single reactions in continuous isothermal reactions
TRANSCRIPT
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Isothermal Reactors
1
CSTR constant densit & SS single reaction
Reversible reaction
Plug Flow 1/r plot concept
Semi-batch reactors
Variable density
Reactors in series CSTR in series
Autocatalytic reactions Reversible Reactions
2
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CSTR Continuous stirred tank reactor (CSTR)
ompos on s e same everyw ere n e reac or an
the exit pipe.o ume =
[L]
o v
CA
-*
3
Exit and tank composition same
Steady-state
Constant Density Reaction
liquid or
ases, no mole chan e, e. . A+2B3C
4
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CSTR Mass Balance:
dNr
dtjjjo += time
Molar flow rates of species j
time
moles
joojo
CvF =*
L mole
5 s L V
CSTR at Steady-State & Const Vol. Flow
= o
=0*
rV
vjjo =ovv =
dCj
Batch
rCC jjjo +=0. .
Flow dj
=
rCC jjjo = rCCjjo=
6j
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.1=AA roducts
rCC AAo =
jjjo
nth order reactionn
AkCr=
rs or er ne cs n=
AAo CC *
kCC += 1AkC
CC Ao=
*
7k+1
r=kCA
Single Reaction, AB AAoB CCC =
CCC AAoB =
CkCCCCC AoAoAoAoAoB
+==
*
8
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.
nA
rjjjo =
=r
o pro uc s
For 2nd order (n=2)
0
411
AA
Ck
C+
=*
02 =+ AoAA CCkC
9
Fractional Conversion X AB Use A as the limiting reactant with stoichiometric
System is at constant densityr=kCA
n
X varies from 0 to 1
= XCCC= XCC AoA )1()(
=XCAo +=
kCC AoA )
11()(
*
=>= )1( XkX +=
k
Ao
1 kkXX =+
Xk = 1 == XCC AoB )(kX +
=1
*
10(X) or X() Cj()
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CkCkr =
AB
=r*
CB=0 =
=CC AAo
*
rrCC jjjo =
AAoAAobAf
CCCCkCk = )(1+
bA kC
AAoAobAbAf
CkCCCkCk
CCCkCkCk
+=++
=+ 1)(0 ++ bfA kkC
11
oo So, CA()
CCC =
bAoAo
kCC =
1)(0 ++=
bfA
A
kkC
)( bAofbAo
fb
kCkkC +As
)( fb kk +*
)( bf
fAo
kk +=
fB kC = (the conc ratio = Keq
12bA
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Same residence time for all molecules Assumptions
Constant Diameter
Steady state
Single reaction
13
ud =
VL ==
3
u (m/s)
v (m3/s)
14z
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Constant Densit Plu Flow
joojo CvF = jj CvF=
2
4
D
v
A
v
u==
z Lz+dz
Mass Balance on j
0)()( =++ rdzdzzFzF jtjj jtjj
Tube Area=
dz
r
dz
dC
dzzCzCjj
=
+lim
tt
udzz
dz
0
15r
dzu j
j= r
dj
j
=uz=
0=++ rdzAdzzFzF
- =
0)()( =++ rdVdVVFVF
rVFdVVF jj
=+ )()(
dF
dv0
16
rdV
j=V V+dV
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For reactant species A, A=-1
AC
AdCVrdCj
=
==AoCo
rvd
r
d
A =
ddCA =
17
First Order in Constant Density PF
=dCj =
C
d
r
dA= dk
CC A
A=
0
kCdC
AA =
kCA =ln
keCC=kd
dCA =
Ao
o
CA
18
than t)
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Exs. 3-1 & 3-3 PF versus CSTR (lst Order)
The reaction AB, r=kCA, occurs with 90% conversion.If k=0.5 min-1 C =2 moles/liter and v=4 liters/min
what residence time and reactor volume will be
re uired? How lon a 2-cm diameter tube would be
required for this conversion and what would be the
fluid velocit ?
Assuming CSTR =CC AAo rCC jjjo =r
*
19
Exs. 3-1 & 3-3 PF versus CSTR (first order)
rdC
j
j=
AssumingPFTR
k
AoA eCC=
*
litersxvVPF 4.1861.44 === ma er an
CSTR
( )mcm
x
DL 6.585860
4/2
.
