Boiler theory question: BTUs lost to replacement combustion air?
Comments

Infiltration is the number you are looking for. Air changes per hour is one of the numbers you need to know. We guess at that number when doing a heat loss calculation. The only real way to know (that I'm aware of) is to have a blower door test on your home. And how accurate that test is depends on the technician doing the test.
Once you know the actual air changes per hour, you can then do the math on that number based on outdoor temperature to indoor temperature. You don't want to limit yourself to the actual combustion air needed to burn the fuel. You also want to account for dilution air, the air going up the draft diverter, and also any exhaust fans in the home, like in the bathrooms and kitchen. By limiting yourself the only combustion air you will miss some of that needed heat.
I guess what I'm saying is that your though process has merit, but the number you are looking for is already accounted for in the load calculation.Edward F Young. Retired HVAC ContractorSpecialized in Residential Oil Burner and Hydronics2 
Not that hard, but does require some arithmetic. First, you need to figure the volume of air involved, including any air entrained in a draught hood or barometric damper. Then you convert that to the mass of air, and it is safe enough to assume that you are working in standard conditions, so you multiply the volume of air in cubic meters by 1.225 to get the mass of the air in kilograms. Then the heat required is 1 kiloJoule per kilogram/degree Kelvin. Conveniently, 1 kiloJoule is almost the same as 1 BTU. 1 degree Kelvin is 1.8 degrees Fahrenheit. 1 cubic meter is, near enough 1,000 cubic feet.
It takes roughly  10 cubic feet of air to burn a cubic foot of natural gas.
So  very roughly  1 therm of gas (1 cubic foot, more or less) will take 10 cubic feet of air. 10 cubic feet of air is, roughly 0.01 cubic meters, and so the weight of that air is around
0.012 kilgrams. So if we suppose that the outside air is coming in at perhaps 30, and is displacing air which is at 70, that's about 20 degrees K, more or less. So 0.012 kilograms of air time 20 degrees K will give you around 0.24 kiloJoules of energy, or about 0.24 BTU per therm of gas.Br. Jamie, osb
Building superintendent/caretaker, 7200 sq. ft. historic house museum with dependencies in New England0 
But the boiler burning conditioned air must increase infiltration (and thus associated heat loss) by some amount, just like turning on an exhaust fan. In alternative scheme where combustion air was directly piped in from outside, presumably infiltration and thus heat loss would be lower, right?0

It does, of course. The increased infiltration due to combustion is, as I noted, 10 cubic feet per curbic foot of gas. In the example I gave above, I assumed that the air infiltrating was at 30, and the conditioned air was at 70. Your numbers may differ.fentonc said:But the boiler burning conditioned air must increase infiltration (and thus associated heat loss) by some amount, just like turning on an exhaust fan. In alternative scheme where combustion air was directly piped in from outside, presumably infiltration and thus heat loss would be lower, right?
And you are correct, using outside air directly from outside will eliminate that loss. However, I'm not at all sure that eliminating that loss  it amounts to one tenth of one percent efficiency  is worth much bother.
Infiltration into the building is far more affected by any wind, and even in no wind conditions any air leaks from inside to outside.Br. Jamie, osb
Building superintendent/caretaker, 7200 sq. ft. historic house museum with dependencies in New England0 
@Jamie Hall  that's what I was going for, although I think some of your math is a little off though.
1 therm = ~100 cubic feet (not 1 cubic foot)
(100 cubic feet of NG) x (10 cubic feet air / cubicftNG) = 1000 cubic feet of air needed
1000 cubic feet of air = 28.3 cubic meters
28.3 m^3 x 1.225 kg/m^3 = 34.6675 kg
If we need 1 kJ/kg per degree K, now we need 34.67 kJ/degree K x (1 degree K / 1.8 degree F)
So we now have 19.26 BTU/degreeF per therm.
At my design temp of 12F, the worstcase should be: (7012) x 19.26 = 1,117 BTU/therm
So 1 therm with a nominal energy of 100,000 BTU x (0.83 efficiency) = 83,000 BTU delivered to the house, and then we're throwing out up to 1,117 / 83,000 = 1.35% by increasing infiltration.1 
I went pretty far down this wormhole a few years ago. When I considered the reduced efficiency of the flame when using cold CA and compared that to the cost of bringing the CA up to room temp, it all came pretty close to washing out. Sorry I did not save the spreadsheet..."If you can't explain it simply, you don't understand it well enough"
Albert Einstein1 
The way I figured it, the CA if pulled from space is really coming from the outdoors and the cost to bring that air to room temp needs to be considered."If you can't explain it simply, you don't understand it well enough"
Albert Einstein0 
I would think you wouldnt want the combustion air sliding in all over the structure and cooling all the rooms down. It would be better to have it dumping directly into the boiler/util room, no ? That way it could warm up a little before being sucking into the burner. But of course, this cools down pipes etc in the util room.
Gotta admit, Im a little concerned about air changes per hour in the building, once we go to sealed/direct venting. Win one battle, lost another.
30+ yrs in telecom outside plant.
Currently in building maintenance.2 
fentonc said:@Jamie Hall  that's what I was going for, although I think some of your math is a little off though. 1 therm = ~100 cubic feet (not 1 cubic foot) (100 cubic feet of NG) x (10 cubic feet air / cubicftNG) = 1000 cubic feet of air needed 1000 cubic feet of air = 28.3 cubic meters 28.3 m^3 x 1.225 kg/m^3 = 34.6675 kg If we need 1 kJ/kg per degree K, now we need 34.67 kJ/degree K x (1 degree K / 1.8 degree F) So we now have 19.26 BTU/degreeF per therm. At my design temp of 12F, the worstcase should be: (7012) x 19.26 = 1,117 BTU/therm So 1 therm with a nominal energy of 100,000 BTU x (0.83 efficiency) = 83,000 BTU delivered to the house, and then we're throwing out up to 1,117 / 83,000 = 1.35% by increasing infiltration.To learn more about this professional, click here to visit their ad in Find A Contractor.0

