Characteristics of Burners and Combustors

Such systems differ broadly in respect of their configurations. The heat release rates, excess air necessities, provisions for ignition, and so on.

Gas Burners:

(a) Burner designs are fundamentally similar to that of a Bunsen burner, with some main air mixed with the gas to yield a rich mixture to give stable flame, and with the secondary air diffusing into the flame from side to entire the combustion.

(b) Turbulence-promotion devices are included in the burners to raise the combustion rate, decrease the flame size, and improve the combustion efficiency.

(c) In order to raise the inlet gas velocity without encountering flame blow-off or blow-out, a flame holder is inserted into the gas stream, therefore its wake gives a low-velocity, highly turbulent area that keeps the flame attached to the burner, the flame holder is generally in the form of a metal ring or a ceramic cup positioned in the centre of the incoming stream of gas – prime air mixture.

Oil Burners:

(a) The rate of vaporization of the droplets is a function of drop dimension. Ambient air temperature, and the volatility of the liquid; the burning time differs from a few ms for iso-octane to some seconds for a heavy fuel oil.

(b) The larger the burner, the more hard the admixture of secondary air from the area surrounding the flame. And the greater the affinity for the flame to blow out. Therefore, the usual process is to use several burners (as, for illustration, in the LPG burners for cooking).

Example:

A certain coal has following elemental composition (eventual analysis):
S: 0.6%; H: 5.7%; C: 79.2%; O: 10.0%; N: 1.5%; Ash: 0%.  When this coal is burned with 30% excess air, compute the AFR on mass basis.  

Solution:

The combustion equation for each of the combustible elements per 100 kg of fuel.  
The molar compositions per 100 kg of fuel become:

Moles of S per 100 kg of fuel = 0.6/32 = 0.02
Moles of H per 100 kg of fuel = 5.7/2 = 2.85
Moles of C per 100 kg of fuel = 79.2/12 = 6.60
Moles of O per 100 kg of fuel = 10/32 = 0.31
Moles of N2 per 100 kg of fuel = 1.5/28 = 0.05

The combustion equation for the combustible elements can now be written; this allows the determination of the hypothetical oxygen needed:

0.02 S + 0.02 O₂ = 0.02 SO₂
2.85 H₂ + 1.425 O₂ = 2.85 H₂O
6.60 C + 6.60 O₂ = 6.60 CO2

Therefore, 8.045 moles of O₂ are needed per 100 kg fuel. As the fuel has 0.31 moles of O₂ per 100 kg fuel, the number of moles of O₂ needed per 100 kg of fuel = 7.735.
Therefore, the theoretical AFR = [7.735 + (7.735 x 3.76) 29.95] / 100
                                                 = 10.63 kg air/kg fuel

Taking the molecular weight of air be 28.95.
For 30% excess air, AFR = 1.3 × 10.63 = 13.82 kg air/kg fuel