SI No
Calculation
1
Heatin time usin !ot water "#on $sot!ermal Heatin %edium&
2 3
Heatin time usin steam "$sot!ermal Heatin %edium& 'oolin time usin non isot!ermal coolin medium
4
Heatin time usin e(ternal Heat e(c!aner ) #on $sot!ermal Heatin
5
Heatin time usin e(ternal Heat e(c!aner ) $sot!ermal Heatin %e
6
'oolin time usin e(ternal Heat e(c!aner ) #on $sot!ermal 'oolin
7
'oolin time usin e(ternal Heat e(c!aner ) $sot!ermal 'oolin %e
8
'alculate t!e Heat *ransfer coefficient for reactors based on different aitator t+pes ) utilit+ flow rate
9
'alculate t!e addition time re,uired based on !eat of t!e reaction*+pe of aitator- %.' ) utilit+ flow rate in a semi batc! reactor
10
Scal Scalee up up of of rea react ctor or from from lab lab to to pla plant nt scal scalee
11 12 13 14 15
'alculate t!e /acuum pump flow rate based on air leaae rate Power re,uired for pumpin Pump affinit+ law Pipe fl flow di distrib ribution Pressure drop across pipe lines
Revision History Rev. 0
Issued to
Downloads section on 5/5/2011
Home
Non Isothermal Heating Reactor Contents Mass M Initial temp t1 Final temp t2 Sp.eat !m #$ea # &! '
72500 120 200 1.05 150 1(0
Heating Fluid Inlet temp Sp.eat Flow $ate
2*0 °F 1.2 "tu/ lb °F 1(000 lb/)$
&1 !w +
lb °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
Time required for heating
ln
&1 , t1 &1 , t2
-
Time reqd
*.30* =
-
e'#/+!
6.4 hrs
+!w M!m
,1
t
Home
Isothermal Heating Reactor Contents Mass M Initial temp t1 Final temp t2 Sp.eat !m #$ea # &! ' Heating Fluid Inlet temp
&1
50000 4( 257 0.5 100 150
lb °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
*20 °F
Time required for heating
ln
Time reqd
&1 , t1 &1 , t2
=
-
.!" hrs
' # time M!
Home
Non Isothermal cooling Mass M Initial temp &1 Final temp &2 Sp.eat !m #$ea # &! '
72500 200 (2 1.05 150 110
!oolin Fluid Inlet temp Sp.eat Flow $ate
72 °F 1 "tu/ lb °F 15500 lb/)$
t1 !w +
lb °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
&ime $e6ui$ed %o$ coolin
ln
&1 , t1 &2 , t2
-
&ime $e6d
2.(33 -
-
e'#/+!
13.11 )$s 114.7( min
+!w M!m
,1
t
Home
Non Isothermal Heating Reactor Contents Mass M Mass Flow $ate w Initial temp t1 Final temp t2 Sp.eat c #$ea # &! '
72500 20000 (2 200 1.05 150 210
Heating Fluid Inlet temp Sp.eat Flow $ate
2*0 °F 1.2 "tu/ lb °F 15500 lb/)$
&1 ! +
lb lb/)$ °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
Time required for heating
ln
&1 , t1 &1 , t2
*
-
* Time reqd
1.21*5( =
-
e'#1/+!,1/wc
"#.#!
hrs
w+! *wc,+!
*,1 M
t
Home
Isothermal Heating Reactor Contents Mass Mass Flow $ate Initial temp Final temp Sp.eat #$ea &!
M w t1 t2 c # '
Heating Fluid Inlet temp
&1
72500 20000 (2 200 1.05 150 210
lb lb/)$ °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
2*0 °F
Time required for heating
ln
&1 , t1 &1 , t2
2
-
2 Time reqd
.(14(3 =
-
e'#/wc
$.4%
hrs
wc Mc
2,1 2
t
Home
Non Isothermal cooling Mass M "atc) %low $ate + Initial temp &1 Final temp &2 Sp.eat ! #$ea # &! '
72500 20000 200 (2 1.05 150 1(0
!oolin Fluid Inlet temp Sp.eat Flow $ate
72 °F 1 "tu/ lb °F 15500 lb/)$
t1 c w
lb lb/)$ °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
&ime $e6ui$ed %o$ coolin
ln
&1 , t1 &2 , t1
-
&ime $e6d
0.4**47 -
-
e'#1/+!,1/wc
1(.13 )$s 1031.42 min
+cw wc,+!
,1 M
t
Home
Isothermal cooling Mass "atc) %low $ate Initial temp Final temp Sp.eat #$ea &!
