Fuels and Combustion

Industrial and Power Plant Engineering is 35% of the MELE, and fuels and combustion sit at the front of it. The good news: most items reduce to a few durable formulas, rearranged. Learn to move between a fuel's composition, its heating value, and the air it needs, and you turn a whole family of questions into fast, reliable points. Every number here is in SI units.

Solid, liquid, and gaseous fuels

Fuels come in three phases. Solid fuels (coal, biomass, coke) are described two ways. The proximate analysis reports moisture, volatile matter, fixed carbon, and ash: the practical burning behavior. The ultimate analysis reports the elemental mass fractions of carbon, hydrogen, oxygen, nitrogen, and sulfur (plus ash): the chemistry you need for combustion math. Fixed carbon is the solid combustible residue left after you subtract moisture, volatile matter, and ash.

Liquid fuels (fuel oil, diesel, gasoline) are graded by properties like the octane number (knock resistance in spark-ignition engines) and cetane number (ignition quality in compression-ignition engines). Gaseous fuels (natural gas, LPG, producer gas) burn cleanly; the principal combustible component of natural gas is methane (CH4).

Heating value: higher versus lower

The heating value is the heat released when a unit mass of fuel burns completely. It comes in two forms:

• Higher heating value (HHV): assumes the water vapor formed during combustion is condensed, so its latent heat is recovered.
• Lower heating value (LHV): assumes that water leaves the products uncondensed, so its latent heat is not recovered.

The HHV therefore always exceeds the LHV by the latent heat of the water formed:

LHV = HHV minus (9 x H) x hfg

Here H is the hydrogen mass fraction, 9 kg of water form per kg of hydrogen burned (2 kg H2 gives 18 kg H2O), and hfg is roughly 2,442 kJ/kg near 25 degrees C.

For a solid fuel of known ultimate analysis, Dulong's formula estimates the HHV in kJ/kg:

HHV = 33,820 C + 144,212 (H minus O/8) + 9,304 S

The term (H minus O/8) is the available hydrogen: oxygen already in the fuel is assumed bound to one-eighth its mass of hydrogen as water, so that hydrogen releases no heat.

Worked example: a coal with C = 0.75, H = 0.05, O = 0.08, S = 0.01. Available hydrogen is 0.05 minus 0.08/8 = 0.04. So HHV = 33,820(0.75) + 144,212(0.04) + 9,304(0.01) = 25,365 + 5,768 + 93 = about 31,200 kJ/kg.

Theoretical air and the air-fuel ratio

Complete combustion needs oxygen for three reactions:

Reaction	O2 needed per kg of element
C + O2 gives CO2	32/12 = 2.667 kg
2 H2 + O2 gives 2 H2O	32/4 = 8 kg
S + O2 gives SO2	32/32 = 1 kg

Air is 23.2% oxygen by mass, so the theoretical (stoichiometric) air-fuel ratio is the oxygen demand divided by 0.232, less any oxygen already in the fuel:

A/F = [2.667 C + 8 H + 1 S minus O] / 0.232

An equivalent, popular form is A/F = 11.5 C + 34.5 (H minus O/8) + 4.3 S.

Worked example: a fuel oil with 85% carbon, 13% hydrogen, 2% sulfur, no oxygen. Oxygen demand is 2.667(0.85) + 8(0.13) + 1(0.02) = 2.267 + 1.040 + 0.020 = 3.327 kg per kg fuel. Theoretical air is 3.327/0.232 = about 14.3 kg air per kg fuel. Pure carbon alone needs 2.667/0.232 = 11.5 kg air per kg; pure hydrogen needs 8/0.232 = 34.5 kg.

The theoretical (stoichiometric) air is the exact minimum air for complete combustion, with none left over.

Excess air

Real burners supply more than the theoretical air so that every fuel particle finds oxygen. The extra is excess air:

percent excess air = (actual air minus theoretical air) / theoretical air x 100

Worked example: a furnace with a theoretical A/F of 15.0 that actually supplies 18.75 kg air per kg fuel runs at (18.75 minus 15.0)/15.0 = 0.25, or 25% excess air. Too little air gives incomplete combustion, so the flue gas carries carbon monoxide from partly oxidized carbon; too much air wastes heat up the stack.

Products of combustion

Track the products by mass. Per kg of each element burned completely:

• Carbon gives 44/12 = 3.67 kg of CO2.
• Sulfur gives 64/32 = 2.0 kg of SO2.
• Hydrogen gives 9 kg of water (so a fuel with 12% hydrogen makes 9 x 0.12 = 1.08 kg of water per kg fuel).

The total wet flue gas mass per kg of fuel is simply the air-fuel ratio plus one (all the air plus all the fuel leave as products, since mass is conserved). A fuel burned at A/F = 14 produces 14 + 1 = 15 kg of flue gas per kg fuel.

Orsat analysis

The Orsat apparatus samples the flue gas and reports the volumetric percentages of carbon dioxide, oxygen, and carbon monoxide on a dry basis (the water vapor is not measured because it condenses in the cooled sample). The nitrogen is found by difference: 100% minus the measured CO2, O2, and CO. From the leftover free oxygen you can back out how much excess air was actually used, which is why the Orsat is the classic tool for tuning a boiler's combustion. Free O2 in the sample means excess air; measurable CO means the fire is starved and fuel is being wasted.

Exam-day strategy

• Keep the oxygen demands memorized cold: 2.667 for carbon, 8 for hydrogen, 1 for sulfur, then divide by 0.232 for air.
• Do not forget to subtract the fuel's own oxygen in the A/F formula; that is the classic trap.
• HHV minus LHV is always the water's latent heat, and 9 kg of water form per kg of hydrogen.
• Flue gas mass per kg fuel is just A/F + 1: mass is conserved.
• Proximate reports moisture, volatile matter, fixed carbon, ash; ultimate reports the elements C, H, O, N, S. Orsat reports dry CO2, O2, CO, with N2 by difference.