PHYSICAL - CHEMICAL TREATMENT OF WATER AND WASTEWATER - CHAPTER 8 ppsx

Advanced Filtration and
Carbon Adsorption

This chapter continues the discussion on filtration started in Chapter 7, except that
it deals with advanced filtration. We have defined

filtration

as a unit operation of
separating solids or particles from fluids. A unit operation of filtration carried out
using membranes as filter media is advanced filtration. This chapter discusses advanced
filtration using electrodialysis membranes and pressure membranes. Filtration using
pressure membranes include reverse osmosis, nanofiltration, microfiltration, and
ultrafiltration.
In addition to advanced filtration, this chapter also discusses carbon

adsorption

.
This is a unit operation that uses the active sites in powdered, granular, and fibrous
activated carbon to remove impurities from water and wastewater. Carbon adsorption
and filtration share some similar characteristics. For example, head loss calculations
and backwashing calculations are the same. Carbon adsorption will be discussed as
the last part of this chapter.

8.1 ELECTRODIALYSIS MEMBRANES

Figure 8.1a shows a cut section of an electrodialysis filtering membrane. The filtering
membranes are sheet-like barriers made out of high-capacity, highly cross-linked
ion exchange resins that allow passage of ions but not of water. Two types are used:

cation membranes

, which allow only cations to pass, and

anion membranes

, which
allow only anions to pass. The cut section in the figure is a cation membrane composed
of an

insoluble matrix

with

water

in the pore spaces.

Negative charges

are fixed
onto the insoluble matrix, and

mobile cations

reside in the pore spaces occupied by
water. It is the residence of these mobile cations that gives the membrane the property
of allowing cations to pass through it. These cations will go out of the structure if
they are replaced by other cations that enter the structure. If the entering cations

came from water external to the membrane, then, the cations are removed from the
water, thus filtering them out. In anion membranes, the mechanics just described
are reversed. The mobile ions in the pore spaces are the

anions

; the ions fixed to
the insoluble matrix are the

cations

. The entering and replacing ions are anions from
the water external to the membrane. In this case, the anions are filtered out from the
water.
Figure 8.1b portrays the process of filtering out the ions in solution. Inside the
tank, cation and anion membranes are installed alternate to each other. Two electrodes
are put on each side of the tank. By impressing electricity on these electrodes, the
positive anode attracts negative ions and the negative cathode attracts positive ions.
This impression of electricity is the reason why the respective ions replace their like
8

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374

ions in the membranes. As shown in the figure, two compartments become “cleaned”
of ions and one compartment (the middle) becomes “dirty” of ions. The two com-
partments are diluting compartments; the middle compartment is a concentrating
compartment. The water in the diluting compartments is withdrawn as the product

water, and is the filtered water. The concentrated solution in the concentrating
compartment is discharged to waste.

8.1.1 P

OWER

R

EQUIREMENT



OF

E

LECTRODIALYSIS

U

NITS

The filtering membranes in Figure 8.1b are arranged as

CACA

from left to right,
where

C

stands for cation and

A

stands for anion. In compartments

CA

, the water is
deionized, while in compartment

AC

, the water is not deionized. The number of
deionizing compartments is equal to two. Also, note that the membranes are always
arranged in pairs (i.e., cation membrane

C

is always paired with anion membrane

A

).
Thus, the number of membranes in a unit is always even. If the number of membranes
is increased from four to six, the number of deionizing compartments will increase
from two to three; if increased from six to eight, the number of deionizing membranes
will increase from three to four; and so on. Thus, if

m

is the number of membranes
in a unit, the number of deionizing compartments is equal to

m

/

2.
As shown in the figure, a deionizing compartment pairs with a concentrating
compartment in both directions; this pairing forms a

cell

. For example, deionizing
compartment

CA

pairs with concentrating compartment

AC

in the left direction and
with the concentrating compartment

AC

in the right direction of

CA

. In this paring
(in both directions), however, only one cell is formed equal to the one deionizing
compartment. Thus, the number of cells formed in an electrodialysis unit can be
determined by counting only the number of deionizing compartments. The number

FIGURE 8.1

Cation filtering membrane (a); the electrodialysis process (b).
CCAA

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375

of deionizing compartments in a unit is m

/

2, so the number of cells in a unit is also
equal to m

/

2.
Because one equivalent of a substance is equal to one equivalent of electricity,

in electrodialysis calculations, concentrations are conveniently expressed in terms
of equivalents per unit volume. Let the flow to the electrodialysis unit be

Q

o

. The
flow per deionizing compartment or cell is then equal to

Q

o

/

(m

/

2). If the influent
ion concentration (positive

or

negative) is [

C

o

] equivalents per unit volume, the total
rate of inflow of ions is [

C

o

]

Q

o

/

(m

/

2) equivalents per unit time per cell. One
equivalent is also equal to one Faraday. Because a Faraday or equivalent is equal to
96,494 coulombs, assuming a coulomb efficiency of

η

, the amount of electricity
needed to remove the ions in one cell is equal to 96,494[

C

o

]

Q

o

η

/

(m

/

2) coulombs
per unit time.

Coulomb efficiency

is the fraction of the input number of equivalents
of an ionized substance that is actually acted upon by an input of electricity.
If time is expressed in seconds, coulomb per second is amperes. Therefore, for
time in seconds, 96,494[

C

o

]

Q

o

η

/

(m

/

2) amperes of current must be impressed upon
the membranes of the cell to effect the removal of the ions. The cells are connected
in series, so the same current must pass through all of the cells in the electrodialysis
unit, and the same 96,494[

C

o

]

Q

o

η

/

(m

/

2) amperes of current would be responsible
for removing the ions in the whole unit. To repeat, not only is the amperage impressed
in one cell but in all of the cells in the unit.
In electrodialysis calculations, a term called

current density

(CD) is often used.
Current density is the current in milliamperes that flows through a square centimeter
of membrane perpendicular to the current direction: CD

=

mA

/

A

cm

, where mA is

the milliamperes of electricity and A

cm

is the square centimeters of perpendicular
area. A ratio called

current density to normality

(

CD

/

N

) is also used, where

N

is
the normality. A high value of this ratio means that there is insufficient charge to
carry the current away. When this occurs, a localized deficiency of ions on the
membrane surfaces may occur. This occurrence is called

polarization

. In commercial
electrodialysis units

CD

/

N

of up to 1,000 are utilized.
The electric current

I

that is impressed at the electrodes is not necessarily the
same current that passes through the cells or deionizing compartments. The actual
current that successfully passes through is a function of the current efficiency which
varies with the nature of the electrolyte, its concentration in solution, and the
membrane system. Call

M

the current efficiency. The amperes passing through the
solution is equal to the amperes required to remove the ions. Thus,
(8.1)
The emf

E

across the electrodes is given by Ohm’s law as shown below, where

R

is the resistance across the unit.
(8.2)
If

I

is in amperes and

R

is in ohms, then

E

is in volts.
IM 96,494 C
o
[]Q
o
η
/ m/2()=
I
96,494 C
o
[]Q
o
η
/ m/2()
M

=
EIR=

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From basic electricity, the power

P

is

EI



=



I

2

R

. Thus,
(8.3)
If

I

is in amperes,

E

is in volts, and

R

is in ohms,

P

in is in watts. Of course,
the combined units of

N

and

Q

o

must be in corresponding consistent units.

