Introduction to Chemical Engineering: Chemical Reaction Engineering

Transcription

1 ========================================================== Introducton to Chemcal Engneerng: Chemcal Reacton Engneerng Prof. Dr. Marco Mazzott ETH Swss Federal Insttute of Technology Zurch Separaton Processes Laboratory (SPL) July 14, 2015 Contents 1 Chemcal reactons Rate of reacton and dependence on temperature Materal balance Converson Energy balance Three types of reactors Batch Contnuous strred tank reactor (CSTR) Plug flow reactor(pfr) Materal balances n chemcal reactors Batch Contnuous strred tank reactor (CSTR) Plug flow reactor(pfr) Desgn of deal reactors for frst-order reactons CSTR PFR Comparson of CSTR and PFR Dynamc behavor of CSTR durng start-up 14 6 Reversble reactons Materal balance for reversble reacton Equlbrum-lmted reactons Thermodynamcs of chemcal equlbrum 17 8 Energy balance of a CSTR The general energy balance Steady-state n a CSTR wth an exothermc reacton Stably of steady-states Multplcty of steady states, gnton and extncton temperatures Adabatc CSTR

2 1 CHEMICAL REACTIONS Equlbrum lmt n an adabatc CSTR Multple reactors n seres Introducton Another mportant feld of chemcal engneerng s that of chemcal reacton engneerng: consderng the reactons that produce desred products and desgnng the necessary reactors accordngly. The desgn of reactors s mpacted by many of the aspects you have encountered n the prevous lectures, such as the equlbrum and the reacton rate, both dependent on temperature and pressure. Whle there s a great varety of types of reactors for dfferent purposes, we wll focus on three basc types: The batch reactor, the contnuous strred-tank reactor, and the plug-flow reactor. 1 Chemcal reactons 1.1 Rate of reacton and dependence on temperature We wll once agan look at the formaton of ammona (NH 3 ) from ntrogen and hydrogen (see secton Chemcal equlbrum of the thermodynamcs chapter). Ths reacton follows the equaton: N 2 + 3H 2 2NH 3 (1) H 0 = 92 kj mol S 0 J = 192 mol K To fnd the Gbbs free energy of formaton at room temperature, recall that G 0 = H 0 T S 0 (2) = 92 kj ( ) kj + (298 K) = 35 kj mol mol K mol Alternatvely, one can also fnd the temperature for whch G = 0, T = H0 S 0 = 479 K = 206 C. At ths temperature the equlbrum favors nether the reactants nor the products. At lower temperatures G s negatve, so the products are favored and the reacton goes forward. At hgher temperatures the equlbrum shfts to favor the reactants, as s expected for an exothermc reacton. We also ntroduced the stochometrc coeffcent ν that descrbes how many molecules of speces react n each occurrence of the reacton. In general, a reacton between speces A and B formng C can be wrtten as ν A A + ν B B ν C C (3) The rate of generaton of each component s then the product of the stochometrc coeffcent and the rate of the reacton, and relates to the rate of generaton of every other component as follows: 2

3 1 CHEMICAL REACTIONS r = ν r (4) r = r = r A = r B = r C ν ν A ν B ν C (5) Remember that the stochometrc coeffcents for reactants are negatve, whle those of products are postve. For systems of multple chemcal reactons the rates can be added to obtan the generaton of component for the whole network of reactons. As an example, take the oxdaton of syngas, a mxture of carbon monoxde and hydrogen gas, where three reactons are to be consdered, each havng reacton rate r j (j = 1, 2, 3): r 1 : H O 2 H 2 O r 2 : CO O 2 CO 2 r 3 : CO + H 2 O CO 2 + H 2 Usng the stochometrc coeffcents, the rate of generaton or consumpton of each component s then gven by: R H2 = r 1 + r 3 R CO = r 2 r 3 R H2 O = r 1 r 3 R CO2 = r 2 + r 3 R O2 = 1 2 r r 2 Note that n these equatons the subscrpt n r j ndcates the reacton, whereas n Equatons 4 and 5 t ndcates the speces. In general then, the rate of generaton of component n a system of reactons j = 1...N r s the sum of the rates of generaton across all reactons: N r N r R = r j = ν j r j = A, B,... (6) j=1 j=1 The rate of each reacton then depends on the concentraton of ts reactants and the temperature, as descrbed by the Arrhenus equaton: r = k(t )c a Ac b B = k 0 e E A RT c a A c b B (7) where a and b are the reacton order wth respect to reactant A and B, respectvely. The overall order of the reacton s n = a + b. 1.2 Materal balance Consder a system of volume V wth a stream enterng and one extng, as shown n Fgure 1. The accumulaton of component n ths system s gven by: dn dt }{{} acc = F n F out } {{ } n out + G }{{} net generaton (8) 3

