1

Reminiscences

2

• Physical data have been supplied to Problem_#4.

• Some physico-chemical data can be found at the web page (E-Tables):

http://uchi.vscht.cz/index.php/en/studium/navody-a-pomucky/e-tabulky

• A lot of data can be found at:

https://www.engineeringtoolbox.com/index.html

3

THREE-STAGE PLANT

8 streams: 8 x 3 = 24 unks. 2 amounts specified + 7 concentrations 7 concentrations can be easily evaluated so 2+14 = 16 data values are known There is 8 unknown values and we can formulate 4x2 balances + 1x sum of concentrations (stream 2) = the problem is overdetermined!!!

7500 m2

50 MMscfd 30% CO2

500 psia

35.4 MMscfd 7.9% CO2

500 psia

56.0% CO2

730 m2

93.1% CO2

30% CO2

14.6% CO2

7500 m2

81.5% CO2

4

Batch process: In a batch process after filling the feed vessel with the starting volume, concentration and/or diafiltration (washing) starts until the final volume or desired washing is reached. Product is either in the vessel or the collected permeate.

cP

cR

cF

V0,c0

Vf, cf

Qw, cw=0

5

ˆ

1

0

0

1

f

W

P

R

f

c V

c V

Q

Q

volumetric flow rate of washing solution

volumetric flow rate of permeate

f … final state 0 … initial state

6

Multicomponent gas separation: stirred to stirred system

,M iP

iJ,F in z ,R in x

,P in y

7

ii R i Pi

M

pP

J x py

1 1 1 11

/

/C C C C C

ik i redi

e

ii

i

i n

jk j r d j

j

i

R i P

i R P i i ii M

j

j

red

n n n n

j R j P j j R j P j j red j

k

j j j jM

k

y

Px

P x P xJ

PJ x P x

S x py

S x p y

P

P

p y pp y p p y

p y p p y pP xp y

1i i iz y x 1, , Ci n

We have 2nC equations for xi and yi supposing feed composition (zi) is given together with the cut () value and permeabilities or selectivities (Pi or Sik).

8

9

10

ULTRAFILTRATION (UF)

11

HISTORY

12

Pressure driven membrane separation process for removal of macromolecular (or colloidal) solutes (103 – 106 Daltons)

Membranes: asymmetric microporous with pore size (approx.) 10 – 100 nm

Separation: due to molecule size (MW) & shape vs. pore size & structure = solute rejection

Applications: separation of proteins, polysaccharides, nucleic acids, thickening of solutions (dairy industry, fruit juices …), waste water treatment …

13

Membrane rejection/retention fundamental factor affecting separation

Solute rejection is a complex process (size × shape)

!

14

Membrane fouling – main problem of UF separations

Fouling – due to small/colloidal particles present in the feed

generally reversible (turbulence, back-flushing …)

generally irreversible (membrane blocking)

15

Turbulence?

Rel u

UF device configuration, channel size, stirring …

feed/retentate stream velocity

Spacers/obstacles

16

BACK FLUSHING – reversal of flow direction

Due to membrane fouling the permeate flux declines with time of operation – membrane cleaning is necessary

BACK FLUSHINGS?

17

Concentration polarization and gel layer formation hydrodynamics of UF modules and from it resulting mass transfer rate is of great importance

Development of concentration polarization layer and (possibly) of gel layer at the membrane feed side with operational pressure

18

polarization modulus

19

Permeate flux vs. applied transmembrane pressure

,

max

,

lni gel

c

i bulk

cJ k

c

membrane permeability

20

MASS TRANSFER COEFFICIENT?

Sh c h

i

k d

D Sherwood number

ic

Dk

definition

Re hu d

Sci iD D

Reynolds number

Schmidt number 4

2

h

h

h

Sd

o

d d

d h

hydraulic diameter of retentate channel

tubes

slits

21

Laminar flow of retentate stream:

Turbulent flow of retentate stream:

3/4 1/3Sh 0.04 Re Sc

1/3

Sh 1.62 ReSc hd

L

4Re 10

40.1 ReSc 10hd

L

22

Permeate flux in ultrafiltration?

23

tot

V

pJ

R

gMtot M g

M g

R R RP P

driving force difference in osmotic pressures

permeate flux density (m/s)

total, membrane, gel layer resistances

pure water flux density (m/s)

M

M M

V

p p PJ

R

evaluation of membrane resistance or permeability using pure water permeation data

24

Membrane area (module size)?

V

PM

VA

J

mass balance (m3/s)

kinetics (m/s)

25

How gel layer concentration can be estimated?