Example 8.1

A brackish water of 378.51 m

3

/day containing 4,000 mg/L of
ions expressed as Nacl is to be de-ionized using an electrodialysis unit. There are
400 membranes in the unit each measuring 45.72 cm by 50.8 cm. Resistance across
the unit is 6 ohms and the current efficiency is 90%.

CD

/

N

to avoid polarization is
700. Estimate the impressed current and voltage, the coulomb efficiency, and the
power requirement.

Solution:

Therefore,

8.2 PRESSURE MEMBRANES

Pressure membranes are membranes that are used to separate materials from a fluid
by the application of high pressure on the membrane. Thus, pressure membrane
filtration is a high pressure filtration. This contrasts with electrodialysis membranes
in which the separation is effected by the impression of electricity across electrodes.
Filtration is carried out by impressing electricity, therefore, electrodialysis membrane

filtration may be called

electrical filtration.
According to Jacangelo (1989), three allied pressure-membrane processes are
used: ultrafiltration (UF), nanofiltration (NF), and reverse osmosis (RO). He states
that UF removes particles ranging in sizes from 0.001 to 10
µ
m, while RO can
remove particles ranging in sizes from 0.0001 to 0.001
µ
m. As far as size removals
are concerned, NF stays between UF and RO, being able to remove particles in
the size range of the order of 0.001
µ
m. UF is normally operated in the range of
100 to 500 kPag (kilopascal gage); NF, in the range of 500 to 1,400 kPag; and RO,
in the range of 1,400 to 8,300 kPag. Microfiltration (MF) may added to this list.
MF retains larger particles than UF and operates at a lesser pressure (70 kPag).
PEII
2
R 3.72 10
10
()
C
o
[]Q
o
η
mM



2
R== =
C
o
[] Nacl[]
4.0
NaCl

4
23+35.45

0.068
geq
L

68
geq
m
3

;=== = = N 0.068=
CD 700 0.068()47.6 mA/cm
2
==
I
47.6 45.72()50.8()
1000 0.9()

122.84 A Ans==
EIR122.84 6() 737 V Ans== =
I
96,494 C
o
[]Q
o
η
/ m/2()
M

96,494 68()
378.51
24 60()60()

()
η
()/400/2()
0.90

==
η
0.77 Ans=
PEII
2
R 122.84
2
6() 90,538 W Ans== = =
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Whereas the nature of membrane retention of particles in UF is molecular screening,
the nature of membrane retention in MF is that of molecular-aggregate screening.
On the other hand, comparing RO and UF, RO presents a diffusive transport barrier.
Diffusive transport refers to the diffusion of solute across the membrane. Due to the
nature of its membrane, RO creates a barrier to this diffusion. Figures 8.2 through
8.4 present example installations of reverse osmosis units.
FIGURE 8.2 Bank of modules at the Sanibel–Captiva reverse osmosis plant, Florida.
FIGURE 8.3 Installation modules of various reverse osmosis units. (Courtesy of Specific
Equipment Company, Houston, TX.)
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The basics of a normal osmosis process are shown in Figure 8.5a. A bag of
semipermeable membrane is shown placed inside a bigger container full of pure
water. Inside the membrane bag is a solution of sucrose. Because sucrose has osmotic
pressure, it “sucks” water from outside the bag causing the water to pass through
the membrane. Introduction of the water into the membrane bag, in turn, causes the
solution level to rise as indicated by the height
π
in the figure. The height
π
is a
measure of the osmotic pressure. It follows that if sufficient pressure is applied to
the tip of the tube in excess of that of the osmotic pressure, the height
π
will be
suppressed and the flow of water through the membrane will be reversed (i.e., it
would be from inside the bag toward the outside into the bigger container); thus,
the term “reverse osmosis.”
Sucrose in a concentration of 1,000 mg/L has an osmotic pressure of 7.24 kNa
(kiloNewtons absolute). Thus, the reverse pressure to be applied must be, theoreti-

cally, in excess of 7.24 kNa for a sucrose concentration of 1,000 mg/L. For NaCl,
its osmotic pressure in a concentration of 35,000 mg/L is 2744.07 kNa. Hence, to
reverse the flow in a NaCl concentration of 35,000 mg/L, a reverse pressure in excess
of 2744.07 kNa should be applied. The operation just described (i.e., applying
sufficient pressure to the tip of the tube to reverse the flow of water) is the funda-
mental description of the basic reverse osmosis process.
FIGURE 8.4 Reverse osmosis module designs.
Perforated
PVC baffle
Spiral wound
Permeate side
backing material
with membrane in
each side and glued around
edges and to center tube
Permeate out
Permeate flow (after
passage through
membrane)
Feed flow
Feed side
spacer
Roll to
assemble
Permeate
Feed flow
Membrane
Porous tube
Large tube
Hollow fine fibers

Product
water
Hollow fibers
Feed
water
Plate and frame
Grooved phenolic
support plate
Cellulose
acetate
membrane
Paper
substrate
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UF, NF, MF, and RO and are all reverse osmosis filtration processes; however,
when the term reverse osmosis or RO is used without qualification, it is the process
operated at the highest pressure range to which it is normally referred. Figure 8.5b is
a schematic of an RO plant. Figure 8.2 is a photograph of a bank of modules in the
Sanibel–Captiva RO Plant in Florida. This plant treats water for drinking purposes.
Take careful note of the pretreatment requirement indicated in Figure 8.5b. As
mentioned before, the RO process is an advanced mode of filtration and its purpose
is to remove the very minute particles of molecules, ions, and dissolved solids. The
influent to a RO plant is already “clean,” only that it contains the ions, molecules,
and molecular aggregates that need to be removed.
FIGURE 8.5 (a) Osmosis process; (b) reverse osmosis system.
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After pretreatment the high-pressure pump forces the flow into the membrane
module where the solutes are rejected. The flow splits into two, one producing the

