Question

Problem 9.38 Carbon Monoxide from Coke x Make sure that you have included all terms in the energy balance. Coke can be converted into Co-a fuel gas-in the reaction CO2(g) + C(s) - 200(9) A coke that contains 84.0% carbon by mass and the balance noncombustible ash is fed to a reactor with a stoichiometric amount of CO2. The coke is fed at 77.0°F, and the Co, enters at 400.0°F, heat is transferred to the reactor in the amount of 5559 Btu/Ibm coke fed. The gaseous products and the solid reactor effluent (the ash and unburned carbon) leave the reactor at 1830°F. The heat capacity of that solid is 0.240 Btu/(lbm °F). 67.52 Calculate the percentage conversion of the carbon in the coke.

43.73% is a wrong answer

Answer 1

CO, fed (n2) Temp (T2) = 400 F Gases out Coke fed Reactor CO2 (g) (n4) CO (8) (n5) Temp (T3) = 1830 F n1 Wt% of C = 84 % Carbon in coke Rest is ash Temp (T) = 77 F Uncombusted coke Coke (n3) Ash Temp (T3 ) = 1830 F Heat Transfer (0) = 5559 Btu/lbm

Basis: Consider 100 lbm of coke fed % of Carbon in coke 84% Carbon in coke 84 100 * Mass of coke 11 0.84 * 100 84 lbm Moles of Carbon in coke Mass of Carbon in coke Molar mass of carbon 84 12 lb-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 *7 7 lb-mol = Carbon in Feed (ni) 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 (9) -110.52 Std. Heat of reaction can be calculated from Std. Heat of formation of components as : AH reaction (Ev AH products Σν ΔH POreactants where v is stoichiometric coefficient, all positive ΔΗ, reaction 2* (AH)co [1 * (AH)co2+1 * (AH) (0) 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 1 mol lb-mol 453.6 .. AH reaction 0.9478 -172.46 * 1 453.6. Btu/lb-mol 74144.36 Btu/lb-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 CO formed * Carbon reacted 2 * 7x = 14 x Carbon unreacted CO2 unreacted 7-7x 7-7x 7*(1-x) 7*(1-x) Thus, the product contains: Component Moles C(s) n3 7*(1-x) CO2 (g) n4 7+ (1 - x) CO (g) n5 14 x Ash 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 400 F Specific enthalpy of CO2 at 1830 F Specific enthalpy of CO at 1830 F 3130 Btu/Ib-mol 20880 Btu/lb-mol 13280 Btu/Ib-mol

Specific Heat capacity of solid (Cp) 0.24 Btu/lbm. F Solid means coke and ash Specific enthalpy of solid in Feed = Cp* (T: - Tres) 0.24 * (77 - 77) 0 Specific enthalpy of solid in Product = Cp* (T3 - Tres) 0.24 * (1830 – 77) 420.72 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 lbm 100 - 84x lbm Inlet-Outlet Enthalpy table is: Component Amount IN Sp. Enthalpy IN Amount OUT Sp. Enthalpy OUT CO2 (g) 7 lb-mol 3130 Btu/lb-mol 7*(1-x) lb-mol 20880 Btu/lb-mol CO (g) 0 14 x lb-mol 13280 Btu/lb-mol Solids 100 lbm 0 100 - 84x lbm 420.72 Btu/lbm Note that units for gases and solids are different.

Extent of reaction based on Carbon can be written as: w Carbon in product-Carbon in feed St.coeff of Carbon in reaction 7*(1-x)-7 (St. coeff is negative for reactants) -1 7x 1b-mol Heat released due to reaction Extent of reaction * Std. Heat of reaction 7x * (74144.36) Btu/lb-mol 519010.52 x Btu/1b-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) Ση, h, over all components Total Enthalpy of Feed 7* 3130 + 100 * 0 21910 Btu Total Enthalpy of Product 7*(1-x) * 20880 + 14x * 13280 + (100 – 84x)*420.72 188232 + 4419.52 x Heat Transfer (9) Rate of Heat Transfer (0) 5559 Btu/lbm q* Mass of coke 5559 * 100 Btu 555900 Btu Substituting all the values in equation (1), 21910 + 555900 188232 + 4419.52 x + 519010.52 x On solving above equation, 0.7443 X х Fractional conversion of carbon in coke % conversion of carbon in coke 100 *x= 0.7443 74.43% (ANSWER)