Experiments, experiments ….

26

Continuous ultrafiltration

SINGLE STAGE

27

TWO STAGES

28

THREE STAGES

29

Batch ultrafiltration

bc

30

1 P

b

cR

c module retention

0

0

( ), when 1

( )

b

b

c t VR

c V t

0

0

( )ln ln , when 1

( )

b

b

c t VR R

c V t

31

DIALYSIS

32

Concentration difference driven membrane separation process for separation of low molecules from large molecules in a solution

Membranes: symmetric, dense (dialysis tubes)

Separation: due to different rates of diffusion in membrane

Applications: desalting of protein solutions, polysaccharides, nucleic acids, recovery of spent acids or alkalis, drinking water …

33

Description

1 1 1,m ipc S c 2 2 2,m ipc S c

Si … distribution/partition coefficients)

1 1 1,

1 2

2 2, 2

n L ip

n m m m

n L ip

j k c c

j k c c

j k c c

FLUX OF A COMPONENT

m

m

Dk

km is (a formal) mass transfer coefficient in the membrane used to describe flux in membrane formally in the same way as mass transfer in liquid films (D is diffusivity)

Partitioning of the component between solution and membrane (equilibrium): solution 1 solution 2 membrane

1c

1,ipc

2,ipc2c

1mc

2mc

nj

,m D

34

1 1 2 2

1 2

1 2

1n

L m L

S c S cj

S S

k k k

when S1= S2=S

1 2 1 21 2

1 2 1 2

1 1 1 1 1n

L m

m

L L L

c c c cj K c c

k S k k k S D k

1 2

1

1 1 1

L m L

K

k S k k

m

m

Dk

and kL1, kL2 are evaluated from Sherwood number

1 2n mn K c c A

total component flux:

m

m

DP kS S

permeability of the membrane

35

BATCH DIALYSIS

1 111 2

2 221 2

dd

d d

dd

d d

V cnK c c A

V cnK c c A

1 1,V c 2 2,V c

when V1=const. and V2=const.

11 2

1

21 2

2

d

d

d

d

c AK c c

V

c AK c c

V

36

0 200 400 600 800 10000.0

0.1

0.2

0.3

0.4

0.5

t sec.

c 1t

,c 2

t

10 1 21 1 2

1 2 1 2

( )( ) exp

c K A V Vc V V

V V VV

10 1 22 1 1

1 2 1 2

( )( ) exp

c K A V Vc V V

V V VV

37

CONTINUOUS DIALYSIS

A

1 1,, inV c 1 1,, outV c

2 2,, inV c2 2,, outV c

1, 2, 1, 2,

1, 2,

1, 2,

ln

in out out in

n meanin out

out in

c c c cn j A K A K c A

c c

c c

n

38

Possible (historical) arrangement of a semi-batch dialysis

FEED SOLUTION

DIALYSIS TUBE

STRIPPING SOLUTION

PERMEATE VESSEL

39

LABORATORY DIALYSIS

40

An Arrangement of Industrial Dialysis

41

Acid Recovery by Diffusion Dialysis

42

MEDICAL DIALYSIS

43

1) An experiment is being conducted to determine the suitability of a cellophane membrane for using it in hemodialysis process. The membrane is 0.025 mm thick. In the experiment, conducted at 36°C using common salt (NaCl) as the diffusing solute, the membrane separates two compartments containing stirred aqueous solutions of NaCl. The bulk liquid phase concentration of diffusing solute in the upstream is 1.0×10-4 mol cm-3 and 5.0×10-7 mol cm-3 in the downstream. The mass transfer coefficients on either side of the membrane is same which calculated as 5.24×10-5 m s-1. Experimental data obtained gave a flux of 8.11×10-4 mol NaCl s-1 m-2 at quasi steady state condition. Calculate:

The membrane permeability in m s-1 and (S D) in m2s-1. The percentage resistance to diffusion in liquid film.

2) Calculate the rate of removal of urea in grams per hour from blood in a cellophane membrane

dialyzer at 37°C. The membrane is 0.025 mm thick and having an area of 2.0 m2. The mass transfer coefficient on the blood side is 1.25×10-5 m s-1 and that on aqueous side is 3.33×10-5 m s-1. The membrane permeability is 8.73×10-6 m s-1. The concentration of urea in blood is 0.02 g per 100 cm3 and that in the dialyzing fluid can be assumed zero.