4 0 L 1 CHEMICAL REACTIONS W Q F n F out V Fgure 1: System of volume V wth a stream enterng and one extng. F n and F out are the mole flows of component nto and out of the system, respectvely. Ẇ s the work done by the systems on ts surroundngs, and Q s the heat flow nto the system. Here, the term G s the net generaton for all reactons over the entre volume consdered. Fndng the net generaton as well as the total amount of a component n the system requres ntegraton over the whole volume: n = G = c dv R dv (9) One assumpton that s frequently made s that the system s homogeneous, at least over certan regons, so n = V c and G = V R. Ths also means that the composton of the extng stream s equal to the composton n the entre volume. Further, the mole flow of a component s often wrtten as the product of the volumetrc flow and the concentraton of the component n the stream, so F = Qc. If one further assumes that only one reacton s takng place, the materal balance becomes 1.3 Converson dn dt = d (c V ) = Q n c n Qc + r V (10) dt The converson of component s the fracton of the reactant that undergoes reacton. It s denoted as X, where X = moles of component that reacted number of moles of component that were fed to the reactor (11) For a contnuous reactor at steady state ths s X = Qn c n Qc Q n c n The desred converson s a key parameter n the desgn of reactors, as we wll see. (12) 4

5 1 CHEMICAL REACTIONS 1.4 Energy balance Consderng the volume n Fgure 1, the energy balance can be wrtten as: de dt = Q Nc Ẇ + N c F n e n F out e out (13) =1 The work n ths equaton conssts of three terms: the so-called shaft work W s, and the volumetrc work done by the enterng stream on the system and by the system on the extng stream. Ẇ = =1 W s + P out Q out P n Q n (14) The shaft work refers to the work done by the strrer, for example, and s typcally neglgble n chemcal systems, so W s 0. The energy n the streams s summed for all N c components, and can also be wrtten n terms of concentratons and volume flow: N c =1 N c F e = Q c e = Q V =1 N c =1 n e = Q V E (15) E s the sum of the nternal energy U, the knetc energy K, and the potental energy E P. The knetc and the potental energy are neglgble n many chemcal reacton engneerng applcatons, so Equaton 15 becomes Q V E = Q V (U + K + E P ) = Q V U (16) we know that U s a functon of the enthalpy, pressure, and volume, so Q V U = Q V (H P V ) = Q V N c =1 N c n h P Q = F h P Q (17) When ths s appled for both streams, the term P Q cancels wth the volumetrc work from Equaton 14, and the energy balance n Equaton 13 becomes du dt = Q ( W s + P out Q out P n Q n) + du dt = Q + F n h n F n h n =1 F out h P n Q n + P out Q out F out h (18) If we are consderng a homogeneous system where only one reacton takes place, G = V ν r, and we can rewrte Equaton 8 by solvng for the flow out of the system: Equaton 18 then becomes du dt = Q + F out = F n F n ( h n + V ν r dn dt ) h V r ν h + h dn dt Note that the sum of the enthalpes of each component multpled by ther correspondng stochometrc coeffcent s the heat of reacton, so V r ν h = V r H r. At the same tme, the dfference n molar enthalpy between the enterng stream and the reactor depends (19) (20) 5

6 2 THREE TYPES OF REACTORS on the temperatures and the specfc heat of each component (assumng that there s no phase change): h n h = h n ( T n ) T n h (T ) = c p, dt ( = c p, T n T ) (21) T Further, the heat transfer nto the reactor s Q = UA (T T a ), where U s the heat transfer coeffcent, A s the heat transfer area, and T a s the ambent temperature or the temperature of the heat transfer flud. The left-hand-sde of Equaton 18 then becomes ( ) du dt = d (H P V ) = d n h P V dt dt = = = V n dh dt + n c p, dt dt + c c p, dt dt + dn h dt d (P V ) dt dn h dt d (P V ) dt dn h dt d (P V ) dt Combnng all ths nto equaton 20, and cancelng the term h dn dt both sdes of the equaton, we obtan (22) that shows up on V ( dt c c p, dt d (P V ) = UA (T T a ) + Q n dt c p, c n ) (T n T ) + V r ( H r ) (23) 2 Three types of reactors 2.1 Batch A batch reactor s a dscontnuous reactor. It s essentally a strred tank that s flled wth the reactants before the reacton starts and empted after t has run to completon (or to the extent that s needed). An example of ths would be the bakng of a cake. All the ngredents are placed n the mold, and then the temperature s ncreased n the oven to the necessary reacton temperature. When the reactons that make up the bakng process have run ther course to the desred extent, they are stopped. One of the dsadvantages of ths type of reactor s that for large producton quanttes the reacton has to be done multple tmes n seres. Ths requres the emptyng and refllng of the reactor, often accompaned by coolng t off frst and heatng t up wth the new batch. Ths large number of steps takes tme and attenton, and thereby reduces the productvty of the reactor. On the other hand, these reactors have the advantage that f multple smlar but dfferent reactons are needed, often the same equpment can be used, and the addtonal effort s comparatvely small. A schematc of a batch reactor can be seen n Fgure Contnuous strred tank reactor (CSTR) A contnuous strred tank reactor s lke a batch reactor n that t conssts of a tank and a strrer, however wth the addton of an nlet and an outlet that allow for a constant flow 6

