Fogler: solucionario de la tercera edición
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About This Presentation
Solucionario de Elementos de ingeniería de las reacciones químicas de Fogler tercera edición
Slide Content
SOLUTIONS MANUAL
Elements of Chemical
Reaction Engineering
THIRD EDITION
by Timothy Hubbard, Jessica Hamman, and
David Johnson
with Kylas Subramanian, Sumate Charoenchaidet, Probjot
Singh, John Santini, H. Scott Fogler, Lisa Ingalls, Abe
Sendijarevic, and Nicholas Abu-Absi
Elements of Chemical
Reaction Engineering
THIRD EDITION
by Timothy Hubbard, Jessica Hamman, and
David Johnson
with Kylas Subramanian, Sumate Charoenchaidet, Probjot
Singh, John Santini, H. Scott Fogler, Lisa Ingalls, Abe
Sendijarevic, and Nicholas Abu-Absi
‘nd Edition, Solution Manual, Chapter 1
Chapter 1
General: The goal of these problems are to reinforce the definitions and provide an
understanding of the mole balances of the different types of reactors. I lays the
foundation for step 1 ofthe algorithm in Chapter 4
PLA. This problem might encourage students to get in the habit of writing down
what they learned from each chapter.
P12, Small open-ended question from which one could choose one or two parts
Parts (a) (b) or (e) are recommended,
P13. This problem use Example 1-3 to calculate a CSTR volume. It is straight
forward and gives the student an idea of things to come in termo of sizing
reactors in chapter 4, An alternative to PI-A1 and PI-L2
PL4. Alternative to PL3, PI-11, and PI-12. See PI-3 above.
Problems P1-5, 1-6, and P1-7 review the definitions given in the chapter.
PLS, This problem can be assigned to just be read and not necessarily to be
worked. It will give students a favor of the top selling Chemicals and top
chemical companies.
This problem will be useful when the table is completed and the students
can refer back to it in later chapters. Answers to this problem can be found
‘on Professor Susan Montgomery s equipment module on the CD-ROM, See
Pua.
Many students like this straight forward problem because they see how CRE
principles can be applied to an everyday example. Iti often assigned as an.
in dass problem and part (gs usually omitted.
Problems FI-11 and PI-12 show a bit of things to come in terms of reactor sizing,
‘Can be rotated from year to year with P1-3 and PI-4. See PI-3 above
PLAB. Asks for details of operation of an industrial reactor.
PIAA. Encourages and requires the student to go outside the text for information
related to CRE. May be a bit early in the text to assign this problem,
PLAS. Encourages and requires using other sources to obtain information.
PLA6. Encourages using other sources to obtain information.
PLZ. 1 strongly recommend this problem be assigned. It can be used in
conjunction with Problem PI. Professor Susan Montgomery has done a
paa
Chapter 1
General: The goal of these problems are to reinforce the definitions and provide an
understanding of the mole balances of the different types of reactors. I lays the
foundation for step 1 ofthe algorithm in Chapter 4
PLA. This problem might encourage students to get in the habit of writing down
what they learned from each chapter.
P12, Small open-ended question from which one could choose one or two parts
Parts (a) (b) or (e) are recommended,
P13. This problem use Example 1-3 to calculate a CSTR volume. It is straight
forward and gives the student an idea of things to come in termo of sizing
reactors in chapter 4, An alternative to PI-A1 and PI-L2
PL4. Alternative to PL3, PI-11, and PI-12. See PI-3 above.
Problems P1-5, 1-6, and P1-7 review the definitions given in the chapter.
PLS, This problem can be assigned to just be read and not necessarily to be
worked. It will give students a favor of the top selling Chemicals and top
chemical companies.
This problem will be useful when the table is completed and the students
can refer back to it in later chapters. Answers to this problem can be found
‘on Professor Susan Montgomery s equipment module on the CD-ROM, See
Pua.
Many students like this straight forward problem because they see how CRE
principles can be applied to an everyday example. Iti often assigned as an.
in dass problem and part (gs usually omitted.
Problems FI-11 and PI-12 show a bit of things to come in terms of reactor sizing,
‘Can be rotated from year to year with P1-3 and PI-4. See PI-3 above
PLAB. Asks for details of operation of an industrial reactor.
PIAA. Encourages and requires the student to go outside the text for information
related to CRE. May be a bit early in the text to assign this problem,
PLAS. Encourages and requires using other sources to obtain information.
PLA6. Encourages using other sources to obtain information.
PLZ. 1 strongly recommend this problem be assigned. It can be used in
conjunction with Problem PI. Professor Susan Montgomery has done a
paa
‘AL Edition, Solution Manual, Chapter 1
great job pulling together the material on real reactors in her equipment
module on the CD-ROM.
1 always assign this problem so that the students will lean how to use
POLYMATH/MatLab before needing it for chemical reaction engineering
problems,
As the WWW becomes more developed, it may be more and more
important to assign this problem.
CDPI-A Similar to problems 3, 4,11, and 12.
CDPIB Points out difference in rate per unit liquid volume and rate per reactor
volume.
Summary
Solution
Assigned Alerts Diffiuliy Time Given
Pra © No
em AA IA, 216) SF ced Yes
“ns AA SATA" FSF Yes
pra AA 3AMIZA ESE Yes
1
PIS Yes
PIS Yes
PLT Yes
ers Read Only Yes
pis Yes
P110 Yes
Le JAMAZA Ye
Pia AMIZA Yes
Pras Yes
PLU No
Plas Partial
Pig Partial
e PL sE No
© mus SF Yes
Piao No
COPLA AMIZA ESP
ps FSF
Assigned
(© = Always assigned, AA = Always assign one from the group of alternates,
O = Often, = infrequently, = Seldom, G = Graduate level
great job pulling together the material on real reactors in her equipment
module on the CD-ROM.
1 always assign this problem so that the students will lean how to use
POLYMATH/MatLab before needing it for chemical reaction engineering
problems,
As the WWW becomes more developed, it may be more and more
important to assign this problem.
CDPI-A Similar to problems 3, 4,11, and 12.
CDPIB Points out difference in rate per unit liquid volume and rate per reactor
volume.
Summary
Solution
Assigned Alerts Diffiuliy Time Given
Pra © No
em AA IA, 216) SF ced Yes
“ns AA SATA" FSF Yes
pra AA 3AMIZA ESE Yes
1
PIS Yes
PIS Yes
PLT Yes
ers Read Only Yes
pis Yes
P110 Yes
Le JAMAZA Ye
Pia AMIZA Yes
Pras Yes
PLU No
Plas Partial
Pig Partial
e PL sE No
© mus SF Yes
Piao No
COPLA AMIZA ESP
ps FSF
Assigned
(© = Always assigned, AA = Always assign one from the group of alternates,
O = Often, = infrequently, = Seldom, G = Graduate level
td Edition, Solution Manual, Chapter 1
alternates
in problems that have a dot in conjunction with AA means that one of the
problem, ether the problem with a dot or any one ofthe alternates are always,
Assigned,
Time
‘Approximate time in minutes it would take a B/B° student to solve the
Problem.
Difficulty
SF = Straight forward reinforcement of principles (plug and chug)
FSE = Fairly straight forward (requires some manipulation of equations or an
intermediate calculation).
IC = Intermediate calculation required
M = More difficult
OE = Some parts open-ended.
‘Note the letter problems are found on the CD-ROM. For example A = CDPI-A.
Summary Table Chl
Review of Definitions and 156789
“Assumptions
Introduction to the CDROM 17184
Make à calculation 10311213
Open-ended 14,1516
Straight forward 20313
Fairy straight forward. 311128
More difficult 10
alternates
in problems that have a dot in conjunction with AA means that one of the
problem, ether the problem with a dot or any one ofthe alternates are always,
Assigned,
Time
‘Approximate time in minutes it would take a B/B° student to solve the
Problem.
Difficulty
SF = Straight forward reinforcement of principles (plug and chug)
FSE = Fairly straight forward (requires some manipulation of equations or an
intermediate calculation).
IC = Intermediate calculation required
M = More difficult
OE = Some parts open-ended.
‘Note the letter problems are found on the CD-ROM. For example A = CDPI-A.
Summary Table Chl
Review of Definitions and 156789
“Assumptions
Introduction to the CDROM 17184
Make à calculation 10311213
Open-ended 14,1516
Straight forward 20313
Fairy straight forward. 311128
More difficult 10
BLL No solucion wil be gives,
e12
(6) Resctantsmigh note bot enough to reset.
(©) Pot Cost vs, Volume on ogg paper. Use thi graph 1 generat an equation for
Cae ación ef ol
ln (Cost) vs. ln (Volume)
y= 02501% + 94932
4 à
In (Volume)
From this we generate the equation: Costa 13,270(¥)°
‘We ean use this equation to find the desired prices:
For a 6000 gallon reactor: Costa 13,270(6000)" = $165,400
Fora 15,000 gallon reactor: Cost 13,270(15,000) = $25,740
Odom, Co 599 4m?
023 min "Doc,
For Constant Pressure
LAN, „ 1d(G¥) _ acy Cav
Va Va a Va
a ent)
ae an) ee)
Hee might oo be able to respond oa malfunction it he/she became injured, and
0 oe would be ete come lo hisher ad
1-4
e12
(6) Resctantsmigh note bot enough to reset.
(©) Pot Cost vs, Volume on ogg paper. Use thi graph 1 generat an equation for
Cae ación ef ol
ln (Cost) vs. ln (Volume)
y= 02501% + 94932
4 à
In (Volume)
From this we generate the equation: Costa 13,270(¥)°
‘We ean use this equation to find the desired prices:
For a 6000 gallon reactor: Costa 13,270(6000)" = $165,400
Fora 15,000 gallon reactor: Cost 13,270(15,000) = $25,740
Odom, Co 599 4m?
023 min "Doc,
For Constant Pressure
LAN, „ 1d(G¥) _ acy Cav
Va Va a Va
a ent)
ae an) ee)
Hee might oo be able to respond oa malfunction it he/she became injured, and
0 oe would be ete come lo hisher ad
1-4
Bind. Fee à CSTR, arten (stm da
À Ego Y = 20 da atn ad à 0827000
A | gute a yo cee
Rector A > 8
(Sotuden by LT. Sani Je)
À Ego Y = 20 da atn ad à 0827000
A | gute a yo cee
Rector A > 8
(Sotuden by LT. Sani Je)
son eg a nr rere
‘The samples mae a eng ag eu for CTA, ar
‘opi aon mnt nana, e io e ve.
De amg matí ng di ai pd ad ae a
JD mt Gael es pr us Cor) of cia) ine
bots neces guy cs o
‘Racor. Au ne gang a propery si e lod y som spe pop ek
sia mene abl a la pend Ot
EN
General Mole Balance:
en a + fra
Foca CSTR, there i no accumulation. Also ame well mixed, o that teo is no spt
vasco ine actor. The mole balance simples to
Bern F0
"Toe rue of ection based on volume is elated to the ae of recio based on catalyst weight by
the bulk ealystdesity py. The reactor volume and catala weight ae also lated az
unten
== bo)
Wen,
‘Combining and rearangin the last thee equations gives te equation fra “hide” CSTR:
re
(Solution by LT. Satine)
1-6
‘The samples mae a eng ag eu for CTA, ar
‘opi aon mnt nana, e io e ve.
De amg matí ng di ai pd ad ae a
JD mt Gael es pr us Cor) of cia) ine
bots neces guy cs o
‘Racor. Au ne gang a propery si e lod y som spe pop ek
sia mene abl a la pend Ot
EN
General Mole Balance:
en a + fra
Foca CSTR, there i no accumulation. Also ame well mixed, o that teo is no spt
vasco ine actor. The mole balance simples to
Bern F0
"Toe rue of ection based on volume is elated to the ae of recio based on catalyst weight by
the bulk ealystdesity py. The reactor volume and catala weight ae also lated az
unten
== bo)
Wen,
‘Combining and rearangin the last thee equations gives te equation fra “hide” CSTR:
re
(Solution by LT. Satine)
1-6
tes tanteo EHE = aus ot acta 3 Ua te nee
a
OS
CEE)
yay foyer
Bis, Nosolatidn willbe sven
a
OS
CEE)
yay foyer
Bis, Nosolatidn willbe sven
fect pray
Given: Los Angeles Basin: A = Basin rea = 2x 10° 6%, H.= Inversion height = 2000,
VAR Of. Basin may be considered “welbmite’
Melo.
i
Wind fom Mojave Desen T= 75 "Fand P= 10atm
Foon = CO emission rom autosFcas = CO in Santa Ana wind
woe mo
Lun CPE,
Se nn
02 molCO/ ot: 3000 sc /r-car 40,00 no) 30.661)
Fooa = 6143410 mob
Fan "Yo VIC
3. vo wind speed= 15 mph W = corrido width = 20 mies
WH, = (Spa Ponies 2001 (5208 me)"
ve 21604107 hr
Ces = (20610 lomo 670107 0)
Feos 3407210 nor
Fos
5. CO Balance: Input to LA Basi rom ar = Fong. from the Santa Ana wind = Fens;
Output = Co. where Co = Concerto of CO i Basi,
a
i
Fi
O
ri
RER, er ee à incorrectas shove
Sco “e Fog att = re
A mue fe.
RENE TEEX na eos rn Le
a
Loos nt
Jeet
co 1
7 "Feo,a°Feo,s ~ "Co
VAR Of. Basin may be considered “welbmite’
Melo.
i
Wind fom Mojave Desen T= 75 "Fand P= 10atm
Foon = CO emission rom autosFcas = CO in Santa Ana wind
woe mo
Lun CPE,
Se nn
02 molCO/ ot: 3000 sc /r-car 40,00 no) 30.661)
Fooa = 6143410 mob
Fan "Yo VIC
3. vo wind speed= 15 mph W = corrido width = 20 mies
WH, = (Spa Ponies 2001 (5208 me)"
ve 21604107 hr
Ces = (20610 lomo 670107 0)
Feos 3407210 nor
Fos
5. CO Balance: Input to LA Basi rom ar = Fong. from the Santa Ana wind = Fens;
Output = Co. where Co = Concerto of CO i Basi,
a
i
Fi
O
ri
RER, er ee à incorrectas shove
Sco “e Fog att = re
A mue fe.
RENE TEEX na eos rn Le
a
Loos nt
Jeet
co 1
7 "Feo,a°Feo,s ~ "Co
PLAO (ta)
seb mme Cog Le rome
Geo = cum 19 nares?
Der: um
Guarico? TREE + 3 40710) EAGLE gut LB jeunes a
; Pa) ES
1
cerro IRA «3 aorazo? Diele co gras ot Dole à à 79 ES
+ SOR ie ets as core,
anotaron? Ti A ne
2 o It Fea = Sosa Th
oa 7 3 LE Fea o a
Proa? AMLO 19 metal 47 = 3000 20 remar
dr mos Coane an for 60)
Era etre
So NO macestas vu
x ‚Fo. = "eno,
cn dina,
seb mme Cog Le rome
Geo = cum 19 nares?
Der: um
Guarico? TREE + 3 40710) EAGLE gut LB jeunes a
; Pa) ES
1
cerro IRA «3 aorazo? Diele co gras ot Dole à à 79 ES
+ SOR ie ets as core,
anotaron? Ti A ne
2 o It Fea = Sosa Th
oa 7 3 LE Fea o a
Proa? AMLO 19 metal 47 = 3000 20 remar
dr mos Coane an for 60)
Era etre
So NO macestas vu
x ‚Fo. = "eno,
cn dina,
PO cont'd)
y Agata,
asomo E zoo) Dale
AT 1
irn = SRE metered à 23700 39
La ut nts ds Lans than Gp ge ++ Tis coneeneracion of MO vi
fot genen 0.4 pe matos LA. o deiriag habits are altered.
putt, Solution
Reaction: A— 3
Constant volumetric flowrate; isothermal; continuous flow
Faae5.0 mal/hr, vont0 dim? /hr —> Cagnd'S mol/ da
Rate Law: ty 2k =0.05 SL
Combine: v= Fa =001F ss „.
Volume of the CSTRA99,0 dm?
y Agata,
asomo E zoo) Dale
AT 1
irn = SRE metered à 23700 39
La ut nts ds Lans than Gp ge ++ Tis coneeneracion of MO vi
fot genen 0.4 pe matos LA. o deiriag habits are altered.
putt, Solution
Reaction: A— 3
Constant volumetric flowrate; isothermal; continuous flow
Faae5.0 mal/hr, vont0 dim? /hr —> Cagnd'S mol/ da
Rate Law: ty 2k =0.05 SL
Combine: v= Fa =001F ss „.
