NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor. F j. T mo Assumptions:

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1 NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flw Reactr T T T T F j, Q F j T m,q m T m T m T m Aumptin: 1. Hmgeneu Sytem 2. Single Reactin 3. Steady State Tw type f prblem: 1. Given deired prductin rate, cnverin and kinetic and ther parameter, determine the required reactr ize, heat duty and temperature prfile. 2. Given reactr ize, kinetic, etc., determine the cmpitin f the exit tream. Let u cnider a ingle reactin υ j A j = 0 (1) with the rate given by r = 0 e E 1 / RT Π C j α j k 20 e E 2 / RT Π C j β j (2) with υ M j / A j υ C j = C A T P A (3) 1+ ε A TP The ma balance in the reactr fr pecie j can be written a: df j dv = υ j r (4) r v = 0 F j = F j (4a) 1

2 F A d dv = ( v A )r = R A (4 ) V = 0 = 0 (4 a) The energy balance baed n (a) negligible change in ptential and kinetic energy and (b) n wrk ther than flw wrk i d dv j=1 F j H j + q v = 0 (5) V = 0 F j H 1 = F j H j (5a) Baed n further aumptin f (c) ideal mixture and (d) ideal gae ne get: j=1 F j C ~ p j dt dv j=1 H ~ j df j dv + q = 0 v (6a) Uing the idea f (e) mean pecific heat which are cntant and (f) cntant heat f reactin, ne get dt (Qρ)C p dv + ( ΔH )r + q = 0 (6) r v Qρ = m tt i the ma flw rate which i cntant J q v i the rate f heat additin per unit reactr vlume m 3 The implet cntitutive relatinhip fr the rate f heat exchange i: q v = Ua v (T m T) (7) m 2 a v - area fr heat tranfer per unit reactr vlume m 3 The equatin t be lved imultaneuly are: dx F A a dv + υ Ar = 0 (A) q dt v QρC pm dv ( ΔH r) r + U av (T T m ) = 0 (B) 2

3 Q m ρ m C pm q dt v m dv U av(t m T) = 0 (C) V = 0; = 0; T = T,, (T m = T m fr ccurrent flw) V = V; (T m = T m fr cuntercurrent flw) (D) and G du dz + dp dz + F = 0 (E) G = ρu - ma velcity P = preure z = V A - axial ditance u = Q A - velcity A cr ectinal reactr area F frictinal le Equatin (E) i the mmentum balance. Hwever thi equatin i uually lved eparately and a mean preure i elected fr evaluatin f ga cncentratin in eq (3). Fr gae the ue f ma fractin, w j, and extent per unit ma, ξ '' i recmmended. (See lecture 1). The equatin can then be written a: '' dξ G dz = r (8) G dt dz = β '' '' r + q v (9) z = 0 ξ '' = 0, T = T (10) β '' = ΔH r C p ; q '' v = q v C p (11) 3

4 where the rate i expreed by: r = 0 e E 1 / RT Π k 20 e E 2 / RT α j m 1+ υ tt j ξ '' F j C j Π C j α j β j m 1+ υ bt j ξ '' F j j=1 T P 1 TP 1+ υ j M av ξ j=1 β j υ j T P 1 TP 1+ ( y)m av ξ "" v j (12) M av - average mlecular weight at feed cnditin m tt = GA ma flw rate m tt F j = M j w j w j ma factin f j in the feed. Fr liquid ne can write dξ dτ = r (13) dt dτ = β r + q v (14) τ = 0 ; ξ = 0 ; T = T (15) β = ΔH r ρc p ; q v = q v '' = Q v ρc p ρ (16) where the rate i given by r = 0 e E 1 / RT Π ( C j + υ j ξ) α j k 20 e E 2 / RT Π ( C j + υ j ξ) β j (17) τ = z u = V - reidence time alng the reactr. Q 4

