Generic Boiler Output Formula? 1

[cfj104]

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

 

[kenvlach]

Approximation:
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 400^oF. The boiler mfr. recommends the stack ID, so from volumetric flow, you get the exhaust velocity.

 

[metengr]

You did not mention the size or type of fuel for your boiler. If you want to calculate the flue gas velocity for a fossil-fueled boiler, the EPA provides and describes 3 test methods. Not very enjoyable reading, but if you wade thru the legal BS, there is some good technical information;

http://www.epa.gov/AIRMARKET/monitoring/flow/rule.pdf

 

[cfj104]

The only information that I was given is 70 gph of #6 fuel oil is fired into a 250 HP boiler (I am assuming that is BHP).

 

[21121956]

Hi cfj104:

You can take a look to the site

http://www.cleaver-brooks.com

Good luck !

 

[kenvlach]

From Perry's Chemical Engineers' Handbook, 7th Edn., pages 27-9 & 27-10, Low sulfur No. 6 F.O. contains 87.26 wt% C and 10.49 wt% H, and weighs about 7.144 pounds/gallon. (I've neglected other components).
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.

 

[kenvlach]

Typo above: The F.O. (presuming low sulfur) weighs about 8.187 pounds/gallon & thus contains 7.144 pounds carbon/gallon.

 

[cfj104]

Do I need to know the molecular weight?

 

[kenvlach]

Not if you only need to find the combustion air/flue gas when firing a fuel of known C & H content at a constant 70 GPH.

But, the composition above is a typical analysis for No. 6 Fuel Oil, of 12.6^o 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.

 

[cfj104]

Thank you for all your help! I found a copy of Perry's Chemical Handbook in the office so everything is much clearer now!

 

[cfj104]

Does 1,600 cu ft/min sound realistic?? I am not familiar with what type of results to expect, just wondering if I was in the ballpark.

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.

 

[kenvlach]

Looks as if you took the fuel's heat content & divided by the heat of formation of CO₂ [(Eq. 9-13) in Perry's 6th Edn.] to get gas volume?

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 32^oF (273.15 K) and 1 atm pressure, 1 lb·mol occupies 359.0 ft³
At final T = 400^oF (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 400^oF.

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.

 

[cfj104]

Hmm, I only have the 5th edition Of Perry's Handbook. I will have to try and find a later addition since this one only goes to chapter 25.

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.