CHEN%204330%20HW%204%20Solutions-scan

CHEN%204330%20HW%204%20Solutions-scan - -er 5-1. (2) For...

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Unformatted text preview: -er 5-1. (2) For batch system, constant volume. ammo):[(1—§)]”[(i-x)]% )5 4N2 amp—133] [i—X]’/2 P3-l 6 (3) Liquid phase reaction '9 assume constant volume Rute Law (reversible reaction): . (1' —r a: k C C ———-C— .4 i A B ] Stoichiometry: CA "‘ CA0 (1 ’X)’ CB " CA0 (1 "X), Cc " CAOX To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for X. CACBKC " CC 1 2 ‘v (“.40 (1" Xe) 'KC " CAoXe l 1 AOKC [rel—(2+ )XeH-O Xe a 0.80 To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric relations. mol mol dm3 dm3 ‘ 1 me] C ac l—X .2” [—0.80 30.4 H A0 ( ) dm} ( ) ‘1m3 g C4 = £po a 2 m", * 0.80 a 1.6 m", i dm3 dm’ CA a CA0 (1_X).. 2 (1 —0.80)== 0.4 P3-16 (b) Stoichiometry: 3-19 8-yA06-(l)(3—-l)-2and0C-0 __&_NA0(1'X)_C (I'X) " V Vo(l+aX) ‘°(1+2x) NC 3NA0X C 3X CC.._.. V Vo(l+sX)- "°(1+2x) Combine and solve for Xe. (I'Xe) 3X4 ch“ (1+ 2X6) C“ (1+ 2X,) KC (1 — Xe )(1+ 211;)2 - 270301133 2 —-(4+ 27C“ )X‘3 +3X¢ +1 - 0 Kc X, -0.58 Equilibrium concentrations: -i-“____10atm -0305fl931. CA0 3 RT" (4OOK)(0.082 ‘1’" mm) d’" moIK (1 -0.58) mol - 0.305 - 0.059— C‘ (1+ 2(o.58)§ dm’ 3(0.58)(0.305) mol -. - 0246—— CC (1+2(0.58)§ dm3 P3-16 (c) Same reaction, rate law, and initial concentration as part (1)) gas phase, batch reaction. Stoichiometry: Combine and solve for X. KCCAO (1" Xe )" (3CA0Xe)3 Xe - 0.39 Equilibrium concentrations 3-20 mu! « ,3 —o. 9 -0.19—-»— (A (0.305)(l 3 ) dm} I ma! (g. a. (0.305)(0.39)- 0.36 2;; P346 (d) (m phase reaction in a constant pressmc, batch reactor Rate law (reversible reaction): ("3. .4"! 3: le'A c 5i!()iL‘iliOlIlctl'yI p, a ylwfi _ (1)(3—1)- 2 and Q. - 0 NA Ego—X) v (l—X) (“J V V0 (1 + ex) “‘0 (1 + 2x) /, NC 3mg X lam» “30 W l“ V i‘},(l+£X) “(1+2x) (Olllhinc and solve for Xe: _ , 3 KCCJO (1- Xe ) 30.40 {a 1+ ZXe A; a 0.58 quliilihrium concentrations: a ‘‘‘‘‘ "WW '5 “m 3(().305)(0.58) mul (7C a. “mmw___._ ., 0 246. 1+2(0.58) ' {15:7 P347 Given: Gas phnsc reaction A t B '9 8C in a batch reactor fitted with a piston such that V 0.1% k = whiff - 1]; mol2 sec «1;, == ijC‘B “AU ’ N30 at i '4 0 v0 0.15 n" |.t()“(‘» 600R (‘unslant 3-21 g Assume that it takes three hours to fill, empty, and heat to the reaction temperature. tf = 3 hours tIolal z “1 + tf tum, : 2. l4hours + 3 hours = 5. l4 hours. Therefore, we can run 4 batches in a day and the necessary reactor volume is l 4400an 4 Referring to Table l—l and noting that 3600 dmJ is about I000 gallons, we see that the price would be approximately $85,000 for the reactor. P4-5 (i) The points of the problem are: I) To note the significant differences in processing times at different temperatures (i.e. compare part (b) and ((3)). 