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Material Balance for FCC Regenerator ProblemBurning the coke deposited on the catalyst particles generates all the heat necessary for catalytic cracking. Therefore, the coke burning rate is a critical parameter to control the rate of cracking. The composition of dry flue gas from the regenerator of an FCC unit is given in vol% as follows:Gas Volume %N2 81.6CO2 15.7CO 1.5O2 1.2
The dry air flow rate to the regenerator is given as 593 SCMM (standard cubic meters per minute). Considering that a significant portion of coke is carbon, you calculated in Exercise 6 the carbon burning rate in the regenerator as 52.6 kg/min. For this exercise, calculate the coke burning rate in kg/min and the hydrogen content (wt%) of the coke burnt. Assume that the coke consists only of carbon and hydrogen.Hint: Use an oxygen balance to determine the missing oxygen which was consumed to burn the hydrogen in coke. Water content of the flue gas is not given because only the dry gas analysis is reported. ------------------------------For Exercise 6 (previously stated) this is how I went about the question:Calculate the mass of carbon burnt/minute Flue gas flow rate?N2 in the combustion air = 0.79(593) = (468.5 m3)/min / (22.4 m3/kgmole) = 20.9 kgmole/minN2 is the tie component, it does not react. =>N2 in the dry flue gas = 20.9 kgmole/minDry flue gas flow rate = 20.9 / 0.816 = 25.6 kgmole/minNow, I can calculate the mass of C in the flue gas (in CO2 and CO)CO2: 25.6 kgmole flue gas/min (0.157 kgmoles of CO2/1kgmole of flue gas)= 4 kgmoles/min of CO2(12 kg C/1kgmole CO2) = 48 kg C /minCO = 25.6 kgmole flue gas/min (0.015 kgmoles of CO/1kgmole of flue gas) =0.4 kgmole/min CO (12 kg of C/1kgmole CO)) = 4.6 kg/min CTotal C burnt = 48+4.6 = 52.6 kg/min
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