HEMISKI Zbirka re{eni zada~i REAKTORI 3

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[esti del. Industriski reaktori Y = 1 10, 52 ( 1 0, 45) ( 1 0, 45) 1, 75 10, 52 150 1, 32 1057, 65 6, 618 0, 0075 3 0, 45 0, 0075 5 2, 782 10 2 1 kg/(m s ) Pa Y 16, 876 . 2 m s kg kat kg kat Bidej}i parcijalnite pritisoci vo brzinskite izrazi se vo atmosferi, vrednosta za Y ja konvertirame vo atm/kgkat: 5 4 Y 16, 876 /( 1, 01310 ) 1, 666 10 atm/kgkat . Ovaa vrednost za Y se zamenuva vo ravenkata (16): dP 4 FT 1, 66610 dW FTo T Po To P atm/kgkat . (17) Re{enieto na sistemot diferencijalni ravenki (11), (13) i (17) e slednoto: POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value W 0 0 2.0E+04 2.0E+04 Fa 500 428.19926 500 428.19926 Fb 250 164.26021 250 164.26021 Fc 0 0 51.864974 51.864974 Fd 0 0 39.871538 39.871538 Fw 0 0 39.871538 39.871538 T 533 508.66839 606.90411 508.66839 P 10 5.7530934 10 5.7530934 k1 3.265E-04 1.139E-04 0.0047697 1.139E-04 k2 4.339E-04 3.592E-04 7.019E-04 3.592E-04 Fto 1750 1750 1750 1750 Ft 1750 1724.0675 1750 1724.0675 Po 10 10 10 10 pa 2.8571429 1.4288712 2.8571429 1.4288712 To 533 533 533 533 Fao 500 500 500 500 pb 1.4285714 0.5481249 1.4285714 0.5481249 r1 0.0011151 1.205E-04 0.0129101 1.205E-04 r2 0.0036143 8.452E-05 0.003818 8.452E-05 ra 0.0047294 2.05E-04 0.0156177 2.05E-04 rc 0.0011151 1.205E-04 0.0129101 1.205E-04 rb 0.0114005 3.138E-04 0.0152947 3.138E-04 rd 0.0072287 1.69E-04 0.0076361 1.69E-04 rw 0.0072287 1.69E-04 0.0076361 1.69E-04 533
vo 7648.55 7648.55 7648.55 7648.55 A 4907.191 124.76435 5625.6392 124.76435 Cpa 65.2496 63.263326 71.017577 63.263326 Cpb 32.497331 32.394968 32.808247 32.394968 Cpc 78.814923 76.027431 86.720499 76.027431 Cpd 49.999704 49.788311 50.641786 49.788311 Cpw 35.275315 34.981997 36.166233 34.981997 Cpn 30.42908 30.337593 30.706961 30.337593 B 7.118E+04 7.007E+04 7.444E+04 7.007E+04 v 7648.55 7648.55 1.25E+04 1.25E+04 X 0 0 0.1436015 0.1436015 ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(W) = -ra [2] d(Fb)/d(W) = -rb [3] d(Fc)/d(W) = rc [4] d(Fd)/d(W) = rd [5] d(Fw)/d(W) = rw [6] d(T)/d(W) = (500*0.0922*(505-T)+A)/B [7] d(P)/d(W) = -1.666*(10^(-4))*(Po/P)*(T/To)*(Ft/Fto) Explicit equations as entered by the user [1] k1 = 1.195*(10^6)*exp(-11737/T) [2] k2 = 0.02252*exp(-2105/T) [3] Fto = 1750 [4] Ft = Fa+Fb+Fc+Fd+Fw+1000 [5] Po = 10 [6] pa = Fa*P/Ft [7] To = 533 [8] Fao = 500 [9] pb = Fb*P/Ft [10] r1 = k1*pa*(pb^0.5) [11] r2 = k2*pa*(pb^3) [12] ra = r1+r2 [13] rc = r1 [14] rb = (r1/2)+(3*r2) [15] rd = 2*r2 [16] rw = rd [17] vo = Fto*0.082*To/Po [18] A = (106000*r1)+(1325000*r2) [19] Cpa = 11.839+0.11967*T-0.000036515*(T^2) [20] Cpb = 30.255+0.004207*T [21] Cpc = -3.201+0.19507*T-0.000077287*(T^2) [22] Cpd = 45.369+0.008688*T [23] Cpw = 28.85+0.012055*T [24] Cpn = 28.425+0.00376*T [25] B = Fa*Cpa+Fb*Cpb+Fc*Cpc+Fd*Cpd+Fw*Cpw+1000*Cpn [26] v = vo*(T/To)*(Po/P)*(Ft/Fto) [27] X = (Fao-Fa)/Fao 534
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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
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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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Page 505 and 506: Zada~a 7 488 Kataliti~ka oksidacija
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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
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 and 534: 516 k k r k k 1 1 1 2 2 2 dF C dV d
Page 535 and 536: So solverot za diferencijalni raven
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: Ravenkite {to treba da se kombinira
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
Page 585 and 586: Zada~a 11 568 Proizvodstvo na akril
Page 587 and 588: na izreagiran propilen (AK/P 0,82-
Page 589 and 590: se dobivaat preku stehiometriskite
Page 591 and 592: Granicite za re{avawe na ravenkite
Page 593 and 594: POLYMATH Results Calculated values
Page 595 and 596: 578 T(V) Grafik 2. Promena na tempe
Page 597 and 598: turata e T 520 K. Vo ovaa presmetk
Page 599 and 600: 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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