NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor

NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor T o T T o T F o, Q o F T m,q m T m T m T mo Aumption: 1. Homogeneou Sytem 2. Single Reaction 3. Steady State Two type of problem: 1. Given deired production rate, converion and kinetic and other parameter, determine the required reactor ize, heat duty and temperature profile. 2. Given reactor ize, kinetic, etc., determine the compoition of the exit tream. Let u conider a ingle reaction υ =1 A = 0 (1) with the rate given by r = k 10 e E 1 / RT Π α C k 20 e E 2 / RT Π β C (2) =1 = 1 with υ M / A x υ A C = C A T o P Ao (3) 1 + ε A TP o The ma balance in the reactor for pecie can be written a: df dv = υ r (4) v = 0 F = F o (4a) 1

or F Ao d dv =( v A) r =R A (4 ) V = 0 = 0 (4 a) The energy balance baed on (a) negligible change in potential and kinetic energy and (b) no work other than flow work i d dv =1 F H +Ý q v = 0 (5) V = 0 F H 1 = F o H o (5a) Baed on further aumption of (c) ideal mixture and (d) ideal gae one get: =1 F C ~ p dt dv =1 H ~ df dv + q ś v = 0 (6a) Uing the idea of (e) mean pecific heat which are contant and (f) contant heat of reaction, one get dt (Qρ)C p dv + (ΔH r )r + q ś v = 0 (6) Qρ = Ý m tot i the ma flow rate which i contant q Ý J v m 3 i the rate of heat addition per unit reactor volume The implet contitutive relationhip for the rate of heat exchange i: q ś v = Ua v (T m T) (7) a m 2 v m 3 - area for heat tranfer per unit reactor volume The equation to be olved imultaneouly are: F ao d dv + υ Ar = 0 (A) 2

QρC pm ś dt 647 q v 48 dv (ΔH ) + U (T T ) r r av m = 0 (B) mq m ρ m C pm ś 647 dt q v 48 m dv U (T T) av m = 0 (C) V = 0; = 0; T = T o,, (T m = T mo for cocurrent flow V = V; (T m = T mo for countercurrent flow) (D) and G du dz + dp dz + F = 0 (E) G = ρu - ma velocity P = preure z = V A - axial ditance u = Q A - velocity A cro ectional reactor area F frictional loe Equation (E) i the momentum balance. However thi equation i uually olved eparately and a mean preure i elected for evaluation of ga concentration in eq (3). For gae the ue of ma fraction, w, and extent per unit ma, ξ '' i recommended. (See lecture 1). The equation can then be written a: '' dξ G dz = r (8) G dt dz = β'' '' r + q v (9) z = 0 ξ '' = 0, T = T o (10) 3

β '' = ΔH r C p ; q '' v = q v C p (11) where the rate i expreed by: r = k 10 e E 1 / RT Π = 1 «α m 1 + υ tot ξ F o C o α =1 T o P 1 TP o 1 + υ M avo ξ =1 υ k 20 e E 2 / RT Π =1 β m Ý 1 +υ bot F o C o ξ '' β T o P 1 TP o 1+ ( y)m avo ξ "" v (12) M avo - average molecular weight at feed condition Ý m tot = GA ma flow rate «m tot F o = M w o w o ma faction of in the feed. For liquid one can write dξ dτ = r (13) dt dτ = β r + q v (14) τ = 0 ; ξ = 0 ; T = T o (15) β = ΔH r ρc p ; q v = q v '' = Q v ρc p ρ (16) where the rate i given by r = k 10 e E 1 / RT Π ( C o + υ ξ) α = 1 k 20 e E 2 / RT Π ( C o + υ ξ) β = 1 (17) 4

