Mass balance in a fixed bed reactor is similar to that of a plugflow reactor (eq. 1.1): dx dv. r F (1.1) Recalling dw = B dv, then. r F. dx dw (2.

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3 Mass balance in a fixed bed reactor is similar to that of a plugflow reactor (eq..): d dv r (.) Recalling dw = B dv, then d dw r B (.)

4 or a reaction: + bb c + dd Species eed Rate hange within Reactor Effluent Rate = = ( ) B B = B b B = ( B b) = c = ( + c) D D = D d D = ( D + d) I I = I I = I I = + (d + c b ) = + OL ELUEN

5 or the above reaction, ( + b) mol of reactants give (c + d) mol of product. In flow systems where this type of reaction occurs, the molar flow rate will be changing as the reaction progresses. Because only equal numbers of moles occupy equal volumes in the gas phase at the same temperature and pressure, the volumetric flow rate will also change. ssuming ideal gas behavior: V N R V N R V N V N (.)

6 Recalling the last row in the previous table: N N N N N N N N = + y he above equation is further simplified by letting: change in total number of moles for complete conversion total number of moles fed to thereactor N N d c b y Equation (.) now becomes: V V (.) (.4)

7 Kalau reaksinya berlangsung sempurna ( = ): N * N N N N erubahan umlah mol kalau reaksinya berlangsung sempurna ( = ): N * N N N N N d c b N d c b N N y

8 o derive the concentrations of the species in terms of conversion for a variable-volume flow system, we shall use the relationships for the total concentration. he total concentration at any point in the reactor is v N V R (.5) t the entrance of the reactor: v N V R (.6)

9 aking the ratio of eq. (.6) to eq. (.5), we have upon rearrangement v v rom the above table: (.7) Substituting eq. (.8) in eq. (.7): (.8) v v v

10 v v y v v v (.9) We can now express the concentration of species for a flow system in terms of conversion: v v v (.)

11 Substituting and in terms of conversion in eq. (.) yields y y (.)

13 Energy balance in a fixed bed reactor is similar to that of a plug-flow reactor (eq..56): d dv Ua r H a p i pi Rx (.56) Recalling dw = B dv, then d dw Ua r H a B p i Rx pi (.)

14 he maority of gas-phase reactions are catalyzed by passing the reactant through a packed bed of catalyst particles. We now must determine the ratio / as a function of volume V or the catalyst weight, W, to account for pressure drop. We then can combine the concentration, rate law, and design equation. However, whenever accounting for the effects of pressure drop, the differential form of the mole balance (design equation) must be used.

15 he equation used most to calculate pressure drop in a packed porous bed is the Ergun equation: d dz G g D c p 5 D p.75g (.) Where : pressure, lb/ft : porosity (volume of void / total bed volume) D p : diameter of particle in the bed, ft : viscosity of gas passing through the bed, lb m /ft h z : length down the packed bed of pipe, ft u : superficial velocity = volumetric flow : cross sectional area of pipe, ft/h : gas density, lb/ft G : u = superficial mass velocity, (g/cm s) or (lbm/ft h)

16 In calculating the pressure drop using the Ergun equation, the only parameter that varies with pressure on the right-hand side of eq. (.) is the gas density,. We are now going to calculate the pressure drop through the bed. Because the reactor is operated at steady state, the mass flow rate at any point down the reactor, m (kg/s), is equal to the entering mass flow rate, m (i.e., equation of continuity), m m v v

17 Recalling eq. (.9), we have v v (.4) ombining eqs. (.) and (.4) gives:.75g D 5 D g G dz d p p c v v (.9)

18 Simplifying yields: d dz (.5) where: G g 5 c D p D p.75g (.6)

19 or tubular packed-bed reactors we are more interested in catalyst weight rather than the distance z down the reactor. he catalyst weight up to a distance of z down the reactor is: weight of catalyst volume of solids density of solid catalyst W c z (.7) where c is the cross-sectional area. he bulk density of the catalyst, B (mass of catalyst per volume of reactor bed), is ust the product of the solid density,, the fraction of solids, ( ) : B (.8)

20 Using the relationship between z and W [eq. (.7)] we can change our equation (.5) into: d dz (.5) d dw c (.9) urther simplification yields: d dw (.) where: c (.)

21 We note that: when is negative, the pressure drop will be less than that for =. When is positive, the pressure drop will be greater than when =.