product water and the other producing the waste discharge. The waste discharge has
one drawback in the use of RO filtration in that it may need to be treated separately
before discharge.
8.2.1 MEMBRANE MODULE DESIGNS
Over the course of development of the membrane technology, RO module designs,
as shown in Figure 8.4, evolved. They are tubular, plate-and-frame, spiral wound,
and hollow fine-fiber modules. In the tubular design, the membrane is lined
inside the tube which is made of ordinary tubular material. Water is allowed to pass
through the inside of the tube under excess pressure causing the water to permeate
through the membrane and to collect at the outside of the tube as the product or
permeate. The portion of the influent that did not permeate becomes concentrated.
This is called the concentrate or the reject.
The plate-and-frame design is similar to the plate-and-frame filter press discussed
in the previous chapter on conventional filtration. In the case of RO, the semipermeable
membrane replaces the filter cloth. The spiral-wound design consists of two flat sheets
of membranes separated by porous spacers. The two sheets are sealed on three sides;
the fourth side is attached to a central collector pipe; and the whole sealed sheets are
rolled around the central collector pipe. As the sheets are rolled around the pipe, a
second spacer, called influent spacer, is provided between the sealed sheets. In the
final configuration, the spiral-wound sealed membrane looks like a cylinder. Water is
introduced into the influent spacer, thereby allowing it to permeate through the mem-
brane into the spacer between the sealed membrane. The permeate, now inside the
sealed membrane, flows toward the central pipe and exits through the fourth unsealed
side into the pipe. The permeate is collected as the product water. The concentrate or
the reject continues to flow along the influent spacer and is discharged as the effluent
reject or effluent concentrate. This concentrate, which may contain hazardous mole-
cules, poses a problem for disposal.
In the hollow fine-fiber design, the hollow fibers are a bundle of thousands of
parallel, self-supporting, hair-like fibers enclosed in a fiberglass or epoxy-coated
steel vessel. Water is introduced into the hollow bores of the fibers under pressure.

The permeate water exits through one or more module ports. The concentrate also
exits in a separate one or more module ports, depending on the design. All these
module designs may be combined into banks of modules and may be connected in
parallel or in series.
8.2.2 FACTORS AFFECTING SOLUTE REJECTION
AND
BREAKTHROUGH
The reason why the product or the permeate contains solute (that ought to be
removed) is that the solute has broken through the membrane surface along with
the product water. It may be said that as long as the solute stays away from the membrane
surface, only water will pass through into the product side and the permeate will
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be solute-free; However, it is not possible to exclude the solute from contacting the
membrane surface; hence, it is always liable to break through. The efficiency at which
solute is rejected is therefore a function of the interaction of the solute and the
membrane surface. As far as solute rejection and breakthrough are concerned, a
review of literature revealed the following conclusions (Sincero, 1989):
• Percentage removal is a function of functional groups present in the
membrane.
• Percentage removal is a function of the nature of the membrane surface.
For example, solute and membrane may have the tendency to bond by
hydrogen bonding. Thus, the solute would easily permeate to the product
side if the nature of the surface is such that it contains large amounts of
hydrogen bonding sites.
• In a homologous series of compounds, percentage removal increases with
molecular weight of solute.
• Percentage removal is a function of the size of the solute molecule.
• Percentage removal increases as the percent dissociation of the solute
molecule increases. The degree of dissociation of a molecule is a function

of pH, so percentage removal is also a function of pH.
This review also found that the percentage removal of a solute is affected by
the presence of other solutes. For example, methyl formate experienced a drastic
change in percentage removal when mixed with ethyl formate, methyl propionate,
and ethyl propionate. When alone, it was removed by only 14% but when mixed
with the others, the removal increased to 66%. Therefore, design of RO processes
should be done by obtaining design criteria utilizing laboratory or pilot plant testing
on the given influent.
8.2.3 SOLUTE–WATER SEPARATION THEORY
The sole purpose of using the membrane is to separate the solute from the water
molecules. Whereas MF, UF, and NF may be viewed as similar to conventional filtration,
only done in high-pressure modes, the RO process is thought to proceed in a somewhat
different way. In addition to operating similar to conventional filtration, some other
mechanisms operate during the process. Several theories have been advanced as to how
the separation in RO is effected. Of these theories, the one suggested by Sourirajan with
schematics shown in Figures 8.6a and 8.6b is the most plausible.
Sourirajan’s theory is called the preferential-sorption, capillary-flow theory. This
theory asserts that there is a competition between the solute and the water molecules
for the surface of the membrane. Because the membrane is an organic substance,
several hydrogen bonding sites exist on its surface which preferentially bond water
molecules to them. (The hydrogen end of water molecules bonds by hydrogen
bonding to other molecules.) As shown in Figure 8.6a, H
2
O molecules are shown
layering over the membrane surface (preferential sorption), to the exclusion of the
solute ions of Na
+
and Cl
−
. Thus, this exclusion brings about an initial separation.

In Figure 8.6b, a pore through the membrane is postulated, accommodating two
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diameters of water molecules. This pore size designated as 2t, where t is the diameter
of the water molecule, is called the critical pore diameter. With this configuration,
the final separation of the water molecules and the solutes materializes by applying
pressure, pushing H
2
O through the pores (capillary flow).
As the process progresses, solutes build and line up near the membrane surface
creating a concentration boundary layer. This layer concentration is much larger
than in the bulk solution and, also, much larger, of course, than the concentration
in the permeate side. This concentration difference creates a pressure for diffusive
transport. The membrane, however, creates a barrier to this diffusion, thus, retaining
the solute and not allowing it to pass through easily. Eventually, however, the solute
will diffuse out and leak to the permeate side.
8.2.4 TYPES OF MEMBRANES
The first RO membrane put to practical use was the cellulose acetate membrane (CA
membrane). The technique of preparation was developed by Sourirajan and Loeb
and consisted of casting step, evaporation step, gelation step, and shrinkage step.
The casting step involves casting a solution of cellulose acetate in acetone containing
an additive into flat or tubular surfaces. The additive (such as magnesium perchlorate)
must be soluble in water so that it will easily leach out in the gelation step creating
a porous film. After casting, the solvent acetone is evaporated. The material is then
subjected to the gelation step where it is immersed in cold water. The film material
sets to a gel and the additive leaches out. Finally, the film is subjected to the shrinkage
step that determines the size of the pores, depending upon the temperature used in
shrinking. High temperatures create smaller pores.
FIGURE 8.6 (a) schematic representation of preferential sorption-capillary flow theory;
(b) critical pore diameter for separation; (c) flux decline with time; (d) correction factor for

surface area of cellulose acetate; and (e) solute rejection as a function of operating time.
Bulk of the
solutions
Pore water
interface
Porous film surface
of appropriate
chemical nature
Film
Film
Demineralized water
Critical pore diameter
(a)
(b)
(c)
(d)
(e)
Flux
Time
Periodic cleaning
of membrane
Feedwater temperature (°C)
0102030
Correction factor,
C
f
1.6
1.4
1.2
1.0

0.8
0.6
1000 2000 3000 7000
100
95
90
85
Percent rejection
Operating time (h)
Divalent rejection
ions
Ca
2+
SO
2-
Na
+
H
2
0Na
+
Cl
-
H
2
0Na
+
Cl
-
H

