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1 The authos and the publishe have taken cae in the pepaation of this book but make no expessed o implied waanty of any kind and assume no esponsibility fo eos o omissions. No liability is assumed fo the incidental o consequential damage in connection with o aising out of the use of the infomation o pogams contained heein. isit us on the Web : opyight 0 Peason Education,Inc. This wok is potected by United States copyight laws and is povided solely fo the use of the instuctos in teaching thei couses and assessing student leaning. Dissemination o sale of any pat of this wok (including the Wold Wide Web ) will destoy the integity of the wok and is not pemitted. The wok and the mateials fom it should neve be made available to the students except by the instuctos using the accompanying texts in the classes. ll the ecipient of this wok ae expected to abide by these estictions and to hono the intended pedagogical puposes and the needs of the othe instuctos who ely on these mateials.

2 Solutions fo hapte - onvesion and Reacto Sizing P-. P-. P-. P-4. P-5. This poblem will keep students thinking about witing down what they leaned evey chapte. This foces the students to detemine thei leaning style so they can bette use the esouces in the text and on the DROM and the web. IMs have been found to motivate the students leaning. Intoduces one of the new concepts of the 4 th edition wheeby the students play with the example poblems befoe going on to othe solutions. This is a easonably challenging poblem that einfoces Levenspiels plots. P-6. Staight fowad poblem altenative to poblems 7, 8, and. P-7. P-8. P-9. P-0. P-. P-. To be used in those couses emphasizing bio eaction engineeing. The answe gives idiculously lage eacto volume. The point is to encouage the student to question thei numeical answes. Helps the students get a feel of eal eacto sizes. Geat motivating poblem. Students emembe this poblem long afte the couse is ove. ltenative poblem to P-6 and P-8. Novel application of Levenspiel plots fom an aticle by Pofesso lice Gast at Massachusetts Institute of Technology in EE. DP- Simila to -8 DP-B Good poblem to get goups stated woking togethe (e.g. coopeative leaning). DP- Simila to poblems -7, -8, -. DP-D Simila to poblems -7, -8, -. Summay ssigned ltenates Difficulty Time (min) P- O 5 P- 0 P- 0

3 P-4 O 75 P-5 O M 75 P-6 7,8, S 45 P-7 S S 45 P-8 6,8, S 45 P-9 S S 5 P-0 S P- 6,7,8 S 60 P- S M 60 DP- O 8,B,,D S 5 DP-B O 8,B,,D S 0 DP- O 8,B,,D S 0 DP-D O 8,B,,D S 45 ssigned = lways assigned, = lways assign one fom the goup of altenates, O = Often, I = Infequently, S = Seldom, G = Gaduate level ltenates In poblems that have a dot in conjunction with means that one of the poblems, eithe the poblem with a dot o any one of the altenates ae always assigned. Time ppoximate time in minutes it would take a B/B + student to solve the poblem. Difficulty S = Staight fowad einfocement of pinciples (plug and chug) S = aily staight fowad (equies some manipulation of equations o an intemediate calculation). I = Intemediate calculation equied M = Moe difficult OE = Some pats open-ended. * Note the lette poblems ae found on the D-ROM. o example DP-. Summay Table h- Staight fowad,,,4,9 aily staight fowad 6,8, Moe difficult 5,7, Open-ended ompehensive 4,5,6,7,8,, itical thinking P-8 P- Individualized solution. P- (a) Example - though - If flow ate O is cut in half.

4 ao/-a v = v/, = O / and O will emain same. Theefoe, volume of STR in example -, If the flow ate is doubled, = O and O will emain same, olume of STR in example -, = /- =.8 m P- (b) Example -4 Now, O = 0.4/ = 0. mol/s, Levenspiel Plot onvesion Table: Divide each tem 0 in Table - by [ O /- ] (m ) Reacto Reacto = 0.8m =. m = ( O /- ) By tial and eo we get: = and = 0.8 Oveall convesion Oveall = (/) + (/) = ( )/ = 0.67 P- (c) Example -5 () o fist STR, at =0 ;

5 0.8m 0 at =0. ;.94 m om pevious example; ( volume of fist STR) =.88 m lso the next eacto is PR, Its volume is calculated as follows O 0.47m d o next STR, = 0.65, O O ( ) m, =.m () Now the sequence of the eactos emain unchanged. But all eactos have same volume. ist STR emains unchanged cst =. = ( 0 /- )* => =.088 Now o PR:, O d By estimation using the levenspiel plot =.8 o STR,

