Generic Boiler Output Formula?
Generic Boiler Output Formula?
(OP)
Hello,
I am looking for a little help. I was asked to find the exhaust flow of a boiler, but I am a little overwhelmed with where to start. I was wondering if there is a general formula to calculate what the exhaust flow would be? If there is not, what would I need to be able to calculate this?
Thanks for any help!
Chrissy
I am looking for a little help. I was asked to find the exhaust flow of a boiler, but I am a little overwhelmed with where to start. I was wondering if there is a general formula to calculate what the exhaust flow would be? If there is not, what would I need to be able to calculate this?
Thanks for any help!
Chrissy





RE: Generic Boiler Output Formula?
Takes 1000 BTU to produce 1 pound of steam.
Natural gas contains 1000 BTU/cu. ft,
boiler about 80% efficent, so requires 1.25 cu. ft NG/lb steam.
NG is mostly methane, CH4, so it burns as
CH4 + 2 O2 = CO2 + 2 H2O
Air is about 21 vol.% O2, and 3% excess air is often used. Thus, combustion air required is
2x (1/0.21) x 1.03 = 9.81 x NG volume.
Since combustion equation has 3 moles of gas on both LHS & RHS, for exhaust gas can use (NG + combustion air) volume. Only need to correct for temperature per gas law. Best to use the actual exhaust T, but if not available, use 400oF. The boiler mfr. recommends the stack ID, so from volumetric flow, you get the exhaust velocity.
RE: Generic Boiler Output Formula?
www.epa.gov/AIRMARKET/monitoring/flow/rule.pdf
RE: Generic Boiler Output Formula?
RE: Generic Boiler Output Formula?
You can take a look to the site http://www.cleaver-brooks.com
Good luck !
RE: Generic Boiler Output Formula?
Calculate the O2 and thus the air to combust 70 GPH of this to the stoichiometric amounts of CO2 and H2.
Then, subtract the O2 and add the CO2 and H2 to the initial air in calculating the volumetric exhaust flow.
RE: Generic Boiler Output Formula?
RE: Generic Boiler Output Formula?
RE: Generic Boiler Output Formula?
But, the composition above is a typical analysis for No. 6 Fuel Oil, of 12.6o API gravity [SG = 0.982]. Always preferable to use hard data if available. The heat content will vary with fuel density per Fig. 27-3 in Perry's, 7th edn. This automatically factors in MW and C/H ratio.
RE: Generic Boiler Output Formula?
RE: Generic Boiler Output Formula?
My Solution:
I took the percentages given in Table 9-11 and inserted them into eq 9-13 to get 35.54.
I then did the following:
70 gph x 150000 BTU/gal x (1/18230 BTU/lb) x 35.54 cu ft/ lb fuel x 1hr/60 min
I am not sure if this is right, but the units worked out...
By multipling and then dividing the heating value, did I end up leave something out? This is the only glaring flag that I can see. My main problem was trying to get the units in cu ft/min, I didn't think I could do a straight conversion since gallons are typically a liquid volume.
RE: Generic Boiler Output Formula?
I came out with a higher flow by the old-fashioned method:
For the combustion of 1 gallon of fuel oil of composition above:
From C: 7.144 lbs/12 = 0.5933 lb·mol C requires 0.5933 lb·mol O2 creating 0.5933 lb·mol CO2
From H: 0.8588lbs/2 = 0.4294 lb·mol H2 requires 0.2147 lb·mol O2 creating 0.4294 lb·mol H2O
Total O2 required is 0.8080 lb·mol, so N2 of air is 0.8080x (79/21) = 3.0396 lb·mol N2. Also, N2 of combustion air = N2 of exhaust.
Molar exhaust flow (CO2 + H2O + N2) = 4.0623 lb·mol per gallon fuel.
Assume ideal gases & vapor.
At 32oF (273.15 K) and 1 atm pressure, 1 lb·mol occupies 359.0 ft3
At final T = 400oF (477.59 K) and 1 atm, 1 lb·mol occupies 627.7 ft3
For 70 GPH, then 70 gal/hr x (4.0623 lb·mol/gal) x (627.7 ft3/lb·mol) = 178,493 ft3/hr
Dividing by 60 gives the exhaust flow as 2975 ft3/min at 400oF.
A small excess of air brings the actual exhaust volumetric flow to 3000 ft3/min.
An exhaust stack cross-sectional area of 2 ft2 gives a linear flow of 1500 ft/min = 25 ft/sec.
RE: Generic Boiler Output Formula?
Thank you for your help, I will try and work through the solution you provided and see if that will work for what I need.