1- : . : - ()

- ()

-

2- : .

1- ( )2- (NOx) ( )3- ( )1- ( )2- (CO) (C)( )

3- 1- (A) :

[Nm3 / kg] (B) : [Nm3/Nm3]

3- 2- (m):(A) :

(B) :A = m Ao n2 : Nitrogen content included in the fuel N2: [Nitrogen (n) included in fuel is to be ignored] nitrogen in the combustion air is 79%,

3- 3- : G = Go + ( m 1 ) Ao G = Go + ( m 1 ) Ao G : G : Go : Go:

Fuel Theoretical Combustion Gas Amount Dry ( Go ) Wet ( Go )Solid &Liquid Fuel(10.21)A0+1.867c+11.2h+0.7s +1.244w+0.8nGaseous FuelG0(h2+2ch4+3c2h6 +2c2h4+4c3h8+5c4h10)1+A00.5(h2+co+c2h62c3h83c4h10)

3- 4- : :(A)Hh, Hl [kcal / kg] [kcal / kg] : (B)Hh=3020CO+3050H2+9520CH4+16850C2H6+15290C2H4+24160C3H8+31790C4H10 [kcal/Nm3] Hl = Hh 480(H2+2CH4+3C2H6+4C3H8+5C4H10) [kcal/Nm3]

3- 5- : :(A)

(CO2)max : (B)

3- 6- : Tb = (Hl + Qa + Qg) / ( Gw x CpG ) Hl: Qa : Qa = Qg: Gw: Gw = (m 1) Ao + Go CpG:

4- :

4- :

4- :

5- : - - Relation between air ratio and CO,O2% in Exhaust Gas 1 -Heat loss by unburned fuel2- Heat loss by sensible heat of exhaust gas3- Total heat loss4- Air ratio minimizing total heat loss

5- The relation between air ratio and exhaust gas composition in a burning Coke Oven Gas

6-

() 1.9 - 4.91.1- 1.33.5 - 6 1.2- 1.4 4.9 - 71.3 - 1.5

: 1. (1)- :

n ; The composition of nitrogen in the fuel ( kg /kg-fue) N2 volume of input air = ( N2 ) G 22.4 [Nm3 / kg-fuel ]And , in the case of complete combustion: N2 volume of excess air = ( O2 ) G [Nm3/ kg-fuel ]Then:

here: ( N2 ), ( O2 ) ; The composition of the exhaust gas ( % ) G ; The volume of actual dry exhaust gas ( Nm3 / kg-fuel l )In the above equation, comparing ( N2 )G with (n/28)22.4, ( N2 ) is about 0.8 for normal fuel, and G is about 10 [ Nm3 / kg-fuel ]. Also n is only 0.02 or less. Therefore, suppose n is small enough to ignore ( n= 0), the equation becomes as follows:

: (2)- : N2 volume of input air = ( N2 ) G n2 [ Nm3 / Nm3-fuel ] N2 volume of excess air = [ Nm3/ Nm3-fuel ]Then:

Suppose n2 ( The composition of nitrogen gas in fuel gas ----Nm3 / Nm3-fuel ) is also small enough to ignore, the equation becomes the same as .That is, is generally the equation used to get m for all kinds of fuel.

Then, in equation , when approximately ( N2 ) = 0.79, it becomes as follows:

: (3)- : [Nm3 / Nm3] Here, ( c ) is the unburned carbon The net excess O2 volume is obtained by subtracting the necessary volume to burn the unburned fuel from the O2 volume of dry exhaust gas. That is, ( O2 ) G, as shown in Fig.- 4.1. Then, when the unburned fuel element is only CO, the equation to get the air ratio ( m ) is as follows:

When the unburned fuel elements include H2, CmHn , and carbon-like soot (c), the above equation becomes as follows; quantity in the exhaust gas: [ kg / Nm3 ] In this case, if ( O2 ) ( O2 )r becomes minus, then m becomes smaller than 1 ( m < 1 ). Anyway, the above equation can be used for both m > 1 and m < 1.

: 2. : In the case where m > 1 and unburned fuel is only CO The above equation becomes, approximately

7- ( )

7- (2)

*********************