2
0
H
2
0Na
+
Cl
-
H
2
0Na
+
Cl
-
H
2
0
H
2
0Na
+
Cl
-
H
2
0Na
+
Cl
-
H

2
0
H
2
0Na
+
Cl
-
H
2
0Na
+
Cl
-
H
2
0
H
2
0Na
+
Cl
-
H
2
0Na
+
Cl
-
H

2
0
H
2
0Na
+
Cl
-
H
2
0Na
+
Cl
-
H
2
0
H
2
0
H
2
0
H
2
0
Monovalent rejection
ions
Porous film surface
of appropriate

chemical nature
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After this first development of the CA membrane, different types of membranes
followed: CAB, CTA, PBIL, and PA membranes. CAB is membrane of cellulose acetate
butyrate; CTA is cellulose triacetate. The PBIL membrane is a polybenzimidazolone
polymer and PA are polyamide membranes. The structure of a PBIL unit is as follows:
Polyethylene amine reacted with tolylene diisocyanate produces the NS-100
membrane (NS stands for nonpolysaccharide). The reaction is carried out as follows:
In this reaction, the H bonded to the N of the n repeating units of polyethylene
amine [(−CH
2
CH
2
NH)
n
−] moves to the N of tolylene diisocyanate [−N=C=O] destroy-
ing the double bond between N and C. The C of the carboxyl group [=C=O] of tolylene
diisocyanate then bonds with the N of the amine. The reaction above simply shows two
of the tolylene molecules participating in the reaction, but in reality, there will be
millions of them performing the reaction of H moving and the C of the carboxyl group
of the tolylene bonding with the N of the amine and so on. The final structure is a mesh
of cross-linked assembly, thus creating molecular pores.
As indicated in the NS-100 product, a closed loop structure is formed. The
ethylene repeating units [−CH
2
CH
2
−] form the backbone of the membrane, and the
benzene rings form the cross-linking mechanism that tie together the ethylene

backbones forming the closed loop. The ethylene units and the benzene rings are
nonpolar regions, while the peptide bonds and the amines [−NH
2
] are
polar regions. In the NS-100, nonpolar regions exceed the polar regions; hence, this
membrane is said to be apolar.
The CA membrane contains the OH
−
and the acetyl groups. The
OH
−
region exceeds the acetyl region in the membrane. OH
−
is polar, while the
acetyl group is nonpolar regions. Since the OH
−
region exceeds the acetyl region,
CA membranes are polar. The polarity or apolarity of any membrane is very impor-
tant in characterizing its property to reject solutes.
N
CO
NH
SO
2
N
C O
NH
O
Polybenzimidazolone unit
CH

3
CH
3
CH
3
NH
N–C
(CH
2
CH
2
N)
n
–CH
2
–CH
2
–N–C–NH
(CH
2
CH
2
N)
n
–CH
2
–CH
2
–N
C

=
O
(CH
2
CH
2
NH)
n
CH
2
CH
2
NH + 2
Polyethylene amine
Tolylene diisocyanate
N
=
C
=
O
N
=
C
=
O
O
O
C
=
O

NH
NS-100 structure
–C–NH–
O
CH
3
CO
2
−
[]
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Polyethylene amine reacted with isophthaloyl chloride produces the PA-100
membrane as shown in the following reaction:
The chloride atom in isophthaloyl chloride is attached to the carboxyl group. In
this reaction, the H in the amine group of polyethylene amine reacts with the chloride
in the carboxyl group producing HCl, as shown by the product over the arrow, and
the PA-100 membrane to the right of the reaction equation. As in the case of the
NS-100, the reaction forms the closed loop resulting in cross-linked structure of the
PA-membrane. Thus, a molecular pore is again produced.
Epiamine, a polyether amine, reacting with isophthaloyl chloride produces the
PA-300 membrane as shown in the following reaction:
Again, in this reaction, the H in the amine group of the amine, epiamine, reacts
with the chloride of isophthaloyl chloride forming HCl. The N, in turn, of the amine
group, from which the H that reacts with the C chloride were taken, bonds with the
C of the carboxyl group of isophthaloyl chloride producing the PA-300 membrane.
As shown in its structure, this membrane is also a cross-linked membrane.
meta-Phenylene diamine reacting with trimesoyl chloride produces the FilmTec
FT-30 membrane according to the following reaction:
C=O

(CH
2
CH
2
NH)
n
CH
2
CH
2
NH + 2
(CH
2
CH
2
N)
n
CH
2
CH
2
N – C
(CH
2
CH
2
N)
n
CH
2

CH
2
N –
C
O
O
Cl
–4HCI
Cl
Isophthaloyl
chloride
Polyethylene amine
C=O
O=C
C
O
PA–100 membrane
C = O
CH
3
CH – O – (CH
2
CH – O)
4
C
O
C
O
O
Cl

C
C– N–CH
2
–CH
2
–N –C
CH
3
CH–O– (CH
2
–O)
4
OO
Isophthaloyl
chloride
Epiamine (polyether amine)
CH
2
CH
2
NH –CH
2
–CH
2
–NH
O
+ 2
– 4HCI
CI
CH

2
O
CH
2
PA-300 membrane structure
C=O
CI
– 3HCI
C=O
C=O
CI
CI
NH
2
NH
2
3
+
meta-phenylene
Trimesoyl chloride
FilmTec FT-30 membrane
NH
H
N–C=O
O
C–N–
–NH
–NH
–C –N–
H

diamine
O
TX249_frame_C08.fm Page 384 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
2-Hydroxy-methyl furan when dehydrated using H
2
SO
4
produces the NS-200
according to the following reaction:
Another membrane formed from 2-hydroxy-methyl furan is Toray PEC-1000.
The PAs in the prefixes for the naming of the membranes stand for polyamide.
Thus, the membranes referred to are polyamide membranes. PA membranes contain
the amide group ; it is for this component that they are called polyamide
membranes. The formulas for the NS-100, NS-300, and the FT-30 membranes contain
the amide group, thus, they are polyamide membranes. The NS-200 is not a poly-
amide membrane.
In the early applications of the reverse osmosis technique, the membranes avail-
able were the polysaccharide membranes such as the CA membrane. As new mem-
branes were developed, they were differentiated from the saccharide membranes by
calling them nonpolysaccharide membranes. Thus, the NS-200 is a nonpolysaccharide
membrane.
8.2.5 MEMBRANE PERFORMANCE CHARACTERIZATION
The performance of a given membrane may be characterized according to its product
flux and purity of product. Flux, which is a rate of flow per unit area of membrane,
is a function of membrane thickness, chemical composition of feed, membrane poros-
ity, time of operation, pressure across membrane, and feedwater temperature. Product
purity, in turn, is a function of the rejection ability of the particular membrane.
Flux decline. Figure 8.6c shows the decline of the flux with time of operation.
This curve applies to a given membrane and membrane pressure differential. The

lower solid curve is the actual decline without the effect of cleaning. The saw-toothed
configuration is the effect of periodic cleaning. As shown, right after cleaning, the
flux rate shows an improvement, but then, it begins to decline again with time. From
experience, the general trend of the curve plots a straight line in a log–log paper.
Thus, empirically, the following equation fits the curve:
(8.4)
where F is the flux, t is the time, m is the slope of the line, and K is a constant.
The previous equation is a straight-line equation between lnt and lnF. The
equation is that of a straight line, so only two data points for lnt and lnF are
required to calculate the constants m and K. Using the techniques of analytic
geometry as applied to straight-line equations, the following equations are produced
( CH2 CH2
–
)
NS-200 membrane
H
2
SO
4
CH
2
OH
,– H
2
O
2-hydroxymethyl furan
OOO
NH
2
−