Volume of the CSTRA99,0 dm?
Pr (cont'd)
BER
Mole Balance: E
mol
Rate Law: tkm
Volume of PFR-29.0 dm?
(6) task Ge with k=0.0001 5
SSIR
Moe Balance in terms of concenendons ve EC)
Rate Law tank 2
acer, (OE
ETC] [ETE] 29 of
vens0éa' *
Combine: Vi
BER
Mole Balance in terms of concentration: Ghia
vr
Rate Law: -taskCa
Combine: [ave ES
q)
v—g[ptgca] ie
$
Ga
=a"
ur
Volume of PFR=128.0 dm?
BER
Mole Balance: E
mol
Rate Law: tkm
Volume of PFR-29.0 dm?
(6) task Ge with k=0.0001 5
SSIR
Moe Balance in terms of concenendons ve EC)
Rate Law tank 2
acer, (OE
ETC] [ETE] 29 of
vens0éa' *
Combine: Vi
BER
Mole Balance in terms of concentration: Ghia
vr
Rate Law: -taskCa
Combine: [ave ES
q)
v—g[ptgca] ie
$
Ga
=a"
ur
Volume of PFR=128.0 dm?
Pret (ont)
2 with ka o E
CERTES
SSIR
Mole Balance in terms of concentration: V=
Rate Law: -ra2kC,?
EN]
a
fee)
(€..-001€,. )
Combine: er ( =
Ss
Volume of CSTR=66,000 dm?
BER
Mole Balance in terms of concentration:
Rate Law: -r42kC,?
"Ta
Volume of PFR=580 dm?
(Solution by JT. Santini, Jr)
2 with ka o E
CERTES
SSIR
Mole Balance in terms of concentration: V=
Rate Law: -ra2kC,?
EN]
a
fee)
(€..-001€,. )
Combine: er ( =
Ss
Volume of CSTR=66,000 dm?
BER
Mole Balance in terms of concentration:
Rate Law: -r42kC,?
"Ta
Volume of PFR=580 dm?
(Solution by JT. Santini, Jr)
Solution
Reaction: A>B4C
Eeiscrmat; wellisixed: Constan volume batch reactor
220 den, Nae=20 moles.
a Aaa, wth 20.865 min!
Mole Balance: (constant volume batch reactor)
a ace ner,
ant nee
(9 Assume [desl Gas Equation Holds
Sempersnare = 127°C = 400K (constant becas
Volume = 20 din? (constand)
Insal Total Moles = 20
Final Toul Males = 40 (reaction goes to completion)
Inka Pressure:
ERT „ (0 moles 0.082 400€)
v am’)
tam
nol Pressure
EEES
D Ga)
Prism th
Reaction: A>B4C
Eeiscrmat; wellisixed: Constan volume batch reactor
220 den, Nae=20 moles.
a Aaa, wth 20.865 min!
Mole Balance: (constant volume batch reactor)
a ace ner,
ant nee
(9 Assume [desl Gas Equation Holds
Sempersnare = 127°C = 400K (constant becas
Volume = 20 din? (constand)
Insal Total Moles = 20
Final Toul Males = 40 (reaction goes to completion)
Inka Pressure:
ERT „ (0 moles 0.082 400€)
v am’)
tam
nol Pressure
EEES
D Ga)
Prism th
pu POR
‘The volumetric Now mei ven wy 0000 mafia TSK and PL = 27 am. Th el as wit
ede amine he volume on cc. au a STP
„az, 200005 +2904 +2Tam
A de
TR” S13K*lam oa 3e00se0 "616
rewind he rence ine of ach meee, we wl sale de a lc
(here Nis heb number of we and AC ro een ara of nh ae).
+4 30,000"
ut. 20 M 59
Whe ~ 2050°589*10- mt "RM = 690
‘Te rien me can sow be calcula ug pt veloc an e eng of ch be
L__2n
nz
= 1158
150,004
Mi
Eure
‘The volumetric Now mei ven wy 0000 mafia TSK and PL = 27 am. Th el as wit
ede amine he volume on cc. au a STP
„az, 200005 +2904 +2Tam
A de
TR” S13K*lam oa 3e00se0 "616
rewind he rence ine of ach meee, we wl sale de a lc
(here Nis heb number of we and AC ro een ara of nh ae).
+4 30,000"
ut. 20 M 59
Whe ~ 2050°589*10- mt "RM = 690
‘Te rien me can sow be calcula ug pt veloc an e eng of ch be
L__2n
nz
= 1158
150,004
Mi
Eure
PAS. Nota wil be give.
ELIE 2 Chemical Matting Repo mica Week
© Chemical Proc Indu (RN. Steve, LA Brisk Je hed New York Mora Hi
Book Co, ne 1977, Indust an Engineering Chem,
‘Api ean racing recor wed pelea eng operas u 44-1020 1030
me RE Steve nd LA Banke ved New York Me
FH Book Cay Ines 1977 Of
niche
No solution wilde given
ELIE 2 Chemical Matting Repo mica Week
© Chemical Proc Indu (RN. Steve, LA Brisk Je hed New York Mora Hi
Book Co, ne 1977, Indust an Engineering Chem,
‘Api ean racing recor wed pelea eng operas u 44-1020 1030
me RE Steve nd LA Banke ved New York Me
FH Book Cay Ines 1977 Of
niche
No solution wilde given
Find number of moles and coocentation
101 33420
De). ae
a
= 15 97.5 = 73.1molesA
Mole Bole 637 moles
volume 200 dm
Determine reaction time
10.731 ln 73.1
1=23 min
9 Determine reaction time
Lunv
A
101 33420
De). ae
a
= 15 97.5 = 73.1molesA
Mole Bole 637 moles
volume 200 dm
Determine reaction time
10.731 ln 73.1
1=23 min
9 Determine reaction time
Lunv
A
det
AO —
A ic
ias
En Gitterentiat atome,
Dr material betas nun,
sary =~ naar
ete GAY = teuesion of xosetor Lone mich te Liquid
capone
Mm ced ta Co ao
More PA Le an 068 Gus à it ane on tte af
Fy + molar flow rate of A (anotes/ase)
AO —
A ic
ias
En Gitterentiat atome,
Dr material betas nun,
sary =~ naar
ete GAY = teuesion of xosetor Lone mich te Liquid
capone
Mm ced ta Co ao
More PA Le an 068 Gus à it ane on tte af
Fy + molar flow rate of A (anotes/ase)
Ard Edition, Solution Manual, Chapter 2
Chapter 2
General: The overall goal of these problems is to help the student realize that if they
the stage
mA.
P22.
(X) they can “design” or size a large number of reaction systems. It sets
@ for the algorithm developed in Chapter 4
This problem will keep students thinking about writing down what they
learned every chapter.
Part (a) is open-ended and encourages the student to do little “OUT OF THE
BOX" thinking.
‘Straight forward rehash of Example 2-7 to calculate reactor volumes.
Uses definition of space time to calculate V. Can be done in 30 seconds.
No calculations necessary for this problem, but does require some thinking.
‘This problem encompasses most all the key points of Chapter 2. That is, if
we are given -ra=f(X) then we can size any number of reactor systems.
Some parts plug and chug, others require more thinking
Good troubleshooting problem. Could ask the students to brainstorm in
groups what could have happened.
Problems P2-8, P2-9, and P2-12 are alternative problems to P2-6 and can be assigned
P20.
itt Poa,
Bis eos
Pas,
an
in different years.
The point is to estimate the sizes of these real reactors. The students can use
the door as a point of reference to estimate the reactor volumes. Could be
used with the ethical dilemma problems. CDP2-B.
Open-ended in that student is faced with decision on how to relax.
In recent years, a number of students have on their own fit a polynomial to
the curves in P2-6, P2-8, P2.9, and P2-12 and then used POLYMATH to solve
the problems.
Problems P2-14 and P2-15 encourage outside reading and help to develop life-long
learning skills by obtaining information on their own.
CDP2A
mx
ors
Similar to 2-9
Good problem to get groups started working together (eg. cooperative
learning,
Similar to problems 2-8, 29, 2-12.
pai
Chapter 2
General: The overall goal of these problems is to help the student realize that if they
the stage
mA.
P22.
(X) they can “design” or size a large number of reaction systems. It sets
@ for the algorithm developed in Chapter 4
This problem will keep students thinking about writing down what they
learned every chapter.
Part (a) is open-ended and encourages the student to do little “OUT OF THE
BOX" thinking.
‘Straight forward rehash of Example 2-7 to calculate reactor volumes.
Uses definition of space time to calculate V. Can be done in 30 seconds.
No calculations necessary for this problem, but does require some thinking.
‘This problem encompasses most all the key points of Chapter 2. That is, if
we are given -ra=f(X) then we can size any number of reactor systems.
Some parts plug and chug, others require more thinking
Good troubleshooting problem. Could ask the students to brainstorm in
groups what could have happened.
Problems P2-8, P2-9, and P2-12 are alternative problems to P2-6 and can be assigned
P20.
itt Poa,
Bis eos
Pas,
an
in different years.
The point is to estimate the sizes of these real reactors. The students can use
the door as a point of reference to estimate the reactor volumes. Could be
used with the ethical dilemma problems. CDP2-B.
Open-ended in that student is faced with decision on how to relax.
In recent years, a number of students have on their own fit a polynomial to
the curves in P2-6, P2-8, P2.9, and P2-12 and then used POLYMATH to solve
the problems.
Problems P2-14 and P2-15 encourage outside reading and help to develop life-long
learning skills by obtaining information on their own.
CDP2A
mx
ors
Similar to 2-9
Good problem to get groups started working together (eg. cooperative
learning,
Similar to problems 2-8, 29, 2-12.
pai
Ard Edition, Solution Manual, Chapter 2
CDP2D Similar to problems 2-8, 2-9, 2-
|
Solution
en
|
'
E
9BCD
9BCD
SBCD
9BCD
Sex 8888 BSsERssusu
waar vor
0000 3 wok
ways assigned, AA = Always assign one from the group of alternates,
ten, I= Infrequently, S = Seldom, G = Graduate level
Pea
P22
P23
P24
P25
e m6
e P27
ms
P29
P210
mu
P212
P213
P214
PRIS
CDP2A
CDP28
CPC
PD
Assigned
.
o
In problems that have a dot in conjunction with AA means that one of the
problems, either the problem with a dot or any one of the alternates are
always assigned.
Time
‘Approximate time in minutes it would take a B/B student to solve the
problem.
SF = Straight forward reinforcement of principles (plug and chug)
FSF = Fairly straight forward (requires some manipulation of equations or an
intermediate calculation).
Intermediate calculation required
OE = Some parts open-ended.
p22
CDP2D Similar to problems 2-8, 2-9, 2-
|
Solution
en
|
'
E
9BCD
9BCD
SBCD
9BCD
Sex 8888 BSsERssusu
waar vor
0000 3 wok
ways assigned, AA = Always assign one from the group of alternates,
ten, I= Infrequently, S = Seldom, G = Graduate level
Pea
P22
P23
P24
P25
e m6
e P27
ms
P29
P210
mu
P212
P213
P214
PRIS
CDP2A
CDP28
CPC
PD
Assigned
.
o
In problems that have a dot in conjunction with AA means that one of the
problems, either the problem with a dot or any one of the alternates are
always assigned.
Time
‘Approximate time in minutes it would take a B/B student to solve the
problem.
SF = Straight forward reinforcement of principles (plug and chug)
FSF = Fairly straight forward (requires some manipulation of equations or an
intermediate calculation).
Intermediate calculation required
OE = Some parts open-ended.
p22
‘and Edition, Solution Manual, Chapter 2
‘Note the letter problems are found on the CD-ROM. For example A = CDPI-A.
Summary Table Ch2
Straight forward 346
Fairly straight forward su
More difficult 57.13
‘Open-ended. 2(a)200)7.101,14,5,
B
Comprehensive 23456
Parameter sensitivity -
Critical thinking EX
‘Note the letter problems are found on the CD-ROM. For example A = CDPI-A.
Summary Table Ch2
Straight forward 346
Fairly straight forward su
More difficult 57.13
‘Open-ended. 2(a)200)7.101,14,5,
B
Comprehensive 23456
Parameter sensitivity -
Critical thinking EX
E
m.
Chapter 2
No solution will be given.
Ploning the given data as Uta vs. U(1-X) produces a linear graph. This
(can be used to extrapolate values for ra for all ofthe conversion values
‘upto the desired 0.98, These values can be used to calculate the desired
volume
Ara ve 1)
De)
‘The following equation was found to ft the data:
1 150.97
312415097
Ex «x
Graphing Ur, vs. 1/(-X) proved to be nonlinear.
(One cannot extrapolate toa higher temperature unless the activation energy is
given
D Scheme A
re Here
ay
TR DE
{2,3 200 +250] +
osu mel ge (20 « 250 SS
sen
Kick. (dei)
ashe rx Xa) ES Jaco,
Vrou = 58.5 + 346.7 = 405.0 dd
at
m.
Chapter 2
No solution will be given.
Ploning the given data as Uta vs. U(1-X) produces a linear graph. This
(can be used to extrapolate values for ra for all ofthe conversion values
‘upto the desired 0.98, These values can be used to calculate the desired
volume
Ara ve 1)
De)
‘The following equation was found to ft the data:
1 150.97
312415097
Ex «x
Graphing Ur, vs. 1/(-X) proved to be nonlinear.
(One cannot extrapolate toa higher temperature unless the activation energy is
given
D Scheme A
re Here
ay
TR DE
{2,3 200 +250] +
osu mel ge (20 « 250 SS
sen
Kick. (dei)
ashe rx Xa) ES Jaco,
Vrou = 58.5 + 346.7 = 405.0 dd
at
Vin Be. (0.3) (300) = 78m
Vas Fo | Sar hf, a _ 4.
IE 7 RES OREA
Vas 3 (25) 1250 + 4279 + 2025) + 4500) + 100]
Vies
Va 2 78 da? 173 de à 251 da
Scheme A
a ys cites Scag.
+B 2000 + 4265) + 550) = 183 da?
Van $ (8-7 (800) = 693 de
Via = 252 dm?
Scheme B
Vı= 2 60650) = 3337 da?
= 22 10 (550 + 200] = 58.5 4m?
Va= 30101550 800) = 58.54
Vow = 392 da?
Vas Fo | Sar hf, a _ 4.
IE 7 RES OREA
Vas 3 (25) 1250 + 4279 + 2025) + 4500) + 100]
Vies
Va 2 78 da? 173 de à 251 da
Scheme A
a ys cites Scag.
+B 2000 + 4265) + 550) = 183 da?
Van $ (8-7 (800) = 693 de
Via = 252 dm?
Scheme B
Vı= 2 60650) = 3337 da?
= 22 10 (550 + 200] = 58.5 4m?
Va= 30101550 800) = 58.54
Vow = 392 da?
En EXA EXE
wol Pesta
Conversion, x
Fen a
+ Gives: ES kes. to achieve X= 0.8 in a CSTR:
vy 21 ata,
23, 50 :
Yevon Gone) CE CR 600 ee? aan
svete lomo. at
P2S
ne = RCA (IX)?
System |
SIR
EAN
TIAS
E
RE ER ax,
(a) GE ae
wol Pesta
Conversion, x
Fen a
+ Gives: ES kes. to achieve X= 0.8 in a CSTR:
vy 21 ata,
23, 50 :
Yevon Gone) CE CR 600 ee? aan
svete lomo. at
P2S
ne = RCA (IX)?
System |
SIR
EAN
TIAS
E
RE ER ax,
(a) GE ae
where a= E
VCE
XX +120
2a Bra (Gras feras
MS a 7
x
For the second root X, 21, which is impossible. Hence,
(2+3)
J: OLX
pda a ras
2
Teather etl”
VCE
XX +120
2a Bra (Gras feras
MS a 7
x
For the second root X, 21, which is impossible. Hence,
(2+3)
J: OLX
pda a ras
2
Teather etl”
P28
conva
2001 joo
MS DEC)
PR» we:
Fer act, Xu ass
From Equation (1)
xo Lean Es
. 2
BER
From Equation 2)
‘System 2 wil give the highest conversion.
Sytem 3 will give the lowest conversion,
conva
2001 joo
MS DEC)
PR» we:
Fer act, Xu ass
From Equation (1)
xo Lean Es
. 2
BER
From Equation 2)
‘System 2 wil give the highest conversion.
Sytem 3 will give the lowest conversion,
rs
Solution
X 0 02 04 06 08 09
ra 10 1687 50 50 125 90
fact 006 00 00 08 alı
(a) To solve this problem, fst plot Ura vs. X trom the chart
above see staches plod. Second, use moe balances as given.
below.
