other information is available (altitude, temperature, pressure drops in system, contribution from combustion products) Having measured the actual input flow to my stove (under standard conditions!) though, we now know actual airflow + any leakage (which is relatively low in my new stove) into the stove from the house, which is what we care about in this thread. Also as stated, air in = air out.

 

[Lake Girl]

I think this is a good question, so I decided to measure my stove's actual input CFM. I did a more accurate measurement of my CPM-10 running full tilt (9/9) which is about 36k BTU/hr input. This time I removed the OAK tube from the back of the stove so I could measure the airflow entering the stove @ a standard temperature (70F) and through an easy to measure rigid pipe.

Input Airflow: 820 LFM
Pipe diameter: 1.75", Cross-sectional Area = 0.0733 sq ft.

Inlet Airflow = 820 ft/min * 0.0733 ft^2 = 60.1 CFM

This is for a stove that is on the low end of BTU output.

Assuming proportional airflow, a large stove like the P61A input probably flows about 100 CFM running full tilt.

You could assume that a typical stove on a cold day is drawing about 60-80 CFM.

Agree/Disagree?

Does your meter tolerate colder conditions? Would be interesting to see what the measurement would be through outside inlet...

 

[woodmakesheat]

Does your meter tolerate colder conditions? Would be interesting to see what the measurement would be through outside inlet...

It can, but it needs a correction factor based on the temperature. Also I realized that I don't seal the inlet tube to the pipe on the stove. The pipe on the stove sticks out like 8 inches and I have the hose slid all the way over that and butted against the back wall of the stove so I'm not worried greatly about air seeping (also stove runs 24/7), but it could affect the flow readings a little. I'm sure there'd be a pressure drop due to the hose and screen, there are mathematical ways to estimate this that would be close enough IMO.

 

[daffonce]

Look you need an oak so the flames dont suck all the oxygen out of the room causing you to suffocate i have thrown all physics out the window now too.

 

[bags]

Look you need an oak so the flames dont suck all the oxygen out of the room causing you to suffocate i have thrown all physics out the window now too.

There is no way anyone will suffocate with nor without the use of an OAK. Homes are not that tight to begin with. That's like saying you have to leave a window or door cracked year round so you get your needed oxygen. Your statement is confusing to me. What is it you are trying to say? The point you are making is?

 

[St_Earl]

i detect sarcasm.