[]
Fln mlnt= ln K F⇒+ Kt
m
=
TX249_frame_C08.fm Page 385 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
from Equation (8.4).
(8.5)
(8.6)
As noted in the previous equations, the summation sign ∑ is used. Although
only two points are needed to determine the constants, the actual experimentation
may be conducted to produce several points. Since the equation only needs two
points, the data are then grouped into two groups. Thus, l is the number of experi-
mental data in the first group in a total of n experimental data points. The second
group would consist of n − l data points.
Equation (8.4) may be used to estimate the ultimate flux of a given membrane
at the end of its life (one to two years).
Example 8.2 A long term experiment for a CA membrane module operated
at 2757.89 kPag using a feed of 2,000 mg/L of NaCl at 25°C produces the results
below. What is the expected flux at the end of one year of operation? What is the
expected flux at the end of two years? How long does it take for the flux to decrease
to 0.37 m
3
/m
2
⋅ day?
Solution:
Time (h) 1 10,000 25,000
Flux (m
3

/m
2
⋅ day) 0.652 0.490 0.45
t (h) 1 10,000 25,000
F (m
3
/m
2
⋅ day) 0.652 0.49 0.45
lnt 0 9.21 10.13
lnF −0.43 −0.71 −0.799
m
∑ Fln
n
l+1
n−l

∑ Fln
l
1
l
–
∑ tln
n
l+1
n − l

∑ tln
l
1

l
–

l∑ Fnl–()∑Fln
l
1
–ln
n
l+1
l∑ tnl–()∑tln
l
1
–ln
n
l+1

==
K
∑ Fm∑ tln
l
1
–ln
l
1
l



exp=
m

l∑ Fnl–()∑Fln
l
1
–ln
n
l+1
l∑ tln
n
l+1
nl–()∑tln
l
1
–

=
K
∑ Fm∑ tln
l
1
()–ln
l
1
l



exp=
TX249_frame_C08.fm Page 386 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
Let l = 1

that is, if the membrane has not broken before this time! Ans
Flux through membrane as a function of pressure drop. The flow of perme-
ate through a membrane may be considered as a “microscopic” form of cake filtra-
tion, where the solute that polarizes at the feed side of the membrane may be
considered as the cake. In cake filtration (see Chapter 7), the volume of filtrate
that passes through the cake in time t can be solved from
(8.7)
where
µ
is the absolute viscosity of filtrate; c, the mass of cake per unit volume of
filtrate collected; , the specific cake resistance; −∆P, the pressure drop across the
cake and filter; S
o
, the filter area; and R
m
, the filter resistance. In RO, c is the solute
collected on the membrane (in the concentration boundary layer) per unit volume of
permeate; and R
m
, the resistance of the membrane. All the other parameters have similar
meanings as explained earlier in Chapter 7.
The volume flux F is Using this and solving the above equation for
(8.8)
Initially neglecting the resistance of the solute in the concentration boundary layer,
µ
c in the denominator of the first factor on the right-side of the equation may
be set to zero, producing
(8.9)
m
1 0.71– 0.799–()2 0.43–()–

1 9.21 10.13+()20()–

−1.51 0.86+
19.34

0.034–===
K
0.43– 0.34–()0()–
1



exp 0.65==
FKt
m
=
F
1year
0.65 365 24()[]
0.034–
0.48
m
3
m
2
⋅ day



Ans==

F
2year
0.65 365 2()24()[]
0.034–
0.47
m
3
m
2
⋅ day



Ans==
0.37 0.65= t
0.034–
t 16,002,176h 1,827 years==
V
t
V

µ
c
α
2 ∆P–()S
o
2

V=
µ

R
m
∆P–()S
o

+
α
V / S
o
.±
V / S
o
± F,=
V
tS
o

F
2
µ
c
α
V 2S
o
µ
R
m
+

∆P–()S

o
==
α
V
V
tS
o

F
1
µ
R
m

∆P–()==
TX249_frame_C08.fm Page 387 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
Now, considering the resistance of the solute, designate the combined effect of
compressibility, membrane resistance R
m
, and solute resistance as . Analogous
to cake filtration, call as specific membrane resistance. Hence,
(8.10)
(8.11)
where s is an index of membrane and boundary layer compressibility. When s is
equal to zero, is equal to , the constant of proportionality of the equation.
Calling the pressure in the feed side as P
f
, the net pressure P
fn

acting on the
membrane in the feed side is
(8.12)
where
π
f
is the osmotic pressure in the feed side. Also, calling the pressure in the
permeate side as P
p
, the net pressure P
pn
acting on the membrane in the permeate
side is
(8.13)
where
π
p
is the osmotic pressure in the permeate side. Thus,
(8.14)
and the flux F is
(8.15)
Table 8.1 shows osmotic pressure values of various solutes. Some generalizations
may be made from this table. For example, comparing the osmotic pressures of
1,000 mg/L of NaCl and 1,000 mg/L of Na
2
SO
4
, the former has about 1.8 times that
of the osmotic pressure of the latter. In solution for the same masses, NaCl yields
about 1.6 times more particles than Na

2
SO
4
. From this it may be concluded that
osmotic pressure is a function of the number of particles in solution. Comparing the
1,000 mg/L concentrations of Na
2
SO
4
and MgSO
4
, the osmotic pressure of the former
is about to 1.4 times that of the latter. In solution Na
2
SO
4
yields about 1.3 more
particles than MgSO
4
. The same conclusions will be drawn if other comparisons are
made; therefore, osmotic pressure depends on the number of particles in solution.
From this finding, osmotic pressure is, therefore, additive.
Determination of and s. The straight-line form of Equation (8.15) is
(8.16)
α
m
α
m
V
tS

o

F
1
µα
m

∆P–()==
α
m
α
mo
∆P–()
s
=
α
m
α
mo
P
fn
P
f
=
π
f
–
P
pn
P

p
=
π
p
–
∆P– P
fn
= P
pn
– P
f
π
f
–()P
p
π
p
–()– P
f
P
p
–()
π
f
π
p
–()–==
F
1
µα

