Design a fluidized bed gasifier. The product gas consist of H 12 m\'/min, CH4- 4
ID: 707460 • Letter: D
Question
Design a fluidized bed gasifier. The product gas consist of H 12 m'/min, CH4- 4 m/min co 15 m min CO2 (stoichiometric basis) and unreacted Air (stoichiometric basis). The gasifier considered as a cylindrical cross section reactor. The actual air to fuel ratio should be 30 % o stoichiometric requirements while 70% exit along with the product gas as unreacted. The ultimate analysis of biomass used for process is as follows: C = 51.7%; H = 7.2% ; 0- gasification rate is 3 GJ/hr-m2, while, 1250 kg of biomass is used for every hour f the 40.53%. The MJ/kgl Density of air 1.2 kg/m; Density of CO, 1.98 kg/m: Heating value of biomass [MW: C-12, H-1, 0-16, N-14, S-32] 18 Total Gas Production Rate Area Gasification rates: vass now rate; Velocity= ass flowrate ; Velocity = Area a. Write empirical chemical formula of fuel. [1 point] b. Write balanced stoichiometric equation. [2 point] c. Calculate air-to-fuel ratio. [1 point d. Calculate the superficial velocity. [6 point]Explanation / Answer
a) C=51.7%, H=7.2%,, O=40.53%
Atomic mass of C=12, H=1, O=16
=> Number of atoms of C in biomass = 51.7/12=4.3 =~4
=> Number of atoms of H in biomass = 7.2/1=7.2=~7
=> Number of atoms of O in biomass = 40.53/16=2.53=~3
=> Formula of Biomass = C4H7O3
b) C4H7O3+1/2O2=>CH4+2CO+CO2+3/2H2
c) Given Air is 30% of stoichiometric requirement
=> Air/fuel ratio = 0.3
d) Given Gasification rate = 3GJ/hr/m2
Mass flowrate = 1250 kg/hr
Heating value of biomass = 18 MJ/kg
=> Mass flowrate = 1250*18*10^-3 GJ/hr
Because, Gasification rate = Mass flowrate/area
=>3=1250*18*10^-3/area
=> Area = 7.5 m2
Molecular weight of Biomass C4H7O3=4*12+7*1+3*16=103 Kg/kmol
=>Molar flowrate of Biomass = mass flowrate/Molecular weight = 1250/103=12.13 kmol/hr
1 mole biomass give 1 mole CO2
=> CO2 flowrate = 12.13 kmol/hr
CO2 molar mass = 44 Kg/kmol
=> CO2 molar flowrate = 12.13*44=533.72 kg/hr
CO2 density = 1.98 kg/m3
=>CO2 volumetric flowrate = 533.72/1.98=269 m3/hr=269/60=4.48 m3/min
=> Total gas production rate = CO2 volumetric flowrate+H2 volumetric flowrate+CH4 volumetric flowrate+COvolumetric flowrate=4.48+4+12+15=35.48 m3/min
Velocity =Gas production rate/area=35.48/7.5=4.73 m3/min/m2
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