2. Selection of Optimal Air Fuel Ratio P M V Subbarao Professor Mechanical Engineering Department Efficient Combustion Requires Sufficient Air…..

3. Model Testing for Determination of important species Air Flow Rate Fuel Flow Rate Water Flow Rate Flue gas Analysis

4. Results of Model Testing. For a given fuel and required steam conditions. Optimum air flow rate. Optimum fuel flow rate. Optimum steam flow rate. Optimum combustion configuration!!!

5. Air Supply Systems

6. Direct Method of SG Performance Analysis Energy balance: Fuel Energy = Steam Enthalpy + Losses. Measurements: –Steam Flow Rate –Steam properties –Fuel flow rate. Difficulties: Measurement of steam flow rate. Measurement of fuel flow rate. Errors in measurements.

7. Selection of Optimal air for Best Performance

8. Results for Best Efficiency

9. Effect of Excess Air on Emissions

10. Selection of Optimal air for Eco-Friendliness

11. Study of Impact of Excess Air

12. Unified Optimization Methods

13. Methods to Apply the Optimization

14. Optimal Excess Air

15. Air Ingression Due to Better Combustion Conditions

16. Effect of Air Ingression on Actual Available Excess Air

18. First Law Analysis of Furnace at Site C X H Y S Z O K +  4.76 (X+Y/2+Z-K/2) AIR + Moisture in Air + Ash Moisture in fuel→ P CO 2 +Q H 2 O +R SO 2 + T N 2 + U O 2 + V CO + W C + Ash Mass of air:  *4.76* (X+Y/2+Z-K/2) *28.89 kg. Mass of Coal: 100 kg. Excess Air:  -1)  4.76* (X+Y/2+Z-K/2) *28.89 kg.

19. C X H Y S Z O K +  4.76 (X+Y/2+Z-K/2) AIR + Moisture in Air + Ash Moisture in fuel→ P CO 2 +Q H 2 O +R SO 2 + T N 2 + U O 2 + V CO + W C + Ash

20. First Law Analysis of Furnace:SSSF Conservation of Mass: m air + m fuel - m fluegas = 0 First Laws for SG in SSSF Mode:  Q + m air h air + m fuel h fuel = m fluegas h fluegas +  W m fuel m air m fluegas Q Q W fans

21. Performance Testing of SG Air Flow Rate Dry Flue gas Analysis Ex. Gas Flow Rate

22. Indirect Method of SG Performance Analysis For every 100 kg of Coal. C X H Y S Z O K +  4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash Moisture in fuel → P CO 2 +Q H 2 O +R SO 2 + T N 2 + U O 2 + V CO + W C + Ash

23. Measurements of Gas Analyser Dry Exhaust gases: P CO 2 +R SO 2 + T N 2 + U O 2 + V CO kmols. Volume of gases is directly proportional to number of moles. Volume fraction = mole fraction. Volume fraction of CO 2 : x 1 = P * 100 /(P +R + T + U + V) Volume fraction of CO : x 2 = VCO * 100 /(P +R + T + U + V) Volume fraction of SO 2 : x 3 = R * 100 /(P +R + T + U + V) Volume fraction of O 2 : x 4 = U * 100 /(P +R + T + U + V) Volume fraction of N 2 : x 5 = T * 100 /(P +R + T + U + V) These are dry gas volume fractions. Emission measurement devices indicate only Dry gas volume fractions.

24. Measurements: Volume flow rate of air. Volume flow rate of exhaust. Dry exhaust gas analysis. x 1 +x 2 +x 3 + x 4 + x 5 = 100 or 1 Ultimate analysis of coal. Combustible solid refuse. nC X H Y S Z O K +  n 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → x 1 CO 2 +x 6 H 2 O +x 3 SO 2 + x 5 N 2 + x 4 O 2 + x 2 CO + x 7 C + Ash

25. nC X H Y S Z O K +  n 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → x 1 CO 2 +x 6 H 2 O +x 3 SO 2 + x 5 N 2 + x 4 O 2 + x 2 CO + x 7 C + Ash x 1, x 2,x 3, x 4 &x 5 : These are dry volume fractions or percentages. Conservation species: Conservation of Carbon: nX = x 1 +x 2 +x 7 Conservation of Hydrogen: nY = 2 x 6 Conservation of Oxygen : nK + 2 n  (X+Y/4+Z-K/2) = 2x 1 +x 2 +2x 3 +2x 4 +x 6 Conservation of Nitrogen:  n 3.76 (X+Y/4+Z-K/2) = x 5 Conservation of Sulfur: nZ = x 3

26. nC X H Y S Z O K +  n 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → x 1 CO 2 +x 6 H 2 O +x 3 SO 2 + x 5 N 2 + x 4 O 2 + x 2 CO + x 7 C + Ash Re arranging the terms (Divide throughout by n): C X H Y S Z O K +  4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → (x 1 /n)CO 2 +(x 6 /n) H 2 O +(x 3 /n) SO 2 + (x 5 /n) N 2 + (x 4 /n) O 2 + (x 2 /n) CO + (x 7 /n) C + Ash C X H Y S Z O K +  4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash Moisture in fuel → P CO 2 +Q H 2 O +R SO 2 + T N 2 + U O 2 + V CO + W C + Ash

27. Specific Flue Gas Analysis For each kilogram of fuel: Air :  4.76 (X+Y/2+Z-K/2) * 29.9 /100kg. CO 2 : P * 44/100 kg. CO : V * 28/100 kg. Oxygen in exhaust : 32 * U/100 kg. Unburned carbon: 12*12/100 kg.

28. Various Energy Losses in A SG Heat loss from furnace surface. Unburned carbon losses. Incomplete combustion losses. Loss due to hot ash. Loss due to moisture in air. Loss due to moisture in fuel. Loss due to combustion generated moisture. Dry Exhaust Gas Losses.