M + &1 &2 ! # '
72500 20000 200 (2 1.05 150 1(0
!oolin Fluid Inlet temp Sp.eat Flow $ate
t1 c w
72 °F 1 "tu/ lb °F 15500 lb/)$
lb lb/)$ °F °F "tu/ lb °F %t2 "&'/)$ %t2 °F
&ime $e6ui$ed %o$ coolin
ln
&1 , t1 &2 , t1
1
-
1 &ime $e6d
*.41725 -
-
e'#/+!
12.77 )$s 744.*7 min
+! M!
1,1 1
t
Home
Calculation of Heat transfer coefficient in reactor Reactor data Diamete$ o% t)e tan8
&
10(
Impelle$ speed
5
$pm
Impelle$ Dia
9 Da
2
in
Impelle$ powe$
p
2
p
#itato$ &:pe
in
3 2700
%t $p)
*.5
%t
#nc)o$,1
;essel
=
1
in
0.0(
%t
ei)t o% li6uid in c:lind$ical section
>
34
in
(
%t
eat capacit:
!p
0.337
Sp $avit:
Bb
1.002(
1002.(
"ul8 ;iscosit:
0.5*4
cp
;iscosit: at wall
? ?w
1.5
cp
&)e$mal conductivit:
8
0.*4(
"tu/)$ %t °F
eat capacit:
!p
1.01
"tu/lb °F
Sp $avit:
Bb
45.2
lb/%t*
"ul8 ;iscosit:
1.5
cp
*.503
lb/)$ %t
;iscosit: at wall
? ?w
1.5
cp
*.503
lb/)$ %t
&)e$mal conductivit:
8
0.**5
"tu/)$ %t °F
=i6uid %low $ate
@
0.22
cu %t / sec
&ro'erties of (atch side fluid "tu/lb °F 42.40
lb/%t*
1.23712 lb/)$ %t
&ro'erties of )ac*et side fluid
Calculation +atch ,ide calculation #itated batc) li6uid $e:nolds numbe$ 9Re #itated batc) li6uid A$andlt numbe$ 9A$
-
Da2 9 Bb / ?
-
1534234
-
!p ? / 8
-
*.51
?/?w
-
0.*70
>/& Da/&
-
0.((3
-
0.*(3
a
1
b
0.47
M )b & / 8 - a 9$e b 9A$
0.1( 1/*
?/?w
M
1gitated liquid Heat Transfer coefficient2h (
=
$#-.4
+tu/hr ft 0F
3ac*et ,ide calculation 9u - #1 9Re0.447 9A$ b u/uw0.1
0.02*
#1
0.0245
b
0. 0.*
9u - )< D< / < D < - 6uivalent c$oss %low diamete$ o% t)e
-
0.447
%t2
+etted - = 2 C + 2 D <
-
14.147
%t
-
0.145
%t
#e6uivalent
-
0.021
%t2
;elocit: ; - @/#
-
10.*00
9Re
-
11*450.5
9A$
-
10.5(
#1
-
0.02*
b
-
0.*
3ac*et side Heat transfer coefficient2 h )
=
!%.-
%t/sec
Fo$ coolin
+tu / hr ft 0F
verall heat transfer coefficient 1/' ove$all - 1/)E C 1/)" C 1/)DM RDM - 1/)DM
-
0.001
1/' ove$all
-
0.0044
5overall
-
"%#.4
+tu / hr ft 0F
*70(2 %t/)$
Home
# $eacto$ 2.0 = capacit: is used to ca$$:out an eot)e$mic $eaction at (0°! &)e $eaction calo$imet$: s)ows )eat o% t)e $eaction o % 15000 "&'/mole o% $eaent >. I% 5 mole o% $eaent > is used calculate t)e addition $ate !oolin wate$ is available at 0°! ' # & @ eat o% $n #ddn Rate
55 "&' / )$ s6 F °F 7 m2 (0 °! 17(0 8cal / )$ 1000 8cal/mole 23.54( lit/)$
24 8cal/)$ m2 °!
