BBL732_Ambika very useful for biotech.pdf

please read to learn

2. 2nd
July
,
U
:
LectureI
Steps
:
1)MassBalances
,
EnergyBalances
2)Equipmentreq
.
For
that
pot(vessel)
[VesselDesign
3
,
sing
thevessel
3)HazardConsiderations
"¥
Spillage
Y
IncorporateFactor
into
Design
Thedesign
exercise
will
use
lotsof
thumbrules
.
Sequenceofprocedureswillbe
imp
,
no
needtorememberdesigneq's
Remember
the
procedureforrussel
design
Easy
course
if
systematic
approach
.

3. HeatExchanger,
Datin
Column
it
e
Only
CourseContents
:3-6
Lectures
:
MBEBalances
↑
VesselDengn
-
-
HeatExchanger
o
o
-aDistillationColumnDesign
/
ProcessSafetyDesign
S
Process
Validation
flowcharts
,
ProcessFlowDiagram
:
Emp

4. TypesofFlowcharts/Diagrams
? streamsElequipments
->
screen
--
Mixer
s
Filter-
labelledbut
?
↑
no
ideaabout
>
Tank ?
-
mass
(quantivesrea.)
->
energyrea
->
enter;equipment
duxillary
equipments
O
+
Quersize
likeconveyerbetts
C
-
Undersize
->
processinstrumentation
automation
,
controldevices

5. Specificuniversally
acceptedsymbolswill
beused
eg
#
:
screen
primitivedrag.
can
beenriched.
mostadvancedflowdesignshaveinfo
aboutprocessinstrumentationEcontrol
danses
->
blueprintof
entire
plant
PNI
diagram
?
P&I

6. EveryWednesday3-3Tutorial
[Practical
Component)
[NoLab] is
the
tutorialitself
M&EBalances
:
Input/output
Models
Issbehaviors
-
arithmetic
(also
us
models)
->
differential(timedimension)
alsospacecoordinates
->
partialdiffer
.
(2
,
t)

7. unusedsubstrateposits
2
problems
-
wasted
raw
material
-
expensivedownstream
no
thumbrules
,
systemspecificsubstraterequirements

8. COURSEPOLICY
:
1)75%attendancepreferred(1
gradedown
if
significantlylowattendance
Nomarksforattendance
2)Minor
38 Agradeabove80
Major
30
(Absolute
Assignments40
-
SurpriseQuizzes
-
some
asignments
3)Textbooks
:
1)
PlantEquipmentDesign(Brownelle
&Young)
2)
Could
?&Richardson(Vol6)

9. Tutsfrom
Week2
-25thJuly
Upstream ProductionDownstream
Screen
toseparatesoud
fractionsBatchReactor
Filter(s-1
separations
Washing
Step
Distillation)-1
separation
Dryer IS-sSep

10. usually
,
20
%
insculumin
anerobicfermentation
Upstream Production
Mi M2 My
M6main
->
Screen
--
Dryer-
Tank
10
,
000L
↓
↓
↓
↓
M3 Ms M7
Mi
=
M2
+M3
WorkingVol
=
6000L
M2
=
My
+M3
f
My
=
M6
+
M7 InoculumVol
=
0
.
2X6000L
=
1200L
or
Mi
=
M3
+
Ms
+
M7
+MG(Input-OutputModel)

11. ↓
solvent/liquid/allowsfastermovement
offiltrate
5)lessretentionoffiltratein
the
filtercake)
In
>
Filter
>
Filter
cake
sturry
(Rawmaterial)
↓
!
Typical
schemeof
a
continuousWe
d
Filtrate
feltration
unit
process
.

12. lossofwash
liquidistypicallynegligible
centrifugealsohas
a
similarscheme
(s-l
separation)
·
DryerMoisture
↑
Moist/
Wet
Call
->
Dryer-> Dry
cake

13. ·
materused
as
solvent
in
any
proces
,heat capareally
i
a
so
,
sequential
separationof
S-l
.
Slurry
-Filtration-wetCake-Dryer-Dry
cake

14. Extraction
:
Liquid-liquid
separation
(based
on
misubility)
Acetone
Water
(mix)
=
Toluene
lorganic
A
+
B
solventL
->
& So
Raffinate
Sigued
Meture-
A
+
B L
+
A
(Process
Liquid)
->
Extract(desired
pat
in
solvent
settles
ofheaverA
:
distributionfactors
%
A
is
more
miscible
un
A
,
Allwill
behigher

15. eg.AntibioticslikePenicillin
can
beextractedby1-1separation
Distillation
:
1-1
separation(based
on
valatility)
steam
(7)
extractingenergy
↑
-
>
more
volatilecomponent
M
Rebater
↑
-
IVIzes
volatilecomponent)
Steamfeed
T feed

16. ·
AdsorptionColumn
:
S-l
separation
or
LiquidMeture(A
+
B)
Similar
schemefor
oCDesorption
"P
Spent
zand(B)

17. Adsorption
of ↑
B
directionreversed
in
case
of
gases
Gases
.T
eg
.Separating02
+
N2
nextureusingPSA
(PressureSwingAdsorption)
:
-
↑
(A
+
B)

18. stripping
&
feedsolvent..
- ↑
↑
-
A
isbeingstripped
outusing
L
mi
-
↑
+
A
A
+
B
Gfeedline

19. QuesMix
2streamsof
sucrose
sol'swithdiff
come.
Feedshaving
mass
fracofsucrose
ofOh
Feed2
"
Wewant100kG
sucrose
so
with0
.
3mas
fraeof
sucrose
.
CalcratioofFeed
1&2fordesired
F
,
(0
.
1)
Song
>
Fout
->
100kg
=
30kg
sucrose
Song
Fu
10
.
3)
RatioofF2toF2
=
1
:
/

20. 50x(0
.
1)
+
50x10
.
3)
=
3
+25
=
30kgin
100
kg
sokg/h
=
0
.
3mas
fraction
~0
.
GkgAlleg
G9RgA/KUORY
-
OhgBlk
-
↑
Gorg/n uug/h&
-> mass
flow
rate
?
a
KgAIng
ausin
"
composition? b
kgB/
kg
0
.
5KgAlkg
------
0
.
5kg
Blkg
30kg/h
0
.
3
kgA/leg
everall
mass
balance
:
100+
30-40-30
0
.
7kgB/leg
-
x
=
0
=>
a
=
60kg/h

21. overall
comp
.
Balance
:
A
:
50+9
-
36-18
-
an
=
0 :outputstream
59
-
54
=
aX60
5
=
60 a 60kgth
a
=
%
=
365ugin
A
35kg/hB
Now
,
componentstream
①&
composition
B
:
50
+
21
-
4
-
12
-
bx
=
0
71
-
16
=
60b
55
=
60b
=
b
=
1
22

24. Eth
August
Enzyme
or
Biomas
catalyse
A
+
B
->
P
+
BP
aCgH1206
+
bNM3
+
cO2
ZdCuHyOz
NoteCa
,
My
,
Oz
,
Ns
,
+
fCO2
+
gHz8
in
a
celllikeE
.
coli
,
95
%
iscontributedbyCHON&5
%
by
otherelements
actualMW
=
Was
,

25. wash
,
centrifugethebromass
,
thendry
it
.
-
murobalance
->
accurateweight
(say
2
.
1
..
g)
#
Biomass
+
On t
&
ICHON)
gas
mixture
afterburning
O+
residual02
TCDThermal
Conductuity
Detectortomeasure
cone
ofNO
↓
NOz
&
marked
(removed

26. #
CO2
,
128
,
No
,
all
are
comingfrom
BM
UtofBiomass-(molesobtained
*
Molweightof
H/C/N]
=
cutof
O
in
BM
Toget
100%MWofBM
weconsider
5-10
%MWcontributedbymicronutrients
totheremaining90-93
%
obtainedbymicronutrients/
elements
landthenbalancethe
ear)
5
%
Ashcontent
*Excess&LuntingReactant

27. ->
Reactor
Et
ResidualReactant
↓
ResidualReactant
?
-
X
=
Reactant
in-Reactant
out
=
Conversionfactor
Reactantin
to
improve
processeconomy
weX&
trytobringitclose
to
1.
SinglePath
System/Mult
Path

28. path
:
howmanytimes
areactorstream
ismoving
thruthesystem
?
ResidenceTime
SemiBatch/Fed
Batch
->
ConstVolume
(add
some
,
removesome
,
but
not
continuously)
->
VariableVolume

29. BatchTime(tb)
Residencetime(T)
(Batch
,
Semi-Batch)
CSTRS
:
BMisshearsensitive
,
we
can't
Heupinabove
a
certainlevel
,
whichmight
not
be
enoughto
keepBMfromsettling
so
sold-lea
separationwill
occur. -
BMsettles
->weremoveit
4)
put
it
for
recycle
->
--
- interacting
-
streams
purge/discardedBM
↓
non-interacting
streams

30. #Selectivity-rateof
desiredpt
a
pat
"¥
Instantaneous
anystream
,
calc.rates
"¥
performance
of
individual
process
units
OverallSelectivity
rateoof
desered/undesired
in
the
final
stream

31. ProcessDiagrams
->
ThemostprimitiveProcess
Diagram
will
haveinput/output
streams
representsprocess
infoconuselybyprowdingallrelevantdata
->
Dataomittedherein
thediagramwill
bepresent
in
theprocesssheet
.
TypesofProcessDiagram
:
1Input
/Output
Diagrams
boxes
2)FunctionalDiagram/flowsheet/chart)
definesfunctionsofindividual
units
3)Operations
Diagram(each
streamhasnumericalvalues
defined
,
whatgoing
standardsymbolswhere)
detaileddescriptionoffloorlines
·
insteadofboxes
unitehave
symbols
:filters
,
reactors
,
pumps
,
heatexchanger
,
etc
all
not

32. 6
-
-
"
=>
-
heatexchanger pump
reactor
-
fetter
distillation
adsorption
unit
unit
4)
Process
InstrumentationDiagram
(Pnl)
/ControlLoops( P&IProcus&InstrumentationDiagram
ControlUnit
:
takes
feedback
fromthe
reactor
intermsofelectricalsignal
measuringsay
,
Do
value
.
ComparesttothesetPointvalue
,
E
takes
an
actionwhenthere's
a
differencebiw

33. Tutorial
I
①E
.
coliisaddedtovigorouslyaerated
mediumcontaining10g/LEtOHandafter
sometime
,
the
ETOH
come.
becomes291Land7
·
5g/Laceticacidis
produced.
Howdoestheobservedyieldof
acetic
acidfrom
EtOH
compare
with
the
theoreticalyield
?
CzHsOH
+
O2
->
CH3COOH
+
H2O
Imal Imol
SgEtOH
-
7
.
5gAcH
"¥
->
Ee
=
5X60=>
7
.
59/60g/mol
7
.
5 x46
-
89/46gimo
=
0
.
718

34. Trueyield
a
CzHzOH
+
bO2-CHzCOOH+at2O
Ye
=
Males
ofActprocess Y
=
&frombalancea
storchometric
ea
usually
<
Yobs
whenonly
one
substratepresent
sometimes
,
when
2subs

35. ①Batchmixing
process
200kgof10
%
wi
Methanol-watersolutionismusedwith100kgof
70
%
wiMethanol-watersolin
a
Batchmixingtank
.
Whatisthe
final
ant
I
compositionofthe
solution?
40
%
of200ng+70%%
of100kg
=
80+
70
=
150kg
Methanol
in
300kgsolution
=
50
%
w/wMethanol-
water
sol
(300ng)

36. Q1000kg
,
batchofpharmaceuticalpowderisdriedin
a
freezedryingoperation.
5%win
moisture
afterdrying
,
90%
ofthewater
is
removed
-
Calc.
Fina
batchcomposition
I cut
ofmoisture
removed
.
5
%Winwater
in
1000
kg=50kgwater
↓
90%
of
this
removed>
45kgremoved
↓
Skyremaining
5kgmosture
in
950hqpowder
=
-X100
=
0
.
52%
955

37. &
Gasleaving
a
fermenterat1atm
Pressureand
25temphasthefollowing
composition.
78
.
2%
N2
19
.2
%Oz
2
.
6%CO2
Calculatemass%offermenteroff-gas
.
If
you
have
a
CO2
absorptiontank
,
thenwhatwill
bethe
mass
ofN2El
②
in
theoutletstreamofabsorption
unit?
let
total
molesbes
0
.
7821
moleN2
~0
.
782xx289
=
20
.
384x&Total
=
27
.
6720g
0
.
192nmoles02
->
0
.
192x
x
329
=
6
.
144xG
I
0
.
026
x
molesCO2
->0
.
026rX44g
=
1
.
144xg

38. ·
allCO2
is
absorbed
:
assy
assuming
new
total
mass
=
20
.
384
nNa
:
38x100
=
13
.
6
I
02
Max 100
=
22
.
20
%
newmass%:
CO2
:M
X100
=
U
.
13
%
Nux
. .