τ = z u = V Q - reidence time along the reactor. From eq (8) and (9) or (13) & (14) we can alway get the following relationhip between temperature and extent or T = T o + β '' ξ '' + 1 G q '' vdz z o (18a) τ T = T o + β ξ + q dτ (18b) v o '' For adiabatic operation (q v = 0, q v = 0 ) thi yield the equation of the adiabatic line, i.e extent and temperature atify the relationhip below at any and every point of the reactor T = T o + β '' ξ '' T = T o + β ξ (19a) (19b) The maximum fractional adiabatic temperature rie i given by the Prater number ut like in the cae of a CSTR. ( = β = ΔH r )C Ao (20) T o ( υ A )T o ρc p ΔT ad max Baic type of problem 1. The temperature in the reactor i precribed a. T(z) = T o iothermal reactor. Integrate (8) or (13) and find extent along the reactor. From eq. (9) or (14) find the heat addition/removal requirement along the reactor and the overall heat duty for the reactor. b. T(z) pecified. Integrate (8) or (13) find ξ (z). Ue ξ (z) and T(z) in eq (9)or (14) to get q v (z) 2. The heat addition (removal) rate i precribed a) Adiabatic operation. T = T o + β '' ξ '' or T = T o + βξ. Subtitute into eq (8) or (13) and integrate 5

b) Heat duty i precribed. q v '' (z) or q v (z) precribed. Simultaneouly integrate (8) or (9) or ubtitute z T = T o + β '' ξ '' + 1 G q '' vdz into (8) and integrate. o 3. Rate of heat addition (removal) controlled by another equation q ś v = Ua v (T T m ) a) T m = cont. Integrate eq (8) and (9) or eq (13) and (14) imultaneouly. Thi i the cae when reactor tube are immered in boiling medium or condening medium. b) T m determined with T and ξ by equation (A) to (E). ( ) dt G m m dz = mκ ' m T m T κ m = Ua v m C pm G m = Q mρ m A m Note: With cocurrent cooling a PFR can be kept iothermal with countercurrent cooling it cannot in the cae of n-th order reaction. Prove that for an exercie. There i alway a unique teady tate in a PFR. Main problem with PFR i: hot pot formation parametric enitivity and temperature runaway. Claical example of temperature runaway preented by Bilou & Amundon (AIChE J., 2, 117 (1956). PFR cooled from the wall t contant T m = T wall. 6

440 342.5 420 337.5 400 T 380 360 335 340 320 300 330 320 310 T m = 300 0 10 20 30 A hot pot i formed due to a very mall change in wall temperature. The ytem how extreme parameter enitivity. τ Reaction runaway i the phenomenon when a mall change in feed concentration, temperature, flow rate or in coolant temperature trigger a dramatic change in he temperature profile and lead to runaway reaction and exploion. Exact criteria for runaway are difficult to develop. Approximate criteria are given on the encloed graph.. Example 1 A reverible firt order reaction (conidered earlier in a CSTR) i now to be per formed in a PFR. A R (liquid phae) k 1 = 5x10 8 e 12,500 / RT (min 1 ) k 2 = 3.4x10 21 e 32,500 / RT( min 1 ) o ΔH r =20,000 cal/ mol ΔG 298 ρc p 2,000 (cal / lit o C) C Ao = 2(mol / lit) =2,500 cal / mol 7

If the feed rate i Q = 100 (lit/min) and the PFR ize i V = 1,500 (lit): a) find final converion in an iothermal reactor operated at 0, 10, 20, 100 C b) determine converion in an adiabatic reactor if the feed i at i) 0 C, ii) 20 C, c) if the maximum permiible temperature i 80 C determine the optimal temperature profile along the reactor neceary to maximize exit converion. d) If the deired converion i 85% find the minimum reactor volume and the deired heat removal rate along the reactor. Permiible temperature range i 0 to 100 C. Solution a) For an iothermal reactor only the ma balance ha to be olved τ = V dx = C A Q Ao o r o A r A = k 1 C A k 2 C R = C Ao [ k 1 (1 ) k 2 ] r A = k 1 C Ao 1 (1e ) in ce k 2 = k 1 e K = k 1 1 e e (r A ) = k C 1 Ao (x x Ae ) e = K Ae 1 + K = k 1 k 1 + k 2 (r A ) = (k 1 + k 2 )C Ao (e ) x 1 A dx τ = A 1 x = ln Ae k 1 + k 2 o e k 1 + k 2 e Solve for converion 1 exp(k 1 (1 + 1 K )τ ) = e 1 exp k 1 τ e τ = 1500 =15 min 100 ( ) 8

