1. No category

Control of Nitrogen Oxides
朱信
Hsin Chu
Professor
Dept. of Environmental Engineering
National Cheng Kung University
1
1. An Overview of the Nitrogen Oxides
Problem


Most of the world’s nitrogen is in the atmosphere
as an inert gas.
In crustal rocks it is the 34th most abundant
element with an abundance of only ≈ 20 ppm.
2


Although nitrogen forms eight different oxides,
our principal air pollution interest is in the two
most common oxides, nitric oxide (NO) and
nitrogen dioxide (NO2).
In addition, we are beginning to be concerned
with nitrous oxide (N2O). It may be a significant
contributor to global warming and to the possible
destruction of the ozone layer.
3
1.1 Comparison with Sulfur Oxides

Fig. 12.1 (next slide) shows part of how nitrogen
moves in the environment, as a result of human
activities.
4

Nitrogen oxides are often lumped with sulfur
oxides as air pollution control problems because
of the similarities between the two:
(1) Nitrogen oxides and sulfur oxides react with
water and oxygen in the atmosphere to form
nitric sulfuric acids, respectively.
These two acids are the principal contributors
to acid rain.
6
(2) Both undergo atmospheric transformations
leading to or contributing to the formation of
PM10 and PM2.5 in urban areas.
(3) Both are released into the atmosphere in large
quantities.
(4) Both are released to the atmosphere by large
combustion sources, particularly coal
combustion sources. Table 12.1 (next slide)
shows the emission sources of NOX in the
U.S. in 1997.
7



We see that vehicles contribute almost half of the
total.
Natural gas appears in this table, but not in the
SO2 table.
The largest contributors to the “All other sources”
are nontransportation internal combustion engines
and residential combustion.
9
There are also major differences between nitrogen
oxides and sulfur oxides:
(1) Motor vehicles are the major emitter of nitrogen
oxides, but a very minor source of sulfur oxides.
(2) Sulfur oxides are formed from the sulfur
contaminants in fuels or the unwanted sulfur in
sulfide ores.
Although some of nitrogen oxides emitted to the
atmosphere are due to nitrogen contaminants in fuels,
most are not.

10
Most of nitrogen oxides are formed by the
reaction of atmospheric nitrogen with oxygen in
high-temperature flames.
(3)The formation of nitrogen oxides in flames can
be greatly reduced by manipulating the time,
temperature, and oxygen content of the flames.
No such reductions are possible with sulfur
oxides.

11
(4)The ultimate fate of sulfur oxides removed in
pollution control or fuel-cleaning process is to be
turned into CaSO4•2H2O, which is an innocuous,
low-solubility solid, and to be placed in landfills.
 There is no correspondingly cheap, innocuous,
and insoluble salt of nitric acid, so landfilling is
not a suitable fate for the nitrogen oxides.
12
The ultimate fate of nitrogen oxides that we wish
to keep out of the atmosphere is to be converted to
gaseous nitrogen and oxygen and be returned to the
atmosphere.
(5) It is relatively easy to remove SO2 from combustion
gases by dissolving SO2 in water and reacting it with
alkali.
 Collecting nitrogen oxides is not nearly as easy this way
because NO, the principal nitrogen oxide present in
combustion gas streams, has a very low solubility in
water.

13


NO must undergo a two-step process to form an acid:
NO + 0.5 O2 ↔ NO2
(1)
3 NO2 +H2O → 2HNO3 + NO
(2)
The first reaction is relatively slow.
It is fast enough in the atmosphere to lead to the formation
of acid precipitation in the several hours or days that the
polluted air travels before encountering precipitation, but
slow-enough that it does not remove significant quantities
of NO in the few seconds that a contaminated gas spends
in a wet limestone scrubber used for SO2 control.
14
1.2 Reactions in the Atmosphere


NO is a colorless gas that has some harmful
effects on health, but these effects are
substantially less than those of an equivalent
amount of NO2.
NO2, a brown gas, is a serious respiratory irritant.
15


Our principal concern with NOX is that nitrogen
oxides contribute to the formation of ozone, O3,
which is a strong respiratory irritant and one of
the principal constituents of urban summer eyeand nose-irritating smog.
The overall reaction is:
NO + HC + O2 + sunlight → NO2 + O3
16



