Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01

Flow rate: 𝑄[!!

" ]; cross section: 𝐴[m#]; void fraction: 𝜖(fluid volume/column volume);

superficial velocity: 𝑢 = $% *

!" +; interstitial velocity: 𝑉 = &

' ; phase ratio: 𝜈 = ()''

Fluid phase concentration of the solute : 𝑐[mol/m*];adsorbed phase concentration: 𝑛[mol/m*]

Phase equilibrium, adsorption isotherm: 𝑛 = 𝑓(𝑐) = +,1+-,: Langmuiradsorption isotherm

Assumptions: isothermal; flux is due only to convection (no diffusion); equilibrium between fluid and solid; solute dissolved in a solvent, which is inert; superficial velocity is constant.

𝐹(𝑐) = 𝑢𝑐𝑀(𝑐) = 𝜖𝑐 + (1 − 𝜖)𝑛

9𝜖 + (1 − 𝜖)𝑓.(𝑐):𝑐/ + 𝑢𝑐0 = 0𝑐 = 𝑐(𝑡, 𝑧)

𝑚𝑜𝑙𝑚#

𝑚#

𝑔𝑔𝑚* =

𝑚𝑜𝑙𝑚*

𝑐1. 𝑐+𝑐

𝑛

𝜎(𝑐) =9𝜖 + (1 − 𝜖)𝑓.(𝑐):

𝑢 =1 + 𝜈𝑓.(𝑐)

𝑉 .𝜎.(𝑐) =𝜈𝑓′′(𝑐)𝑉 < 0

𝑓(𝑐) =𝐻𝑐

1 + 𝐾𝑐.𝑓.(𝑐) =

𝐻(1 + 𝐾𝑐)#

. 𝑓..(𝑐) < 0. 𝜆(𝑐) =1

σ(𝑐)=𝑑𝑧𝑑𝑡

𝑧 = 0𝑧 = 𝐿

DESORPTION (𝑐1 = 0), simple wave

𝑧 = 0𝑧 = 𝐿

ADSORPTION (𝑐1 = 0)

𝑧 = 0𝑧 = 𝐿

Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.

Conservation law in finite form9𝑀(𝑡 + Δ𝑡) − 𝑀(𝑡):Δ𝑧 = 9𝐹(𝑧) − 𝐹(𝑧 + Δ𝑧):Δ𝑡

N9𝜖𝑐(𝑡 + Δ𝑡) + (1 − 𝜖)𝑛(𝑡 + Δ𝑡): − 9𝜖𝑐(𝑡) + (1 − 𝜖)𝑛(𝑡):O Δ𝑧 = 9𝑢𝑐(𝑧) − 𝑢𝑐(𝑧 + Δ𝑧):Δ𝑡 = 𝑢Δ𝑡(𝑐+ − 𝑐1)= 9(𝜖𝑐+ + (1 − 𝜖)𝑛+) − (𝜖𝑐1 + (1 − 𝜖)𝑛1):Δ𝑧

Δ𝑡Δ𝑧

= 𝜎P(𝑐1, 𝑐+) =1 + 𝜈Δ𝑛Δ𝑐

𝑉=1𝑉Q1 + 𝜈

[𝑓][𝑐]R 𝜎(𝑐) =

1𝑉(1 + 𝜈

𝑑𝑓𝑑𝑐)

[𝑓][𝑐]

=1

𝑐+ − 𝑐1Q

𝐻𝑐+1 + 𝐾𝑐+

−𝐻𝑐1

(1 + 𝐾𝑐1)R =

𝐻(1 + 𝐾𝑐+)(1 + 𝐾𝑐1)

𝑑𝑓𝑑𝑐

=𝐻

(1 + 𝐾𝑐)#

1. Langmuir isotherm, anti-Langmuir, BET 28-29, 322. Chromatographic cycle 29-303. Pair of equations (32-38)4. Sedimentation 38-39 5. Constant pattern/shock layer 30-32 (19.05)

Residue Curve Maps applied to batch distillation (26.05)

𝑐

𝑛

𝑓(𝑐) =𝐻𝑐

1 − 𝐾𝑐

𝑓(𝑐) = 𝑎𝑐 + 𝑏𝑐#

Adsorption: simple wave Desorption: shock

Adsorption:

𝑓1 − 𝑓2𝑐1 − 𝑐2

= 𝑓′(𝑐2)

BET isotherm

𝑡+ = 𝜎(𝑐+)𝐿.𝑡1 = 𝜎(𝑐1)𝐿𝑡(𝑐) = 𝜎(𝑐)𝐿. 𝑐(𝑡). 𝑡+ ≤ 𝑡 ≤ 𝑡1𝑉𝑡𝐿= 1 + 𝜈

𝐻(1 + 𝐾𝑐)2

. (1 + 𝐾𝑐)2 =𝜈𝐻

𝑉𝑡𝐿 − 1

𝑡3: pulse

𝑡2 =𝐿𝑉Q1 +

𝜈𝐻1 + 𝐾𝑐+

R

Breakthroughtimedependson𝑐4556 = 𝑐+

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡3: pulse

𝑡7 = 𝑡3 +𝐿𝑉(1 + 𝜈𝐻)

Regenerationtimedoesnotdependon𝑐4556

𝑡8 =function(𝑐8)

𝑢𝐴𝑐9𝑡3 = 𝑢𝐴h 𝑐(𝑡)𝑑𝑡 =:

7

𝑢𝐴𝐿𝜈𝑉

h 𝑐𝑓..(𝑐)𝑑𝑐 =⬚

7

,"

=𝑢𝐴𝐿𝜈𝑉

[𝑐𝑓. − 𝑓],"7 =

𝑢𝐴𝐿𝜈𝑉

9𝑓(𝑐8) − 𝑐8𝑓.(𝑐8):Singleimplicitequationin𝑐8

𝑡(𝑐) = 𝜎(𝑐)𝐿

𝑑𝑡 = 𝐿𝑑𝜎 =𝐿𝑑𝜎𝑑𝑐 𝑑𝑐 =

𝐿𝑉 𝜈𝑓

..(𝑐)𝑑𝑐

𝐾𝑐8 =𝛽

1 − 𝛽. 𝛽 = √

𝑐9𝑡3𝑉𝐾𝐿𝜈𝐻

Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01