6 O STR = => =.6 0.m () The wost aangement is to put the PR fist, followed by the lage STR and finally the smalle STR. onvesion Oiginal Reacto olumes Wost angement = 0.0 = 0.88 (STR) = 0. (PR) = 0.60 = 0.8 (PR) = 0.5 (STR) = 0.65 = 0.0 (STR) = 0.0 (STR) o PR, = 0. 0 O d Using tapezoidal ule, O = 0., = 0. O f O f m 0.m o STR, o = 0.6, O O.m, = o nd STR, =.(0.6 0.) = 0.5 m o = 0.65, O m, = 0. m P- Individualized solution. P-4 Solution is in the decoding algoithm given with the modules. P-5

7 O /- (m ) =.6 m P-5 (a) Two STRs in seies o fist STR, = ( o /- ) => = 0.5 o second STR, = ( o /- ) ( ) => = 0.76 P-5 (b) Two PRs in seies o o d d 0 By extapolating and solving, we get = 0.6 = 0.84 P-5 (c) Two STRs in paallel with the feed, O, divided equally between two eactos. NEW /- = 0.5 O /- = (0.5 O /- ) Solving we get, out = 0.68 P-5 (d) Two PRs in paallel with the feed equally divided between the two eactos. NEW /- = 0.5 O /- By extapolating and solving as pat (b), we get out = 0.88 P-5 (e)

8 STR and a PR ae in paallel with flow equally divided Since the flow is divided equally between the two eactos, the oveall convesion is the aveage of the STR convesion (pat ) and the PR convesion (pat D) o = ( ) / = 0.67 P-5 (f) PR followed by a STR, PR = 0.50 (using pat(b)) = ( o /- -STR ) ( STR PR ) Solving we get, STR = 0.70 P-5 (g) STR followed by a PR, STR = 0.44 (using pat(a)) PR STR O d By extapolating and solving, we get PR = 0.7 P-5 (h) m PR followed by two 0.5 m STRs, o PR, PR = 0.50 (using pat(b)) STR : = ( o /- -STR ) ( STR PR ) = 0.5 m STR = 0.6 STR : = ( o /- -STR ) ( STR STR ) = 0.5 m STR = 0.7 P-6 Exothemic eaction: B + (mol/dm.min) /-(dm.min/mol) P-6 (a)

9 To solve this poblem, fist plot /- vs. fom the chat above. Second, use mole balance as given below. STR: Mole balance: 00mol / min 0.4 5mol / dm.min 0 STR => => STR = 4 dm PR: Mole balance: PR 0 0 d = 00(aea unde the cuve) PR = 7 dm P-6 (b) o a feed steam that entes the eaction with a pevious convesion of 0.40 and leaves at any convesion up to 0.60, the volumes of the PR and STR will be identical because of the ate is constant ove this convesion ange. PR d d P-6 (c) STR = 05 dm Mole balance: STR 0

10 05dm 0.5dm min/ mol 00mol / min Use tial and eo to find maximum convesion. t = 0.70, /- = 0.5, and /- = 0.5 dm.min/mol Maximum convesion = 0.70 P-6 (d) om pat (a) we know that = Use tial and eo to find. Mole balance: 0 Reaanging, we get t = 0.64, onvesion = 0.64 P-6 (e) om pat (a), we know that = Use tial and eo to find. Mole balance: PR 7 d t = 0.908, = 00 x (aea unde the cuve) => = 00(0.4) = 7dm onvesion = d

11 P-6 (f) See Polymath pogam P-6-f.pol. P-7 (a) S 0 S S0 = 000 g/h t a convesion of 40% Theefoe S dm h 0.5 g 0.5 (000)(0.40) 60 dm P-7 (b) t a convesion of 80%, S0 = 000 g/h Theefoe S dm h 0.8 g 0.8 (000)(0.80) 640 dm P-7 (c) PR S 0 0 d S om the plot of /- S alculate the aea unde the cuve such that the aea is equal to / S0 = 80 / 000 = 0.08 = % o the 80 dm STR, 80 dm S 0 /- s = om guess and check we get = 55% S