m

P
f
P
p
–()
π
f
π
p
–()–[]=
1
µα
mo
∆ P–()
s

P
f
P
p
–()
π
f
π
p
–()–[]=
1
µα

mo

∆P–()
1−s
=
αα
αα
mo
µ
F()ln 1 s–()∆P–()
α
mo
ln–ln=
TX249_frame_C08.fm Page 388 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
This equation needs only two data points of ln(
µ
F) and ln(−∆P) to determine the
constants and s. Assuming there are a total of n experimental data points and
using the first l data points for the first equation, the last n − l data points for the
second equation, and using the techniques of analytic geometry, the following
equations are produced:
(8.17)
(8.18)
Example 8.3 The feedwater to an RO unit contains 3,000 mg/L of NaCl,
300 mg/L of CaCl
2
, and 400 mg/L of MgSO
4
. The membrane used is cellulose

acetate and the results of a certain study are shown below. What will the flux be
if the pressure applied is increased to 4826.31 kPag? Assume that for the given
concentrations the osmotic pressures are NaCl = 235.80 kPa, CaCl
2
= 17.17 kPa,
and MgSO
4
= 9.93 kPa. Also, assume the temperature during the experiment is
25°C.
TABLE 8.1
Osmotic Pressures at 25°°
°°
C
Compound
Concentration
(mg/L)
Osmotic
Pressure (kPa)
NaCl 35,000 2758
NaCl 1,000 76
NaHCO
3
1,000 90
Na
2
SO
4
1,000 41
MgSO
4

1,000 28
MgCl
2
1,000 69
CaCl
2
1,000 55
Sucrose 1,000 7
Dextrose 1,000 14
Applied Pressure (kPag) Flux (m
3
/m
2
⋅⋅
⋅⋅
day)
1723.68 0.123
4136.84 0.187
α
mo
s 1
l∑
µ
F()nl–()∑
µ
F()ln
l
1
–ln
n

l+1
l∑∆P–()ln
n
l+1
nl–()∑ ∆P–()ln
l
1
–
–=
α
mo
1 s–()∑ ∆P–() ln
l
1
∑
µ
F()ln
l
1
–
l



exp=
TX249_frame_C08.fm Page 389 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
Solution:
Therefore,
Effect of temperature on permeation rate. As shown in Equation (8.15), the

flux is a function of the dynamic viscosity
µ
. Because
µ
is a function of temperature,
the flux or permeation rate is therefore also a function of temperature. As temper-
ature increases, the viscosity of water decreases. Thus, from the equation, the flux
P
1
(kPag) −∆P (N/m
2
) ln (−∆P) Flux (m
3
/m
2
⋅⋅
⋅⋅
day)
µµ
µµ


F ln(
µµ
µµ


F)
1723.68 1,460,780
a

14.19 0.123 9.56 2.26
4136.84 3,873,940 15.17 0.187 14.54 2.68
a
(P
f
− P
p
) − (
π
f
−
π
p
) = − ∆ P = [(1,723,680 +101,325) − 101,325]
F
1
µα
mo
∆P–()
s

P
f
P
p
–()
π
f
π
p

–()–[]
1
µα
mo

∆P–()
1−s
==
s 1
l∑
µ
F()nl–()∑
µ
F()ln
l
1
–ln
n
l+1
l∑∆P–()ln
n
l+1
nl–()∑ ∆P–()ln
l
1
–
–=
α
mo
1 s–()∑ ∆P–()ln

l
1
∑
µ
F()ln
l
1
–
l



exp=
µ
9.0 10
4–
()kg= /m s⋅ 77.76 = kg/m d⋅
π
f
235.80 17.17 9.93++ 262.9 ==kN/m
2
262,900= N/m
2
P
p
101,325 N/m
2
= ;
π
p

0, assumed.=
262,900 0–[]–1= ,460,780 N/m
2
s 1
2.68 2.26–
15.17 14.19–
– 1 0.43– 0.57===
α
mo
1 0.57–()14.19 2.26–
1



exp 46.6==
F
1
µα
mo

∆P–()
1 s–
1
µα
mo

∆P–()
1−s
==
∆P– at 4826.31 kPag P

f
P
p
–()
π
f
π
p
–()–[]=
4,826,310 101,325+()101,325–[]262,900 0–[]–= 4,563,410 N/m
2
=
F
1
77.76 46.6()

4,563,410()
1 0.57–
0.20
m
3
m
2
⋅ day

Ans==
TX249_frame_C08.fm Page 390 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
is expected to increase with increase in temperature. Correspondingly, it is also expected
that the flux would decrease as the temperature decreases. Figure 8.6d shows the

correction factor C
f
for membrane surface area (for CA membranes) as a function
of temperature relative to 25°C. As shown, lower temperatures have larger correction
factors. This is due to the increase of
µ
as the temperature decreases. The opposite
is true for the higher temperatures. These correction factors are applied to the
membrane surface area to produce the same flux relative to 25°C.
Percent solute rejection or removal. The other parameter important in the
design and operation of RO units is the percent rejection or removal of solutes. Let
Q
o
be the feed inflow, [C
o
] be the feed concentration of solutes, Q
p
be the permeate
outflow, [C
p
] be the permeate concentration of solutes, Q
c
be the concentrate outflow,
and [C
c
] be the concentrate concentration of solutes. By mass balance of solutes,
the percent rejection R is
(8.19)
The index i refers to the solute species i.
Figure 8.6e shows the effect of operating time on percent rejection. As shown,

this particular membrane rejects divalent ions better than it does the monovalent
ions. Generally, percent rejection increases with the value of the ionic charge.
Example 8.4 A laboratory RO unit 152.4 cm in length and 30.48 cm in
diameter has an active surface area of 102.18 m
2
. It is used to treat a feedwater with
the following composition: NaCl = 3,000 mg/L, CaCl
2
= 300 mg/L, and MgSO
4
=
400 mg/L. The product flow is 0.61 m
3
/m
2
⋅ day and contains 90 mg/L NaCl,
6 mg/L CaCl
2
, and 8 mg/L MgSO
4
. The feedwater inflow is 104.9 m
3
/day. (a) What
is the percent rejection of NaCl? (b) What is the over-all percent rejection of ions?
Solution:
8.3 CARBON ADSORPTION
Solids are formed because of the attraction of the component atoms within the solid
toward each other. In the interior of a solid, attractive forces are balanced among the
various atoms making up the lattice. At the surface, however, the atoms are subjected
to unbalanced forces—the ones toward the interior are attracted, but the ones at the

R
∑Q
o
C
oi
[]∑Q
p
C
pi
[]–
∑Q
o
C
oi
[]

100()
∑Q
c
C
ci
[]
∑Q
o
C
oi
[]

100()==
a()R

∑Q
o
C
oi
[]∑Q
p
C
pi
[]–
∑Q
o
C
oi
[]

100()
104.9 3,000()0.61 102.18()90()–
104.9 3,000()

100()==
98.2%= Ans
b()R
104.9 3,000 300 400++()0.61 102.18()9068++()–
104.9 3,000 300 400++()

100()=
388,130 6,482.30–
388,130
= 100()98.3%= Ans
TX249_frame_C08.fm Page 391 Friday, June 14, 2002 4:35 PM

