HEMISKI Zbirka re{eni zada~i REAKTORI 3

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POLYMATH Results Calculated values of the DEQ variables [esti del. Industriski reaktori Variable initial value minimal value maximal value final value W 0 0 6600 6600 X 0 0 0.273805 0.273805 tetaI 0.2222 0.2222 0.2222 0.2222 To 565.55032 565.55032 565.55032 565.55032 CAo 72.33633 72.33633 72.33633 72.33633 T 565.55032 565.55032 798.53776 798.5377 K 2.7381906 0.7532669 2.7381906 0.7532669 k 5.266E-08 5.266E-08 1.949E-06 1.949E-06 R 2.756E-04 2.756E-04 0.0021888 0.0021888 FAo 20 20 20 20 Xeq 0.6233164 0.4646422 0.6233164 0.4646422 v 0.2764862 0.2764862 0.3903891 0.3903891 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = R/FAo Explicit equations as entered by the user [1] tetaI = 0.2222 [2] To = (625+tetaI*298)/(1+tetaI) [3] CAo = 5*(10^4)/To/(1+tetaI) [4] T = To+(1040/(1+tetaI))*X [5] K = 145.3*exp(-8.395*(T-298)/T) [6] k = 0.0125*exp(-7000/T) [7] R = k*(CAo^2)*((To/T)^2)*(((1-X)^2)-((X^2)/K)) [8] FAo = 20 [9] Xeq = (K^0.5)/(1+(K^0.5)) [10] v = (FAo/50000)*(1+tetaI)*T Komentar za dobieniot rezultat: So dodavawe laden vozduh na vlezot vo reaktorot se izbegnuva prethodno ladewe na reaktantot vo toplinski razmenuva~ – edna operacija i eden ured pomalku! Se pokaùva deka na vlezot vo reaktorot strujata na reaktantot treba da se me{a so struja laden vozduh so protok FI = FAo·0,2222 = 20·0,2222 = 4,444 mol/s. So ova razreduvawe vleznata koncentracija na reaktantot se namaluva (72,336 mol/m 3 nasproti 94,835 mol/m 3 za ~ist reaktant so prethodno ladewe), {to implicira namaluvawe na brzinata na reakcijata! No, od druga strana, vleznata temperatura e povisoka (To = 565,55 K nasproti To = 527,23 K), so {to pak se zgolemuva brzinata na reakcijata! 455
Zna~i, so ovaa varijanta razreduvaweto se kompenzira so izveduvawe na reakcijata vo podra~jeto na povisoka temperatura. Se dobivaaat slednive rezultati (sporedi gi site podatoci so podatocite dobieni so adijabatska rabota so ~ist reaktant): 456 I = 0,2222; FI = FAo·I = 20·0,2222 = 4,444 mol/s; W = 6600 kg; To = 565,55 K; CAo = 72,336 mol/m 3 ; X izlez = 27,4%, T izlez = 798,54 K. Zabele{ka: Ako istiov problem go re{ime so zanemaruvawe na promenata na volumenskiot protok so promenata na temperaturata, odnosno so zemawe deka To/T = 1 (se pojavuva vo izrazite za molskite koncentracii na reaktantot i produktot i konsekventno vo brzinskiot izraz – vidi ja stehiometriskata tablica 2), rezultatite bi bile slednive: I = 0,2627; FI = FAo·I = 20·0,2627 = 5,254 mol/s; W = 6600 kg; To = 556,97 K; CAo = 71,1 mol/m 3 ; X izlez = 27,4%, T izlez = 782,83 K. 3) Vo ovaa varijanta na zada~ata ladeweto na reaktantot }e se izveduva so ramnomerno dodavawe laden vozduh so temperatura od 298 K po celata dolìna na reaktorot. Rabotata na reaktorot i vo ovaa varijanta e adijabatska. Sega zada~ata ja postavuvame vaka: Strujata od ~ist reaktant so temperatura od 625 K i so molski protok FAo = 20 mol/s se dodava samo na vlezot vo reaktorot. Sniùvaweto na temperaturata na reaktantot se ostvaruva so postepeno i ramnomerno dodavawe na ladniot vozduh niz dovodi za vozduh postaveni na ednakvi rastojanija po dolìnata na reaktorot. Se opredeluvame za 10 dovodi na vozduh. Prviot dovod e na vlezot vo reaktorot. So vakva raspredelba na dovodite i pod pretpostavka za klipno te~ewe na reakcionata smesa, reaktorot moème da go razgleduvame kako da e konfigurirana serija od 10 mali cevni reaktori (segmenti), sekoj so dolìna Li = 7/10 = 0,7 m i so masa katalizator Wi = 6600/10 = 660 kg. Iako segmentite ne se fizi~ki odvoeni, bidej}i te~eweto e klipno, sekoj moè da se razgleduva kako mal idealen reaktor so fiksen sloj katalizator. Presmetkata }e zapo~ne so prviot reaktor (segment) so pretpostaven molski protok na ladniot vozduh! ]e
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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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Re{avawe na ravenkata (8) zna~i pre
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Analizata za razli~ni vrednosti na
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Re{enie: 1) Ravenkata za dizajn na
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[8] Po = (1-X)*P/(2-X) [9] Ph = (1-
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1 ( rO ) f ( W ) ili D f ( X )
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380 f 9, 375 ( 1, 75 0, 0585 ?) 9
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Za da zaklu~ime dali ovoj volumen e
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pritisokot i konverzijata. Zavisnos
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k 3.194E-05 3.194E-05 3.194E-05 3.1
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[esti del Industriski reaktori
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392 - vlez vo reaktorot: T C o 650
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Izleznata konverzija normalno raste
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gramata se vnesuvaat ovie dve raven
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Toplinskiot bilans vo ovoj slu~aj }
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400 X 0.5 0.4 0.3 0.2 0.1 1 2 3 4 5
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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 431 and 432: Za vrednosti na Re-brojot pome|u 30
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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
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Page 461 and 462: enzen izvedena pod dadenite uslovi
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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 475 and 476: segment }e treba da se presmetuva p
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Page 479 and 480: Explicit equations as entered by th
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Page 483 and 484: Sedmiot segment: POLYMATH Results C
Page 485 and 486: Explicit equations as entered by th
Page 487 and 488: 470 X 0.4 0.32 0.24 0.16 0.08 0 500
Page 489 and 490: Zada~a 5 472 Proizvodstvo na glukon
Page 491 and 492: 30 24 / 144 5 . Zna~i, na po~etok
Page 493 and 494: T a b e l a 1 Po~etok na ciklus 10
Page 495 and 496: 478 Vmax / K m 1 5 3 k 0, 016 h
Page 497 and 498: te~nata reakciona smesa se odrùva
Page 499 and 500: 482 r r , r , 1 r , 2 r , 3 C
Page 501 and 502: azot za vkupno sozdadena kiselina (
Page 503 and 504: 486 CB(t) CM(t) t (s) t (s) t (s) A
Page 505 and 506: Zada~a 7 488 Kataliti~ka oksidacija
Page 507 and 508: 490 dX dW ( rA ) F1 ( X , T ) (2)
Page 509 and 510: Podolu se izve{taite i grafi~kite p
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Page 513 and 514: Reakcioniot plan pretstavuva grafik
Page 515 and 516: 498 Podatoci potrebni za re{avawe n
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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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