4/ 22====
min/12705860
cmL
u ===
May be
too LONG
2061.4
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Exs. 3-2 & 3-4 (2nd Order)
The reactionAB r=kCA2 occurs with 90%. . , A0
and v=4 liters/min, what residence time and reactor
*
literslitersxvVCSTR 360904 ===
21
Exs. 3-2 & 3-4
PFTR rdCj
= r=kCA2
=AC
AdC
AoC r
min9111111
2= = ==
AdCAo
A
PF ==
22
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PF versus Batch Reactor In PFTR, time t that a molecule spends in the reactor
is z/u. Time for molecule to leave reactor is L/u.
Residence time in a constant-density, constant cross-
u
L
v
VPFTR ==
By using udzdt / d We can convert Batch Reactor to PFTR Design
e uation:r
ddzu
dt
AAA ==
a) Aproductsdzdt,PFTR
23
Completely mixed inside 1 slice
z
L
(Constant Density)
nth order kinetics in a batch reactor
111)1(1 nnAoAoA tkCnCC +=
nth order kinetics in a PFTR is same
[ ] )1/(11)1(1 nnAoAoA kCnCC+=
batchPFTR t=
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(FIRST order irreversible reactions with constant density)
AA
C
A
batchPFTR C
C
kC
C
kC
dC
kt
A
0
ln
1
ln
11
==== rd
dC
PFA
=:Ao
AA CC = 0AkC
1= XCA
1
1
1
0
0
=
C
C
A
A
1/ XCCCCV AAoAAoCSTRCSTR =
=
==
A
25
00 AAAAAPFTRPFTR
Com arison between 3 reactors
[ ])1/(1ln)1()/ln(1/
)/ln( 0
0
0
0
XX
X
CC
CC
CCC
CC
V
V
AA
AA
AAA
AA
PFTR
CSTR
PFTR
CSTR
=
=
==
X=(CA0-CA)/CA0PFTR
CSTR
0.0 1.0
. .
0.9 3.91
0.95 6.34
0.99 21.5
0.999 145
26
high conversion for these kinetics.
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(FIRST order irreversible reactions with constant density)
AB
- a c or w g ve e same convers on
- CSTR gives a lower conversion for the same reaction
27
me a c or res ence me con nuous
Difference between PF and CSTR becomes larger as
the order of reaction becomes larger.
CC r=kCA2
AA
AAPFTR
CkC 0
0= 2A
CSTRkC=
1C
)1( XCAPFTR ==
28
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10.00
12.00
Fn=2
6.00
8.00
R/Tau_
P
2.00
4.00
Tau_
CS
0.00
0 0.2 0.4 0.6 0.8 1
X
29
Com arison between 3 reactors
th
are the same.
Negative order the CSTR requires a smaller volume.
Not true for non-isothermal.
Positive order the CSTR requires a larger volume
30
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(FIRST order irreversible reactions with constant density)
Batch CSTR PFTRReactor Size for a given
conversion
+ - +
Simplicity and cost + + -
Continuous operation - + +
Large throughput - + +
Cleanout + + -On-line analysis - + +
Product certification + - -
31
1/r Plot Aproducts
1AAo CCCC
=
=
A = -1*
1C
AA rr
AdC
C
CA
APFTR Cr
A
Ao
=CSTRrd
j
=
)( 0 AC
A
A
PFTR CCdCr
Ao
=
PF
= ydx*
(area under curve)
XA=0 XA=1
CSTR very inefficient
for high X (For first
32d(CA0-CA)=-dCA
orders)
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1/r Plot for r=kC n
r is monotonically decreasing for n (+)
r increases if n
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A 3B r=kC 2
At complete conversion, v=3voRT
nv =
, o
In Batch, we use X instead of C for variable
35
Mass Balance dNj
dtjjjo
=
jjo
X X
- jjjo
Molar flow rate varies with conversion
AoA
XV
=*
XV
AA0
36XrFA=
0
1=A
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XV A products
XrFA0No inert
)1( Xvv o +=Since X is constant throughout,
the volumetric flow rate is
A(g)3B(g)+C(g)e.g. 3113 =+== j
*
rrvv
A
o
A
o
00 ===
37
PFTR Variable Densit For variable density and cross-sectional area.