7 2 THREE TYPES OF REACTORS c products c, V reactants t Fgure 2: Schematc of a batch reactor and typcal evoluton of the concentraton of reactants and products n a batch reactor c n, Q n c steady-state products c, V c, Q reactants t Fgure 3: Schematc of a contuous strred tank reactor (CSTR) nto and out of the reactor. Once the reactor s started up and reaches steady-state, t s usually assumed to have a constant volume as well as constant and homogeneous temperature, pressure, and composton. Whle contnuous processes don t have the varablty of batch processes, and durng start-up wll produce product that does not meet specfcatons, they have a number of advantages that make them attractve to use. For one, contnuous reactors don t have to be cooled off, empted, cleaned, reflled, and then heated to operatng temperature. For another, f a reacton produces heat and the reactor needs to be cooled, the coolng duty for a CSTR s constant, and can be tuned as needed. For a batch reactor the coolng duty needed would vary wth the reacton rate, and nsuffcent coolng can lead to a runaway reacton. Addtonally, the product from one reactor s often used n subsequent steps for other reactons. If multple steps are done n seres n batch reactors, and each step takes a dfferent amount of tme, the ntermedate products need to be stored n buffer tanks. These tanks can be elmnated or greatly reduced n sze f each reactor produces a steady stream that can be fed to the next reactor. If a process has to be done n batches, several reactors are often used n parallel, shfted n tme to gve a contnuous stream from the group of reactors. See Fgure 3 for a schematc representaton of a CSTR. 2.3 Plug flow reactor(pfr) Another type of contnuous reactors s the plug flow reactor, or PFR. It s a tubular reactor, meanng that t conssts of a long cylndrcal ppe through whch the reacton mxture s 7

8 2 THREE TYPES OF REACTORS c n, Q n Dx A c, Q 0 L Fgure 4: Schematc of a plug-flow reactor (PFR) flowng steadly. Typcally the assumpton s made that the temperature, pressure, and composton do not vary radally wthn the ppe, creatng a plug that flows through the reactor. As the reactants flow through the PFR, they are consumed, creatng a concentraton profle along the length of the ppe. Whle these reactors can have a heatng or coolng duty requrement that vares along the reactor, the reactor volume necessary to reach a partcular converson s lower than for a CSTR, whle keepng the advantages of a contnuous process. 8

9 3 MATERIAL BALANCES IN CHEMICAL REACTORS 3 Materal balances n chemcal reactors 3.1 Batch A batch reactor has no flow nto or out of the reactor: Q n = Q = 0 (24) Ths reduces the general mole balance from equaton 10 to V dc dt + c d (c V ) = r V dt dv dt = r V (25) Often, the reactor volume n a batch process s nearly constant. In ths case, the equaton reduces even further, and the rate of change n concentraton s smply the rate of reacton. If ths s not the case, one can stll rewrte equaton 25. Both cases can be seen here: dv dt = 0 dv dt 0 dc dt + c dc dt = r (26) d ln V dt Calculatng the converson X for a batch process s relatvely straghtforward. It s the dfference between the number of moles of reactant ntally n the reactor and those left at the end of the reacon dvded by the total number at the begnnng. It can then be related to the reactor volume and the reacton rate: = r X = n0 n n 0 (27) dn = N 0 dx (28) dx = r V dt n 0 (29) dx dt = r c 0 (30) where c 0 s the ntal concentraton of reactant. 3.2 Contnuous strred tank reactor (CSTR) A CSTR, as mentoned earler, has a feed stream enterng the reactor and a product stream extng. It s usually assumed to be well-mxed, gvng t a constant temperature, composton, and reacton rate throughout ts entre volume. It s almost always operated at steady state, meanng that after start-up s complete, the pressure, temperature, composton, and reacton rate no longer vary n tme. Once steady-state s reached, the number of moles of any gven speces no longer changes, and the flow out of the reactor matches the feed flow. dn dt = 0 and Qn = Q (31) 9

10 3 MATERIAL BALANCES IN CHEMICAL REACTORS Ths allows us to smplfy the mole balance from equaton 10 as follows: dn dt = Qn c n Qc + r V 0 = Qc n Qc + r V c c n = r V Q = r τ (32) Here we ntroduced the varable of space-tme, τ = V Q. The converson can be calculated form the concentraton of component n the feed and product stream as such: X = cn The flowrate, nlet concentraton, desred converson, and reacton rate relate to the reactor volume n ths way: c c n (33) 3.3 Plug flow reactor(pfr) Qc n X = r V (34) X τ = r c n Whle a PFR s assumed to be perfectly mxed radally, there s assumed to be no mxng along the length of the ppe. The reacton rate s therefor dependent on the poston, and the mole balance has to be wrtten as follows: dn dt = Qn c n Qc + V (35) r dv (36) As the reactor s assumed to be well-mxed radally, the reacton rate s only dependent on the poston along the length of the reactor, x. If we look at a slce of the reactor of cross-secton A and thckness x, we can wrte the mole balance for component for that secton as: dn dt = Qc (x) Qc (x + x) + r A x (37) dc dt A x = Q (c (x + x) c (x)) + r A x (38) If we let the thckness of the slce go to zero, we obtan: A c t = Q c x + r A (39) c t = υ c x + r (40) where υ = Q A s the flud velocty n the reactor. As ths s a partal dfferental equaton, we need the ntal and boundary condtons. These are 10