Mote Baer Vi = Fe} SE oro ets énonce 72602
04 a6
x
(D) For a feed stream that eners the reaction with a previous
Conversion of 040 and leaves a any conversion up 0 00, he
Solar of the PER and CSTR will be identical because the rate
Le constant over this conversion range. (ee Below)
nm
ey ox
va de Se
(Note: The reactors) before and up to achieving a conversion of
‘040 must be in Seres with no side reams)
2-1
Solution
X 0 02 04 06 08 09
ra 10 1687 50 50 125 90
fact 006 00 00 08 alı
(a) To solve this problem, fst plot Ura vs. X trom the chart
above see staches plod. Second, use moe balances as given.
below.
Mote Baer Vi = Fe} SE oro ets énonce 72602
04 a6
x
(D) For a feed stream that eners the reaction with a previous
Conversion of 040 and leaves a any conversion up 0 00, he
Solar of the PER and CSTR will be identical because the rate
Le constant over this conversion range. (ee Below)
nm
ey ox
va de Se
(Note: The reactors) before and up to achieving a conversion of
‘040 must be in Seres with no side reams)
2-1
cont'd
(© Van =10.5 dm?
First pick a conversion. Second, find 1/-14 from the plot. Third.
check to see if X/-ra is equal to 0.035. [fit hen the correct
conversion has been chosen.
1
m x27, 005, mode ni
Maximum conversion = 0.70
@
From part (a), we know that X,=0.40. Use trial and error to
find Xe,
(© Van =10.5 dm?
First pick a conversion. Second, find 1/-14 from the plot. Third.
check to see if X/-ra is equal to 0.035. [fit hen the correct
conversion has been chosen.
1
m x27, 005, mode ni
Maximum conversion = 0.70
@
From part (a), we know that X,=0.40. Use trial and error to
find Xe,
B26 conta
ya Ful%=%)
Er
Mole Balance:
rearranging, we get
E
“ox, Fag 300
Conversion = 9.64
%
From part (a), we know that X120.40. Use trial and error to
find Xe.
xx
x
Mole Balance: vane [220]
At, = 0.908 v= 30 Aa Under conve] 3000242 72
Conversion = 908
ya Ful%=%)
Er
Mole Balance:
rearranging, we get
E
“ox, Fag 300
Conversion = 9.64
%
From part (a), we know that X120.40. Use trial and error to
find Xe.
xx
x
Mole Balance: vane [220]
At, = 0.908 v= 30 Aa Under conve] 3000242 72
Conversion = 908
2:6 cont'd
(D See plots below.
Conversion asa Function of Rate of Reaction asa Function of
PER Volume PER Volume
ate extn
ee SEER ER SES
IA
FER Volume (en
Questions from Sgt. Ambercromby:
Could a dent cause the string blade not to tur?
Ts the temperature the same?
‘What causes the whoosh sound?
How much does the dent reduce the volume?
Possible reasons forthe drop:
1. Temperature ofthe feed.
“The rate of reaction depends on temperature.
2. Flow rate
Residence time depends on the flow rate, If low rate T, residence time 4,
hence conversion .
3. Level of liquid in the CSTR
TE liquid level | effective volume of CSTR J, hence conversion L.
4. Concentration of Ain feed
Rate of reaction depends on the concentration of A in the feed.
5. Performance of the sirer in the CSTR
Y bere is some malfunctioning inthe stir, the CSTR will no longer be
wellstirred and there will be some ‘dead volume". This decrease inthe
effective volume would cause a decrease in conversion,
(D See plots below.
Conversion asa Function of Rate of Reaction asa Function of
PER Volume PER Volume
ate extn
ee SEER ER SES
IA
FER Volume (en
Questions from Sgt. Ambercromby:
Could a dent cause the string blade not to tur?
Ts the temperature the same?
‘What causes the whoosh sound?
How much does the dent reduce the volume?
Possible reasons forthe drop:
1. Temperature ofthe feed.
“The rate of reaction depends on temperature.
2. Flow rate
Residence time depends on the flow rate, If low rate T, residence time 4,
hence conversion .
3. Level of liquid in the CSTR
TE liquid level | effective volume of CSTR J, hence conversion L.
4. Concentration of Ain feed
Rate of reaction depends on the concentration of A in the feed.
5. Performance of the sirer in the CSTR
Y bere is some malfunctioning inthe stir, the CSTR will no longer be
wellstirred and there will be some ‘dead volume". This decrease inthe
effective volume would cause a decrease in conversion,
9 01 020304 05 66 07 05 09
Conversion, x
Initial Operating Conditions
X=0, Xı=064, X:=082
0 OF 02 03 04 05 06 07 08 09
Conversion, X
Final Operating Conditions:
Verein = 38°0.17 = 65 dm”
Decrease in volume = 1.5 dm?
2-15
Conversion, x
Initial Operating Conditions
X=0, Xı=064, X:=082
0 OF 02 03 04 05 06 07 08 09
Conversion, X
Final Operating Conditions:
Verein = 38°0.17 = 65 dm”
Decrease in volume = 1.5 dm?
2-15
E2-7 cont'd
This 20% decrease in volume could not have been caused by a small dent. The
most likely explanation is thatthe ster fll off, which caused “dead zones" and a
decrease in the effective volume.
“To increase the conversion to above 0.50, switch the order of reactors. By placing |
the PER before the CSTR, the necessary conversion can be achieved. 1]
X1=0.16
Vemestay= 65 dm? => X2= 062
o G1 02 03 04 05 06 07 08 09
Conversion, X
Comarsion. x
Figure 2-8.
2-1#
This 20% decrease in volume could not have been caused by a small dent. The
most likely explanation is thatthe ster fll off, which caused “dead zones" and a
decrease in the effective volume.
“To increase the conversion to above 0.50, switch the order of reactors. By placing |
the PER before the CSTR, the necessary conversion can be achieved. 1]
X1=0.16
Vemestay= 65 dm? => X2= 062
o G1 02 03 04 05 06 07 08 09
Conversion, X
Comarsion. x
Figure 2-8.
2-1#
LS cont'd
0) For am intecnodiete convertion of 0.3, Figsze Pacte shove har a RZ
7ieide the malloae volume, since for the PFE ve se the aros under the enero,
Asintoce volume is also achieved by following the FFB vith «CST. la tale
Cto she ares considered vould be the rectangle bounded ty 190.3 and 100.7
THB # Relghe equal to the Clo/-zy vale at 220,7, which is 1
under the curve.
D ve 50 Lan
Verlorene
Ts ares considered ia past an
Loro e
A +
rit
is ain
MO Va v, Zn (50 L/mta) (15 min) = 750 1 = 750 ax.
Fhe saeliese ares can be achieved by using only one CSTR vith Chis system,
¢
Zero aos = 10.5 ste
A x00.7
Se Vey, 1 = (50) Unia (20.8) min 3231
Ye vould farther seduce the total volume dy using a PER at Fiese ap co che
conversion thet gives the same Cyo/-2, as 260.7.
To obtain equal CSTR and PFR volumes the aren ander the carve
So the area of the rectancle up to the specified conversions
By trial and error ve see chat X=0.45 is a solution. For the CSTR,
€
Em (0.45-0) Mn (o.45-0)(37 = 16.65 mia
Ro.
So Ver To (30 1/mia)(16.65 mia) = 632.5 1
> 15
0) For am intecnodiete convertion of 0.3, Figsze Pacte shove har a RZ
7ieide the malloae volume, since for the PFE ve se the aros under the enero,
Asintoce volume is also achieved by following the FFB vith «CST. la tale
Cto she ares considered vould be the rectangle bounded ty 190.3 and 100.7
THB # Relghe equal to the Clo/-zy vale at 220,7, which is 1
under the curve.
D ve 50 Lan
Verlorene
Ts ares considered ia past an
Loro e
A +
rit
is ain
MO Va v, Zn (50 L/mta) (15 min) = 750 1 = 750 ax.
Fhe saeliese ares can be achieved by using only one CSTR vith Chis system,
¢
Zero aos = 10.5 ste
A x00.7
Se Vey, 1 = (50) Unia (20.8) min 3231
Ye vould farther seduce the total volume dy using a PER at Fiese ap co che
conversion thet gives the same Cyo/-2, as 260.7.
To obtain equal CSTR and PFR volumes the aren ander the carve
So the area of the rectancle up to the specified conversions
By trial and error ve see chat X=0.45 is a solution. For the CSTR,
€
Em (0.45-0) Mn (o.45-0)(37 = 16.65 mia
Ro.
So Ver To (30 1/mia)(16.65 mia) = 632.5 1
> 15
cont'd
Fee the PPL
rf" 3
a a
sing flapson's me
1+ QUE quo + cas) + 20) = aan 2660 a © 2050 +
un + 2048) «m
PER quo» aes + 39 + 48 + 48) 2020 à 48e 30 2 49) + D + 364
So Va yE = (50 1/aia)(15.72 aia) = 7862 6% aitterence, pretty
There ie also a solution at an 0.7
Te 1-0
For the CSTR
T= (028-0039) = 26.4 min
For the PFE
o + 26) + aan) «200 + an + am
“aus + a3
2 MA (20 + 420 + $0 + 32 + 45) #2043 + 48 + aT) em 0 ate
Tey 190.19
For the CSTR
Ze 0.799650) = 23.7 mín
Se Verde (0 t/nin)(23.7 min) = 3285 1 (Om)
Fee the PPL
rf" 3
a a
sing flapson's me
1+ QUE quo + cas) + 20) = aan 2660 a © 2050 +
un + 2048) «m
PER quo» aes + 39 + 48 + 48) 2020 à 48e 30 2 49) + D + 364
So Va yE = (50 1/aia)(15.72 aia) = 7862 6% aitterence, pretty
There ie also a solution at an 0.7
Te 1-0
For the CSTR
T= (028-0039) = 26.4 min
For the PFE
o + 26) + aan) «200 + an + am
“aus + a3
2 MA (20 + 420 + $0 + 32 + 45) #2043 + 48 + aT) em 0 ate
Tey 190.19
For the CSTR
Ze 0.799650) = 23.7 mín
Se Verde (0 t/nin)(23.7 min) = 3285 1 (Om)
RZ: cont'd
30 23,9 - 2 ¢0.8-0.199(30095) = 29.9 - 0.313 = 23.58 nía
So Ve y Ln (50 1/m10)(23.58 mia) = 1179 1
0.58 difterence
So/-rats
ain) 20 8.6 150 17.2 26.0 10:2 20.8 26.4
»
ta (min) 13
een
Ve
2 on 02 03 04
2
es os 07
Figure Pa:
For ose perticntas case
+ Be ate
Tee Graph yields eases possible atendy states
E + 0.285, 1, = 0.535, and x, 0.130,
2-17
30 23,9 - 2 ¢0.8-0.199(30095) = 29.9 - 0.313 = 23.58 nía
So Ve y Ln (50 1/m10)(23.58 mia) = 1179 1
0.58 difterence
So/-rats
ain) 20 8.6 150 17.2 26.0 10:2 20.8 26.4
»
ta (min) 13
een
Ve
2 on 02 03 04
2
es os 07
Figure Pa:
For ose perticntas case
+ Be ate
Tee Graph yields eases possible atendy states
E + 0.285, 1, = 0.535, and x, 0.130,
2-17
AMC
‘Data taken at 1013 kPa (10 atm) and 227°C (500.2 K)
YaP (03330)
1 = 0.08113 gmol/ám”
RT” 0.082502) ”
33 Cu
.000010]0.0000030.00000270.000001]
5 | 02 | 04 | 08
911277 [10225 34]20563.84]81127.8]
(2) 30% conversion in PFR:
a (sss fe mtimin)n 155m
mia ,
> Vom = (12.1 s] [2 m’/min) m’
ln (oe) 105.64
© Toal Volume:
Vyas 21555 +405 = 56141m°
(® 60% conversion in PFR:
8X 2028195 = Vin
>18
‘Data taken at 1013 kPa (10 atm) and 227°C (500.2 K)
YaP (03330)
1 = 0.08113 gmol/ám”
RT” 0.082502) ”
33 Cu
.000010]0.0000030.00000270.000001]
5 | 02 | 04 | 08
911277 [10225 34]20563.84]81127.8]
(2) 30% conversion in PFR:
a (sss fe mtimin)n 155m
mia ,
> Vom = (12.1 s] [2 m’/min) m’
ln (oe) 105.64
© Toal Volume:
Vyas 21555 +405 = 56141m°
(® 60% conversion in PFR:
8X 2028195 = Vin
>18
P29 (cont'd)
80% conversion in PFR:
1, is not known for X>0.60 ~ cannot do.
Coa xin) = 1014.1
© 501060% conversion in CSTR.
\em’min)2104m°
os,
Rate of Reaction vs. Volume Conversion vs. Volume
=
0) Critique
Answers are Valid:
Constan Temperature and Pressure
No heat effects
No pressure drop
Single interpolation to X, = 0.15, 0.30, 0.45, and 0.50 allowable
Huge volume (he size ofthe LA Basin)! Raise T? Raise P?
3-19
80% conversion in PFR:
1, is not known for X>0.60 ~ cannot do.
Coa xin) = 1014.1
© 501060% conversion in CSTR.
\em’min)2104m°
os,
Rate of Reaction vs. Volume Conversion vs. Volume
=
0) Critique
Answers are Valid:
Constan Temperature and Pressure
No heat effects
No pressure drop
Single interpolation to X, = 0.15, 0.30, 0.45, and 0.50 allowable
Huge volume (he size ofthe LA Basin)! Raise T? Raise P?
3-19
Problem 2-10 involves estimating the volume of tree reactors from a picture.
“The door on the side ofthe building was used as areference. It was assumed to
bes ft high.
The following estimates were made:
CR
hessk a=9n
Ve ne = 14:50 (56 D = 3562 1 = 100,865 L
BER
Length of ove segment = 23
Length of entre reactor = (23 f) (12) (11) = 20368.
D=i
‘Verne = 10.5 12036 1) = 2384 = 67,507 L
‘Answers will vary for each individual
B2:L1 No solution necessary.
Pre
‘The smallest amount of catalyst necesary to achieve 80% conersion in a CSTR and
PBR connected in series and conaining equal amouats of catalyst can be calculated
from tbe figure below.
Comers sm
Some ran
‘The fit shaded area with the lines going up from the right to he left denotes the
CSTR while the area withthe lines going up from right to left denotes the PBR.
“This figure shows that che smallest amount of catalyst used is obtained when the
(CSTR is used before the PBR,
2-20
“The door on the side ofthe building was used as areference. It was assumed to
bes ft high.
The following estimates were made:
CR
hessk a=9n
Ve ne = 14:50 (56 D = 3562 1 = 100,865 L
BER
Length of ove segment = 23
Length of entre reactor = (23 f) (12) (11) = 20368.
D=i
‘Verne = 10.5 12036 1) = 2384 = 67,507 L
‘Answers will vary for each individual
B2:L1 No solution necessary.
Pre
‘The smallest amount of catalyst necesary to achieve 80% conersion in a CSTR and
PBR connected in series and conaining equal amouats of catalyst can be calculated
from tbe figure below.
Comers sm
Some ran
‘The fit shaded area with the lines going up from the right to he left denotes the
CSTR while the area withthe lines going up from right to left denotes the PBR.
“This figure shows that che smallest amount of catalyst used is obtained when the
(CSTR is used before the PBR,
2-20
P22 contra
One can calculate the amount of catalyst needed
80% conversion using a single CSTR by dete
10 carry out the same reaction tp
rectangle below.
inning the area ofthe shaded
s 10
Conversion, x
‘he wea ofecangle canbe found 1 equal apprit 22.4 k ty
(© Tie ESTR esa weigh: necessary to achieve 40% conversion canbe obtained by
‘alculaing the area ofthe reactangle shown in toe
shaded area on the figure below,
2 4 Peso
Conversion, x
‘The amount of catalyst W. needed can be found o be about 8 kg
9-21
One can calculate the amount of catalyst needed
80% conversion using a single CSTR by dete
10 carry out the same reaction tp
rectangle below.
inning the area ofthe shaded
s 10
Conversion, x
‘he wea ofecangle canbe found 1 equal apprit 22.4 k ty
(© Tie ESTR esa weigh: necessary to achieve 40% conversion canbe obtained by
‘alculaing the area ofthe reactangle shown in toe
shaded area on the figure below,
2 4 Peso
Conversion, x
‘The amount of catalyst W. needed can be found o be about 8 kg
9-21
22:12 cont'd
(@ The caulyst weight necessary 10 achieve 80% conversion fora PBR can be
determined by calculating the shaded area undemeah the curve ia te figure below
(6g Cast)
so
ee am
Conversion, x
‘The necessary catalyst weight can be found to be approximately 30.