39. CsHi20s
+
aO2
+
bNHs
>
Cy.uH7
.
3 01
.
2
No
.
86
S
X
+
dH20
+
eCO2
conventionally
,
yieldsfor
gaseous
products
:
mol/mal
yields
forsolide/bq.
products
:
glg
OGiven
,
2/3rdofGlucose
'C'
goes
toBroman
Calculate
a
,
b
,
c
,
d
,
e
and
Yxs
,
YY/02
If
BMcontains
10%
ash
,
findits
actualMW
.
W
(obtainedfor
a

40. stoichiometric
balances
:
C-balance
:
6
=
4
.
4c
+e
:
Erdof
C
fromGlu
-
B
=>1
.
4c
=
4
=c
==
and
e
=
2
H-Balance
:
12
+
3b
=
7
+
ad
2d
-
3b
=
5
.
36
-
D

41. Obalance
:
6
+2a
= 1
.
2
+
d
+
2e
↑
6
-
1
-
4
=
d
-
2a
3
.
07
0
.
909
=
d
-
2a
-
②
2a
=
3
.
07
-
0
.
909
-
=
1
.
08

42. Nbalance
:
36
=
0
.
86x1
1-2(6
=
0
.
26
putting
inQ
2d
-
3(0
.
26)
=
5
.
36
2d
=
6
.
14-
=
3
.
07

43. AvailableelectronbalanceE
Degrees
ofReduction
specific
toCcontainingempoe
+
4
+
1
-
2
CgHi20s
+
aO2
+
bNHs
-
>
Cy
.
u
H7
.
3 01
.
2
No
.
86
↓
max
oudiedstate
Ginammonia
120
+
e
CO2
+
y
-
4
↓
a
.
einc
=
4
V
=
available-percarbonatom
:
CH1206
=
@
+
2X1
+
6X-2
=
G
=(degreeof
reduction
a

45. ->
0(NH3]
->
O
>
CwHnOyNz
+
aO2
+
bHgOuNicCHrOBNs
+
ACO2
+
substrate N-source
Biomass↓
eH2O
+
fCjHuOeNm
W
↓
of
1
.
organicpat
Vs
=
4w
+u
-
2y
-
3z GeneralBiochemicalRxn
VB
=
u+x
-
2B
-
38
-
ta
C
Availablee-balance
:
wUs-Ya
=
cVB
+
fjUp
TheoreticalOuygenDemand
a
=
wVs-cVB-fjUp
of
NH3istheN-source U

46. ManyfacultativeorganismsproduceCHy(methane)from
C
inlow
EN2O fromN
O2
conc.
CHy-18X
greenhouse
potentialthanCO2
N20
->
270x
Also
,
Ratof
available
e inreactant
i
wVs
-
Ya
=
1
CVB
+
fjUp

47. substrated
wrs
=
cVB
+
fiUp
+
a
1
=
(c)
+
(+)
forMaxBiomassproduction Theoretical
negligibleproductionofotherpots MaxBromassYield
CaxUB
=
1
=
Cmax
=
W
VB

48. Theoretical
MaxPotYield(considerBM&02Fraction
zero)
frax
=
WUs
:
Up
Butinreality
,
some
CO2 I
water
will
be
produced(someO2willbe
consumed)

49. Oues)BiomassconvertsGlu
->
CO2
+
120
al
metabolic
rxnof
Gluis
the
common
resperation
rxnin
presence
of
O2
.
Metabolic
vexn
for
glucose
Thecell
compositionis
CH200
.
55
No
.
2
+
5/ash
.
Yield
of
BMfromsubstrateis
0
.
5919
If
notmentioned
theoreticalobserved
NH3
as
Nitrogen
source
CalculateO2demandwithgrowthElcompare
withdemandwithoutgrowth
.im ca
justresperation
CoMizO6
+
aO2
+
bNH3
e
cCH200
.
35
No
.
2
+
dCO2teH2O
-180gsubstrate 12
+
2+
16(0
.
55)
+
14(0
.
2)
"¥
gogBM
=
236
=
26
.
9glmoa
=>26
.
9xc
=
90 VB
=
y
+
2
-
2(0
-
55)
-
3(02)
=
4
.
3
c
=
3
.
34

50. TheoreticalO2demandwithgrowth
a
=
ebe
Upo
U
a
=
24
-
14
.
362
=
2
.
4095mal
Y
TheoreticalO2demandw/Ogrowth
/just
respiration(
CoH1206
+
602
->
6CO2
+
6120
↓
a
=
6
mal

51. 8thAugust
-
usually
-
Stainlesssteel
Iwithglasslining
or
highgradeforinertness
316L
-
Production
vessel
woodenvesselsdeliberatelyused
inbrabies
toenhancetaste
,
aroma
disposablereactors
can
alsobeused
eg
forMammalian
cellproduction.
(but
very
expensive

52. airinlet
tempbrobe
DOProbl
PH
->
HeadPlate
-
rubbergasket
almostall
- accessoriesimpellers
are
placed
on
thehead
plate
Callholes
on
headplateto
Li
not
create000
,
00
-
S
unnecessarypressure
on
reactorwalls)
GPressureunbalance

53. toarrest
deflectionof
a
longshaftrodintall
reactors
,
a
ballbearingis
addedatthebottom
.
but
unnecessary
shaftweight(in
theemptyworkingvolheadspace
creates
load
on
themotor
.
..
forlargescalesystems
:
BOTTOMmountedshafts
arepreferred
.
with
mechanicalsealtopreventleakage
-largescale
-
designconsiderations noglassjacket(uneven/draste
change
in
tempblu
instead,
&steam
coil
or
electric
la
en
reactor
us
jacket
can
create
thermal
B
000,
00
-
S
heaterstress
that
breakstheglass
Jacket(
·
drain
or
"¥
Bottommountedor
coolantcoil
discardvalue
w
mech.
seal

54. PH
,
tempEDOsensors
atmultiplelocations
Log
mean
DO
emp
Do
-
one
atthetop
,
one
DO
?
largescalereactoratthebottom
,
cm
their
log
mean
diff.
Pressureat
PPh
:very
tallliquid
height
Temp
·
)
pro
so
Dowill
besignificantlyaffected
Control
:
log
mean
Dowillbecalculated.
signal
receiver
<
comparator
-
actuator/activator

55. DO
control
actuatoractivateseither
:
increase
motorspeed
orflow
rateofO2thrusparger
pHcontrol
actuatoractivates
either
:
alkal
pump
it
pump
Tempcontralactuatoractivateseither
:
heatingcoil
orcoolingjacket
depending
on
themag/sign
ofthe
error
value

56. Drawafunctionaldiagramfor
YeastProductionin
a
large
scalefermenter
:
eachstreamis
&)
Deniallin
welldefinedwith
numbers
Efunction
Inoculum
-
Offgas
Clarfer-
>
Sterilizer
-
1
fermenter
Brothfilter
filtrate
->
unit
names
Slurry
->mass
streams
Molasses ↓ labelled
↑ Biomass
->
energy
can
be
Af omitted

57. Now
,
convertthe
functionaldiagramto
an
OperationalDiagram
"¥
mass
Elenergy
inputsspecified
numerically?
Inoculum
-
Affgas
/min
Clarfer-
>
Sterilizer
-
fermenter
brothfilter
filtrate
1 orSlurry
Malaseslgth regin
↓
↑ Biomass
Air4/min

58. Youshould
beableto
make
FunctionalDiagrams
&Operational
wirBiomas
usgiven
ofeach
unit=90%, $
2tonnesof40%.
conc
Slurry
needstobeeg
f Spreparedperday
,
make
an
operationaldiagramfor
thisprocess
ofyeart
&functional production
.

59. 12thAugust
Processflowsheet
is
Itspresentationmustbeaccurate
,
comprehensive
a
complete
similar
to
a
logical
diagram
.
BlockDiagramis
thesimplestformofrepresentationof
a
processwhereeach

60. blockrepresents
a
single
pieceofequipment
Incertain
cases
,
a
singleblock
canbeusedtodescribe
a
complete
operational
stage
.
Usefulfor
representing
simple
processes
.
Butforcomplex
processes
,
itsutilityis
limited.
III
OperationalDiagram/Pictorial
detailed
flow
chetudfordesign
operationwhere
up
drawn
s
BS1553
-or
ANST/AmericanNationalStandardInstitute)
British
Standard
for
set
of
symbols

61. -
-
-
-
L
-
C
O
weight
ElectricalLoadingSpray Rotary
G
devicedevice device device movement
6
↑
H
Boundary
zine-
Stirring
Discharge
to
Equipment
device AtmosphereBranch

62. or
3
- -
I =
HE
------
------
T
-------
------
+S
~ ~
>
Pressure TrayColumn
↓ Vessel
Shell
&TubeHE

63. Fluch
- Contact
a
=
#
1x
E
->
-
-
It
/
-=
~
-
T
↓
↓ Absorption
simplecolumn Russel
withpackedbed
representation

64. -
-
↑
>
↓ ↓
00
OpenTank
Clarir
or
SettlingTank&
-
DometypeGasHolder
CentrifugalPump Rotary
Pump
orRotary
compressor?

65. ElectricMotor
ReaprocatingPump
RollCrusher
.....
x
Filter Filter
Press
RibbonBlender
-
RotaryFilter

66. T
#
1111%
->
"¥ ↓
Centrifugal
compressor Cyclone
separator
BeltConveyor
ScrewConveyor

67. "¥
-
↓
-
-
-
-
=-
Bowl
Turbine
/
Centrifuge

68. ↑
3
T
Elevator
-~ ~
Reactor

69. -
S
a ----
streame
information
in
boxes
y
compulsory/mandatory
+
optionalunfo
one
typeofduignpractice

70. anothertypeofdengnpractice
:
2
-1
S----
y
12345
E

71. EssentialInformation OptionalInfo
·
StreamComposition
Mass
·
Malar%Composition
·
FlowRateof
eachcomponent(mostlyrg1h)
·
PhysicalPropertydata
Oor wocanty
,
I
,
surface
tension
StreamComposition
as
mightfraction
·
StreamEnthalpy
·
TotalStreamFlowRate
(mostly
Mass
flow
Rate
kg/h)StreamName
StreamTemperature
·
StreamOperatingPressure/Required
OperatingPressure

72. #
Whilegivingphysical
propertydata
,
we
report
mean
valuesofphysicalproperties
StandardPractices
indrawing
:
2.
Equipmentshould
bedrawnapproximatelytoscale
2
.
Principal
Equipment(usuallyreactor)should
bedrawn
in
itscorrectposition
la
proportionalityfactorforsizeshould
be
followed
across
equipments)
3.Connectorsto
beusedbluseparatesheets
1000-
lon
next
page) 10

73. markedproperlywithhighprecision
E
basiccalculationshouldshowproper
mass
&energybalances(accurate)
Eltimedependentparametershould
be
rational
&
common
foralloperatinglines.
similaruntconventionsacrosssupplylines
,
backedby
proper
rationale
assumptionsshouldbelisted
atthebottomof
the
table
.
->
for
continuousoperation
,
Imaynotbementored(can
be
calculated
~
butforBatchoperation
,
to
mustbementionedatthebottom/
in
thetable
->
servicewhitesmaynotbedescribed
in
the
main
sheet
-
Symbolsforthemeg
SUI
can
beexplained
in
a
separatesheet
.

74. Abbreviations
forSymbols
:
Reactor
-
R
HeatExchanger
-
HE
Column
-
C
Felter-F
uncommon
abbreviations
can
beexplainedatthebottomofthesheet
.