© 2003 by A. P. Sincero and G. A. Sincero
surface are not. Because of this unbalanced nature, any particle that lands on the surface
may be attracted by the solid. This is the phenomenon of adsorption, which is the
process of concentrating solute at the surface of a solid by virtue of this attraction.
Adsorption may be physical or chemical. Physical adsorption is also called van
der Waals adsorption, and chemical adsorption is also called chemisorption. In the
former, the attraction on the surface is weak, being brought about by weak van der
Waals forces. In the latter, the attraction is stronger as a result of some chemical
bonding that occurs. Adsorption is a surface-active phenomenon which means larger
surface areas exposed to the solutes result in higher adsorption. The solute is called
the adsorbate; the solid that adsorbs the solute is called the adsorbent. The adsorbate
is said to be sorbed onto the adsorbent when it is adsorbed, and it is said to be
desorbed when it passes into solution.
Adsorption capacity is enhanced by activating the surfaces. In the process using
steam, activation is accomplished by subjecting a prepared char of carbon material such
as coal to an oxidizing steam at high temperatures resulting in the water gas reaction:
C + H
2
O → H
2
+ CO. The gases released leave behind in the char a very porous structure.
The high porosity that results from activation increases the area for adsorption.
One gram of char can produce about 1000 m
2
of adsorption area. After activation,
the char is further processed into three types of finished product: powdered form
called powdered activated carbon (PAC), the granular form called granular activated
carbon (GAC), and activated carbon fiber (ACF). PAC is normally less than 200
mesh; GAC is normally greater than 0.1 mm in diameter. ACF is a fibrous form of
activated carbon. Figure 8.7 shows a schematic of the transformation of raw carbon

to activated carbon, indicating the increase in surface area.
8.3.1 ACTIVATION TECHNIQUES
Activation is the process of enhancing a particular characteristic. Carbon whose
adsorption characteristic is enhanced is called activated carbon. The activation
techniques used in the manufacture of activated carbons are dependent on the nature
and type of raw material available. The activation techniques that are principally
used by commercial production operations are chemical activation and steam acti-
vation. As the name suggests, chemical activation uses chemicals in the process and
FIGURE 8.7 Raw carbon material on the left transforms to the carbon on the right after
activation.
TX249_frame_C08.fm Page 392 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
is generally used for the activation of peat- and wood-based raw materials. The raw
material is impregnated with a strong dehydrating agent, typically phosphoric pen-
toxide (P
2
O
5
) or zinc chloride (ZnCl
2
) mixed into a paste and then heated to tem-
peratures of 500–800°C to activate the carbon. The resultant activated carbon is
washed, dried, and ground to desired size. Activated carbons produced by chemical
activation generally exhibit a very open pore structure, ideal for the adsorption of
large molecules.
Steam activation is generally used for the activation of coal and coconut shell
raw materials. Activation is carried out at temperatures of 800–1100°C in the pres-
ence of superheated steam. Gasification occurs as a result of the water–gas reaction:
(8.20)
This reaction is endothermic but the temperature is maintained by partial burning

of the CO and H
2
produced:
(8.21)
(8.22)
The activated carbon produced is graded, screened, and de-dusted. Activated
carbons produced by steam activation generally exhibit a fine pore structure, ideal
for the adsorption of compounds from both the liquid and vapor phases.
8.3.2 ADSORPTION CAPACITY
The adsorption capacity of activated carbon may be determined by the use of an
adsorption isotherm. The adsorption isotherm is an equation relating the amount of
solute adsorbed onto the solid and the equilibrium concentration of the solute in solution
at a given temperature. The following are isotherms that have been developed:
Freundlich; Langmuir; and Brunauer, Emmet, and Teller (BET). The most commonly
used isotherm for the application of activated carbon in water and wastewater
treatment are the Freundlich and Langmuir isotherms. The Freundlich isotherm is
an empirical equation; the Langmuir isotherm has a rational basis as will be shown
below. The respective isotherms are:
(8.23)
(8.24)
X is the mass of adsorbate adsorbed onto the mass of adsorbent M; [C] is the
concentration of adsorbate in solution in equilibrium with the adsorbate adsorbed;
n, k, a, and b are constants.
CH
2
OH
2
CO+→ 175,440++kJ/kgmol
2CO O
2

2CO
2
→ 393–+ ,790 kJ/kgmol
2H
2
O
2
2H
2
O→ 396,650 –+ kJ/kgmol
X
M

kC[]
1/n
=
X
M

ab C[]
1 bC[]+

=
TX249_frame_C08.fm Page 393 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
The Langmuir equation may be derived as follows. Imagine a particular exper-
iment in which a quantity of carbon adsorbent is added to a beaker of sample
containing pollutant. Immediately, the solute will be sorbed onto the adsorbent until
equilibrium is reached. One factor determining the amount of the sorbed materials
has to be the number of adsorption sites in the carbon. The number of these sites

may be quantified by the ratio X/M. By the nature of equilibrium processes, some
of the solutes adsorbed will be desorbed back into solution. While these solutes are
desorbing, some solutes will also be, again, adsorbed. This process continues on,
like a seesaw; this “seesaw behavior” is a characteristic of systems in equilibrium.
The rate of adsorption r
s
is proportional to the concentration in solution, [C],
(at equilibrium in this case) and the amount of adsorption sites left vacant by the
desorbing solutes. Now, let us determine these vacant adsorption sites. On a given
trial of the experiment, the number of adsorption sites filled by the solute may be
quantified by the ratio X/M, as mentioned previously. The greater the concentration
of the solute in solution, the greater this ratio will be. For a given type of solute and
type of carbon adsorbent, there will be a characteristic one maximum value for this
ratio. Call this (X/M)
ult
. Now, we have two ratios: X/M, which is the ratio at any
time and (X/M)
ult
, which is the greatest possible ratio. The difference of these two
ratios is proportional to the number of adsorption sites left vacant; consequently, the
rate of adsorption r
s
is therefore equal to k
s
[C][(X/M)
ult
− (X/M)], where k
s
is a
proportionality constant.

