solution manual chemical reaction engineering 3rd Octave Levenspiel solution manual chemical reaction engineering 3rd Octave Levenspiel solution manual chemical reaction engineering 3rd Octave Levenspiel Bài giải bài tập kỹ thuật phản ứng
Trang 1SOLUTUONS MANUAL, accompany)
Trang 2SOLUTIONS MANUAL
to accompany
CHEMICAL REACTION ENGINEERING
THIRD EDITION
| Includes Solutions to All 228 Odd-Numbered Problems |
OCTAVE LEVENSPIEL
Chemical Engineering Department Oregon State University
Corvallis, OR 97331-2702 Telephone 541-737-3618 Fax 541-737-4600 E-mail levenspo @ peak.org
9
JOHN WILEY & SONS, INC
Trang 4Copyright © 1999 by John Wiley & Sons, Inc
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Requests for permission or further
information should be addressed to the Permissions
Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012
ISBN 0-471-3 1478-1
Printed in the United States of America
10987654321
Trang 5"` Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter 20 21 22 23 24 25 26 27 29 30 oO nN DH Ww Rh Q2 He * , * CONTENTS
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\ OMNES
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vers oc? # sal be,
Trang 71
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Trang 74— soma epost ễ i V\Oc 6€ _ oe vn Ww Se ie 4X š 4 va sae ` “
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Trang 85Suy” \á- 34 oy ` # he
22 WWE OVE W An € Ss a đà = + & Ỹ
Trang 86
A
Trang 91{74 W=loo € LOND Toke) {~~ | Coo 3 Ì 33 (bypass) q 3 a gi )+ b ý +- a loo 64 200 * 200 67 100 ` 100 @ LAD b —) 1330 «| 137 _ d ; heat “ K=£ 100 ee inert 3 stream IDO men ⁄X^^ — ta om eh — — #8 2á
joo § ' joo OF soos a) 200 | 100 200 200
Trang 92cm Ệ th Lope Sư E Xp ⁄ \ ‘ M 4 4 Ậ 4 ‘ A “tr N ] Tớ + Sứ 6.85 -~-~—~ — — X ⁄ 0 os Plug flow 0-85 w= fe | a =(!00) (a 1a4) = 212-4 keg
Heat added ahead of the reactor
O.= Eh, Cp AT = (100) (40) (700-300) z2-4niP TZ aaa
xa
Heat removed from the reactor
Qa = Ty XA Í-AHr) + FA Cp lav)
Trang 93
CONES 20% A- 80% inert, adiabatic reactor A slope = —e _ #+0x5 _ _ |
~TAHc 90ooo “09 Let us construct = VS Xa graph , From this graph we can See that recycle
flow is best amd
0.5 0 + &= 0.†6-0.51 ” CA =2~ | = anes 8 T wj= E Ao a (io0x0.2) (0.75) (22.8) ~h = 342 kg t Rate 1S Vis of
Heat duty “Ta that Given in
_ the ch Q= For Cp AT Dà,Xa (-Anr) & chart
|
= (100)(40)( 825-300) —(20)(0-75) (P0000) số số si ; ¬ nh `
il Ix18 T/S Cheating) 0 Of 02 3 oO4 OF O06 07 108
Venn ee A 825 K R25kK _ Xa=o.T5 Aock am 525k T30k & Fa,=20 wel | Q=700 kw R=2125 fe : YO Tha
fo ( Sự” f vi 3 O ag I amir ett ees Fs emit ttn psn `
Trang 95(4.14 202 Ậ- §04B, Xu=o-$5
slope = —CP_ _ = #oxs = |
—AHr g0000 400
Plug flow with recycle is best
et rhe opt oF rectangles 2sp Reaction rate is Wo of that
Ỳ SUaa#4E whee to locate in the chart, I
‘ hag ⁄
: 4
Xa JOO +
đ(e & A_ = ÍTT.3 ka'S/mị ` a R= OLD
_W NX
50+ 4 / ` Uy : 2
4 = 4b kgs '
Wi= Fa,"Aceal = @o) (464) / Sẻ L23 “+5 đe `
=42.8ka " _ We = Fao areat=(20) (11.3) Oo L—- Pa >1 ———+— + = 24 kg 0 0.2 0.4 0.6 0-8 | | XA Ro i Ri io Heat duty r* >< - 6.6% 0,85 o 0.32 059 Qi = Fao Cp (AT = (20)(200) (5@-300)
= lL#xto?° 3% Cheating) \ [ Slope “in Qa = FaoCp (ar) = (20)(200) (670-930) p.3St eer xã
=-O84 x10 T/s5 (cool ing ) /
Note that this solution was obtained after a Few trials
Aook 2A 650K 058 -< ' { l \ ‡ ( í ‘ f ! oO i Bed, >> 3oo + 650: 880
đo ¿4o - T8o
