HEMISKI Zbirka re{eni zada~i REAKTORI 3

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Petti del. Kataliti~ki reaktor so fiksen sloj katalizator (PBR) Na pritisok P = 30 atm i temperatura T = 314 o C = 587 K vrednostite na konstantite vo brzinskiot izraz se: 5 k 7, 0 kmol/(kgkat h atm ); 1 1 K1 1, 73atm ; K 2 0, 3atm . Ovaa reakcija se izveduva vo ceven reaktor so fiksen sloj katalizator pod uslovi za koi e dobien brzinskiot izraz. Vleznata smesa vo reaktorot e so stehiometriski odnos na reaktantite, so molski protok na jagleroddioksidot FAo = 100 kmol/h. Da se presmeta potrebnata koli~ina katalizator za 20% konverzija na reaktantite. Re{enie: Najnapred izrazot za brzinata na reakcijata daden preku parcijalnite pritisoci }e go transformirame preku konverzijata. Go izbirame jagleroddioksidot kako baza za presmetki. Stehiometriskata ravenka ja pi{uvame vaka: A 4B C 2D i ja sostavuvame slednava stehiometriska tablica: A (CO2) B (H2) C (CH4) D (H2O) F ( X ) i Fio Fi() Fi(X) pi yi P P FT ( X ) ( 1 X ) ( 5 2X ) FAo FA = FAo– FAo(1–X) pA yAP P FBo= 4FAo FTo=5FAo FB = = 4FAo–4 4FAo(1–X) 4( 1 X ) pB yB P P ( 5 2X ) X 0 FC = FC = FAoX pC yCP P ( 5 2X ) 2X 0 FD = 2 FD = 2FAoX p D y D P P ( 5 2X ) FT = = FAo(5 – 2X) P vk = pi = P FAo FA B FBo / FAo 4 / 1 4; X A X ; FAo X ; P = 30 atm. F Ao 363
364 Za brzinski izraz preku konverzija se dobiva: ( 4 k pCO p 2 H2 rCO ) kmol/(kg 2 5 kat ( 1 K1 pH K 2 pCO ) 2 2 4 h) ; pi ( atm); ( 1 X ) 4( 1 X ) k P ( 5 2 ) P ( 5 2 ) ( ) X X r A . (1) 5 4( 1 X ) ( 1 X ) 1 K1 P 2 ( 5 2 ) K P ( 5 2 ) X X Izrazot (1) moè da se razvie i sredi! Po taa procedura i so vnesuvawe na numeri~kite vrednosti za konstantite se dobiva slednava forma: 10 4, 35456 10 ( 1 X ) ( rA ) 5 ( 221, 6 218, 6X ) Ravenkata za dizajn na PBR, kmol/(kgkath) . (2) W F Ao X 0, 2 0 5 dX , (3) ( r ) ja kombinirame so brzinskiot izraz (1) i ja re{avame za 20% konverzija na jagleroddioksidot, odnosno vodorodot. Se izbira metod za re{avawe na ravenkata (3): Simpsonovoto ednotretinsko pravilo ili druga integraciona formula, ili solver za diferencijalni ravenki. Ako se izbere solverot za diferencijalni ravenki od POLYMATH, toga{ se re{ava diferencijalnata forma na ravenkata (3), A Ao A dX (r ) . (4) dW F Za zadadeniot vlezen molski protok na jagleroddioksidot FAo=100 kmol/h i 20% konverzija na reaktantite se dobivaat slednive rezultati:
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HEMISKI REAKTORI 3 Zbirka re{eni za
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Skopje, 2010
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Predgovor . . . Hemiskiot reaktor e
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lot da se stekne so ve{tina da gi k
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~ite }e mu ovozmoì preku niv da go
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5. TOPLINSKI BILANSI ..............
