HEMISKI Zbirka re{eni zada~i REAKTORI 3

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Petti del. Kataliti~ki reaktor so fiksen sloj katalizator (PBR) Za re{avawe na ravenkite (3) e potreben i podatokot za vlezniot molski protok na reaktantot A: F o To RT P o o F Ao C o Ao ; 3 82, 05(cm atm)/(molK) 427, 5K 65, 9mol/h ; 1atm 2, 31110 cm /h ; o 6 FAo 2, 31110 5 10 1, 156 mol/h. Ravenkata za dizajn ne }e moè da se re{i sama. Preku brzinskiot izraz taa, pokraj konverzijata, ja vklu~uva i temperaturata. Zatoa e potrebna i ravenkata na toplinskiot bilans. Ovaa bilansna ravenka vo sebe }e gi vklu~i: toplinata na reakcijata, toplinata {to se odzema od reakcionata smesa (razmena na toplina) i toplinata {to se nosi so reakcionata smesa (razlika na entalpiite na vlezot i izlezot od sistemot). Toa e ravenkata (106), dT UaV ( Ta T ) ( H r )( rA ) . (106) dV N F C i1 Ovaa ravenka prisposobena za uslovite vo ovaa zada~a }e izgleda vaka: dT ( H r )( rA ) FT CP UaV ( T Ta ) . (4) dV Vo ravenkata (4) se zamenuva brzinskiot izraz (2) i site zadadeni numeri~ki vrednosti. Za sekoj ~len od ravenkata se dobiva: 6 i 7 P, i ( H r )( rA ) 152000185, 712exp( 2958/ T )( 1 X ) 7 0, 578 0 2, 82310 exp( 2958/ T )( 1 X ) T Ua a V Ua V ( T T ) 8, 67 a ( D) L 4 2 ( D / 4) L D V a ( T 4 3 ( T T ) 11, 557( T 427, 5) ; a V 3 427, 5) 1, 333 cm 2 /cm 0, 578 3 , 578 T ; 0, 578 343
344 dT dV dT dT dT FT CP 65, 9 6, 9 454, 71 . dV dV dV Na krajot se kombiniraat site ~lenovi: 62080exp( 2958/ T)( 1 X ) 0, 578 T 0, 578 0, 0254( T 427, 5) . Potrebniot set diferencijalni ravenki e gotov: ravenkite (3) i (5) se re{avaat vo kombinacija so brzinskiot izraz (2) i so vlezen uslov definiran so vleznata struja vo reaktorot. Integraciite se izveduvaat do dolìna na reaktorot L = 20 cm, 2 3 odnosno V 0, 785 3 20 141, 35 cm . Se opredeluvame za solverot za diferencijalni ravenki od POLYMATH. So cel da se ovozmoì grafi~ko pretstavuvawe na zavisnostite X(L) i T(L), vo programata se vnesuva i eksplicitniot izraz, V V L 0, 1415V. 2 2 0, 785 D 0, 785 3 Se dobivaat slednive rezultati: POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 141.35 141.35 X 0 0 0.9936627 0.9936627 T 427.5 427.5 563.74864 491.44326 L 0 0 20.001025 20.001025 FAo 1.156 1.156 1.156 1.156 R 0.0055354 6.74E-04 0.0135492 6.74E-04 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = R/FAo [2] d(T)/d(V) = (62080*exp(-2958/T)*((1-X)^0.578)*(T^(-0.578)))-(0.0254*(T-427.5)) Explicit equations as entered by the user [1] L = 0.1415*V [2] FAo = 1.156 [3] R = 185.712*exp(-2958/T)*((1-X)^0.578)*(T^(-0.578)) (5)
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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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Page 347 and 348: 330 Fio Fi() Fi(X) pi = yiP = (Fi/F
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Page 351 and 352: Ako reakcijata se odviva vo pove}ec
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Page 359: Re{enie: Rabota so razmena na topli
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Page 369 and 370: 352 W (kg) W (kg) (-rA) = f(W) W (k
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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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