75. follow
theclassandpracticeregularly
.
14thAugust2024
Flowsheet
calculations
Streamflowsandcompositioncalculatedfrommaterialbalance&diff
designequs
obtainedfrom
the
process&equipmentdesignconstraints
②typesofdeugnconstraints
:
externalEinternal
designconstraints
Withinexternalconstraints
we
have
·
potspecifications
,
safety
considerations
affectdegn
nations
->
greennorms,
effluent"
(
000
effluentis
environmentallysafeIegproducing
Penicillin G
outofallotherpenicillins

76. directlylinkedtoproductionprocess
-Withininternal
constraints
:
-
process
stoichiometry
-
exnconversions
I
yeeld
-
chemicalequilibria
-
Physical
"
-
energybalanceconstraints
Elgenerallimitations
on
equipmentdesign
->
authoritiesaudit
process
plants
[GMOnorms)
O
external
constraintsneedtobedealt
with
even
thetheymaynotaffectthepot
yeld/
qualitydirectlybut
whothe
process
maynotbe
allowedtobesetup

77. effluenttreatmentmayneedtreatment&equipments
willneedtobedesigned
orrecyling
leventho
no
directlinkoftheeffluentto
productioncapacity)
more
more
constrainespopupateachscale-upstep
eg
.
mass
transfer
,
heattransferconstraints
selection
ofparameters/crucial
us
auxillary

78. weakly
dependent
Strongly
dependentparamete
equipmentde
aparameters
or
streams
are
the
ones
associatedwith
equipmentdesignwhose
performancecanbe
assumed
,
or
approximatedw/oentroducing
ugnificant
error
in
theflow
sheet
Geg
.
assumingdensityfor
dilutea
solutionsto
beclose
to
thatofwater
:
assumptions
inP
of
liquidthat'sunchanging
won'tbe
problematic

79. Calendaryear
us
ProductionYear
I
calendaryear
-
dauntime
operationaldays
,
operationalhours
Capacity
per
hourI
alculatedfromhere
/flow
rates
etc)
some
unplannedshutdownsmaybe
accountedforin
a
contigencymargin
C
(emergency/contingencyfactor

80. CapitalDepreciation
Payingyour
Expenditure
inInvestmentemployees
evenCapital
on
unproducturedays
:.
Bare
minimumdowntime
,
only
what
isrequiredfor
maintenance
.
yield
or
molt
per eg
overall
50%
yield
&paper
molRawmaterial
=>
forevery1kgpot
,
2kg
raw
materialneeded
alsocalledscale-upfactor?
scalingfactor

81. scalingfactor
canbe
usedto
estimate
raw
materialrequirementfor
scale-up
or
scale-down
Eachscaleneedstobe
optimisedseparately
El
a
realisticvalueforyieldiscalculated
say,
atlabscale toget
10
,
000kg
antibiotic?
20
,
000kgraw
material
zkg
raw
materi
aa
->
"¥
,
10
,
000ng
antibiotic
0
.
5 kg/kg
yeld
say
now
,
0
.
4
ge
optimise

82. usuallyyield
goesdownasme
scale
up
BUT
largerscalesystems
can
becontrolled
betterwithcontrolsystems
C5)
y
thepot
is
favored
ina
controlledsituationthat
canbebetteroptimised
at
a
larger
scalethenthat
might
givehigheryield
egproductionof
ETOHby
a
microderophile
easier
in
largerscalewhere
low
DO
can
bebetter
controlledrelate
toshaes
a
Softwares
:
ASDEgoodformakingflow
sheets
Chemdraw,

83. P&IDiagram
:
Piping&InstrumentationDiagram
criteriaforselectingpiping
material(costlyinvestment
·
mechanicalstrength
·
chemicalproperties(shouldbe
inert
Wi
easy
maintenance
·
typical
case
of
flowlines
so
sedimentation
,
scaleformation
,
frictionallossshould
beminimumtoreducepumpingcasts
pumps
,
values
,
otheraccessoriestocontrolflow
:
instrumentation

84. a
P
&IDiagramshouldhave
:
all
process
equipmentidentifiedby
an
equipmentnumber
,
and
equipment
drawnmustbe
roughlyin
proper
locationof
nozzles
,
values
,
pumps
shouldbeshown
000allpipesshouldbe
identifiedby
a
specificlinenumber
O
-
o
Repesize
,
materialofconstructionEl
diff.accessoriesofpipes
shouldbe
oO
undicated
(union
,
elbow
,
T/Yjoints
,
sockets
T
Jointhassuddendirectionchange
(909
-
max
pressuredrop
-
chancesofprecipitationofhighwsoety
leg
=>
chancesofclogging
..
Yfountpreferred
over
T

85. accessoriesofpiping
:
socket
me
en
a
-
um
union
preferred
if
it
needs
tobeopenedfrequently
↑
--
>
"¥
itispreferrednottoposition
a
valuehere
butcreationofbypass
line
can
bedonetocreatebetter
control
on
thedeliverylinewo
excess
pressure
on
thepump

86. delivery
line
>
XD
->
bypass
line
failureofpower
willleadtovaluesto
not
workelectricvalues
depending
on
theneed
d
~
failto
openvalues
solenoidvalues
use
->
fail
toclose electromagnets

87. #allvalues
with
an
identification
number
,
type
type
can
beshowedwith
a
symbol
orbe
includedin
thecodeusedfor
identification
ancillaryfittingslikesideglasses
(watch
glass
,
strainertofeltersolidmaterials
fromthefeed
,
moisturetrap
,
steamtrap)
"
,,,,1 can
be
numbered
I
describedbelow
/ANA
,
ANZ...
#Pumpsshouldbeidentified
,
w
their
respective
symbol/codenumber.
AllcontrolloopsEltheirinstrumentsIf
dealing
in
simple
processes
,
keep
utility
Elservice
lines
in
thePnI
drag.
for
complex
processes
,
service dutilitylines
can
have
aseparatedragbuttheir

88. connectionsmustbeshown
on
theequipmentwithnumbers.
Samenumbering
conventions
must
beused
across
difflevelsofcomplexdrags
1
.
c
.
same
numbering
scheme
forblock
-
functionalsoperational
-
Pn7drag.
N FailsOpen
-
Power
Failure-
Value ~ FailsShut

89. -
=
MaintainPosition
LocallyMountedPanelMounted
/InstrumentationControlUnit
I
when
asp
.
levelofvalueneedsto ControlUnit
mounted
remain
open
allthe
time
,
solenoid
mounted
locally)
somewhere
valuescan'tbeused
or
needpowerbackup)
on
thesystemelse
itself

90. MechanicalparameterstoconsiderforPipingDesign
:

91. BBL732
ASSIGNMENT
AMBIKABATRA
2021BB10329

92. Set
BBLT32ProblemstoPractice
AmbikaBatra
2021BB10329
@
Gasleaving
a
fermenterat
...
1atm
25
°
C
comportion
:
18
.
2%
Na &
19
.
2%Oz
mole
%
2
.
6%CO2

93. fromIdeal
GasLaw
,
PV
=
nRT
P
=
1atth
&van
R
=
0
.
0821LatmK
"mel'
T
=
298K
considering1m3ofoff-gas
:vol
%
will
beproportionalto
mole
.
in1ms
off-gas, andin
100molesofgas
:
0
.
782
m3
:
Na 78
.
2moles
:
N2
0
.
192m3
:
Oz 19
.
2moles
:
02
0
.
026m3
:
Coa 2
.
6
moles
:
CO2

94. a)
MassComposition
:
calculating
mass
ofcomponents
: mass
N2
:10
.
2x
98
=
2189o
2189
.
6*100
=
75
.
0%
2918
.
4
02
:
19
.
2X329
mee =
GlUUg
X100
=
2%
CO2
:
2
.
6X44glmal
=
114
.
4g
Y
Y
X100
=
3

95. b)
mass
of
CO2
in
each
[m
ofoffgas
CO2
:
Vol
xno
of
molesof
gas
(PV
=
nRT)
In1m3of
gas
,
volofCO2
=
0
.
026
m
·:
PV
=
nRT
wecan
calc
.
I
(no
ofmolesofCO2
in
thisvolume
1
atmx
0
.
026
X1031
=
n
10
.
0821
(am)
28
n
=
2
=
1
.
06males
0
.
0821X298
Xyng/mel
=
4675g
CO2

96. #
2
CoHi20
+
NH3+
1
.
502
->
CgHaNOu
+
CO2
+
3H28
Glucose L-Glutamicacid
↓
160
+9+14+
64)glma
=
147
glma
massof02requiredtoproduce1479Glutanueand
=
1
.
5X32g
:"
"
159
Glutamicand-18
XI5
147
=
4
.
89g02

97. Ans3
CoHizO6
->
2CHzOH
+
2002
Glucose EtOH
log/L 3
.
29/L
"¥
1gt
Bases
:
11
Ima
Glu-
>
2 malEtOH(theoretically
alractional
conversion
X
we'llfirst
calculatehowmuchglucese
was
directedtowards
ECOHbiosynthesis

98. massofGlurequiredfor
3
.
2gofETOH
2
3)2gEtOM
=
0
.
878molEWH
46glma
from
Stachometry
,
0
.
078malETHrequires
00
=
0
.
033mal l
=
0
.
035X180glad
=
6
.
39Glu
1theoretically
butexperimentally
,
logGlu
was
presentperIL

99. 6.3gglu
was
usedforZWH
SynthesisEl10gwas
available
:
X
=39
=
0
.
63
109(available)
But99
ofGlu
was
consumedfromthesalh
:
X
=
639
=
0
.
78
aglactuallyconsumed

100. b)YETOH/
Gle
99Glu-
>
3
.
29EOH
per
:
Yuld
=
32g
GO
=
0
.
33g
o
And
30
%water
-225%ETOH
15%MeOH byweight
12%
Glyaro
10%acetic
acid
8
%
benzaldehyde

101. In1009
solk
,
30g
water
=
30
=
1
.
67mal logActuaced
=
10
=
0
.
17mal
CH3COON
Go
239EtOH
=
2
=
0
.
34modto 8gbenzaldehyde
=
8
=
0
.
073mol
GH60
106
159CHzOH
=
E3
=
0
.
47mal
32
12gGlyceral=
=
0
.
13mal
C3H803 a2
total
moles
=
3
.
055males

102. Molefraction
:
water
:
0
.
55 glyceral
:
0
.
042
ETOH
:
0
.
18 ACH
:
0
.
056
MeOH
:
0
.
15 Benzaldehyde
:
0
.
024
stachiometry
foraverallsynthes
:
167CH1286
+
2NH3
+0
.
502
+
HaSOU
+
CHgOa
Glucose
->
GoMsOuNaS
+
2CO2
+
9H20
Penicillin

103. a)Maxtheoreticalyieldfrom
glucose
=
smal
Penicilling
glmal
-
1
.
67malGlu
=
0
.
6mol/mal
=
111
gig
b)from
the
1
.
67
molgluavailable
,
only
6
%
(0
.
06X1
.
67)
of
it
was
usedfor
0
0
Penicillin
=
0
.
1
malglu
=
189Glu-
> 1
.
11X18
1
.
67
mol
-
=
19
.
989
Penicillinfrom
300
.
69Glu

104. =
1998
g
a
=
0
.
066
glg/actualyield-
3
.
591
m
50g/LGlu
Yg/LPAA

105. i)Limiting
Reagent
1
.
67melGlu-1mo Penicillin
1malPAA- 1malPenicillin
Yeeldcale
·
un
(b)
=
0
.
066
glgPenicilm/Glu
perlitre
:066
g
Penicillin
X44
.
59Glu(consumed)
g
Gl
per L
=
937 g
Penicillinproduced
=
0
.
009mol
Peneluo

106. perL
:9
PAApresent
=
0
.
02
ma
perL
:
6%
ofGlaavailableforPersullenprod.
=
6
%
of=o
gimal
=
0
.
0167male
every
mal
ofPenicillinrequires
(molofPAA
1
.
67molofGlu