3) A liquid containing dilute solute at a concentration of 3×10-2 kmol m-3 is flowing rapidly by a

membrane of thickness 3.0×10-5 m. The distribution coefficient of the solute equals 1.5 and diffusivity is 7.0×10-11 m2 s-1 in the membrane. The solute diffuses through the membrane and its concentration on the other side is 0.50×10-2 kmol m-3. The mass transfer coefficient kL1 is large and can be considered as infinite and kL2 = 2.02×10-5 m s-1. Calculate the flux and the concentrations at the membrane interfaces. Sketch the concentration profile also.

DIALYSIS - 1 An experiment is being conducted to determine the suitability of a cellophane membrane for using it in hemodialysis process. The membrane is 0.025 mm thick. In the experiment, conducted at 36°C using common salt (NaCl) as the diffusing solute, the membrane separates two compartments containing stirred aqueous solutions of NaCl. The bulk liquid phase concentration of diffusing solute in the upstream is 1.0×10-4 mol cm-3 and 5.0×10-7 mol cm-3 in the downstream. The mass transfer coefficients on either side of the membrane is same which calculated as 5.24×10-5 m s-1. Experimental data obtained gave a flux of 8.11×10-4 mol NaCl s-1 m-2 at quasi steady state condition. Calculate: 1) The membrane permeability in m s-1 and DAB K in m2s-1. 2) The percentage resistance to diffusion in liquid film.

44

DIALYSIS - 2 Calculate the rate of removal of urea in grams per hour from blood in a cellophane membrane dialyser at 37°C. The membrane is 0.025 mm thick and having an area of 2.0 m2. The mass transfer coefficient on the blood side is 1.25×10-5 m s-1 and that on aqueous side is 3.33×10-5 m s-1. The membrane permeability is 8.73×10-6 m s-1. The concentration of urea in blood is 0.02 g per 100 cm3 and that in the dialyzing fluid can be assumed zero.

45

DIALYSIS - 3 A liquid containing dilute solute at a concentration 3×10-2 kmol m-3 is flowing rapidly by a membrane of thickness 3.0×10-5 m. The distribution coefficient is 1.5 and diffusivity 7.0×10-11 m2 s-1 in the membrane. The solute diffuses through the membrane and its concentration on the other side is 0.50×10-2 kmol m-3. The mass transfer coefficient kc1 is large and can be considered as infinite and kc2 = 2.02×10-5 m s-1. Calculate the flux and the concentrations at the membrane interfaces. Sketch the concentration profile also.

46

U3.3

An aqueous solution of nonpurified proteins was filtered using an experimental ultrafiltration

membrane. A dependence of the intensity of volumetric permeate flux (JV) on applied

transmembrane pressure difference (p) was measured. The results are given in the table:

p [kPa] 0 13 27 36 45 55 62 81 105 118 136

JV [10-3

m s-1

] 0 0.046 0.105 0.150 0.200 0.249 0.295 0.350 0.365 0.370 0.363

Using these data judge whether a gel layer was formed at the membrane surface and, eventually,

estimate limiting value of the permeate flux intensity.

Results: The gel layer has formed, limiting intensity of permeate flux is about 0.365 × 10-3 m s-1.

47

U3.4

The retentate stream in an ultrafiltration membrane module flows inside membrane fibers with

inner diameter of 0.1 cm and length of 100 cm. The mean velocity of the flow in fibers is 3 m s-1.

The retentate stream contains a dissolved protein. Its diffusivity equals 9 × 10-11 m2s-1. The viscosity

of the protein solution is 0.0012 Pa s and its density is 1100 kg m-3. The permeate flux intensity

equals 0.045 m h-1. Determine a value of the polarization modulus during the ultrafiltration and

discuss significance of membrane polarization. How would situation change if fibers of inner

diameter of 0.025 mm are used?

Results: The polarization modulus in 0.1 cm fibers is 14.35 and the membrane polarization is

therefore very strong. The modulus equals 1.325 in larger fibers and the membrane polarization is

therefore insignificant.

48

U3.6

It is known that the proteins of the blood plasma form a gel layer at the membrane surface during

their ultrafiltration. The mass fraction of proteins in the gel layer equals 0.2. The molecules of the

proteins can be considered as spherical particles of 30 nm in diameter. A filtration experiment with

pure water yielded value of the intensity of volumetric flux of the permeate 2.5 × 10-5 m s-1 at the

transmembrane pressure difference p = 68.96 kPa. When the blood plasma was filtered at

p = 103.44 kPa the intensity of volumetric permeate flux equaled 0.416 × 10-5 m s-1. The filtration

is performed at 20o C. Auxiliary experiments yielded permeability of the gel layer of 3.1 × 10-18 m2.