5 Frm eq (8) and (9) r (13) & (14) we can alway get the fllwing relatinhip between temperature and extent r T = T + β '' ξ '' + 1 G z q '' v dz (18a) τ T = T + β ξ + q v dτ (18b) '' Fr adiabatic peratin (q v = 0, q v = 0 ) thi yield the equatin f the adiabatic line, i.e extent and temperature atify the relatinhip belw at any and every pint f the reactr T = T + β '' ξ '' T = T + β ξ (19a) (19b) The maximum fractinal adiabatic temperature rie i given by the Prater number jut like in the cae f a CSTR. ( = β = ΔH r)c A (20) T ( υ A )T ρc p ΔT ad max Baic type f prblem 1. The temperature in the reactr i precribed a. T(z) = T ithermal reactr. Integrate (8) r (13) and find extent alng the reactr. Frm eq. (9) r (14) find the heat additin/remval requirement alng the reactr and the verall heat duty fr the reactr. b. T(z) pecified. Integrate (8) r (13) find ξ (z). Ue ξ (z) and T(z) in eq (9)r (14) t get q v (z) 2. The heat additin (remval) rate i precribed a) Adiabatic peratin. T = T + β '' ξ '' r T = T + βξ. Subtitute int eq (8) r (13) and integrate b) Heat duty i precribed. q v '' (z) r q v (z) precribed. Simultaneuly integrate (8) r (9) r ubtitute z T = T + β '' ξ '' + 1 G q '' vdz int (8) and integrate. 5

6 3. Rate f heat additin (remval) cntrlled by anther equatin q v = Ua v (T T m ) a) T m = cnt. Integrate eq (8) and (9) r eq (13) and (14) imultaneuly. Thi i the cae when reactr tube are immered in biling medium r cndening medium. b) T m determined with T and ξ by equatin (A) t (E). dt G m m dz = κ ' m T m T ( ) κ m = Ua v m C pm G m = Q mρ m A m Nte: With ccurrent cling a PFR can be kept ithermal with cuntercurrent cling it cannt in the cae f n-th rder reactin. Prve that fr an exercie. There i alway a unique teady tate in a PFR. Main prblem with PFR i: ht pt frmatin parametric enitivity and temperature runaway. Claical example f temperature runaway preented by Bilu & Amundn (AIChE J., 2, 117 (1956)). PFR cled frm the wall t cntant T m = T wall T T m = τ 6

7 A ht pt i frmed due t a very mall change in wall temperature. The ytem hw extreme parameter enitivity. Reactin runaway i the phenmenn when a mall change in feed cncentratin, temperature, flw rate r in clant temperature trigger a dramatic change in he temperature prfile and lead t runaway reactin and explin. Exact criteria fr runaway are difficult t develp. Apprximate criteria are given n the encled graph. Example 1 A reverible firt rder reactin (cnidered earlier in a CSTR) i nw t be per frmed in a PFR. A R (liquid phae) = 5x10 8 e 12,500 / RT (min 1 ) k 2 = 3.4x10 21 e 32,500 / RT ( min 1 ) ΔH r = 20,000 cal /ml ΔG 298 ρc p 2,000 (cal /lit C) C A = 2 (ml /lit) = 2,500 cal /ml If the feed rate i Q = 100 (lit/min) and the PFR ize i V = 1,500 (lit): a) find final cnverin in an ithermal reactr perated at 0, 10, 20, 100 C b) determine cnverin in an adiabatic reactr if the feed i at i) 0 C, ii) 20 C, c) if the maximum permiible temperature i 80 C determine the ptimal temperature prfile alng the reactr neceary t maximize exit cnverin. d) If the deired cnverin i 85% find the minimum reactr vlume and the deired heat remval rate alng the reactr. Permiible temperature range i 0 t 100 C. Slutin a) Fr an ithermal reactr nly the ma balance ha t be lved τ = V dx = C A A Q r A r A = C A k 2 C R = C A [ (1 ) k 2 ] r A = C A 1 x (1 x ) A Ae ince k 2 = e K = x 1 Ae e ( r A ) = k C 1 A (e ) e = K e 1+ K = + k 2 7 ( )

