Chemical Reaction Engineering. Lecture 7

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1 hemical Reaction Engineering Lecture 7

2 Home problem: nitroaniline synthesis the disappearance rate of orthonitrochlorobenzene [ ] d ONB ra k ONB NH dt Stoichiometric table: [ ][ ] 3 hange Remaining* oncentration** ONB F -F F (1-) (1-) NH 3 B F B -F F (Θ B -) (Θ B -) N F F F (Θ +) (Θ +) NH 4 l D F D F F (Θ D +) (Θ D +) *) ONB is a limiting reagent (from stoichiometry) **) reaction in liquid, thus the volume is constant

3 Home problem: nitroaniline synthesis the disappearance rate of orthonitrochlorobenzene in terms of conversion [ ][ ] ( )( ) r k ONB NH k Θ 1 a 3 B rate of the reaction at 188º, 5º, 88º when.9: kt ( ) e E 1 1 E / RT R T T1 kt ( ) kt ( 1) e k(188 ) cal / mol J / cal k(5 ) k(188 ) e 1 r a r a k(88 ) r a

4 Home problem: nitroaniline synthesis find the reactor volume for STR at.9: 5º 188º 88º Reaction rate 3.7*1-7 3.*1-4.9*1-3 F v r r 1. m m 3 find the reactor volume for PFR at.9: at 88º:.15 m 3 at 5º: 1.14*13 m 3 v d r a Levenspilplot, T a olume (.9) 8.6*1 3 m Levenspilplot, T5 a F/-ra dv( x) F/-ra dv( x) x.9 x.9 onversion, onversion,

5 Our starting point: ; ; Fitting the data Hippo s stomach 1.45m 3 flow rate 4 kg/day, with density of grass 365kg/m 3, volumetric rate.13 m 3 /day

6 Hippo s digesting problem The hippo has picked up a river fungus and now the effective volume of the STR stomach compartment is only. m 3. The hippo needs 3% conversion to survive. Will the hippo survive? STR: F r PFR:.1.5. ( ).6.75 ( (1 x) dx 1 x _pfr( 1).15.1 if pfr.15 m3 as before, 1~.31, so the hippo survives

7 Hippo s digesting problem The hippo had to have surgery to remove a blockage. Unfortunately, the surgeon, Dr. No, accidentally reversed the STR and PFR during the operation. Oops!! What will be the conversion with the new digestive arrangement? an the hippo survive? PFR: ( ) ( (1 x) 1 x dx if pfr.15 m 3 as before, 1 ~.11 _pfr( 1).1 STR: v3() x.4.3 ( ) F 1 r if pfr.45 m 3 as before, ~ x.45

8 Design structure for isothermal reactors

9 Example: batch operation alculate time necessary to achieve given conversion for irreversible nd order reaction. Mole balance: N d dt r B Rate law: r k Stoichiometry: ( ) d dt 1 ombining: k ( ) 1 t 1 k ( 1 )

10 Typical reaction times 1 st order reaction: t 1 1 ln k 1 ( ) nd order reaction: t 1 k ( 1 )

11 STR single STR mole balance τ ( r ) ( v r ) F exit exit rate law (1 st order) r k Stoichiometry: ( ) 1 1 ombining τ or k 1 1 +τ k Dahmköler number r Dahmköler rate of the convective number is transport a ration of at the the rate entrance of the reaction to the Da F For the 1 st order: For the nd order: Da Da r F k v r F k v τ k τ k

12 STR in Series concentration flowing to the nd reactor design equation the nd reactor so, ( 1+ τ k )( 1+ τ k ) if the reactors have the same size and F F r τ k 1 1 ( ) v 1 k temperature: ( 1+ τ k )

13 STR in Series in terms of Damköler number for n reactors conversion: ( ) ( 1+ Da) n ( 1+ Da) 1 ; 1 n n ( 1+ Da) ( 1+ Da) 1 when the Damköler number is above 1, a high conversion is achieved in few reactors

14 STRs in parallel Iet s consider identical reactors with the feed equally distributed, than conversion factors and the reaction rates are the same i Fi r or F n n r So, the situation is identical to a single reactor with the size equal to the total volume of all reactors.

15 Tubular reactors: liquid phase Design equation F d r d in the absence of pressure drop F d r Let s consider a nd order reaction products rate law: r k stoichiometry: ( ) 1 F v k k d 1 ( 1 ) ( ) τ k Da 1+ τ k 1+ Da

16 Tubular reactors: gas phase For T, P constant, the concentration is a function of conversion F ( 1 F ) stoichiometry: v v 1+ ε 1+ ε ( ) ( ) rate law: combining: r k F k ( 1+ ε ) ( 1 ) d if e<: i.e. number of moles : flow rate, the residence time, so if e>: i.e. number of moles : flow rate, the residence time, so

17 Pressure drop in Reactors in the liquid phase pressure drop doesn t lead to any significant volume and therefore concentration changes can be neglected. in the gas phase can be an important factor volumetric flow: ( 1 ) P T v v +ε P T j F ( η ) ( η ) F Θ + Θ + j j j j j v P T v ( ) ( 1 ) 1+ ε + ε P T PT P T

18 Pressure drop in Reactors to account for a pressure drop we have to use differential form of the equation rate law: F d stoichiometry: r dw r k for isothermal operation: r ( 1 ) ( 1+ ) ( 1 ) ( 1+ ) P T ε P T k P T ε P T ( ) ( ε ) ( ) ( ε ) d 1 1 P T 1 k P k dw F 1+ P T v 1+ P

19 Ergun equation pressure drop in a packed porous bed is described by Ergun equation: dp G 1 φ 15( 1 φ ) µ 1.75G 3 + dz ρgcdp φ Dp dominant for laminar flow dominant for turbulent flow φ D p G g c ρu porosityvolume of void/total volume particle diameter superficial mass velocity conversion factor, 1 for metric system in terms of catalyst weight: W (1 φ) zρc

20 for the flow in pipes: Pressure drop in a pipe dp du fg G dl dl ρd where: u average velocity; f Fanning friction factor; G- mass flow rate. PdP dp fg ρ G + P dl PdL D P P P L P G f + ln ρ D P P P 1 4 fg ρ DP c 1/

Chemical Reaction Engineering

Chemical Reaction Engineering Lecture 6 hemical Reaction Engineering (RE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 6 Tuesday 1/9/13 Block