We get the following reult: T K k 1 x ae x a 273 1494 0.0498 0.999 0.526 283 407 0.112 0.998 0.813 293 121 0.239 0.992 0.965 303 40 0.486 0.975 0.974 313 13.5 0.943 0.931 0.931 323 5.0 0.755 0.833 0.833 333 2.0 3.149 0.662 0.662 343 0.81 5.46 0.448 0.448 353 0.35 9.17 0.262 0.262 363 0.16 15.0 0.139 0.139 373 0.08 23.8 0.071 0.071 Same a equilibrium converion The reactor pace time i o large that above 50 C practically equilibrium converion i obtained. a) The adiabatic operating line i T = T o + β A C Ao β A = ΔH r A = 20,000 ρc p 2,000 =10 lit o C mol C Ao = 2 mol lit T = T o + 20 Subtitute thi relationhip into the ma balance and integrate: dx C A Ao dτ =( k 1 + k 2 ) C Ao ( e ) = k 1 C Ao ( k 1 + k 2 ) C Ao τ = 0 = 0 k 1 = k 10 e E 1 / RT ad = k 10 e E 1 / R(T o + 20 ) k 2 = k 20 e E 2 / R(T o +20 ) e = K 1 + K = k 1 k 1 + k 2 Thu integrate numerically d dτ = k 10e E 1 / RT(T o + 20 x ) A k 10 e E1 / R( To+20 ) + k20 e E 2 / R(T o +20 ) τ = 0 = 0 9

12, 500 d dτ = 5x108 1.987(T e o +20 x ) A 5x10 8 e τ = 0 ; = 0 Deired reult i obtained at τ = 15. 12, 500 1.987(T o +20 ) + 3.4x10 21 e Alternatively we could olve by trial and error the following integral: We find: τ = 15 = o 32,500 1.987(T o +20 dx 5x10 8 e 12, 500 1.987(T o + 20 x ) 5x10 8 e 12,500 1.987(T o +20 x ) + 3.4x10 21 e 32,500 i) T o = 0 C = 273 K = 0.78 ΔT adiabatic = 15.7K = 16K T = 289 K ii) T o = 20 C = 293 K = 0.94 = e ΔT adiabatic = 18.8 = 19K T = 292K 1.987(T o +20 x ) x c) In order to maximize converion at given pace time we hould follow the line of maximum rate. T m = ( E 2 E 1 / R) ln k 20E 2 k 10 E 1 + ln 1 = 10, 065 x 30.51 + ln A 1 Since maximum permiible temperature i 80 C (353 K) we have to preheat the feed to 33 K, cool the reactor and keep it iothermal a 353 K until the locu of maximum rate i reached and then run along the locu of maximum rate. The interection of the iothermal line T = 353 K and the T m line determine up to which point the reactor ha to be run iothermally. T = 353 = T m = 10,065 x 30.51+ ln A 1 10,065 353x30.51 exp 353 = = 0.119 10,065 353x30.1 1+ exp 353 10

τ = 1 k 1 + k 2 dx 1 = (e ) (k 1 + k 2 ) ln e e 0.119 o At 80 C (353 K) from the table given earlier τ = 1 9.17(1+ 1 ln 0.35 ) 0.262 0.262 0.119 = 0.017(min) The iothermal operation hould occur in the very entry ection of he reactor. After that the T m line hould be followed. d dτ = 5x108 e 12,500 1.987T m (1 ) 3.42x10 21 e 32,500 1.987T m 10,065 T m = x 30.51 + ln A 1 τ = 0.017 = 0.119 Deired reult at τ = 15 =0.988 T exit = 288K Really one hould preheat only to adiabatic line. Adiabatic line hould end at T = 353 K, = 0.119. Hence, the fluid mut be preheated up to T o = T β A C Ao = 353 20x0.119 = 350K The graphical repreentation of part (a-c) ha the following form: e a. Iothermal. Solid line are operating line for τ = 15 min T 11