2) That the reaction is so fast at 77°C that a batch reactor is not appropriate. One minute to react and 180 to fill and empty. 3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black. = 3600alm3 P4—6 (a) O emu-in: Farm. “3 $MLE’O “Fa 3 k CA Ca / $1: m“), Fin) 4-9 Ed}! I .~. A. m ’r,‘ “w .3: M ) .1», W < (26 J3 ,_ ‘ "‘c " X L" . F(,, 4 :2 1a,; m / yr . 31AM): OptfiSAIDH F “no”? X 12:45 llbmol M~w~~ a .. 195:»; mc: yr 304 24m 3“ 2781b 20'” n: 3 ’ in o e l t VCSTR m 3313C” £21 x ‘"”}'7f‘_r: 133.7 ft} 7.48 g1; P4-6 (b) To increase conversion, use PFR, higher temperature, or use better catalyst. P4-6 (c) I»! ‘5 P4-6 (d) P’i‘ii Damian Eguamm ,1 Y w y r If! {1‘3 FM, ( ~ J}; \ A j t .22" ‘; w. 4-10 :: L a' 5 f “M WWW» , ‘ 4, ,. IQ) t Lu:1'1:1ti011 :xirsées the reaction time ‘wtr :2 finch and unit m an (m! x _ , in}: that hush, ‘! inn: is. a {nuicmfi'f' i'r muzzm‘: ihih conversion and few h mth and low convmnmu bm many lszzzu'l‘xes per day. What «,v’mv; ’ ’ “ ‘ :u‘ualloStnumiw' :21';:.;¢::t01’:~:'f N m ' E411.“ ‘ Cm ‘ (TIM, : 0.2"bnmll f1“ 1‘5 H .‘5 ‘ ‘ Uzr » 1" w X24 N j: H T: I . w- 1 v7 » U m—ww " ' {’M“;\ 3‘ i Y) thunk {rm +3 W wlwwl/éf’ (Fifi «l 3 . . ‘ 1 ‘ ‘ «Cm /‘ " 9101)] ; t 3H m" " {I'd} II é ’? 1 /‘ c \k / i I ~\\\‘ / I \“““a~ ' r" > . 1 . J {3 0. 0,4 1 X.-..\> I‘M: ‘ _;,,},-, '{\';’f- 3'; (s1 .1 lief 1‘4)’;z_w‘)i3212§'é i:“\ (“é K; ik’, Mfg”) P4-6 lmlividlmlized solution [)4-6 Individlmlizml solution [)4-6 (h) lndix‘ iduulized so luliun E ~~-,“,.MM.MWMM 4-H l =l———-~—=.526 l+rkCA0 2.lll So, while calculating PFR conversion they considered reaction to be lst order. But actually it is a second order reaction. P4-8 (d) A graph between conversion and particle size is as follows: Originally we are at point A in graph, when particle size is decreased by l 5%, we move to point B, which have same conversion as particle size at A. But when we decrease the particle size by 20%, we reach at point C, so a decrease in conversion is noticed. Also when we increase the particle size from position A, we reach at point D, again there is a decrease in the conversion. => X tip ----—----I- ~_\\K P4-9 A ¢> B To = 300K KCO (300K)= 3.0 V = l000gal = 3785.4 dm’ Mole balance: V = FAUX -w " rA 2 C; Rate law: —rA =k0 C, —-E Stoichiometry: CA = CAD (I — X) and C5 = CBOX => V = i X . 2 kUCAU (I _ X)? _ (Ki KC. 2 V (l _ X)2 _ Xm 2 Z = .40 = K} = 3785.4 0 7 _ 0.4 kUsz X 0.4 3 Z = 2902.2dm3 , 2= i =exp[£(i_i)] k R 7;, T (1 4-20 ) X: I-X M —-~~w K} [( ) 113.] k ex E I I I 6‘ Z=M= " ““~ am whne ko P R To T AH , KI‘ = Kw exp my" "’L‘i ‘ ' R T” T Solving using polymalh to get a table ofvalucs ofX Vs T. 800 Polymath program P4~9,pol. POLYMATH Results I $ v.1rjnble Value f(x) Ini GUESS x 0‘4229453 3.638E~12 0.5 3 To 300 i T 305 . 