1

MIE516 Assignment 3 Solutions

NOx formation

1. Identify the key radical in the conversion of NO to NO2 in combustion systems. Why does NO2 not

appear in high-temperature flame regions?

Solution:

The following mechanism is responsible for the conversion of NO to NO2 in the combustion products:

2 2NO + HO NO + OH (formation) (N.12)

2NO + H NO + OH (destruction) (N.13)

2 2NO + O NO + O (destruction) (N.14)

From these reaction steps, it is apparent that the HO2 radical is the key radical in the conversion process.

NO2 does not appear in high-temperature regions since HO2 only forms in low-temperature regions,

meaning NO2 is not formed in high-temperature regions, and the NO2 destruction reactions are more

active at high temperatures, resulting in the destruction of any NO2 that is transferred to the high-

temperature regions.

2. For the conditions are:

2100KT

30.167 kg/m

28.778MW

5

O, 7.6 10 (mole fraction)eq

2

3

O , 3.025 10 (mole fraction)eq

2N , 0.726 (mole fraction)eq

14 3

1 1.82 10 exp[ 38,370 / ( )] with units of cm /gmol-sfk T K

10 3

2 1.8 10 exp[ 4,680 / ( )] with units of cm /gmol-sfk T T K

a) How long does it take to form 50 ppm of NO?

b) Calculate the value of the reverse reaction rate coefficient for the first reaction rate (i.e.,

2O N NO N, for a temperature of 2100 K.

c) Determine the steady state numerical values for [N] and .

d) For the first reaction, determine the ratio of the forward reaction rate versus the reverse reaction

rate.

2

Solution:

a) d[NO]/dt can be evaluated from

1 2

d[NO]2 [O] [N ]

df eq eqk

t

Converting the mole fraction of O and N2 to molar concentrations

O,

O,

5

7 3 10 3

[O]

0.1677.6 10

28.778

4.41 10 kmol/m 4.41 10 gmol/cm

eq eq

u

eq

P

R T

MW

2

2

2 N ,

N ,

3 3 6 3

[N ]

0.1670.726

28.778

4.21 10 kmol/m 4.21 10 gmol/cm

eq eq

u

eq

P

R T

MW

The reaction rate of NO is

1 2

14 10 6

9 3

d[NO]2 [O] [N ]

d

2(1.82 10 )exp[ 38,370 / 2100](4.41 10 )(4.21 10 )

7.846 10 gmol/cm -s

f eq eqkt

Converting the mole fraction of NO to molar concentrations

NO

NO

6

7 3 10 3

[NO]

0.16750 10

28.778

2.902 10 kmol/m 2.902 10 gmol/cm

u

P

R T

MW

So the time takes to form 50 ppm of NO is

10

9

2.902 100.0369 s

7.846 10

3

b) The forward and reverse rate coefficients are related to the equilibrium constant, Kp, via:

1

1 1

1

( )( ) ( )

( )

f

c p

r

k TK T K T

k T

Thus, to find k1r, we need to evaluate k1f and K1p at 2100 K. K1p can be evaluated as:

1 exp ,o

Tp

u

GK

R T

where

22100K ,N ,NO ,O ,N2100K

339,735 63,782 115,023 0

288,494 kJ/kmol

o o o o o

f f f fG g g g g

8

1

288,494exp 6.679 10

8.315(2100)pK

The forward rate coefficient of the first reaction at 2100 K is

14

1

6 3 3 3

38,3701.82 10 exp

2100

2.113 10 cm /gmol-s 2.113 10 m / kmol-s

fk

So,

31 10 3

1 8

1

2.113 103.164 10 m / kmol-s

6.679 10

f

r

p

kk

K

c) The forward rate coefficient of the second reaction at 2100 K is

10

2

12 3 9 3

4,6801.8 10 (2100)exp

2100

4.07 10 cm /gmol-s 4.07 10 m / kmol-s

fk

With the steady-state approximation,

2

2

1 N O1 2

2 2 2 O

3 5

9 3

11 3

[N ][O][N]

[O ]

