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 )