NO, emitted during the morning commuter rush,
is oxidized in the atmosphere to NO2 over a
period of several hours.
The NO2 thus formed then reacts with HC to form
O3.
The O3 peak occurs after the NO2 peak.
17
1.3 NO and NO2 Equilibrium


The most important reactions for producing NO
and NO2 in flames are:
N2 + O2 ↔ 2NO
(3)
and Eq. (1).
For any chemical reaction at equilibrium, the
Gibbs free energy is at a minimum for the
reaction’s temperature and pressure.
18

From that condition it follows that
G o
ln K  
RT

(4)
where ∆Go = standard Gibbs free energy change
K
= equilibrium constant
R
= the universal constant
T
= absolute temperature (K or oR)
Using the published Go values for the reactions in Eqs. (1)
and (3) we may construct Table 12.2 (next slide).
19



Here we show Kp, the equilibrium constant based
on taking the standard states as perfect gases at 1
atm pressure, and expressing gas concentrations
as partial pressures.
Because the number of mols does not change in
Eq. (3), its Kp is dimensionless.
However, in Eq. (1) the number of mols decreases
by ½, so that Kp has the dimension (atm)-0.5.
21
Example1


Calculate the equilibrium concentrations of NO
and NO2 for air that is held at 2000 K = 3140oF
long enough to reach equilibrium.
Assume that the only reactions of interest are Eqs.
(1) and (3).
22

Solution:
The definitions of the two equilibrium constants are:

[ NO ]2
K3 
[O2 ][ N 2 ]
(5)
[ NO2 ]
K1 =
[ NO ][O2 ]1/ 2
(6)
Solving Eq. (3) for the equilibrium concentration of NO,
we find
[ NO]   K3[ N 2 ][O2 ]
1/ 2
23

Substituting values, including the value of K3 from Table
12.2, into this equation, we see
[ NO]   0.0004[0.78][0.21]
1/ 2

 0.0081  8100 ppm
Solving Eq. (6) for [NO2], we find
[ NO2 ]  K1[ NO][O2 ]
1/ 2
24

Substituting values into this equation, we see
[ NO2 ]  0.0035[0.0081][0.21]1/ 2  1.3 105  13 ppm

#
Following the same procedure as in Example 1, we may
make up Table 12.3 (next slide), which shows the
calculated equilibrium concentrations of NO and NO2 at
various temperatures.
25


The starting gas that is 78% nitrogen, 4% oxygen
is more representative of combustion gases in
which most of the oxygen has been consumed by
the combustion.
The values from Table 12.3 are shown in Figs.
12.2 (next slide) and 12.3 (second slide).
27

From Table 12.3 and Figs. 12.2 and 12.3 we see:
(1) If the atmosphere were at equilibrium (at a
temperature near 300 K = 80oF), it would
have less than a part per billion of NO or NO2.
• The concentrations of NO and NO2 observed
in cities in the world often exceed these
equilibrium values, so that equilibrium alone
is not a satisfactory guide to the presence of
NO and NO2 in the atmosphere.
30
(2)The equilibrium concentration of NO
increases dramatically with increasing
temperature.
The rapid increase begins at about
2000~2500oF (1367-1644 K).
(3)At low temperatures the equilibrium
concentration of NO2 is much higher than that
of NO, whereas at high temperatures the
reverse is true.
31
(4) We are likely to get to the temperatures where
NO and NO2 are formed from atmospheric N2
and O2 only in flames and in lightning strikes.
• Lightning strikes are a major global source of
NOX, but combustion in our vehicles and
factories is the main source of NOX in heavily
populated areas.
32
1.4 Thermal, Prompt, and Fuel NOX


Thermal nitrogen oxides, which are generally the
most significant, are formed by the simple heating
of oxygen and nitrogen, either in a flame or by
some other external heating.
Prompt refers to the nitrogen oxides that form
very quickly as a result of the interaction of
nitrogen and oxygen with some of the active
hydrocarbon species derived from the fuel in the
fuel-rich parts of flames.
33


