It's still a problem with the TO furnace
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No matter how I calculate it, the amount of air entering is much, much higher than the chemical oxygen demand – by many times over. All this extra air needs to be preheated using heat from combustion. What I want to ask is: according to the literature, sometimes very little or no gas heating is needed when the design is proper; how is that achieved? Haha, I’m going to take into account the heat released during combustion as well
Is it the flow meter that’s inaccurate? Is something burning? Have you tried calculating it using nitrogen or some other inert gas? In reality, there’s definitely an excess of air, but it seems wrong by many, many times over
Organic waste gases, xylene, and things like that – do you work with these as well? What is the standard concentration used to measure waste gases in your area?
On my end, it’s inorganic waste gas – actually, it’s the blowing air from fixed-bed batch gasifiers used for ammonia synthesis. The concentrations of this waste gas are based on actual measurements; using empirical values would result in significant deviations
Could you show me your data and the calculation process?
I used the specified maximum concentration, which is 25% of the lower explosive limit; as a result, the amount of air introduced is many times higher than the theoretical value
1# Use exhaust air to recycle 15 gas stoves; operate two D500 fans and two D700 fans； Based on the D500 fan, the air volume is: 30,000 Nm3/h ； The fan outlet pressure is set at 2500 mmH2O, the fan efficiency is set at 76%, and there are 4 blast furnaces. Components of the purge gas and vent gas: CO2, CO, H2, N2, CH4, O2. For the purge gas: 14.8% , 7.9% , 1.4% , 75.4% , 0.9% , 0.5%; for the vent gas: 0, 0.3% , 17.0% , 40.9% , 41.5% , 0.3%. 2. Calculation process: The volume of Vent Gas No. 1, V1, is 1300 m3/h. The air supply volume V2 = 30000*0.79*28*15*0.76*(9800+2500)/(9800*0.754*120) = 104939 m3/h. (Calculated on a nitrogen basis). According to the blow-air theory, the air supply volume V3 = (V2*7.9%/2 + V2*1.4%/2 + V2*0.9%*2)/0.21 = 35878 m3/h (based on the oxygen requirements of CO, H2, and CH4). The theoretical air supply volume according to the exhaust gas theory is V4 = (V1*0.3%/2 + V1*17.0%/2 + V1*41.5%*2)/0.21 = 5674 m3/h (based on the oxygen requirements of CO, H2, and CH4). The theoretical flue gas volume is V5=V1+V2+V3+V4-V2*7.9%/2-V2*1.4%/2-V1*0.3%/2-V1*17.0%/2=144098 m3/h. In practical operation, with an excess of air and about 8% residual O2 in the flue gas, the excess air volume V6 satisfies the following relationship: V6*0.21 = (V5 + V6)*8%, which gives V6 = 87876 m3/h. The actual flue gas volume V7 = V5 + V6 = 230,674 m3/h. The actual air volume V8 = V3 + V4 + V6 = 129,428 m3/h. If you don’t know the concentrations of the various components, the calculation is meaningless