11 3 MATERIAL BALANCES IN CHEMICAL REACTORS c = c 0 for t = 0 and 0 < x < L (41) c = c n for t > 0 and x = 0 (42) When consderng the plug flow reactor n steady-state, the dependence on tme dsappears, and we get where ϑ = x υ r = υ dc dx = dc dϑ = Q dc dv s a resdence tme dependent on the poston along the reactor length. (43) 11

12 4 DESIGN OF IDEAL REACTORS FOR FIRST-ORDER REACTIONS 4 Desgn of deal reactors for frst-order reactons In ths secton we wll see how to use the prncples above to desgn CSTR and PFR reactors under sothermal condtons and only consderng a sngle, rreversble, frst-order reacton: A products (44) Under the assumed condtons we can wrte the rate of reacton as r = kc A or r A = kc A (45) Recall that because A s a reactant, ts stochometrc coeffcent ν s negatve (n ths case -1). 4.1 CSTR Applyng ths equaton n the rate for a CSTR, we can rewrte equaton 32 as c A c n A = r A τ = kc A τ c n A = c A (1 + kτ) c A = cn A 1 + kτ here, the product kτ s also known as the frst Damköhler number, denoted as Da. It can be used to gve a rough estmate of the converson that can be expected gven a known rate constant and resdence tme: (46) X A = cn A c A c n A = cn A cn A 1+kτ c n A = kτ = kτ 1 + kτ Ths way t s farly smple to estmate that for a frst Damköhler number of 0.1, the converson s less than 0.1, whle for a value of 10 t s over 0.9. Equaton 46 can also be rewrtten to render the volume of the reactor as a functon of the flowrate, reacton rate constant, and converson: ) (47) ( V c n A c A = kc A τ = kc A Q V = Q ( c n A c ) A kc A V = Q ( c n ) A QX A 1 = k c A k (1 X A ) (48) 12

13 0 L 4 DESIGN OF IDEAL REACTORS FOR FIRST-ORDER REACTIONS c products reactants Fgure 5: concentraton profle along the length of a plug flow reactor x 4.2 PFR Smlarly, the reacton rate can be substtuted nto equaton 43 for a PFR to yeld Q dc A dv = kc A and snce c A = c n A for x = 0, c A c n A c A = c n A e k Q V (49) = 1 X A = e kτ X A = 1 e kτ (50) the assumptons made n the desgn of the PFR cause each dfferental volume of the reactor to behave lke a batch reactor as t moves through the ppe. As a result, the concentraton profle along the length of a PFR looks much lke the concentraton profle n a batch reactor over tme, as seen n Fgure 5. One can also solve for the volume of reactor necessary to acheve a desred converson, startng from equaton 49: c n A V = Q k ln ( c n A c A = e k Q V c A ) = Q k ln ( 1 1 X A ) (51) 4.3 Comparson of CSTR and PFR In general, reactons tend to exhbt knetcs of postve order, meanng that as the reactants are consumed the rate of reacton decreases. As a CSTR s already at the composton of the product, the reactants are already consumed, and ther concentraton s low. Consequently, the CSTR typcally has a larger volume than a PFR that reaches the same converson. For the reacton consdered above, Fgure 6b compares the volumes of both types of reactors. 13

14 5 DYNAMIC BEHAVIOR OF CSTR DURING START-UP kv Q CSTR kv Q CSTR PFR PFR 1 c n A (a) ca X A (b) 1 Fgure 6: Comparson of reactor volume for CSTR and PFR for varous cn A ca (b) (a) and X A 5 Dynamc behavor of CSTR durng start-up So far we have looked at the steady-state operaton of reactors. But when a CSTR s started up t undergoes a transtonal perod untl t reaches that steady-state. How does the concentraton n the reactor behave durng ths tme perod, and how long does t take to get reasonably close to the steady-state? To answer ths, we go back to the mole balance from equaton 32, but we no longer assume the number of moles of component to be constant. We stll assume, however, that the reactor s well-mxed and sothermal, and that the volume s constant (so Q n = Q). Consderng the reacton from before: we obtan A products, r A = kc A (52) dn A dt = V dc A = Qc n A Qc A + r A V dt dc A = 1 dt τ cn A 1 τ c A kc A (53) where τ s constant. Equaton 53 s a lnear nhomogeneous ordnary dfferental equaton. ( dc A dt + k + 1 ) c A = 1 τ τ cn A (54) The ntegral of the homogeneous part and the partcular soluton are, respectvely, c h A = Ae (k+ 1 τ )t (55) c p A = cn A 1 + kτ By applyng the ntal condton c A (0) = c 0 A we obtan ( ) c A = c 0 A cn 1+kτ A e ( τ )t + cn A 1 + kτ 1 + kτ Note that f you take ths soluton and fnd the lmt as t you arrve at equaton 46, the steady-state condton. See Fgure 7 for an example of the evoluton of concentraton n a CSTR durng start-up. (56) (57) 14