(©) Te amount of catalyst necessary for a single PBR 10 buin 40% conversion can be
calculated by finding the asa ofthe shaded region below.
69 catayst
20
conversion, x
this region can be found 10 equal approxima
22
(@ The caulyst weight necessary 10 achieve 80% conversion fora PBR can be
determined by calculating the shaded area undemeah the curve ia te figure below
(6g Cast)
so
ee am
Conversion, x
‘The necessary catalyst weight can be found to be approximately 30.
(©) Te amount of catalyst necessary for a single PBR 10 buin 40% conversion can be
calculated by finding the asa ofthe shaded region below.
69 catayst
20
conversion, x
this region can be found 10 equal approxima
22
2-12 (cont'd)
o
sa and X vs. Catalyst Weight
For different (+) vs. (X) curves, reactors should be arranged so thatthe smallest
amount of catalyst is needed to give the maximum conversion. This can be done
by minimizing the area that is occupied by a given reactor.
(One useful heuristic is that for curves with a negative slope, itis generally better
to use a CSTR. Similarly, when the curve bas a postie slope, itis generally
better to use a PBR,
2.13 No solution will e given.
2:14 No solution willbe given.
22:18 No solution will be given.
o
sa and X vs. Catalyst Weight
For different (+) vs. (X) curves, reactors should be arranged so thatthe smallest
amount of catalyst is needed to give the maximum conversion. This can be done
by minimizing the area that is occupied by a given reactor.
(One useful heuristic is that for curves with a negative slope, itis generally better
to use a CSTR. Similarly, when the curve bas a postie slope, itis generally
better to use a PBR,
2.13 No solution will e given.
2:14 No solution willbe given.
22:18 No solution will be given.
Tee eenstion A 23
= 02 ii
rs
3) Tati ede point vaere = UE, in independent of Tye des:
CRETE EE
eus
SES
em 0
a 3
serene 1, 0.5. Fence = 2/5, + ano" Sit = enol dea
DEEE
sad ence conversión = X= 311071
da
= 02 ii
rs
3) Tati ede point vaere = UE, in independent of Tye des:
CRETE EE
eus
SES
em 0
a 3
serene 1, 0.5. Fence = 2/5, + ano" Sit = enol dea
DEEE
sad ence conversión = X= 311071
da
oa mu ("a
net 7, + 2000
mo Ta), hig woop | a
‘The integral on the RES. stag che graph de
321014521024 2.
GR sonny o eas eue le
So the pine flor renstor volen = 2,3 2 104 a} (Sine of LAL Sasi
Kot)
a
Fao = 1000 molo/mia
x, -0.10
1, = 0.0
“Mey + Ceesetion ater? ag
sm = 1 2 (0.9 = 0.7) 2 1200
RS ot Lk main.
©) For a baron sonetos
Aone ots
(Uitte panes Venen
net 7, + 2000
mo Ta), hig woop | a
‘The integral on the RES. stag che graph de
321014521024 2.
GR sonny o eas eue le
So the pine flor renstor volen = 2,3 2 104 a} (Sine of LAL Sasi
Kot)
a
Fao = 1000 molo/mia
x, -0.10
1, = 0.0
“Mey + Ceesetion ater? ag
sm = 1 2 (0.9 = 0.7) 2 1200
RS ot Lk main.
©) For a baron sonetos
Aone ots
(Uitte panes Venen
CDPZA cont'd
Stace for X vetveen O and
32204 voie
de en oo e coa ann“ e Gato = aa at
A z
+ SAL zen A resction viese progress can be followed by senerstions!
Crique: This is ridiculously small ate of reaction
"No solution will be given
a = (Ares)
M 12008
AMET
From the graph we canse at X,
For the PFR
EX,
240
Y, . 500
a under curve = Mt.» 60
Are =.
From the graph we can see hat X, = 0.80
Stace for X vetveen O and
32204 voie
de en oo e coa ann“ e Gato = aa at
A z
+ SAL zen A resction viese progress can be followed by senerstions!
Crique: This is ridiculously small ate of reaction
"No solution will be given
a = (Ares)
M 12008
AMET
From the graph we canse at X,
For the PFR
EX,
240
Y, . 500
a under curve = Mt.» 60
Are =.
From the graph we can see hat X, = 0.80
CDP2-p
400LCSTRand 1ODLPFR Feeds 41% A, 41%B, and 18%
P=l0am T=227C=500K
Sun 10 am
ORT" (0082 L-aummol- KXS00 K)
Che =041C, = 0410244 mol) =0.1 moUL.
E, = 0,Cu =1 L4(0.1 moUL) =0.1 mols = 6 molinin
=0244 mol.
(@) There are two possible arrangements of the system:
1. CSTR followed by the PER.
2. PER followed by the CSTR
Case 1: CSTR PER
CSTR: V,=F,,(Area)
Area 29 66.67
rs 6
From the graph - X, = 0.36
V, =F, (Area under curve)
100
Area under curve = Vi = 100
6
From the graph - X, =0.445
Case 1: CSTR => PER Case 2: PER => CSTR
00
04 06 08 10 00 02 04 06 oF 10
Case 2: PFR + CSTR
PFR: Area under curve = 16.67
From the graph - X,
Area = 66.67
From the graph + X,
400LCSTRand 1ODLPFR Feeds 41% A, 41%B, and 18%
P=l0am T=227C=500K
Sun 10 am
ORT" (0082 L-aummol- KXS00 K)
Che =041C, = 0410244 mol) =0.1 moUL.
E, = 0,Cu =1 L4(0.1 moUL) =0.1 mols = 6 molinin
=0244 mol.
(@) There are two possible arrangements of the system:
1. CSTR followed by the PER.
2. PER followed by the CSTR
Case 1: CSTR PER
CSTR: V,=F,,(Area)
Area 29 66.67
rs 6
From the graph - X, = 0.36
V, =F, (Area under curve)
100
Area under curve = Vi = 100
6
From the graph - X, =0.445
Case 1: CSTR => PER Case 2: PER => CSTR
00
04 06 08 10 00 02 04 06 oF 10
Case 2: PFR + CSTR
PFR: Area under curve = 16.67
From the graph - X,
Area = 66.67
From the graph + X,
CDP2-D (cont'd)
o
‘Two 400 L CSTR's in series,
CSTRI: V=F, (Area)
‘Area = 6667
From the graph - X,
Area = 6667
From the graph - X, = 0.595
(0) Two CSTR la Series (©) Two CSTR’ la Parallel
soo 00
400 00
200 ¿20
00 02 04 06 08 10 00 02 04 0608 10
x x
‘Two 400 L CSTR’s in parallel
To each CSTR goes half of the feed.
F,,=62=3 molmio
VF, (Area)
y 400
am
LLama
From the graph: X=052
PFR: V=F, (Area under curve)
From the graph we can find the area under the curve for a conversion of 0.60:
a = (260100), 9
V= (2 molmis (90) = 180 L
o
‘Two 400 L CSTR's in series,
CSTRI: V=F, (Area)
‘Area = 6667
From the graph - X,
Area = 6667
From the graph - X, = 0.595
(0) Two CSTR la Series (©) Two CSTR’ la Parallel
soo 00
400 00
200 ¿20
00 02 04 06 08 10 00 02 04 0608 10
x x
‘Two 400 L CSTR’s in parallel
To each CSTR goes half of the feed.
F,,=62=3 molmio
VF, (Area)
y 400
am
LLama
From the graph: X=052
PFR: V=F, (Area under curve)
From the graph we can find the area under the curve for a conversion of 0.60:
a = (260100), 9
V= (2 molmis (90) = 180 L
CDP2-D (con's)
(4) Single PER
Pressure reduced by a actor of 10.
A decrease in pressure would cause a decreas in the overall concentration which
‘would in tum cause a decease in Cu, and Fu, By looking atthe design equation:
EX
AE appart tt compensa or te decreas in Freud be and increase
Use the graph of Ui, vs. X to find values for all volumes. (Assume a Now rate of
Timovinin) Generat the following able and grap
(4) Single PER
Pressure reduced by a actor of 10.
A decrease in pressure would cause a decreas in the overall concentration which
‘would in tum cause a decease in Cu, and Fu, By looking atthe design equation:
EX
AE appart tt compensa or te decreas in Freud be and increase
Use the graph of Ui, vs. X to find values for all volumes. (Assume a Now rate of
Timovinin) Generat the following able and grap
sition, Solution Manual, Chapter 3
Chapter 3
See P4:1, page 205 of the text for guidelines.
(&, the point is that the rate law (Eqn. E3-21) is not valid at very low
trations of NaOH (i.e. the rate law predicts a rate of reaction even i
this problem motivates the students as they se applications of CRE uti
This pre een plans Surprisingly, he acivation energie a
Stay the same À
Short problem from DuPont about competing effects corrosion rate 4
Function of concentration and temperature that involves reasoning, A
calculations.
Temperature, that the students learned in chemistry. Part (0) is a dis
version of P33 à
Short problem that einfores the specific reaction rate k, must be tid
ett Piguticular species This problem, which helps lay the foundation MI
MIE resetions in chapier 6, can be given atthe same lime as P3-0.
lems P37, P36 at setting 9
stoichiometric table 1 flow sheets are given for P3-8 and Figs
D Gress problems can be alternately assigned from year to year. My favor
i Po &
This problem begins to prepare the students for those situations (ef
ne reactors, multiple reactions) where they cannot use conversi
as a variable
This problem reinforces the concept that at or near equilibrium, the r
must reduce to give the equilibrium conversion (and the concentra!
that are derived from thermodynamics
See ab
See above.
See above
Reinforces the point that the equilibrium conversion will
flow system and a constant volume batch system.
Chapter 3
See P4:1, page 205 of the text for guidelines.
(&, the point is that the rate law (Eqn. E3-21) is not valid at very low
trations of NaOH (i.e. the rate law predicts a rate of reaction even i
this problem motivates the students as they se applications of CRE uti
This pre een plans Surprisingly, he acivation energie a
Stay the same À
Short problem from DuPont about competing effects corrosion rate 4
Function of concentration and temperature that involves reasoning, A
calculations.
Temperature, that the students learned in chemistry. Part (0) is a dis
version of P33 à
Short problem that einfores the specific reaction rate k, must be tid
ett Piguticular species This problem, which helps lay the foundation MI
MIE resetions in chapier 6, can be given atthe same lime as P3-0.
lems P37, P36 at setting 9
stoichiometric table 1 flow sheets are given for P3-8 and Figs
D Gress problems can be alternately assigned from year to year. My favor
i Po &
This problem begins to prepare the students for those situations (ef
ne reactors, multiple reactions) where they cannot use conversi
as a variable
This problem reinforces the concept that at or near equilibrium, the r
must reduce to give the equilibrium conversion (and the concentra!
that are derived from thermodynamics
See ab
See above.
See above
Reinforces the point that the equilibrium conversion will
flow system and a constant volume batch system.
3nd Edition, Solution Manual, Chapter 3
135, Uses the rate law and stoichiometric table to give practice at expressing
A=/(X) In Chapter 2 the student saw that once one has -r,=/(X) a number
P36. Straigh blem to set up a stoichi table and then substitute
numbers 10 cal
P347. Alternate
P3:18, This problem to expr
straight forward. Reinforces the {
function
hapter
P59, Requires the student 4 book for ation and
practice life-long
P3.20. Usually assigned at graduate le Land
Problems P3-21, P3-22, and P3-23 a
reactions with phase €
CDP3A Similar to th
CDP3.B Similar to problem
Points out that
Alternative
lation
(psc) Yes
Yes
135, Uses the rate law and stoichiometric table to give practice at expressing
A=/(X) In Chapter 2 the student saw that once one has -r,=/(X) a number
P36. Straigh blem to set up a stoichi table and then substitute
numbers 10 cal
P347. Alternate
P3:18, This problem to expr
straight forward. Reinforces the {
function
hapter
P59, Requires the student 4 book for ation and
practice life-long
P3.20. Usually assigned at graduate le Land
Problems P3-21, P3-22, and P3-23 a
reactions with phase €
CDP3A Similar to th
CDP3.B Similar to problem
Points out that
Alternative
lation
(psc) Yes
Yes
jon, Solution Manval, Chapter 3
CDP+A
Cor
cDP3C
CDF+-D
CDP3E
alculation required
CD-ROM. For ex
«ans that one of the
of the alternates art
and chug) d
equations orá
ple A = COPLA.
CDP+A
Cor
cDP3C
CDF+-D
CDP3E
alculation required
CD-ROM. For ex
«ans that one of the
of the alternates art
and chug) d
equations orá
ple A = COPLA.
Chapter 3
P3:1 No solution wil be gi
nd b will tay be same while the reaction rates will increase by a
Rate law should à from an experimental observion
= KC aa the reaction rate is zero-order wih respect to NaOH,
I nechanisms »C (low temp)
Cake wil burn on he ou
at transfer so reaction tempera
the outside
BAL ives: Frequency of Hashing ot fire!
P3:1 No solution wil be gi
nd b will tay be same while the reaction rates will increase by a
Rate law should à from an experimental observion
= KC aa the reaction rate is zero-order wih respect to NaOH,
I nechanisms »C (low temp)
Cake wil burn on he ou
at transfer so reaction tempera
the outside
BAL ives: Frequency of Hashing ot fire!
9 Plot Speed otre Honey bee
From the grap
4 Kal
Tiesto a 40°C,
M = -13 37203 73
Ve bases
Ausc,
Bm
Va 0005 cave
Vafomunately forthe bee, iu
From the grap
4 Kal
Tiesto a 40°C,
M = -13 37203 73
Ve bases
Ausc,
Bm
Va 0005 cave
Vafomunately forthe bee, iu
B33 cont'd
From the graph
En 77613 kcalimole
Bump. Te clases it ee
ants and bees have à
round the mic of the clu
BAS ira cence
From the graph
En 77613 kcalimole
Bump. Te clases it ee
ants and bees have à
round the mic of the clu
BAS ira cence
Bs, Solution
following esos
following esos
p36 cont'd
le will be the
or flow reactor,
le will be the
or flow reactor,
3-2 cont'd
Gas Phase, constant
BETEN
eno (+1
vv (190
Gas Phase, constant
BETEN
eno (+1
vv (190
B3-Z cont'd
Gas Phase, Isothermal and no pressure drop
Tide
(=x) _ (0.092.1-x)
DT: Ox
Gas Phase, Isothermal and no pressure drop
Tide
(=x) _ (0.092.1-x)
DT: Ox
P28 cont'd
E
yan =
em 748 =0. af:
The inert Nitrogen would also change the values of ya and €.
2) AB — eC
Mole Balance for a CSTR:
Fao-Fa= HV
E
yan =
em 748 =0. af:
The inert Nitrogen would also change the values of ya and €.
2) AB — eC
Mole Balance for a CSTR:
Fao-Fa= HV
Steichionem
8 POLYMATH or similar sotovas p
8 POLYMATH or similar sotovas p
ii
PER
LY MATH Oc similar software package.
PER
LY MATH Oc similar software package.
539.1)
Consequently, we know thatthe rte law most reduced to Equation (PS3-9.1) when a = 0
‘Therefore
tant [Ca
2] 28392)
At equilibrium «ra = 0 and equation (PS3-9.2) becomes upon
Ka Pa + Ka Po
Consequently, we know thatthe rte law most reduced to Equation (PS3-9.1) when a = 0
‘Therefore
tant [Ca
2] 28392)
At equilibrium «ra = 0 and equation (PS3-9.2) becomes upon
Ka Pa + Ka Po
P3-10cont’a
High termperarures
AS —_Kr |
ASTRA PA Ro Pi
eau 24 4983 2C+
Ex Sab Farma Tocata
Tapiz SR
reg FT
Oxygen 3 MEX Fi) Eso (da 90 P Ta
ose) FT
Pathalic Aabyérido Box Fox Box pre
mo (+80 POT
Carbon Dioxide 210 X 2EoX PT
sao (1900 POT
water 0 20x
CUS]
V2 ye=00 = Yas"5 = 0039-112 = 00175
FART/P, 125007
Co = Cantey 8099
don
ra (+ €X) = Fae (1484) (+ OD RT.
7
High termperarures
AS —_Kr |
ASTRA PA Ro Pi
eau 24 4983 2C+
Ex Sab Farma Tocata
Tapiz SR
reg FT
Oxygen 3 MEX Fi) Eso (da 90 P Ta
ose) FT
Pathalic Aabyérido Box Fox Box pre
mo (+80 POT
Carbon Dioxide 210 X 2EoX PT
sao (1900 POT
water 0 20x
CUS]
V2 ye=00 = Yas"5 = 0039-112 = 00175
FART/P, 125007
Co = Cantey 8099
don
ra (+ €X) = Fae (1484) (+ OD RT.
