1 CE 583 – Control of Nitrogen Oxides Jeff Kuo, Ph.D., P.E.

2 Content Overview of the Nitrogen Oxides Problems Comparison with sulfur oxides Reaction in the atmosphere NO and NO 2 equilibrium Thermal, prompt, and fuel NO x Thermal NO Prompt NO Fuel NO Non-combustion Sources of Nitrogen Oxides (~3%) Control of Nitrogen Oxide Emission By combustion modification By post-flame treatment

3 Overview of Nitrogen Oxides Problem Oxides of Nitrogen NO, nitric oxide NO 2, nitrogen dioxide N 2 O, nitrous oxide, a greenhouse gas NO 3 N 2 O 3, N 2 O 4, N 2 O 5 NO x [VOC] [NO x ] [O 3 ] = 0.30 ppm [O 3 ] = 0.20 ppm [O 3 ] = 0.10 ppm [O 3 ] = 0.05 ppm Denver Isopleth

4 Overview of Nitrogen Oxides Problem NO x Emission Vehicles: ~50% Fuel combustion: 1/3 Others: IC engines Residential Combustion Industrial/agricultural

5 Overview of Nitrogen Oxides Problem - Similarities with SO x Both SO x and NO x react with oxygen and water to form acids (sulfuric and nitric). both are the principal components to acid rain. both are the primarily removed by rain. Both SO x and NO x emissions lead to significant PM 10 and PM 2.5 formation in urban area. Both are regulated pollutants (NAAQS) and cause respiratory problems at high doses. NAAQS for NO 2 = ppmV (100  g/m 3 ) – annual average. Both are emitted by large combustion sources.

6 Overview of Nitrogen Oxides Problem - Differences from SO x Motor vehicles are the primary emission source of NO x, but play a minor role in SO 2 emissions. If we remove S in fuel, we can eliminate SO 2 emissions, while removal of N in fuel would only result in 10-20% reduction in NO x emissions. NO x emissions can be greatly altered by changing the combustion conditions (time, temperature, and oxygen content). Not true for SO 2 emissions.

7 Overview of Nitrogen Oxides Problem - Differences from SO x Ultimate fate of SO x  CaSO 4  2H 2 O  landfill. But no comparable cheap insoluble salt of nitric acid. Much harder to get NO into water than SO 2. NO must undergo two steps to form an acid: NO + ½ O 2  NO 2 (slow) 3 NO 2 + H 2 O  2 HNO 3 + NO

8 Overview of Nitrogen Oxides Problem - Reaction in the atmosphere NO is colorless, not as harmful as NO 2 (brown gas). Principal concern of NO x : ozone formation. NO + HC + O 2 + sunlight  NO 2 + O 3 VOC needs high T, t, and turbulence; NO x control is opposite.

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10 Overview of Nitrogen Oxides Problem - NO and NO 2 equilibrium If the atmosphere were in equilibrium, it would have less than 1 ppt of NO and NO 2. Equilibrium concentration of NO increases dramatically with increasing T (begin at ~ 2000 – 2500 o F). NO O 2 NO 2 K 2 = [NO 2 ] [NO][O 2 ] 1/2 N 2 + O 2 2NO K 1 = [NO] 2 [N 2 ][O 2 ]

11 Overview of Nitrogen Oxides Problem - NO and NO 2 equilibrium At low T, the equilibrium concentration of NO 2 is much higher than that of NO, whereas the reverse if true at high T (due to the different natures of the two K’s). Lightning are a major global source of NO x, but combustion is the main NO x source in populated area. NO O 2 NO 2 K 2 = [NO 2 ] [NO][O 2 ] 1/2 N 2 + O 2 2NO K 1 = [NO] 2 [N 2 ][O 2 ]

12 Overview of Nitrogen Oxides Problem - Thermal, prompt, and fuel NO x Thermal NO x : formed by simple heating of O 2 and N 2 (most significant at high T, but negligible below 1300 o C) Prompt NO x : O 2 + N 2 + active HC species derived from fuel in fuel-rich part of the flame (such as CH 3 ). Fuel NO x : formed by conversion of some N originally present in the fuel to NO x.

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14 Thermal NO – Zeldovich Kinetics Thermal NO is the most important of the three at high T. If air is heated to 2000K, NO = 8100 ppm at equilibrium. The intermediate steps involve highly energetic species, free radicals such as O, N, OH, H, CH 3, CH 2. Zeldovich mechanism: most quoted kinetic equations. Used to predict NO concentration at different T & t.

15 Thermal NO – Zeldovich Kinetics N 2 + O NO + N k +1 k -1 N + O 2 NO + O k -2 k +2 N + OH NO + H k -3 k +3 r NO = k +1 [N 2 ][O] – k -1 [NO][N] + k +2 [N][O 2 ] – k -2 [NO][O] + k +3 [N][OH] – k -3 [NO][H]

16 Thermal NO – Heating and cooling times How much thermal NO is formed and the extent converted back to N 2 and O 2 as the gases cool depend greatly on how fast gases heat and cool in flames. For a peak T of 2400 o F and combustion time of s, the rate can be estimated as 5 x 10 5 o C/s (huge!) T-t curve is often used with Zeldovich mechanism to calculate the NO concentration. [NO] = 5.2x10 17 e -72,300/T y N 2 y O 2 t McKinnon Global Model

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18 Prompt NO During the first part of combustion, carbon-bearing radicals from the fuel react with nitrogen by CH  + N 2  HCN + N  The N  produced attack O 2 to increase the amount of NO; HCN partly reacts with O 2 to form NO and partly with NO to produce N 2. Production of prompt NO is only weakly dependent on T and average about 30 ng/J (Fig 12-4).