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XIV êtvrti del IDEALEN CEVEN REAKT
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Zada~a 4. Adijabatski PBR so pretho
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1. STEHIOMETRIJA 1.1. Prost reakcio
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Prv del. Definicii, koncepti, stehi
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2. TOPLINA NA REAKCIJA I HEMISKA RA
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3. IZRAZI ZA BRZINA NA REAKCIJA 3.1
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Molskite koncentracii vklu~eni vo b
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kako ravenkata (52). Razlikata e vo
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Ravenkite za dizajn na cevniot reak
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Ravenkata za dizajn na ceven reakto
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Za presmetka na brojot na reaktorit
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28 Po~eten uslov za re{avawe na sis
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Toplinski bilansi za adijabatska ra
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32 T H ( r ) V r A a, vlez Ta, vle
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Toplinski bilansi za izotermna rabo
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Osnovnata ravenka na toplinskiot bi
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38 H / L Q q dz Ua ( T T ) dz ;
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Vtor del [arèn reaktor
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Re{enie: a) Za izotermna rabota na
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POLYMATH Results Calculated values
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Explicit equations as entered by th
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Jasno e deka vremeto potrebno za re
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Zada~a 2 64 [arèn reaktor i CSTR s
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Rezultatot e deka so eden {arèn re
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Vo ravenkata za dizajn (48) gi zame
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a) Brojot na {arìte koj }e go obez
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Kone~nata forma na brzinskiot izraz
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76 Mi % (w/w) 1 3 Cio smesa kmol/m
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Re{enie: Samo so primena na ravenka
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data e so sè pomali temperaturni r
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) Za varijantata na razmena na topl
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( H r )( rA ) V TW , vlez T
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Vkupnata koli~ina toplina {to }e se
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) Za presmetka na vremeto na reakci
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( 1 X ) To C A C Ao ; (1) ( 1 0, 5
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11620 X T To . (6) 38, 73 12, 91X
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96 t r t r C P dP dP C Ao ; 0, 5(
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a) Od kineti~kite podatoci da se op
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100 C C W n C Wo / V Wo W rastvor
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A B ; ( rA) k1C A k2C B ; 7 1 k
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) Vremeto na reakcija za 80% konver
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presmetki za izotermna rabota na re
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Zada~a 1 Hidrogenacija na etilen vo
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Tret del. Proto~en reaktor od rezer
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1 Tret del. Proto~en reaktor od rez
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POLYMATH Results Tret del. Proto~en
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F F X X A1 B1 1 1 Tret del. Proto~e
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C B2 0, 1056 F F X 2 A2 B2 2 Tre
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Zada~a 8 Tret del. Proto~en reaktor
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Explicit equations [1] R = 0 [2] T
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X * XMB XEB X*; XMB; XEB 1 0.8 0.6
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XMB XEB 1 0.8 0.6 0.4 0.2 Zada~a 10
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( C C A1 ( C C A2 ( C C A3 A Tret d
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1 Tret del. Proto~en reaktor od rez
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Zada~a 15 Tret del. Proto~en reakto
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C C C C Bo Bo Wo Wo Tret del. Proto
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dT dt Tret del. Proto~en reaktor od
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Zada~a 1 Endotermna reakcija vo gas
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êtvrti del. Idealen ceven reaktor
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Re{enie: êtvrti del. Idealen ceven
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A (NO) B (O2) C (NO2) I (N2) êtvrt
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X 0 0, 2 êtvrti del. Idealen ceve
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Re{enie: a) Izotermna rabota: êtvr
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T (K) k (T) izraz (4) êtvrti del.
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( r êtvrti del. Idealen ceven reak
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0,22 êtvrti del. Idealen ceven rea
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Zada~a 16 êtvrti del. Idealen ceve
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V PFR k F Ao êtvrti del. Idealen c
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Page 330 and 331: Ropt=1,5 Vmin=1884litri êtvrti del
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Page 341 and 342: Ponatamu podgotvuvame brzinski izra
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Page 347 and 348: 330 Fio Fi() Fi(X) pi = yiP = (Fi/F
Page 349 and 350: 332 A FAo B Bo Fio Fi() Fi(X) F 2F
Page 351 and 352: Ako reakcijata se odviva vo pove}ec
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Page 359 and 360: Re{enie: Rabota so razmena na topli
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Page 363 and 364: Kako {to moè da se vidi od dobieni
Page 365 and 366: Komentari: 1) So adijabatska rabota
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Page 379: POLYMATH Results Calculated values
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Page 385 and 386: 368 Sega go pi{uvame bilansot na ak
Page 387 and 388: Re{avawe na ravenkata (8) zna~i pre
Page 389 and 390: Analizata za razli~ni vrednosti na
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Page 399 and 400: Za da zaklu~ime dali ovoj volumen e
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Page 406: [esti del Industriski reaktori
Page 409 and 410: 392 - vlez vo reaktorot: T C o 650
Page 411 and 412: Izleznata konverzija normalno raste
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Page 415 and 416: Toplinskiot bilans vo ovoj slu~aj }
Page 417 and 418: 400 X 0.5 0.4 0.3 0.2 0.1 1 2 3 4 5
Page 419 and 420: 402 1060 1032 T(K) X 1004 976 948 9
Page 421 and 422: Zada~a 2 404 Piroliza na etan do et
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Page 427 and 428: T 1053 930.72495 1053 930.72495 FAo
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Za vrednosti na Re-brojot pome|u 30
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416 V (m 3 ) P(V) V (m 3 ) X(T) T (
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eaktorot: dali toa }e bide izotermn
