HEMISKI Zbirka re{eni zada~i REAKTORI 3

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[esti del. Industriski reaktori (toplinata {to se nosi so strueweto na reakcionata smesa, toplinata {to se osloboduva so slu~uvaweto na reakciite i toplinata {to se razmenuva preku yidovite na cevkite so mediumot za ladewe). Toplinskiot bilans vo diferencijalna forma e slednata ravenkata: dT dV Ua V ( T a T ) i1 R2 j1 N F C i ( H P, i r, j )( r A, j ) . (6) Vo ravenkata (6) se zamenuvaat numeri~kite vrednosti za koeficientot na prenos na toplina i za temperaturata na vodata za ladewe, potoa temperaturnite zavisnosti za toplinskite kapaciteti (kako {to se zadadeni), dodeka za specifi~nata povr{ina na toplinska razmena, za sumata na toplinite na reakciite i za sumata na proizvodite FiCP,i se zamenuva kako {to sleduva: 2 j1 6 i1 a V ( H i r, j n n )( r cevki cevki A, j D ( D cevka 2 cevka ) ( H ( H L / 4) L D r, 1 r, 1 )( r )( r ) 4 cevka ) ( H 212000 r 441667 r ; 1 A, 1 B, 1 1 A, 1 B, 1 4 0, 039 )( r ( H 2 r, 2 r, 2 2 106000 r ( 1325000 / 3) r 2 102, 564 m A, 2 ) )( r B, 2 ) F C F C F C F C F C F C F A, 2 B, 2 P, i A P, A B P, B C P, C D P, D W P, W M P, M So site zameni ravenkata (6) }e izgleda vaka: dT dV 633102, 546( 405 T ) ( 212000 r1 441667 r2 ) . (7) 6 i1 F i P, i C 1 ; C . 517
So solverot za diferencijalni ravenki od POLYMATH se re{ava sistemot diferencijalni ravenki (2) zaedno so diferencijalnata ravenka na toplinskiot bilans (7). Kon ovie ravenki se dodavaat site drugi potrebni ravenki i izrazi. Kako krajna granica za re{avawe na ravenkite se zadava volumenot na katalizatorskiot sloj, Vfinal = V kat.sloj = 42,5 m 3 . Izve{tajot od koristeweto na softverot (programa i rezultati) i grafi~kiot prikaz na site Fi(V), T(V), r1(V) i r2(V), se dadeni podolu. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value V 0 0 42.5 42.5 Fa 5835 4301.3729 5835 4301.3729 Fb 1435 0.0561717 1435 0.0561717 Fc 0 0 1266.375 1266.375 Fd 517 517 1051.5042 1051.5042 Fw 0 0 534.50422 534.50422 T 405 405 862.11706 418.52804 k1 0.0048133 0.0048133 2.241E+04 0.0122718 k2 3.742E+07 3.742E+07 5.881E+08 4.426E+07 Fto 1.27E+04 1.27E+04 1.27E+04 1.27E+04 Ft 1.27E+04 1.207E+04 1.27E+04 1.207E+04 Po 16.65 16.65 16.65 16.65 To 405 405 405 405 vo 2.533E+04 2.533E+04 2.533E+04 2.533E+04 Fao 5835 5835 5835 5835 v 2.533E+04 2.487E+04 5.237E+04 2.487E+04 Ca 0.2303288 0.087784 0.2303288 0.1729236 Cb 0.0566447 1.838E-06 0.0566447 2.258E-06 r2 1566.5472 8.814E-11 1566.5472 8.814E-11 r1 2.639E-04 3.189E-06 161.93359 3.189E-06 rc 5.277E-04 6.378E-06 323.86718 6.378E-06 ra 522.18292 6.378E-06 522.18292 6.378E-06 rb 1566.5474 3.189E-06 1566.5474 3.189E-06 rd 1044.3648 5.876E-11 1044.3648 5.876E-11 rw 1044.3648 5.876E-11 1044.3648 5.876E-11 Cpm 41.82898 41.82898 63.825271 42.479895 Cpa 54.315977 54.315977 87.870868 55.528075 R 8.314 8.314 8.314 8.314 Cpc 63.12535 63.12535 107.53105 64.903232 Cpb 31.958835 31.958835 33.88207 32.015747 Cpd 48.88764 48.88764 52.85937 49.005172 Cpw 33.732275 33.732275 39.243234 33.895356 B 5.936E+05 5.936E+05 9.031E+05 5.994E+05 A 6.919E+08 0.6760932 6.919E+08 0.6760932 X 0 0 0.2628324 0.2628324 518
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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
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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
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
Page 511 and 512: K2 13.126279 5.2903007 13.126279 5.
Page 513 and 514: Reakcioniot plan pretstavuva grafik
Page 515 and 516: 498 Podatoci potrebni za re{avawe n
Page 517 and 518: nosno ravenka za dizajn (preku konv
Page 519 and 520: 0, 01094 kmol ( rA ) k 0, 425 P ; P
Page 521 and 522: - proizvod od koeficientot na preno
Page 523 and 524: 506 - Drugite podatoci se: sloj 5
Page 525 and 526: 508 X(W) W (kg) Zavisnost konverzij
Page 527 and 528: temperatura dava podobar efekt i vo
Page 529 and 530: Bidej}i reakcijata e egzotermna so
Page 531 and 532: 514 - Kinetika r ( r k r k 1 1 2 2
Page 533: 516 k k r k k 1 1 1 2 2 2 dF C dV d
Page 537 and 538: na koncentracijata na kislorodot. T
Page 539 and 540: 522 Varijanta 2. Proces so kislorod
Page 541 and 542: Ravenka za dizajn na PBR: Ravenkata
Page 543 and 544: Sistemot diferencijalni ravenki (11
Page 545 and 546: 528 FA(W) FB(W) FC(W) W (kg) FD(W);
Page 547 and 548: POLYMATH Results Calculated values
Page 549 and 550: Ravenkite {to treba da se kombinira
Page 551 and 552: vo 7648.55 7648.55 7648.55 7648.55
Page 553 and 554: 536 r2(W) r1(W) Grafik na zavisnost
Page 555 and 556: Y = 1 7, 077 ( 1 0, 45) ( 1 0, 4
Page 557 and 558: 540 r2(W) r1(W) W (kg) Grafik na za
Page 559 and 560: 1 ( rA ) 1 r1 k1 p A pB kmol/(kg
Page 561 and 562: 544 Za specifi~nata povr{ina se zam
Page 563 and 564: na toplina i za protokot na vodata
Page 565 and 566: Cpa 44.257982 44.257982 65.187219 5
Page 567 and 568: Na prviot grafik ne e pretstavena z
Page 569 and 570: 552 Da se presmetaat i analiziraat
Page 571 and 572: FMo 9, 07 3 CMo 0, 169 kmol/m .
Page 573 and 574: povtorno treba da se re{at ravenkat
Page 575 and 576: da se odviva taka vleznata temperat
Page 577 and 578: 560 ~ , i iC P 35 12, 518 0, 05
Page 579 and 580: Vremeto na nestacionarniot period n
Page 581 and 582: 564 C C C Ao Bo Mo F F Ao o Bo
Page 583 and 584: 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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