· web viewsherwood number n sh = convective mass transfer coefficient diffusive mass transfer...
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Liquid or solid mixture Gas mixture
c= nVcA=
nA
VcB=
nB
Vx A=
cAc
=n A
nxB=
cBc
=nBn
c A+cB=c x A+xB=1
c A=PA
RTcB=
PB
RTy A=
PA
PyB=
PB
P
c A+cB=c=PA
RT+PB
RT= PRT
y A+ y B=1
ρA=c AM A y A∨xA=
ωA
M A
ωA
M A+ωB
M B
mass fractionωA=xAM A
xAM A+x BM B∨
y AM A
y AM A+ yBMB
massaverage velocity v=∑i=1
n
ρi v i
ρmolar average velocity V=
∑i=1
n
c iV i
c
Fick’s law for molecular diffusion
J Az=−DAB
d c A
dz=−c DAB
d xA∨d y A
dz=DAB (c A1−c A 2)
z2−z1=c DAB (x A 1−x A2 )
z2−z1
c A=nA
V=PA
RTJ Az=
DAB (PA1−PA 2 )z2−z1
ConvectionN A=J A+cA vM
J A+J B=0cA+cB=c
N=J A+c A vM+J B+cBvM=c A vM+cB vM=cvM=N A+NB vM=N A+N B
c
N A=J A+c A
c (N A+N B )
N A=−c DAB
d xAdz
+c A
c (N A+N B )N B=−c DBA
d xBdz
+c B
c (N A+N B )
N A=kc (cL 1−c Li )
K=c Li
c i
Equimolar counterdiffusion in gasesJ Az=−J Bz
c=cA+cBd c A=−d cB
J Az=−DAB
d c A
dz=−JBz=DBA
d cBdz
DAB=DBA
N A=−c (DAB+εM )d x A
dz+ xA (N A+N B)
If A is diffusing in stagnant, nondiffusing B, NB=0
N A=−c DAB
d xAdz
+c A
cN A=−D AB
d cAdz
+x A N A
N A=−DAB
1−x A
dc A
dz=
−c DAB
c−c A
d cAdz
N Adz=c DAB
c−c Ad c A N A ( z2−z1 )=c DAB ln
c−c A2
c−c A1=c DAB ln
1−x A2
1−x A1
N A=−DAB
RTd PA
dz+PA
PN A N A (1− PA
P )=−DAB
RTd PA
dz
N A=DAB P
RT ( z2−z1 )lnP−P A1
P−P A2P=PA 1+PB1=PA2+PB2
PBM=PB2−PB1
lnPB 2
PB 1
=PA 1−PA2
lnP−PA2
P−PA1
N A=DABP
RT ( z2−z1 )PBMPA1−PA 2
Diffusion from a sphere
N A=N A
Area
N A=−4 π r2 DdCA
dr
dN A
dr=0=D
r 2ddr (r2 d CA
dr )=0
ddr (r2 d CA
dr )=0BC :r=RC A=CAR r=∞C A=0
r2 dC A
dr=A1
dC A
dr=A1
r2 CA=A2−A1
rA1=
−CAR
RA2=0
C A=C ARRr
N A@ R=−DdC A
dr R=−DCAR R∗− 1
r2R
=CARDR
dM=−N∗A∗dtρdV=−N∗A∗dt
ρ 4 π r2dr=C ARDr
∗4 π r2∗dt
∫R
0
ρrdr=CARDr
t t= ρ R2
2CARD
Diffusion through a conduit of nonuniform cross-sectional area
r=r2−r 1
z2−z1z+r1N A=
N A (kgmols
)
π r2 =−DABd PA
RT (1− PA
P )dr=
N A
π ( r2−r1
z2−z1z+r1)
2
ℜ= inertial forcesviscous forces
= ρLvμ
Sc= viscousdifusion ratemolecular diffusion rate
= μρDAB
Sherwood number N Sh=convective mass transfer coefficientdiffusivemass transfer coefficient
=kc ' LDAB
=kc yBM LDAB
=k x ' Lc D AB
= f ( ℜ , Sc )
Equation for diffusion in liquids
equimolar counterdiffusionN A=DAB (c A1−c A 2)
z2−z1=DAB c Av ( xA 1−x A2 )
z2−z1
diffusion A thorughnondiffusing BN A=DAB c Av (x A 1−xA 2 )
( z2−z1 )xBM=DAB (c A 1−c A2 )
z2−z1xBM=
xB2−xB1
lnxB2
xB1
Diffusion in solids