107. &(basis
:
amtofglucoseconsumedforpenicillinproduction
=
6
%
of
49
.
5
=
2
.
679
=
1
.
48X10-2
moles
amtofPAA
availableforpemallinprod
=
4g
=
49
132g/mal
=
2
.
qu
X102
moles
-
[X1
.
67
:
4
.
91X10-2 mol
>amtof
gluavailable
4
.
91X10-2
molesGlurequiredfor
Glucoseis
theLimitingReagent itscomplete
conversion

108. ]
11basis
50g
-
3
.
59/left)
Glu Glu
un
100L
Glucommmed
=
44
.
59-
>
4450gGlucoseconsumed
↓
24
%ofthisusedforgrowth
=
1068gAutal
Glucoseusedfor
growth

109. Actual
ofPenicillin(b)
=
0
.
066
glg
=
0
.
066X
44
.
5X100L
↓Total
Glu
consumed
=
296g
is)final
come.
Of
PAA
.per
L0
.
009molesofDemallin
were
produced
=>
"I
"PAA
were
consumed
=
0
.
009X132g
=
1
.
188g/L
PAAconsumed
=>
1911-1
.
188glL
=
2
.
812
gllleft

110. Ans
MFfor
cells
:
Cy
.
th
.
301
.
2
No
.
so
cubstrate
:
C6HzU
CoHzn
+16
.
2802
+
1
.
42NH3
->
1
.
65
Cynth
.
301
.
2
No
.
so
+8
.
74CO2
+
13
.
11
H20
a)checkingofbalancedof
IHS
=
RHS
16
=
7
.
26+
8
.
74-
yes
,
balanced3
3
+4
.
26
=
12
.
045
+26
.
22v
32
.
36
=
1
.
98
+
8
.
74X2
+13
.
11-
N142
=
1
.
65x8
.
86
-

111. b)
mal
Citzu-1
.
65moles
Cynth
.
301
.
2
No
.
so
(12x16
+
34)g
-> 1
.
65(12X4
.
4+7
.
3+
16x1
.
2+14x0
.
86)
226g
-
=
150
.
7g
130
.
79
cells
226ghexadecane
=
0
.
67gig
C)
"¥
.
T
galls
=
0
.
29
glg
16
.
28X32g0a

112. d)finally
:
2
.
5kgofalls
=
2500gcells
=
2009
=
274mala
i)
mun
amtofLenadecane
=
1
X
molesofcellsneeded
(for
100%Conv
.
)
1
.
65
=
16
.
6moleshexadecane
=
16
.
6X
2269
=
3751
.
6g
=
3
.
75kg

113. i
min
come.=
6g
=
125
ii)
air
at
20
,
1atm
27
.
4molesofcells
require
28X274moles
of
=
270
.
35moles02
PV
=
nRT

114. assuming
a
is
21%O2
=>
total
molesofairrequired
=
n
=
2035
X10
=
1287
.
36moles
air2xn
=
270
.
35
PV
=
nRT
latmxV
=
1287
.
36X0
.
0821Latm
X29317
-
Kmol
V
=
30
,
9671
=
30
.
9m3

115. An
Corn
steep
liquor
Beetmolasses
2.5%invertsugars 50
%sucrose
50
%
water
1
%
invertsugars
18%
water
by
Joe
e
#Okg
+x
->
2%
W/Winvertsugars
2.5
%
of125+11.of45
=
2%of1170
+
x)
3.373
=
0
.
02
/170
+
x)
=0
.
024
=
3
.
575
-
3
.
4=n
=
0

116. sucrose
50
%ofU5kg
=
5%
of(170
+
8
.
75)
22
.5
=
2x178
.
73
100
=>
S
=
12
.
6%
AM8EIOH-AcH
↓
->
LughAch
CHOH
->
CHaCOOH
+
02
+
H20
cone
ofAct
Air-
=
12%
200mal/h

117. per
hourbasis
:
a)
20009Act
produced
=
glmal=
33
.
33
moles/h
Ima
ETOH->
ImalAct/under
100%
conversion)
..min
.
33
.
33moles
ECOH
required
=
1533
.
33g
ELOH
b)
20009AcHmustbeat12%come.
inside
reactorxu
=
Gu
=
200
=
1667kg
of so
=67kgwataIG
-
2kgACH

118. c)Comportionoffermenteroff gas
33
.
33molesAcHrequire33
.
33moles02perhi
input
rate
=
200ma/kAir
assuming21%O2
79
%
Na
=>
Input
:
42moles02
,
158molesNa
consumed
:
33
.
33moles
Output(Offgas)
:
(42-33
.
33molsO2
,
158molesN2)
Offgas
=
5
.
2%
02
,
94
.
8%
Na

119. AmaXanthur
gum
Experimentally
,
Yoz/e
=
-
0
.
23
glg YH20/m
=
0
.
13glg
YNH3/Gl
=
-
0
.
01g/g
ugum/Glu
=
0
.
75glg
Yals/Gm
=
0
.
09glg
YCO2/Ge
=
0
.
27 glg

120. T
Guen
gumque
=
0
.
75 g1g
20
,
000L =>0
.
75
=
700-
130kgoffers
X
cut
in
20
,
000Lwater x
=
10X
final
gum
conc
.
=
3
.
5
% 3
(20
,
000
kg)
=
933
.
33g
35x20
,
000
=
700kggum
finally
produced
3
X
(20
,
000+
x)
=
x
700
+
0
.
035x
=
x
-
700
=
0
.
965nTh
=

121. a)amtofGlu&NH3
required?
amtofGluconsumedtoreachthefinal
come
ofgum
=
933
.
33g
amtofNHsconsumed/gofGluConsumed
=
0
-
01glg
=>
totalNH3consumed
=
9
.
33 g
-
b)whatpercentageexcess
airis
provided?
genen
Y0z/Gm
=
-
0
.
23919
=
0
.
23glgX
933
.
33g
=
214
.
7g
02consumed
Ycoz/glu
=
0
.
27 g/g
=
amtofCO2
=
0
.
27g/gX
933
.
339
Glu
=
252
q
CO2formed

122. 200rymair-
ProductDig
3.5%
Xanthangum
=
0
.
035Pkg
20
,
000L
M0
.
767AlgN2-
-
> 30kgoffgas
"¥
>Air
final
gum
conc
.
=
3
.
3
%
wit
Akg
moles
[L23
t

123. PreliminaryCalculations
:
Airis
21
%
O2
79
%
Na
In1moleair
0
.
21mole02
=
21x32glmo
=
6
.
72g
0
.
79moleN2
=
0
.
79X28
glma=212g
228
.
849
ut%:02
=
62X100
=
23
.
3%
N2
=
x100
=
76
.
7

124. SinceNaremainsunreacted,
76
.
7%
ofAkg
remains
unchanged
=
0
.
767Akg
in
offgas
.
MassBalanceTable
:
(kg)
In
StreamGluO2N2CO2 Gum
CellsNH3H2O
Total
Feed ?000 00?
20
,
000
F
Air
00
.
233A0
.
767A
O A
Pot
-
---880
O
offgas
-
-
-
- - -
g
Total F
+
A

125. OUT
StreamGluO2N2CO2 Gum
CellsNH3H2O Total
Feed
-
-Air
i
-
Pot P
offgasO
?
0
.
737A?
O00 O 1258
Total
0
.
737A 0
.
0359 1250
+
Pkg

126. Yoz/ge
=
-0
.
23
g/g
YNH3/Gl
=
-
0
.
01g/g
final
gum
conc.
in
Pot
=
3
.
5%
=
0
.
035Pkg
ugum/Glu
=
0
.
75glg
:
0
.
35ggum
is
produced
per
1g
Gluconsumed,
Yals/Gm
=
0
.
09glg
YCO2/Ge
=
0
.
27 glg
YH20/glu
=
0
.
13glg
=>5
g
Glucomumed
=
0
.
0467P
hgGluIn
OverallMB
:
F
+
A
=
P+
Affgas-Q
NH3consumed
=
-0
.
01gigX0
.
0467PRgGlu
=
0
.
000467PkgNH3
F
=
20
,
000kgwater
+
0
.
0467PGlu
+0
.
000467PNH3

127. ①becomes
20000+0
.
04674
+
0
.
0004674
+
A
=
p+1258
18730
+
A
=
0
.
953P
-
②
O2comumed
=
0
.
2391gX0
.
0467Pkg
=
0
.
01074P
CO2produced
=
0
.
27glgX0
.
0467P
=
0
.
0126
P
=>Offgas
:
1230kg
=
(0
.
233A-0
.
01074p)
+
10
·
767A)
+
10
.
016p)-
Oleft
N2
same CO2produced

128. :
1250
=
A
-
0
.
010744
+0
.
0126P
1250
=
A
+0
.
00190
-
&'
/
18750
=
-
A
+
0
.
9534
-
2
Adding
1
+
20000
=
0
.
9549P
=>
p
=
20
,
944
.
6hg
..
O2consumed
=
0
.
01074 p
leg Gluconsumed
=
224
.
94
=
0
.
0467P
hg
=
978
.
11
kg
NH3consumed
=
0
.
000467
x
20
,
941
.
6
=
9
.
78
lg

129. b)What
encers
airis
provided?
02provided
=
0
.
233A
=
281
.
98
O2required
=
0
.
01074P
=
2249Ukg
from②
1250
=
A+0
.
0019
9
1250
-
0
.
0019/20
,
944
.
6)
=
A
A
=
1210
.
205leg
%excess
provided
=
281
.
98
-
224
.
94X100
224
.
94
=
25
.
35
%

130. #10Cozy
+
a02
+
bNHz - CH16600
.
27
No
.
2+
dCO2
+
et2O
genRQ
=
0
.
43
=
d
=
0
.
43
Individual
elementbalances
:
a
=
d/0
.
43
CBal
:
16
=
c+
d
-
②
c
=
16
-
d
-?
HBal
:
34+
36
=
1
.
66c
+
2
-
③
O
Bal
:
2a
=
0
.
27c
+
2d
+e
- ③
becomes
:
NBal
:
b
=
0
.
2c
-
⑤
34+
3/0
.
2c)
=
1
.
66c
+
2e
3410
.
6116
-
d)
=
1
.
66(16
-
a)
+
2e
-
⑤
⑧
becomes
:Ed
=
027(16-d)
+
ad
+
e-x

131. 34+0
.
6/16
-
d)
=
1
.
66(16
-
d)
+
2e
-
⑤d
=
0
.
27(16-d)
+
ad
+
e-x
-
9
.
3d
=
-
0
.
54(16-d)
-
1d-2e
- "
3+9
.
6
-
0
.
6d
=
26
.
36
-
1
.
66d
+
e
-
'
43
.
6
-
9
.
9%
=
26
.
56
-
8
.
96
+
0
.
54d
-
4d
-
1
.
66d
43
.
6
-
9
.
99
=
17
.
6
-
5
.
12d
26
=
9
.
9d
-
5
.
12d
26
=
4
.
78d
d
=
5
.
44
,
a
=
=
126
,
c
=
16-5UU
,
b
=
02XO
=
10
.
56
=
2
.
11

132. from
⑤
43
.
6
-
26
.
56+
(
.
66
-
0
.
6)d
=
2e
e
=
11
.
40
cells
ANDCoHi2O6
+
602
-
GCO2
+
6128(duringgrowth)
allcomposition
:
CH1
.
84 00
.
53
No
.
2
+5%ash
MWx
=
12
+
1
.
84
+0
.
55(16)
+0
.
2/14)
=
26
.
78glmoh
0
.
95
YXs
=
0
.
5g1g
-
0
.
59
X
=
3
.
36
gmo
I MolarYuld
-
26
.
789/mol
-
19/1809/molGlu

133. TheoreticalOuygenDemand
a
=
wVs-cVB-fjUp
of
NH3istheN-source U
MolaryeldofBM
is
thestouch-coefffor
BMofstorch
-
Coeff
ofsubstrate
=
1
VBforCH1
.
800
.
53
No
=
u
+
184
-
2(0
.
53)
-
310
.
2)
=
4
.
14
2
a
=
G(u)
-
3
.
36(4
.
14)
=
2
.
52
Y
..Ofdemandwithgrowth
=
2
.
52gmal02/gmd
Glu

