Question

3. A process stream of 100 mol/s of n-Hexane vapor at 70.0 °C and 1 atm is fed into a furnace along with a process stream of 50.0% excess dry air (20% O2, 80% N2) at 70 °C and 1 atm. In the furnace, all the hexane is consumed, and the production rate of 60 mol/s CO is obtained. The reaction schemes for complete and incomplete combustion of n- hexane are given as: (1) C6H14 (9) + 5.02(g) – 6CO2(g) + 7H20(g) (2) C6H14(9) + 202(g) – 6C0(g) + 7H20(g) The furnace is well insulated, meaning the exiting gas is at the adiabatic flame temperature and 1 atm. This flue gas contains only 02, N2, CO2, CO and H20. The heat capacity of each species are approximated as constants, as follows: Ĉr Smolcc) C6H14 (9):0.137,02(9): 0.029, CO2(g): 0.036, CO(g): 0.035, H20(g): 0.034, N2(g): 0.029 a. Draw a label a flow chart for this system. b. Using the standard heat of formation and/or combustion values can be found in the table B.1 excerpt provided, determine the standard heats of reaction for complete and incomplete combustion, ΔΗ), and ΔΗ92. c. Solve the material balance for the inlet and outlet molar flow rates of each species. Determine the extents of reaction, Ši and $2. d. Write the relevant form of the energy balance. Use the reference state given for each species: C6H14 (9, 25°C, 1atm),O2(g, 25°C, 1atm), CO2(g, 25°C, latm), CO(g, 25°C, latm), H20(g, 25°C, 1atm), N2(g, 70°C, latm). Make an enthalpy table. e. Calculate the value or function of Tf representing each specific enthalpy variable in your enthalpy table. f. What is the temperature of the exiting flue gas?

Answer 1

Solution: (1 6H14 g +19 O₂ (g) 19 0₂ (g) 60₂6 60, (q) + 7H₂O(g) (2) GH14 (q) + 13 02 (9) 6 Col + 7H₂O(g) ca 100 molestials 70°c Gomol cols na mol cools) nadmol H₂0/) ng (mot Orls) n (mol Nels) 50% excess air 0.2 mol O2/molair 0.8 mol Nel molair I 70c (b). Heat of formations from Table B. 1; (A ff cottin _167.2 kJ/mol 2 0 (@flori (Aflfcon (6h foco -39.5 kJ/mol 2 kJ Imol -110:52

- 2ч 1, 93 ET/ms) (дн) H₂O lofts no ад дні, с (лн., * 100м, 34, -ев 2:1 (від «х(-343-s) take41-82) - (-167-2)- 14 хо Altro - 388 б. 61 kJ 13 (дн), allir 6 Catýlco. +7(Auflino - (Até chin's е хо г Сх (-0.52 )++(- 2 1 - 03) – (-1,7-2) - 13 Гі. Altro - 210. +3 KI. mol.

(2) will use equation ni anio + Z0; }; та - ) - {-}. Tбъ * - 25 2 loo ( *,**, 2 и п= 61, 10 molls " put in 0 2 90 molls о) — quo1 з , с Islom no corls ons z а. 2 Схо т noi 6%, 2. in Ho +, ++, + 1xqo + al Xt т 7to mol Hools | lен H2O

(now theoretical 2 two molgtmls 19.5 motor I mol & Hiu (15 95 6 mot Orls air 4] set. ๕๕ (44 0.2 ทอ (055) 5:( - mol airls ht 28 - °น : (0) 17 s w61 044 53 50 51 NM4 Now, now 2 (76, 2 - 4 , - 2 1ฯy - 11x90 - 13 % -sonli C 5 D5 nor mol Oals 2 AN น 2 (ทม) 5700 mol Nels C

cd) halance Energy equation; à - AH + QEp + AEK -0 (Adiabatic) (No shaft work) W, z 2 D ZO AEp = AEpz soi ди со 1, 2 , s Enthalpy Tables References; thu lgs, Orug), co.(g), colgs, H2O(g), N₂ (g) at 25°C, lath ý out fin Substance nin Icmolls) nout Ismolls) (kJ/mol (kJ/mol 100 H fu C Hig O, Å 505 Irurs 540 cor 60 Но AL fr co too HO 5700 Hg NA 5700 î 3