15 6 REVERSIBLE REACTIONS c A c 0 A c n A 1+kτ c 0 A Fgure 7: concentraton of reactant A n a CSTR durng start-up t 6 Reversble reactons 6.1 Materal balance for reversble reacton Untl now we have only consdered rreversble reactons, where reactants form products, but not the other way around. In realty, many reactons can go both ways, even f one sde of the equaton mght be strongly favored over the other one. Let us consder now a general reversble reacton: B (58) k2 A k 1 where the forward reacton s governed by the rate constant k 1 and the reverse reacton by the rate constant k 2. The overall reacton rate s then the dfference between the rate of consumpton of A and the rate of ts producton. r = k 1 c A k 2 c B (59) Gven enough tme ths system wll settle n an equlbrum that depends on the two rate constants. At ths pont there s no net reacton, so the concentraton of the reactant and the products can be related to each other through the two rate constants: r = 0 (60) c B c A = k 1 k 2 = K (61) where K = k 1 k 2 s the equlbrum constant for ths reacton under these condtons. You may recall ths constant from earler n the semester when t was found through the Gbbs free energy of formaton G 0 r n the thermodynamcs scrpt. To fnd the concentraton of the reactant n a CSTR for a reversble reacton, start at Equaton 32: 15

16 6 REVERSIBLE REACTIONS c A c n A = r A τ = ( k 1 c A + k 2 c B ) τ (62) and c B c n B = r B τ = ( k 1 c A k 2 c B ) τ Further, snce mass s conserved, Rewrtng Equaton 62, we get c A + c B = c n A + c n B c B = c n A + c n B c A (63) and smlarly for component B c A (1 + k 1 τ) k 2 τc B = c n A c A (1 + k 1 τ) = c n A + k 2 τc B = c n A + k 2 τ ( c n A + c n ) B c A c A (1 + k 1 τ + k 2 τ) = c n A + k 2 τ ( c n A + c n ) B c A = cn A + k 2τ ( c n A + ) cn B 1 + τ (k 1 + k 2 ) c B = cn B + k 1τ ( c n A + ) cn B 1 + τ (k 1 + k 2 ) For a reversble reacton, there s a lmt to the converson, set by the equlbrum. If you had an nfntely large CSTR, τ, and c A = k ( 2 c n A + c n ) B = cn A + c n B (66) k 1 + k K c A, and as a result the converson X, depends on both the nlet concentraton of component A and B. Assumng that you are only feedng your reactant (and c n B = 0), the lmt to the converson s found to be (64) (65) X = cn A c A c n A = cn A cn A 1+K c n A = K = K 1 + K (67) 6.2 Equlbrum-lmted reactons The converson at equlbrum depends on the equlbrum constant, whch n turn depends on the temperature: K = e G0 (T ) RT (68) Fgure 8 shows how the equlbrum converson changes wth temperature. Especally for exothermc reactons ths s a concern. To acheve hgh reacton rates a hgh temperature s needed, but ths can severely lower the converson. Not only does the equlbrum pose a lmt to the converson, even gettng close to t takes a hgh toll on reacton rate. There are dfferent attempts to crcumvent ths, such as usng hgher pressures, as we wll see n the next secton, or usng tanks n seres, as we wll see later. 16

17 7 THERMODYNAMICS OF CHEMICAL EQUILIBRIUM 1 1 X eq X eq T T (a) (b) Fgure 8: Equlbrum converson dependency on temperature for an edothermc (a) and an exothermc (b) reacton. 7 Thermodynamcs of chemcal equlbrum Consder a sngle reacton occurrng n a batch reactor. The rate of change of the amount of any component n the reactor s gven by dn dt = ν rv dn = ν rv dt = ν dλ (69) where dλ = rv dt s the extent of reacton. For multple reactons then, the dfferental change of the amount of s gven by the sum over all reactons j, dn = j ν j dλ j (70) If we now consder a system at a gven temperature T and pressure P, then and at equlbrum dg = ( ) µ dn = µ ν dλ (71) dg = 0 µ ν = 0 (72) Now take an deal gas mxture n ths system, n whch a reacton s takng place, e.g. A + B C + D. Then ( ) ( ) ν T, P, y = µ P (T, P r ) + RT ln (73) 0 = µ P r ( ) ν P ν µ ( ) T, P, y = ν µ (T, P r ) + RT ln P r = ν µ (T, P r ) + RT ln (P y ) ν P ν r [ ( ) P = G 0 ν (T, P r ) + RT ln P r y ν ] (74) 17

