HEMISKI Zbirka re{eni zada~i REAKTORI 3

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[esti del. Industriski reaktori C6H 5C 2H 5 H 2 C6H 5CH3 CH 4 ( E H T M ) k ( T ) p p 3 kmol/(m s) ; p , p ( atm) (3) rT , 3 3 E H sloj E H ( r rE ) 3 ( rH ) 3 rT , 3 rM , 3 11000 3 k3 ( T ) 1282, 2exp0, 2961 kmol/(msloj s atm T b) Podatoci za vleznata smesa: – temperatura: To = 800; 900; 930; 1100 K – pritisok: Po = 2,4 atm – odnos vodna parea/etilbenzen: 3 2 ) ; T ( K) W/E = FW/FEo = W = 8,72; 14,5; 20; 58,82 kmol/kmol. v) Toplina na reakcija i specifi~ni toplini: C C C C P, E P, H P, Et P, T 30 H H H o R, 1 o R, 2 o R. 3 299 kJ/(kmol kJ/(kmol 90 kJ/(kmol 249 kJ/(kmol 118000 kJ/kmol 105200 kJ/kmol E E 53900 kJ/kmol H Et E T K); K); K); K); C C C C P, S P, W P, B P, M E 273kJ/(kmol 40 g) Podatoci za katalizatorskiot sloj: 3 2137 kg/m ; 0, 4. cvrsto Re{enie: 1) Adijabatski PBR kJ/(kmol 201kJ/(kmol S W 68 kJ/(kmol Koga dehidrogenacijata na etilbenzen vo stiren se razgleduva kako prost reakcionen sistem, toga{ nejzinata celosnost se prikaùva so edna stehiometriska ravenka i eden brzinski izraz, ravenkata (1), zaedno so temperaturnite zavisnosti za brzinskata i ramnote`nata konstanta. Ova sugerira za B K); M K); K); K). 421
e{avawe na zada~ata da se koristi priodot preku stepenot na konverzija na edinstveniot reaktant. Metodologijata na ovoj priod e poznata: edna ravenka na molski bilans, edna ravenka na toplinski bilans i stehiometrija! No, bidej}i sakame re{enie i za slu~ajot koga reakcioniot mehanizam e sloèn (so site tri reakcii), zada~ata }e ja re{avame so priodot so site molski bilansi, za site u~esnici vo reakcijata, zaedno so toplinskiot bilans. Ravenka za dizajn – molski bilansi: Obi~no ravenkata za dizajn na kataliti~ki reaktor se izrazuva preku masata na katalizatorot. Vo ovaa zada~a }e koristime ravenka za dizajn na PBR preku volumenot na reaktorot (vo koj e smesten katalizatorot), odnosno preku volumenot na fiksniot sloj katalizator, bidej}i brzinskiot izraz e daden vo takva edinica za reakcionen prostor, kmol/(m 3 slojs). Ravenkata za dizajn, odnosno molskiot bilans na u~esnikot i vo PBR: dFi ri 0 , (4) dV }e ja primenime na site u~esnici vo reakcijata (1): dFE dFE – etilbenzen, E: ( rE ) r1 (5) dV dV 422 dF S – stiren, S: 1 r S dF rS (6) dV dV – vodorod, H: 1 r dFH dFH rH (7) dV dV Za ovoj sistem od tri diferencijalni ravenki da se re{i, potrebno e: 1) da se zameni brzinskiot izraz (1); 2) bidej}i vo brzinskiot izraz se prisutni parcijalnite pritisoci na site u~esnici vo reakcijata, treba da se najde vrskata pome|u niv i molskite protoci; 3) bidej}i rabotata na reaktorot e adijabatska, temperaturata }e se menuva nadol` reaktorot, pa }e bide potrebna dopolnitelna ravenka. Toa e toplinskiot bilans; 4) bidej}i diferencijalnite ravenki se obi~ni od prv red, }e tre-
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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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ili 318 - diferencijalen oblik, kak
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Od izve{tajot i od rezultatite od p
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Grafi~ka prezentacija na site rezul
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Ponatamu podgotvuvame brzinski izra
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Zada~a 3 326 Reakcija vo gasna faza
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Iako brzinskiot izraz e prili~no go
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330 Fio Fi() Fi(X) pi = yiP = (Fi/F
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332 A FAo B Bo Fio Fi() Fi(X) F 2F
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Ako reakcijata se odviva vo pove}ec
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Isto taka, dobro e da znaeme kolku
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Prvo, brzinskiot izraz se pi{uva vo
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340 Yexp X Ycalc 6087E-12 0.105 8.0
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Re{enie: Rabota so razmena na topli
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344 dT dV dT dT dT FT CP 65, 9 6,
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Kako {to moè da se vidi od dobieni
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Komentari: 1) So adijabatska rabota
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za KC(T) se dobiva sledniov izraz:
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352 W (kg) W (kg) (-rA) = f(W) W (k
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354 - toplina na reakcijata: Hr = 1
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356 dT dV ( H r )( r C A ) smes
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358 Potrebniot broj reaktorski cevk
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360 C TS r a, T kT pT Cv K
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364 Za brzinski izraz preku konverz
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366 0, 2 0 h f ( X ) dX 3 0, 2 0
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368 Sega go pi{uvame bilansot na ak
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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 }
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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 425 and 426: 408 X 1 0.8 0.6 0.4 0.2 0 900 950 1
Page 427 and 428: T 1053 930.72495 1053 930.72495 FAo
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Page 431 and 432: Za vrednosti na Re-brojot pome|u 30
Page 433 and 434: 416 V (m 3 ) P(V) V (m 3 ) X(T) T (
Page 435 and 436: eaktorot: dali toa }e bide izotermn
Page 437: 2) Da se povtori re{enieto pod 1),
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Page 443 and 444: Ramnoteà: Za da se kompletira slik
Page 445 and 446: Explicit equations [1] Feo = 0.0034
Page 447 and 448: POLYMATH Results Calculated values
Page 449 and 450: 432 T (K) Presek na ramnote`na i ad
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Page 453 and 454: 436 - vodorod, H: - benzen, B: - et
Page 455 and 456: Za da se re{i sistemot od 7+1 difer
Page 457 and 458: 440 FE(V) Promena na molskite proto
Page 459 and 460: 442 F F F F F F F E FEo 1) 1 ( 1
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Page 463 and 464: - toplinski kapaciteti na reaktanto
Page 465 and 466: dX (rA ) . (3) dW FAo Ravenkata (3
Page 467 and 468: FAo 20 20 20 20 T 527.23 527.23 812
Page 469 and 470: 2) Novata varijanta na adijabatskat
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Page 473 and 474: Zna~i, so ovaa varijanta razreduvaw
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Page 479 and 480: Explicit equations as entered by th
Page 481 and 482: Toi 650.66442 650.66442 650.66442 6
Page 483 and 484: Sedmiot segment: POLYMATH Results C
Page 485 and 486: Explicit equations as entered by th
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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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