12 P-7 (d) To achieve 80% convesion with a STR followed by a STR, the optimum aangement is to have a STR with a volume to achieve a convesion of about 45%, o the convesion that coesponds to the minimum value of /- s. Next is a PR with the necessay volume to achieve the 80% convesion following the STR. This aangement has the smallest eacto volume to achieve 80% convesion. o two STR s in seies, the optimum aangement would still include a STR with the volume to achieve a convesion of about 45%, o the convesion that coesponds to the minimum value of /- s, fist. second STR with a volume sufficient to each 80% would follow the fist STR. P-7 (e) ks s and 0. S 0 S K M S s k S 0. S 0 K M S S 0.00 s k S K 0. M S 0 S S 0.00 Let us fist conside when S is small. S0 is a constant and if we goup togethe the constants and simplify then K M S s ks k s since S < K M K M s ks k s lage and as S gows deceases). Now conside when S is lage ( is small) which is consistent with the shape of the gaph when is lage (if S is small is s S gets lage appoaches 0: 0. S 0 S 0.00 and S S 0

13 If k S s then K M S s K M k S S s S gows lage, S >> K M nd s k S S k nd since is becoming vey small and appoaching 0 at = 0, /- s should be inceasing with S (o deceasing ). This is what is obseved at small values of. t intemediate levels of S and, these diving foces ae competing and why the cuve of /- S has a minimum. P-8 Ievesible gas phase eaction See Polymath pogam P-8.pol. + B P-8 (a) PR volume necessay to achieve 50% convesion Mole Balance 0 d ( ) olume = Geometic aea unde the cuve of ( 0 /- ) vs ) = m P-8 (b) STR olume to achieve 50% convesion Mole Balance 0 ( ) = 50000m 0.5

14 P-8 (c) olume of second STR added in seies to achieve 80% convesion = 50000m ( ) ( ) ( ) P-8 (d) olume of PR added in seies to fist STR to achieve 80% convesion PR ( PR = 90000m ) ( ) P-8 (e) o STR, = m (STR) Mole Balance 0 ( ) ( = ) o PR, = m (PR) Mole balance = 0.4 d ( ) 0 ( ) d P-8(f) Real ates would not give that shape. The eacto volumes ae absudly lage.

15 P-9 Poblem -9 involves estimating the volume of thee eactos fom a pictue. The doo on the side of the building was used as a efeence. It was assumed to be 8 ft high. The following estimates wee made: STR h = 56ft d = 9 ft = π h = π(4.5 ft) (56 ft) = 56 ft = 00,865 L PR Length of one segment = ft Length of entie eacto = ( ft)()() = 06 ft D = ft = π h = π(0.5 ft) (06 ft) = 84 ft = 67,507 L nswes will vay slightly fo each individual. P-0 No solution necessay. P- (a) The smallest amount of catalyst necessay to achieve 80 % convesion in a STR and PBR connected in seies and containing equal amounts of catalyst can be calculated fom the figue below.

16 The lightly shaded aea on the left denotes the STR while the dake shaded aea denotes the PBR. This figue shows that the smallest amount of catalyst is used when the STR is upsteam of the PBR. See Polymath pogam P-.pol. P- (b) alculate the necessay amount of catalyst to each 80 % convesion using a single STR by detemining the aea of the shaded egion in the figue below. The aea of the ectangle is appoximately. kg of catalyst. P- (c) The STR catalyst weight necessay to achieve 40 % convesion can be obtained by calculating the aea of the shaded ectangle shown in the figue below. The aea of the ectangle is appoximately 7.6 kg of catalyst.

17 P- (d) The catalyst weight necessay to achieve 80 % convesion in a PBR is found by calculating the aea of the shaded egion in the figue below. The necessay catalyst weight is appoximately kg. P- (e) The amount of catalyst necessay to achieve 40 % convesion in a single PBR can be found fom calculating the aea of the shaded egion in the gaph below. The necessay catalyst weight is appoximately kg.

18 mao/-am P- (f) P- (g) o diffeent (- ) vs. () cuves, eactos should be aanged so that the smallest amount of catalyst is needed to give the maximum convesion. One useful heuistic is that fo cuves with a negative slope, it is geneally bette to use a STR. Similaly, when a cuve has a positive slope, it is geneally bette to use a PBR. P- (a) Individualized Solution P- (b) ) In ode to find the age of the baby hippo, we need to know the volume of the stomach. The metabolic ate, -, is the same fo mothe and baby, so if the baby hippo eats one half of what the mothe eats then ao (baby) = ½ ao (mothe). The Levenspiel Plot is shown: utocatalytic Reaction Mothe Baby onvesion

