Question

How to calculate theoritical NOx and SO2 (mg/Nm3) in gases at coal fired power plant, if known :

Coal Calorific Value = 4500 kcal/kg
Fuel consumption = 180 t/h
Excess air = 18%
Oxygen in exhaust gases = 5%
Flue Gas Temperature in exhaust gases = 135 deg C

Carbon Content = 30.110
Hydrogen Content = 3.610
Oxygen Content = 9.850
Sulfur Content = 0.330
Nitrogen Content = 0.700
Ash Content = 5.740
Moisture Content = 32.530

Thank you!

Answer 1

Theoritical No. and so (mg/Nm') in gases in wal fired power plant. given: tetonnes : fuel consumption 180 t/hr 180x10° leg Ihr excess air - 18%. or in the gas -5%. gases = 1358 in exhaust Temperature carbon content = 30.110 wty. Montent = 3.610 wt % o content : 9.050 wty. s content : 0.330 wty Ni wntent = 0.700 wty. moisture content= 32.530 wty.

(Aus) The ne jure flue gas gas will contain CO, CO2, 4,0, 50,, Nz., NOXexcess Q (54.) reactions are stO₂- SO2 C+ to. - 2H+ + O HO c+ O₂ Wr. Nota: Nox formation is complex, however in fuel beomes Nox of N₂ we assume 20% so, calculation : giris Imole so .. s - Imole s (kmol) where i so, formed : 32.06 wal sulphur in s= mass of sulphur 32.06- mol.wt of

giveni s content = 0.330 mass of s in wal = 0.330 x 180x10² kg/hr : 59.4*10'legiur. So, formed thmol)= 59.48103 kg Ihr 32.06 C ly I lemon) :1.85 XIO' Icmol hr gas all formed in find the lomol of - Now flue gas. amt of Nein fuel. + h ic + & . 12.01 2.016 + s + h - 32.06 28.02 (som formed) Cmnogen formed) Com CH forned) m + x 3,77 0₂ that is 18.016 nitrogen in air not rearted. in the water

how carbon content: Kalhr C: 30.110 x 180X10%, 5419.8 XIo Hydrogen content ! h = 3.610x 180 x 10² = 649.8 XI0² kg/hr content: - Nitrogen 3. 126x103 kg Ihr h: 0.700x180x103. 126 X103 content moisture 1.5855.4 x 10² kg Ihr m. 32.530 X 180 810 = 5855.4 x10 al ratio a= equivalence 18% excess air arvin wal is burned with 100+18y. theoritical air wal is burned with 118 118 IDO uivalence rato :

The N₂ in in coming air that is unreacted the theoritical amt of 2 depends on il needed for combustion : complete clcmol) -o 32 Oric + 0.5 h + 12.01 2.016 32.06 1 in fuel o oxygen untent where 9.850 x 180 X 10² = 1773x10² leg/hr 0. 5x 0:5419.8x10 3 0.50 649.8x10° + 1.85% 103 2.016 12 ott 1773 X10 32 O. 558.87 X10° kamol these

amt of flue gas in kmol + x 3.77 0₂ c 12.01 + h 2.016 + s 32.06 + n 28.02 + m 18.016 - = 5419.8 %10% + 649,8X10% + 1.05x10'+ 12 6.X10° + 28.02 12.01 2.016 5855.4X10% + 1.18 * 3.77x558.87X10 19.016 flue gases. { flowrate of 1.14x10 kmol/hr Now we have 5%. Og in exhaust gas. I needed for complete combustion - 558.87 X 10² lumol/hr

5% of this is 5 x 558.87X10²: 27.94 X 10² kmol/hr excess o gas (including - the Amt of = 3591.14010² + 27.94X103 lcmol/hr Flowrate of - 3619. Ilmol/ hra) free gas. ideally flue gas kehares Annume at 0°c = 22.7 Nm3 kmol The molar volume

: Flowrate of flue gas (2) Q- 3619 kmol x 22.7 Nm3 : 82 153.19 Nm's hr lemol lcmol . Now, amt of so = 1.85*10 ² lemol/hr. i 1.85 % 103 lemol 82153.19 Nm3/mol as mol. wt of so₂ = 32.06 lcg llcmol amt of so, in (mg) = 1.85010² kmolx 32.06 leg x 10 (mg / kg) - limol 82 153.19 Nm3 72 1956,14 mg /Nm3

Amt of NOx's assuming 20%. of N in wal.is converted to Nox. wal = 126 x 10² lcg. Amt of N₂ in to Nox this converts now 20% of 0.20 x 12 66103 = 25.2 lg Nox. NOx = amt of 25.2 x 10' mg 82 153.19 Nm 3 t of NOx (mg) Nms = 306.744 mg / Nm3