c A=solubility(m3 solute per m3)∗PA
22.414( kgmol Am3 solid
)
N A=D ABS (PA1−P A2 )22.414(z2−z1)
Y=c1−cc1−c0
c=c0t=0 x=x1Y=1c=c1 t=t x=0Y=0c=c1 t=t x=2 x1Y=0
Heat transfer Mass transfer
Y=T1−TT 1−T o
X=αtx1
2
x2√αt
m= kh x1
hk √αt
n= xx1
Y=c1/K−cc1/K−co
X=DAB tx1
2
x2√DAB t
m=DAB
K kc x1
K kcDAB
√DABt
n= xx1
Initial conditions (IC)
C A ( t=0 , x , y , z )=costant
Boundary conditions (BC)
C A ( z=zs , x , y ,t )=costant
Boundary conditions (BC)dC A
dz=costant ,∨constant=0
−DdC A
dz=k (CA ∞−C As)
Different types of fluxesMass flux (kg/s/m^2) Molar flux (kg/s/m^2)
Relative to fixed coordinates nA=ρA v A N A=c Av A
Relative to molar average velocity vm jA=ρA (v A−vM ) jA=c A (v A−vM)Relative to mass average velocity v jA=ρA (v A−v) jA=c A (v A−v )
Relations between the fluxes aboveN A+N B=c vM N A=nA /M A N A=J A+cAVJ A+J B=0 J A= jA /M A nA+nB=ρvjA+ jB=0 N A=J A+cA vM nA= jA+ρAV
J A=−c DAB
dx A
dzjA=−ρD AB
dwA
dz
General mole balance ∇⃑ ∙ N⃑+∂CA
∂t=R A
N A=−DAB
dc A
dz+(x A∨ y A) (N A+N B )
Spherical coordinates (typical 1D transport) ∇⃑ ∙ N⃑= 1r2∂ (r2 N )∂r
N A=−DAB
d cAdr
+(xA∨ y A) (N A+N B )
If dilute or if counterdiffusive N A=−DAB
dc A
dr
If no reaction, ∂C A
∂t=−∇⃑ ∙ N⃑=−1
r2∂ (r2N )∂r
= 1r 2
∂(r2DAB
d cAdr )
∂r=DAB
r2 [2 rd c A
dr+r2 d
2 cAd r2 ]
If steady state with reaction, ∇⃑ ∙ N⃑=R A if steady state without reaction, ∇⃑ ∙ N⃑=0
Typical ICs, t=0, C A=0Typical BCs, r=R, C A=C A0 r=0 ,d CA
dr=0
Cartesian coordinates (can be any dimension) ∇⃑ ∙ N⃑=∂ N∂ z
N A=−DAB
d cAdz
+(x A∨ y A) (N A+N B )
If dilute or if counterdiffusive N A=−DAB
dc A
dz
If no reaction, ∂C A
∂t=−∇⃑ ∙ N⃑=−∂ N
∂z=∂(DAB
d c A
dz )∂ z
=DAB
d2 cAd z2
If steady state with reaction, ∇⃑ ∙ N⃑=R A if steady state without reaction, ∇⃑ ∙ N⃑=0
Typical ICs, t=0, C A=0Typical BCs, z=something, C A∨x A=something
Cylindrical coordinates (can be 1D radial, or 2D axial) ∇⃑ ∙ N⃑=∂ N∂ z
+1r∂(rN )∂r
N Ar=−DAB
d c A
drN Az=c Av z
If no reaction, ∂C A
∂t=−∇⃑ ∙ N⃑=−∂ N
∂z−1r∂ (rN )∂r
If steady state with reaction, ∇⃑ ∙ N⃑=R A if steady state without reaction, ∇⃑ ∙ N⃑=0=∂ N∂z
+ 1r∂(rN )∂r
Typical ICs, t=0, C A=0
Typical BCs, z or r=something, C A∨x A=something∂C A
∂r=0at r=0 , symmetry
Equimolar counterdiffusion
N A=k y' ( y AG− y Ai)=k x
' (x Ai−x AL)=K y' ( y AG− y A
¿ )=K x' (x A
¿ −x AL)
−k x'
k y' =
( y AG− y Ai)(x AL−xAi )
y AG− yA¿ =( yAG− y Ai )+( y Ai− y A
¿ )
m'=y Ai− y A
¿
xAi−x AL
1K y
' =1k y' +
m'
kx'
If m’ small, gas phase controlling.
