HEMISKI Zbirka re{eni zada~i REAKTORI 3

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Petti del. Kataliti~ki reaktor so fiksen sloj katalizator (PBR) Odnosno, ovaa analiza zna~i deka se presmetuva kolkava e potrebnata koli~ina katalizator za 1 kmol/h procesiran reaktant A, za razli~ni vrednosti na pritisokot, za izlezna konverzija na reaktantot od 50%. Za ovie presmetki, pokraj ravenkata za dizajn, e potrebna i stehiometriska tablica poradi izrazuvawe na brzinata preku konverzijata i vkupniot pritisok. Stehiometriska tablica: F ( X ) i Fio Fi() Fi(X) pi yi P P FT ( X ) ( 1 X ) ( 2, 5 X ) A FAo FA = FAo– FAo(1–X) p A y AP P ( 1 2X ) ( 2, 5 X ) B FBo = FAo FB=FAo+2 FAo(1+2X) pB yB P P 0, 5 I FI = 0,5FAo FI = 0,5FAo FI = 0,5FAo pI yi P P ( 2, 5 X ) FTo = 2,5FAo FT= FAo(2,5+X) P vk = pi = P B F Bo / F Ao 40 / 40 1; F / F FAo F X F Ao A I ; F I Ao Ao X . 20 / 40 0, 5; Brzinski izraz preku konverzijata i pritisokot: 0, 001 10 p A ( rA ) kmol/(kgkath); pi ( atm); ( 1 10 p p ) A B ( 1 X ) 0, 001 10 P ( 2, 5 X ) ( rA ) f ( X , P). (7) ( 1 X ) ( 1 2X ) 1 10 P P ( 2, 5 X ) ( 2, 5 X ) ( ) vo kmol/(kg h); P( atm). r A kat Brzinskiot izraz (7) se zamenuva vo ravenkata za dizajn: W F Ao 0, 5 0, 5 dX dX kg kat . (8) ( r ) f ( X , P) kmol/h 0 A 0 369
Re{avawe na ravenkata (8) zna~i presmetka na vrednosta na integralot, odnosno na koli~nikot W/FAo. Bez razlika na izbraniot metod, vrednosta na integralot }e se presmetuva za razli~ni vrednosti na pritisokot. Sekoja presmetka zna~i integriraweto da se izveduva do X = 0,5! Primena na solver za diferencijalni ravenki sekako e upatna, bidej}i mnogu re{enija se dobivaat za kuso vreme, a za sekoja promena na pritisokot programata e ista! Analizata za razli~ni vrednosti na pritisokot vo intervalot od P = 0,5 atm do P = 10 atm gi dade slednive rezultati (tabelarno i grafi~ki): 370 P = 1,8 atm P (atm) X W/FAo (kgkat/(kmol/h)) 0,5 0,5 1056 1,0 0,5 978 1,5 0,5 960 1,8 0,5 958 2,0 0,5 959 2,2 0,5 960 2,5 0,5 965 3,0 0,5 973 3,5 0,5 984 4,0 0,5 990 5,0 0,5 1010 7,0 0,5 1060 10,0 0,5 1135 P (atm) W/FAo = f(P)
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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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Ropt=1,5 Vmin=1884litri êtvrti del
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Petti del Kataliti~ki reaktor so fi
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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
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Page 379 and 380: POLYMATH Results Calculated values
Page 381 and 382: 364 Za brzinski izraz preku konverz
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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
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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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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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