What is the thickness of the gel layer at the steady state, i.e., when limiting intensity of the

permeate flux is achieved? Consider (for simplification) that the plasma viscosity and the permeate

viscosity values are the same and equal to water viscosity at the temperature of the ultrafiltration.

Result: The gel layer thickness is 68.5 m.

49

U3.7

An aqueous colloidal solution is concentrated in an ultrafiltration membrane module. The inet

concentration of the dissolved substance is 50 kg m-3, the outlet concentration would increase to

200 kg m-3. The retention factor of the membrane used for filtration equals one (an ideal

membrane). The volumetric flow rate of the feed is 3.6 m3 h-1. The dissolved substance creates a gel

layer at the membrane surface where its concentration equals 300 kg m-3. The module consists of

tubes with the inner diameter of 5 mm and length of 1250 mm. The retentate stream flows inside

the tubes. Evaluate the intensity of the permeate volumetric flux and the total membrane area

needed for the filtration at mean velocity of the retentate stream within the tubes of 1 m s-1. The

density of the retentate stream is 1260 kgm-3, dynamic viscosity is 2.12 × 10-3 Pa s and the

diffusivity of the dissolved substance in water is 2.75 × 10-9m2 s-1.

Results: The intensity of permeate volumetric flux is 2.858 × 10-5 m s-1 and the membrane area of

26 m2 is necessary.

50

U8.10

A salt is removed from an aqueous solution of a protein by dialysis. The protein solution is tightly

closed within a dialysis tube and its volume is 50 cm3. The tube surface area available for dialysis

equals 110 cm2. The dialysis tube is made of a membrane whose thickness is 120 m. The diffusivity

of the salt within the membrane is 7.21×10-10 m2s-1. The initial salt concentration was set to

0.37 mol dm-3. The mass transfer coefficient at the inner (retentate) side of the membrane is

1.6×10-5 m s-1. The mass transfer coefficient at the outer (permeate) side of the tube equals

7.8×10-5 m s-1. The tube is immersed in pure water whose volume is many times larger than the

volume of the protein solution. Determine at what time the salt concentration in the solution

within the tube will reach 10% of its initial value.

Result: 2530 s.

51

U8.11

The phenol contaminates a solution of macromolecular substances. A continuous dialysis is

suggested to remove the phenol from the solution. The dialysis equipment consists of a glass pipe

(inner diameter: 8 cm) within which 20 membrane dialysis tubes made of a copolymer of a

polycarbonate and polyether are placed. The outer diameter of the dialysis tubes is 10 mm, the

tube wall thickness is 50 m. The phenol diffusivity in the tubes walls equals 1.05 × 10-9 m2s-1. The

contaminated solution flows at flow rate of 0.25 m3h-1 within the dialysis tubes, the inlet phenol

concentration is 0.028 mol dm-3. The outer surfaces of the dialysis tubes are washed with water

that enters the pipe at zero phenol concentration. The water flow rate is 1.525 m3h-1. Determine

the length of the dialysis tubes necessary to reduce the phenol concentration in the treated

solution to 10% of its inlet value. Suppose counter-current operation of the device. The mass

transfer coefficient at the inner surface of dialysis tubes equals 5.6 × 10-4 m s-1 and at the outer

surface equals 1.37 × 10-4 m s-1.

Result: The length of tubes is 14.2 m.

52

53

U8.1 A permeation experiment using pure hydrogen has been performed to determine permeability of a paladium based membrane for the hydrogen. 1.526 moles of H2 permeated the membrane within 50 seconds when the steady state was reached with the membrane of the thickness of 1 mm, of total surface area of 9 cm2 and at 30°C. The pressure difference across the membrane was kept constant at 1.44 MPa. Determine the membrane permeability.

Result: Permeability of the membrane for hydrogen equals 2.35510-8 mol s-1Pa-1m-1.

U8.2 Permeability of a polyethylene foil for oxygen equals 2.31510-12 kg m-1s-1Pa-1 (at 30°C). The foil of

thickness of 150 m is used to wrap perforated paper box with fruits. The metabolic activities of the fruits in the box result to oxygen consumption of 2.1 g of oxygen per hour. The box dimensions are

0.50.60.3 m. The resistance of the box itself to the oxygen transport can be neglected. What will be the oxygen partial pressure within the box in a steady state if the box is surrounded with the air at 98 kPa and 30°C? Result: The oxygen partial pressure value in the box is 20.55 kPa.