My numbers were back of the envolope. Sorry. The error was, of course, in the conversion from cubic feet to cubic meters. Still, it is a small fraction.
@Dave Carpentier has a very good point  and an excellent comment in the last line. With echoes from my experience with the furor the Arab oil embargo some years back (most of you are, I think, too young to remember that!). There was a terrific effort at sealing up draughts in some of the most unlikely places, and in some cases (such as schools) reducing outside air replacement to very low levels. The end result was, quite predictably, truly horrible air quality  much worse indoors in some buildings than outdoors. Even now many buildings  particularly retrofitted and upgraded residences  are below any recommendation. Most sources will quote 4 to 6 air changes per hour for residences  other occupancies vary considerably. In occupancies which have ducted air handling, the impact on heating can be substantially reduced with heat recovery ventilators, of course (I, at least, do not recommend energy recovery ventilators which also recover latent heat, as they do not reduce most indoor air pollutants, which is the objective of the exercise after all). Too tight a building has quite an interesting array of adverse health effects.Br. Jamie, osb
Building superintendent/caretaker, 7200 sq. ft. historic house museum with dependencies in New England0 
And we need to know this because it will make a difference where?
Remember the words of Gil Carlson. "For a difference to be a difference it needs to make a difference."
I have always understood the need of a load calculation. But when I measure a room that is 14'7.5" x 12'3.1" and wright down 15 x 12 for that room size on the calculation, (Which is what we all do) how accurate do you need to be? The boilers and furnaces don't come in custom sizes. There are thousands of BTU/h difference between each size. You are trying to get as close as you can without going under. Just the opposite of The Price Is Right!
Just thinking of how much time I wasted watching Paint Dry today.
Tomorrow we will watch the grass grow!
Then Sunday, we rest!Edward F Young. Retired HVAC ContractorSpecialized in Residential Oil Burner and Hydronics1 
I had a blower door test done a few years ago (part of a rebate program). I was surprised that the tech didnt want to cover the barometric damper on my boiler vent.EdTheHeaterMan said:The only real way to know (that I'm aware of) is to have a blower door test on your home. And how accurate that test is depends on the technician doing the test.
It wasnt sealed for either test (before my house envelope sealing efforts and after), so at least it's a consistent reading. Im just not sure I can use the number at all, for any kind of reference, once I go to a direct vent boiler.
30+ yrs in telecom outside plant.
Currently in building maintenance.0 
When I do my energy studies, I estimate between 12 and 15 parts of air for every part of gas depending on the burner. That would equal 2050% excess air. If the boiler is a million btuh it would consume 1,000 cubic feet of gas per hour at high fire. That would be 16.6 cubic feet per minute which would make the air 199 to 249 CFM. I then use a 50% load factor figuring the boiler will run roughly 50% of the time during the heating season. You can calculate the sensible heat by using this formula Hs = 1.08 x CFM x Delta T. The off time calculations are a bit trickier due to the chimney effect. Hope this helps
RayRay Wohlfarth
Boiler Lessons2 
I was just interested in understanding this aspect of my boiler's operation better  it sounds like the worstcase for my location would be between 10002000 BTU/therm (at 12F), so around 12% or so. Thanks for the explanations!0


Typical atmospheric boilers have much higher excess air than that.RayWohlfarth said:When I do my energy studies, I estimate between 12 and 15 parts of air for every part of gas depending on the burner. That would equal 2050% excess air. If the boiler is a million btuh it would consume 1,000 cubic feet of gas per hour at high fire. That would be 16.6 cubic feet per minute which would make the air 199 to 249 CFM. I then use a 50% load factor figuring the boiler will run roughly 50% of the time during the heating season. You can calculate the sensible heat by using this formula Hs = 1.08 x CFM x Delta T. The off time calculations are a bit trickier due to the chimney effect. Hope this helps
Ray
To learn more about this professional, click here to visit their ad in Find A Contractor.0 
What I find interesting is that nearly all the older cast iron burner equipped atmospherics I've tested have excess air around 30% ( after a good heat exchanger cleaning even) All the newer "more efficient" atmospherics have about 60% excess air. So much for real efficiency gains with newer equipment.To learn more about this professional, click here to visit their ad in Find A Contractor.0

the higher efficiency is from the latent heat extracted from the flue gas, not by minimizing the excess air to limit high stack temperatures as with atmospheric equipment. that is an excessive number that can be brought down to the 30% range. whether its high efficiency or atmospheric your combustion numbers are same. combustion is combustion0
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