5 2 2902 2 1 v 3785.4 g H 1.5E+04 3 R 2 y 1.5684405 3 Ken 3 er ~2.SE+04 5 KC 1.4169064 NLE Report (safenewt! 0443 7 Nonlinear equations . ...,,V J f(X) : (Uy)*X/((1~X)"2 — XAZ/Kc) Av : 0 “2 “m '37 ““L “N “ ”“ 041 _.___.._‘__# I. . _ ‘__~‘_“ M“ Explicit equations ,, ’ fl "1 To : 300 0'4 A ~«« ~~~ ~-~ a ’1 1 T : 305.5 K , I ' Z : 2902.2 039 r —- — —r-— Aw w v “~— 3 I V 1 3785.4 033 ‘ -“fi __~“.. ““ V -E“‘- ‘ I E = 15000 ' R I 2 037 w I» ~ H AWN-3 —l—_ i Y : exp(E/R*(1/TO~1/T)) 0.35 ,, V‘ ,.4,_-1______1._1‘_‘____V-,:‘_“,,j l KOO : 3 295 300 305 310 315 320 ” ' er : ‘25000 Temperature 'III KC 1 KCO*eXp(HrX/R*(1/TO‘1/T)) 4—21 We get maximum X = 0.4229 at T = 305.5 K. P4-10 (a) For substrate: Cell: VOCC = rgV Fso—F5+T,V=0 C “max '0‘ (Cso— Cs)Vo = 1’; VYS/c = VYS/C (. K M + CS ' P4-10 (b) CC = YC/S [C50 ‘ CS] ,u C. (C50 ‘ CS )vo ‘ VYS/C [#61] = 0 M S _\ v C ’1 v 3 (C50 ‘ Cs )vo ‘ VYC/S (C50 ‘ C5 )= 0 M s 05sz O—C —5 .3 (3 S)5 2 x0 ><[5+CS (30—CS)=0 Solving we get Cs = 5.0 g/dm3 or 30 g/dm3. if C5: C30 no reaction has occurred so the only valid answer is C5 = 0.5 g/dm’. P4-10 (c) CC = YC/s(Cso ' C5) = 0.8(30 — 5.0)g/dm3 = 20 g/dm3. P4-10 (d) vnew = vo/2 = 2.5 dm3/h Using equation from above, we get Cs = L67 g/dm3 and CC = 22.67 g/dm3 4-22 P4-10 (e) vnew = vo/3 = 25/3 dm3 Using equation from above, we get Cg = 3.0 g/dm3 and CC = 21.6 g/dm3 P4-10 (1) For batch reactor: Cso = 30 g/dm3 Cm : 0.] g/dm3 CC = CU) + YC/s(Cso A Cs) v = IO dm3 See Polymath program P4,, 104’. pol. POLYMA‘I‘II Results Calculated valugg of the DES! ygriablgg Variable initiaL value minimal value maximal value final. value t 0 O 15 15 CS 30 0.0382152 30 0.0382152 C80 30 30 30 30 YCS 0.8 0.8 0.8 0.8 Km 5 5 5 5 Umdx 0.5 0.5 0.5 0.5 CCO 0.1 0.1 0.1 0.1 CC 0.1 0.1 24.069428 24.069428 rg 0.0428571 0.0428571 5.4444349 0.0912841 rs 70.0535714 76.8055436 70.0535714 ~0.1141052 nega.t,ive_ 0.0535714 0.0535714 6.8055436 0.1141052 ODE ReporHRKF4Si ; Differential equations as entered by the user a 1 J d(Cs)/d(t) = rs Explicit equations as entered by the user :1] Cso=30 12] ch=0.8 § 1Km=5 ' 2} Umax = 0.5 1 C00 = 0.1 I CC = Cco+ch*(Cso—Cs) i l rg = (Umax*Cs/(Km+Cs))*Cc ‘J rs = —(1chs)*rg J negative_rs = —rs 'r) 30.0 7.0 21.0 5.6 18.0 _A 4.2 ' (‘s 12.0 2.8 6.0 1.4 0.1m [ 9.0" 13.0" 15.0" 0.0” 3.00 6.1)" t ‘10" I210” '51"! IN] MM 0.00 J_mi L 4-23 .1—1: P4-10 (g) Graphs should look the same as part (0 since reactor volum vnlnme hatch reactor, 6 is not in the design equations for a constant ...
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CHEN%204330%20HW%204%20Solutions-scan - -er 5-1. (2) For...

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