7
Pen CRT» Cod: 92X) PRT
100 Po
cet-2m0 »
des Pe
van (le 00 PoP
ET
ftom part a
1-0
from pared kE (5820020. re)
dso
tank Coca
from par a amd * AU 20? + (92-92%
Gs
from par rank "CEA (97.9030. ES
1-00
By asin pare oxygen instead far, e concentration ofthe rar ace higher and the reaction
rat ie dherefove fs,
No answer willbe given
Ret,
100 Po
cet-2m0 »
des Pe
van (le 00 PoP
ET
ftom part a
1-0
from pared kE (5820020. re)
dso
tank Coca
from par a amd * AU 20? + (92-92%
Gs
from par rank "CEA (97.9030. ES
1-00
By asin pare oxygen instead far, e concentration ofthe rar ace higher and the reaction
rat ie dherefove fs,
No answer willbe given
Ret,
T= 227€ = 500€
Ho
3-11
Ho
3-11
Palacenta
DMA
Mole balance fora PER.
de
wv
ER
Mole balance fora PER.
de
wv
ER
Pa-J3conva
Combine wih equaco 340 1943-48
a, 3 (ti
(Er +)
cs predicts the reacting species in a reversible
lated by the equation given below.
Plug Flow Reactor (PFR):
O 0 =x)
ro de
ne CEE A
ar
Combine wih equaco 340 1943-48
a, 3 (ti
(Er +)
cs predicts the reacting species in a reversible
lated by the equation given below.
Plug Flow Reactor (PFR):
O 0 =x)
ro de
ne CEE A
ar
P2eL4coat’d
The equations above were solved using POLYMAT The
POLYMATH notation are shown below.
The equations above were solved using POLYMAT The
POLYMATH notation are shown below.
jilibrium Conversion asa
Function of Entering Inert Mole
rhum conversion highrise PER because number of mole ol qu increases ding a
non lnresing volume. Sie buch is conta volume: change mes fan de
Ánfluence the system. ù
Mmamber of moles of pas says corsa PER an con
Sucio conversion. If number of moles gas
clave high equiitrium conver cn.
Pais Hy + Br; > 2HBr, Given that yur
Yi=0.50, Po 10 am, To =
The kinetic express
Function of Entering Inert Mole
rhum conversion highrise PER because number of mole ol qu increases ding a
non lnresing volume. Sie buch is conta volume: change mes fan de
Ánfluence the system. ù
Mmamber of moles of pas says corsa PER an con
Sucio conversion. If number of moles gas
clave high equiitrium conver cn.
Pais Hy + Br; > 2HBr, Given that yur
Yi=0.50, Po 10 am, To =
The kinetic express
0.0097, 0- Xi (motes
THE
THE
P3-1Seont'd
(2) Constant Volase (sas next pago)
Ce * Geo (1-2) : 6, 3
7 7 co
Pe = CRT = Cogir GR
DEC = CgghT X= Fp
ART (x) SER
TR CogRTO-X)+ Ky CogRTX
ar
an
T+ Coût RL A ae]
Consider the general ense LÉ which both P and T ace varying
Sept
“EIA
Ceoktt1-1
OTRO
Siailacly for B and P
Geo
CCS
AT
M SE
RI
(2) Constant Volase (sas next pago)
Ce * Geo (1-2) : 6, 3
7 7 co
Pe = CRT = Cogir GR
DEC = CgghT X= Fp
ART (x) SER
TR CogRTO-X)+ Ky CogRTX
ar
an
T+ Coût RL A ae]
Consider the general ense LÉ which both P and T ace varying
Sept
“EIA
Ceoktt1-1
OTRO
Siailacly for B and P
Geo
CCS
AT
M SE
RI
The above expression ca be ateplified further by cotter that
or
or
Pä-Léconta
Stotenione
A)
»
‘noo * ¥40
= xe,
acto
2 cadmio? ama
+ 149007? -
demo
np (LODO ca
Stotenione
A)
»
‘noo * ¥40
= xe,
acto
2 cadmio? ama
+ 149007? -
demo
np (LODO ca
Pa-l6convd
Pron Eqsation (13.120).
ELIL Solution
5 lid
Rate Cave Le
1 the equilibrium co
law, and solve for Xe
à re > assume constant volume
HG = Cle XX Ce ACK
aversion, set-ra = 0, combine stoichiometry and rate the rate
meer)
(139 eme A
220 3 wet 0 ma
Steichlomeny:
keys“
3-29
Pron Eqsation (13.120).
ELIL Solution
5 lid
Rate Cave Le
1 the equilibrium co
law, and solve for Xe
à re > assume constant volume
HG = Cle XX Ce ACK
aversion, set-ra = 0, combine stoichiometry and rate the rate
meer)
(139 eme A
220 3 wet 0 ma
Steichlomeny:
keys“
3-29
30.514009,
(20030)
same reaction, rate law, and inal concent
Seichlomeuy:
Cam Naf Ve Nu (l-20/Vo= Cua
Cen Ne/V a 3NoX/Vo=30X
Combine and Sole for Xe:
KeCao(l Xe) = (Cao Xe?
Xe2039
3-50
(20030)
same reaction, rate law, and inal concent
Seichlomeuy:
Cam Naf Ve Nu (l-20/Vo= Cua
Cen Ne/V a 3NoX/Vo=30X
Combine and Sole for Xe:
KeCao(l Xe) = (Cao Xe?
Xe2039
3-50
De equiirium concent
Cas (0.3051 -039) = 019
Ces = (09050392036
js phase remedo a const prestas; bee
O]
Stoichiometry
En ya bag = Zand De O
Cam Nal V Nu (301 ONE REN
Nel Ve INA! (Yo (+ EI = Cn SKC 2X)
Combine and tlve fee Xe
(Cae Ke u +2)
Fad equiitiam con
Cum 0305 (1-058) {1 + 21058) #0059 mol fee?
= 30305) (0.58) /(1 +20.58) = 0.246 mal dr?
(Solusons by JT. Santi 363
23-18 Given: Gas phase reaction À + B > C in a Batch reactor fitted with a piston such
V=0.1Pg k= 10 (bci
SA=kCA Cp N, =
T= 140°C = 600°R = Constant
8-81 e=roba3
Teel Peto
Ova OO or Ve Po=10Vo
Se
Cas (0.3051 -039) = 019
Ces = (09050392036
js phase remedo a const prestas; bee
O]
Stoichiometry
En ya bag = Zand De O
Cam Nal V Nu (301 ONE REN
Nel Ve INA! (Yo (+ EI = Cn SKC 2X)
Combine and tlve fee Xe
(Cae Ke u +2)
Fad equiitiam con
Cum 0305 (1-058) {1 + 21058) #0059 mol fee?
= 30305) (0.58) /(1 +20.58) = 0.246 mal dr?
(Solusons by JT. Santi 363
23-18 Given: Gas phase reaction À + B > C in a Batch reactor fitted with a piston such
V=0.1Pg k= 10 (bci
SA=kCA Cp N, =
T= 140°C = 600°R = Constant
8-81 e=roba3
Teel Peto
Ova OO or Ve Po=10Vo
Se
Na=Nao(1-X1 nos Mao=10
KOENNEN Naa =UnaPoBT Vo
Bros
[ZueRY xy
LAT ) (+)
LO (> / (Emo) hee) (0.75 76 UTR GOR? AR
do
“428.03 + 10% (Lx mot / ace
ess
Va Voti 09
2015 (1 + 320)
10% pmol ee
‘There will be no solution given.
There will be no solution given.
AU get, + cay ce, + ae
Tease. ds Pr 950 hPa = 9.39 ate
SL, vapor pressure o = 094 ate = Py
3-35
KOENNEN Naa =UnaPoBT Vo
Bros
[ZueRY xy
LAT ) (+)
LO (> / (Emo) hee) (0.75 76 UTR GOR? AR
do
“428.03 + 10% (Lx mot / ace
ess
Va Voti 09
2015 (1 + 320)
10% pmol ee
‘There will be no solution given.
There will be no solution given.
AU get, + cay ce, + ae
Tease. ds Pr 950 hPa = 9.39 ate
SL, vapor pressure o = 094 ate = Py
3-35
anbseitacing Ce
Aires conde
Creplentor of
ea) cy = Cf
$CuXC4 SVS
Cat NES 9-0:
CASOS
15 und Eg = 0.0658 gaol/ı
0
3-3
Aires conde
Creplentor of
ea) cy = Cf
$CuXC4 SVS
Cat NES 9-0:
CASOS
15 und Eg = 0.0658 gaol/ı
0
3-3
Bä:21contà
EEE)
Fe = Mor
Eon
Pre)
Ac
EEE)
Fe = Mor
Eon
Pre)
Ac
jpecies Symbol b After condensation Py =
Pe remaining
FR)
BEd IX
o FX TO
Pe remaining
FR)
BEd IX
o FX TO
B3-22cont’a
E —
Fa AIN
RL) ZII INGA
Ark
ELIO.
PUT
E —
Fa AIN
RL) ZII INGA
Ark
ELIO.
PUT
Pa22conva
Cho
where Fag = 0.106 4 tho = 0.212 anol’
After condensation
Note exar at X = 0,609 (veginaise of condensation) €, = Ci
Cho
where Fag = 0.106 4 tho = 0.212 anol’
After condensation
Note exar at X = 0,609 (veginaise of condensation) €, = Ci
Efi]
TEEN
Yu =05
Es pa5=0 + 12-3-6)23
TEEN
Yu =05
Es pa5=0 + 12-3-6)23
P3.23cont'd
Cra ÇA
(=x)
1728
16810 Gees
3
2.06" 10* x
1738
Cra ÇA
(=x)
1728
16810 Gees
3
2.06" 10* x
1738
{Ch + 3H > 281 (5) + TH+
Take $, HC, as bass
Si +:
Take $, HC, as bass
Si +:
CDPLB cont'd
Sees Symbol Er Gms Leming
SiHOS @ Fo Fux Fama)
Fe) Fao" @sFro -FaoX Fa = Fao (On - x)
£ o E Few E
o rex F=lrox
o Brox
Pressure, Neglect the vapor pressure of S()
5 only involves the changes in gasp
Cao = 0.0088 moe?
28 RG 5,4
lat Par (a)
Sees Symbol Er Gms Leming
SiHOS @ Fo Fux Fama)
Fe) Fao" @sFro -FaoX Fa = Fao (On - x)
£ o E Few E
o rex F=lrox
o Brox
Pressure, Neglect the vapor pressure of S()
5 only involves the changes in gasp
Cao = 0.0088 moe?
28 RG 5,4
lat Par (a)
UPI Given sks B=)
Morros
e
Morros
e
NX 1
ANAL Nasl 2%)
Nak Kader
Nas —Naolly*X) Nan
Final cons. of smncsin =
[Po $toiehiometrie table for « Stow system eaing ammonia as the dasis
Species Symbol Eneieg Change
Armonía B
Benzo chlor
Denzylamde .c
Ammonium Che "D
Concentration
Fao FX Fate) Fi
A Go FuX2 Fox X2) Faolôs Ps
ScFao FX? Fao(®c+X2) Fuori.
Domo Fuck Faul@oeX/2) Fanta ig
BONO toe the Flor ayston esta ammonia an the baste is difte
Pisten in car fetter
Fy
De.
»
Molar flow rates considered setter than suaber of soler.
Plmerie r
ANAL Nasl 2%)
Nak Kader
Nas —Naolly*X) Nan
Final cons. of smncsin =
[Po $toiehiometrie table for « Stow system eaing ammonia as the dasis
Species Symbol Eneieg Change
Armonía B
Benzo chlor
Denzylamde .c
Ammonium Che "D
Concentration
Fao FX Fate) Fi
A Go FuX2 Fox X2) Faolôs Ps
ScFao FX? Fao(®c+X2) Fuori.
Domo Fuck Faul@oeX/2) Fanta ig
BONO toe the Flor ayston esta ammonia an the baste is difte
Pisten in car fetter
Fy
De.
»
Molar flow rates considered setter than suaber of soler.
Plmerie r
rd Edition, Solution Manual, Chapter 4
chi
General he h
students to re
the algorith
\e purposes ol
text and lecture
u
¿ho
P41, This problem gives
problems. If you think ot
vg
parts (0) through (i). The hat
Shows a CRE application in the foo
Many of our stud
Jed this,
pro
action in which par
Liquid phase
Pan tli opereended and
me actes fs O, er
"Dry,
Realliguid phase resction and dat,
‘This problem is somewhat
figure why the pressure con
conversion. Ask the students
hi able, I would only assig
assigned at the sam
4 do
the student
ines should be ad
aps (i) or (9) and then
ssigned coulé
god ind:
to give an example of re
forward
CRE pr
Pad. Pa.
is GREATER
brainstorming
part (a), especially if part (e)
is to encourage the
his chapter reinfomy
to develop o
sd send me an
to the next pri
kinetics in th
time is ch
straight forward MN
tions, Part AN
3 and Pi
than the
Depending o
peri
chi
General he h
students to re
the algorith
\e purposes ol
text and lecture
u
¿ho
P41, This problem gives
problems. If you think ot
vg
parts (0) through (i). The hat
Shows a CRE application in the foo
Many of our stud
Jed this,
pro
action in which par
Liquid phase
Pan tli opereended and
me actes fs O, er
"Dry,
Realliguid phase resction and dat,
‘This problem is somewhat
figure why the pressure con
conversion. Ask the students
hi able, I would only assig
assigned at the sam
4 do
the student
ines should be ad
aps (i) or (9) and then
ssigned coulé
god ind:
to give an example of re
forward
CRE pr
Pad. Pa.
is GREATER
brainstorming
part (a), especially if part (e)
is to encourage the
his chapter reinfomy
to develop o
sd send me an
to the next pri
kinetics in th
time is ch
straight forward MN
tions, Part AN
3 and Pi
than the
Depending o
peri
Pio,
point out
à Kepner Tre
Prentice Ha
and 10)
Strat
problem could be alternated with P4-4 a
Problems P4-11, P4-12, Pa
Pas,
Paz.
Problems 74-18,
Paas,
Pars,
Pe
Pen
Pan.
for the
still may be true
toughest
Ther
hour per problem
This problem requir
st use a model balance L
This problem also s
CRE principles are ap
PLAS, 7120
ng pressure d
Pia
Pas
Trial and error solution using POLY
constant, k, or the pressure drop p:
Trial and T
of the analysis rec
or using POLYMA
red in PS-21
Parts (a) and ) orward, plus
through (g) requires
depending on how ma
some anal
Fa ight forward problem using POLYM:
what could
Pechap
he
and i
it not th
to pá
lat
lem because
of ECOLOGIC
web module
shorter versi
Parts (o)
can be quite long
to P4-20,
point out
à Kepner Tre
Prentice Ha
and 10)
Strat
problem could be alternated with P4-4 a
Problems P4-11, P4-12, Pa
Pas,
Paz.
Problems 74-18,
Paas,
Pars,
Pe
Pen
Pan.
for the
still may be true
toughest
Ther
hour per problem
This problem requir
st use a model balance L
This problem also s
CRE principles are ap
PLAS, 7120
ng pressure d
Pia
Pas
Trial and error solution using POLY
constant, k, or the pressure drop p:
Trial and T
of the analysis rec
or using POLYMA
red in PS-21
Parts (a) and ) orward, plus
through (g) requires
depending on how ma
some anal
Fa ight forward problem using POLYM:
what could
Pechap
he
and i
it not th
to pá
lat
lem because
of ECOLOGIC
web module
shorter versi
Parts (o)
can be quite long
to P4-20,
ard Edition, Solution Manual, Chapter 4
eral times to plol
mum catalyst size
se CDPA-1, and CDP.
7929. Apı unsteady. state
POLYMATH to solve three ¢
P430. Membrane reactor. There should ime for the
out part (4) parameter var id ask “What if...” quest
PLL, Membrane reactor problem that is an alternative to 14-30.
CDP4-A Good problem concerning b d on the web
CDPEB Alternative to problem P4-17.
CDP4C Reversible batch reaction
COPED Alternative to CDP4-13
CDPLE California p
CDPLF Very straight fo
CDP4G Optimization problem -
CDPHH Very
corns
COPE} Most straight forward problem in
Problems CDPS-K through CDP4-M a
CDPA-N Radial flow reactor. N
CDPLO Bacteria growth ir
eral times to plol
mum catalyst size
se CDPA-1, and CDP.
