FUELS and COMBUSTION Fuel -any material that can be burned to release thermal energy. -in all fuels the two basic combustible elements are carbon and hydrogen. Classification of Fuel 1. Solid Fuels - Are coal, chiefly anthracite anthracite and coke. -Coal contains varying amounts of oxygen, hydrogen, nitrogen, sulfur, moisture, and ash.
Analsis of Coal 1. "roximate "roximate Analysis Analysis - %etermines the the &ercentage of moisture, moisture, volatile matter matter,, fixed carbon, and ash. - 'n some cases cases the &ercentage &ercentage of sulfur is also also obtained. (. )ltimate Analysis - ultimate analysis analysis divides all the remaining &art of the coal into the elements carbon, hydrogen, oxygen, sulfur, sulfur, and nitrogen in &ercentages by weight. - *he ash content of the coal is inde&endent of the ty&e of analysis and is therefore the same for both.
2. Liquid fuels - a mixture mixture of numerous numerous hydrocarbons and are obtained from crude oil by distillation. distillation.
3. Gaseous Fuels -As natural gas, oil derivatives !"#$, acetylene, manufacture gas from coal or oil residue$ and biogas from manure or sewage$.
Basis of !e"o#tin$ Analsis Analsis - *he +ureau of ines re&orts coal in the following manner 1. As eceived eceived or As Fired Fired (. %ry or oisture Free /. oisture and Ash Free or Combustible Combustible 0. oisture, Ash, Ash, and Sulfur Free
Sam&le "roblem 1. Convert the &roximate analysis of Cherokee County, ansas coal given in *able 2-1 to a.$ dry and b.$ to moisture and ash free. "roximate analysis of coal as fired$ oisture 3 2.456 7olatile atter 3 /0.086 Fixed Carbon 3 2(.096 Ash 3 8.5:6 (. Convert the ultimate analysis of Cherokee County, ansas coal given i n *able *able 2-1 to a.$ dry b.$ to moisture and ash free c.$ to moisture, ash, and sulfur free bases. )ltimate analysis of coal Carbon 3 81.916 ;ydrogen 3 2.(/6 =xygen 3 14 14.126
Sulfur 3 /./06 =xidi?er
Balancin$ C&e%ical Equations -Chemical [email protected]
are balanced on the basis of the conservation of mass principle, which states that the total mass of each element is conserved during a chemical reaction.
1. ethane C; $ is burned with stoichiometric amount of air during a combustion &rocess. Assuming com&lete combustion, determine the airfuel and fuelair ratios. 0
(. "ro&ane fuel C ; $ is burned in the &resence of air. Assuming that the combustion is theoreticalBthat is, only nitrogen ;( 2
3. Octane (C H ) is burned burned with dry air. The volumetric volumetric analysis of the the prod produc ucts ts on a dry dry basi basis s is 9.21 9.21 perc percen entt CO , 0.61 0.61 percent CO, 7.06 percent O , and 83.12 percent N . Determine (a ) the air-fuel ratio and ( b ) the percentage of theoretical air used. 8
Properties of Fuels & Lubricants: 1. Density and Specific Gravity
4. A coal from Illinois which has an ultimate analysis (by mass) as 67.4 67.40 0 perc percen entt C, 5.31 5.31 perc percen entt H , 15.1 15.11 percen percentt O , 1.44 1.44 perc percen entt N , 2.36 2.36 perce ercent nt S, and 8.38 8.38 perce ercent nt ash ash (no (noncombustibles) is burned with 40 percent excess air. Calculate the mass of air required per unit mass of coal burned and the apparent molecular weight of the product gas neglecting the ash constituent. 2
Specific Gravity =
where: Density of water = 1000 kg/m 3 = 9.1 !"/m 3 = 1 kg/#i = $%.& #'/ft3
%. o ()* + o ,(-./ -nits: o API
!" ( )!" *+, %, %, ,3%0 ,3%0 #$% "!&'
3. !eatin" #alue of Fuels: a. eating va#e of so#id f0e#s 2 $ % : ?
$ % = ))(45*6 + '&&(5'5/! 7 0 + ,()*&S( 1231" or
!" 45"!&' 6789:)!"
! ! " ( )!" *+, -,./0 #$% "!&'
Specific gravity with correction factors de to temperatre effect: SG t =
#$"!&'; ( ! < % ( "! & = >
'. eating va#e of #i30id f0e#s: $ % = &'(')* + '),-. / o API0( 1231" 1igher heating va#0e 4 va#0e 4 the the heating heating va#e o'tained o'tained when when water in the prodct prodct of com'stion is in the #iid state. 5ower 1eating va#0e 4 va#0e 4 the heating heating o'tained when the water in the prodct of com'stion is in the vapor state. 6a#orimeter 7 7 is se to measre the heating va#e va#e of fe#. fe#.
8%eoretical Air7Fuel 9atio for Soli Fuels : 7 = ''-;6 + )&-;/! 7 0 + &-)S( 1" air31" fuel ?
(ct0a# (ir480e# (ir480e# 9atio:
= /8%eoretical 0/' + e0
Given the fo##owing #timate ana#ysis: 6 = 0;