19 baby ao.6 * m Since the volume of the stomach is popotional to the age of the baby hippo, and the volume of the baby s stomach is half of an adult, then the baby hippo is half the age of a full gown hippo. 4.5 yeas ge.5 yeas ) If max and m ao ae both one half of the mothe s then m o m 0 mothe M M and since m M vmax K then M v max M baby M KM m o M o m o mothe m M baby M mothe M mothe will be identical fo both the baby and mothe. ssuming that like the stomach the intestine volume is popotional to age then the volume of the intestine would be 0.75 m and the final convesion would be 0.40 P- (c) stomach = 0. m o om the web module we see that if a polynomial is fit to the autocatalytic eaction we get: m0 = M

20 m0 nd since stomach =, M solve = = 0. m stomach =.067. o the intestine, the Levenspiel plot fo the intestine is shown below. The outlet convesion is Since the hippo needs 0% convesion to suvive but only achieves 7.8%, the hippo cannot suvive. P- (d) PR STR PR: Outlet convesion of PR = 0.

21 STR: We must solve = 0.46 = (-0.)( ) =0.4 Since the hippo gets a convesion ove 0% it will suvive. P- o a STR we have : = 0 f 0 So the aea unde the vesus cuve fo a STR is a ectangle but the height of ectangle 0 coesponds to the value of at = f 0 But in this case the value of is taken at = i and the aea is calculated. Hence the poposed solution is wong. DP- (a) Ove what ange of convesions ae the plug-flow eacto and STR volumes identical? We fist plot the invese of the eaction ate vesus convesion. Mole balance equations fo a STR and a PR:

22 STR: 0 PR: 0 d Until the convesion () eaches 0.5, the eaction ate is independent of convesion and the eacto volumes will be identical. i.e. d PR d 0 0 STR DP- (b) What convesion will be achieved in a STR that has a volume of 90 L? o now, we will assume that convesion () will be less that 0.5. STR mole balance: DP- (c) 0 v m 0 v 0 0 m mol 8 m. s s mol m This poblem will be divided into two pats, as seen below: The PR volume equied in eaching =0.5 (eaction ate is independent of convesion). v m The PR volume equied to go fom =0.5 to =0.7 (eaction ate depends on convesion).

23 inally, we add to and get: DP- (d) tot = + =. x0 m What STR eacto volume is equied if effluent fom the plug-flow eacto in pat (c) is fed to a STR to aise the convesion to 90 % We notice that the new invese of the eaction ate (/-) is 7*08. We inset this new value into ou STR mole balance equation: STR DP- (e) v m If the eaction is caied out in a constant-pessue batch eacto in which pue is fed to the eacto, what length of time is necessay to achieve 40% convesion? Since thee is no flow into o out of the system, mole balance can be witten as: Mole Balance: dn dt

24 Stoichiomety: N N ( ) 0 ombine: N 0 d dt om the stoichiomety of the eaction we know that = o(+e) and e is. We inset this into ou mole balance equation and solve fo time (t): 0 ( ) N 0 d dt t dt d ( ) fte integation, we have: t 0 ln( ) Inseting the values fo ou vaiables: t =.0 x 0 0 s That is 640 yeas. DP- (f) Plot the ate of eaction and convesion as a function of PR volume. The following gaph plots the eaction ate (- ) vesus the PR volume: Below is a plot of convesion vesus the PR volume. Notice how the elation is linea until the convesion exceeds 50%.

25 The volume equied fo 99% convesion exceeds 4*0 m. DP- (g) itique the answes to this poblem. The ate of eaction fo this poblem is extemely small, and the flow ate is quite lage. To obtain the desied convesion, it would equie a eacto of geological popotions (a STR o PR appoximately the size of the Los ngeles Basin), o as we saw in the case of the batch eacto, a vey long time. DP-B Individualized solution DP- (a) o an intemediate convesion of 0., igue above shows that a PR yields the smallest volume, since fo the PR we use the aea unde the cuve. minimum volume is also achieved by following the PR with a STR. In this case the aea consideed would be the ectangle bounded by =0. and = 0.7 with a height equal to the 0 /- value at = 0.7, which is less than the aea unde the cuve.

26 DP- (b) DP- (c) DP- (d) o the PR,

27

28 DP- (e) DP-D

29 DP-D (a) DP-D (b) DP-D (c) DP-D (d)

30 DP-D (e) DP-D (f) DP-D (g) DP-D (h)

31 DP-E DP- (a) ind the convesion fo the STR and PR connected in seies. - /(- )

32 DP- (b)

33 DP- (c) DP- (d)

34 DP- (e) DP- (f) DP- (g) Individualized solution