Similarity of PF and CSTR (one is integral form)z z+dzj
zFj(z+dz)
=X
AA
r
dXFV 0A=-10)()( =++ rdzAdzzFzF
jtjj
==X
AA
dXC
V0AA
dVrdzzFzF +=+ )()(
jtjj0
1
)1(
AA
AAA
CX
XCC
=
=
o rv 0
ACA r
dF =
0
AA
A
C
dCdX =
==AC
A
o rv0
ConstdensityXFF )1( =AdX =
38AAA dXFdF 0=AA
dV0
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Ex. 3-6 CBA + 2 r=kCA2
Find the expression for the reactor volume V for specified and
feed flow F for the reaction amon ideal ases with no
diluent in a CSTR and PFTR.
2)21( XXFX Ao +22 )1( XkCXr Ao
AoCSTR)1(0 XFF AA =
= = Xr
dXFV
X
AoPF
00
0
XFF AC
AB
=
+=
dX
XFV
X
A
PF
)21(
2
2
2
0
0A +=
)21( Xvv +=
++=
=
XXFA
XAo
9)1ln(12940
0
*
39
Ao
Ex. 3-6 CBA + 2
+ XX )21( 2 2+ +2- +4
( )= XX
X 102 4x2- 8x+4
12x-3
( )
+= =dX
XX 14
0
214
+= BAX
+= dX
X
XX
1
1434
2)1(14
22
+=
BXAX
+
+=
=
dX
X3
34
34)1(14 +=
BXAX
= XXX
90 3,1
==BX
40 =
X1,
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Ex. 3-7 Variable Vol An B , r=kC Find the reactor volume V required to obtain 90%
CSTR and in a PFTR with no diluent fornB=2, 1 and-1 A0 , . , o
liters/min. (high pressure reactor, p=49 atm).
2=Bn*
=CA
)1(0 XFF AA =
)1(0
0
XFFF ABA
AB
+=+
=
41
=r)1( Xvv o +=
Ex. 3-7 Variable Vol A2BCSTR
AkCr=
XXvXXFXFV oAAoCSTR
)1()1(0
+=
+==
Ao
)9.01(9.04 +CSTR .
)9.01(5.0
42
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Ex. 3-7 Variable Vol A2BAkCr=PFTR
dXXFdXFV
X
A
X
AoPFTR )1(0+
== -
Xv
r
X
Ao
1
00
+
-x+1x
-
XkPFTR
)1(0
=+1
dX
XXk
vV oPFTR
)1(
11
1
1
+
=
[ ] litersXXv
VPFTR 6.29)1ln(20 ==
43
Ex. 3.7 (Continued) Variable Vol A1B
1=Bn 00F
C AA =AA XFF = 0 )1(
)1( XkCkCr AoA
o
==
ABA
AB
FFF =+
=
0
0
o =
litersX
X
k
v
X
X
kC
F
Xr
XFV o
Ao
AACSTR 72
1.0
9.0
5.0
4
11
00 ==
=
==
LXk
v
X
X
kC
F
Xr
XFV o
Ao
AAPFTR 4.1810ln81
1ln1
0
0
0
0 ===== 44
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Ex. 3.7 Continued2/1=n [AB]
)1()1(0 XCXF
C A
=
=)1(0 XFF AA =
)1(
)1()1( 2121
X
XXo
o
)1( 2102
1
XFFF
XFF AB
=+
=
)1( 21 XAoA
==)1( 21 Xo =
Xk
v
XkCXrFV o
Ao
AACSTR
11
2200
=
==
liters6.399.01
)45.01(9.0
5.0
4=
=
[ ]XXv
dXXFdX
FV oX
A
X
APF )1ln()1( 210
0 +=
==
45L
Ao
8.12)9.010(ln4
00
=+=
- AB. V is larger than with no density change if the
react on pro uces more mo es nB= ecause t s
dilutes the reactant, while V is smaller if the
react on re uces t e num er o mo es nB=An B
AkCr=nB CSTR PFTR
2 136.8 29.6*
*
39.6 12.8
46
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, B Fig 3-6,
47
, B
valid when density varies with conversion.
Constant-density used normally as an
approximation when:
E ual moles of reactants and roducts
Liquids PFBatch* u e gases
EstimationsConstant
48Variable
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Reactor residence time defined based on the
inlet conditions V *
vo=
STV
vSV o ==
In variable density reactions, the velocity
c anges w convers on. ang ng an
along reactor can also change v.
