6 replies to this topic

[balakfs123]

Hello

I am trying to calculate the stoichiometric air requirement for coal combustion.

 

I have a small question regarding combustion of coal. 

 

Ultimate analysis for Coal as fired on wt basis%

C                     73%       

H                     5.4

O                     10.5%

N                     1.5%

S                     0.6%

Ash                 8%

Moisture         1%

 

Stochiometric reactions

C    +   O2 ------>CO2

12kgc   32kgO2      44kg CO2

1kg C  need 2.67kg O2

73kg c need 194.91 kgO2

 

2H  +  1/2O2------> H2O

2 kg H   16kg O2          18 kg H2O

1kg H needs 8kg O2

5.4 kg H need 43.2kgO2

 

S    +   O2  -------> SO2

32kg S + 32 kg O2     64kg SO2

1kg S needs 1 kgs O2

0.6 kg S needs 0.6Kg O2

 

Total stoichiometric O2 required by per kg of coal = 11.67kg

Total actual O2 = Theoretical value - O2 already in coal

                         =  (238.71-10.5)/100

 

                         =   2.281 kg of O2 needed per kg of coal

 

N from coal 1.5% = no reactions it comes out as inert and will be added to total air requirement calculation.

 

Questions:

1. From the ultimate analysis of coal C, H ,N,O if i take reaction 2 (2H+0.5O2--->H2O) do I need to consider it as H2 or H? If I use H or H2 what will be the O2 requirement, the difference will be half the value?

 

 

2. For O, do i need to consider O2 or O in the calculation?

If i take that as O2 = what should i use in the below formula for "O=10,5%". what will be the value to deduct in the below formulae?

      Total actual O2 = Theoretical value - O2 already in coal

                         =  (238.71-10.5)/100

3. similarly what should I do with N2 ? is it only N or N2? what will be the value for N2 for a value of  N=1.5% -coal ultimate analysis?

 

 

 

Please let me know what is the agreed procedure for this calculation. I hope It is solvable from experienced person from industry. Different books use different way to calculate(for H2/H) so I would like to get it correct.

 

Thanks

 

[PingPong]

There is no H₂ or O₂ or N₂ present in coal. So use H, O, N so as not to look foolish.

(It makes no difference though, because the burning of 1 kg of H and then burning of 1 kg of H₂ both require the same kgs of O₂)

 

 

Burning of Coal can better written as: C_xH_yO_zN_pS_q + ....... O₂ --> ...................

 

in which x, y, z, p and q you can determine from the given composition.

Edited by PingPong, 12 January 2015 - 11:35 AM.

[balakfs123]

Thanks for the clarity. sorry I assumed few things in a different way.

IS there any changes required for the following calculations. If I know  O as 10.5% in the Coal

Total actual O2 = Theoretical O2 required - Oxygen already in coal

                         =  (238.71-10.5)/100

Do I need to change this 10.5% as different value if it is only O not O2?

[PingPong]

That 10.5 % is WEIGHT %

 

1 kg O is equivalent to 1 kg O₂

[balakfs123]

Thanks for your answers.

[P.K.Rao]

1. Whether it is H or H2, stoichiometrically, 2 gm of hydrogen reacts 16 gm of oxygen.
2. For all reactions, go on weight percent basis after considering equation. For example,

H2 + O =  H2O

2  gm of hydrogen reacts with 16 gm of oxygen

Or, 1 gm of hydrogen reacts with 8 gm of oxygen

So, 5.4 gm of hydrogen (present in 100 gm of coal) reacts with 5.4 x 8 = 43.2 gm of oxygen

 

C + O2 = CO2

12 gm of carbon reacts with 32 gm oxygen

So, 73 gm of carbon will react with 32 x 73/12 = 194.7 gm of oxygen

And so on

Then combine all oxygen required and subtract oxygen present in coal. This would be oxygen required by 100 gm of coal for stoitiometric combustion. Convert the result obtained to air equivalent.

Add 20% more to the requirement (excess air) for complete combustion. Otherwise part of carbon will be converted to carbon monoxide instead of carbon dioxide. (There by some energy would be lost in the gases due to incomplete combustion)

1. Nitrogen present in the coal is not present as free nitrogen. It is present as organic nitrogen compound. So it will not be present as free nitrogen in the combustion products also. It will be present as NO2. Please take Nitrogen dioxide also as a reaction product.

Hydrogen and nitrogen can also be present in the volatile matter of the coal 

 

Hope this answers your queries.

Edited by P.K.Rao, 14 January 2015 - 12:33 AM.

[balakfs123]

Thanks for your detailed reply Mr.Rao

 

I was adding N from coal on to the N2 from air. 

I will consider NO2 as 1.5% into my combustion calculation.

 

Thanks