22 d dw r B r' r B (.) r H d Ua a Rx (.) dw B p i pi d dw (.)

23 EMLE. alculate the pressure drop in a 6 ft length of /-in. schedule 4 pipe packed with catalyst pellets /4-in. in diameter when 4.4 lb/h of gas is passing through the bed. he temperature is constant along the length of pipe at 6. he void fraction is 45% and the properties of the gas are similar those of air at this temperature. he entering pressure is atm. SOLUION d dw

24 d dw d dw W d W dw W W W W

25 W c c z W c z W z W z

26 c G g 5 c D p D p.75g or ½-in. schedule 4 pipe, =.44 ft G 4.4 lbm h.4 ft lb 78. h.ft m or air at 6 and atm: =.67 lb m /ft.h =.4 lb m /ft

27 D p = ¼ in =.8 ft g c lb lb m f.ft.h lb f ft atm 64. ft 44 in 4.7 lb in f atm ft ka m

28 z atm 7.5 atm

29 EMLE. Ethylene oxide is made by the vapor-phase catalytic oxidation of ethylene with air: O H 4 + ½ O H H We want to calculate the catalyst weight necessary to achieve 6% conversion. Ethylene and oxygen are fed in stoichiometric proportions to a packed-bed reactor operated isothermally at 6. Ethylene is fed at a rate of. Ib mol/s at a pressure of atm. It is proposed to use banks of ½-in.-diameter schedule 4 tubes packed with catalyst with tubes per bank. onsequently, the molar flow rate to each tube is to be 4 Ib mol/s. he properties of the reacting fluid are to be considered identical to those of air at this temperature and pressure. he density of the a ¼-in.-catalyst particles is lb/ft and the bed void fraction is.45

30 he rate law is: r ' k B lbmol lb cat.h k lb mol.4 atm.lb cat.h at 6 SOLUION Reaction: + ½ B Mole balance: d dw r '

31 Rate law : B B B ' kr R R k k r Stoichiometry: gas phase, isothermal v v v v (.9) (.) or an isothermal gas phase reaction:

32 Ethylene and air are fed to the reactor where ethylene and oxygen are in stoichiometric proportions: B I B I.5 79 B.88 B.5 y N N /.5.48

33 B B B.5.5

34 B ' r kr '.5.5 kr r '.5 kr r kr.6 r ' W.6kR r ' ' W k' r where k =.6 kr =.6 k

35 d dw d dw r ' k' W k' d W W dw k' ln W W k' ln (a)

36 w W W dw W W W W

37 arameter evaluation per tube (i.e., divide feed rates by ): Ethylene = -4 lbmol/s =.8 lbmol/h Oxygen B =.5-4 lbmol/s =.54 lbmol/h Inert (N ) Jumlah I mol lbmol s mol N O = lbmol/s =. lbmol/h = + B + I =.65 lbmol/h y y y. atm. atm

38 k'.6k.6 lbmol.4 atm.lb cat.h atm lbmol.66 h.lb cat or 6% conversion, eq. (a) becomes W k' ln W. k'

39 m lbmol.8 h lb 8 lbmol.4 lb h m B lbmol.54 h lb lbmol 7.8 lb h m I lbmol. h lb 8 lbmol lb h m lb 4.4 h m G c 4.4 lb h.44 ft lb 78. h.ft

40 or air at 6 and atm: =.67 lb m /ft.h =.4 lb m /ft D p = ¼ in =.8 ft g c lb lb m f.ft.h G g 5 c D p D p.75g

41 lb f ft atm 64. ft 44 in 4.7 lb in f.775 atm ft c lb cat

42 W. k'.66. lb cat lbmol.66 lb cat.h.66 lb cat lbmol.8 h = 45.4 lb of catalyst per tube = 45,4 lb of catalyst total

43 W.66 lb cat 45.4 lb cat. 496 =.496 = 4.96 atm = = 4.96 = 5.4 atm

Chemical Reaction Engineering. Lecture 7

Chemical Reaction Engineering. Lecture 7 hemical Reaction Engineering Lecture 7 Home problem: nitroaniline synthesis the disappearance rate of orthonitrochlorobenzene [ ] d ONB ra k ONB NH dt Stoichiometric table: [ ][ ] 3 hange Remaining* oncentration**