1K y
' =1k y'
x A¿ − xAL=(x A
¿ −xAi )+(x Ai−x AL )
m' '=y AG− y Ai
x A¿ −x Ai
1K x
' =1
m' ' k y' +
1k x'
If m’’ big, gas phase controlling.
1K x
' =1k x'
A diffusing through stagnant B
N A=k y' ( y AG− y ℑ)(1− y A )ℑ
=k x' (x ℑ− xAL )( 1−x A )ℑ
=k y ( y AG− y Ai )=k x (x Ai−x AL)
N A=K y
' ( yAG− y A¿ )
(1− y A )¿M=K x
' (x A¿ −x AL)
(1−x A )¿M=K y ( y AG− yA
¿ )=K x ( xA¿ −x AL)
−k x
k y=
−k x'
(1−xA )ℑk y'
(1− y A )ℑ
=( y AG− yAi )(x AL−xAi )
k y=k y'
( 1− y A )ℑ
(1− y A )ℑ=(1− y Ai)−(1− y AG)
ln1− y Ai
1− y AG
k x=kx'
(1−x A )ℑ
(1−x A )ℑ=(1−x AL )−(1−x Ai)
ln1−x AL
1−x Ai
K y=K y
'
(1− y A )¿M
(1− y A )¿M=¿¿
1K y
'
(1− y A )¿M
= 1k y'
(1− y A )ℑ
+ m'
k x'
(1−x A )ℑ
K x=K x
'
( 1−x A )¿M
(1−x A )¿M=(1−xAL )−¿¿
1K x
'
(1−xA )¿M
= 1m' ' k y
'
(1− y A )ℑ
+ 1k x'
(1−x A )ℑ
Stripping (transfer of solute from L to V)
Pi=H i x i y i=H i
Ptotx i
L'=L0 (1−x0 )
L' x0
1−x0+V ' yn+1
1− yn+1=L' xn
1−xn+V ' y1
1− y1
V= V '
1− yn+1
N=
ln [ x0−yN+1
m
xN−yN+1
m
(1−A )+A]ln 1
A
Absorption (transfer of solute from V to L)
Lmin' x L
1−xL+V ' y1
1− y1=Lmin
' x1
1−x1+V ' y2
1− y2
V '=V 1(1− y1)
N=ln [ y N+1−mx0
y1−mx0(1− 1
A )+ 1A ]
ln A
A=√ LN L0
m2V N+1V 1
Plate absorption towers L' x0
1−x0+V
' yN +1
1− yN +1=L' x N
1−xN+V
' y1
1− y1 Packed column
L' x1−x
+V 'y1
1− y1=L' x1
1−x1+V ' y
1− y
For stripping (transfer of solute from L to V)
N=
ln [ x2−y1
m
x1−y1
m
(1−A )+A ]ln 1
A
For absorption (transfer of solute from V to L) N=
ln [ y1−mx2
y2−mx2(1− 1
A )+ 1A ]
ln A
A= LmV
dA=aSdz
a is the interfacial area in m2 per m3 volume of packed section S is the cross-sectional area of the tower.d (Vy )=d (Lx )
z=∫y2
y1 VdyK y
' aS(1− y )¿M
(1− y ) ( y− y¿ )=∫
x2
x1 LdxK x
' aS(1−x )¿M
(1−x ) ( x¿−x )
z=∫y2
y1 Vdyk y' aS
(1− y )ℑ(1− y ) ( y− y i)
=∫x2
x1 Ldxkx' aS
(1−x )ℑ(1−x ) (x i−x )
if dilute
z=[ VK y
' aS(1− y )¿M
(1− y )]∫y2
y1 dy( y− y¿)
=[ LK x
' aS(1−x )¿M
(1− x )]∫x2
x1 dx( x¿−x )
z=[ Vk y' aS
(1− y )ℑ(1− y )
]∫y2
y1 dy( y− y i )
=[ Lkx' aS
(1−x )ℑ(1−x )
]∫y2
y1 dx(x i−x )
(1− y )ℑ(1− y )