18 7 THERMODYNAMICS OF CHEMICAL EQUILIBRIUM where G 0 (T, P r ) = ν µ (T, P r). Rearrangng the equaton gves us ( ) P ν P r y ν } {{ } Q(P,y) ( ) = exp G0 (T ) RT }{{} K eq(t ) The rght-hand-sde of ths equaton s the equlbrum constant for the reacton, as we have seen before. Whle these quanttes are equal n equlbrum, they are not when the system s not n equlbrum: (75) Q(P, y) = K eq (T ) dg = 0 no reacton Q(P, y) > K eq (T ) dg > 0 A + B C + D Q(P, y) < K eq (T ) dg < 0 A + B C + D Consder now once agan the reacton of ntrogen and hydrogen formng ammona (Equaton 1). Applyng Equaton 75 gves us the followng relatonshp: P 2 y 2 K eq (T ) = Pr 2 NH 3 P 4 y N2 yh 3 2 = P 2 r P 2 y 2 NH 3 y N2 y 3 H 2 (76) Ths shows that whle the equlbrum constant K eq s a functon of temperature, the pressure of the system also affects the composton at the equlbrum. In the above reacton, NH 3 s desred, so t s of nterest to ncrease the mole fracton of t at equlbrum. As we saw earler n Secton 1.1 a low temperature would favor ammona. Ths would, however, severely lower the reacton rate. To get a hgher converson despte the hgh temperature, then, the pressure can be ncreased. P y NH3 good! P y NH3 bad! 18

19 8 ENERGY BALANCE OF A CSTR 8 Energy balance of a CSTR 8.1 The general energy balance Consder agan a CSTR n steady state, meanng all dervatves n tme are zero. Ths reduces the energy balance from Equaton 23 to 0 = UA (T T a ) + Q n ( c n c p, ) (T n T ) + V r ( H r ) (77) Ths can be rewrtten as (note that we consder Q n = Q out = Q) UA (T T a ) + Q n ( c n c p, ) (T T n ) = V r ( H r ) c n ( ) UA (T Q (T T a) + c n c p, T n ) = τr ( H r ) c p, [ ] UA Q ( cn c ) (T T a ) + T T n = τr ( H r ) (78) p, We now defne two new varables: UA β = Q ( cn c ) p, (79) T c = T n + T a β 1 + β (80) Ths allows us to rewrte the above as c n c p, [ T (1 + β) T n T a β ] = τr ( H r ) c n c p, [(1 + β) (T T c )] } {{ } heat removed R(T ) = τr ( H r ) }{{} heat generated G(T,c A ) (81) Note that the two new varables are not arbtrary: β relates the nfluence on the reactor temperature of the heat exchanger to that of the enterng feed, and T c s an ntermedate temperature between the feed temperature and the temperature of the heat transfer flud,.e. t s a weghted average of the two. The two sdes of Equaton 81 are the heat removed from the reactor and the heat generated by the reacton, respectvely. They represent the energy balance for a CSTR at steadystate, as a functon of three groups of parameters: The feed condtons (c n, T n ), the reactor coolng characterstcs (T c and β), and the resdence tme (τ = V ). The heat Q n removed R (T ) (left-hand sde) depends only on two parameters, namely β and T c ; the effect that each of these parameters has on the removed heat s shown n Fgure 9. The heat generated s not only a functon of the reactor temperature, but also of the reacton rate, whch depends on the reactant concentraton at steady state. 19

20 8 ENERGY BALANCE OF A CSTR R(T ) R(T ) β T c T T a T n T (a) (b) Fgure 9: The heat removal rate from a CSTR changes wth T c, the ntermedate temperature of the heat transfer flud and the feed, as seen n (a) and on the value of β as seen (b) Reversble Reactons As we saw n Secton 6, chemcal reactons are often reversble, meanng they can go n both drectons: B r = k 1 c A k 2 c B (82) k2 A k 1 Alternatvely, they can be consdered as two entrely separate reactons as such: A k 1 B r forward = k 1 c A (83) B k 2 A r back = k 2 c B (84) If the heat of reacton for these two reactons are H and H back, respectvely, then H back = H. Ths leads to the followng expresson for the net rate of heat generaton [ J m 3 s] : r forward H + r back H back = (r forward r back ) H (85) = (k 1 c A k 2 c B ) H (86) Here t s obvous that the reacton rate for the combned reactons s the same as the one for the reversble reacton (Equaton 82). Ths means that wth regard to the energy balance of a reactor (see Equaton 81), all the consderatons concernng a sngle, rreversble reacton can be appled to a reversble reacton, provded that the approprate rate expresson from Equaton 82 s used. Further, recall from Equaton 63 that c B can be expressed as a functon of the ntal concentratons and c A, so the net heat generated depends on c A only: (k 1 c A k 2 c B ) H = ( k 1 c A k 2 ( c n A + c n B c A )) H = G(T, ca ) τ (87) 20