Home
&wo assumptions , Aowe$ pe$ unit volume is constant , &ip speed is constant ,cale u' 7a( scale data8 Dt1 0.10( m 1 0.10( m ;ol 0.001 m* Da1 0.0* m 91 00 $pm Densit: 1*00 8/m* ;iscosit: 1000 !A Scale up times
1.00
=it
4.47
$ps
1.0000
8 / m s
1000 times
1ssum'tion2 &o9er 'er unit volume is constant 92 91 1/R powe$ 2/* Ratio 92 9p
10
-
1. $ps 5 #ssume
Da2 Dt2 ;2 9$e
0.**41 m 1.0(0*5 m 1000.00 = *51.04(7
A
235.2272 E/sec 0.515142 )p
(4.2
0.*0 + *0 G =oss
1ssuming Ti' s'eed is constant 92 91 / R 92 0.444447 $ps 9$e
142.3517 E/sec
A
23.52272 E/sec 0.051514 )p
;2 ;1 $pm
1/*
0.0
)p
0.0
)p
0 $pm
0.0* + *0 G =oss
: The assum'tion is also (ased on 'o9er 'er unit volume as c onstant
Home
Designing Based on pressure drop test ample &otal S:stem ;olume S:stem vacuation to Final A$essu$e D$op test pe$iod Desi$ed ope$atin p$essu$e
; Ai A% t Ao
-
Pump capacity required
@
-
*50 2 * 10 1
%t* in #bs in #bs min in #bs
(Rise in pressure) x Syste time t Desi$ed ope$atin p$es
@
-
*5
!FM
Designing Based on pump evacuation time &otal s:stem volume Initial p$essu$e Final p$essu$e vacuation time
; Ai A% t
-
Pump capacity required
@
-
100 740 50 2.25
%t* mm abs mm abs min
V x ln(Pi/Pf) t
@
-
120.3
#!FM
Hnce t)e pump sie is selectedJ we must $ecalculate t)e evacuation time b: usin t)at pumps a capacit:.
Designing Based on Air Leakage Rate Assuming the inlet gas composition is only air at 75°F Fall in vacuum &otal S:stem volume &empe$atu$e Initial A$essu$e Hpe$atin A$essu$e #i$ =ea8ae $ate
; & Ai Ao
-
*00 5000 75 740 100 2.*(
mm /)$ lit °F mm #bs mm #bs 8/)$
&)e capacit: in #!FM can be calculated usin #!FM - m / Mwt *73/40 Initial p$essu$e / Hpe$atin A$essu$e 40C&/520 #!FM
S
-
(.(2
!o$$ection %acto$
Fc
-
0.(5
#!FM
!o$$ected capacit:
Sc
-
10.*7
#!FM
"ased on t)isJ %$om t)e pump manu%actu$e$ cu$ve $e6ui$ed !FMJ )p K pm o% wate$ %o$ se$vice c
Designing or air ! solvent vapour !omposition o% vapou$s a$e #i$ C Met)anol ;apou$ C wate$ vapou$
Fall in vacuum &otal S:stem volume &empe$atu$e Initial A$essu$e Hpe$atin A$essu$e Met)anol vapou$ +ate$ vapou$ #i$ =ea8ae $ate &otal
-
*00 5000 10 740 50 15 5 2.2* 22.2*
@t: 15.0 5.0 2.2
G wei)t 47.( 22.3 10.02
-
27.00
S
-
(5.7
!o$$ection %acto$
Fc
-
0.(5
!o$$ected capacit:
Sc
-
100.(7
; & Ai Ao
!ompound Met)anol +ate$ #i$ !alculatin ave$ae molecula$ wei)t #!FM
mm /)$ lit °F mm #bs mm #bs 8/)$ 8/)$ 8/)$ 8/)$ Mwt *2.0 1(.0 23.0
#!FM
#!FM
volume (V) u$e Ao
e$ae
0. 5 234.3
atm m*
5.1(
lb/)$
n be obtained
0.*37*7 atm 5 m* *1*.0
.31
lb/)$
Home
&o9er Required for 'um'ing Flow $ate
@
*00 m*/)$
Densit: Aump )ead $e6ui$ed
B
1000 8/m* 25 m
Aump e%%icienc:
L
0.75
&o9er Required
&
$$$. *g m / sec
0.0(** m*/sec
!$.#4 h'
(*.** 8/sec
Home
@J Flow
1ffinity 7a9s A$opo$tional
9
J ead
A$opo$tional
92
A J Aowe$
A$opo$tional
9*
;
91 @1 1 A1
1500 00 25 3.
$pm m*/)$ m )p
Final RAM
+y ad)usting 92
100
$pm
Final %low Final ead Final Aowe$
Result @2 2 A2
*7* 21.( 0.1
m*/)$ m )p
": $educin $pm b: ead $educed b: Aowe$ $educed b:
4.47 12.(3 1(.70
G G G
&i'e flo9 distri(ution
ata ;iscosit:
B ?
