This preview shows pages 1–11. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: An Calculator
NF ORMA T I ON SHEET PROVIDE D A18436 . Graph paper available Calculators may be used in this examination but must not be used to store text.
Calculators with the ability to store text should have their memories deleted
prior to the start of the examination. THE UNIVERSITY OF BIRMINGHAM
Level C
School of Engineering Department of Chemical Engineering. First Examination. CHEMICAL ENGINEERING 04 17043
(ChE 1CBP) Chemical and Biochemical Processes Time Allowed: Two Hours May/June, 2005 Answer ALL questions in Section A. To record your answers to section A, use the
separate sheet provided and submit it at the end of the examination.
BE CAREFUL NOT TO FOLD OR CREASE THE ANSWER SHEET. Answer T W0 questions chosen from Section B and record each of your answers in
separate answer books. The ﬁgures in brackets at the end of each question or part of a question are an
indication of the marks available. Sections A and B each count 50% of the total marks for this paper PLEASE TURN OVER... Page 1 of 9 A18436 An Calculator
NF ORMA T I 0N SHEET PROVIDE D SECTION A: Answer ALL questions on the grid provided. Each
question carries the same mark. The total of the marks for this
section is 50. A reaction 2A —> B is to be scaled up for industrial production. Lab tests
have shown that the conversion obtained in a batch reactor is 80 % aﬁer
2 hours. Assuming the same behaviour at full scale, how many moles of
A need to be supplied to the reactor to produce 88 moles B in 2 hours? a 55 moles b 88 moles c 110 moles d 176 moles e 220 moles At 25°C a rate constant has a value of 2 x 10'3 s". The activation energy
is 50 kJ mol". What is the value of the rate constant for the same reaction at 85 °C? b 0.059s‘l c 5.9x 10‘3s" a2x10‘3s" d 4><10_3 s" e 0.059 m3mol" s" For a second order reaction 2A —> Products, the rate constant is to be
deduced from a set of reaction kinetics data. A table of values of
conversion versus time is available and all initial conditions in the
reactor, such as compositions, are known. What variables should be
plotted to give a graph with a slope equal to the rate constant, k? a 1n(1X)vst b lanst c In M—X vs(M1)CA0t
M(1X)
Note M: Cgo/CAO
d ( M—X ) vst X
1" e vsC t
M(1—X) 1_X A0 Note [kl= Cgo/CAO A CSTR reactor of volume 2 m3 is fed with A at a rate of 3 x 10'3 m3s". If the conversion at the reactor outlet is 60 %, calculate the ﬁrst order rate constant.
a 2.25 x 10'3 s'1 b 1.25 x 10'2s'1 c 1.25 x 10%" d 5 ><10'3s‘l e 3 x10'3s'] PLEASE TURN OVER Page 2 of 9 An Calculator
FORMATION SHEET PROVIDE D _~_—_—_,___T_____
5. A plug ﬂow reactor has a volume of 4 m and operates with a volumetric ﬂow rate of 0.5 m3 min'l. The rate constant for the reaction A ——> B,
which is carried out in the PFR is 0.05 min". Calculate the conversion,
assuming a liquid phase reaction without volume change. a 0.2 b 0.330 c 0.419 d 0.5 e 0.735 6. A 3 batch reactor has a volume of 1 In and can be ﬁlled at a rate of
0.1 m3 min'1 and emptied at a rate of 0.05 m3 min'l. The rate constant is
0.02 min'1 and the conversion of the only reactant is 70%. 1 hour is
allowed for cleaning. What is the batch cycle time? a 1.33 hours b 147 min c 2.5 hours d 134.6 min e 45 min Three CSTRs are to be operated in series. The reactors have a
volumetric throughput of 0.5 m3 s'1 and a rate constant of 0.1 s'1 in each
reactor. The overall conversion after the third tank is 85 %. Assuming
the tanks to be of equal volume, calculate the volume of one tank. a 16.7 m3 b 25 m3 c 4.41m3 d 0.176 m3 e 12 m3 A gas phase reaction A + B a P is to be operated in a PFR without
