thermal efficiency

[zekeman]

I was wondering how the tech who measures my "efficiency" comes up with it based only on stack temp and CO2. I made the calculation based on complete combustion and X excess air (absenting sulphur and nitrous oxides), as follows. From data I obtained, #2 oil is approximately 85% carbon and 12 % hydrogen, I wrote and balanced the chemical formula (air is 3.76/1 nitrogen to oxygen) as follows
C+.818H2+1.409O2+1.409*3.76N2 yields CO2+.818H2O+1.409*3.76N2+1.409(4.76)X
Now if the fraction of CO2 measured is Z, I get
Z=1/(1.818+1.409*3.76+1.409*4.76X =1/(7.115+6.706X)
Now the tech measured my CO2=0.1, so from this I get the excess air,
X=0.43
It also looks like I get 13% co2 for X=0, the stoichiometric value, which looks low compared with data I have seen. To get the efficiency , I first get the stoichiometric ratio lb air/ lb fuel =1.409*(32+3.76*28)/(12+2*.818)=14.18
The stack temp was measured at 500 deg, ambient 65 so corrected stack was 435 deg The heat loss for lb fuel is then
(15.18+.43*14.18)*.25*435=2313 BTU
Assuming 7.5 lb /gal this becomes
2313*7.5=17,354 BTU/gal
If the heating value of oil is 140,000BTU/gal, my calculated efficiency is
1-17,354/140,000= 87%
The tech said it was 78.75%, a large discrepancy.
Where have I gone wrong?

 

[AbbyNormal]

you got the enthalpy of 'superheated steam' going up the stack?

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[DrRTU]

Stack temp and O2 would be more accurate than CO2. This is a good site on efficiency. testing.

http://www.bacharach-training.com/Combustion Zone/Comb/O2CO2/advantages_of_measuring_o2_vs_co.htm

 

[AbbyNormal]

Probably all that measures accurately CO2 these days will be a fyrite bottle or some other orsat.

The electronic analyzers just estimate the CO2.

They have the full efficiency calculation procedures in the ASHRAE HOF.

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

The tech most likely has a chart that works the percentage CO2 and a net stack temperature.

The chart has it worked out then based on the typical composition of the fuel.

The only reason measuring O2 is anymore accurate than just measuring CO2, is you can have the same CO2 reading with less than enough air for complete combustion as well as when you have excess air.

To be sure you need the O2, so Bacharach et al will keep costs down and just make one with an O2, CO and perhaps NOx sensor. Cost more to acutally have the CO2 sensor as well.



Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[Bubbleheadski]

Wow... Stoichyometry? Some people have way to much time on their hands! LOL
I loved balancing equtions in chemistry clas and advanced chemistry way way back in High School.
But to see fractions instead of whole numbers just blew my mind. Do engineers really sit back and do all these calculations?

Bubble "overwhelmed" head

 

[zekeman]

Thanks to all od the responses.
Abbey,
The steam enthalpy is taken care of in the heat equation where 0.25 is used for the average specific heat; steam is about 0.5 but only o few % of the exhaust is steam
To Bubbleheadski,
Yes and yes
I just like to know how those socalled charts are developed and the accuracy of these measurements. I have come away very skeptical of oil companies sending out techs with poor equipment to make these "efficiency" calculations which is sometimes the basis of a HO decision to scrap an existing furnace for a new one. Also, I have another bone to pick with the industry. First, they call this "efficiency", combustion efficiency which should NOT include stack temperature; the way they do it,that efficiency excludes radiation and leakage losses. The true measure of efficiency is the overall efficiency meaning the amount of heat that goes into your supply registers divided by the BTU heat of combustion of the oil flowing into the burner which would mean getting accurate readings of both oil flow and air flow in the ducts and delta temperatures accross the heat exchanger. But that would be too much to ask on a tuneup, so they give you that half assed "efficiency" that is meaningless even with their fancy Bachrach equipment and thet know it.
And yes you dont have to deal with whole numbers in balancing chemical equations even though they teach you that way in high school and it is true that you can't have a fraction of o molecule react with a fraction of another molecule.