7929. Apı unsteady. state
POLYMATH to solve three ¢
P430. Membrane reactor. There should ime for the
out part (4) parameter var id ask “What if...” quest
PLL, Membrane reactor problem that is an alternative to 14-30.
CDP4-A Good problem concerning b d on the web
CDPEB Alternative to problem P4-17.
CDP4C Reversible batch reaction
COPED Alternative to CDP4-13
CDPLE California p
CDPLF Very straight fo
CDP4G Optimization problem -
CDPHH Very
corns
COPE} Most straight forward problem in
Problems CDPS-K through CDP4-M a
CDPA-N Radial flow reactor. N
CDPLO Bacteria growth ir
Ard Edition, Solution Man
corer
Solutio
corer
Solutio
3rd Edition, Solution Manual, Chap
© COPIE
CDP4K
core
CDPIM
CDPEN
corso
Cops
corsO
Assig
Always assigned, AA = Always assign one from the group of alter
Often 1 jently, $ = Seldom, G = Graduate level
Difficulty
ht fo a
te cal )
IC = Intermediate calculation require:
M
© COPIE
CDP4K
core
CDPIM
CDPEN
corso
Cops
corsO
Assig
Always assigned, AA = Always assign one from the group of alter
Often 1 jently, $ = Seldom, G = Graduate level
Difficulty
ht fo a
te cal )
IC = Intermediate calculation require:
M
‘xd Edition, Solution Manual, Chapter 4
res
| Gas Phase
Straight Forward
Fairly Straight
More Difficult
Intermediate Calculations |
Parameter Variation
res
| Gas Phase
Straight Forward
Fairly Straight
More Difficult
Intermediate Calculations |
Parameter Variation
paz
(a) Cooking food (effet oft
concentration), dissolution
(6) Accept catalyst
1) X is small ar
Reject east
1) Pressure drop now too high
2) Catalyst nor as active as il
(© Bad idea for 100 rubes in series. Pressu
mass flow ra
POLYMATH and the set of equations given in ex
function of exalyst weight
Wein
run it so that they do not
there is
(a) Cooking food (effet oft
concentration), dissolution
(6) Accept catalyst
1) X is small ar
Reject east
1) Pressure drop now too high
2) Catalyst nor as active as il
(© Bad idea for 100 rubes in series. Pressu
mass flow ra
POLYMATH and the set of equations given in ex
function of exalyst weight
Wein
run it so that they do not
there is
42 cont'd
H program given la the
example and make Fag af io action of the inert and use
the equation
PATA
to find the conversion, Record the results and graph the two conversions
as a function of mole fraction:
Conversions vs. mole percent inert
‘The advantages of the inert are obvious by the graph; as more
added the conversion ine S that the more inert
5 of you will create
um value of Fa decreases, w
H program given la the
example and make Fag af io action of the inert and use
the equation
PATA
to find the conversion, Record the results and graph the two conversions
as a function of mole fraction:
Conversions vs. mole percent inert
‘The advantages of the inert are obvious by the graph; as more
added the conversion ine S that the more inert
5 of you will create
um value of Fa decreases, w
42 cont'd
3. KK. Ko ao, it can be determined that keeping
0 high would opimize the process geting the
Ke has lite effect on the resction as a whole. Keepin
Fo low also» the volume down, but on the other band it keeps it
From having too much reacted. Ifthe temperature were raised the volume
d would
Fiswrate
3. KK. Ko ao, it can be determined that keeping
0 high would opimize the process geting the
Ke has lite effect on the resction as a whole. Keepin
Fo low also» the volume down, but on the other band it keeps it
From having too much reacted. Ifthe temperature were raised the volume
d would
Fiswrate
maximum conceatation of Cand D in Example 4
profit + so. 04*3600"
[o.04+.425+30,06+ 360024365]
[o.0:3+0.0736 *3600+24*365]-5,000,000
[200,000,000 0:36]
hey wil aot make as
Kooks unfavorable, However, etiylene oxide
will be mi ame location
profit + so. 04*3600"
[o.04+.425+30,06+ 360024365]
[o.0:3+0.0736 *3600+24*365]-5,000,000
[200,000,000 0:36]
hey wil aot make as
Kooks unfavorable, However, etiylene oxide
will be mi ame location
42 cont'd
removing con!
protective clothing
y gogales, prevent contact with warm surface.
removing con!
protective clothing
y gogales, prevent contact with warm surface.
Mole balance:
Rate law and stoichiometry:
Specific
Combine:
x =108
Ca=Caoti-X)
nd molecular weight are constant the equation can be rewriven
maot1-X)
TU = myo (1-0.108)
myo = 7287 10
co
Rate law and stoichiometry:
Specific
Combine:
x =108
Ca=Caoti-X)
nd molecular weight are constant the equation can be rewriven
maot1-X)
TU = myo (1-0.108)
myo = 7287 10
co
) KGO°C)
0.0048 weeks
2 be rewiten a5
0.0048 weeks
2 be rewiten a5
= 0.07 dm’/molmin
En = 20,000 calmo!
Rate Law: ra=kCaCo
a CSTR 2200 am”
Using te Anis equon atthe CSTR x k the new specific
(200m? Ras
X=
From te quadratic equate
PBR’ -V=800 dm?
Design Equation
En = 20,000 calmo!
Rate Law: ra=kCaCo
a CSTR 2200 am”
Using te Anis equon atthe CSTR x k the new specific
(200m? Ras
X=
From te quadratic equate
PBR’ -V=800 dm?
Design Equation
P4-S (conta)
b) BatchReacior V=20 X=0.90 Assume Iso
Desig
Find the specific reaction rate a P 73 K using the Actbenius
Equation
K=2.51x10
(200x0)
Ga x 101200)
d) Athigher temperan action may be extremely high, If he react
extremely exolhermi
e) Assume for es
temperature. The
batches ad
In one day. the reactor p
D The points
(1) To ro the significa
(3) Not io be co
red or black
b) BatchReacior V=20 X=0.90 Assume Iso
Desig
Find the specific reaction rate a P 73 K using the Actbenius
Equation
K=2.51x10
(200x0)
Ga x 101200)
d) Athigher temperan action may be extremely high, If he react
extremely exolhermi
e) Assume for es
temperature. The
batches ad
In one day. the reactor p
D The points
(1) To ro the significa
(3) Not io be co
red or black
nr te
ñ
D
AS
Fan (FM) Ee
ñ
D
AS
Fan (FM) Ee
ici (= YS-x)
F
The above equation relates th
achieved during that bsteh, There iss Ih conversion and few
batches and low conversion but many batches per day. What conversion will
result inthe
MX 24
(52x
F
The above equation relates th
achieved during that bsteh, There iss Ih conversion and few
batches and low conversion but many batches per day. What conversion will
result inthe
MX 24
(52x
The minimum occurs at X = 0.82 and e
X= 090%
IR
= %)=
rar
ations in the POLYMATH
IR
= %)=
rar
ations in the POLYMATH
PAT (con's)
Swichiomes
a.
e=2 and €,
Kumosz
X=(090)x,, «047
Swichiomes
a.
e=2 and €,
Kumosz
X=(090)x,, «047
Benzene
PRR
Design Equation
Rate Law:
Stoichiometry
PRR
Design Equation
Rate Law:
Stoichiometry
4-10 (cont'd)
Combine
(one:
3032)
(x-1145 )
Equation
Stoichiometry
Combine:
Combine
(one:
3032)
(x-1145 )
Equation
Stoichiometry
Combine:
P4.10 (ont)
& bours. Assun
A day there can be 6 runs per day,
(© — E=30202 butmol
forX=0,
and other down time assume that
abe used twenty-four Bours
& bours. Assun
A day there can be 6 runs per day,
(© — E=30202 butmol
forX=0,
and other down time assume that
abe used twenty-four Bours
Pati (coat
: 1 28.000 ext
as er 1
ÓN
y,
E
J
: 1 28.000 ext
as er 1
ÓN
y,
E
J
PA12 (conri)
P4-13 (cont'd)
Stace 7, = 1.0, 0790 5
Stace 7, = 1.0, 0790 5
P4-13 (cont'd
Pé17 (cont'd)
@) Evaporation
anta" ax)
Alsa
Quí2.110=3)-0.5
faoccaorve
evaporation
rays 2 0)
evaporation
and condensation
ra ve 2 0)
and condensation
@) Evaporation
anta" ax)
Alsa
Quí2.110=3)-0.5
faoccaorve
evaporation
rays 2 0)
evaporation
and condensation
ra ve 2 0)
and condensation
4) Making the ratio QC, smal postive number immediately pus the
conversion close o 1, Coaverly making that tio any negative number makes the
Conversión go down very quickly.
©) No solution will be given
2) Assume isothermal and e=0
therefore, P=Po(l-aW’ 110
Wa 995,
te from X=0 10 X=.05
Last 5% conversion integrate from X=35 to X=90
Wal0.85
4-37
conversion close o 1, Coaverly making that tio any negative number makes the
Conversión go down very quickly.
©) No solution will be given
2) Assume isothermal and e=0
therefore, P=Po(l-aW’ 110
Wa 995,
te from X=0 10 X=.05
Last 5% conversion integrate from X=35 to X=90
Wal0.85
4-37
P4-19 (conta
D Since ted
HINDERN
By using tal and enoris POLYMAT
[zen
por
1e
Thal and exo by POLYMATI gas uste following
Guess ae MX 006 y= 123
Brlations;
BG) 140.) ane (2-1) trop»
WW) au = (1093050
Alb» 000946
8.000101
wi
vr,
D Since ted
HINDERN
By using tal and enoris POLYMAT
[zen
por
1e
Thal and exo by POLYMATI gas uste following
Guess ae MX 006 y= 123
Brlations;
BG) 140.) ane (2-1) trop»
WW) au = (1093050
Alb» 000946
8.000101
wi
vr,
P4-20cont'd
We will make the approximation that c'X will be neglected with respect to 1 and
then use the above equations. k il increase as D, will decrease as per Chapter 12
and tha alpha increases as D, decreases so that y decreases and there isa
competition forthe effet on also decreases as G
e) No solution willbe give
We will make the approximation that c'X will be neglected with respect to 1 and
then use the above equations. k il increase as D, will decrease as per Chapter 12
and tha alpha increases as D, decreases so that y decreases and there isa
competition forthe effet on also decreases as G
e) No solution willbe give
pez,
a PR
D PER
MB:
Rate Law
Combine
‘is inversely proportion to cross sectional area and therefore the pipe
Sine this stn
ameter so ineressing the ameter wil
internal difusion, the specific reaction rate is inversely proportional to
ce the conversion itis wise t ne
a PR
D PER
MB:
Rate Law
Combine
‘is inversely proportion to cross sectional area and therefore the pipe
Sine this stn
ameter so ineressing the ameter wil
internal difusion, the specific reaction rate is inversely proportional to
ce the conversion itis wise t ne
erage
iter way tsa goodies
44
iter way tsa goodies
44
P4-23 (cont)
Combine
Be:
no 2200100
Kö
ea,
os
Combine
Be:
no 2200100
Kö
ea,
os
Mole Balance
2. Ratelaw 1 =-kCipc(l-9)
Where pe ¿and @ are unknown constants, They willbe grouped into one unknown,
constant ky.
k. =k pelle)
(ex)
ex)’
cy
3. Stoichiomeiny: Cy =Cy
4. Pressure Deo (+)
az
when neglecting the tube isgiventy
EOS
Lrspie |
1 ofthe bracketed variables are unknown constant
The mass flow rates also and unknown consta canbe writen asa constat,
B,overeross - sectional area.
so-e) ul
Where B is a function ofthe mass flow r properties, and catalyst propertes—all
of which are constant for par (a). Note aly toa tubular
PBR and spherical PBR. To find B and ko we must mod
POLYMATH by entering the above equations in édition
Ae = 2 Cam} =516dm
B and kz must be arbitrarily hos “Then, depending on whether POLYMATH
gives high or low values of y and X, one can converge on the true values of ke and B. In
Order to do so efficiently, however, one must make use of the following rende: (Note
thatthe following table is completely true only when € > 0.)
4-45
2. Ratelaw 1 =-kCipc(l-9)
Where pe ¿and @ are unknown constants, They willbe grouped into one unknown,
constant ky.
k. =k pelle)
(ex)
ex)’
cy
3. Stoichiomeiny: Cy =Cy
4. Pressure Deo (+)
az
when neglecting the tube isgiventy
EOS
Lrspie |
1 ofthe bracketed variables are unknown constant
The mass flow rates also and unknown consta canbe writen asa constat,
B,overeross - sectional area.
so-e) ul
Where B is a function ofthe mass flow r properties, and catalyst propertes—all
of which are constant for par (a). Note aly toa tubular
PBR and spherical PBR. To find B and ko we must mod
POLYMATH by entering the above equations in édition
Ae = 2 Cam} =516dm
B and kz must be arbitrarily hos “Then, depending on whether POLYMATH
gives high or low values of y and X, one can converge on the true values of ke and B. In
Order to do so efficiently, however, one must make use of the following rende: (Note
thatthe following table is completely true only when € > 0.)
4-45
P4-24 cont'd)
POLYMATHy [POLYMATHR] Remedy [Effect ong] Eifec oa
150 RIGH | too HIGH T Towers X
100 HIGH WoLOW | mise ES
co HIGH | vert, Towers
so LOW | TowerB | ay] ax]
(Once the comect values of an a are found, one can proceed with the standart
spherical algo
Y Fonrocrari
BayP = 973"1500=1459.5
is fats)
amo.
Facade
ict
Elo
wen
aly
Scones
Ce
Bin.
POLYMATHy [POLYMATHR] Remedy [Effect ong] Eifec oa
150 RIGH | too HIGH T Towers X
100 HIGH WoLOW | mise ES
co HIGH | vert, Towers
so LOW | TowerB | ay] ax]
(Once the comect values of an a are found, one can proceed with the standart
spherical algo
Y Fonrocrari
BayP = 973"1500=1459.5
is fats)
amo.
Facade
ict
Elo
wen
aly
Scones
Ce
Bin.
P4-24 cont'd)
xd
1) From POLYMATH
maxim Nowrate= 1750001
(0) Fao = 1750
xd
1) From POLYMATH
maxim Nowrate= 1750001
(0) Fao = 1750
PA.24 (ond)
9) From POLYM
Minimum presare 80 kPa
ews
9) From POLYM
Minimum presare 80 kPa
ews
ont)
(©) Using the same POLYMATH equations, and changing values for V and ve
Change ve from 0,050 drs from 0.025 dm
0.025
(©) Using the same POLYMATH equations, and changing values for V and ve
Change ve from 0,050 drs from 0.025 dm
0.025
P4-26 (cont'd)
Change V from 200 dm? o 100 dm!
4-26 (e) V=100
Change V from 200 dm? o 100 dm!
4-26 (e) V=100
4.26 (cont'd)
(€) Evaporadon 0f0.5 mals changes the mole balance for spec
EA
Simulations show that Ny is always negative, Ther
sx,
where Ny
See POLYMATH solution below.
The equations:
arar!
‘dtt)8, 05810, 99-rv01
reel
ng 00,0
xt Ont -5) XEna val}
sasSt4-n07/1544
x=. 05810, 31 -n99/4,05)
Initial values:
age 0.0
Final value
(€) Evaporadon 0f0.5 mals changes the mole balance for spec
EA
Simulations show that Ny is always negative, Ther
sx,
where Ny
See POLYMATH solution below.
The equations:
arar!
‘dtt)8, 05810, 99-rv01
reel
ng 00,0
xt Ont -5) XEna val}
sasSt4-n07/1544
x=. 05810, 31 -n99/4,05)
Initial values:
age 0.0
Final value
P4-26 (cont'd)
Using the same equations in POLYMATH
varying vo and V.
Vo 100 dm? and va = 0.025 dns
4-26 (e) V=100 and vo = .025
Using the same equations in POLYMATH
varying vo and V.
Vo 100 dm? and va = 0.025 dns
4-26 (e) V=100 and vo = .025
4.26 (@) V=3
4.26 (e) V=300 and vo = -100
4.26 (e) V=300 and vo = -100
D Mo = Cyog = Bate
P4:27 (cont)
MU = = ote + à
atico -
150200)
MU = = ote + à
atico -
150200)
Hain tata)
200
308
500
350
“oo
Fremen
200
308
500
350
“oo
Fremen
‘This problem was solv
a,b, and care given
(cat )/d(t)=({(voa/uo)*cao-ca)/tau)-keeal **2
dlca2)/diU=(cat-ca2)/tau)-krcaze"2
úlca3)/d(Ua((ca2-ca3)/tau)-k*cas*
tau=10
K=0.025
cao=2
Voa=18
vo=28
N={unascao-uo*ca5)/(uoarcao)
tg=0
falg=o
fa2y-0
tasg- 0
tr= 78
»
From POLYMATH, the steady state conv
sion of A is approximately
039,
(0:959(0.6110)=(0.605)
A concentration of 0.605 mol/dm:
approximately 610 minutes.
