Fuels & Combustion Calculations
FUELS & COMBUSTION CALCULATIONS
Unit 5
Prabir Basu
TYPES OF FUELS
FOSSIL FUELS
Solid fuels (COAL) Liquid fuels (OIL) Gaseous fuels (NATURAL GAS)
NUCLEAR FUELS RENEWABLE FUELS (BIOMASS) WASTE FUELS (MUNICIPAL WASTES)
SOLID FUELS PROPERTIES
HEATING VALUE ULTIMATE ANALYSIS PROXIMATE ANALYSIS ASH DEFORMATION POINTS
Initial deformation temp. Softening temp. Hemispherical temp Fluid temp.
LIQUID FUELS PROPERTIES
HEATING VALUE FLASH POINT IGNITION POINT (SELF, FORCED) VISCOSITY POUR POINT SULFUR ASH
Refinery process
Distillation
GASEOUS FUEL PROPERTIES
HEATING VALUE COMPOSITION DENSITY
BASIS OF ANALYSIS
AS RECEIVED Ultimate C +H +O +N +S +A +M =100 Proximate VM +FC +M +A = 100 AIR DRY [100C/(100-Ma)] DRY ASH FREE [100C/(100-M-A)]
HEATING VALUE
HIGHER HEATING VALUE (GROSS) LOWER HEATING VALUE (NET) LHV = HHV – LH of steam (9H/100+M/100)
II-1 CHEMICAL REACTIONS
Combustion C + O2 = CO2 + 32,790 kJ/kg of carbon,
Heat of formation at 25C is 393.7 kJ/mol [Perry p-2-188]
mCn Hm + (n +m/4)O2 = nCO2 + m/2 H2O + Q S + O2 = SO2 + 9260 kJ/kg of sulfur Calcination CaCO3 = CaO + CO2 – 1830 kJ/kg of CaCO3gCO3 = MgO + CO2 – 1183 kJ/kg of MgCO3. Sulfation CaO + SO2 + 1/2 O2 = CaSO4 + 15141 kJ/kg S.
Basic Stoichiometry
C + O2 = CO2 + q 1 kmol of carbon combines with 1 kmol of oxygen to produce 1 kmol of carbon dioxide and release q amount of heat. 1 kmol of reactant = M kg of the reactant when M is the molecular weight of the reactant. So mass of one kmol of oxygen (O2) is 2x16 = 32 kg 1 kmol of a gas occupies 22.4 nm3 at 00C 1 atm
BASIC EQUATION
1.
2.
3. 4.
C + O2 = CO2 kJ/kg carbon H2 + ½ O2 = H2O S + O2 = SO2
Adding oxygen requirements of above eqns and subtracting the oxygen in fuel we get the total oxygen required VO2= (1.866C + 5.56H + 0.7S - 0.7O) Nm3/kg
Since air contains 21% oxygen by volume,
Unit 5
Prabir Basu
TYPES OF FUELS
FOSSIL FUELS
Solid fuels (COAL) Liquid fuels (OIL) Gaseous fuels (NATURAL GAS)
NUCLEAR FUELS RENEWABLE FUELS (BIOMASS) WASTE FUELS (MUNICIPAL WASTES)
SOLID FUELS PROPERTIES
HEATING VALUE ULTIMATE ANALYSIS PROXIMATE ANALYSIS ASH DEFORMATION POINTS
Initial deformation temp. Softening temp. Hemispherical temp Fluid temp.
LIQUID FUELS PROPERTIES
HEATING VALUE FLASH POINT IGNITION POINT (SELF, FORCED) VISCOSITY POUR POINT SULFUR ASH
Refinery process
Distillation
GASEOUS FUEL PROPERTIES
HEATING VALUE COMPOSITION DENSITY
BASIS OF ANALYSIS
AS RECEIVED Ultimate C +H +O +N +S +A +M =100 Proximate VM +FC +M +A = 100 AIR DRY [100C/(100-Ma)] DRY ASH FREE [100C/(100-M-A)]
HEATING VALUE
HIGHER HEATING VALUE (GROSS) LOWER HEATING VALUE (NET) LHV = HHV – LH of steam (9H/100+M/100)
II-1 CHEMICAL REACTIONS
Combustion C + O2 = CO2 + 32,790 kJ/kg of carbon,
Heat of formation at 25C is 393.7 kJ/mol [Perry p-2-188]
mCn Hm + (n +m/4)O2 = nCO2 + m/2 H2O + Q S + O2 = SO2 + 9260 kJ/kg of sulfur Calcination CaCO3 = CaO + CO2 – 1830 kJ/kg of CaCO3gCO3 = MgO + CO2 – 1183 kJ/kg of MgCO3. Sulfation CaO + SO2 + 1/2 O2 = CaSO4 + 15141 kJ/kg S.
Basic Stoichiometry
C + O2 = CO2 + q 1 kmol of carbon combines with 1 kmol of oxygen to produce 1 kmol of carbon dioxide and release q amount of heat. 1 kmol of reactant = M kg of the reactant when M is the molecular weight of the reactant. So mass of one kmol of oxygen (O2) is 2x16 = 32 kg 1 kmol of a gas occupies 22.4 nm3 at 00C 1 atm
BASIC EQUATION
1.
2.
3. 4.
C + O2 = CO2 kJ/kg carbon H2 + ½ O2 = H2O S + O2 = SO2
Adding oxygen requirements of above eqns and subtracting the oxygen in fuel we get the total oxygen required VO2= (1.866C + 5.56H + 0.7S - 0.7O) Nm3/kg
Since air contains 21% oxygen by volume,