2.11 10 (7.6 10 )(0.726) 0.167

4.07 10 3.025 10 28.778

5.487 10 kmol/m

ff

ss

f f

kk

k k MW

11 3

N,

28.778[N] 5.487 10 9.45 10 ppm

0.167SS

MW

4

d) The forward and reverse reaction rate of the first reaction is

2

2

1 2 1 N O

2

3 5

6 3

[N ][O]

0.167(2.113 10 )(7.6 10 )(0.726)

28.778

3.926 10 kmol/m -s

f fk kMW

1 1 NO

10 11 6

7 3

[NO][N] [N]

0.167(3.164 10 )(5.487 10 )(50 10 )

28.778

5.037 10 kmol/m -s

r rk kMW

Thus, the ratio of the forward reaction rate versus the reverse reaction rate is

61 2

7

1

[N ][O] 3.926 107.79

[NO][N] 5.037 10

f

r

k

k

3. A designer of a power plant wonders how long it will take for NO in the post flame region to reach

100 ppm. To simplify, assume that the post flame region has a temperature of 1700K, a pressure of

1atm and a gas composition of the major products of methane/air combustion (50% excess air).

a) What is the mole fraction of oxygen (O2) in the post flame region?

b) Calculate this time.

Solution:

a) For 50% excess air, the methane/air combustion equation can be written as

4 2 2 2 2 2 2CH 3(O 3.76N ) CO 2H O O 11.28N

Thus, the mole fraction of oxygen in the products is

2

2

O

O

mix

16.54%

1 2 1 11.28

N

N

Similarly,

2

2

N

N

mix

11.2873.8%

1 2 1 11.28

N

N

2N 73.8% .

b) d[NO]/dt can be evaluated from

1 2

d[NO]2 [O][N ]

dfk

t

5

Converting the mole fraction to molar concentrations

NO

4

7 3 10 3

[NO]

101,3251.0 10

8315(1700)

7.17 10 kmol/m 7.17 10 gmol/cm

u

P

R T

22 N

3 3 6 3

[N ]

101,3250.738

8315(1700)

5.29 10 kmol/m 5.29 10 gmol/cm

u

P

R T

The concentration of O can be evaluated through equilibrium approximation,

2O 2O

The concentration of O2 is

22 O

2 4 3

[O ]

101,3256.54 10 4.69 10 kmol/m

8315(1700)

u

P

R T

The equilibrium constant is

2,O ,O1700K

(2) (1)

(2)(141702) (1)(0)

283404 kJ/kmol

o o o

T f fT

G g g

9G 283,404exp exp 1.96 10

8.315(1700)

o

Tp

u

KR T

Thus,

1/2 1/29

4

2

8 3 11 3

1.96 10 101325[O] [O ] 4.69 10

8315 1700

8.12 10 kmol/m 8.12 10 gmol/cm

o

p

u

K P

R T

The forward rate coefficient at 1700 K is

14

1

4 3

38,3701.82 10 exp

1700

2.869 10 cm /gmol-s

fk

6

The reaction rate of NO is

1 2

4 6 11

11 3

d[NO]2 [O][N ]

d

2(2.869 10 )(5.29 10 )(8.12 10 )

2.465 10 gmol/cm -s

fkt

So the time takes to form 100 ppm of NO is

10

11

7.17 1029.1s

2.465 10

4. Consider an adiabatic combustion of gasoline (i-C8H18) and air at Φ = 0.9 proportions. Assume no

dissociation and constant and equal specific heats for reactants and products of 38 kJ/kmol-K. Assume

a fuel heating value of 5×106 kJ/kmol. The air and fuel are initially at 300 K and atmospheric pressure.

a) If 20% (by moles) of the reactant mixture is replaced by 20% (by moles) of the exhaust gas

(which is about 800 K) to reduce the exhaust temperature, by how would the NO rate of

formation change {i.e., (d[NO]/dt)2/(d[NO]/dt)1}?

Note that this replacement does not change the exhaust gas species concentrations. Assume the

exhaust gases are isothermal. Assume the NO levels are much smaller than equilibrium levels. The

system is at 1 atm. A diagram is shown below to help clarify the combustor set up.

