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Rotary kiln pulverized coal combustion and kiln burner

Date:2019-04-26 15:33     writer:admin     Views:

 The key to kiln combustion is 3T, temperature, time, turbulence, which is temperature, time and mixing. At present, there is usually no problem in achieving the above three main burners. Here, the decomposition furnace will be described.
The influence of temperature includes the tertiary air temperature and the tail coal feeding position. For example, if the tail coal is directly fed into the tertiary air, the temperature is obviously higher (such as the DD decomposition furnace). Although the separation from the tertiary air helps reduce the NOx production, it will inevitably affect the combustion. It is not easy to burn from the raw material too close to nature. Time is the pulverized coal residence time, which is also the purpose of most enterprises to expand the furnace. The mixing situation is not taken seriously by everyone. In fact, the flow field in the decomposition furnace is very uneven. Some areas have high oxygen concentration, some areas have low oxygen concentration, some areas have more coal powder, and some have less coal powder. This is why SNCR The reason why the NOx concentration in the different positions of the spray gun is very different is that if the mixing is not good, the volume of the large decomposition furnace is limited. On the contrary, the good mixing effect can achieve the kiln output under the condition that the volume of the decomposition furnace is constant. In addition, the problem of mixing may be one of the fundamental reasons for the significant difference in the classification of different companies.

Don't underestimate the existence of CO. Once CO is not completely burned, it means that almost half of the calorific value of pulverized coal has not been used. If you use the 6000 kcal coal now, if it produces CO, it means that you coal. The calorific value is only 3,000. It is difficult to react with oxygen after CO is produced, and the reaction rate is very slow, but if some hydrogen-containing groups (such as common water) at this time contribute to the complete combustion of CO. (Of course, the amount of water added needs to take into account the effects of water vaporization and heat absorption.) In addition to the hydrogen-containing groups, mixing (as already mentioned above) is also the key to limiting the complete combustion of CO.
We have been emphasizing the momentum of the main burner, ie 6 N/MW is a minimum limit, usually 9-11 N/MW. The greater the momentum, the faster the mixing of pulverized coal and air (and the mixing problem), the more favorable it is to form a short, concentrated flame. However, when the momentum is too large, it may mean that the wind volume of the primary wind is large, which is not conducive to the thermal efficiency. The burner momentum can be visually evaluated by the reflux of the flue gas, that is, moderate flue gas recirculation means that the pulverized coal and air are well mixed. On the contrary, if there is no flue gas recirculation, it means that the momentum is too small, which is bound to be caused by The secondary air entrainment is not enough to cause incomplete combustion of the coal. The figure below shows a schematic diagram of a burner with a small momentum and a burner with a large momentum.
The calculation of the momentum of the burner is: burner momentum (N / MW) = primary air mass flow (kg / s) at the exit of the burner × primary wind speed (m / s) / heat of coal (MW), which requires attention The primary air volume and the primary air speed do not refer to the overall primary air volume and speed, but the weight of the axial wind, the swirling wind, the central wind and the corresponding speed are weighted; the heat of the coal powder is generally represented by a large card, which needs to be converted into MW, ie 1 large card = 1000 * 4.186 J = 4186 J = 4186 × coal powder per second mass (kg / s) / 1000000 (MW), such as head coal per hour coal feeding amount of 12 t / h, calorific value For the 5500 kcal, the conversion to MW is =12000/3600*5500*1000*4.186/1000000=76.74 MW.

Cylinder diameter(mm) Cylinder length(mm) Speed(r/min) Yield(t/h) Motor power
1900 39000 0.31-2.82 1.6-3.4 37
2500 50000 0.68-1.45 6.2-7.4 55
2500 55000 0.79-2.38 7.3-9.8 55
2800 20000 0.8-1.45 15-18 75
3000 28000 0.8-1.45 18-23 75
3000 60000 0.49-2.41 16-20 90
3200 26000 0.8-1.45 17-23 90
3200 55000 0.1-1.52 22-27 110
3500 29000 0.16-1.38 28-40 110
3500 60000 0.65-2.25 28-40 125
4000 30000 0.38-3.52 52-70 160
4000 60000 0.38-3.52 55-78 315
4200 62000 0.38-3.52 80-105 315
4500 65000 0.38-3.52 110-140 420
4700 72000 0.38-3.52 150-190 550

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