Prompt NOX are not observed in flames of fuels
with no carbon, e.g., H2.
Fuel nitrogen oxides is formed by conversion of
some of the nitrogen originally present in the fuel
to NOX.
34


Coal and some high-boiling petroleum fuels
contain significant amounts of organic nitrogen;
low-boiling petroleum fuels and natural gas
contain practically none.
Fig. 12.4 (next slide) shows estimates of the
contribution of the thermal, fuel, and prompt
mechanisms to the NOX emissions from coal
combustion.
35
2. Thermal NO

From Fig. 12.4, the thermal mechanism is the most
important of the three ways of making NO at the highest
temperatures.
2.1 The Zeldovich Kinetics of Thermal NO
Formation


Example 1 showed that a specific equilibrium
concentration of NO could be reached at certain
temperature.
Here we inquire how rapidly the mixture approaches that
equilibrium value.
37



The reactions shown in Eqs. (1) and (3) do not
proceed as written in those equations.
Rather, they proceed by means of intermediate
steps involving highly energetic particles called
free radicals.
The free radicals most often involved in
combustion reactions are O, N, OH, H, and
hydrocarbons that have lost one or more
hydrogens, e. g., CH3 or CH2.
38


These materials are very reactive and energetic
and can exist in significant concentrations only at
high temperatures.
In principle they can be formed by equilibrium
reactions like the following:
N2 ↔ 2N
O2 ↔ 2O
H2O ↔ H + OH
39

However, the reactions are conventionally written
with an M on both sides of the equation to
indicate that another molecule, which is not
chemically changed in the reaction, must collide
with the N2 or O2 molecule to supply or remove
energy for the reaction to occur, e.g.,
N2 + M ↔ 2N +M
40



The concentration of M, which can be any other gas
molecule (e.g., O2 or N2 or H2O) does not influence the
equilibrium but does influence the rate of the reactions.
The most widely quoted mechanism for the thermal NO
formation reaction is that of Zeldovich.
It assumes that O radicals attack N2 molecules by
O + N2 ↔ NO + N
and that N radicals can form NO by
N + O2 ↔ NO + O
41

If one assumes that the O radicals are in equilibrium with
O2, that the concentration of N radicals is not changing
significantly with time, and that one term is small
compared with the others, one can simplify the resulting
kinetic equations to:
2
d [ NO]
[
NO
]
 k f [ N 2 ][O2 ]1/ 2  kb
dt
[O2 ]1/ 2
(7)
where kf and kb are the forward and backward reaction
rate constants.
42

Then one can observe that the equilibrium value of the NO
concentration, [NO]e, is given by Eq. (5), and that the
equilibrium constant is related to the two rate constants by
K3 

kf
kb
One may show this by setting the rate of change of
concentration in Eq. (7) to zero and comparing the result
to Eq. (5).
43

Making this substitution and rearranging in Eq. (7), one
finds:
kb
d [ NO]

dt
2
2
1/ 2
[ NO]e -[ NO] [O2 ]
(8)
which may be integrated and rearranged (with an
assumption of zero NO at time zero) to
[ NO] 1  exp( t )

[ NO]e 1  exp( t )
(9)
where
2[ NO]e kb

[O 2 ]1/2
(10)
44
Example 2

Estimate the concentration of NO in a sample
containing 78% N2 and 4% O2 that is hold for one
second at 2000 K = 3140oF, according to the
Zeldovich thermal mechanism.

Solution:
From Table 12.3, we know the equilibrium
concentration, [NO]e, is 3530 ppm.
45


Seinfeld suggests a value for kb of 4.1  1013 exp (91600/RT), for T in K, R in cal/mol/K, t in seconds, and
concentrations in (mol/cm3).
At 1.0 atm and 2000 K, the molar density of any perfect
gas is 6.1  10-6 mol/cm3, so that
6
8 mol
[ NO]e  6.110  0.003530  2.15 10
cm3
mol
[O2 ]  6.1106  0.04  2.44 10 7
cm3
46

kb  4.1  10
13
 -91600 
 mol 
exp 
  4003  3 
 1.987  2000 
 cm 
1 2
1
 
s
2  2.15 10-8  4003 0.349


-7 1/ 2
(2.44 10 )
s

and
[ NO] 1  exp[(0.349 / s)(1s)]