inerts. In the feed 1 mol A is added for every 1 mol B. What is the value
of the expansion coefﬁcient? Assume all of A reacts. a 0.5 b 0.25 c 1 d 4 e 2 A yeast, Pichia pastoris, is widely forecast to be a useful future
expression system for the production of recombinant protein products. In
earlier experiments the value for the maximum speciﬁc growth rate umax
for this culture has been found to be 0.5 h'1 and the saturation constant Ks
value for glucose was 2 kg m'3. It is proposed to use a 1 m3 bioreactor
with a continuous feed containing glucose as the limiting substrate at a
concentration of 40 kg m‘3 at a flow rate of 0.3 m3 h". The glucose
concentration in the outlet would be: a 1kgm'3 b 2kgm'3 c 3kgm'3 d 4kgm'3 e 5kg1n'3 ___——______—_1___
I 10. A18436 Page 3 of 9 f the Yield Constant Yx/S in the last question is 0.55 kg cells (kg g1ucose)’ ,
the cell concentration in the reactor outﬂow would be: a 35 kgm'3 b 16.25 kgm'3 c 18.5 kgm‘3 d 20.35 kgm'3 e 12.15 kgm'3 PLEASE TURN OVER An Calculator
NF ORMA T I 0N SHEET PR 0 VIDE I Figure 1 below is the pressuretemperature diagram for a pure substance. Refer to
Figure 1 to answer questions 11 and 12: Figure l.
11.
a.
b.
c.
d.
e.
12.
a.
b.
c.
d.
e.
13.
a.
b.
c.
d.
e.
A18436 Page 4 of 9 Line AB Pressure Temperature
Regions A, B and C represent the following:
A = solid, B = liquid, C = vapour.
A = solid, B = vapour, C = liquid.
A = vapour, B = liquid, C = solid.
A = solidvapour mixture, B = liquidvapour mixture, C = solidliquid mixture. A = boiling region, B = melting region, C = sublimation region. Lines AB, BC and AC represent the following: AB = boiling points line, BC = melting points line, AC = sublimation line.
AB = melting points line, BC = sublimation line, AC = boiling point line.
AB = triple points line, BC = boiling points line, AC = critical line. AB = melting points line, BC = boiling points line, AC = sublimation line. AB = evaporation line, BC = critical points line, AC = triple points line. In the design of a fractionation column, if the slope of the top operating
line (TOL) is at maximum (i.e. coincides with the 45° line), the column is
then operating under Minimum reﬂux conditions. Total reﬂux conditions.
Very low pressure.
Very high vapour and liquid ﬂow rates. The column is shut down. PLEASE TURN OVER An Calculator
NF ORMA T 1 ON SHEET PRO VIDE D 14. On each tray of the distillation column, the statements below describe
what takes place. Which of these statements is FALSE? :1. The vapour stream rises and mixes thoroughly with the falling liquid
stream. b. Some of the more volatile component evaporates to join the rising
vapour, and some of the less volatile component condenses to join the
falling liquid. c. The vapour stream becomes richer in the less volatile component, and the
liquid stream becomes richer in the more volatile component. d. Concentration of the more volatile component reaches maximum at the
top tray. e. Concentration of the more volatile component is minimum at the bottom tray.
15. Below is the feedline equation, whose slope describes the condition of
the feed:
q xF
—  — x + —
yq 1 q q 1q where yq and xq are the mole fractions of the more volatile component
(MVC) in the vapour and liquid at the feed plate, respectively, and xp is
the mole fraction of the MVC at the feed. In which way does the value of the slope of the q line describe the feed condition?
a. Liquid at b.p = negative, partially vapour = O, totally vapour = 00.
b. Liquid at b.p = positive, partially vapour = 00, totally vapour = 0.
c. Liquid at b.p = 0, partially vapour = 1, totally vapour = 00.