49
PFTRs in series A Products
1 2 N
X
ith
1 PFTR: XdXV
1 2 N-1
==A
o rv 00
Consider N plug flow reactors connected in series. Let
X X X be the fractional conversion of A leavin
reactor 1,2,..N. Based on the material balance for ith
reactor
==iX
Ai
i
dXCV 0
50iX 10
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For N reactors in series
nXXX dXdXdXV 21
=nXXXA
rrrF110
...0
If no change of composition of intermediate streams,
*
then PFTRs in series is same as 1 large PFTR
nX
=rF
51
0
CSTR in Series lst Order Kinetics*
1 3n
2
01
CC AA =111110 AAAA kCCrCC ==
1
222221 AAAA kCCrCC ==
012
CCC AAA ==
52
212
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s r er ne cs
333332 AAAA r ==
CC
)1)(1)(1()1( 32133
kkkkCA
+++=
+=
==
AnA CC 0)1(
nnnnnn
=
+=+= n
Ankk )1()1(
53
,
have the same residence time ( )
o a res ence men
1
oncen ra on rom n reac or n erms o A0is given by C
nn
An kC
)1( +=
54
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-
AAo CC 1AAo
AA
CSTRCrCr
qua res ence me = same rec ang e area
55A series of CSTR has a performance close to PFTR
PFTR+CSTRFirst order kinetics
A1
then CA2.
o owe yAoC 1AC 2AC
1
12
1 kC AA+=
1keCC=
1
2
0
2
12
11
kAAA ekkC
+=+=
56
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PFTR+CSTR first order kinetics CSTR followed by PFTR
0AC 21AC 2AC
1
0CA=
22 0 kAkC
1
11 k
A+
=
1
121
AAk+
kinetics, the expressions are identical for bothconfigurations. So which reactor is first is
57
nconsequen a .
high) then PFTR to minimize total reactor
.
en
CSTR
en
PFTR
58
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. The reaction AB, r=kCA occurs in n equal volume
s n ser es, eac w t res ence t me w t
90% overall conversion. Ifk=0.5/min, CA0=2mo es ter, an v= ters m n, w at res ence t mes
and reactor volumes will be required forn=1,2, and 4?
/1 n
AoCnnAo
An
CC =
CSTR
Total
residence
time
An
n
Ckn
nAn k
C += 1 n
AoCPF
A kC
C=1ln
Cn 1
59n
AoC+=
-.min18
2.00.2)110(
11
1 01 =
=
==
==
CCC AAAo
...
22 2/12/1
==
== AoC
.
5.0 ACk3/1
min92.6)110(5.0
133/1
3 ==
==A
Ao
Ck
min22.6)110(4
14
4 4/14/1
4 ==
== Ao
C
.A
11
60
.10
n5.0=
=PF
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.For n=1, we have single CSTR.
as n to approach PFTR for which =4.61min
Total Reactor volumes:
n=1, V1=72 L, (=4 x 18)vV =
n=2, V2=34.6 L, (=4 x 8.65)
n=3, V3=27.7 L, (=4 x 6.92)
n=4, V4=24.9 L, (=4 x 6.22)
The total reactor volume decreases as n.
61
BAr= ,
a e o orwar reac on s en ance yconcentration of a product. Can also be written by
CkCrBBA =+ ,2
B . , B , , B .
CA0CA0CB0=0
62
A0
CB0=0CAfCBf0 (TANK reaction happens)
B0
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Autocatal tic Reactions Rate )( 00 ABAABAA CCCkCCkCCr +== B0= ,
)(0 AAA
CCkCr =
r versus CA0-CA is a parabola. PF requires timesince r=0 at feed (pure A). CSTR is feasible.
63
Autocatal tic Reactions PFTR
==AC
AdCV
+=
C
A
CCCkC
dCA
AoCo
rv
+=C
dCdCA
A
111
0
+
++ ABAC ABA
CCCC
CCCCCCkA
0000
1
0
( ) +
+=
ABAABA CCCCCCk 000000nn
( ) ++= ABAA
A
B
BA CCC
C
C
C
CCk 0000
00ln
1
64
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Partial fractions 00
1)( AABA BCCCCA =++
C
AdC
A
=
00 ,1)(,0 BAA CCACLet =+=
AA
C
C ABAA
dCB
dCAA
A
+=
000
1
00
1
BA CCA+
=
C
AABA
ABAC A
dCBCCCCAA
A
++=
+0
00
00
)(1
00 ,1, ABAA BCCCCLet =+=
C ABAACCCCk
A+
000
00
1
BA CC
B+
=
AA
C
dCdCA
+= 111
65
ABAC ABA A0000
0
00 AAAACCCC
=
=)( 00 ABAAA CCCkCCr +
oss e
Tubulareac ors
66
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Reversible Reactions Rate goes to 0 near. For PF and CSTR, as
.