=(1− y )¿M
(1− y )=
(1−x )ℑ(1−x )
=(1−x )¿M
(1−x )=1
VS ( y1− y2 )=k y
' az ( y− y i )MLS (x1−x2)=k x
' az (x i−x )MVS ( y1− y2 )=K y
' az ( y− y¿)M
LS (x1−x2)=K x
' az ( x¿−x )M
Concentrated solutions, stagnant B
z=HOG∫y2
y1 (1− y )¿M dy(1− y ) ( y− y¿ )
=HOGNOGHOG=V
K y' aS
= VK ya (1− y )¿M S
z=HOL∫x2
x1 (1−x )¿M dx(1−x ) (x¿−x )
=HOLNOLHOL=V
K x' aS
= VK x a (1−x )¿M S
z=HG∫y2
y1 (1− y )ℑ dy(1− y ) ( y− y i )
=HGNGHG=V
k y' aS
= Vk ya (1− y )ℑS
z=H L∫x2
x1 (1−x )ℑ dx(1−x ) (x i−x )
=H LNLH L=V
kx' aS
= Vk x a (1−x )ℑ S
Dilute solutions
(1− y )ℑ(1− y )
=(1− y )¿M
(1− y )=
(1−x )ℑ(1−x )
=(1−x )¿M
(1−x )=1
z=HOG NOG=HOG∫y2
y1 dy( y− y¿ )
NOG=y1− y2
( y− y¿)M
z=HOLNOL=HOL∫x2
x1 dx(x¿−x )
NOL=x1−x2
(x¿−x )M
z=HGNG=HG∫y2
y1 dy( y− y i )
NG=y1− y2
( y− y i )M
z=H LNL=H L∫x2
x1 dx(x i−x )
NG=x1−x2
(x i−x )M
NOG=1
1− 1A
ln [ y1−mx2
y1−mx2(1− 1
A )+ 1A ]
NOL=1
1−Aln [ x2− y1/m
x1− y1/m(1−A )+A ]
NOG=N∗ln A
1− 1A
HETP=HOG
ln 1A
1−AA
Vapor-liquid separationRaoult’s law
pA+ pB=P
PA x A+PB (1−xA )=P
y A=pA
P=PA x A
PPA∧PBare the vapor pressuresof A∧B
Relative volatility α AB=
y A
x A
yBx B
=
y A
x A
1− y A
1−x A
=PA
PBy A=
α AB x A
1+(α−1 ) x A
Flash distillation FxF=Vy+Lx=Vy+(F−V ) x
xL= (x−dx ) (L−dL )+ ydL=xL−xdL−Ldx+dxdL+ yDL
dLL
= dxy−x∫L2
L1 dLL
=lnL1
L2=∫
x2
x1 dxy−x
L1 x1=L2 x2+(L1−L2 ) yav
McCabe-Thiele MethodV n+1+Ln−1=V n+ Ln
V n+1 yn+1+Ln−1 xn−1=V n yn+Ln xn
n-1V n , yn ↑ ↓
↑ ↓Ln−1 , xn−1
nV n+1 , yn+1 ↑ ↓
↑ ↓Ln , xnn+1
Enriching sectionF=D+W
FxF=DxD+WxWV n+1=Ln+D
V n+1 yn+1=Ln xn+D xD
yn+1=Ln
V n+1xn+D
xDV n+ 1
= RR+1
xn+xD
R+1
R=Ln
DEnriching line will intersect y=x line at xD, y=xD
Draw from xD, y=xD with the slope Ln
V n+1∨ RR+1
Stripping sectionV m+1=Lm−W
V m+1 ym+1=Lm xm−W xW
ym+1=Lm
V m+1xm−
W xWV m+1
Enriching line will intersect y=x line at xW, y=xW
Draw from xD, y=xD with the slope Lm
V m+1
Two lines intersect at a point, connect that point to xF, y=xF to form q line
q=H V (dew point )−HF (entrance conditions )HV (dew point )−H L (bubble /boiling point )