8 ( r A ) = ( + k 2 )C A (e ) x 1 A dx τ = A 1 x = n Ae + k 2 e + k 2 e Slve fr cnverin 1 exp( (1+ 1 K )τ) = e 1 exp k 1 τ τ = 1500 =15 min e 100 We get the fllwing reult: T K x ae x a Same a equilibrium cnverin The reactr pace time i large that abve 50 C practically equilibrium cnverin i btained. a) The adiabatic perating line i T = T + β A C A β A = ΔH r A = 20,000 ρc p 2,000 =10 lit C ml C A = 2 ml lit T = T

9 Subtitute thi relatinhip int the ma balance and integrate: dx C A A dτ = (k + k )C (x x ) = k C (k + k )C x 1 2 A Ae A 1 A 1 2 A A τ = 0 = 0 = 0 e E 1 / RT ad = 0 e E 1 / R(T +20 ) k 2 = k 20 e E 2 / R(T +20 ) e = K 1+ K = + k 2 Thu integrate numerically ( ) ; τ = 0 = 0 d dτ = 0 e E 1 / RT (T +20x ) A 0 e E1 / R( T+20xA ) + k20 e E 2 / R(T +20 ) d dτ = 5x108 e 12, (T +20x ) A 5x10 8 e 12, (T +20x ) A x10 21 e 32,500 Deired reult i btained at τ = 15. Alternatively we culd lve by trial and errr the fllwing integral: 1.987(T +20 x ; τ = 0 ; A = 0 τ =15 = We find: dx 5x10 8 e 12, (T +20x ) 5x10 8 e 12, (T +20x ) + 3.4x10 21 e 32, (T +20x) x i) T = 0 C = 273 K = 0.78 ΔT adiabatic =15.7K =16K T = 289 K ii) T = 20 C = 293 K = 0.94 = e ΔT adiabatic =18.8 =19K T = 292K c) T maximize cnverin at given pace time we huld fllw the line f maximum rate. (E T m = 2 E 1 /R) 10,065 n k E n x = A x n A 0 E 1 1 x A 1 Since maximum permiible temperature i 80 C (353 K) we have t preheat the feed t 33 K, cl the reactr and keep it ithermal a 353 K until the lcu f maximum rate i reached and then run alng the lcu f maximum rate. 9

10 The interectin f the ithermal line T = 353 K and the T m line determine up t which pint the reactr ha t be run ithermally. 10,065 T = 353 = T m = x n A 1 10, x30.51 exp 353 = = , x exp τ = + k 2 dx 1 = (e ) ( + k 2 ) n e e At 80 C (353 K) frm the table given earlier τ = 9.17(1+ 1 n = 0.017(min) 0.35 ) The ithermal peratin huld ccur in the very entry ectin f he reactr. After that the T m line huld be fllwed. d dτ = 5x108 e 12, T m (1 ) 3.42x10 21 e 32, T m 10,065 T m = x n A 1 τ = = Deired reult at τ = 15 =0.988 T exit = 288K Really ne huld preheat nly t adiabatic line. Adiabatic line huld end at T = 353 K, = Hence, the fluid mut be preheated up t T = T β A C A = x0.119 = 350K 10

11 The graphical repreentatin f part (a-c) ha the fllwing frm: e T a. Ithermal. Slid line are perating line fr τ = 15 min e b. Adiabatic. Adiabatic line with τ = 15 T T m e T max c. Operating alng the lcu f maximum rate d) Permiible temperature range i 0 C t 100 C. We want minimum reactr ize fr = Preheat t 100 C, run alng the lcu f maximum rate 11