(1. Liquid at b.p = 1, partially vapour = 0, totally vapour = positive. e. Liquid at b.p = 00, partially vapour = negative, totally vapour = 0. A18436 PLEASE TURN OVER Page 5 of 9 An Calculator
NF ORMA T ION SHEET PROVIDE D SECTION B: Answer TWO questions. All questions carry equal
marks. 1. The reaction A + B —> C + D has been investigated experimentally in a
batch reactor, to determine the reaction kinetics. Values of the
concentration of A, CA, at various times, t, are available below for a temperature of 25 °C. I. 3000 4000
00213 By making the appropriate graphical plot, show that the reaction is
second order and determine the rate constant, k. Note that C130 = 0.1449 mol dm'3.
[25 marks] 2. In this question, constant density conditions can be assumed and the
reactions are ﬁrst order: the moles reacted per unit time in unit volume,
r = kC, where k is the rate constant and C is the concentration of
substance which reacts. (a) A plug ﬂow reactor of volume 1 m3 is being used for the ﬁrst order
reaction of substance A. The feed rate of A to the reactor is 0.2 kmol s'1
and the concentration of A in the feed is 3 kmol m'3. The conversion of
A is 70 %. Calculate a value for the rate constant, k. [8 marks] (b) For the reaction in part (a), the frequency factor in the Arrhenius
equation is 1.4 x 107 s'1 and the energy of activation is 50,000 J mol".
At what temperature must the reactor be operated to achieve the rate
constant calculated in part (a)? [7 marks] (c) A CSTR of volume 1 m3 is to be added after the PFR, to which it is
connected in series. What is the reactant concentration after the CSTR, assuming operation of the CSTR at 50 °C?
[10 marks] A18436 PLEASE TURN OVER Page 6 of 9 An Calculator
NF ORMA T I 0N SHEET PR 0 VIDE I 3. (a) Bacteria, mammalian cells and whole plants each have characteristics
that make them suitable for bioproduct manufacture. List the principal
advantages of each and indicate in each case some typical products in current or projected manufacture.
[7 marks] (b) There is more to a bioreactor than just a ‘stirred tank chemical reactor’.
In what ways (e. g. process inputs and outputs, control loops, etc) is a bioreactor different?
[8 marks] (c) The simplest model for the dependency of microbial speciﬁc growth rate
u on the concentration of a limiting substrate S is given by Monod as: What is the signiﬁcance of umax and Ks, and how may these parameters
be determined using a continuous completelymixed stirred tank bioreactor? Derive any relationships required in answering the question.
[10 marks] 4. The feed to a fractionating column operating at atmospheric pressure
contains 55 mole per cent of hexane and 45 mole per cent of toluene
(methylbenzene). It is 65 mole per cent vapour. The top product
(distillate) contains 90 mole per cent of hexane. (a) If the minimum number of theoretical stages is three, show on the
vapourliquid equilibrium diagram on either page 8 or 9 how the composition of the bottom product is found. What is that composition?
[4 marks] (b) Determine the minimum reﬂux ratio, Rm, using the vapourliquid equilibrium diagram on either page 8 or 9.
[4 marks] (c) The column is to be operated with a reﬂux ratio of 1.5 to obtain a top
product containing 90 mole per cent of hexane and a bottom product
containing 10 mole per cent of hexane (this is not the answer to (a) l).
The top product is produced at a rate of 45 kmol h'l. (i) Determine the feed rate and rate of production of the bottom
product.
[4 marks]
(ii) Determine the number of theoretical stages required to obtain
the desired products.
[10 marks]
(iii) Which is the feed stage?
[3 marks] VAPOUR—LIQUID EQUILIBRIUM DATA FOR HEXANETOLUENE ARE ON
PAGES 8 AND 9 (THE SAME ON EACH). It these diagrams are used gut your registration number on them and attach them to your answer book.