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2) Da se povtori re{enieto pod 1),
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e{avawe na zada~ata da se koristi p
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ni ravenki, (5), (6) i (7), zavisno
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Ramnoteà: Za da se kompletira slik
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Explicit equations [1] Feo = 0.0034
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432 T (K) Presek na ramnote`na i ad
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cionen mehanizam koj, pokraj glavna
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436 - vodorod, H: - benzen, B: - et
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Za da se re{i sistemot od 7+1 difer
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440 FE(V) Promena na molskite proto
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442 F F F F F F F E FEo 1) 1 ( 1
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enzen izvedena pod dadenite uslovi
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- toplinski kapaciteti na reaktanto
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dX (rA ) . (3) dW FAo Ravenkata (3
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FAo 20 20 20 20 T 527.23 527.23 812
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2) Novata varijanta na adijabatskat
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454 C P, A F C i H r C P, i P, B
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Zna~i, so ovaa varijanta razreduvaw
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segment }e treba da se presmetuva p
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dodeka izrazite za molskite koncent
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Toi 650.66442 650.66442 650.66442 6
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Sedmiot segment: POLYMATH Results C
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470 X 0.4 0.32 0.24 0.16 0.08 0 500
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Zada~a 5 472 Proizvodstvo na glukon
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30 24 / 144 5 . Zna~i, na po~etok
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T a b e l a 1 Po~etok na ciklus 10
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478 Vmax / K m 1 5 3 k 0, 016 h
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te~nata reakciona smesa se odrùva
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482 r r , r , 1 r , 2 r , 3 C
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azot za vkupno sozdadena kiselina (
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486 CB(t) CM(t) t (s) t (s) t (s) A
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Zada~a 7 488 Kataliti~ka oksidacija
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490 dX dW ( rA ) F1 ( X , T ) (2)
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Podolu se izve{taite i grafi~kite p
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K2 13.126279 5.2903007 13.126279 5.
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Reakcioniot plan pretstavuva grafik
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498 Podatoci potrebni za re{avawe n
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nosno ravenka za dizajn (preku konv
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0, 01094 kmol ( rA ) k 0, 425 P ; P
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- proizvod od koeficientot na preno
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506 - Drugite podatoci se: sloj 5
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508 X(W) W (kg) Zavisnost konverzij
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temperatura dava podobar efekt i vo
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Bidej}i reakcijata e egzotermna so
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514 - Kinetika r ( r k r k 1 1 2 2
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516 k k r k k 1 1 1 2 2 2 dF C dV d
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So solverot za diferencijalni raven
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na koncentracijata na kislorodot. T
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522 Varijanta 2. Proces so kislorod
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Ravenka za dizajn na PBR: Ravenkata
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Sistemot diferencijalni ravenki (11
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528 FA(W) FB(W) FC(W) W (kg) FD(W);
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Ravenkite {to treba da se kombinira
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vo 7648.55 7648.55 7648.55 7648.55
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536 r2(W) r1(W) Grafik na zavisnost
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Y = 1 7, 077 ( 1 0, 45) ( 1 0, 4
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540 r2(W) r1(W) W (kg) Grafik na za
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1 ( rA ) 1 r1 k1 p A pB kmol/(kg
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544 Za specifi~nata povr{ina se zam
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na toplina i za protokot na vodata
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Cpa 44.257982 44.257982 65.187219 5
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Na prviot grafik ne e pretstavena z
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552 Da se presmetaat i analiziraat
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FMo 9, 07 3 CMo 0, 169 kmol/m .
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povtorno treba da se re{at ravenkat
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da se odviva taka vleznata temperat
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560 ~ , i iC P 35 12, 518 0, 05
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Vremeto na nestacionarniot period n
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564 C C C Ao Bo Mo F F Ao o Bo
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Kako {to se gleda od dobienite rezu
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Zada~a 11 568 Proizvodstvo na akril
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na izreagiran propilen (AK/P 0,82-
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se dobivaat preku stehiometriskite
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Granicite za re{avawe na ravenkite
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578 T(V) Grafik 2. Promena na tempe
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turata e T 520 K. Vo ovaa presmetk
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smesa moè da se izrazi preku edins
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inerti i amonijak. Koga vtoroto e s
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586 k1 pC 2 R 0 K P ( atm ) ; (
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duvanata reakcija. Ovaa operacija t
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I dvata grafika, 2 i 3, pokaùvaat
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( r A 592 F C F F F F To T T F
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Sledniot ~ekor e da se sostavat mol
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na i specifi~nata povr{ina na topli
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[5] k1 = 1.79*(10^4)*exp(-20800/1.9
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Stepen na konverzija na azot Stepen
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602 1) UaV = 25000 kJ/(m 3 ·h·K)
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LITERATURA 1. Levenspiel, Octave, C
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ISBN 978-608-65044-3-4
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HEMISKI Zbirka re{eni zada~i REAKTORI 3

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