0.(275
Flow $ate Aipe Dia
@ D
Densit:
#ccele$ationJ
1000
8/m* cp
0.000(* 8/m sec
500 10
m*/)$ in
0.1*((3 m*/sec 0.25
3.(
m/s2
# ; 9Re
0.05 2.7
m2 m/s
(1775.2
%
0.0023((
Calculation #$ea ;elocit: Re:nolds 9umbe$ F$iction Facto$
ia m8 0.20
7ength m8 *0
0.15
40
0.10
0
Flow $ate # m*/sec 0.1*(3 Calculation 0 &i'e #a" #b" #c" ,olver !ons,1 !ons,2 !ons,*
ia m8 7engthm8 f >2m/s 0.20 *0 0.00233 2.35*4* 0.15 40 0.00233 1.(0(72 0.10 0 0.00233 1.(0(72
0.00000 0.00000 0.00000
ead loss ac$oss t)e pipe line is e6ual
;quations 5sed ;quation of Continuity@
hf 0.73(0 0.73(0 0.73(0
? m!/sec8 ? m!/hr8 0.0327 ***.(( 0.0*13 115.01 0.012 51.11 %##.##
;quality of &ressure dro's@ DelA due to %$iction in Aipe #a" - DelA due to % $iction in Aipe #b" DelA due to %$iction in Aipe #a" - DelA due to %$iction in Aipe #c"
'sin t)ese * sets o% e6uations a$e %o$medN &)ese e6uations a$e o% non, linea$. &o solve t)eseJ t)e solver add:in is bein used.
&)is p$o$am will $un wit) Mac$os. Set :ou$ mac$os settin to low and t)en p$oceed t)e calculation.
"hese values are arrived rom solving the constraints '% ( and ) ,y -olver Add./n
"arget #ell "his cell is to have the value o $% in such a &ay that the constraints '%(% and ) are *ero+
Home
# *5o #AI distillate is bein t$ans%e$$ed %$om a sto$ae tan8 at 1 atm absolute p$essu$e to a p$essu$e vessel at 50 psi b: means o% t)e pipin a$$anements s)own in %iu$e. &)e li6uid %lows at t)e $ate o% 2*100 lb/)$ t)$ou) * inc) Sc)edule 0 steel pipeN t)e lent) o% t)e st$ai)t pipe is 50 %eet. !alculate t)e minimum )o$sepowe$ input to t)e pump )avin an e%%icienc: o% 40 pe$cent. &)e p$ope$ties o% t)e distillate a$eO viscosit: - *. cAJ densit: - 52 lb/%t *. The following are the data for the pipe and fittings: Fo$ * inc) Sc)edule 0 9ominal pipeJ HD - *.5 inc)N &)ic8ness - 0.214 inc) Flow coe%%icients %o$ t)e %ittins a$eO Pate valve - 0.25N 30 o elbow - 0.3N !)ec8 valve - 10
F$iction %acto$ can be calculated %$om "lasius e6uation. #ccount %o$ ent$: and eit losses also.
Conversion Factors 1 %eet 1 lb 1 inc) 1 centipoise 1 atm
0.*0( 0.5 0.025 0.001 1.7
m 8 m 8/m.sec psi
1.01C05 9/m2
1 atm
3.(12 m/sec2
ata given@ Mass %low $ate Densit: ;iscosit:
2*100 lb/)$ ρ µ
52 lb/%t* *. cA
Converted data@ 2.31*147 8/sec -
(**.70(7 8/m* 0.00* 8/m.sec
Aipe HD Aipe t)ic8ness Aipe lent) ;e$tical )ei)t Aump e%%icienc: in %$action
= 1,2
*.5 inc) 0.214 inc) 50 %eet 70 %eet 0.4
=oss coe%%icient o% Pate ;alve =oss coe%%icient o% elbow coe%%icient o% c)ec8 valve ;alve
-
1*7.14 m
-
21.**4 m
0.25 0.3 10
Aipe ID
D
*.04( inc)
-
0.077327 m
A$essu$e at 2
A2
50 psi *.01*41
-
*.5C05 9/m2 *01*.41
;olumet$ic %low $ate ;elocit: Re:nolds 9umbe$ F$iction %acto$
@ v 9Re %
Calculations@ 0.00*3 m*/sec 0.7*24 m/sec 1*333 0.00724
)% o% pipe
1.*3(5 m
v2/2 )% o% Pate valve
0.027*5 m
)% o% 2 numbe$ o% elbows
0.032* m
)% o% !)ec8 valve
0.27*51 m
)% o% sudden cont$action at inlet
0.0103 m
)% o% sudden epansion at outlet
0.027*5 m
&otal %$ictional )ead
1.7442 m
Aump )ead
22.541 m
Minimum power for the pump
0.004( m
1074.81 Watt