134. Inthe
absenceof
growth
,
no
Glu
is
consumedforBNsynthesisItallofitis
oxdisedto
CO2
+
H20
aspertheresperation
eqh
:.
O2demandW/Ogrowth
=
GymolO2/gma
Glu
b)
EtOH
as
substrate
fractionalallocationoffrom
subs
to
subs
TBM
subs
to
of a=
WVs
MaxpombleBM
yieldis
whenallesgofromsubs
2
BM
1
.
e
.
In
absenceof02
=>1
=
C
&wopatformation
WVs

135. CaM30H
:.
Smax
=
=
=
6
forEtOH
,
Vs
=
6 forGlu
,
Vs
=
&
UB
=
4
.
14
VB
=
U
.
14
W
=
2 w
=
6
Cmax
=
u
Cmax
=x
u
=
5
.
80grnd/gmd
=
2
.
9gmd/gmd=
26
.
78g
=
1
.
69g
o
=
32678
=
0
.
80g

136. onmass
base
.
YSH
*
2) YXsal)
And
CH3COOM
+
NH3
-> Bromass
+ T
H20
+
CH
ICH
1.
400
.
u
No
.
2)
Y
CO2/s
=
0
.
67kg/kg
=
0
.
67gly
=
167g
=
0
.
911small goal
ugimal
Togimal

137. CH3COOH
+
bNHz
->
cCHinOo
.
uNo
.
2
+
dCO2
+
eHzO
+
fCHu
↓
0
.
914
performingelementalbalances
:
c
=
2
=
c+0
.
914
+
f
-
0
-
f
=
1
.
086
-
C
i
Hi n+3b
=
1
.
4c
+2e
+
If
-
②
u
+0
.
6c
=
14c
+
2e
+
4(1
.
086
-
c)
0
:
2
=
0
.
4c
+
2(0
.
9(4)
+e
-
-
2
-
1.828
=
0
-
4c
+e
-
N
:
b
=
0
.
2
-
&:
0
.
6c-1
.
4c
-
2e
+
yc
=
4
.
344
-
48
:
04c
+e
=
0
.
172
3.20
-
2e
=
0
.
344
-
②"
I=
0
-
172

138. f
=
0
.
91 =>
YCHy/X
=
0
.
914qmal/goal
b
=
0
.
0344
e
=
0
.
1032
MaxPossibleYeldofMethane
is
whenallfromsubs
-
pot
orn
=
fi
j
=
1
CHU
Up
=
4+4
=
S
Tfmax
=
CH3COOH
Us
=
8
+
4
-
4
==
gmo/gid
=
..91
.
4%of
max
yield.
=
4

139. Ans13
:
all
Crubs-
>
BM
↑
0-96-6 012
.
76
CH2OG
+
bNHz
-,
Chis6005yNo16
+
d
+
eH2O
VB
=
y
+
1
.
56+0
.
54(2)
+0
.
161
-
3)
=
4
MWofcells
=
12
+
1
.
36+0
.
54(16)
+0
.
16
/14)
=
25
.
73
glad
0
.
95
Performingelemental
balances
:
Actual
YXs
=
G gmal/gmal
6
=
C
I
=
0
.
86
g19
b
=
0
.
16c
=
0
.
16x6
=
0
.
96
6
=
0
.
54(6)
+e
=
e
=
2
.
76 (max
=
w
=
=
Ggmalgn
a
&

140. :
actualyeld
ofBM
=
Maxpossible
yield
i)
AnotherSystem
:
CHzOH
+
aO2
+
bNH3
->
cCH16800
.
36
No
.
22+
dCO2
+
eH2O
MWofBiomass
=
12+
1
.
68
+0
.
36(16)
+
0
.
22(14)
=
23
.
96
glmal
0
.
94
VB
=
U+
1
.
68
-
210
.
36)
-
310
.
22)
I
Us
=
Y
=
4
2
=
6
=
14.3

141. ElementalBalances
:
1
=
c+
d
->
d
=
1
-
C
4
+
3b
=
1
.
68c+2
-
4
+0
.
66c
=
1
.
68c
+2
1+
2a
=
0
.
36c
+
2d
+e
=
1
+2a
=
0
.
36c
+
2
-
2
+
e
b
=
0
.
22c
Cmax
in
the
case
=w
=
means
1
.
03g
C-malyieldfrom
glucess
isIgmal/gmo
Reasonforincreasedyield
:
highVsforMethandthan
Glucose.

142. i)
quen
,
actualyield(c
=
0
.
42X
1
.
40
gma/gmal)
=
0
.
59gena/gmot
for
no
potformation
,
O2demand
a
=
1/wVs-
CVB)
=
1x6
-
0
.
59X4
.
30
=
0
.
87gma/gmot
Y
=
87g/gMWforbath
same

143. A
->Of
gas
(47LO2
,
I5LCO2
u
10
day
14239
3%
Glu
1
.
75
%NH3
->
Draineda
9
5
.
25%
H20
M
1110
g
0
.
063%
Glu
latene
,
25
°C 1
.
7%NH3
22c/minfor
10
days
CH1206
+
aO2
+
bNHz-cCHaOpNg
+
dCO2
+
eH2O

144. Total
are
spargedintoreactor
in
10days
=
22cm3
X
10&X24h
X
60 mu
Tur
V
=
3
.
17x105am3
:
PV
=
nRT
n
=
PV
RT
=
latix3
.
17X10L
=
12
.
95gmd
0
.
0821X298K
Au
:
21
%
O2
,
79%
N2
=>molesof02
un
=
0
.
21X
12
.
95
=
2
.
72gud
=
87
.
04g
=>
" "N2und
out
=
10
.
23mel
=
286
.
49

145. Total
mase
of
our
in
=
A
=
373
.
449
offgas
:
UTLO2
,
15LCO2S -
n
=
PV
=
0
.
613gmal
-
n
=
PV
=
1
.
92qual RT
=
26
.
979
RT
=
Glung
=>
man
of02consumed
=
25
.
60g
&Total
massofoffgas(G)
=
374
.
81g(02
+CO2+
N2)

146. Total
man
balance
:
1798
.
44gtotal
mas
in
=
(1110
+G
+
B)
out
=>B
=
313
.
639
a)
:
thedry
biomassproduced
=
1X313
.
63g
=
20
.
92g
IU
El
mass
of
water
in
biomass
=
(4X313
.
639
=
292
.
T1g
1+14
Glubalance
:
42
.
79
in
=
0
.
699gout
+consumed
Gluconsumed
=
12
.
05g
NH3balance
:
24
.
94g
=
18
.
87g
+
consumed
NH3cons
.
=
G
.
07g

147. b)Reactants Pats
go
=
=
On a
Hoo
=
2583
=
1
.
433good
UN
=
G07
=
0
.
357
and
NBM
=
20
.
92
MWB
these
are
storhometriccoefficients

148. 0
.
233
(6H1206
+
0
.
802
+0
.
357
NH3-2092CHROBNs
+O
.
GB
O
+1
.
435H28
=
0
.
233
(for
per
quo
glu
(M1206
+3
.
4302
+
1
.
53NH3
-T9
CHOpNs
+2
.
63C02
+
61
ElementalBalances
:
C
:
6
=
0979
+
23
H
:
12
+
3 (1
·
53)
=
3
.
37a
+
2(66)
=x
=
1
.
27

149. 0
:
6
+
2(3
.
43)
=
3
.
37B
+
2(2
.
63)
+6
.
16
=
7
B
=
0
.
43
N
: 1
.
53
=
3
.
378
=
8
=
0
.
45
:
CHaOpNs
:
CH
,
2700
.
43
No
.
13
c)
MalesofGluIn
=
0
.
24gial
7
02In
=
2
.
72
gid
"
NH3In
=
1
.
47gial
O
,
E,
1
:
bySC)
"¥
smallestGlu
iszinting
.

150. d)YX/s
=
2929
BM
=
0
.
5g/g(dryweight)
42
.
059Glu
Aus15GlycerolC3HsOMW
=
92g/mal
C3H8O
+
a02
+
bNH3
->
c CHaOpNs
+
dCO2
+
eH2O
from
PaulineDoranTable1
.
3
,
wecan
taketheorganum'sMolecularFormula
as
CH1
.
7300
.
43
No
.
22
VB
=
4+
1
.
75
-
2(0
.
43)
-
310
.
22)
=
4
.
23
MWBiomass
=
12+
1
.
75
+0
.
43(16)
+0
.
22
(14)
=
25
.
8
glma
0
.
92

151. YX1squen
=
0
.
4g/g
Glyc
O2requirement
a
=
1WVs-cUB)
=
(3x4
.
67-1
.
93X423)
=
1
9gl a
gund
=
11x3
An16
anerobegrowthElpatformation
=
0
.
69glg
CoHi206
+
bNH3
-
>
CHi800
.
3
No
2
+
dCO2
+
eH2O
+
fCcHsO

152. YXs
=
0
.
11g1gforyeast
,
0
.
05g1gforbacteria
for
yeast
,
c=
MWs=g
X180
=
0
.
81
gmol
na
for
bacteria
c
=
5 X88
=
0
.
37geollgno
a
ElementalBalances
(Yeast]
C
:
6
=
c+
d+
2f
=
0
.
81
+
d
+
2f
=d
=
5
.
18
-
2f
H
:
12+3 b
=
1
.
8c
+
2e
+
6f
->
10.54
+
3b
=
2e
+
6f
0
:
6
=
0
.
3c
+
2d
+
e+
f
->
5
.
595
=
2a
+e+
f
N
:
b
=
0
.
2c
=
0
.
16

153. Bacteria
:
10
.
54
+
310
.
16)
=
2e
+
6f
C
:6
=
c+
d
+
2
e
=
5
.
51
-
35 Hi1810
.
37)
+2e+
6f
-
11
.
33
+
3b
=
2e
+
67
68
=
10
.
2939
0
:
0
.
310
.
37)
+
[d
+e+
f
->
5
.
815
=
2d
+e+
f
Ni0
.
210
.
37)
=
0
.
074
f
=
here
e
=
5
.
78-3f
:
Yps
=
1
.
72gind/smal6f
=
11
.
225
f(pat)
=
1
.
87
YPIs
=
1
.
87
gmol/gmot

154. b)fruax
=
s=
=
2amoloma
·
yeart:86
ofthereticala
is
An
assuming
no
pots
:
CH1206
+
a02
+
bNH3
->
c CH1
.
1900
.
36
No
17
+
dCO2
+
eHcO
c
=
Y
MWsubs
=
2
.
65gmalan
a Y
XIS
=
0
.
37glg
MWcells
=
25
.
16glia

155. ..
2
.
65gmdcells/gmal
Gluconsumed
O2demand
:
0
.
88g1g
=
0
.
69
gmal/gmal
butobserved Odemand
=
qu
(26gma l=
VB
=
(xh
+
1
.
79X
1
-0
.
56(2)
-
0
.
17(3)
=
4
.
16
TheoreticalO2demand
:
a
=
1wVs-cUB)
=
3
.
24
Cegnificantlyhigher
thanobservedformation
ofotherpotsacting
as
-acceptors
likely

156. Ansi8MW(NHula
Son
=
132glial
MWBM
=
26
.
16
glial
14
Bases
:
Prodof25galls
->
23
=
0
.
956
gendllo
1gendcells
-
8
.
25
gidN
0
.
956""
-0
.
239gidN(from
N
source
Nsource
(1gmal)
-
2 N
:
@
andN
sourcereq
.
=
O
.
12X132glma
=
g
to
:.nun
conc
=
15
.
9
glL

157. Ans19Guun
,
Proteinpat
:
CH1
.
33 00
.
31
No
.
2
E
.
coll(frombook)
:
CH1
.
7700
.
ya
No
.
zu
CGH206
+
a02
+bNHz
->
c
Ch
1.
7700
.
na
No.
zu+
dCO2
+
eH28
+
f
CH1
.
3500
.
31
No
.
25
MWBM
=
25
glma
MWpdt
=
22
.
83gima
YPIs
=
0
.
2X0
.
48
=
0
.
096g1g
given
YMS
=
0
.
48glg
f
=
YXMWs
=
Ozagnd
c
=
08
x180
=
3
.
46
gmd/gmo smal

158. a)
NBalance
:
b
=
0
.
24c
+
0
.
25f
=>b
=
0
.
24(3
.
46)
+
0
.
25/0
.
79)
b
= 1
.
03
NH3rea
=
1
.
03
gmd/giol
Glu
b)Up
=
U+
1
.
55
-
0
.
31(2)
-
0
.
23(3)
=
1
.
18
Ozdemand
a
=
WVs-cVB-fiVp
=
24
-
3
.
46(4
.
07)
-
0
.
79(4
.
18)
& &
= 1
.
63qmo/gma
Glu

159. c)
If
f
=
0
but
c
=
3
.
46
NBalance
:
b
=
0
.
24c
=
0
.
83
:.me
cutE
.
coleNH3requirementdecreasesfrom
1
.
83to0
.
83
gia/gmd
Orrequirement
:
a
=
wV-cUB
=
24
-
3
.
46(4
.
07)
=
2
Y
O2
demandAses
in
cutE
.
cole
from
1
.
65to2
.
48gmol/gmol

160. Ans
20
CH60
+
a02
+
bN13
->
c CH1800
.
3
No
.
2+
dCO2
+
eH2O
+
fCaHyOa
MWBm
=
24
.
63
glad
YXIs
=
0
.
149
=
MX46
=
0
.
26
gmol/gat
i
YP/S
=
0
.
92glg
=
2 x 46
=
0
.
11gmd/gd
a
Us
=
6
,
Up
=
4
,
Ve
=
4
+
1
.
8
-
0
.
5(2)
-
0
.
2(3)
=
4
.
2

161. O2demand
:
a
=
wVs
-
cUB-fjUp
=
12-0
.
26(4
.
2)-0
.
71x2XY
& H
=
1
.
31gmol/gmol
WithGrowth
:1
.
31gmalO2viaper
gmaGlu
WithoutGrowth
:
Igna
O2
reapergiatEl
=>39
%increaseun
02
demand
.

