HEMISKI Zbirka re{eni zada~i REAKTORI 3

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1 Tret del. Proto~en reaktor od rezervoarski tip so idealno me{awe (CSTR). Polu{arèn reaktor Grafik 1 Moè samo u{te da se konstatira deka istite re{enija kako dobienite so tabelarniot prikaz se dobivaat pobrzo i istovremeno i na grafi~ki na~in! b) Rabota na reaktorot so razmena na toplina I za ovoj na~in na rabota na reaktorot, za presmetka na stacionarnite to~ki }e bidat potrebni ravenkata za dizajn i toplinskiot bilans, odnosno relaciite X – T od dvete bilansni ravenki. Prvata relacija, XMB–T, e ve}e izvedena. Toa e ravenkata (3), ( 1 X ) X X MB 750 k( T ) . (3) ( 1 X ) Vtorata relacija, konverzija–temperatura, se dobiva od toplinskiot bilans, vo ovoj slu~aj toa e ravenkata (120): 0 2 XMB(T) XEB(T) T (K) N Q Ao i P, i o r A i1 ~ F C ( T T ) ( H )( r ) V . (120) So soodvetna zamena vo sekoj ~len od ravenkata (120) se dobiva slednovo: ~ 0 ( T T ) F ( C , )( T T ) ( H ) F X . (7) UA a Ao A P A o r Ao Ravenkata (7) ja izrazuvame eksplicitno vo odnos na konverzijata, gi zamenuvame poznatite numeri~ki vrednosti i stignuvame do baranata relacija XMB–T, 3 117
118 ~ UA( T Ta ) FAoC P, A( T To ) X X EB ; F ( H ) 0, 05( T 373, 2) 0, 001120( T 325) X X EB ; 0, 001 70000 Ao r X X EB 0, 0024 T 0, 824 . (8) Ako ja sakame relacijata temperatura–konverzija, od ravenkata (8) }e go dobieme slednovo: T 343, 23 416, 6 X . (9) Vo ravenkata (9) T = 343,23 K e prese~nata to~ka so X = 0, T = TC! Ne e vlezna temperatura kako vo slu~ajot so adijabatska rabota! Na krajot simultano se re{avaat ravenkata (3) ili nejzinata eksplicitna forma (6) i ravenkata (8). Koj na~in za re{avawe }e se izbere e seedno. Ako go izbereme na~inot so tabelirawe na rezultatite (postapkata e kako prikaànata za adijabatska rabota na reaktorot), povtorno }e se dobijat tri stacionarni to~ki: X X X 1 2 3 0; 0, 55; 1, 0; T T T 1 2 3 343, 23 K 575K 760K Istiov rezultat moè da se dobie pobrgu, to~no i istovremeno i vo grafi~ka forma, so primena na solver za diferencijalni ravenki. Ako se primeni softverskiot paket POLY- MATH, vo solverot za diferencijalni ravenki se pi{uva diferencijalna ravenka so koja se zadava ~ekorot na promena na temperaturata, potoa se vnesuvaat ravenkata (3) ili (6) i ravenkata (8) zaedno so temperaturnata zavisnost na brzinskata konstanta. Programata, izve{tajot i grafi~kiot prikaz (grafik 2) se slednite:
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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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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
Page 91 and 92: Explicit equations as entered by th
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Page 95 and 96: Re{enie: Samo so primena na ravenka
Page 97 and 98: data e so sè pomali temperaturni r
Page 99 and 100: POLYMATH Results Calculated values
Page 101 and 102: ) Za varijantata na razmena na topl
Page 103 and 104: ( H r )( rA ) V TW , vlez T
Page 105 and 106: Vkupnata koli~ina toplina {to }e se
Page 107 and 108: ) Za presmetka na vremeto na reakci
Page 109 and 110: ( 1 X ) To C A C Ao ; (1) ( 1 0, 5
Page 111 and 112: 11620 X T To . (6) 38, 73 12, 91X
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
Page 126 and 127: Zada~a 1 Hidrogenacija na etilen vo
Page 128 and 129: Tret del. Proto~en reaktor od rezer
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Page 160 and 161: F F X X A1 B1 1 1 Tret del. Proto~e
Page 162 and 163: C B2 0, 1056 F F X 2 A2 B2 2 Tre
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Page 170 and 171: Explicit equations [1] R = 0 [2] T
Page 174 and 175: X * XMB XEB X*; XMB; XEB 1 0.8 0.6
Page 178 and 179: XMB XEB 1 0.8 0.6 0.4 0.2 Zada~a 10
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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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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
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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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