 0.173
[ NO]e 1  exp[(0.349 / s)(1s)

Therefore,
[ NO]  3530 ppm  0.173  610 ppm #
47


This is a simplified version of the Zelodvich
simplification of the kinetics of the thermal NO
formation reaction.
Using this simple relation, one may readily make
up plot like Fig. 12.5 (next slide), which show the
expected time-temperature relation for one
specific starting gas composition.
48


More complex versions of the Zeldovich
mechanism add another equation to the reaction
list,
N + OH ↔ NO + H
with a resulting increase in mathematical
complexity.
The result cannot be reduced to any simple plot
like Fig. 12.5.
50


The Zeldovich mechanism shown here makes
reasonably accurate predictions of the rate of
formation of NO from N2 and O2 in the highest
temperatures observed in flames, and in other
high-temperature situations.
It predicts much lower NO concentrations than
those observed in low-temperature flames.
51

Before we discuss the prompt and fuel NO
mechanisms that predominate at low temperatures,
we explore the temperature and times in flames,
to help understand what kinds of flames produce
significant thermal NO and what kinds do not.
52
2.2 Heating and Cooling Times


How much thermal NO is formed in a flame as
well as how much is then converted back to N2
and O2 as the gases cool is a strong function of
how fast the gases heat and cool in the flames.
Fig. 12.6 (next slide) shows some possible
heating and cooling patterns for flames.
53



The square wave of Fig. 12.6a leads to the easiest
calculations, as shown in Example 2.
That example implicitly assumed that cooling
after the flame is instantaneous, so that none of
the NO formed in the flame is converted back to
N2 and O2.
Nothing in this world is truly instantaneous!
55
Example 3



Estimate the heating and cooling rates in the flame of an
ordinary gas stove.
The peak flame temperature is about 2400oF and the
combustion time about 0.005 s.
Solution:
If we assume that the temperature-time pattern is the
triangle wave shown in Fig. 12.6b, we can say that the
heating and cooling rates should be about the same, and
o
o
dT T 2400o F  68o F
5 F
5 C


 9.3 10
 5.2 10
dt
t
0.0025 s
s
s
56
#


The assumption of a symmetrical triangle temperature-time pattern is
closer to reality than the square wave and still leads to relatively simple
calculations.
If we take the viewpoint of a person riding with a parcel of gas through
the flame (The Lagrangian view), we can say that:
V  CP
dT
 heat added by combustion- heat lost to surroundings
dt
=rhcombustion  UA(T flame  Tsurroundings )
(11)
where r is the rate of combustion and UA the product of the overall heat
transfer coefficient and the applicable surface area.
57


For the rising part of T-t pattern on Fig. 12.6c, the
observed behavior is that the rate starts slowly,
increases as the the temperature rises, and then
slows to a stop as the last of the fuel is consumed.
After the peak is reached, the decline must be
steeper than the rise was because the rise was
proportional to the difference between the two
terms on the right of Eq. (11) but the decline is
proportional to the cooling term alone.
58
Example 4

Repeat Example 2 for the strong simplifying
assumptions that
(1) the gas heats from 293 to 2000 K in 0.5 s with
a linear temperature increase, then cools from
2000 to 293 K in 0.5 s with a linear
temperature decrease (Fig. 12.6b) and
(2) the initial NO concentration is zero.
59

Solution:
Here we use the simple zeldovich mechanism and Eq. (8),
which we know to be only approximate.

