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Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01
Flow rate: π[!!
" ]; cross section: π΄[m#]; void fraction: π(fluid volume/column volume);
superficial velocity: π’ = $% *
!" +; interstitial velocity: π = &
' ; phase ratio: π = ()''
Fluid phase concentration of the solute : π[mol/m*];adsorbed phase concentration: π[mol/m*]
Phase equilibrium, adsorption isotherm: π = π(π) = +,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.
πΉ(π) = π’ππ(π) = ππ + (1 β π)π
9π + (1 β π)π.(π):π/ + π’π0 = 0π = π(π‘, π§)
ππππ#
π#
πππ* =
ππππ*
π1. π+π
π
π(π) =9π + (1 β π)π.(π):
π’ =1 + ππ.(π)
π .π.(π) =ππβ²β²(π)π < 0
π(π) =π»π
1 + πΎπ.π.(π) =
π»(1 + πΎπ)#
. π..(π) < 0. π(π) =1
Ο(π)=ππ§ππ‘
π§ = 0π§ = πΏ
DESORPTION (π1 = 0), simple wave
π§ = 0π§ = πΏ
ADSORPTION (π1 = 0)
π§ = 0π§ = πΏ
Mean of the slopes (through integration)1.Red line splits triangle in two equal parts2.
Conservation law in finite form9π(π‘ + Ξπ‘) β π(π‘):Ξπ§ = 9πΉ(π§) β πΉ(π§ + Ξπ§):Ξπ‘
N9ππ(π‘ + Ξπ‘) + (1 β π)π(π‘ + Ξπ‘): β 9ππ(π‘) + (1 β π)π(π‘):O Ξπ§ = 9π’π(π§) β π’π(π§ + Ξπ§):Ξπ‘ = π’Ξπ‘(π+ β π1)= 9(ππ+ + (1 β π)π+) β (ππ1 + (1 β π)π1):Ξπ§
Ξπ‘Ξπ§
= πP(π1, π+) =1 + πΞπΞπ
π=1πQ1 + π
[π][π]R π(π) =
1π(1 + π
ππππ)
[π][π]
=1
π+ β π1Q
π»π+1 + πΎπ+
βπ»π1
(1 + πΎπ1)R =
π»(1 + πΎπ+)(1 + πΎπ1)
ππππ
=π»
(1 + πΎπ)#
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)
π
π
π(π) =π»π
1 β πΎπ
π(π) = ππ + ππ#
Adsorption: simple wave Desorption: shock
Adsorption:
π1 β π2π1 β π2
= πβ²(π2)
BET isotherm
π‘+ = π(π+)πΏ.π‘1 = π(π1)πΏπ‘(π) = π(π)πΏ. π(π‘). π‘+ β€ π‘ β€ π‘1ππ‘πΏ= 1 + π
π»(1 + πΎπ)2
. (1 + πΎπ)2 =ππ»
ππ‘πΏ β 1
π‘3: pulse
π‘2 =πΏπQ1 +
ππ»1 + πΎπ+
R
Breakthroughtimedependsonπ4556 = π+
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘3: pulse
π‘7 = π‘3 +πΏπ(1 + ππ»)
Regenerationtimedoesnotdependonπ4556
π‘8 =function(π8)
π’π΄π9π‘3 = π’π΄h π(π‘)ππ‘ =:
7
π’π΄πΏππ
h ππ..(π)ππ =β¬
7
,"
=π’π΄πΏππ
[ππ. β π],"7 =
π’π΄πΏππ
9π(π8) β π8π.(π8):Singleimplicitequationinπ8
π‘(π) = π(π)πΏ
ππ‘ = πΏππ =πΏππππ ππ =
πΏπ ππ
..(π)ππ
πΎπ8 =π½
1 β π½. π½ = β
π9π‘3ππΎπΏππ»
Equilibrium chromatography (isothermal adsorption)Tuesday, 5 May 2020 13:01