Question

Problem 9.26 Benzaldehyde Production Benzaldehyde is produced from toluene in the catalytic reaction CH.CH3 + O2 + CHECHO + H20 Dry air and toluene vapor are mixed and fed to the reactor at 350.0 °F and 1 atm. Air is supplied in 200.0 % excess. Of the toluene fed to the reactor, 13.0 % reacts to form benzaldehyde and 1.70 % reacts with oxygen to form CO2 and H20. The product gases leave the reactor at 379 °F and 1 atm. Water is circulated through a jacket surrounding the reactor, entering at 80.0 °F and leaving at 105 °F. During a four-hour test period, 34.3 lbm of water is condensed from the product gases. (Total condensation may be assumed.) The standard heat of formation of benzaldehyde vapor is -17,200 Btu/Ib-mole; the heat capacities of both toluene and benzeldehyde vapors are approximately 31.0 Btu/(Ib-mole °F); and that of liquid benzaldehyde is 46.0 Btu/(Ib-mole.°F). Physical Property Tables Volumetric Flow Rates of Feed and Product Calculate the volumetric flow rates (f/h) of the combined feed stream to the reactor and the product gas. 103 ft3/h Vout = x 10 ft/

Answer 1

Basis: Consider 100 lb-mole of CH-CH: entering the reactor Thus, C.HSCH; in n = 100 lb-mole Reactions are : CH3CH2 + O2 + CHECHO + H2O (Main) (1) CH3CH3 + 902 → 7CO2 + 4H20 (Side) (2) Calculating Theoretical And Excess Amount of air entering Theoretical O2 is only based on main reaction i.e. reaction forming Benzaldehyde i.e reaction (1) Based on stoichiometry, Theoretical Amount of O2 reqd for reaction 1 = 1 * C.HSCH; in ni = 1 * 100 100 lb-mole Air contains 79 mole % N2 and 21 mole % 02 Thus, Theoretical Amount of air required Theoretical Amount of O2 reqd Mole fraction of O2 100 476.19 lb-mole 0.21 But 200 % excess Air is used. Thus, Actual air used (1 + 200 100 Theoretical Amount of Air 3 * 476.19 1428.57 lb-mole Moles of air entering (n2) 1428.57 lb-mole Thus, O2 in actual air Thus, N2 in actual air 0.21 * 1428.57 0.79 * 1428.57 300 1b-mole 1128.57 lb-mole Calculating Composition of Outlet Product Out of total toluene fed to reactor, 13 % reacts to form Benzaldehyde and 1.7 % reacts to form CO2 and H20. 13 Thus, C.HSCH; reacted by reaction (1) * C.HSCHz in ni 0.13 * 100 100 13 lb-mole

C6H-CH3 reacted by reaction (2) 1.7 100 * CH CHz in ni 0.017 * 100 1.7 lb-mole By stoichiometry, O2 reacted by reaction (1) * CH CH3 reacted by reaction (1) = 13 lb-mole 1 CH CHO formed by reaction (1)= * CH:CH3 reacted by reaction (1) 13 lb-mole 1 H2O formed by reaction (1) * CGH-CH3 reacted by reaction (1) 13 lb-mole 1 O2 reacted by reaction (2) *CH:CH; reacted by reaction (2) 15.3 lb-mole 1 7 CO2 formed by reaction (2) * CH-CH3 reacted by reaction (2) 11.9 1b-mole 1 H2O formed by reaction (2) **CH CH3 reacted by reaction (2) = 6.8 1b-mole 1 Total CH CH3 reacted Total O2 reacted Total H2O formed 13 +1.7 13+ 15.3 13 + 6.8 14.7 lb-mole 28.3 lb-mole 19.8 lb-mole CH=CH; unreacted O2 unreacted 100 - 14.7 300-28.3 85.3 lb-mole 271.7 lb-mole Thus, Composition of product is: Component Moles (Ib-mole) C.HCH 85.3 C.HSCHO 13 02 271.7 N2 1128.57 CO2 11.9 НО 19.8 Total 1530.27 Now, we need to calculate Volume of Feed and Products for 100 lb-mole as basis which we considered. Ideal gas Law is given as: V n+R+T P (where n is in lb-mole, T is in Rankine, P is in atm, V is in ft3) R=0.7302 ft.atm/lbmol.R

For Feed, Vfeed (100+300+1128.57) 0.7302 + (350+460) 1 ft = 904091.07 ft For Product, Vprod 1530.27 0.7302 (379+460) 1 ft 937501.25 ft First, need to find Specific Enthalpy of each component at Initial and Final Temp to do Energy Balance Heat of Formation for all components involved are: And to convert kJ/mol to Btu/Ib-mole, multiply by 430.28 Std. Heat of Formation (KJ/mol) 50 Component C.HSCH; (g) C.HECHO (9) 02 (9) N2 (8) CO2 (g) HO (9) 0 0 -393.5 -241.83 Std. Heat of Formation (Btu/lb-mole) =50 * 430.28 - 21514 -17200 0 0 =-393.5*430.28 = -169315.18 =-241.83*430.28 = -104054.61 Here, heat of formation is given at 25 °C. This is the reference temperature. Thus, Tref 25 °C 77 °F Also, Initial Temp (T) 350 F Final Temp (T2) 379 °F Constant specific Heat capacity of CH$CH3 and CoH$CHO are: 31 Btu/lb-mole °F CPCHỊCH (8) CPCH, CHO (8) 46 Btu/lb-mole °F Specific Enthalpy of component at given temperature for constant Cp is given as: h AH; 25 c + Cp* (T – Tref) where T is Temperature of that stream Let 1 and 2 denote inlet and outlet streams. For Inlet Toluene, 21514+31 * (350 - 77) h1C.HSCH 29977 Btu/Ib-mole For Outlet Toluene, 21514 +31 * (379-77) h2CH CH3 30876 Btu/Ib-mole