163. AugustMaterialsofConstruction
forProcess
Equipment

164. loss
of
material
dueto
erosion
productioncost
1
.EffectofStress
2.
Erosion
3.
High
TempOxidation Operatingtemp
Clowalloysteelshowhightempoudation
rapidly
->
<500
°
whereas
Crcontainingalloys
can
withstandhighertemp
>500%
chromealloys

165. 4.HydrogenEmbrittlement
-lossofdubblyof
a
material
caused
by
absorption
Hydrogenassistedembrittlement
majorissueswithmetalhealth.
SelectionofMaterialforCorrosionResistance
:
factors
:
1)Temp 5.Oxidation/Reduction
2)
Pressure
S
.
Slurry/PureLiquid
,
3)pH 7.Mixingof
liquid
ina
reactor
4)Purityofalloy
,
impurities
present8
.
Contaminationligorousmixing
O
willcreate
abberations
on
wall)

166. MaterialofConstruction
:
1)
Tron
2)Steel
->
difftypesbased
on
compositionI
c content
i)LowCarbonSteel
(MildSteel/MS)
easy
transformationwelding
,
bending
strength
,
ductility
,
workability
.
g
notresistanttocorrosion
can
beused
inindustrialenvironmentswithlesscorrosivematerials
/solvents
eg.organsolvents
ii)
StainlessStul(SS)
mostcommonly
used
inindustry
.
Crcontent12%
resistscorrosion
inadd"tobeingstrong
,
Ihaving
good
workability
,
ductility
it
is
alsoresistantto
corrosion
.
YangCr
%Usesresistanceto
corrosionbutshouldn'tbeTsedbeyond
20%
otherwisemechanicalproperties
are
compromisedIstrength
,
duckling

167. Add"ofNi-
>
Resistancetocorrosioninreducingenvironment
Add"ofCr
>
"
""Oxdiing"
MengC%
.
beyond
a
point
-
buttleness
C%maintained
<1%
confers
strength
Cry"
>12
%*
20%
Nix
.
"
<10
%
Crackpropagation
in
brittlematerials

168. Ferritic
->
13-20%
Cr No
Nickel
10
·
1
%C
RestFe
Austentic
->
18-20
%
Cr
NoCarbon
mostcommonlyused >7%
Ne used
inmildCorrosue
env.
Mortensitic12-10?%Op
esNi
22
%
No
Type
sav
also1818steelSS
0
.
2-0
.
4%C
SS301
or
18/8SS

169. Type
316Steel
:
Molybdenumadded
eg.dilute
SS316
(Mo) higheramtofCI
more
resistancetoreducing
corrosiveenr
CorrosionResistanceofSS
SSTypes
:
306 304L 321 316 316L318
Ranking
& 1
.
D 1
.
1 1
.
25 1
.
3 1
.
6
(CorrosionResistance
Scale)
selectappropriatetype
ofSSbased
on
corrosion
resistance
needs

170. Deignstress
shouldnotexceed
Mechanical
stress
StrengthofSS
(N/mm)
Temp
is
a
crucialfactorin
strength
Temp
(oc)
300 400 500 600
MS 77
6
e
-SS304
or
18/8108 62(
prometoThermalstress/

171. Monalmetal?
Ni-CuAlloywithN2
:
Cu
=
2
:
1
:
significantimprovement
in
corrosionresistance
food
industrystillwes
e
containers
alloyreatoras
->
C
gettleachedintofood&YesNututionalvalue?
->
Antibacterialproperties
Cu-Su
:
Bronze
Scannotdeal
w
ammoniacalenvironment
Su-In
:
BrassdelH2SOuproblematic
can
handlealkalineconditions

172. Allining/based
alloys
are
highlycorrosionresistantbutlowtensilestrength
Al
basedalloys
can
beusedin
:
TextileIndustries
I
strengthnotvumpbutanti-corrosion
Food
StorageofReactureMaterials
Duralumin
:
improvedtensilestrengthE
also
corrosion
resistant

173. Lead(Pb)
strongresistantto?
Titanium
goodresistancetoredying
enr
but
experie
lunert)
biocompatiblematerials(prosthetics
HeatExchangers
(cyclicthermal
stresstolerated
Plasticsplasticlivingforante-corrosion
wheremuchmechanicalstrengthnotrequired

174. 1)
ThermoplasticPVC
,
etcPTFE(Teflon)
,
Polyethylene
softerdueto↑se
in
tamp
2)Thermosetting
regedmaterials
,
crosslinked
Polyester
,
epoxy
resins
pipes
,
living
ofvessels
Composite
materials
ofpolymersfellers
&
plastasescan
semech.properties
can
havemechanicalstrength
,
good
corrosionresistanceetc
00
-
pump
impellers
,
pipe
fillings
,
values

175. FRP
:
FiberReinforcedPlastics
Tease
or
Cfiber
Ases
mechanicalstrengthofThermosettingplastics
(asmuch
as
MS
advantage
:
cheap
.
piping
,
flow
ofadnesseliquid
,
high
pressureliquid
Disadvantages Advantages
->
plasticsare
flammable
I
-
GoodResistancetocorrosion
->
interact
withorganic
solvents
socan'tbeused
-
Mech
.
Strength
can
beusedby
O
withthem
(
can
dissolve
in
someorganic
reinforang
with
fibers
solvents)
Is
easilyhandledilmineral
acide,alkalis

176. EpoxyResins(expensive)but
goodresistancetoAlkalis
->
strength
~
MS
~
muchhigher
corrosion
resistance.
Gaskets
Rubber
,
metal
,
asbestes
softmaterialthatshowsgoodstretchability
Naturalrubber
canalsobeusedforpiping
->
and
,
alkalitolerated
->
but
notresistantto
cone
HNO3
or
organeacids

177. CeramicsElGlass
:
Brittle
but
vv
resistant
corrosion
"¥
Borosilicate
glassvV
commonlyusedun chemical&Brochemicalindustries
(Glasslined
SSwesels)
resistantto
:
and
,
alkale
,
salts
,
oxidying
,
reducingem
,
organeacids

178. PackingMaterials
->
enert
->
light
->
provdeSA(forbroflm
form")
->
durable/mech.strength)
stonewareused
intrickling
felters
,
packedbeds
Asbestesusedforliving
Einsulation
3utilitylines
Glasswool
Refractorymaterialswhendealingwith
v
hightemp

179. heatresistantbricks(refractorymateriallined)
es
Pyrolysesreactor800-900
SS/MS/Titanium
vesselwith
inner
wallmadeofandresitantbrucks
or
-externally
asbestos(refractorymaterial)
providesMech
.
Strength
Coating
:
barrier
biwthe
materialI
outside
en
When
we
requireprotectionfromand
,
heat
,
light
,
ete

180. Microwave
over
cavityisprotectedwith
a
ceramicpaint?
thermalpaint
provides
thermalstability
stableat
1100
%
C
PowdercoatedMStoprotectMSfromCorrosion
Spraypainting
g
Hughtempspraypainting decreasescrack
Metalexposedtohightempthenspray
painted propagation

181. intermediate
compounds
(Primers)firstpainted
(Binder
1-
kindbendtopaint
tometal
Thus
iseadhesion
of
thepainttothe
metal
.

182. "¥
ItAugust
MechanicalDesignof
ProcessEquipment
StaticLoad
W
W
STRESS
W
&
↓
f
=
wA
↓
near
stre a
at area
↑
W
↑ ↓
+
s
=
W
A
W ofchange
in
dimension
occurs
CompressionTension
duetofs
,
thenextentof
deformation
is
measuredbystrain

183. STRAIN
=
*
dimension
dimension
THERMALSTRESS
-1-E
E
metal
rodfixedin
placebut
temp
dimensioncan'tchangebut
stress
buildingup.
↑
Temp=coffof
linear
panin
a f
=
ExXt
↓
f
=
Thermal
Stress
Modulusofelasticity

184. if
freerod
:
ItwillfreelyexpandXL
4)
=
LXOt
stram
=
=
XXt
E
=
stre tram
ot
+
E
=
-f
=
Edit
beam
------
neutral
axes
umaginaryplanewheretensile
&
compr
.
stress
is
zero
neutralplane

185. Stress
causedbybendingEn---o
M
=
M
=
f
=
f
M
=
Bendingmoment
=
Momentofresistanceatthecross-sectionbeingconsidered
I
=
Momentof
inertia/second
momentof
area
of
cross-sectionabout
the
neutralaxis
f
=
Tensile/Compressure
bendingstressat
a
distance'Y'from
the
neutralaxis
z
=
Z
=
Modulusofsection

186. STRUTS? failuredueto
↓
h
compres
a
Crippling
or
Buckling
longitudinal
compressionmay
cause
Structe
5
CripplingLoad
or
BucklingLoad
Short
struts
-
failureofthe
material
duetodirectload
p
=
fo
A
↓
↓
->
A
=
xnal
area
Itoload
LoadCompressive
stress-
mod
ofelasticity
P
=
a
<El-Mol
Fuler'sformula
"¥
lengthof
body/Light
ofchamber

187. LongStruts
↓
↓d
C= Sufferingrunge
preventbuckling
-
-
-
our
ultunate
objective
is
tocale-Thicknessof
materialbytakingcumulative
stressexperiencedby
a
body
(I
can
withtand
it
+
IPYincperyearcorrosion
value
Isafety
factor

188. E
-applied
torque
=
Torsional
stress
I
=
"¥G
L
Ip
E
T =
appliedtorque
Tp
=
polarsecondmoment
,
of
area
fs
=
shearstressatradius
i
G
=
modulusof
rigidity
T
=
-I
O
=
angleof
twist
overthe
length's'
=
fsZp Ip
=
Ep
=
polarmomentofsection
M

189. P R
Todd
17
z
T
+H
...
--H2
fz=Spy<R(3m
+
)
- (m]
fu
=
3by((3m+
(R]

190. fz
and
fo
are
maximum
when
yismaximum
El
n
is
min
.
fz
=
fu
=3
33MR R
the
=
poisson'sratio
Crateofthestrainin
transversedirtothatof
longitudinaldirection

191. Uniformlyloaded
and
filled
at
theedges
:
stressatthecentre
oftheplate
:
fu
=
fz
=
M=
=
parson'srai
a
stressatthearcumference
:
fr
=(fr=

192. Perforated
plate
Maxi
stress
in
perforatedplate
g
Sp(pitch)
feax
=
max"
stress
insolid
00
ligament
effeciency
-
Rectangularplate
unformlyloaded
,
supported
atthe
perimeter.
-max
=
]
+
a

