Home-exercise 3) From 1 ton of wood with a higher heating value of 20 MJ/kg and lower heating value of 19.3 MJ/Kg we obtain 500 kg of CO and 80 kg of H2. Utilize the physical properties Hf etc. given in the previous exercise in the calculations. The density is approximately 0.082 kg/m3 for hydrogen gas and 1.145kg/m3 for carbon monoxide gas at 25 °C and 1 bar. 1) Calculate mass yield from 1 ton of wood 2) Calculate higher heating value by mass (MJ/Kg) for CO and H2 by combustion reactions and through standard heat of formation for the components 3) Calculate the lower heating value for hydrogen, what is the lower heating value of CO? 4) Calculate the volumetric heating value of hydrogen and CO and compare them with natural gas with a heating value of 38 MJ/m3. 5) What is the energy yield by higher and lower heating value from 1 ton of wood for the given gas mixture? 6) If we assume that we can produce the gas mixture containing 4.62 kmol CO and 13.23 kmol of methane from 1 ton of wood. What would be the energy yield by higher and lower heating value? 1) Mass yield = (500+80) kg/1000 kg = 58 % 2) CO+1/2O2->CO2 = ΔHr =HFproduct -Hfreactants :(-393,5kJ/mol*1)-(0.5*0+-1*110,5kJ/mol) =-283 kJ/mol HHV =-ΔHr/Mw(CO) : 283 kJ/mol/28.01g/mol =10.1 kJ/g or MJ/kg H2+(1/2)O2->H2O ΔHr = HFproduct -Hfreactants: (-1mol*285.33kj/mol-0.5*0-1*0) =-285.33 kJ/mol HHV =-ΔHr/Mw(CO) : 285.33 kJ/mol /2.016 g/mol =141.5 kJ/g or MJ/kg 3) LHV = LHV =HHV –HlatenH2O*mH2Oinfluegas = HHV –HlatenH2O*(mH2/MH2)*(nH2O/nH2)*MH2O =141.5 MJ/kg -2.443 MJ/kg *(1kg/ 2.016 kg/kmol*(1/1)*18.018 kg/kmol)) =119.7 MJ/kg 4) 141.5 MJ/kg*0.082kg/m3 = 11.6 MJ/m3 for Hydrogen and 10.1Mj/kg*1.145 kg/m3= 8.8 MJ/m3 for CO. The volumetric heating values are much lower than for methane. 5) By higher heating value = HHV energy yield = HHV products /HHV raw material =(500kg *10.1 MJ/kg + 80kg*141.5)/(20 MJ*kg*1000kg) ~82 % By lower heating value LHV energy yield = LHV products /LHV raw material =(500kg *10.1 MJ/kg + 80kg*119.7MJ/kg)/(19.3 MJ/kg*1000kg) ~76 % 6) First we calculate the heating value for methane from heat of formation of the components: CH4+ 2O2-> 2H2O+ CO2 ΔHr =HFproduct -Hfreactants = (-2*285.83kJ/mol+-1*393.5KJ/mol)(2*0kj/mol+1*-75kJ/mol) = -890.16 kJ/mol MW CH4 =16.042 g/mol HHV =-ΔHr/Mw(CO) : --890.16 kj/mol/16.042 g/mol = 55.5 kJ/g or MJ/kg LHV HHV –HlatenH2O*mCH4/MCH4*(nH2/nCH4*MH2O) =55.5 MJ/kg -2.443 MJ/kg*(1/16.042 kg/kmol*(2/1)*18.018kg/kmol) =50.0 MJ/kg No we can determine the energy yield HHV energy yield = (mCO*HHVCO + mCH4*HHVCH4)/HHVrawmterial (4.62 kmol*28.01kg/kmol*10.1 MJ/kg+ 13.23 kmol *16.042 kg/kmol*55.5 MJ/kg) /(20 MJ/kg*1000kg) ~ 65 % and LHV energy yield LHV energy yield = (mCO*LHVCO + mCH4*LHVCH4)/LHVrawmterial (4.62 kmol*28.01kg/kmol*10.1 MJ/kg+ 13.23 kmol *16.042 kg/kmol*50.0 MJ/kg) /(20 MJ/kg*1000kg) ~ 62 %
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