Air required for combustion per pound of coal By Er. Laxman Singh Sankhla B.E.Mech., hartered Engineer !odhpur, "ndia Mail "#$ laxman%%%&''()yahoo.co.in A bituminous coal has the following ultimate analysis: C = 71.6 N = 1.3 A = 9.1 H = 4. ! = 3.4 " = 3.# $ = 6.3 Calculate the total ai% %e&ui%e' fo% com(lete combustion (e% (oun' of coal. )he assume' +,-. )he theo%etical ai% %e&ui%ement is calculate' as follows: " ta = 11.#3C 34.36 /H 0 $ 2 4.3+! $% " ta = 11.#3 ,.716 34.36 /,.,45,.,63 2 4.3+ ,.,34 $% " ta = 9.7 lb lb of coal. Now calculate the total ai%: *cess ai% (e%centage = /" a 0 " ta 2 " ta 1,, $% ,.+ = /" a 0 9.72 9.7 1,, $% " a = 11.74 lb lb of coal.
A bituminous coal has the following ultimate analysis:C = 71.6N
= 1.3A = 9.1H = 4.8S = 3.4W = 3.5O = 6.3Calculate the total air
required for complete combustion per pound of coal. The excess air
is assumed 20%.
The theoretical air requirement is calculated as follows:Wta =
11.53C + 34.36 (H O/8) + 4.32SOr, Wta = 11.53*0.716 +
34.36*(0.048-0.063/8) + 4.32*0.034Or, Wta = 9.78 lb/lb of coal.Now,
calculate the total air:Excess air percentage = (Wa Wta)/ Wta
*100Or, 0.2 = (Wa 9.78)/9.78 *100Or, Wa = 11.74 lb/lb of coal.