Below is a plot of
function of time
4-60
a,b, and care given
(cat )/d(t)=({(voa/uo)*cao-ca)/tau)-keeal **2
dlca2)/diU=(cat-ca2)/tau)-krcaze"2
úlca3)/d(Ua((ca2-ca3)/tau)-k*cas*
tau=10
K=0.025
cao=2
Voa=18
vo=28
N={unascao-uo*ca5)/(uoarcao)
tg=0
falg=o
fa2y-0
tasg- 0
tr= 78
»
From POLYMATH, the steady state conv
sion of A is approximately
039,
(0:959(0.6110)=(0.605)
A concentration of 0.605 mol/dm:
approximately 610 minutes.
Below is a plot of
function of time
4-60
com POLYMATH, we found the con
9956 of the steady st
(€.99)(0.70)=(.693
‘That corresponds to 48 minutes
Below is a plot of concentration of exiting the three CSTRs asa function of tine,
©. Varying any of the parametem effets the results rather dramatically, Increasing ve
1050 caused the coaverson 1 goto 69 and decreasing ito decreased te conversion fo
12, Increasing Vt 1000 dm creased the conversa 18.87 while Gesmasing 10 20
dim caused the conversos to decrease to 0088, Increseing kt 1 caused he Conversion
¡0 increase to 0.67 while decreasing Kio 000
A person can ss Wat 208 V hana most
4-1
9956 of the steady st
(€.99)(0.70)=(.693
‘That corresponds to 48 minutes
Below is a plot of concentration of exiting the three CSTRs asa function of tine,
©. Varying any of the parametem effets the results rather dramatically, Increasing ve
1050 caused the coaverson 1 goto 69 and decreasing ito decreased te conversion fo
12, Increasing Vt 1000 dm creased the conversa 18.87 while Gesmasing 10 20
dim caused the conversos to decrease to 0088, Increseing kt 1 caused he Conversion
¡0 increase to 0.67 while decreasing Kio 000
A person can ss Wat 208 V hana most
4-1
P4-30 (con's)
(Use the same equations fort
plot
ie IMRCF in POLYMATH to generate the desired
(Use the same equations fort
plot
ie IMRCF in POLYMATH to generate the desired
Peso (eons)
By varying one parameter
and malay flow fe ru more a
Lowering the wranspon coulent (k) causes an increase in bth Cy and Fy, which
By rising the equim conta
case de molar ow ate
(cause a decease la À aná Cy and an increase in
{A iniicaı decrease a tmperan for an exotermic reacio would cause
ease in he rate ofthe forward reaction. Ths would cave op X aed Cle
By varying one parameter
and malay flow fe ru more a
Lowering the wranspon coulent (k) causes an increase in bth Cy and Fy, which
By rising the equim conta
case de molar ow ate
(cause a decease la À aná Cy and an increase in
{A iniicaı decrease a tmperan for an exotermic reacio would cause
ease in he rate ofthe forward reaction. Ths would cave op X aed Cle
Conveaona PER
(0) Membrane Restor
(©) Conveatoaal PER
(0) Membrane Restor
(©) Conveatoaal PER
copie
(4) Use the same equations in (a) except vo values cl
V = 15000000-10000:
o = 80000
The reason the gr
evaporated, bt water with a
¡ÓN
Beene
V = 15000000-10000:
o = 80000
The reason the gr
evaporated, bt water with a
¡ÓN
Beene
Find the number of moles of receptors
pacs ai, pars, | mole receptors _
ral. ell 6022107 recepiors
166% 10% + OLL = 1.66% 10 moles
Design equation:
Rate Law:
KC,C,
Stoichiometry
C.=Call-X)
Cuel8 X)
Where
“Total number of moles:
1.6610"? +1107 = 1.00210
bence
156x100 0917
1.00210
DA =0.998
700210
0.998
007
‘Combining and solving
1.6610" M
pacs ai, pars, | mole receptors _
ral. ell 6022107 recepiors
166% 10% + OLL = 1.66% 10 moles
Design equation:
Rate Law:
KC,C,
Stoichiometry
C.=Call-X)
Cuel8 X)
Where
“Total number of moles:
1.6610"? +1107 = 1.00210
bence
156x100 0917
1.00210
DA =0.998
700210
0.998
007
‘Combining and solving
1.6610" M
od apy
Design Equas
Rate Law
Stoichiometry
Design Equas
Rate Law
Stoichiometry
He 3.49 x 20! gu/(iimie - Le)
i, (280 pele) = Cp 6500 pui
®
i, (280 pele) = Cp 6500 pui
®
CDP&F 4
CDP#F (cont'd)
oom
«sono 3-0
Vs 38.06 00)
Sanrio te?
na Eo
oom
«sono 3-0
Vs 38.06 00)
Sanrio te?
na Eo
CDP4-F (cont'd
[CR aro Te Eur Er)
(150100.58110.06 E
ñ + asco.
an sana
[CR aro Te Eur Er)
(150100.58110.06 E
ñ + asco.
an sana
CDP4F (cont'd)
73031075 /001/8.325/e01/E(L/373-1/328) apra
At msitibetes
(0.033) 1/2
FER"
FR x = 0.70 (0.664) = 0.465
fa
pad, =
Be sites, |" usa
Panecieni integration yields a
Un 2.559 an?
ER x = 0.050.664) = 0.564
73031075 /001/8.325/e01/E(L/373-1/328) apra
At msitibetes
(0.033) 1/2
FER"
FR x = 0.70 (0.664) = 0.465
fa
pad, =
Be sites, |" usa
Panecieni integration yields a
Un 2.559 an?
ER x = 0.050.664) = 0.564
cppa.G
(2) Plot tra e Xto find the
0.0 010202 040506070009 1.9
From the ch » o minimize volume the best system would bea
CSTR followed
0 find the volume ofthe CSTR, the area ofthe rectangle drawn in the
Chart and mulúply kin a low rate of À:
2075 and 030.
Rate Law
Steichiomeuy
in POLYMATH to generate Ihe plot of X, Cy, an
4-3!
(2) Plot tra e Xto find the
0.0 010202 040506070009 1.9
From the ch » o minimize volume the best system would bea
CSTR followed
0 find the volume ofthe CSTR, the area ofthe rectangle drawn in the
Chart and mulúply kin a low rate of À:
2075 and 030.
Rate Law
Steichiomeuy
in POLYMATH to generate Ihe plot of X, Cy, an
4-3!
DRA ao dk Vo:
COTE
£06 #0640)
COTE
£06 #0640)
CDP4-H (cond)
a OE ae
Ba PSE a,
Ki = 00857
bos
X=
CDP4-1
ZA+B + Cap
Nu = 0.005 kino, €,
yee = 003 kon”
Feed stopped when V = 0.53 a?
0334 4x10
0 min
dh
0.005(1~x,)
a OE ae
Ba PSE a,
Ki = 00857
bos
X=
CDP4-1
ZA+B + Cap
Nu = 0.005 kino, €,
yee = 003 kon”
Feed stopped when V = 0.53 a?
0334 4x10
0 min
dh
0.005(1~x,)
CDP4- (cone)
Phugrng that into POLYMATH we get be folio
(xe) /4(6) #6000" (2, 20-8
(Cbu0 005" (2-8) /(.33+¢0-3"E)
can (1.2e-éte 02942) / (.33446-3€)
ce .008xb/(.33448-3%€
Nu = Co V = (572109) kml (053)0
Na = Ca V a (40.2104) Emol (0.33)
NEEND 384100 malus OS
Now Ais the lilting macan
Phugrng that into POLYMATH we get be folio
(xe) /4(6) #6000" (2, 20-8
(Cbu0 005" (2-8) /(.33+¢0-3"E)
can (1.2e-éte 02942) / (.33446-3€)
ce .008xb/(.33448-3%€
Nu = Co V = (572109) kml (053)0
Na = Ca V a (40.2104) Emol (0.33)
NEEND 384100 malus OS
Now Ais the lilting macan
zu
= 0143x712
À FOLYMATI ete os u
mation
ote) 181015 1039 malen.
Be ose
ar 5
Ms —
tn
Ba
98992605
0-39678108
0.99a97206 aa
DS6108:87
Now we teed 1 combine he co
Paris the seria operation x,
Xie And Inc,
= 0143x712
À FOLYMATI ete os u
mation
ote) 181015 1039 malen.
Be ose
ar 5
Ms —
tn
Ba
98992605
0-39678108
0.99a97206 aa
DS6108:87
Now we teed 1 combine he co
Paris the seria operation x,
Xie And Inc,
DPA (cont'd)
During th semibateh stage
where N, is the total amount of A introduced in the reactor by the time Land N, is
the amount of A left in the reactor a hat Game, so
MAA
x
Cant
During the batch stage
Caps GY
€
ere à is te time at which the flow of feed was stopped. (Le: 50 minutes)
|
caros
ce “102
Denn
During th semibateh stage
where N, is the total amount of A introduced in the reactor by the time Land N, is
the amount of A left in the reactor a hat Game, so
MAA
x
Cant
During the batch stage
Caps GY
€
ere à is te time at which the flow of feed was stopped. (Le: 50 minutes)
|
caros
ce “102
Denn
COPA (cont'd)
oh o.sonases
310.0
02.4] 6.00999526]
65.5| 0.9909
68.6 0.90905
b) From the table X, = 0.97 att =9.71 min
©) From the table X, = 0.59 att = 49.42
© New reaction ate
[-2f1_
Gi
sal 35 ja
MT E TE
ho kexpl
oh o.sonases
310.0
02.4] 6.00999526]
65.5| 0.9909
68.6 0.90905
b) From the table X, = 0.97 att =9.71 min
©) From the table X, = 0.59 att = 49.42
© New reaction ate
[-2f1_
Gi
sal 35 ja
MT E TE
ho kexpl
CDP#I (coat)
etc
5
{(1.2*10%r-0.01X, (0.00 )
ef (EZ 1071-0013, 00051 Ky
a
x
\
Recall that inthe semibatch reactor stage Xy
“Tue equation forthe conversion of A remains the same as
g for Xy, X, and Xp we obtain
-1915°%,
etc
5
{(1.2*10%r-0.01X, (0.00 )
ef (EZ 1071-0013, 00051 Ky
a
x
\
Recall that inthe semibatch reactor stage Xy
“Tue equation forthe conversion of A remains the same as
g for Xy, X, and Xp we obtain
-1915°%,
CDPé- (cont'd)
ol |
| | Bi
|
bbls
ele
HER
|
ol |
| | Bi
|
bbls
ele
HER
|
copra.
BATCH
Stichionezy
A E
Eos), ax "Eis
Ion iegrls canbe found in Appendix A-10, 9
BATCH
Stichionezy
A E
Eos), ax "Eis
Ion iegrls canbe found in Appendix A-10, 9
CDPS.K(con'
2. Rave Law
Fox
Fett
Parameter evaluaion
2. Rave Law
Fox
Fett
Parameter evaluaion
CDP (cour
CNE]
15.000 do
CNE]
15.000 do
E = Fy (l= X)= Fell + Ro) JOA)
vale) ete Rex +e, x)
{ 1+R0-x,) )
sil 3)
ar)
(pee)
Baum Fast Eu + RFu(1— Xe)
Combining:
(Bot Eu = XML REX) pr
ACER ES (4%) y
We also know tha X,
E, [ De
i (A
Two equations with two unknowns is solvable and the answer is
Enps-L,
vale) ete Rex +e, x)
{ 1+R0-x,) )
sil 3)
ar)
(pee)
Baum Fast Eu + RFu(1— Xe)
Combining:
(Bot Eu = XML REX) pr
ACER ES (4%) y
We also know tha X,
E, [ De
i (A
Two equations with two unknowns is solvable and the answer is
Enps-L,
CDPSL cont'd
vavfi+alirex)fi+e,x
Fe Fall+ A
rt aX) Fale
+ R(L= Xo) |
EL
Fe
x
Plugging into POLYMATH
vavfi+alirex)fi+e,x
Fe Fall+ A
rt aX) Fale
+ R(L= Xo) |
EL
Fe
x
Plugging into POLYMATH
conta
Vers)
Vers)
pese design equation:
eB hos tr arar
El
“Then make ty
eB hos tr arar
El
“Then make ty
EN ca
D Now wit press
Plugging tis ino POL
aston
pe
tomer
decreases de coaverion. Inc
and decreasing I wi increase the
Inconversion,Incressig De Beit e
en Inercia R dere
conven, WeresingR, wil increase
D Now wit press
Plugging tis ino POL
aston
pe
tomer
decreases de coaverion. Inc
and decreasing I wi increase the
Inconversion,Incressig De Beit e
en Inercia R dere
conven, WeresingR, wil increase
A yy
ae, avt
en becomes ve
ae, avt
en becomes ve
CDP SON"
CDP4-0 (con)
From Ego (4) part dl,
Erg
REIFEN
USES AI TGS,
gw attest E
(ous cime =
Solving vith the guaeatie equation
c= 963, 7579
allovable flow rate (part ©)
AA
= 15.19 Ake
From Ego (4) part dl,
Erg
REIFEN
USES AI TGS,
gw attest E
(ous cime =
Solving vith the guaeatie equation
c= 963, 7579
allovable flow rate (part ©)
AA
= 15.19 Ake
CDP4-0 (cont)
Ax his sebstrate concentration, the steal growth cate de
Mo aproximacion OME à 2 y © 288 AE a a
ES
Ths muz. cell concentration Le 13 corremos
Ue. 500)
0 LS [RENTE] a
DTO Tom DONT toad an CATED
Soiree for x,
Ken. un
Seen
Ax his sebstrate concentration, the steal growth cate de
Mo aproximacion OME à 2 y © 288 AE a a
ES
Ths muz. cell concentration Le 13 corremos
Ue. 500)
0 LS [RENTE] a
DTO Tom DONT toad an CATED
Soiree for x,
Ken. un
Seen
CDP4O (conve)
From Ea. (4) part 0),
o Pa MSE
NAS y IG ARE)
Su At Fe 75.79 Uhr, the cell concentration is maxima.
11 concent Be catealaced weion <=
aus ST Th
aan
15.79 ake
From Ea. (4) part 0),
o Pa MSE
NAS y IG ARE)
Su At Fe 75.79 Uhr, the cell concentration is maxima.
11 concent Be catealaced weion <=
aus ST Th
aan
15.79 ake
DPEP Given: à 8
CDP4P («
mote balance for HO,
Concentration
For Concentration
Concentration
For Concentration
coreg.
Specs Bt
Specs Bt
3nd Edition, Solution Manual, Chapter 5
General: In Chapters 3 and 4, the rate law. The problems la
this chapter reinforce how to 0 om experimental da
1. An oper to create and original
problem and solution.
P52 "What pr
Problems P5-3, P5-5, P5-7, and P5-18 all invo) .actor experiments to find
the reactor order and specific reaction rate. The students can use differeil|
techniques to differential the data or can use regression. These problems cf
be alternated from year to year
‘ny go blindly aha
This problem usually trips up
c ere taken in
and differentiate the data ( r the data
CSTR at steady state
Differentiate the data down a PER to fi 0 eters, Note ti
reaction is reversible. Quite tricky me consuming,
California Exam Problem where one first needs to determine in the rate ll
Part (a) is quite straight forward ba
a. Requires thinking and al
ight forward
Good problem involving regression AND thinking, May be i
consuming ifthe st st got the hint and see iti the sum of two TH
laws, zero and first o
Very the % decomposition 1)
la
same for different
determine the reaction order,
straig
reaction rate. Sho
industry
Problems P5-13, P5-14, P 1 forward prob
using nonlinear re ar to year.
nis is used to md
P5-16, Shows an important concept ¥
first and second order reactions.
General: In Chapters 3 and 4, the rate law. The problems la
this chapter reinforce how to 0 om experimental da
1. An oper to create and original
problem and solution.