A18436 PLEASE TURN OVER Page 7 of 9 An Calculator
FORMA T I 0N SHEET PROVIDE I If this diagram is used for the calculations, But your registration number in the sgace
below and attach it to {our answer book. Candidate' Number VAPOUR  LIQUID EQUIUBRIUM DATA FOR HEXANE  TOLUENE
AT ATMOSPHERIC PRESSURE 1.013 BAR 1 .0 _ l—ll'l III'I I— l m ,4.
ﬁnnﬂngﬁaﬁw "" =:..:= .. .a:::=§§qmnm"m::ﬁ=.2F_?:' Mﬁ—gﬁﬁéﬁﬁﬁiiﬁﬁi'miﬁﬁﬁﬁﬁﬁmﬁﬁﬁéw 0 9 ' Wﬂﬁgﬁwﬁaﬂh“ﬁﬁ=nﬁﬁfg _ '5" .. mﬁaﬁé—gﬁgﬁﬁwmnﬁﬁqmy 4:"... 9&ﬂw. Eiwaﬁmi‘zﬁﬂ" 08 mﬁﬁw ymgwgﬁmwﬁﬁ
ﬁwﬁ%=:::a—'m wﬁﬁmmﬁ :.'§£§:E$E=ﬂ “mm =. II . IIIIIl mmme$a.ﬂ%==mﬁﬁiamﬁ=ﬂa=
mem .='._:'" :ma .0
\l 9
.p. Mole Fraction of Hexane in the Vapour Phase
.0
m .0
m Mole Fraction of Hexane in the Liquid Phase (The smallest scale division is a mole fraction of 0.01) A18436 PLEASE TURN OVER Page 8 of 9 An Calculator
F ORMA T ION SHEET PROVIDE D If this diagram is used for the calculations, gut [our registration number in the sgace
below and attach it to your answer book. Candidate' Number VAPOUR — LIQUID EQUIUBRIUM DATA FOR HEXANE  TOLUENE
AT ATMOSPHERIC PRESSURE {1.013 BAR) III I. ml! Inm .1.
.... ::: ::a:::::.:..:..:a:=. ..:.=:%5::":..=: ﬁrmﬂ
I... nﬂ II II I: .
.........._=_§“§=.........:..:. =::§E..::ﬁ_igﬁﬁ:ﬁi_=_ﬁ§==ﬁirnﬂ§ﬁﬁ§v .3.
03 gammaaaﬁﬁgm” hag? ....:.'.!'.:*'.:::§§gang?":g
mnlnllmmmnuumlnmnlunlullmill ‘I Inwln M mﬁﬁﬁﬁm" =='=''==w=.a§== “:4: m ' mEam.w = m II II... .‘lllI—I Ill :9 I... nnm:. .' J. = V
IllllI—Illlll—llIlllllll II? . ﬂ III III’I _
umunmlnnmmﬂglnwmlm lm=Imm=§nm=l$ELmaﬁ WME—IIIII“ Ell"IIIIIIlIlﬂl'laa— =I mam
IIIIII...” Iulnllnﬂllum==gﬁﬁﬁﬁi:a 0.7 .0
m .0
.p. Mole Fraction of Hexane in the Vapour Phase
.0
m .0
N Mole Fraction of Hexane in the Liquid Phase (The smallest scale division is a mole fraction of 0.01) A18436 END Page 9 of9
PTO. A18436
Page 1 of 2 CHEMICAL ENGINEERING 04 17043 (1 CBP)
INFORMATION SHEET FOR REACTION ENGINEERING (Dr J Wood) NOTATION
. . . Glossary of Terms and Deﬁnitions
L Rm.“ W W) or (m . _
_
_
_ (—rA) indicates that reactant A being usedup
(r3) indicates that B is the product 2. The rate of reaction is expressed as rateof change of me] A with respect to unit
volume or some other appropriate property of the system. a) Reaction volume (V) . 8.314 J mol‘ K' (_ _ i «INA moi m—s (1
V dt
In a constant volume system, NA = CA V dNA = d(CA V)
Fractional convetsron of species 1 l d(CA V) d(CA ) 3 1
. , _ =___.___~__._mol ‘t
x' “3 X" (es Species A) — ( r") V dt d m NB. For liquid phase reactors V at V9, the reactor .volume but
for gases V = Visince all of the reactor is occupied by reacting phase. we mm A
_
V, Volume of reaction system
(eg. P V’ = N.R'I') Total mols in a system N mols of species i M 01
‘ (eg. species A) b) 2Phase Reactor
The reaction occurs at the interphase between two phases  ldNA 2 1
_r __ molm t
( A) S dt
S = interfacial area c) Heterogeneous Catalytic Reaction _ There are two possible ways of expressing rate (_ Q): _ 1 “NA mol m'2 t“
.._____ S dt
, . ii . as a function of catal st mass
_ ) y (w)
1 dN l l
. . . . . . (—r )=—__—A mol(gcat) t — A w at
x mole extent of reaction
(eg. CMXA = Amotmt of product formed) Oi Page 2 of 2 3. Order of Reaction 4. 5. The rate of reaction is a function of reactant (and sometimes product) concentration.