T  293K 
1707 K
K
t  293K  3414 t
0.5s
s
for 0 < t < 0.5 s
and
1707 K
(t  0.5s )
0.5s
K
 2000 K  3414 (t  0.5s) for 0.5 s < t < 1 s
s
T  2000 K 
60



If we consider the time interval 0.0 ≤ t ≤ 0.01 s, we
see that the initial temperature is 293 K and the final
temperature is 293 + 34 = 327 K.
For this time interval the average temperature is 310
K.
Using that value, we may estimate the equilibrium
constant as
43400 

30
K P  21.9 exp  
  5.5 10
 1.987  310 
61


For simplicity we will proceed as if the oxygen content
were the 4% in that example, independent of time.
Substituting this value of Kp in Eq. (5) and solving, we
find
[ NO]e  (5.5 1030 [0.78][0.04]) 0.5  4 10 16

At this temperature and one atmosphere pressure, the
molar density is (4.46  10-5) (273 / 310) = 3.93  10-5
mol/cm3, so that in concentration units [NO]e = 4 10-16
(3.93  10-5) = 1.6  10-20 mol/cm3.
62

At this temperature, using the equation from Example 2,
we have
91600 

51
kb  4.110 exp  

1.1

10

1.987

310


13

d [ NO] kb ([ NO]e2  [ NO]2 ) 1.11051 ([1.6 1020 ]2  [0]2 )


1/ 2
dt
[O2 ]
[0.04  3.93 105 ]1/ 2
 2.2 1088
mol / cm3
s
63


Multiplying this by ∆t = 0.01 s, we conclude that
during the first 0.01 s the concentration of NO
went from zero to 2  10-90 mol/cm3.
This value is negligible, but we now proceed for
the remaining 99 time steps, each of the same size,
using a spread-sheet (e.g., Excel), and find the
results shown in Fig. 12.7 (next slide). The final
calculated NO concentration is 20 ppm. #
64


From Fig. 12.7, we see that the calculated
equilibrium concentration goes from 5.5 10-24
ppm (≈ 0) to the 3530 ppm in Table 12.3, as the
temperature rises, and then back to zero as the
temperature falls.
The reaction rate is practically zero, and hence the
amount of NO actually formed is practically zero
until about 0.44 s (T ≈ 1780 K), after which the
rate of NO formation becomes significant.
66


By the time the Equilibrium and calculated curves
cross, the temperature is low enough that the
reaction rate is practically zero, so that the NO
concentration remains practically constant during
the further cooling.
One speaks of the concentration being “frozen” at
its higher temperature value.
67
3. Prompt NO


During the first part of combustion, the carbonbearing radicals from the fuel react with nitrogen
by:
CH + N2 ↔ HCN + N
and several similar reactions involving the CH2
and C radicals.
The N thus produced attacks O2 to increase the
amount of NO formed; and the HCN partly reacts
with O2, producing NO, partly with NO,
producing N2.
68


There does not appear to be any simple theoretical
treatment like the Zeldovich mechanism.
But it appears certain that the amounts of NO
found in low-temperature flames, which are much
larger than one would predict from the Zeldovich
thermal mechanism, are actually mostly prompt
NO.
69
4. Fuel NO



Some coals as well as the hydrocarbons derived
from oil shale can contain up to 2% N.
Most of the fuel nitrogen is converted in the flame
to HCN, which then converts to NH or NH2.
The NH and NH2 can react with oxygen to
produce NO +H2O, or they can react with NO to
produce N2 and H2O.
70


Thus, the fraction of the fuel nitrogen that leaves
the flame as NO is dependent on the NO/O2 ratio
in the flame zone.
Keeping the oxygen content of the gases in the
high temperature part of the flame low
significantly lowers the fraction of the fuel
nitrogen converted to NO.
71


The fraction of the fuel nitrogen that appears as
NOX in the exhaust gas is estimated to be
typically 20% to 50%, depending on furnace
conditions and, to some extent, the chemical
nature of the N in the fuel.
Fig. 12.8 (next slide) shows one schematic view
of the interactions of the there mechanisms of NO
formation.
72
5. Noncombustion Sources of Nitrogen
Oxides


The production and utilization of nitric acid lead
to emissions of NO and NO2, as do some other
industrial and agricultural processes, e.g., silage
production.
Table 12.1 shows that such sources account for
about 3% of NOX emissions in the U.S.
74
6. Control of Nitrogen Oxide Emissions

There are two possible approaches to controlling
NOX in combustion gases:
(1) Modify the combustion processes to prevent
the formation of NOX.
(2) Treat the combustion gas chemically, after the
flame, to convert the NOX to N2.
75