193. Generalequforthicknessofflatplate
t
=
CDP/fmax
Y
thickness
C
=
constant
;depends
on
edgesupport
(when
M
=
0.3typicallyfor
steel
of
eagesare
completely
rigid
c
=
0
.
43
ofedgesarefreetorotate
-
c
=
0
.
56
D
=
Effectureplatediameter
f
=
Maxmallowable
stress
(designstress)
1
.
e
-
calcCumulativestressX
safetyfactor

194. andSeptember
,
2024
1)CrumferentialStress/Hoop
Stress
2)RadialStress
3)
AvalStress/Longitudinal
stress
Thumbrule
:
Ratio
ofthicknesstointernal
diameter
is
less
thanI
,
then
it
may
be

195. assumedthatthehoop
stressandlongitudinal
stressareconstoverthe
thicknessofthe
cylender
theRadial
stress
issmallI
can
beneglected.
ThinCylinder
:
1)HoopStress
-internal
pressure
HoopStress
Ap
=
Pe
internal
diameterIt
thicknessof
cylinder
2)
Aval
stress
fa
=
P=
f

196. for
a
thin
sphericalShell
:
fp
=
fa
=
b
thin
sphericalshellunder
internal
pressure.
for
a
thick
cylinderunder
internal
pressure
innerouterdiasRadialstrus
fr
=
A-BR2
A
,
Bcounts
00g
Ovarysignificantly R
:
anyradues
Hoopstress
fp
=
A
+
BR
Axial
stress
fa
=
p
Ri
:
internaldia
R2-R&
Re
:
externaldia

197. A
=
PiRR-boR2
IR2
-
RY)
Po
B
=
/Po
-
pi)RiR
E
Pir
(R?-Ri)
thehighest
strees
Obtained
is
consideredthedaughstress

198. For
THICK)
spherical
shell
e
as
ind
once
internal
pressure
RadialStree
Fr
=
A
isfixed
.
A&B
will
beconstants
for
a
system
fp
=
A
+3
90'bend
in
pipe
:
suddenchange
in
flowdirleadstomuchhigher
amt
ofstresscome.
-

199. Frregularityin
stress
distribution
duetoabrupt
changesofformiscalled
stress
concentration
Themax
valueofstressatsuchpaintisgiven
by
a
stressconcentrationfactork.
K
=
ratio
of2difffactors
=
highestvalueofthestressat
a
specific
point
nominal
stressgivenby
elementary
ea
forminimal
crossection
eg
=
outputport
g
-riveted
plate
for
watchglass

200. -thermal
stress
absent
----
UPV
Unfired
PressureVesselFPV
FiredPressureVesselb
agnificantthermal
stress
superheatedsteam/electricalcoil
/
heated
diquid
abruptchange
indir
creates
high
strust one~
domeshaped watertanks
CS
flammable
gases
feel
tanksfor
domeshaped
closerslowerstress
come.

201. ->> transient
,
conditions
startup/
maintenance
/
shutdownperiods
:
normalconditions
uss
temp
,
pressure
operatingconditions
-
environmental
conditions
extremeconditions(tornadoes
,
earthquakes
,
floods
Environmentalinfluences
are
alsopartofoperatingconditions
eg
.
brittleness
,
corrosioninduced
by
flow
etc
#ExternalLoading
(deadweightofaccessories
Ripingsystems
,
pumps
,
values
,
baffles
,
stirrers
attachedtothevessel
theyalsocreatestress
,
bothinnormalI
operatingcond

202. StandardsforReigningPressureVessels
:
standard
Standard
code
forUPVisIS2823andspecificationforformedendandtanks
UnfredPressureVessel
andnormalpressureVessel
,
standardcode
is[S4049
SpecificationforFlangeforvesselandequipmentis254869and154870
StandardforManholeandInspectionOpeningDesignforChemicalEquipment
:
2S3133
CodeofPracticeforDeign
,
FabricationElErectionofVerticalMilestellCylindrical
Welded
Ou
StorageTank
:
75803

203. SpecificationsofShell&TubeTypeHeatExchanger[S4503
Theexerciseofdesigningstandards
istohour
a
failsafe
reference
DesignCriteriadepend
on
ElasticAnalys
assumptions
:
materialissotropic
homogeneous
workingcond"
are
withinthe
clastic
range(extendingbeyondto
plasticlimitcreatesfadure
valueoftheloadforwhichmaterial
goes
totheplasticregion(beyond
clastic
lunt)
=
collapseload
or
bursting
pressure

204. of
a
materialis
under
certaincyclicloading
,
we
needto
ensurewe
don't
go
beyondclasticbut
.
Cyclicloadingshould
betested
in
such
a
waytomakesure
elasticlimit
Corrosionallowance
:
extrathicknessaddedto
counter
corrosion
forC-steel&CastFrom
:
1
.
5
mm
extreme
cases
ofdealingwithsupercorrosue
:
2X1
.
5
=
3
mmuquid
forhighalloysteel&non-ferrousmaterial
:
nocorrosion
allowancerea.

205. of
thickness
30mmthenalsocorrosionallowanceis
necessary.
5
componentsofvesseldesign
Shell
&
Head/Cover
Vessel
-
>
Nozzle
Daugh
Tat
short
->
"¥
mostcommon
in
vessels
Rivetedjoints
,
flangejoints
,
weldingjoints
00000
0

206. Radiography
EfficiencyofJoints
is
heckedbyX-RayforWeldedJointsElof
nogape
/
cracks
are
foundthroughoutthe
vessels
thenjointefficiencyconsidered
100
%
of
checked
ona
fewspots
by
Radiography
:
n
=
85%
1
11
by
nakedeye
:
50-85
%
Inoradiography)
forRivetedJoint
:n
=
70
-
85%
u
=
50-70%
=
IIIIII
ButtJoint LapJoint

207. 5thSeptember
,
2024
.
Rivetedjoint
lesseff
.
thanwelded
joint
?
rivets
E
y
=
75
-
80% n
=
50 - 70%
ButtJaunt LapJoint
-
-
(lessefficient
,
-
-
there
can
be
00sheet
slipping
E

208. Howisjoint
efficiencycalculated?
O
based
on
stressbearingcapacity
JointEfficiency
X-RayInspection
can
monitorminute
cracks
··
certifies
100%
efficiency
whenthere's
no
crackfound.X-Raymontoring
tosubstantiate
the
claimof
100
%
y
,
X-Raycheckingneedstobe
done.
J
:
Jointefficiency(1
ofX-Rayconfirms100%,ofninualinspection
based
efficiency
is
considered
,
we
can'tstandardise
it
Ihastoconsidered
1.
Weldedjoints
>
RivetedJoints

209. sometimesthe
,
some
materialsnot
suitableforweldingImayneedtosufficewith
Riveting
.
Cylindricalallsubjectedtointernal
pressure
stresses
:
fp
=
H fa
=
PD(t)
4t
Hoop) (Axal)
Common
convention
:
tensile
stress(+
)
O
compressur
(-)
Ap
=
2fa
:Apishigher
,
then
it'll
be
considered
as
the
max
possible
stressusedtocalculate
thickness.

210. fp
=
f
r
design
stress
(maxstress
CYLINDER
Jointyt=p
=
p
Di
+
t
=
Do
f
=
y
=
P
=
Pi
=
pDo SPHERICALSHELL
4f]4f]-p
-
ness
considerable

211. Letofcontent
,
any
externalload
,
piping
I
deadut
.
bending
dueto
wind
Additional
factors(
1)InternalPr
2)Witofcontents
,
externalload
3)Piping
4)Wind
Stressesbecome
:
->
only
dueto
internalpressure
as
before
?
fp
=
p(Di
+
t)(t) fa
,=
p
(t)
dusto
internal
pressure
2t
fac
=
W
(t)
whycompressive?
7)
t(Di
+
t)
on
cylinderwalls
,
not
base
plate
witofcontent
,
vessel

212. ->
bendingmoment
.
loadI
axs
ofvessel
fas
=
it onesectionexperiences
O
tencile
,
othercompressive
Total
fa
=
faitfaz
+
fas
to
ethis
mind
>
tensile
-scomp
particularlyformind
:
tooffectthewindstress
weneedto
createcertainatof
torque
.ts
=
1
torque
TItDi/Ditt)
stressthat
will
offsetstressdueto
wind
Isis
effective/netstress
in
bothradual
&analdirn
.
instead
,
fasonlyaxial

213. ↓
f+
&
fp
aresame
ShearStrain
EnergyTheory:
BrownelledYoung
Failure
of
thespecimensubjected
to
any
,
combinationofloadwhenthe
max
O
shearing
stressat
anypointreachesthe
failurepointequaltothatdeveloped
O
atyieldingin
an
axial
,
tensile
orcompressuretestofthe
same
material
Cateriaforthistheory
Fr
calculation
fre
=
Ifp-Epfa
+
fa
+
3fs)
Isheaustra
Energy
They
is
fr(Tensile)
Apit(permissible)

214. fa(Tensile)
Sp
,
/Permissible
fa/compressive)
<
fc
(permissible)
tfo(permissible)
=
-
-M2)
(Dolt)
forAlloySteel
E
:
Elasticity
U
:
passon'sratio

215. SS316
t
-t
-
·
Forrosion
allowance
↓
to
-> 10
↓next
higheravailable
thickness
ofto
comes
outtobeour
degn
thickness
thatsatufes
stress
requirement
C
and
us
<30mm
(then
corrosionallowanceadded
ofdealingwith
corrosuemat
.
->vacuum
NEGATIVEPRESSUREinthevessel/orexternal
pressure
can
createdeformationofthebody
->
eitherMethicknessfurther
or
->
·
-
rigsatweakpoints
(calledstiffeningrings)
I
use
=
to
avoidlobeformation
(buckling/crippling)
material
lesscost
lobeformation
canoccur
by
excessivestress

216. -
-
critical
pressure
forbuckling/lobeformation
L
V
Pc
=
+((m
=1)
+
emM
n
=
no
·
oflokes

217. it
0
.
1
I 18
L/D
forPNID
:
eachstudent
will
have
a
uniqueprodcapacity
for
entereprod.processof
yeart/select
one
outofthe3
yearts)definedprod"capacity
Stateassumptions

218. &Septem
2024
POIDs
ITableto
explainindividualequipment
names
1Tabletoexplain
mass
&energy
input
,
outputs
symbolstorememberfromPerry'sHandbook
:
Utility
Line
Reactor
eg
.37
%
C
run
temp
,
but
felter
,
centrifuge ambienttemp28%,calc.energy
distillationcol
,
mixingtank requirements(
+
energyloss
O
heat
enchanger
,dryer,chilles assumptions
or
state
if
neglected
centry
,
reciprocatingpump
Boiler
,
diff.values
,
nomenclatureofdiffcontrollers

219. stirring
energy
requirementsfor
deffreactorvolumes
1fonne/day
capacity(YeastBiomass)
->
Reactorcapacitychosensay
:
10
,
0001
lookupfromtheatternetthe
kindof
energy
requirementsof
stirrerfor10kLreactor saccordingly
scale
upor
down
in
exam
I
stateassumption
-
similarly
,
pressuredroprequiredto
filter
a
5%-10%BMSlurry.
Energy
Requirementsforspecificunits
you'reexpectedtoknowballparksofenergyreg
.
fordiffunitsforreference
values
2)thenscaleup
I
downin
exam
[internet
sources]

220. 1000Lreactor
usuallyrequires1Hpmotor
"¥
10
,
000Lreactormayrequire7-20HP
motor
/reasonable)
but
100HP(unreasonable)
so
ofyoudon'tknowclosetoaccuratevaluesfor
some
reference
,
youwantbeabletoscaleup/down
with
any
accuracy
.
you
could
assumeeffeciency
ofallequipmentstobe
88%
oranme
it
accessory
-
cue
DONOT
givevariablesto
giveidea
aboutpump
capacity

221. Pumpsonlyneededonce
(to
rasse
feedtofeedtank)
entire
process
flowafterreactor
can
bemanaged
bygravity
OnechartforNomenclature OneChartfor
equipment
specifications
PRFeedPump PR10HP
P2AndAlkaliPump
,
etc
P21HP
,
etc
· :