21 8 ENERGY BALANCE OF A CSTR 8.2 Steady-state n a CSTR wth an exothermc reacton Consder now an exothermc, rreversble reacton wth a sngle reactant takng place n a cooled CSTR. A B, r = kc A (88) The energy and mass balances for a CSTR at steady state are then gven by: R (T ) = G (T, c A ) = τ ( H) r (T, c A ) c n A c A = τr(t, c A ) = τc A k (T ) (89) Note that the second equaton s Equaton 32, and that by solvng t wth respect to c A, one obtans Equaton 46. Therefore the system (89) can be rewrtten as: c A = cn A 1+τk(T ) R (T ) = G (T, c A ) = ( H) c n A ( ) τk(t ) 1+τk(T ) = G (T ) Therefore the concentraton of A can be expressed as a functon of T through k(t ), and the heat generated s gven as solely a functon of temperature, wth parameters τ and H. The effect of these parameters on the heat generaton curve s shown n Fgure 10. (90) G(T) G(T) τ H T T (a) (b) Fgure 10: The generaton of heat from the reacton depends strongly on the temperature of the reactor, the resdence tme (as seen n (a), and on the heat of reacton (seen n (b)) Stably of steady-states For the reactor to be at steady-state, R(T ) = G(T ). Fgure 11 shows three heat removal lnes together wth a heat generaton lne, and hghlghts the steady-states. At any other temperature, the behavor of the reactor depends on the relatonshp between the two lnes: R(T ) > G(T ) R(T ) = G(T ) R(T ) < G(T ) reactor cools off steady-state reactor heats up 21

22 8 ENERGY BALANCE OF A CSTR G(T) IV V R(T) III I II T Fgure 11: The roman numerals I-V denote steady-states. The crcles are stable, whle the square s an unstable steady-state. In Fgure 11, for example, f the reactor s operated between ponts III and IV usng the mddle value of T c, the heat generaton s hgher than the removal, leadng the reactor to heat up. Once the temperature passes pont IV, however, more heat s removed than s generated, causng the reactor to cool off. Ths s why pont IV s consdered a stable steady state: A small devaton n temperature n ether drecton wll cause the system to self-correct and return to the steady-state. Pont III, meanwhle, s an unstable steady state. If the reactor s operated at that temperature, t wll nether heat up nor cool off, and s at steady state. If, however, there s a small dsturbance that warms the reactor a lttle bt, the heat generated wll outwegh the removed heat, and the reactor wll heat up further and further. A small dsturbance to a lower temperature wll have the opposte effect: The now prevalent heat removal wll cool the reactor off more and more. What t comes down to s that f the dervatve of the heat removal lne at the steady state s hgher than the heat generaton lne, the steady state s stable; f t s lower, the steady state s unstable Multplcty of steady states, gnton and extncton temperatures As you can see n Fgure 11, the mddle heat removal lne allows for three dstnct steady states. Ths s called a multplcty of steady states. As III s an unstable steady state, the reactor would not reman there for a long perod of tme. Whether the reactor runs at pont II or at IV depends on ts startng pont. At any temperature below III t wll settle on pont II, at any hgher temperature t wll end up at IV. If T c changes a lttle, the reactor temperature wll change accordngly, but reman n the same regon. Only f T c changes past the gnton temperature wll the reactor be forced to go to the hgh temperature. Conversely, f the temperature drops below the extncton temperature, the reactor drops to cool temperatures, as seen n Fgure 12 22

23 8 ENERGY BALANCE OF A CSTR G(T) T ss R(T) T (a) T c T ext T gn (b) Fgure 12: (a) shows a curve of heat generaton along wth fve possble heat removal lnes. 1 and 5 have only one steady-state. 3 has two stable steady states. Whch one the reactor s at depends on the ntal temperature. 2 and 4 are the extncton and gnton temperatures, respectvely. (b) shows the steady state reactor temperature for a range of T c, ncludng the fve temperatures seen n (a), clearly showng the regon wth two possble operatng condtons 8.3 Adabatc CSTR Equlbrum lmt n an adabatc CSTR Reactons are also frequently carred out n a vessel that s nether heated nor cooled, wth the heatng/coolng takng place ether upstream or downstream of the reactor. As a result, these reactors are adabatc and modeled as such. We wll now look at an adabatc CSTR n whch an exothermc, reversble reacton s takng place. ( B ( H r ) > 0 r = k 1 c A k 2 c n A + c n ) B c A k2 A k 1 We have seen n secton 6.2 that ths exothermc reacton wll be lmted by ts equlbrum. In secton 8.1 then we saw that the temperature n the reactor depends on the heat generated and the heat removed. The energy and mass balances are repeated here: (91) heat removed R(T ) heat generated { }}{ G(T,c A ) {}}{ c n c p, (1 + β) (T T c ) = τr(t, c A ) ( H r ) c n A c A = τr(t, c A ) = X A c n A (92) where we explot the defnton of converson X A,.e. X A = cn A c A. In an adabatc CSTR, c n A β = 0 and T c = T n. As a result, substtutng the second equaton nto the frst yelds: ( ) (T c n c p, T n ) = c n A X A ( H r ) (93) In ths last equaton, the left-hand sde s the heat removed through the coolng effect of the ncomng feed, and the rght-hand sde s the heat produced by the reacton. Equaton 93 descrbes a lnear relatonshp between the converson and the temperature n the reactor. 23

