OCR Text |
Show These values are shown in Figure 8 for different adiabatic flame temperatures. 3.12 3.1 3.08 ~ 3.06 c: ::l3.04 j §. 3.02 u:- 3 -; 2.98 E 2.96 10 ~ 2.94 ~ 2.92 ~ 2.9 ~ ~ 2.88 2.86 2.84 2.82 3.8 4 4.2 Natural Gas 21 % Oxygen .!i Air 10% Excess Air 4.4 4.6 Adiabatic Flame Temperature (F) (thousands) FIGURE 8, PEAK TEMPERATURE VS ADIABATIC FLAME TEMPERATURE, The NOx formation is modeled using the Zeldovich mechanism. The reactions that describe the Zeldovich mechanism are: (0) + N2 -k1- -k-1- NO +(N) (R1) (N) +02 -k2- -k-2- N + (0) (R2) O2 + (M) -k3- _k_3_20 + (M) (R3) in which: [0] is a free oxygen radical [N] is a free nitrogen radical [M] is a third body (k] is the specific reaction rate constant Before obtaining an expression for the rate of formation of NOx the following must be noted: 1. It has been observed in combustion that NO is formed after ° is formed. 192 2. It has also been noticed that at the flame,O and 02 are near the1, r equ1'l '1b r1'u m concentra-tion when NO is first detected. This can be represented in equation R3. Since ° and ~2 are assumed to be in equilibrium, the equ1- librium constant is 3. Reaction(R1) is an endothermic reaction with a high value of activation energy (+75 Kcal) which means that this reaction will set the rate in the chain. 4. Reaction(R2) is an exothermic reaction; since the activation energy of the reverse reaction is +32 kcal(endothermic reaction) the presumption that the activation energy of reaction( R2) is -32 Kcal is valid. 5. Reaction(R2) is faster than reaction(R1) so the N produced in reaction(R1) will be immediately consumed in reaction(R2), so N can be assumed to be under steady state condition. 6. In order to calculate the time it takes NO to reach equilibrium, it can be assumed that N2 is already in equilibrium. From reactions R1, R2 and R3 the rates of formation of NO and N will be given by the following equations (1) and (2) and the equilibrium (0) can be expressed in terms of (02) by equation (3) d(NO) = k1(0)(N2) + ~(N)(02) - k-1(NO)(N) - k-2 (NO)(O) (1) dt deN) = k1(0)(N2) - k-1(0)(N) + k-2 (NO)(O) - k2(NX02) (2) dt (0)2 = Keqo(02) (3) Note that the symbols of the elements in parenthesis represent concentrations of the different elements in mol/cm3 . Since the radical concentrations of N , NO and ° are very small, there products (NO)(O) and (NO)(N) are still smaller and can be neglected. Further, since N was assumed to be under steady state conditions, its rate of change with time is zero and the rate of change of NO with time can be expressed by equation(7) which reduces to equation(8) expressed in mole fractions(Note that ni = xiP/RT) below: |