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Basic calculations on fuels & combustion


Fuel:
It is a substance which releases heat energy on combustion. The principal combustible elements of each fuel are carbon and hydrogen.

Fuels classification:

Primary Fuels: These fuels directly available in nature, Ex: Wood, Peat, Lignite coal, Petroleum and Natural gas.
Secondary Fuel: These are prepared fuels, Ex: Coke, Charcoal, Briquettes, Kerosene, fuel oil, petroleum gas, producer gas etc.

Fuels are also classified as Solid fuels, liquid fuel and Gaseous fuel.

Different types of coals: Peat, Lignite, Bituminous coal, Anthracite coal and Coke.


Gaseous fuels:Natural gas, Coal gas, Coke oven gas, Blast furnace gas, producer gas, Water gas and Sewer gas.

Energy producing elements in fuel: Carbon, oxygen, hydrogen and Sulphur.

Formation of Charcoal & Coke:

Charcoal:It is obtained by destructive distillation of wood.

Coke is formed by destructive distillation of certain types of coal.

Calorific value of fuel: Calorific value (GCV) is the amount of heat released on complete combustion of unit quantity of fuel.It is measured in kcal/kg.

GCV is also known as Higher Calorific Value (HCV),it is given by following formula

GCV or HCV = (8084 X C% + 28922 X (H2%–O2%/8) + 2224 X S%)/100…kcal/kg

Where C, H2, O2 and S are percentage of Carbon, hydrogen, oxygen and Sulphur respectively in fuel.

Net Calorific Value (NCV) or Lower Calorific Value (LCV):


The total heat released by fuel during combustion is not completely utilized. Some heat is taken out by water vapour which is produced during combustion of hydrogen. Such heat value taken by considering heat taken away by water vapour is called NCV or LCV.

LCV = HCV – (9 X H2% X 586), Where H2 = Hydrogen% in fuel and 586 is latent heat of steam

in kcal/kg.

Useful heat Value (UHV) of coal:

NCV = GCV - 10.02 X Percentage of total moisture.

Ultimate analysis:Ultimate analysis indicates the various elemental chemical constituents such as Carbon, Hydrogen, Oxygen, Sulphur etc.

Proximate analysis of coal:Proximate analysis determines fixed carbon, Volatile matter, moisture and percentages of ash.

Volatile Matter (VM) & is its significance:

VM is generally a composition of methane, hydrocarbons, hydrogen and carbon monoxide and other incombustible gases like CO2 and Nitrogen. It is the indication of presence of gaseous fuel in the fuel.


Significance:

  • Helps in easy ignition of coal by increasing the flame length
  •   Sets minimum limit on the furnace height and volume.
Properties of Coal:
·         Caking Index: Indicates binding property of coal.
·         Swelling Index: Indicates caking capacity of coal.
·         Slacking Index: Indicates the stability of coal when exposed to open atmosphere.
·         Grinding Index: It gives the idea of ease of grinding of coal.
·         Abrasive Index: Indicates the hardness of coal.


Combustion:

Combustion is the rapid oxidation of fuel accompanied by the production of heat

Oxygen is the major element on earth, making up to 21% (by volume) of our air. Carbon, hydrogen and Sulphur in the fuel combine with oxygen in the air to form carbon dioxide, water vapour and Sulphur dioxide releasing tremendous heat.

Oxygen is the major element on earth, making up to 21% (by volume) of our air. Carbon, hydrogen and Sulphur in the fuel combine with oxygen in the air to form carbon dioxide, water vapour and Sulphur dioxide releasing tremendous heat.

Basic requirements of combustion:Fuel, Oxygen and 3T’s

C + O2 = CO2+ Heat 8084 kcal/kg of Carbon.

2C + O2 = 2CO + Heat 2430 kcal/kg of Carbon.

2H2 + O2 = 2H2O 1 Heat 28922 kcal/kg of Hydrogen.


S + O2 = SO2 + Heat 2224 kcal/kg of Sulphur.

Products of Combustion: CO2, CO, O2, SO2 and ash.

Spontaneous combustion: is a phenomenon in which coal bursts into flame without any external ignition source but by itself due to gradual increase in temperature as a result of heat released by combination of oxygen with coal.

Major contents of ash:
  • Silica (SiO2)
  • Aluminum oxide (AlO3)
  • Iron Oxide (Fe2O3)
  • Sodium Oxide (Na2O)
  • Potassium Oxide (K2O)
  • Calcium Oxide (CaO)
  • Magnesium Oxide or Magnesia (MgO)
Fly ash & Bottom ash :

Fly ash 70–80%, Bottom ash 20–30%.