H- enthalpy of feedLm=Ln+qF
V n=V m+ (1−q )F
y= qq−1
x+xFq−1
q line will intersect y=x line at xF, y=xF
(1) H at entrance condition > H at bubble point (and dew point), q < 0, slope < 1(2) H at entrance condition = H at dew point, q = 0, slope = 0(3) H at entrance condition < H at bubble point, 0<q<1, slope < 0(4) H at entrance condition = H at bubble point q=1, slope = infinite(5) H at entrance condition < H at dew point, q>1, slope > 1
Reboiler is an equilibrium stage where there is both liquid and vapor leaving. Condenser is not because full condensation occurs.
x’ and y' represent pinch point, where the number of
stages required becomes infinite, Rm
Rm+1=xD− y '
xD−x '.
Typically want R = 1.2 Rm ~ 1.5 RmR = infinity, slope of the enriching section = 1
Nm=log
xD
1−xD
1−xWxW
logα av
Overall efficiency EO=number of idealtraysnumber of actual trays
Murphree tray efficiency EM=yn− yn+1
yn¿− yn+1
if the operating line is straight with slope m
EO=log1+EM (mVL −1)
log mVL
H=HOGNOG
HETP=
HOG ln mLV
mLV
−1=Tray spacingT
EO
Liquid-liquid and fluid-solid separationsEquilibrium relationships - 3 isotherms for adsorption
Freundlich is favorable, Langmuir is strongly favorable.
qFM+c FS=qM+cS
q=Kc q=Kcn q=q0 cK+c
WashingAnalogous to stripping where U (underflows) O (overflows)
recovery=1−x N (outlet )x0 (inlet )
=x0−xNx0− yN+1
=( 1A )
N+1
− 1A
( 1A )
N+1
−1A=
LmV
=UO
V1, x1←
→L0, N0, y0, B
V2, x2←
→L1, N1, y1, B
L0+V 2=L1+V 1=ML0 y A0+V 2 x A 2=L1 y A1+V 1 xA 1=M x AM
B=L0 N0=L1 N1=N MM
N A=k1 (C1−C1 i )=DSL (C1 i−C2i )=k2 (C 2−C2 i )
D (diffusivity )∗S(solubility )=P( permeability)
C1−C1 i=N A
k1
C 1i−C2 i=N A LP A
C2i−C2=N A
k2
N A=C1−C2
1k 1
+ LPA
+ 1k2
V A
A=V P yPA
=PA'
L (Phx0−P l y p )V B
A=V P (1− y p )
A=PB'
L (Ph(1−x0)−P l(1− y p))
α ¿=PA'
PB' Lf=L0+V pθ=
V p
L f
y p
1− y p=
α¿ [ x0−plph
y p]
(1−x0 )−plph
(1− y p)
y p=−b±√b2−4ac
2a
a=1−α¿b=−1+α ¿+phpl
+x0 ph
p l(α¿−1 )c=
−α ¿ x0 php l
Lf x f=L0 x0+V p y p x0=x f−θ y p
1−θy p=
x f−x0 (1−θ )θ
A=θ L f y p
PA'
L ( ph x0−p l y p )