CSTR needs a longer (like irreversible reaction)
r=kCACSTR
r=kfCA-kbCB PF
67
Ex. 3-9 Reversible Reactions
BbAf CkCkr =AB
For the reaction, find the residence times for 50%
conversion in a CSTR and in a PFTR ifkf=0.5 min-1,
kb=0.1 min-1, CAO=2 moles/liter, v=4 liters/min, andCBo=0.
For the CSTR
*
68
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Ex. 3-9 Reversible Reactions
For the PFTR
C
AdCA
=C AAobAf
CCkCkAo
)(
AobAbf
Aof
bf
PFTRCkCkkkk ++
=)(
ln
25.01 .21.016.06.0
69
Ex. 3-9 continuedFor this reaction calculate the residence time for90%
AAo CC =AAobAf CCC 2.00.2=
=
Ck1
)2.00.2(1.02.05.0 18.010.0 =
AobAbfbf
PFTRCkCkkkk ++
=)(
n
)ln(21.02.06.0
.
ln6.0 =
=
7008.0=
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Ex. 3-9 (continued)
For this reaction calculate the residence time for90%
= in either reactor. We go beyond the equilibriumconvers on, w c s
CCCk
,
,
,
, ===eqAeqAb CCk
so a A,eq= . mo es er, B,eq= . mo es er, an
the equilibrium conversion is 83%.
This is the maximum conversion obtained for these
71
ne cs n any s ng e reac or
-
Assume spatially uniformFjo
dNj =
Fj
dto
Constant Density (AProducts) Constant Volume
VrCCdCV AAoA = 0
72
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Semi Batch Reactors A may be added continuously but nothing
removed to iverVFF
dt
jjjo
j+=
o ume o reac or ncreases near y w me
tvVV += We obtain
rVFdC
VdV
CdVC AA
+=+=dtdtdt
o
dCA =
dCA
dtjoAooA
73dt
ooAAoo 0
-
Bioreactors
Solids processing
74
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Durin startu and shutdown
Mass-balance equation for species A ssume reac or vo ume an ow ra e are
constant and density is unchanged 1=A
A CVrFFdN
= AdC
AdC
t
AAAodt
=
AAAodt
75CSTR MASS BALANCE
Transients in Continuous Reactors
Solvent replacement (No Reaction)=
Initially t=0
CA0 is inlet concentration
C is concentration within tank
Ai
AdC =
CAi is initial concentration within tank
Let u=C -C
AAAo
dt
= AAoA CC
dC)( AAi
AA
i
t
CC
CC
u
u=
=
0
0lnln
=tC
A dtdCA 1
t
AA eCC
CC =
0
C AA
CCAi 00
tu
t
=
76
=u
dtu
i 0
i
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Transient Reaction in CSTR Tank initially filled with pure solvent.At t=0, feed
A0 A
dCkdu
CkCu AA
+=
+=
1
)1( 0 kCCCdt
AAAoA = )(
dudCA =1
t
kCCCAAAo
A =
dtdu
=1
dt
kCC
dCA =
+ 1 uk+1
77
Transient Reaction in CSTRdtdu
tu
=1 t
k
u
u)
1exp(
+=
tu
uu i
=1
0
tkCkC AA )
1exp(
)1( 0 +=+
uk i+1
k1 +
AAi 0
tAiAAA exp11 00
+=+
1
)exp()1(00
k
tkCCC
CAiAA
A+
+=
At long times0CA SS lst order in CSTR
78)1( k
A+
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te.g. A is salt
AiAAA eCCCC = 00CAo
CA
CAi=0 at t=0
)1(
0
kC AA+=
CA With reaction
CAi=0 at t=0= , A= Ai=
t=, CA Pure solvent
79
dV
productsA
ccumu a on = n u + enera on
dNA =
dVrdVVFVFdC
dVdV
CVCd
dt
AA +=+=)(
( dVrdVVFVFdC
dV
dtdtdt
A += )(
rFC
t
A =
rC
uC AA =
80
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-
dN1 dFVariable Density
rdtV
A= rAA = 0 rdVA=
VXFA = 0
dVdXF AA = 0
rA rA
Constant Densit
rdC
j
j= rCC jjjo = r
dCj
j=
81
THE END