19 Fuel NO N in fuel  HCN  NH, NH 2 NO/O 2 ratio determines fate of fuel N Keep [O 2 ] low in in high temperature zones (fuel oil contains % N and coal % N) Fraction of fuel nitrogen that appears as NO x in the exhaust gas is estimated to be 20 – 50%, depending on furnace conditions and chemical nature of N in the fuel. O2O2 NO NO + H 2 ON 2 + H 2 O

20 NO x Control Options Three possible approaches Modify the combustion process Treat flue gas to convert NO x to N 2 Scrubbing with NaOH and KMnO 4 (for small industrial sources only – nitric acid plants) Combustion modification Reduce temperature in flame Reduce gas residence time in flame Reduce oxygen concentration in flame Flue Gas Treatment Catalyzed or non-catalyzed reduction Adsorption - experimental

21 NO x Control Options – Combustion modification Recall, r NO initial = k[N 2 ][O 2 ] 1/2 k = k 0 exp (- E/RT), k 1600 K /k 1000 K ~ 10 6 Minimize time spent at high T and high oxygen C. 2-staged combustion (reburning): mix part of combustion air w/ the fuel, then add remaining air to finish combustion. In the first stage, the max T is lowered because not all the fuel is burned, and the max T is reached when all oxygen is used up  not enough oxygen to form NO.

22 NO x Control Options – Combustion modification In stage 2, enough of heat has released in Stage 1, now the T is low enough even with excess oxygen. In some schemes a small amount of additional fuel such as methane is added for the Stage 2. It is cheap, but needs a larger firebox and incomplete burning in Stage 2.

23 NO x Control Options – Combustion modification Low-NOx burners (LNB): The incoming gas is thoroughly mixed w/ air and recycled combustion products (Flue Gas Recirculation, FGR) with ~15% excess air  prevents any fuel-rich part of the flame, in which prompt NO can form. The gas-air-FGR mixture flows outward around an internal recirculation zone (The zone stabilizes the flame). The NO x is reduced to 10 ppm from ~ 100 ppm.

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25 NO x Control Options – Post-flame treatment Add reducing agents (CO, CH 4, HC, NH 3 ) to combustion gas stream to take oxygen away from NO. In auto exhaust:

26 NO x Control Options – Post-flame treatment For large furnaces, o F) – Selective Catalytic Reduction (SCR)

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28 NO x Control Options – Post-flame treatment Or directly in gas o F – Selective Noncatalytic Reduction (SNCR) At > 1800 o F, the reaction below dominants (no good!)

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30 Summary At high T thermal NO forms quickly, but not long enough to reach equilibrium. The hot combustion gases are cooled so rapidly that the flames does not have time to revert to N 2 and O 2. Prompt NO is the major NO X for kitchen appliances (low T). 10 – 50% N in fuels is converted to NO x when fuel is burned. Most of NO x in flames is NO (NO:NO 2 = 90%: 10%).

31 Summary NO reacts rapidly with O 3 to form NO 2. Under sunlight, NO, NO 2 and O 3 are in equilibrium. HC forces it in the direction of NO 2 and O 3. NO 2 react with water to produce nitric acid  PM 10, PM 2.5, and acid rain problems. NO formation can be reduced by minimizing peak T, oxygen or time at peak T. NO in flue gas can be converted back to N 2 by a reducing agent: CO in IC and NH 3 in power plant. The reduction can be thermal or catalytic.

32 Sample Question – PE Exam A preliminary dispersion modeling analysis of NO x emission from gas-fired engines at a pipeline compressor station reveals that the ambient air exceedences occur outside the station’s property line. To obtain an air quality permit for the station, the analysis must show no modeled exceedences. What would be the best alternative? (a) Increase the stack height (b) Increase engine exhaust stack diameter (c) Reduce air/fuel ratio of engines (d) Quench the discharge

33 Sample Question – PE Exam A fan turning at 90 rpm supplies 725 cfm of air. Determine the fan speed (rpm) required to move 1,000 cfm of air. N 2 = N 1 (Q 2 /Q 1 ) = 950(1,000/725) = 1,310 rpm

34 Sample Question – PE Exam The following information applies to the reaction of methane: CH 4 + 2O 2  CO 2 + 2H 2 O What is the total heat of reaction (J/mol)? = (-394,088) + 2(-242,174)-(-74,980) = 803,456

35 Sample Question – PE Exam The EPA NSPS for a large utility boiler is 0.2 lb NO x (as NO 2 ) per 10 6 Btu heat input. The boiler burns natural gas. Determine is the boiler is in compliance. Natural gas fuel rate = x 10 3 scfh Average NO x concentration = ppmV Stack velocity = fps Stack diameter = 92 inch Stack pressure = in-Hg Stack temperature = 355 o F Moisture in stack gas = 12.25% 1 ppm NO 2 = x lb NO 2 /dscf stack gas (how?) Gas Btu value = 1,000 Btu/scf

36 Sample Question – PE Exam