combustion calculations

COMBUSTION CALCULATIONS

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Solid & Liquid Fuels:

where

C = weight fraction of Carbon in Solid/Liquid fuel

S = weight fraction of sulfur in Solid/Liquid Fuel

Gaseous Fuels:

where

CO = weight fraction of Carbon Monoxide Gaseous fuel

HHV's (Higher Heating Value) for API fuel oils:

HHV = 17887 + 57.5*°API - 102.2*Swhere°API = Degree API of Fuel Oil (measure of specific gravity)S = weight percent of sulfur in fuel oil

MM Btu method of combustion:

Examples for calculation of thoretical/stoichiometric Air:

Example 1:A Fuel oil has the following characteristics:Degree API 28 °APIweight percent C = 87.5 %weight percent H = 12.5 %weight percent S = 0 %

0 %

Astoi = 11.53*C + 34.34*(H2 - (O2/8)) + 4.29*S

Astoi = Dry stoichiometric air required for combustion, lb/lb of fuel

H2 = weight fraction of Hydrogen in Solid/Liquid fuel

O2 = weight fraction of Oxygen in Solid/Liquid fuel

Astoi = 2.47*CO + 34.34*H2 + 17.265*CH4 + 13.3*C2H2 + 14.81*C2H4 + 16.12*C2H6 + 15.69*C3H8 + 15.44*C4H10 - 4.32*O2

Astoi = Dry stoichiometric air required for combustion, lb/lb of fuel

H2 = weight fraction of Hydrogen in Gaseous fuel

CH4 = weight fraction of Methane in Gaseous Fuel

C2H2 = weight fraction of acetylene in Gaseous Fuel

C2H4 = weight fraction of ethylene in Gaseous Fuel

C2H6 = weight fraction of ethane in Gaseous Fuel

C3H8 = weight fraction of propane in Gaseous Fuel

O2 = weight fraction of Oxygen in Gaseous fuel

This method for combustion is based on the concept that the weight of air required in the combustion of a unit weight of any commercial fuel is more nearly proportional to the unit heat value than to the unit weight of the fuel. Hence the quantity of air required is expressed in lb per million btu (lb/MMBtu) fired.

weight percent O =


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Calculate the amount of theoretical air required per million Btu fired for above fuel oil


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Calculations:HHV = 19497 Btu/lb

14.38 lb/lb fuel

1 MM Btu fired requires = 51.29 lb fuelHence air required per 1 MM Btu fuel = 737.6 lb

Example 2:Natural gas is used as fired fuel for boiler. It has the following volumetric analysis:

83.4 %

15.8 %

0.8 %

Calculate the amount of theoretical air required per million Btu fired for natural gas

Calculations:

Component volume % Molecular wt weight % wt fraction

83.4 16 72.89 0.7289 23876 yellow colored cell are inputs

15.8 30 25.89 0.2589 22320

0.8 28 1.22 0.0122 -

16.76 lb/lb fuelHHV = 23181.0 Btu/lb1 MM Btu fired requires = 43.14 lb fuelHence air required per 1 MM Btu fuel = 722.9 lb

Total Atmospheric Air required for Combustion in fired Heaters:

where

total atmospheric air, lb/MM Btu

Dry stoichiometric or theoretical air required, lb/lb fuelHHV = Higher Heating Value, Btu/lbH= absolute humidity of atmospheric air expressed as a fraction - lb of water vapor per 1 lb of dry air

excess air expressed as a fraction (10% excess air will be entered as 0.1, 15% = 0.15 etc.)

Astoi =

CH4 =

C2H6 =

N2 =

Combustion Constants or HHV, Btu/lb

CH4

C2H6

N2

Astoi =

Atot = Astoi*(1000000/HHV of fuel)*(1+ H)*(1 + Aexc)

Atot =

Astoi =

(to be calculated from % Relative Humidity (RH) & Dry Bulb Temperature using psychrometric chart, example: at 105°F & 100% RH, H = 0.0507 from Humid Air chart)

Aexc =


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Common Combustion Reactions & the Heats of Combustion:

Heating Value (Btu/lb)HHV LHV

61100 51600

14093 14093

4440 4440

4345 4345

10160 10160

23885 21500

22263 20370

21646 19929

21293 19665

Note that where no water is formed during combustion reaction the HHV & LHV are the same.

Heating Value (LHV(net) & HHV(gross)) @ 60°F, 14.696 psia (Source: Fig 23-2, GPSA Engg. Data Book)

Compound

Heating Value, 60°FNet Gross

Methane 909.4 - 1010 - -Ethane 1618.7 20277 1769.6 22181 65869Propane 2314.9 19757 2516.1 21489 90830Isobutane 3000.4 19437 3251.9 21079 98917n-Butane 3010.8 19494 3262.3 21136 102911Isopentane 3699 19303 4000.9 20891 108805n-Pentane 3706.9 19335 4008.9 20923 110091Neopentane 3682.9 19235 3984.7 20822 103577n-Hexane 4403.8 19232 4755.9 20783 115021Hydrogen 273.8 - 324.2 - -

320.5 - 320.5 - -

Prepared by: Ankur SrivastavaChemical Engineere-mail: [email protected]

H2 + 1/2O2 = H2O

C + O2 = CO2

C + 1/2O2 = CO

CO + 1/2O2 = CO2

S + O2 = SO2

CH4 + 2O2 = CO2 + 2H2O

C2H6 + 3.5O2 = 2CO2 + 3H2O

C3H8 + 5O2 = 3CO2 + 4H2O

C4H10 + 6.5O2 = 4CO2 + 5H2O

Btu/ft3, ideal gas,

14.696 psia

Btu/lbm Liquid

Btu/ft3, ideal gas,

14.696 psia

Btu/lbm Liquid

Btu/gal Liquid

Carbon Monoxide

Disclaimer: The information and methods included within this spreadsheet are presented for combustion air calculations. It is intended to be used by technically skilled persons at their own discretion. I do not warrant the suitability or accuracy of these methods.

C114

asrivas1: Temp. is above Critical Point

E114


F114


C124


E124


F124



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+ 15.44*C4H10 - 4.32*O2


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: The information and methods included within this spreadsheet are presented for combustion air calculations. It is intended to be used by technically skilled persons at their own discretion. I do not warrant the

combustion calculations

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