For Outlet Benzaldehyde gas, h2 C,H, CHO -17200+ 31 * (379 - 77) -7838 Btu/1b-mole Specific Enthalpy of O2, N2, CO2 and H2O can be directly calculated using tabulated values. From standard table, Specific Enthalpies in Btu/lb-mole of given components at required temperature with respect to Reference temperature of 77 F are: + Component 02 N2 CO2 HO T= 300°F 1602 1558 2108 1818 T=400°F 2342 2265 3129 2652 T= 350°F 1972 1911.5 2618.5 2235 T=379°F 2186.6 2116.53 2914.59 2476.86 Enthalpies at 350 F and 379 F are not available directly. So assuming linear interpolation, we calculate H at 350 F and 379 F using H at 300 F and H at 400 F By Linear Interpolation, H350 H300 + (H400-H300 (350 – 300) 400-300 H379 H300 + H400-H300 400-300 (379 – 300) Above enthalpies in table are enthalpy with respect to 77 F. Actual Enthalpy is given as: ht AH; 77 p + HT For Inlet Oxygen, 0+ 1972 1972 Btu/lb-mole мо, For Inlet Nitrogen, 0+ 1911.5 MN 1N2 1911.5 Btu/lb-mole For Outlet Oxygen, 0+2186.6 h2oz 2186.6 Btu/lb-mole For Outlet Nitrogen, H2N2 0+2116.53 2116.53 Btu/lb-mole For Outlet CO2, -169315.18 + 2914.59= hzco, -166400.59 Btu/b-mole For Outlet H20, -104054.61+ 2476.86 = h2H₂O -101577.75 Btu/Ib-mole

+ Molar flowrate and Specific Enthalpy table is: Component ni (lb-mole) h (Btu/1b-mole) C.HSCH; (g) 100 29977 C.HSCHO (g) 0 O2 (g) 300 1972 N2 (8) 1128.57 1911.5 CO2 (g) 0 H:0 (g) 0 12 (lb-mole) 85.3 13 271.7 1128.57 11.9 19.8 h2 (Btu/lb-mole) 30876 -7838 2186.6 2116.53 -166400.59 -101577.75 Now, we need to do Energy Balance Total Energy IN Total Energy OUT Total Enthalpy of Feed + Heat Transfer Total Enthalpy of Product (1) Total Enthalpy = Sum of (Molar flowrate * Specfic enthalpy of that component at given Temp) Ση,h, over all components Total Enthalpy of Feed 100 * 29977 + 300 * (1972) + 1128.57 * 1911.5 5746561.56 Btu Total Enthalpy of Product 85.3 * 30876 +13 * (-7838) + 271.7 * 2186.6 +1128.57 * (2116.53) + 11.9* (-166400.59) + 19.8* (-101577.75) 1523173.81 Btu Substituting all the values in equation (1), 5746561.56 + Heat Transfer 1523173.81 Heat Transfer - 4223387.7 Btu Negative sign indicates this much heat is being removed. This heat is removed by water in jacket. Initial Temp of water (T1w) 80 F Final Temp of water (T2w) 105 F Cp of liquid water (Cpw) 1 Btu/lb. F Assuming no heat loss to surroundings, Energy gained by water - 1 * Heat Transfer 4223387.7 Btu

Let mw be mass of water. Then, Energy gained by water mw * Cpw * (T2w - Tiw) 4223387.7 = mw*1*(105-80) mw 168935.508 lb SCALING: and 34.3 lbm water is Here, it is said that complete condensation of water in product is do obtained in 4 hours Actual Rate of condensation of water lbm/hr = 34.3 8.575 lbm/hr 4 In our product gas, we had 19.8 lb-mole of water. Moles of water in product = 19.8 lb-mole Mass of water in product 19.8 * 18.016 lb Thus, water that would have been condensed 356.72 lbm 356.72 lbm This is because we considered 100 lb-mole of Toluene as basis. Now we need to scale everything for given rate of water condensation. Scaling Factor Actual Rate of condensation of water hr Water condensed for given basis 8.575 356.72 0.024 hr! Actual Volumetric flowrate of feed is: Scaling Factor * V feed = 0.024 * 904091.07 ft/h 21698.19 ft3/h 21.698 * 10 ft/h (ANSWER) Vout Actual Volumetric flowrate of product is: Scaling Factor * V prod = 0.024 * 937501.25 ft/h 22500.03 ft /h 22.5 * 10 ft/h (ANSWER) Actual Heat Transferred = Scaling Factor * Energy gained by water calculated before 0.024 * 4223387.7 Btu/h = 101361.31 Btu/h 101.36 * 103 Btu/h (ANSWER)

Required cooling water Scaling Factor * mw 0.024 * 168935.508 lb 4054.45 lb/h Density of water 62.4 lb/ft? Vol. flowrate of water required = Mass flowrate of water Density of water = 4054.45 62.4 ft/h 4054.45 7.48052 II gal/min 62.4 60 (1 ft3 = 1 hr 7.48052 gallon 60 min) 8.1/gal/min (ANSWER)