P52 "What pr
Problems P5-3, P5-5, P5-7, and P5-18 all invo) .actor experiments to find
the reactor order and specific reaction rate. The students can use differeil|
techniques to differential the data or can use regression. These problems cf
be alternated from year to year
‘ny go blindly aha
This problem usually trips up
c ere taken in
and differentiate the data ( r the data
CSTR at steady state
Differentiate the data down a PER to fi 0 eters, Note ti
reaction is reversible. Quite tricky me consuming,
California Exam Problem where one first needs to determine in the rate ll
Part (a) is quite straight forward ba
a. Requires thinking and al
ight forward
Good problem involving regression AND thinking, May be i
consuming ifthe st st got the hint and see iti the sum of two TH
laws, zero and first o
Very the % decomposition 1)
la
same for different
determine the reaction order,
straig
reaction rate. Sho
industry
Problems P5-13, P5-14, P 1 forward prob
using nonlinear re ar to year.
nis is used to md
P5-16, Shows an important concept ¥
first and second order reactions.
Ad Edition, Solution Manual, Chapte
P5-17. A problem on designing experiments.
F518. Can be used alternatively with P
CDPS-A Alternative to problems P5-3, P
CDPSB Alternative to P5-6, but si
CDPSC Fairly difficult problem. Usually a
DISD Alternative to P549
CDPSE More difficult to alterna
Summary
Alternates
78,18,A, BCE
378,18,4,B,C
10,14,18,19,202
10131819,
3578, BCE
101314,
101314,
P5-17. A problem on designing experiments.
F518. Can be used alternatively with P
CDPS-A Alternative to problems P5-3, P
CDPSB Alternative to P5-6, but si
CDPSC Fairly difficult problem. Usually a
DISD Alternative to P549
CDPSE More difficult to alterna
Summary
Alternates
78,18,A, BCE
378,18,4,B,C
10,14,18,19,202
10131819,
3578, BCE
101314,
101314,
3rd Edition, Solution Manual, Chap!
gned
e = Always assigned, AA = Al
O = Often, I = Infrequently, S = Seldom, G = G:
mm the group of alternates,
nat
In problems that have a
problems, either the problen
Jot or any one of the alternates are
Time
‘Approximate time in minutes
problem.
+ would take a B/B' student to solve the
Difficulty
SF = Straight forward reinforcement of and chug)
ESF = Fairly straight ion of equations or af
intermediate
IC = Intermediate calculation required
M = More difficult
OE = Some parts ope
“Note the letter problems are found on the CD-ROM. For example A = CDPI-A.
Summary Table CI
ea Method o
Reasoning rica Half Line
Straight Forward 1218
Fairly Straight Forward
More Difficult
Critical Thinking
gned
e = Always assigned, AA = Al
O = Often, I = Infrequently, S = Seldom, G = G:
mm the group of alternates,
nat
In problems that have a
problems, either the problen
Jot or any one of the alternates are
Time
‘Approximate time in minutes
problem.
+ would take a B/B' student to solve the
Difficulty
SF = Straight forward reinforcement of and chug)
ESF = Fairly straight ion of equations or af
intermediate
IC = Intermediate calculation required
M = More difficult
OE = Some parts ope
“Note the letter problems are found on the CD-ROM. For example A = CDPI-A.
Summary Table CI
ea Method o
Reasoning rica Half Line
Straight Forward 1218
Fairly Straight Forward
More Difficult
Critical Thinking
Chapter 5
B51. No solution will be given.
28:2 No solution will be given,
PS3
025
175 007
Pot of og Cyl v og Ca shows a
de
B51. No solution will be given.
28:2 No solution will be given,
PS3
025
175 007
Pot of og Cyl v og Ca shows a
de
Acc
LS 20
Pas Paz
Task (. Rewrite the design equation (Le.
variables. Recall Y = AcZ,th
eX ertade
az
LS 20
Pas Paz
Task (. Rewrite the design equation (Le.
variables. Recall Y = AcZ,th
eX ertade
az
5-6 conta
For isothermal operation and ao pressure drop,
Ca = PART
(1+ eX Pa = Pao
7
+ Pan
ait -PaPao] vil mole balance
variables Pa and
Now we have the diffee
in terms of the meas
Task 2, Look for simplifications.
A) See if volume change ean be neglected
y = [gx 104)-0
©) We also see that for runs Land 3 hat B isin excess and that for excess B
57
For isothermal operation and ao pressure drop,
Ca = PART
(1+ eX Pa = Pao
7
+ Pan
ait -PaPao] vil mole balance
variables Pa and
Now we have the diffee
in terms of the meas
Task 2, Look for simplifications.
A) See if volume change ean be neglected
y = [gx 104)-0
©) We also see that for runs Land 3 hat B isin excess and that for excess B
57
Pas = 3.0 or and Pass
Kaos Ph 2
KPaot Poy Fat
2 Bestie
Den mean
We know y = | because of thermodynamic consistency,
weaning Pm E]
Task 5. Evaluate k’ and K,
From AH of 3.0 or we ses equilibrium is reached at Pag 0.01
Pce =0.129-001 = 119
Tao
P woe
Om meo
Kaos Ph 2
KPaot Poy Fat
2 Bestie
Den mean
We know y = | because of thermodynamic consistency,
weaning Pm E]
Task 5. Evaluate k’ and K,
From AH of 3.0 or we ses equilibrium is reached at Pag 0.01
Pce =0.129-001 = 119
Tao
P woe
Om meo
cont'd
weh ö=3-1 const. vol
arranging
(tex) at
Le La evident from the plot that
weh ö=3-1 const. vol
arranging
(tex) at
Le La evident from the plot that
Hated
tom) 24
(a) Mote Balance: constant Y
ts A tee + tag,
Kain)
Stein)
[Re]
ac
2% oon,
7 on 0.051 35 -0.026 0,018
tom) 24
(a) Mote Balance: constant Y
ts A tee + tag,
Kain)
Stein)
[Re]
ac
2% oon,
7 on 0.051 35 -0.026 0,018
Cat] 0082
Inc aC a/a) | 2.501
In Ca 0895
Using linear regression: a = 1.0
Ink=-3.3864 > k=0;
wating by equal area
0.81
508 Ton [ass [ao]
3.10 [3507 | 426
7 10.128 0821
Im
429 En
a
> &-1
Inc aC a/a) | 2.501
In Ca 0895
Using linear regression: a = 1.0
Ink=-3.3864 > k=0;
wating by equal area
0.81
508 Ton [ass [ao]
3.10 [3507 | 426
7 10.128 0821
Im
429 En
a
> &-1
259
(@) Method of Half Lives
Linear Regress
0025
00133
TOT
To
0075
007
Plot In Cao vs. In 11/2
Regression
-1.0128x - 2,9529
(@) Method of Half Lives
Linear Regress
0025
00133
TOT
To
0075
007
Plot In Cao vs. In 11/2
Regression
-1.0128x - 2,9529
25:9 conta Using POLYMATH we can
I caomio?
+ tal ac2rCalp= 19-19 cemCalD:
8.99104
©
Using the equation
Solve for eat 110°C,
ke 110Cis20
LG 100'Cis 10.52
From these values of k we can calculate the activation energy (E).
nr
kı
I caomio?
+ tal ac2rCalp= 19-19 cemCalD:
8.99104
©
Using the equation
Solve for eat 110°C,
ke 110Cis20
LG 100'Cis 10.52
From these values of k we can calculate the activation energy (E).
nr
kı
Oy + wall > loss of Os
O5 + alkene > products
Lan
Un FOLYMATI noie pion we a fd th le oc, nd
5-14
O5 + alkene > products
Lan
Un FOLYMATI noie pion we a fd th le oc, nd
5-14
S-11 Given: Por of percent Decomposition of NO, va. V/F)
4 Decompositión of HO,
same that e uch
wim aso ree Gy
Ets tinear wish WEjgy #4 figure to Che cÍght shows
5 no
cesction ix urn andee- amt
PEZ 500, + GP > BASI + 28,0
+ notes of $10; = Hae
aotecuise weight of $402 = 60.0
depen of si
ee of 16 + TB
ft potecular mess of PF 20.0
4 Decompositión of HO,
same that e uch
wim aso ree Gy
Ets tinear wish WEjgy #4 figure to Che cÍght shows
5 no
cesction ix urn andee- amt
PEZ 500, + GP > BASI + 28,0
+ notes of $10; = Hae
aotecuise weight of $402 = 60.0
depen of si
ee of 16 + TB
ft potecular mess of PF 20.0
Can EE tee
+ vere pa (o Y
pe were D rl]
E apta
13.09
aan
+ vere pa (o Y
pe were D rl]
E apta
13.09
aan
wy, = 60 s/usst
rar nan à 106
Cuasatook of Censstez and Faysies, St obs pe DRSS?
rro
ao
wr, 20 a/eeot
3
Ve 0.5 de = 0.507 a
9 = 2298600
OPA 105
à (69 ty Eds
10.2 basas ts
ons
Volum ‘sor 104 10*10%{10X1000) = 1*10%n?
Moles of SiO, removed = 1*10%232*10'60 = 0.386 mol
Moles of HF used = 6(0.386)
Moles of HF available = 0.5*107/5*1*10°20=5 moles
Hansanı (HPL © 0.2 (ganes!
rar nan à 106
Cuasatook of Censstez and Faysies, St obs pe DRSS?
rro
ao
wr, 20 a/eeot
3
Ve 0.5 de = 0.507 a
9 = 2298600
OPA 105
à (69 ty Eds
10.2 basas ts
ons
Volum ‘sor 104 10*10%{10X1000) = 1*10%n?
Moles of SiO, removed = 1*10%232*10'60 = 0.386 mol
Moles of HF used = 6(0.386)
Moles of HF available = 0.5*107/5*1*10°20=5 moles
Hansanı (HPL © 0.2 (ganes!
IE,
| Er
a = data 22007 alt 0.775,
gr Gone MS A
Tes as aa © 2000 à
0.10.7737 393%
EA 350 mio
E
aw
aw =05
Using POLYMATH's non linear regression, the following results were found:
68 a=t B=2
FancxPproi ono?”
Pe 600214 ber = 9.201097
Ste = 120305;
IE & Poncivo recauats, 2 negative resicuals. Sun ci
2 5-18
| Er
a = data 22007 alt 0.775,
gr Gone MS A
Tes as aa © 2000 à
0.10.7737 393%
EA 350 mio
E
aw
aw =05
Using POLYMATH's non linear regression, the following results were found:
68 a=t B=2
FancxPproi ono?”
Pe 600214 ber = 9.201097
Ste = 120305;
IE & Poncivo recauats, 2 negative resicuals. Sun ci
2 5-18
re ro eo m 7»
site ns am
ram os 0.8 3 005 0.
0. os 0.5 1 016 0.6
2 0. Ds 06 ds 06 02
oe me 1 0.10
0.6 0.4 0.6 0.05
A
PC RICART) Pal)
® ».-c0=c.rro.-x)="l0,-x)
EN
Da 2
Fak _ Fe bo
oy an A
ner )eint-amnp, — Pla Pa
Y= A AX AX
0.015
ass
0.420
0.320
0.500
ous
0.015
0.468
0.420
0.520
0.500
0.00
à
instelkerks
1.0625
1.0500
120
9.600
0.9378
2.0025
site ns am
ram os 0.8 3 005 0.
0. os 0.5 1 016 0.6
2 0. Ds 06 ds 06 02
oe me 1 0.10
0.6 0.4 0.6 0.05
A
PC RICART) Pal)
® ».-c0=c.rro.-x)="l0,-x)
EN
Da 2
Fak _ Fe bo
oy an A
ner )eint-amnp, — Pla Pa
Y= A AX AX
0.015
ass
0.420
0.320
0.500
ous
0.015
0.468
0.420
0.520
0.500
0.00
à
instelkerks
1.0625
1.0500
120
9.600
0.9378
2.0025
B5-14cona
Seumteg over ‘a’ dete pointe
Ina)
uni
Baur
blog ene a ls
6A, 42698 Ay = 4.602 à = 0,4403 a = 0.097
UBS A, + 3.8837 Ay + 3.5133 Ay
4.695 A, + 3.5735 Ay + 1.6858 A = 0.5066
see poole
aoe Ik 0.016 ees
foo um 3 Er
spe
Seumteg over ‘a’ dete pointe
Ina)
uni
Baur
blog ene a ls
6A, 42698 Ay = 4.602 à = 0,4403 a = 0.097
UBS A, + 3.8837 Ay + 3.5133 Ay
4.695 A, + 3.5735 Ay + 1.6858 A = 0.5066
see poole
aoe Ik 0.016 ees
foo um 3 Er
spe
AD oo
= Comp (CET, = TIT (Cys
la Crying) = EUR GT, = UTA) + a da CEIC
rate us) EC
GT, - way
HR ne 1098.27
a
te, 4
AS pod a Ca
Be pod ka
mak Ca He
IS)
C=C, + CAO = Cy (L-Xq}+ Cd Xa)
Cyt) = CuK #Op Ky) = Cu Ka Ma)
ac,
HE Cully Kr,
LE à Gus taka Hh )
E = Cf, X, +075 00
= Comp (CET, = TIT (Cys
la Crying) = EUR GT, = UTA) + a da CEIC
rate us) EC
GT, - way
HR ne 1098.27
a
te, 4
AS pod a Ca
Be pod ka
mak Ca He
IS)
C=C, + CAO = Cy (L-Xq}+ Cd Xa)
Cyt) = CuK #Op Ky) = Cu Ka Ma)
ac,
HE Cully Kr,
LE à Gus taka Hh )
E = Cf, X, +075 00
BilGcont’d
mot
CACEET
rte = Cp)
ee) mated vee
nny
au ane must aot
nu. a cons? um D 1.000 ea
Ab/aeIta y 90090
mot
CACEET
rte = Cp)
ee) mated vee
nny
au ane must aot
nu. a cons? um D 1.000 ea
Ab/aeIta y 90090
Plan to find the rate aw forthe hydrogenation af eyelopentane on a
Since this i a sable catalyst we don’ have to worry about catalyst decay and an
‘Integral Reactor willbe used.
Perform several different runs, holding C,, and W constant while E, is varied
rom ua to run,
Plot Xu v5. WR, for al runs.
Fita curve though all points which pases through che origin. The slope at any
point is the reaction ale. Record the slope and corresponding C,, for many
different X, values. These data can be used to deienmine the at aw
Experimental Plan to find the rae law forthe Liquid phase production of met
bromide from an aqueous solution of methyl anne and bromine cjanide:
1. Fora liquid phase reaction without a catalyst, sea bath recto.
2. While running the reaction record both C, and Cy at equal time intervals
3. Repeato ensure accurate data,
Experimental Plan o find the ate law for the aid-ctalyaed produc
ol form an aqueous soluion of ethylene oxide:
is an aqueous solution with a soluble catalyst, «batch reactor will be
For each run record C, at equal ime intervals, Use this data o determino the
effect of C, on the ca law
Since this i a sable catalyst we don’ have to worry about catalyst decay and an
‘Integral Reactor willbe used.
Perform several different runs, holding C,, and W constant while E, is varied
rom ua to run,
Plot Xu v5. WR, for al runs.
Fita curve though all points which pases through che origin. The slope at any
point is the reaction ale. Record the slope and corresponding C,, for many
different X, values. These data can be used to deienmine the at aw
Experimental Plan to find the rae law forthe Liquid phase production of met
bromide from an aqueous solution of methyl anne and bromine cjanide:
1. Fora liquid phase reaction without a catalyst, sea bath recto.
2. While running the reaction record both C, and Cy at equal time intervals
3. Repeato ensure accurate data,
Experimental Plan o find the ate law for the aid-ctalyaed produc
ol form an aqueous soluion of ethylene oxide:
is an aqueous solution with a soluble catalyst, «batch reactor will be
For each run record C, at equal ime intervals, Use this data o determino the
effect of C, on the ca law
SEM 92g
015
(4C.) 131-4095: +07
FialPoin; (a) „121420323007,
Cat), 4 m
Graphical Method
02st
0x4
TS
Poly. (Senos!)
m1
lie ers the edges ofthe bars we wil ge or vales or Cade
oca whee the wen
5-0)
015
(4C.) 131-4095: +07
FialPoin; (a) „121420323007,
Cat), 4 m
Graphical Method
02st
0x4
TS
Poly. (Senos!)
m1
lie ers the edges ofthe bars we wil ge or vales or Cade
oca whee the wen
5-0)
5-18 cont'd
Polynomial Fit
10588 - 0.3847
Using that we can find ad graph the denvañive of C, a any time
Graphs ofthe natural log plot ofthe derivative are given below
the plot made by the Numerical ® ent the Gi
02 04
ye ss - 10173
y= 16970 = 19221
In deere
Sty = arm - 22409
Polynomial Fit
10588 - 0.3847
Using that we can find ad graph the denvañive of C, a any time
Graphs ofthe natural log plot ofthe derivative are given below
the plot made by the Numerical ® ent the Gi
02 04
ye ss - 10173
y= 16970 = 19221
In deere
Sty = arm - 22409
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