Fly ash at Economiser : 7–8%, APH: 10–12% and ESP: 80–82% of total fly ash.


Oxygen & Nitrogen present in the air :By weight Oxygen 23% and Nitrogen 77% and by volume Oxygen 21% and Nitrogen 79%.

Stoichiometric air fuel ratio: A mixture of air and fuel, which contains sufficient amount of oxygen for complete combustion.

Rich mixture: Mixture with deficiency of air
Lean mixture: Mixture with excess air


Theoretical air of combustion:

Minimum amount of air that supplies the sufficient amount of oxygen for the complete combustion of all carbon, hydrogen and any other elements in the fuel that may oxidize is called theoretical air.
Theoretical air required for combustion of carbon:
We know that Carbon on oxidation with Oxygen forms Carbon dioxide.

C + O2 = CO2
12 + 32 = 44 (Molecular weights of Carbon and Oxygen are 12 and 16 respectively)
1 + 2.67 = 3.67
So, 1 kg of Carbon requires 2.67 Kg of Oxygen for complete combustion into 3.67 kg of carbon dioxide.

Similarly
Hydrogen on oxidation forms Water
2H2 + O2 = 2H2O

4 + 32 = 36 (Molecular weights of Hydrogen and Oxygen are 1 and 16 respectively)
1 + 8 = 9
So, 1 Kg of Hydrogen requires 8 Kg of Oxygen for its combustion & forms into water.

Excess air: Amount of extra air given to ensure complete combustion is called excess air.

3T’s of combustion:

  • Temperature: High enough to maintain the ignition of the fuel.
  • Turbulence: Is the mixing of fuel and air.
  • Time: Sufficient enough for combustion.
Specific heat Cp and Cv:

Specific heat is the amount of heat in kcal needed to raise the temperature of 1 kg of substance by 1 °C. Cp and Cv are the specific heat at constant pressure and constant volume of gas.

Solved examples:

 Example-1:A Biomass (Bagasse) contains 23% of carbon, 22% of oxygen, 3.5% of Hydrogen and 0.05% of Sulphur, then calculate the theoretical air required for its combustion.

We have, Theoretical air requirement = (11.6 X %C + 34.8 3 (%H2 - %O2/8) + 4.35 X %S)/100… 

Kg/kg of fuel.

Therefore, Th air requirement will be = (11.6 X 23 + 34.8 X (3.5 – 22/8) + 4.35 X 0.05))/100
                                                             = 2.9 kg of air/kg of fuel.

Example-2:What is the quantity of excess air, if O2 measured in the Boiler outlet flue gas is 6%?

We have, Excess air = (O2%/( 21 - O2%)) X 100

                                 = (6/(21 – 6)) X 100 = 40%

Example-3:A complete combustion requires 2.9 kg of theoretical and 20% of excess air, then calculate the total air consumed for complete combustion per kg of fuel burnt.

Actual quantity of air supplied = (1 + Excess air %/100) X theoretical air

                                                   = (1 + 20/100) X 2.9 = 3.48 kg of air/kg of fuel.

Example-4: A flue gas has 10% of CO2 and theoretical calculated CO2 is 12%, then calculate percentage of the excess air.

                                         % of Excess air = ((Theoretical CO2%/Actual CO2%) - 1)) X 100

                                                                   = (12/10) - 1) X 100 = 20%

Example-5: A Coal sample contains 10% of ash, coal required is 300 MT/day, assuming 100% combustion calculate the mass of ash generated in a day.

                                                   Mass of ash generated = (300 X10)/100
                                                                                        = 30 MT

Example-5:A Indonesian coal contains 58% of Carbon, 4.2% of Hydrogen, 11.8% of Oxygen and 0.5% of Sulphur, also needs 20% of excess air for its complete combustion. Calculate the Total air required for complete combustion and O2% in flue gas.

We know that,

Theoretical air required for combustion is = (11.6 X 58 + 34.8 3 (4.2 - 11.8/8) + 4.35 X 0.5)/100

                                                                    = 5 7.7 Kg/Kg of Coal
                                                      Total air = (1 + EA/100) X Theoretical air
                                                                    = (1 + 20/100) X 7.7
                                                                    =  9.24 Kg/Kg of Coal

                               Also we know that EA = (O2 %/(21 - O2%)) X 100

                                                               20 = (O2/(21 - O2)) X 100
                                                               O2= 3.5%



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