Solution:

The combustion equation can be written as

8 18 2 2 2 2 2 2

12.5C H (O 3.76N ) 8CO 9H O 1.389O 52.22N

0.9

No EGR:

, ,1

prod

( 300)c i p i fh N c T

7

,1

prod

6

300

5 10300 2163.5 K

38 (70.61)

c

f

p i

hT

c N

With EGR:

EGR

prod

0.2 iN N

, EGR , ,2

prod prod

, EGR

prod

,2

prod

6

0.8 0.2 ( 300) ( 300)

0.8 0.2 ( 300)

300

0.8 (5 10 ) 0.2 (38) (70.61) (800 300)300

38 (70.61)

1890.8 K

c i p i i p i f

c i p i

f

p i

h N c T N c T

h N c T

Tc N

The ratio of the NO formation rate is

1 ,22 2 2 2

1 1 ,1 2 1 1

( )(d[NO] / d ) [N ] [O]

(d[NO] / d ) ( ) [N ] [O]

f f

f f

k Tt

t k T

The molar concentration of N2 and the rate coefficient can be evaluated from

22 N[N ]u

P

R T

14

1 ( ) 1.82 10 exp[ 38,370 / ]fk T T

The molar concentration of O2 can be evaluated through equilibrium approximation,

2O 2O

2

1/2

O O[O] p

u u

P PK

R T R T

Thus,

1/2

,2 ,2 ,222

1 ,1 ,1 1 ,1

exp[ 38,370 / ] 1 / 1 /exp( G / )(d[NO] / d )

(d[NO] / d ) exp[ 38,370 / ] 1 / exp( G / ) 1/

exp[ 38,370 /1890.8] 1/1890.8 exp( 256768 / 8.315(18

exp[ 38,370 / 2163.5] 1/ 2163.5

of f fT u

o

f f T u f

T T TR Tt

t T T R T T

1/2

90.8)) 1/1890.8

exp( 223352 / 8.315(2163.5)) 1/ 2163.5

0.07747 (1.3092) (0.1411)

0.01431

8

Emissions

5. Using textbook chapters 5 and 15 if needed, explain the differences among the nitric oxide formation

mechanisms denoted:

Simple thermal (Zeldovich) mechanism

Superequilibrium mechanism

Promt Fenimore mechanism

Solution:

Mechanism 1: Thermal or Zeldovich mechanism

The Thermal mechanism consists of the following two reaction steps and is important at high

temperatures (above approx. 1800 K). Comparatively, the Thermal mechanism is rather slow and is

generally considered to be important in the post-flame gases, although superequilibrium o-atom

concentrations in the flame zone may result in rapid NO formation rates.

2O + N NO + N

2N + O NO + O

Mechanism 2: Prompt superequilibrium mechanism

As mentioned above, superequilibrium O and OH radicals result in the rapid formation of NO through the

thermal mechanism reactions in the flame zone. Thus, the superequilibrium mechanism is closely related

to the Thermal mechanism.

Mechanism 3: Prompt Fenimore mechanism

The prompt mechanism for NO formation is associated with the hydrocarbon oxidation process.

Hydrocarbon radicals react with molecular N2 to form amines and cyano compounds plus a free N radical

which reacts with OH to form NO. The prompt Fenimore reaction results in rapid NO formation in flame

zones and is faster than the Thermal mechanism.

2CH + N HCN + N

2C + N CN + N

HCN + O NCO + H

NCO + H NH + CO

2NH + H N + H

N + OH NO + H

6. Using the CHEMKIN equilibrium code to determine the equilibrium composition of the products of

CH4-air combustion at phi = 1 (considering dissociation) for the following conditions: (i) T = 2000 K

and P = 10 atm, (ii) T = 925 K and P = 1 atm.

a) Calculate the CO mole fraction at both conditions.

9

b) Assuming condition i approximates a point early in the SI engine expansion process and

condition ii approximates conditions in the exhaust pipe, what are the implications of the

calculations you made in part a.

Solution:

a) (i): 3

CO,1 1.4448 10 (ii): 9

CO,1 3.9511 10

b) We see that the CO mole fractions reduce greatly as the gases expand and move from the inside of the

SI engine cylinder into the exhaust pipe. However, due to the expansion being slow and chemical reaction

rates reducing with lower temperatures, the mixture cannot fully equilibrate, thus the actual CO emissions

will be somewhere between i) and ii).

