1.0 Introduction
We have already
explained this in details in Coagulation & Flocculation
TABLE__1
|
Chemicals
|
Aluminum
sulfate
|
Ferrous
sulfate
|
Ferric
sulfate
|
Chlorine
|
Hydrated
Lime
|
Sulfuric
acid
|
|
Formula
|
Al2(SO4)3.18
H2O
|
FeSO47H2O
|
Fe2(SO4)3
.2H2O
|
Cl
|
Ca(OH)2
|
H2SO4
|
|
Reduction in Alkalinity as CaCO3
|
0.45
|
0.36
|
0.75
|
1.40
|
-
|
0.95
|
|
Increase in
Alkalinity as CaCO3
|
1.26
|
|||||
|
Reduction in
CO2 asCO2
|
1.11
|
|||||
|
Increase in
CO2as CO2
|
0.40
|
0.31
|
0.66
|
1.30
|
0.84
|
|
|
Increase in
SO4 as CaCO3
|
0.45
|
0.36
|
0.75
|
0.95
|
||
|
Increase in
Cl as CaCO3
|
1.4
|
|||||
|
Increase in
Hardness as CaCO3
|
1.26
|
1.1 Adding coagulants
The amount of coagulants
required to be added for effective coagulation is found out by JAR TEST.
In coagulation & Flocculation we
have already mentioned the minimum coagulant dosages used For example Let alum be the
coagulant. The minimum dosage is 34ppm.
By the use of following formulae
we can find out – How much additional alum, acid or alkali must be added to
obtain an Optimum pH of 6.8. The pH /alkalinity /CO2 table
shows that for a pH of 6.8 the ratio of 3 Alkalinity to 1CO2 is
required.
Let us assume that after
adding the minimum alum dosage the pH is low. To raise the pH we add either
soda ash (sodium carbonate) or hydrated lime (calcium hydroxide)
1.2 Using soda ash
Let X be the amount of
soda ash required (Purity of soda ash 99.16%)
We require a ratio of
3Alkalinity to 1CO2
Alkalinity + 0.94X
3 =
------------------------------
(refer table1)
CO2
- 0.41X
3CO2 -- 1.23X
= Alkalinity + 0.94X
3CO2 -- Alkalinity =
2.17 X
3CO2 -- Alkalinity
X = ----------------------------------
2.17
1.3 By using
hydrated lime
Let Y be the
amount of lime required
Alkalinity + 1.26 Y
3 =
-------------------------------
CO2 -- 1.11Y
3CO2 - 3.33Y =
Alkalinity + 1.26Y
3CO2 – Alkalinity = 4.59 (say 4.6)
3CO2 –
Alkalinity
Y =
-------------------------------
4.6
If pH is
high after adding the minimum alum dosage we require adding more
alum to get the optimum pH. This can be found out as shown Let’s say that
additional alum required is Z ppm.
This will lower the pH.
As mentioned earlier we
require a ratio of 3 Alkalinity to 1 CO2.
Therefore
Alkalinity
- 0.45Z
3 = ---------------------------
CO2 + 040 Z
3CO2 +1.20 Z
= Alkalinity - 0.45Z
Alkalinity - 3 CO2 = 1.65Z
Alkalinity – 3CO
Or
Z
= -------------------------------
1.65
At times in larger
plants adding additional alum to lower pH becomes highly uneconomical. Therefore to bring the
pH to optimum level sulfuric acid is used.
Let’s say Acid
required is R ppm. Again to keep the ratio
of 3 Alkalinity to 1 CO2
Alkalinity – 0.95 R
3 = --------------------------------
CO2 + 084 R
3 CO2 + 2.52 R = Alkalinity - 0.95R
Alkalinity – 3 CO2 = 3.47 R
Alkalinity – 3CO2
R = ------------------------------------
3.47
The formulas above are
when alum is used for Turbidity reduction.
Example
Coagulation for
turbidity reduction. Let the alkalinity & CO2 of raw water
be 45 & 5 ppm
respectively.
34 ppm of alum is added.
this is the minimum dosage.
There will be change in
Alkalinity and CO2 because of alum addition
Alkalinity will decrease and there will be increase in CO2
Reduction in
alkalinity
= 34* 0.45 = 15ppm
(When alum is added)
Change in
alkalinity
= 45 – 15
= 30 ppm
Increase in
CO2
= 34 * 0.40 =
13 ppm
Change in CO2 of raw
water = 5 + 13
= 18 ppm
We require a
pH of 6.8. This requires a ratio of 3 . Ratio alkalinity to CO2 = 30/18 =
1.67
This results in a
pH of 6.5
3CO2 - Alkalinity
By using equation for
soda ash X= ---------------------------------------
2.17
3*18 -
30
54 -30
X= ----------------------
=
--------------------
2.17 2.17
= 24/2.17 = 11.04
Alkalinity after adding
soda ash = 11 *0.94 = 10.4 ( say 11 )
CO2 after adding soda
ash = 11*
0.41 =4.5 (say 4 )
Alkalinity = 30 +11 = 41 ppm
CO2
= 18- 4 = 14 ppm
Ratio = 41/ 14 = 3
This results in a pH of 6.8
Color Removal
Now let’s
calculate alum required for Color Removal
For color removal the pH
required is 5.6 Let alkalinity of raw
water be 30ppm andCO2 be 3ppm Adding 43 ppm of alum
Alkalinity
= 30 - ( 43*0.45) = 30
- 19 = 11
CO2
= 3 + ( 43 *0.40) =
3 + 17 = 20
Alkalinity /
CO2 = 11/20
= 0.55
Results in pH of 6.05.