54

U8.3 A gas releasing from a bioreactor is transferred to an analyzer by means of a tubing made of the silicone rubber; the tubing outer diameter is 5 mm and wall thickness is 0.25 mm. The length of the tubing is 5.2 m. The gas pressure in the tubing is 125 kPa, the outer pressure is 99 kPa. The gas from the bioreactor contains 4.5 vol.% of carbon dioxide, 16.5 vol.% of oxygen and 4.3 vol.% of water

vapour; the rest up to 100 % is nitrogen. The permeabilities of the silicone rubber are: 2.710-11

m3(m s Pa)-1 for oxygen, 1.7810-10 m3(m s Pa)-1 for carbon dioxide, 2.3810-9 m3(m s Pa)-1 for water vapour – the permeabilities were measured under the same conditions as are those of the permeation from the tubing. What amounts of O2, CO2 and H2O vapour will permeate through the tubing wall? Will it cause significant change in gas composition prior the gas enters the analyzer?

Results: Oxygen permeates into the tubing at the rate 1.4610-6 m3s-1, 3.2710-4 m3s-1 of carbon

dioxide escape from the tubing together with 4.1810-3 m3s-1 of water vapour. The composition of the gas will change significantly.

U8.4 In a single-stage membrane separator the air is to be enriched with oxygen in such a manner that oxygen concentration in the permeate is 30 vol.%. The retentate side pressure is kept at 1 MPa, the permeate side is kept at 0.2 MPa. The permeate is leaving the separator at 40°C. The permeability of

the membrane for the oxygen is 4.510-15 m2s-1Pa-1 (at normal temperature and pressure). The

thickness of the active layer of the membrane is 10 m. The selectivity of the membrane for oxygen with respect to nitrogen equals 2.2. The separator would produce 10 m3h-1 of the enriched air (at temperature and pressure of the permeate). Determine composition of the retentate stream, the area of the membrane necessary for the enrichment and the stage cut value. Consider ideal mixing of the gas at both sides of the membrane. Results: The oxygen concentration in the retentate will be 19 mol.%. The necessary membrane area is 2.5 m2. The stage cut equals 0.179.

55

U8.5 An experimental reverse osmosis membrane has been used for treatment of a waste water containing dissolved sodium carbonate (Na2CO3). The intensity of pure water flux through the membrane was 1.5 m3 per 1 m2 of the membrane area per 1 day (laboratory experiment value) at the pressure difference of 3 MPa at 30°C. The full scale waste water treatment will be performed at the same temperature but at the pressure difference of 5 MPa. The required sodium carbonate concentration in the retentate is 0.05 kmol m-3. The streams at both membrane sides can be considered as ideally mixed. Determine the intensity of volumetric flux of the permeate through the ideal membrane supposing that concentration polarization of the membrane can be neglected. Determine also the membrane area required for treatment of 125 m3 of waste water per day; the water contains 0.027 kmol m-3 Na2CO3.

Results: The intensity of volumetric flux of the permeate is 2.67510-5 m s-1. The membrane area is 24.9 m2.

U8.6 The reverse osmosis (RO) is used (RO) to treat an aqueous solution of low molecular substances that comes to the RO module at the rate of 26 m3 per day. The concentration of the substances equals 0.1 kmol m-3 and the osmotic pressure of the solution equals 0.46 MPa. We can suppose that the osmotic pressure of the solutions is directly proportional to concentration of dissolved substances concentration. The concentration polarization of the membrane can be neglected. The RO is performed at the pressure difference of 3 MPa and the stage cut equals 0.4. The retention factor value

of the module is 0.9. An experimentally determined permeance of the membrane (Pm=P/m) for pure

water equals 510-13 m s-1 Pa-1. Determine volumetric flow rates of the retentate and the permeate and concentrations of dissolved substances in both streams. Determine also the required membrane area. Results: The permeate volumetric flow rate is 10.4 m3/day, the retentate flow rate is 15.6 m3/day. The retentate contains 0.1563 kmol m-3 of dissolved substances, the permeate contains 0.01563 kmol m-3 of the dissolved substances. The membrane area is 103 m2.

56

U8.7 The air is enriched with the oxygen in a membrane separator with the membrane made of silicone rubber. The permeate containing 27 mol.% of the oxygen has to be produced. The membrane selectivity for oxygen with respect to the nitrogen under the conditions of the separator equals 2.1. The reduced pressure value equals 2.86. Determine the stage cut value necessary to achieve goals of the separation. Result: The stage cut value has to be set to 0.233.