12 τ = = 0.85 dx 5x10 8 e 12, T m 5x10 8 e 12, T m + 3.4x10 21 e 32, T m x 10,065 with T m = x n A 1 τ =1.8min Thu with Q = 100 lit/min we need nly V = 160 liter The deired temperature prfile alng the reactr i preented in the encled graph. The heat remval per unit vlume i q Q = ρc p (T T) + ( ΔH r )C A = 2,000(100 T) + (20,000x2) Thi curve i al preented in the figure. The ttal heat denity i: q Q tt = 2,000(100 70) + 40,000x0.85 =1.56x10 5 (cal /lit) With Q = 100 lit/min q tt =1.56x10 7 (cal /min) Fr cmparin, if cling failed and reactr ran adiabatically with T = 100 C ne wuld get adiabatic = 0.068,T exit =126 C The adiabatic temperature prfile i hwn al n the encled figure. 12

13 Extenin t Multiple Reactin υ ij A j = 0 i =1,2,...R (1) j=1 r R df j dv + υ r ij i = 0,2,...R (2) R i=1 υ ij i=1 d X i dv + d X i dv + r i = 0 ( H j ) d F j dv R υ ijr i = 0 i=1 + q v = 0 (3) (2a) V = 0 ; F j = F j ( X i = X i ) ; H j = H j With the uual aumptin made abut the energy balance (ee the lecture n CSTR) ne get: dt F j C ~ p j dv + j=1 R i=1 ( ΔH r Ti ) r i + q v = 0 (4) The equatin t be lved fr a et f multiple reactin are: d X i dv + r i = 0 i =1,2...R (A) ρc p Q dt R dv + ( ΔH r i )r i + q v = 0 i=1 V = 0 ; X i = X i T = T ρq = cnt r i = k i10 e E 1i / RT Π C j α ij k i20 e E 2i RT Π C j β ij (B) (C) with 13

14 1+ ρt C j = C j ρ T 1+ R υ ij i=1 F j R X i υ ij i=1 j=1 F bt X i (D) The cntitutive relatinhip fr q v i: q v = U av (T m T) a) T m = cnt b) T m i gverned by anther D.E. ρ m Q m C pm dt m dv q v = 0 (E) V = 0 T m = T m (ccurrent flw) V = V T m = T m (cuntercurrent flw) Prblem Cnider the reactin intrduced in the lat lecture A R R= C A -k 2 C R (ml/lit ) = exp 7 83,700 x10 3 ( -1 ) RT k 2 exp , ( -1 ) RT ΔH r = 80,000 (J /ml) C ~ p = 40(J /ml K) (Activatin energie given in jule) 1. The abve reactin ccur in liquid phae! Permiible temp range f peratin i 300<T<900 K. Feed cnditin: Q = 100 (lit/) ; T = 300 K ; C A = 1(ml/lit) Yu have a V = 100 liter PFR. Hw wuld yu perate thi reactr if the nly bjective i t maximize the prductin rate f R. 14

15 a) What i maximum F R. b) What are final and ΔT. c) What i the prfile f heat additin r remval fr every 10% f reactr vlume. d) What i the verall heat duty fr the reactr and any heat exchanger preceding it. e) Sketch yur ytem. 2. The abve reactin ccur in ga phae. The ga feed ate i Q =100(lit /) at T = 300K, P = 24.6 atm The feed i 50%A, 50% inert. Permiible temperature range i 250< T < 900 K. Preure i cntant in the reactr. Gae tart t cndene belw 250 K. Deired cnverin i 85%. a) What reactr vlume i needed if yu perate alng the lcu f maximum rate? b) What i the ditributin f heat duty alng the reactr? c) What i the prductin rate f R? 3. Fr the abve prblem what wuld F R and be if yu had a reactr (PFR) f V = 100 liter available? 4. Suppe that the reactr can nly be perated adiabatically and the deired cnverin i 85%. Minimize the required reactr ize. a) What reactr type d yu recmmend? b) What feed temperature wuld yu ue? c) What i the heat duty? 15