e b. Adiabatic. Adiabatic line with τ = 15 T T m e T max c. Operating along the locu of maximum rate d) Permiible temperature range i 0 C to 100 C. We want minimum reactor ize for = 0.85. Preheat to 100 C, run along the locu of maximum rate τ = = 0.85 o dx 5x10 8 e 12,500 1.987T m 5x10 8 e 12,500 1.987T m + 3.4x10 21 e 32, 500 1.987T m 10,065 with T m = x 30.51 + ln A 1 τ = 1.8min Thu with Q = 100 lit/min we need only V = 160 liter The deired temperature profile along the reactor i preented in the encloed graph. The heat removal per unit volume i x 12

q ś Q = ρc p(t o T) + (ΔH r )C Ao = 2, 000(100 T) + (20, 000x2) Thi curve i alo preented in the figure. The total heat denity i: q Ý = 2,000(100 70) + 40,000x0.85 Q tot = 1.56x10 5 (cal / lit) With Q = 100 lit/min q Ý tot = 1.56x10 7 (cal / min) For comparion, if cooling failed and reactor ran adiabatically with T o = 100 C one would get ad = 0.068,T exit =126 o C The adiabatic temperature profile i hown alo on the encloed figure. Extenion to Multiple Reaction υ i =1 A = 0 i =1,2,...R (1) or R df dv + υ ir i = 0 = 1,2,...R (2) R υ i i =1 i=1 d X Ý i dv + d X Ý i dv + r i = 0 ( H ) df dv R υ r = 0 i i i=1 + q Ý v = 0 (3) (2a) V = 0 ; F = F o ( Ý X i = Ý X i o ) ; H = H o With the uual aumption made about the energy balance (ee the lecture on CSTR) one get: 13

F o =1 C ~ p dt dv + R i =1 ( ΔH r Ti ) r i + q Ý v = 0 (4) The equation to be olved for a et of multiple reaction are: d X Ý i dv + r i = 0 i =1,2...R (A) ρc p Q dt R dv + ( ΔH r i )r i + q Ý v = 0 i=1 V = 0 ; X Ý i = X Ý io T = T o ρq = cont (B) r i = k i10 e E 1i / RT Π α C i k i 20 e E 2i RT Π =1 = 1 C β i (C) with 1+ ρt C = C o o ρ o T 1+ R υ Ý i X i i =1 F o R υ Ý i X i i =1 =1 F bot o (D) The contitutive relationhip for Ý q v i: Ý q v = U av (T m T) a) T m = cont b) T m i governed by another D.E. mρ m Q m C pm dt m dv Ý q v = 0 (E) V=0 T m = T mo (cocurrent flow) V = V T m = T mo (countercurrent flow) 14

Problem Conider the reaction introduced in he lat lecture A R R=k 1 C A -k 2 C R (mol/lit ) k 1 = exp 7 83,700 RT x10 3 ( -1 ) k 2 exp 18 167,400 10 3 ( -1 ) RT ΔH r =80,000 (J / mol) C ~ p = 40(J / mol K) Activation energie given in oule. 1. The above reaction occur in liquid phae! Permiible temperature range of operation i 300 <T < 900 K. Feed condition: Q o = 100 (lit/) ; T o = 300 K ; C Ao = 1(mol/lit) You have a V = 100 liter PFR. How would you operate thi reactor if the only obective i to maximize the production rate of R. a) What i maximum F R. b) What are final and ΔT. c) What i the profile of heat addition or removal for every 10% of reactor volume. d) What i the overall heat duty for the reactor and any heat exchanger preceding it. e) Sketch your ytem. 2. The above reaction occur in ga phae. The ga feed ate i Q o =100(lit / ) at T o = 300K, P o = 24.6 atm The feed i 50%A, 50% inert. Permiible temperature range i 250< T < 900 K. Preure i contant in the reactor. Gae tart to condene below 250 K. Deired converion i 85%. 15

a) What reactor volume i needed if you operate along the locu of maximum rate? b) What i the ditribution of heat duty along the reactor? c) What i the production rate of R? 3. For the above problem what would F R and be if you had a reactor (PFR) of V = 100 liter available? 4. Suppoe that the reactor can only be operated adiabatically and the deired converion i 85%. Minimize the required reactor ize. a) What reactor type do you recommend? b) What feed temperature would you ue? c) What i the heat duty? 16