Premixed Flames

7. Using the single-step reaction rate data for methane, how would the laminar flame velocity change

with pressure? How does this compare to the other fuels?

Solution:

The pressure dependence of SL is

( 2)/2n

LS P

10

where n is the reaction order. Reaction order for CH4 is m + n = 1.3 + (–0.3) = 1.0 (see Table 5.1).

Therefore,0.5

,methane .LS P The experimental results of Andrews and Bradley verifies that:

0.5

,methane 43 .LS P

Other fuels in Table 5.1 have reaction orders close to 2 and therefore, the effect of pressure on their

laminar flame speed is small. For example:

0.125

ethane ,ethane1.65 0.1 1.75 Ln S P

0.125

octane ,octane1.5 0.25 1.75 Ln S P

0.125

ethanol ,ethanol1.6 0.15 1.75 Ln S P

8. For a laminar premixed flame burning a stoichiometric gaseous ethane (C2H6)-air mixture with an inlet

reactant temperature of 300K, a pressure of 1 atm and adiabatic flame temperature given in Table B.1.

Assume complete combustion and use the fluid properties of air.

a) Estimate the mean reaction rate using the global one-step reaction rate.

b) Estimate the laminar flame speed.

c) Estimate the flame thickness.

d) Estimate the characteristic time of the flame.

e) How does the flame speed compare to the values of propane and acetylene as discussed in class?

f) Using CHEMKIN, calculate the laminar flame speed and flame thickness. How does it compare

to the calculations on part b & c.

Solution:

a) For 1 mole of C2H6, the combustion equation at stoichiometric condition can be written as

2 6 2 2 2 2 2C H 3.5(O +3.76N ) 2CO 3H O 13.16N

So the initial mole fraction of C2H6 and O2 is

2 6C H ,i

10.05662

1 3.5(4.76)

,

2O ,i

3.50.1982

1 3.5(4.76)

The mean reaction rate using the global one-step reaction rate

0.1 1.652 62 6 2

d[C H ]exp( )[C H ] [O ]

d

a

u

EA

t R T

A mean temperature is chosen to evaluate the reaction rate as

1 1( ( ) ) 1769 K

2 2r u b bT T T T

11

The mole concentration used to evaluate the reaction rate is

2 6

4 3 7 3

(0 0.05662) (0 0.05662) 101325[C H ]

2 2 8315 1769

1.950 10 kmol/m 1.950 10 gmol/cm

u r

P

R T

2

4 3 7 3

(0 0.1982) (0 0.1982) 101325[O ]

2 2 8315 1769

6.826 10 kmol/m 6.826 10 gmol/cm

u r

P

R T

So the mean reaction rate is

0.1 1.652 62 6 2

12 7 0.1 7 1.65

3 3

d[C H ]exp( )[C H ] [O ]

d

150981.1 10 exp( )(1.950 10 ) (6.826 10 )

1769

3.091 10 gmol/s-cm

a

u r

EA

t R T

b) For air, 31.1614kg/mu at 300KT

The mean temperature over the entire flame thickness is

1( ) 1279.5 K

2u bT T T

For air at 1279.5 KT

0.0808 W/m-Kk , =1186 J/kg-Kpc

So

5 20.08085.866 10 m /s

1.1614(1186)u p

k

c

2 2

2 6

O N

C H

3.5MW 13.16MW16.016

1MW

A

F

m

m

392.73 kg/s-mF Fm MW

Thus,

1/2 1/252(1 ) 2(1 16.016)(5.866 10 )

(92.73) 0.40 m/s1.1614

L F

u

S m

c) The flame thickness

52 2 5.866 100.293 mm

0.40LS

12

d) The characteristic time of the flame

c

0.0002930.00073 s 0.73 ms

0.40LS

e) Flame speed of ethane is less than the values of propane and acetylene.

f) Using CHEMKIN, the laminar flame speed is 0.411 m/s and the flame thickness is around 0.3 mm.

The CHEMKIN results are very similar to the calculations on part b & c. The difference is mainly due to

the global one-step reaction assumption.