Trang 96Injection of cold feed slo e=x 40x20 — t p #0000 100 Ậ 4000 + A @ 0945 - ott ccc TT” 0.-83ƑF TT—~~~~-~~~~—=——— 0.65- - - - moray 3000 r ⁄ ⁄ X ⁄ P ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ a ⁄ A 2000+ 0 ⁄ J i 4 | 300 S00 —Y?/ T (000E- AYêa 2=22!I
Fraction of feed A that
goes to the first reactor
0-65 — areal=84 0.83 - ot 8S D/1/002 CTT kaa Me 0 0.2 0.4 0.6 0.3 LO Xa 329 kg | TA Hos kg ị VI CO SỞ _ Wi = Fao @real = (5x0.7831)( 84) = Wa = fa, area2 =($)(221) =
A= Fay, Cp (AT) = (5x 0.7831) (800) (699 -300)= [22x10 I< (heating) Neke: Only one heaf axchan ger i> Needed wm crs desig he
&)
€9ok Xa=0.83ÿý XA=0.6sf
T=113k €'To k TOOK
Trang 109I2.&6%Lo #9
Trang 111
29:1 Fist letus see if ether extreme well ‘ths Sima 9 Lf i @QrLOUS apely
“` coustenut Hhvougouk /
(0 3250
a= Th —=—
K 200 Even ve to Xez®4 we have the
Ar whet Cg, = 400 4Ô 7 eevee (ke Cg Ca, Lebus vse ©
the ex pre: So cous & hats Qe EUWE loa
guess XgS 40, Cy = AOA fw? J hon check Ib at the ect
ÊvaluaF€ quauhities needed to {24 the rate
a = Os - 695%) pag ait), A» 5 xo
Se R a - (3.86% 17> )( bab) = 0, O2LF ao | th 5 “197? (714007 -5MìoO, a M+ = Lb | a Satter G dt \SO0) = 2 bu là ia se 6 & Atle % tg? ? New ky the cake 24 vabum,
7 \ 100325
on Va on
i | “Gee
Bbooo (tot) 6.02 0.0269 (AMF IO} (1G00) (01D OSS) = , we +50 43701 + 416.4 ⁄ | 94229” = p.022 bo
vốn va de - VWOMA C£@usÊf1e€
Qui eơnelav# Pa throughout ne trrelele bed and will, the rake
inde penderk as <8 the weaFerval balance becomes vevy aimale, as cacy
Fay A = Teo Xe =¢ hy Vr
Ves = Ủy Coo -Ípx TU_C = 0,084\/s
## và sen aR NEESER NETTIE: eas ea ee
cự
+ Xe = $ «) Va (0.0224 (5 €O.") = onl ~ > IS ha 5.08 _——— A~~~
Trang 112`
som €F _ *
“em Soh fobs, iA a © Sot Rcd `" 2
Trang 113
2245 Fist let vs see “if sec extreme ap plies
a cồuA sau” i IO A SOOT nese cope 4 ram A \ot22E Z C, = TALL \025 - 3555 mol / 2 H, 28500 _ /vs
At stewt : Cay = 1097 MIA
At end run — Ca ge = 104.7 “0U,
Coweoav ` ise ‹2@ Heol
Cam Ca
the vghout tho vu,
Ose the equations Fow tle Ce >> Cy echeme
ofs 6 (6255 we Q 2 “If = cS==~ - #8ooo ^* ¢ lg avwœtt mộ ka Ge = lo? (S000) = 0.05 57! 2owcGgig13i0g0m03n0502014 9210/đ8 TT” ¬— (7%
Maz / Kes „ 2x0 0.05 (750) te sa Q
v øt 6 2/25 xi0 t9 (0.6 lọ 0.02944
We ave new ready to wribe the rake 2nOreS SOW
củ \ D225 mm \ + | 28500 0.01 0.05 o.D05 (I52)(n.01+43s,2s
98 + 20 4 11% , (3-5 a s6) =O0OTT wo) Bg? reading S
t in ally qo by the
V _ ae wes
“he Xp = a are \ = (- Oy 3V
4
Ugeles sales = use these teyvms to wheavahe
maternal loalance prt "4 ‘i Ene A
Trang 128ne a oe AO + Ệ on &
tm Wily VŒ%S$ tái đWe@
Trang 129
7) pm | VN Test the 2 possible shrinks = —= CĐr© mechanisms to See Wheel eS
Consistent with the Followma naparted date
Xp =O.Q1S at t=thy | +
sor Xg = 2 af =e OS
Xa = 1.000 at tér\= 2h +
Since the shape of the solids is not specified assume to start
wil Lat Theu are sphencal , Then From “Table 1
masse
IF ma đ(foseen controls: 4 = Xe
2 l
but from the data: OS £0375 - this doestt a vee 50 Wwe reyect ths
mechanim
Pp 2/
\F ash diffusion controls: = = |- 3(i-X,) 7+ 2(-Xe) 2 |
again From thedata OS = \- 3(b)% 420%) = OS - this
Carers
New before Wwe feel that all is solved let us try the thud
mechansin , Who kywdws maybe “oe Too will tt, So
IF surface veastvbn conhdls = \— (—vv)$
& aaw from the data: OF = |= (Ye \’3 Iv
H
wee Aaa this also
ay “0?
So hove Wwe have Cr comcidonce where loath it
` i
a , : Ƒ ~
reaction & ash Atfoscin mechanisms it the data |
Note ~ Figuce FG oc JO shows that the data Were taken at
the precise powt wheve the asl Affusion Curve
cyosses the reactiin controlling, Cure ,
~\F we try the 2quations for Flat plate or cylindrical