![CO2 fed (n2) Temp (T2) = 300 F Gases out Coke fed Reactor CO2 (g) (n4) CO (B) (n5) Temp (T3 ) = 1630 F n1 Wt% of C = 84 % Car](//img.homeworklib.com/questions/e8a63ab0-0c76-11eb-86b7-374927491463.png?x-oss-process=image/resize,w_560)
![Basis: Consider 100 lbm of coke fed % of Carbon in coke 84 % Carbon in coke 84 100 * Mass of coke = 0.84 * 100 Il 84 lbm Mole](//img.homeworklib.com/questions/e90b9f60-0c76-11eb-bd53-3b0f55d311ad.png?x-oss-process=image/resize,w_560)
![We need to convert kJ/mol to Btu/lb-mole 1 kJ 0.9478 Btu 1 mol = lb-mol 453.6 0.9478 :: AH reaction -172.46* Btu/lb-mol 53.64](//img.homeworklib.com/questions/e988f5c0-0c76-11eb-9a78-05e5630341bd.png?x-oss-process=image/resize,w_560)
![Specific Heat capacity of solid (Cp) 0.24 Btu/lbm. F Solid means coke and ash Specific enthalpy of solid in Feed Cp* (T1 - Tr](//img.homeworklib.com/questions/e9fd5460-0c76-11eb-bccb-b58550d9249f.png?x-oss-process=image/resize,w_560)
![Extent of reaction based on Carbon can be written as: Carbon in product-Carbon in feed St.coeff of Carbon in reaction 7*(1-x)](//img.homeworklib.com/questions/ea6e5ac0-0c76-11eb-9f2a-99420260dfa9.png?x-oss-process=image/resize,w_560)
CO2 fed (n2) Temp (T2) = 300 F Gases out Coke fed Reactor CO2 (g) (n4) CO (B) (n5) Temp (T3 ) = 1630 F n1 Wt% of C = 84 % Carbon in coke Rest is ash Temp (T) = 77 F Uncombusted coke Coke (n3) Ash Temp (T) = 1630 F Heat Transfer (Q) = 5259 Btu/lbm
Basis: Consider 100 lbm of coke fed % of Carbon in coke 84 % Carbon in coke 84 100 * Mass of coke = 0.84 * 100 Il 84 lbm Moles of Carbon in coke Mass of Carbon in coke Molar mass of carbon 84 12 1b-mol 7 lb-mol Reaction is: CO2(g) + C(s) +200 (g) It is said that stoichiometric amount of CO2 is fed. .. CO2 fed * Moles of Carbon in coke 1 1*7 7 lb-mol Carbon in Feed (n.) n2 7 lb-mol 7 lb-mol First we need to find heat of reaction at standard tempeature i.e. 77 F Heat of Formation for all components involved are: Component Std. Heat of Formation (KJ/mol) Carbon (s) (g) 0 CO2 (g) -393.5 CO (8) -110.52 Std. Heat of reaction can be calculated from Std. Heat of formation of components as : (Ev AH products (Ev AH )reactants where v is stoichiometric coefficient, all positive 2* (AHA)co [1 * (AH)co2+1 * (AH) (0] AH reaction AH reaction 2* (-110.52) - [1 * (-393.5) +1 * 0] 172.46 kJ/mol
We need to convert kJ/mol to Btu/lb-mole 1 kJ 0.9478 Btu 1 mol = lb-mol 453.6 0.9478 :: AH reaction -172.46* Btu/lb-mol 53.64 74144.36 Btu/Ib-mol Standard heat of reaction = AH reaction 74144.36 Btu/lb-mol Let x be fractional conversion of C and CO2 : Carbon reacted CO2 reacted 7*x 7*x 7x 7x 2 CO formed = * Carbon reacted 2 * 7x = 14 x 1 Carbon unreacted CO2 unreacted 7-7x 7-7x 7*(1-x) 7(1-x) Thus, the product contains: Component Moles C(s) 7*(1-x) CO2 (g) n4 7*(1-x) CO (g) 14 x Ash n3 n5 Now, we need to find specific enthalpy of all components at given temperatures. Let reference temperature for C(s), CO2 (g), CO (g), ash be 77 F Specific enthalpy of gases can be directly calculated from table From table B9 in Feldar, Specific enthalpy of CO2 at 300 F 2108 Btu/Ib-mol Specific enthalpy of CO2 at 1630 F 18177 Btu/1b-mol Specific enthalpy of CO at 1630 F 11652 Btu/lb-mol
Specific Heat capacity of solid (Cp) 0.24 Btu/lbm. F Solid means coke and ash Specific enthalpy of solid in Feed Cp* (T1 - Tref) 0.24 * (77 - 77) Specific enthalpy of solid in Product = Cp* (T3 - Tres) 0.24 * (1630 - 77) 372.72 Btu/lbm We need to calculate mass of solids OUT in product Mass of solid in Feed 100 lbm (given) Ash in solid in feed 100 - Carbon in solid 100 - 84 16 lbm Ash remains unreacted. Thus, amount of ash in feed and product will be same Mass of solid in Product Unreacted carbon + Ash 7*(1-x) * 12 +16 100 - 84x lbm Ibm + Inlet-Outlet Enthalpy table is: Component Amount IN Sp. Enthalpy IN CO2 (g) 7 lb-mol 2108 Btu/lb-mol Amount OUT Sp. Enthalpy OUT 7*(1-x) lb-mol 18177 Btu/lb-mol CO(g) 0 14 x 1b-mol 11652 Btu/lb-mol Solids 100 lbm 0 100 - 84x lbm 372.72 Btu/lbm Note that units for gases and solids are different.
Extent of reaction based on Carbon can be written as: Carbon in product-Carbon in feed St.coeff of Carbon in reaction 7*(1-x)-7 (St. coeff is negative for reactants) 7x 1b-mol Heat released due to reaction Extent of reaction * Std. Heat of reaction 7x * (74144.36) Btu/1b-mol 519010.52 x Btu/lb-mol Energy Balance is given as: Total Energy IN Total Energy OUT Total Enthalpy of Feed + Total Enthalpy of Product Rate of Heat Transfer Heat change due to reaction + (1) Total Enthalpy = Sum of (Molar flowrate * Specfic enthalpy of that component at given Temp) Enihi over all components Total Enthalpy of Feed 7* 2108 + 100 * 0 14756 Btu Total Enthalpy of Product 7*(1-x) * 18177 + 14x * 11652 + (100 – 84x)*372.72 164511 + 4580.52 x Heat Transfer (9) Rate of Heat Transfer (0) 5259 Btu/lbm q* Mass of coke 5259 * 100 Btu 525900 Btu Substituting all the values in equation (1), 14756 + 525900 164511 + 4580.52 x + 519010.52 x On solving above equation, 0.7184 Fractional conversion of carbon in coke % conversion of carbon in coke X 100 *x 0.7184 71.84 % (ANSWER)