U8.8 The methane is purified in a membrane separator. The traces of carbon dioxide is removed from the methane at 35°C and 2 MPa at the retentate side of the membrane. The permeability of the

membrane (at conditions of the separation) for carbon dioxide is 1.12510-16 m2s-1Pa-1, for the

methane is 3.610-18 m2s-1Pa-1 (the permeabilities were evaluated at conditions at the retentate side of the membrane). The gas entering the separator contains 90 mol.% of CH4 and 10 mol.% of CO2. The pressure at the permeate side is 0.11 MPa, the stage cut equals 0.5. The thickness of the membrane

active layer is 1 m. Determine values of the membrane selectivity for CO2 against CH4, the compositions of both permeate and the retentate and the intensities of volumetric flow rates of both outlet streams. Determine separator production rate (measured by the permeate volumetric flow rate) if the membrane area equals 200 m2? Results: The membrane selectivity is 31.25. The retentate contains 1.68 vol.% of the carbon dioxide, the permeate contains 18.3 vol.% of the carbon dioxide. The volumetric flow rate of the permeate equals 108 m3h-1 (at the conditions of the permeate side).

57

U7.1

Experiments yielded equilibrium adsorption data listed in table:

AX [mg/g] Ac [mg/cm3]

0,02 3,2010-9

0,08 3,2810-6

0,10 1,0010-5

0,50 3,1210-2

0,70 1,6810-1

1,00 1,00

Determine parameters of Freundlich and Langmuir isotherms using the above data.

Results: Freundlich isotherm: 0,201,00A AX c . Langmuir isotherm does not fit the data.

58

U7.4

Clear fermentation broth after separation of cells contains 810-5

mol dm-3

of

immunoglobulin G. 90% of the immunoglobulin present in the solution has to be separated by

an adsorption to a synthetic nonpolar adsorbent. Adsorption equilibrium can be described by

the isotherm: 0,3555,5 10A AX c , where AX is the equilibrium concentration in the adsorbent

(mol cm-3

) and Ac is the equilibrium concentration in the liquid (mol dm-3

). Compute

minimum amount of the adsorbent required for treatment of 2 m3 of the broth in a single

stirred adsorption stage.

Result: 0,16 m3 of the adsorbent.

U7.5

Into the vessel filled with a bed of activated carbon a solution of the acetic acid was poured in

such a way that void volume between the particles has been just filled up. The acid adsorbs to

the carbon particles at 60°C. The vessel performs as an equilibrium stage. The equilibrium is

expressed by relation 0,4119

m 3,019A ALX c , where XmA is equilibrium concentration of acetic

acid in the adsorbent (moles CH3COOH per 1 kg of carbon) and ALc is equilibrium

concentration of CH3COOH in liquid (mol dm-3

). Total volume of the bed of the adsorbent is

Vb = 1,34 m3, porosity of the empty bed is e = 0,434, inner porosity of the carbon particles is

p = 0,570, density of nonporous adsorbent is C = 1820 kg m-3

. Determine amount of carbon

necessary for adsorption, volume of the acetic acid solution within the bed and concentration

of acetic acid in the liquid after reaching equilibrium. Initial concentration of acetic acid in

solution is 0,25 mol dm-3

.

Results: 594 kg of the adsorbent, volume of the solution 0,582 m3, concentration of acid

0,0022 mol dm-3

.

59

U7.6

The same final concentration of the acetic acid as in the problem U7.5 has to be reached by

the adsorption in two consecutive stages. Determine amount of the adsorbent necessary in

each stage if volume of the liquid and other conditions will remain the same as in problem

U7.5.

Výsledek: The first stage: 156 kg of adsorbent, the second stage 87 kg.

U7.7

Albumin in a solution is adsorbed in a stirred one-stage adsorber with ion-exchange resin.

Adsorption vessel is filled with 60000 kg of the solution containing 0,0012 mass % of the

albumin. 10 kg of dry clean ionex is used for the adsorption. the equilibrium is described by

the Langmuir isotherm 254000y

X1+30790 y

AA

A

, where XA is relative mass fraction of the albumin

within the adsorbent and yA is the mass fraction of the albumin in the solution. Determine

values of albumin concentrations in both the liquid phase and in the adsorbent. What fraction

of the albumin will be transferred from the solution to the adsorbent?

Results: Relative mass fraction in the adsorbent: 0,07 and mass fraction in the solution:

2,7810-7

. 97,6% of the albumin will be transferred to the ionex.