For smaller industrial sources, like nitric acid
plants, other control techniques are used, e.g.,
scrubbing with solutions of NaOH and KMnO4.
The strongly oxidizing KMnO4 quickly converts
NO to NO2, allowing it to be captured by the
alkaline NaOH.
This process requires an expensive
electrochemical regeneration of the KMnO4; it is
not suitable for large-scale combustion sources.
76
6.1 Nitrogen Oxide Control by
Combustion Modification


The formation of thermal NO is increased by
increasing any of the following three variables:
peak temperature, time at high temperature,
oxygen content at the high temperature.
Two-stage combustion or reburning as shown in
Fig. 12.6d is a good way for combustion
modification.
77


In the first stage, the maximum temperature is
lowered because not all the fuel is burned, and the
maximum temperature is reached when all the
oxygen has been used up, so that there is not
enough oxygen to form NO.
In the second stage, enough of the heat released in
the first stage has been removed that the
maximum temperature reached-in the presence of
excess oxygen-is low enough that the formation
of NO is small.
78



The advantage of this approach is that it is cheap.
The disadvantages are that it requires a larger fire
box for the same combustion rate and that it is
difficult to get complete burning of the fuel in the
second stage.
Fig. 12.9 (next slide) shows an alternative
approach to designing a low-NOX burner.
79



In normal operation the incoming gas is thoroughly mixed
with air and recycled combustion products (Flue Gas
Recirculation, FGR), with about 15% excess air.
This thorough mixing, together with excess air, prevents
there ever being any fuel-rich part of the flame, in which
prompt NO can form.
Fig. 12.10 (next slide) show some test results for this
burner, burning natural gas with 15% excess air and
varing degrees of air preheat and FGR.
81



The dilution with recycled flue gas simply lowers
the temperature at every part of the flame.
As the figure shows, it is possible to reduce the
NOX to 10 ppm if one is willing to bear the extra
cost of pumping 20 to 40% of the flue gas back
into the burner.
However, compared to other methods of
achieving 10 ppm–i.e., catalytic treatment of the
flue gas-the cost of FGR is very reasonable.
83



This burner is a moderately low-NOX burner
when it is burning fuel oil.
Ultra-low NOX levels are not achieved since most
fuel oils contain bound nitrogen that is oxidized in
the flames to NOX.
Also, because there will be a fuel-rich region at
the surface of evaporating oil droplets, there will
be prompt NO formation.
84
6.2 Nitrogen Oxide Control by
Postflame Treatment


Most post flame treatment processes add a
reducing agent to the combustion gas stream to
take oxygen away from NO.
Almost any gaseous reducing agent can be used,
e.g., CO, CH4, other HCs, NH3, various
derivatives of NH3.
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In modern auto engines the reaction is
platinum rhodium catalyst
2 NO  2 CO 
 N2  2CO2
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
For power plants and other large furnaces a variety of
reducing agents are used, of which the most popular is
ammonia.
The desired reaction is:
6 NO + 4 NH3 → 5 N2 + 6 H2O
(12)
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However, there is always some oxygen present,
which can lead to reactions like:
4 NO + 4 NH3 + O2 → 4 N2 + 6 H2O
(13)
is which one ammonia molecule is required for
each molecule of NO.
The NO2 is reduced by:
2 NO2 + 4 NH3 + O2 → 3 N2 + 6 H2O
(14)
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
Reactions (12) – (14) can be carried out over a variety of
catalysts, of which zeolites and supported TiO2 are the
most popular, at about 700oF (370oC), or simply in the gas
stream in the part of the furnace where the temperature is
between 1600o and 1800oF (870o – 980oC).
Below 1600oF the reaction rate is too slow. Above 1800oF
the dominant reaction becomes
NH3 + O2 → NO + 3/2 H2O
which increases rather than decreases the NO content of
the gas.
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
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The catalytic processes are called selective
Catalytic Reduction, or SCR.
The high-temperature ones, without catalysts, are
called Selective Noncatalytic Reduction, SNCR.
There are experimental processes under
development that adsorb both SO2 and NO onto
an activated carbon adsorbent.
They are then stripped off one at a time and
treated, making S and N2 the final products.
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