 

[AbbyNormal]

I guess I missed the enthalpy part, the latent heat

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

as compared to being 100% with water leaving a condensate drain at the combustion air temp

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[DickRussell]

Zekeman, your final calculation is somewhat dependent on the value used for heat of combustion. I suspect your 140,000 BTU/lb is a gross heat of combustion, because the net heat for a typical no. 2 fuel oil is more like 125,000 to 130,000 I believe. Also, your density (7.5 lb/gal) seems high; that would give a specific gravity of 0.90, vs. one source I saw saying east coast no. 2 oil is more like 0.86. Finally, Even with revised properties, your efficiency comes to 86-87%, considerably above what the tech guy got. To refine the calcs more you would have to put your equation into a simulator and have it tweak air to give the measured CO2, then compare heat removed to 77 F vs. to the actual stack temperature to get efficiency. That still would be dependent on the C/H ratio of the actual fuel. Hardly worth it, unless you really do have too much time on your hands.

As a rough rule of thumb, for approximating efficiency vs. stack temperature of a fired heater, I have used 92% less 3% for every 100 F over 280 F. Your 500 F stack temp would give 85.4% using that approximation, not all that far from your calculation. If the tech guy really did used his chart correctly and came up with 78%, then I share your skepticism. I wonder if your "%" and the tech guy's "%" mean the same thing. Maybe someone who routinely uses such charts can post something.

 

[AbbyNormal]

A "higher heating value" of #2 is somewhere around the 140,000 per gallon, but a "higher heating value" includes the latent heat of water made as a product of combustion.

Lower heating values to not include this latent heat. You seem to be only dealing with the specific heat of the steam then.

You burn a pound of methane and you make 2.25 pounds of water, so there is a couple thousand Btu in latent heat right there.

Be some heat locked away in water vapour when you burn a gallon of oil too.



Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[zekeman]

Abbey , Dick and others
You are probably quite right about my error in not including heat of vaporization of H2O. If I include it, I get about 1lb of H2O for each lb of fuel; so my calc that shows a loss of 2313 BTU/lb should be
2313+1000=3313 BTU/lb
where at low pressure the heat of vaporization is roughly 1000 BTU/lb and my overall efficiency would be
1-.13*3313/2313=0.814, closer to the tech's answer of 78.75% I suppose.

Dick quote "...... unless you really do have too much time on your hands."
Apparently,yes, otherwise I would be out somewhere making an honest living.
Also, I got my density and ultimate analysis of "furnace oil" from the Handbook of Chemistry and Physics, 23rd edition, 1939 which I inherited somewhere in my travels. Do you think it has changed substantially, though oil prices have gone from $.05 to $2.50/gal.
As a footnote, I noticed that the hancbook sold new for $4.00 in 1939, probably worth $200 today, in about the same ratio as fuel oil.
So much for inflation which is another forum.

 

[quark]

Good advice by AbbyNormal and DickRussel. The huge difference seems to be from the latent heat of vapor. Further, you should check whether you and your technician are using the same calorific value (either GCV or NCV).

For Furnace Oil (No.6 or Bunker C), your calorific value seems to be high as pointed out by DickRussel. The NCV is 9500 kcal/kg. The density value is correct.

Further, 1 lb of water vapor for 1 lb of fuel is too high. The moisture content of furnace oil is 1%v/v max. If the combustion air is at 65F and 75% RH, the water vapor content is 0.0099lb/lb. So you will get maximum 0.1496lb of vapor per lb of fuel burned with air to fuel ratio of 14:1 (0.0099*14 + 0.133gal*0.01*3.785lit/gal*1kg/lit*2.2lb/kg)

Secondly, check both of you didn't consider CO.

You can use the following equations to estimate NCV of fuel oils.

NCV (MJ/kg) = 51.5 - [5.95/(1.53-sg)] where sg is specific gravity of fuel

NCV (kcal/kg) = 9380 + 26x[sup]0[/sup]API

 

[zekeman]

Quote from Quark:
"Further, 1 lb of water vapor for 1 lb of fuel is too high. The moisture content of furnace oil is 1%v/v max."
Two things. First I get my 1lb/1lb ratio by looking at my combustion equation which shows 1mole of carbon +.818 moles of hydrogen yielding .818 moles of H2O. Then the weight ratio of water product to fuel is simply
(.818(2+16))/(1*12+.818*2)= 1.079
Second the moisture in air is already vaporized so you DO NOT add its heat of vaporization to the equation.
Finally my pet peave. I can't understand why the British who gave us those lousy ft/lb/gal units abandoned them and left us stuck with them so we Americans (as well as you) have to constantly do these conversions so we can talk the same talk.

 

[AbbyNormal]

Leave the brits alone, that was an American gallon you were deailing with

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[AbbyNormal]

http://i32.photobucket.com/albums/d41/a_bee_normal/Bacharach_oil.jpg

Take the "V" out of HVAC and you are left with a HAC(k) job.