(—rA)=kf(CA,CB,Cc.....)
= kc; ,C3 ,C2 where a, b, c are the respective orders of reaction with respect to CA , CB , Cc . . . . andn = a+b+c+. . . .. = Overallreactionorder Thus if,
(—rA) = k CA reaction is FIRST ORDER
(4,.) = k C1 reaction is SECOND ORDER
(—rA) = k CA CB reaction is SECOND ORDER (ie. First order in cA and First order in CB) NOTE: a) Reaction order must be detemiined experimentally but most simple reactions ' are ZERO, FIRST or SECOND order.
b) Complex reactions can have. simple kinetics (eg chain reactions or
heterogeneous reactions) and cases of their kinetics can lead to 0, V2,
1%, 2 order reactions but in the case of very complex kinetics, for example:
(_r )= k cA cB
A (1+irA cA + chB)2
an overall reaction order cannot be stated The above equation is an example of
heterogeneous kinetics. Arrhenius equation: [Sr] k = (frequency factor) e Variable Volume Batch Reactors
These are, in a sense, similar to plug ﬂow reactors because the volume changes to maintain constant pressure. Thus for a nth order reaction,
, N n
(" 1'A) = k where, NA = number of moles of A
V = Volume of the system
NA = NAo (1 — XA)
XA = fractional conversion of A
__ NA0 — NA Initial mols — Final mols X __ —— =
" NM Initial mols and V= V0 (1 + sXA)
A18436 6. The volume of a system is a linear function of the number of moles and hence the
conversion. Thus, 6 _ Final mols — initial mols _
Initial mols dXA _ k 05;“)(1—XA)“ . dt (1 + .9 X A )" “
When a = 0 (liquid phase, and gas phase where there is no mole change) (if: = k Caro—l) (1 _ XA)D
FIRST ORDER (n= 1)
dX .
A = k(1 — XA)
dXA
=kdt =>=> —ln(1—XA)=kt
(1 — XA)
SECOND ORDER (n= 2)
dX
th = k CA0(1_ XA)2
IfCAo =Cno:
dXA X
———=kC dt =>=> 4=kcot
(l—XA)2 AD (l—XA) A
If CM ¢ Cg. (ie. when Cr;o = M CM)
1,, _M_'_X_A__ =k(M1)th
MO—XQ
Design Equations
X
A dX
a) Batch t =CA0 I A
0 (— rA)
XA
b) Pl g Flow (PFR) V' dXA
U =
FAo o (— I.A)
) CSTR V' X“
c — =
FAo ( 1'A) \l ...
View
Full
Document
This note was uploaded on 06/06/2011 for the course CHEM 3040 taught by Professor Reddy during the Spring '10 term at Taylor's.
 Spring '10
 reddy

Click to edit the document details