HEMISKI Zbirka re{eni zada~i REAKTORI 3

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X X 1 0.8 0.6 0.4 0.2 Vtor del. [arèn reaktor 0 560 580 600 620 640 660 T 1 0.8 0.6 0.4 0.2 Q = 0 Operacioni linii X(T) Q * T=const. 0 0 400 800 1200 1600 2000 t Zavisnost X(t) g) Sporedba na rezultatite Za sporedba na vremeto potrebno za postignuvawe 70% konverzija pri razli~nite toplinski protoci, vklu~uvaj}i gi i adijabatskata i izotermnata rabota na reaktorot, }e go analizirame grafikot X(t). * Q T (K) t (s) Q = 0 61
Jasno e deka vremeto potrebno za reakcija raste vo nasokata na adijabatskata rabota. Kolku e toplinskiot protok povisok, vremeto na reakcijata se namaluva, duri stanuva i pokratko otkolku za izotermna rabota. Ova se slu~uva za toplinski protoci Q > 25 kcal/s. Eve gi vremiwata potrebni za postignuvawe 70% konverzija (pro~itani od grafikot): 62 Q (kcal/s) t (s / min) 0 (adijabatska rabota) 2267/37,783 1,261 1380/23 8 600/10 12,61 (*) 449/7,48 20,61 330/5,5 25,61 280/4,67 30,61 246/4,01 37,95 210/3,5 Izotermna rabota 301/5 Visinata na toplinskiot protok se nagoduva so pravilen izbor na goleminata na greja~ot. Za goleminata na greja~ot }e odlu~uva promenata na temperaturata ili vremeto na reakcijata. Na slednava slika se pretstaveni operacioni linii samo za izotermna, adijabatska i rabota so razmena na toplina so izbran toplinski protok od Q = 12,61 kcal/s: X 1 0.8 0.6 0.4 0.2 Q = 0 0 560 572 584 596 608 620 T T (K) Sporedba na rezultatite T = const. *
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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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Prv del. Definicii, koncepti, stehi
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Page 35 and 36: Molskite koncentracii vklu~eni vo b
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Page 41 and 42: Ravenkata za dizajn na ceven reakto
Page 43 and 44: Za presmetka na brojot na reaktorit
Page 45 and 46: 28 Po~eten uslov za re{avawe na sis
Page 47 and 48: Toplinski bilansi za adijabatska ra
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Page 51 and 52: Toplinski bilansi za izotermna rabo
Page 53 and 54: Osnovnata ravenka na toplinskiot bi
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Page 70: Vtor del [arèn reaktor
Page 73 and 74: Re{enie: a) Za izotermna rabota na
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Page 77: Explicit equations as entered by th
Page 81 and 82: Zada~a 2 64 [arèn reaktor i CSTR s
Page 83 and 84: Rezultatot e deka so eden {arèn re
Page 85 and 86: Vo ravenkata za dizajn (48) gi zame
Page 87 and 88: a) Brojot na {arìte koj }e go obez
Page 89 and 90: Kone~nata forma na brzinskiot izraz
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Page 93 and 94: 76 Mi % (w/w) 1 3 Cio smesa kmol/m
Page 95 and 96: Re{enie: Samo so primena na ravenka
Page 97 and 98: data e so sè pomali temperaturni r
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Page 101 and 102: ) Za varijantata na razmena na topl
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Page 105 and 106: Vkupnata koli~ina toplina {to }e se
Page 107 and 108: ) Za presmetka na vremeto na reakci
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Page 113 and 114: 96 t r t r C P dP dP C Ao ; 0, 5(
Page 115 and 116: a) Od kineti~kite podatoci da se op
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Page 119 and 120: A B ; ( rA) k1C A k2C B ; 7 1 k
Page 121 and 122: ) Vremeto na reakcija za 80% konver
Page 123 and 124: presmetki za izotermna rabota na re
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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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POLYMATH Results Calculated values
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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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Explicit equations as entered by th
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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
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Zada~a 2 404 Piroliza na etan do et
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406 C A C Ao ( 1 X ) ; ( 1 0, 6X )
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408 X 1 0.8 0.6 0.4 0.2 0 900 950 1
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T 1053 930.72495 1053 930.72495 FAo
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sokot, potoa podatokot za konstante
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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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