24 8 ENERGY BALANCE OF A CSTR 1 X eq = K(T) 1+K(T) X τ T n Fgure 13: Converson and temperature n the reactor have a lnear relatonshp. The only nfluence that the resdence τ has s that t determnes where along the lne the reactor s operated. T Whle t does not provde the steady-state condtons of the reactor drectly, t descrbes a path along whch the operatng pont of the reactor can be found for any gven resdence tme τ. One can solve the system 92 numercally to fnd the temperature and converson of the reacton for any gven resdence tme. Consder now a reactor where the feed s pure A (so c n B = 0). Substtutng the reacton rate gven n Equaton 91 nto the mass balance n system 92 yelds Equaton 64 for c A c A = cn A + k 2τ ( c n ) A (94) 1 + τ (k 1 + k 2 ) Ths equaton can be substtuted for c A n the second equaton n (92), whch when substtuted nto the frst equaton, yelds the relatonshp between the temperature n the reactor and the resdence tme as ( ) (T c n c p, T n ) = c n k 1 (T )τ A ( H r ) 1 + τ (k 1 (T ) + k 2 (T )) (95) Here agan one can use the defnton of the converson X A = cn A c A τk = 1 c n 1+τ(k A 1 +k 2 ), where k 1 = k 1 (T ) and k 2 = k 2 (T ). The converson n the reactor can be taken to two lmts, nfnte τ and nfnte temperature, wth the followng results: X A K(T ) 1+K(T ) when τ τk 1, 1+τ(k 1, +k 2, ) when T where k 1, and k 2, are the lmts of the two rate constant for nfnte temperature. As was shown before, the converson does not go to 1 for an nfnte τ, but approches the equlbrum lmt, as s shown n Fgure 13. (96) 24

25 8 ENERGY BALANCE OF A CSTR n c A n T j 1 T j-1 j 1 c A n T j T j j c A j-1 j Fgure 14: Several CSTRs n seres. The product from each reactor s cooled to move away from the equlbrum lmt, and then fed to the next reactor. As reactors often contan a catalyst wthout whch the reacton does not progress, t s typcally assumed that the feed enterng reactor j s of the same composton as the product from j 1 (.e. c n,j A = cj 1 A ) Multple reactors n seres As you could see n Fgure 13, the equlbrum lmt for converson mght stll be relatvely low. By coolng the reactor a lower temperature can be mantaned for hgher conversons, however t s often more effcent to use a dedcated heat exchanger to cool the streams. If the product from the reactor s cooled, t can be fed to another reactor, where t can contnue to react. See Fgure 14 for a schematc. In ths system, c n A and cn B refer to the concentratons of A and B n the feed enterng the frst reactor. The concentratons enterng all subsequent reactors are equal to the outlet concentratons from the prevous reactor, so c n,j A = cj 1 A and c n,j B = cj 1 B (97) For obvous reasons, the converson n reactor j s not calculated on the bass of the feed enterng t, but on the ntal feed. As a result, as c A decreases throughout the reactor cascade, the converson ncreases wth each reactor: X A = cn A cj A c n A (98) XA 1 < XA 2 <... < X j 1 A < X j A <... (99) For each reactor n the cascade, Equaton 93 becomes ( ) (Tj c n c p, Tj n ) = τj r j ( H r ) ( cn c ) p, ( H) c n A ( = = c n A c j 1 A cj A ( X j A Xj 1 A ) ( H r ) ) ( H r ) ( Tj Tj n ) = X j A Xj 1 A (100) So for each reactor, the converson and temperature are lnearly related, and both a functon of τ. A seres of CSTRs wth nterstage coolng can acheve hgher converson than a sngle reactor, whle remanng at reasonably hgh temperatures. A schematc of the converson for a fve reactor cascade s llustrated n Fgure 15, where fve reactors n seres wth ntermedate coolng produce a sawtooth wave pattern to hgh converson. Note 25

26 8 ENERGY BALANCE OF A CSTR 1... X eq = K(T) 1+K(T) X T 2,X 2 A ntermedate coolng T 1,X 1 A T n Fgure 15: Several CSTRs n seres. The product from each reactor s cooled to move away from the equlbrum lmt, and then fed to the next reactor. T that Equaton 100 does not gve the reactor condtons at steady state, but descrbes a lne along whch the operatng condtons depend on τ. In the desgn of the reactor cascade one has to consder the nherent trade-off: A hgher converson n each reactor requres a larger τ, however as the equlbrum lmt s approached the ncrease n converson becomes smaller for an ncrease n τ (and therefore reactor sze). Interstage coolng can be used to move away from the equlbrum lmt, allowng the reacton to proceed further n the next reactor. Ths allows a step-wse approach to hgher converson at the expense of the amount of necessary equpment (reactors, heat exchangers, etc.). 26