For the desorption process, as the ratio (X/M) forms on the adsorbent, it must
become a driving force for desorption. Thus, letting k
d
be the desorption proportion-
ality constant, r
d
= k
d
(X/M), where r
d
is the rate of desorption. The process is in
equilibrium, so the rate of adsorption is equal to the rate of desorption. Therefore,
(8.25)
Solving for X/M produces Equation (8.24), where a = (X/M)
ult
and b = k
s
/k
d
.
Note that in the derivation no mention is made of how many layers of molecules
are sorbed onto or desorbed from the activated carbon. It is simply that solutes are
sorbed and desorbed, irrespective of the counts of the layers of molecules; however,
it is conceivable that as the molecules are deposited and removed, the process occurs
layer by layer.
The straight-line forms of the Freundlich and Langmuir isotherms are, respectively,
(8.26)
(8.27)
Because the equations are for straight lines, only two pairs of values of the
respective parameters are required to solve the constants. For the Freundlich iso-

therm, the required pairs of values are the parameters ln(X/M) and ln[C]; for the
Langmuir isotherm, the required pairs are the parameters [C]/(X/M) and [C].
k
s
C[]
X
M



ult
X
M



– k
d
X
M



=
X
M

ln k
1
n

C[] straight-line formln+ln=
C[]
X/M

1
ab

1
a

C[] straight-line form+=
TX249_frame_C08.fm Page 394 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
To use the isotherms, constants are empirically determined by running an
experiment. This is done by adding increasing amounts of the adsorbent to a sample
of adsorbate solution in a container. For each amount of adsorbent added, M
i
, the
equilibrium concentration [C
i
] is determined. The pairs of experiment trial values
can then be used to obtain the desired parameter values from which the constants
are determined. Once the constants are determined, the resulting model is used to
determine M, the amount of adsorbent (activated carbon) that is needed. From the
derivation, the adsorption capacity of activated carbon is a = (X/M)
ult
. From this
ratio, the absorption capacity of activated carbon is shown as the maximum value
of the X/M ratios. This ratio corresponds to a concentration equal to the maximum

possible solute equilibrium concentration.
The value of X is obtained as follows: Let [C
o
] be the concentration of solute
in a sample of volume before adsorption onto a mass of adsorbent M. Then
(8.28)
8.3.3 DETERMINATION OF THE FREUNDLICH CONSTANTS
Using the techniques of analytic geometry, let us derive the Freundlich constants in a
little more detail than used in the derivation of the constants in the discussion of reverse
osmosis treated previously. As mentioned, the straight-line form of the equation requires
only two experimental data points; however, experiments are normally conducted to
produce not just two pair of values but more. Thus, the experimental results must be
reduced to just the two pairs of values required for the determination of the parameters;
therefore, assuming there are m pairs of values, these m pairs must be reduced to just
two pairs. Once the reduction to two pairs has been done, the isotherm equation may
be then be written to just the two pairs of derived values as follows:
(8.29)
(8.30)
The index l is the number of data points for the first group. These equations
may then be solved for n and k producing
(8.31)
(8.32)
V
XC
o
[] C[]–()V=
X
M

ln

1
l
∑
k
1
n

C[]ln
1
l
∑
+ln
1
l
∑
lk
1
n

C[] for the first pairln
1
l
∑
+ln==
X
M

ln
l+1
m

∑
k
1
n

+ln C[]ln
l+1
m
∑
ml–()k
1
n

C[] for the second pairln
l+1
m
∑
+ln=
l+1
m
∑
=
n
l∑ C[]ln
m
l+1
ml–()∑
m
l
C[]ln–

l ∑
l+1
m
ln
X
M

ml–()∑
1
l
X
M

ln–()

=
k
∑
X
M

1
n

∑ C[]ln
l
1
–ln
l
1

l




exp=
TX249_frame_C08.fm Page 395 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
8.3.4 DETERMINATION OF THE LANGMUIR CONSTANTS
As was done with the Freundlich constants, the Langmuir equation may be manipulated
in order to solve the Langmuir constants. From the m pairs of experimental data
(8.33)
(8.34)
Solving for the constants a and b produces
(8.35)
(8.36)
Example 8.5 A wastewater containing [C
o
] = 25 mg/L of phenol is to be
treated using PAC to produce an effluent concentration [C ]
eff
= of 0.10 mg/L. The
PAC is simply added to the stream and the mixture subsequently settled in the
following sedimentation tank. The constants of the Langmuir equation are deter-
mined by running a jar test producing the results below. The volume of waste
subjected to each test is one liter. If a flow rate of Q
o
of 0.11 m
3
/s is to be treated,

calculate the quantity of PAC needed for the operation. What is the adsorption
capacity of the PAC? Calculate the quantity of PAC needed to treat the influent
phenol to the ultimate residual concentration.
Solution:
Test PAC Added M (g) [C] (mg/L)
1 0.25 6.0
2 0.32 1.0
3 0.5 0.25
4 1.0 0.09
5 1.5 0.06
6 2.0 0.06
7 2.6 0.06
C[]
X/M

l
1
ab



1
a

C[]
1
l
∑
+=
1

l
∑
C[]
X/M

l 1+
m
∑
ml–()
1
ab



1
a

C[]
l+1
m
∑
+
a
l∑ C[] ml–()∑C[]
l
1
–
m
l+1
l∑

l+1
m
C[]
X /M

ml–()∑
1
l
C[]
X /M

–




=
b
l
a∑
1
l
C[]
X /M

∑ C[]
l
1
–

=
a
l∑ C[] ml–()∑C[]
l
1
–
m
l+1
l∑
l+1
m
C[]
X /M

ml–()∑
1
l
C[]
X /M

–




=
b
l
a∑
1

l
C[]
X /M

∑
1
l
C[]–

=
TX249_frame_C08.fm Page 396 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero
Neglect tests 6 and 7, because the additional values of 0.06’s would not conform to
the Langmuir equation.
Let l = 3; m = 5
Total phenol to be removed = 0.11(0.025 − 0.0001) = 0.00274 kg/s.
Therefore,
The lowest concentration of phenol is 0.06 mg/L; at [C] = 0.06 mg/L,
Total phenol to be removed to the ultimate residual concentration = 0.11(0.025 −
0.00006) = 0.00274 kg/s.
Therefore,
Test No. PAC Added M (g) [C ] (mg/L) X/M [C ]/X/M
1 0.25 6.0 0.076 78.95
2 0.32 1.0 0.075 13.33
3 0.5 0.25 0.0495 5.05
4 1.0 0.09 0.0249 3.61
5 1.5 0.06 0.0166 3.61
a
3 0.09 0.06+()2 6.0 1.0 0.25++()–
3 3.61 3.61+()2 78.95 13.33 5.05++()–

0.45 14.5–
21.66 194.66–

0.081===
b
l
a∑
1
l
C[]
X /M

∑ C[]
l
1
–

3
0.081 78.95 13.33 5.05++()6.0 1.0 0.25++()–

==
3
7.88 7.25–

4.76==
X
M

ab C[]

1 bC[]+

0.081 4.76()0.10()
1 4.76()0.10()+

0.026
kg phenol to be removed
kgC

== =
PAC required
0.00274
0.026

60()60()24()9,105== kg/d Ans
X
M



ult
a 0.081==
kg phenol to removed
kgC

Ans
X
M

0.081 4.76()0.06()

1 4.76()0.06()+

0.018
kg phenol
kgC

==
PAC required
0.00274
0.018

60()60()24()13,152 kg/s== Ans
TX249_frame_C08.fm Page 397 Friday, June 14, 2002 4:35 PM
© 2003 by A. P. Sincero and G. A. Sincero