222. SetPoint
Valuesnot
requiredfor
controllers
butsp
.
symbolsshould
Stirrer
controller Jbutcontrolloopsshouldbecloud&equipment
pHcontrollermustbeconnectedtorespectivecontrols
ElectricHeatingCoils
easer
for
energybalances
(higher
energy
efficiencybut
some
usHeatExchangers safety&cleaning
concerns
↓
localisedhotspotsmaygive
lower
energyeff-but
safer problems
w
nutrientmedia
qual
&heats
more
evenly scale
form"
oncalls
hard
to
clan
.
~
someunitsmayrequireheat
,
somemayneed needsproperearthing.
heat
extraction
:
heat
recoveryunitpossible
i
heretenchangersbutnotwithheatingcals
(so,
,
processeconomymisebetter

223. BrowniePointsforestimatingheatrequirements
usingHeatExchanger
.
Rational
guess
regardingaeration
rea
eg10
,
000Lreador
->
calclea
vol
(working)
say
,
9800L
&
rum
->
soyou
knowvolof
air
neededtobepumped
"¥
Energyveeq
forblower
to
supply&X9800L
Typically
,
ofratio
=
1forCSTRtypereactor
20
%emptyhead
spaceextra
for
acroble
vxns

224. sayworkingrol
=
10
,
000L
20%
head
=>
reactorvol
=
10
,
0001
+2000L
I
12
,
000L
-
=
1
calculatelength
-
>
then
use
itforovercominghydrostate
head
&
cale
.Blowercapacity
YeastPrep
:
Inoculumtransferrequired
atmultipleplaces
howdoyou
decidewhichpup
us
used?
->
10-200Xmore
expensive
CentrugalPump
cheaper
.ButDiaphragmPump
or
PeristalticPumpgoodforsterity
(BasseReg.

225. Staterationale
,
justificationforequipmentselection
Dewaivingmostlydonebyfelt/centrifugforBM
(yeasts
dryer
atmax
10-50
%
c
:
Pandrying
byevaporation
or
ethyleneoxidegas/SO2to
kill
bacteria
,
I
contamination.
the
amtofethyleneoxide
we'reusing
toI
shelf
life
ofspies
-
manycountries
banningIndian
spires
.ElEthylene
oudeis
a
known
carcinogen.

226. notrationalto
connder80-85
%
yforfiltration/separationunits
.
Consider
closeto95-96
%.
Scrapingsolidcatewentrecove100%
BM
.
Multistage
filtrationmayachieve
-
100%
If
felter
care
is
highly
compressible(eg
.
biomass)
Pressuredropdyffrom
uncompressible
cake
.
felteraidunparts
some
uncompressibilitytocompr
.
Materials(hard
mat.
as
felterand
If
we
addbetsofgrains(eg
.
tiles
,
diatomaceous
earth
,
glassbeads
,
smallparticles
that
createchannelsthat
and
filtration
.
O
You
can
improve
filtrationeasilyusingfilteraidif
youpotofinterest
is
feltrate
I
enterested
in
BM
:
felterand
(heavierswillneedtobeseparated
density
diff
-
settles?
I
separatesout
.

227. #
choosingequipment-wiseefficiency
:
BROWNIEPoints
The
more
detailIthoughtyouputintoyourP&ID
,
thebetter
.
Minor
20POID
(5
Func
,
5OperationalDiag
,
10P&
I
30marks
-
10marks
O
->
1 Ques
from
PuressMass&EnergyBalances
(likeassignment
~
High
Pressure
VesselDesign
eachsymbol
wdneedsnomenclature
typeofvalueusedneeds
tobementioned
.
Pipediameters
can
beomitted.
Iflowcontrolvalues
or
ON/OFFvalues

228. consideroverallboxesformultiunit
operations
(in
M&EBalances
(
startMEBcalc.onlyafter
drawing
blockdiagrams(don't
jumpinto
cales,
mastere-balances
,
stoichiometry
,
degreeofreductions
calc
.
consider
what
can
beconsidered
overall
insteadofcomponent
balances)

229. POST-MINORS

230. Wednesday
,
23thSept

231. Thursday
,
26thSept.

232. &
Design
a
shell/vessel)
andheadplate
.InternalDiaofshell
=
1200mm11
.
2 m)
G
&Material
ofConstructions
&
%Cr
18%
Ni
Permissible
stressof
thismaterial130 N/mm2
If
InternalPrusure
=
0
.
3
N/mm
?
Jountn
=
85
%
notofvessel
=
3200kg
torque Calculatethickness
I
due
tooffect
,
piping
,
mind
=
900Nm

233. HeadFlange
ExternalDra
=
1200mm
GrownRadius
=
1200mm
&verzea
a
KnuckleRadius
=
72mm
MOC
=
SS10
.
5
%Cr
.
18
%
Ni)
Jointy
=
100%
(Welding;thoroughlyinspectedbyRadiography
Bare
minimum
safetyfactor
=
10%
O..
ofvessel
internal
pressure
=
0
.
3N/mm
:
internal
designpreseurs
=
P
+0
-
IP
=
0
.
33
N/mm

234. Thicknessofshellcalculation
-
internaldesignpressurst
=
pi
internaldia
2f]-p
Jointy
design
or
permissible
stress
un
gett
=
0
.
33
N/mm2X1200
mm
2X130N/mm2X0
.
85
-
0
.
33N/mm2
t
=
X100
mm
=
1797
mm
/but
not
a
standards
a

235. vessel
dia(m)
MinimumVesselThicknessRecommendation(mm)
A
"¥
↓
1
-
2
->
curt
=
1
.
8mm
<
7mm
ours
2
-
2
.
5
1
~but
minimumrequired
is
2
.
5
-
3 forhigh
pressurevessel
1200mm3-3
.
5 12 design
80
1
.
2
m
soour
+
=
7mm
-If
we'reusing
this
vesseljust
inch
peryear
?forstorage&no
p:requirement
from
datasheetyoucouldcheckIPV
?
then
we
couldsafelydoublethe
yearlycorrosion calculated&
use
It
=
Imm
CorrosionAllowance
-
30%
so
t
=
+0
.
3t
=
7mm
+G
.
Imm
=
9
.
Imm
->
liberty
tochoose
amm
:
alreadysafety

236. -had
it
been
9
.
5mm
go
for10mm
for
SS
,
nocorrosion
allowancerequired
so
Imm
straightawayisshellThickness
Usingthist
,
cheek
if
allconditions
are
satisfied
I fr
criteriasatisfied
ofnot
-go
fornextthickness(9mm)I
checkagain
.

237. 3rdOctober
,
2024
metrictonne
=
1000 kg
USton
=
907kg
(ContinuedSolution
forprevproblem)
ft
=
p(Di
+
+)
2t
fe=
Difai
+
t)
+
3
=
0
(extpr)
designpressure

238. fs
=
pt
:)Di
+
t)
fa
=
fi
+
fz
+
+3
fr
=
[fR-ftfa
+
fa
+
3fs2]'2
ofquestionmentionsSSMoC
-
nocorrosion
allowance
rea
7 1
"some
otherMoC
-
corrosionallowanceforthatmaterial
""does
NOT
mentionMoC-state
assumptionsabout
yourMOCthen
consider
its
CA

239. first
calc.fr
and
fo
thencompare
,
ofbelowacceptablestressthen
elsego
for
next
higherthickness.
shellthickness
comes
outtobe9mm?
G
Head">=shellthickness
HeadThicknesscalc
:
(Ans-14
.
37
mm-
then
chooseclosestnext
levelfromstandard
table)
th
=
pu
w=
1/3+
Refe
numrasead
#Safety
factorfor
thehead
is
the
same
(10
%
)
:
same
designpressure
is
used
=
0
.
33N/mm2
x
(1200mm)
x
1
.
771 ↓
13
+d
2X
130
N/mm2
X1
=
2
.
697
mm
↑(3
+
10)
=
1
.
7706

240. #
if
mentionedthevessel/headisexperiencing
entpressurebutnotmentioned
Nowmuch
,
thenusuallyconsidered
1
.
6-1
.
7times
the
internal
pressure
#y
mentionedthoroughly
checked
byradiography
,
then=I
11
checkedat
some
places thenJ
=
85%
11
l
"by
nakedeyethen-50-85
%
-Sir's
Solution
pipe
jointsea
undersse
->
but
pressurecanmake
these
slip

241. for
pressures
upto20kg/cm2
Fe
tocounteract
interprese
either
riveting
Beyondthis
,
or
welding
Standardsfor
piping
joints
[S6392-64??
1S4864-4869(Standardsforflanges)

242. Flat
RingGasket
.
0
.
5 mm-3mm
eg
in
blenders
,
miners
,
grinders thickness
lunhioningtoprevent
abrasion)
Below120°
>
Rubber
,
Paper
upto350
-
Asbestos
About
350
-
Metals(soft)
Temp
,
settingstress
are
factorsto
choosegasket
material

243. ServatedGasket
M
- ↑serrationto
maintain
roughness
-
-
Sufficientlockingto
notgetleakage
- ·"smoothnessis
good
,
but
roughnessisbetter"
-
Zia(2024)
LaminatedGasket
sandwichedmaterials
duetochange
inelastic
properties
in
thegasket
-
metal
tolerant
tovariablepressures
/highlow
-
>
asbestos(diffelasticity)
O
-
met
al
more
flexible

244. Corrugated
GasketConetypeofselfsealinggasketwhich
can
work
in
highertemp
a
flexible
pressure
Eltemperatureranges(both
corrugated
&serrated
Self
-sealing
ket

245. IUthOct
,
2024
FLANGEDeign
-
Design
Steps
1)Selectionofgasket
2)Flangefaung
3)Betting
4)HubProperties
3)
Flange
with
a
eness
Maindesignconsiderations
are
O
1)
to
improvethecontactpressureatthegasket-flangeinterfall
under
all
serviceconditions
&topreventleakage

246. 2)to
create
lightly
enforced
wooverstressingthebalt
3)to
ensure
structuralintegrity
oftheflangeIminimise
deflectionofflanges
1)elasticReaction
:
withinelasticlimits
gasket
onlythegaret
surface
willtry
to
fill
theurregularitiesoftheflangefaces
2)
PlasticReaction
gasket
is
overstrued
I
itsmassbecomesplastic

247. 3)SpringyReaction
GasketiscompletelyenclosedwithintheflangeIbehaves
like
confinedfluid
for
estimating
sezingofgaskets&determiningthe
no
E
diaofbatts
1)
AtmosphericPressure
2)
Operating
Pressure
1)
Underatmosphericconditions
gasketreaction
Wa
=
Ag
-
sealing
stress
(seatingstuss)

248. Wp
=
AgYp
+
Anp
We
use
Hooke'sLawtogive
us
relationshipsblumaterial
properties
Xy
=
Ya
-
Yp
XT
=
T
"¥
-
Ya-Yp
->
modulus
UT
=
0L
+
2Δf ofelasticity
↓
>
changein ofgasketY
change
change
in
in
deflection
weam
forclasue
or
gasketbalt
offlange springy
rxn
(d
not
plastic
Micknesslength rxn)
I
dueto
lightening)

249. "¥
lengthofbalt
"¥
change
inboltload
,
varieswith
pressure
0
=
1W
T
modulusofelasticity
ofbelt
XArea
ofcrossectofbolt
&W
=
Wp-Wa
change
W
Y
under
un
baltunder
atmpressure
loadoperating
=LAW
pressure
&W
=
Wp-Wa
Of
=
c&W
=
AgYp
+
Anp-AgYa
↓
=
Ag(Yp-Ya)
+
Anpconst
(deflectionfactoroffraa
OW
=
Anp-OYAg

250. -
relatedtogasket
nowwe
haveDW
ito
DY
wecan
replaceIW
in
expressions
its #
wecan
geteverything
expressed
its
Y
.
Yaisalways
greaterthan
Yp
.is
significantforrubber(
:Egsmall)aT
u
large
formetallicgaskets,
I
can
be
neglected
(i
Eg
vhighformetals
aturns
is
small

251. G
small
~
significant
formetals
forrubber

252. Ch11Perry'sHandbook