This pH has to be depressed to 5.6
So more alum is required
Let’s say X is
additional alum required
By using the
formula we can calculate additional required
Alkalinity - 0.2
CO2
11 - 0.2 ( 20)
X
= -------------------------------
= --------------
0.53
0.53
X
= 11 - 4 / 0.53 = 7/ 0.53 =
13.2 ppm ( say 13 ppm)
This additional alum
will further reduce Alkalinity in water and increase CO2
Reduction in
Alkalinity = 11 - (0.45 * 13) = 11 - 6 = 5 ppm
Increase
in CO2 = 20 +
(0.40 * 13 ) = 20 + 5 = 2
Ratio Alkalinity/
CO2 = 5 / 25 = 0.2 . Results in pH of 5.6
Total alum required is
= 43ppm(minimum dosage) + 13ppm (additional dosage)
We have suggested
a minimum dosage of various coagulants but we repeat that best
method to arrive at optimum coagulant dosage is by conducting a JAR TEST.
In the example for color
removal we gave the formula for pH > 5.6
If pH is less than an
alkali must be added . for color removal caustic soda is used for
increasing pH .
Let’s say C is the
required amount of caustic
0.2 CO2 - Alkalinity
C = ---------------------------
1.45
Note :- these
calculations are available in Excel also. There you have to just feed the user
Entry data’s and calculation will be done automatically.
2.0 Introduction
Chlorine is one
substance which finds wide utilization in water treatment. It is Because chlorine is a
powerful oxidizing agent. Chlorine is used for Disinfecting water, for
ammonia removal , color reduction, for microorganism control, taste & odor
control, in iron & Manganese removal ,H2S Removal
& destruction of organic
matter Chlorine is a hazardous
chemical and should be handled carefully. As mentioned above Chlorine is
for variety of application and chlorine dosage for each may be different. Chlorine is available in
all three
forms solid ,
liquid & gas.
We limit our discussion
to Chlorinating by hypochlorites. The two hypochlorite’s used are Calcium
hypochlorite and Sodium hypochlorite . Free chlorine combines
with water to form hypochlorus and hydrochloric acid
Chlorine +
water- Ã
Hypochlorus acid + hydrochloric acid
Cl2
+ H2O
Ã
HOCl + H+ + Cl—
HOCL
à H+
+ OCl--
Distribution Of HOCl And
OCl at different pH and temperature
|
|
|
PH
HOCl
OCl-
|
4
100
100
0
0
5
99.9
99.9
0.1
0.3
6
98.5
97.4
1.5
2.6
7
86.9
79.2
13.1
20.8
7.5
67.9
54.7
32.1
67.9
8
40.1
27.6
59.9
72.4
8.5
17.4
10.8
82.6
89.2
9
6.3
3.7
93.7
96.3
10
0.7
0.3
99.3
99.6
11
0.07
0.03
100.0
100.0
Before we discuss
further let us understand few terminology associated with
chlorination.
2.1 Chlorine demand
Chlorine when added to
water containing organic and inorganic materials, will combine with these
impurities to form chlorine compounds. At some point reaction with
organic and inorganic material will stop, This point is known as chlorine
demand. In other words the DEMAND is the amount of chlorine consumed by
oxidation and substitution reaction with organic and inorganic
materials.
Chlorine which is left
after the chlorine demand has been met is free residual chlorine. This free
Residual chlorine is the available residual chlorine to kill the microorganism
present in water.
Chlorine dose mg/liter =
Chlorine demand mg/liter + Chlorine residual mg//liter
Reaction with water
Chlorine reacts with
water to form hypochlorous and hydrochloric acid
Cl2 +H2O ®
HOCl + HCl
Hypochlorous acid being
a weak acid further dissociates ,depending on the pH to
Hydrogen ions and
hypochlorite ions
HOCl ®
H+ + OCl-
A chlorine
solution therefore may contain elemental chlorine, hypochlorous acid or
hypochlorite ions depending upon the pH
At pH<3 all
chlorine remains has elemental
Cl2
At
pH>3<5 all chlorine remains has hypochlorous acid
At pH >5<10
all chlorine is in form of HOCl and OCl-
At pH >10 all
chlorine is form of OCl--
As it can be seen from
above pH is of vital importance in chlorination In dilute chlorine
solution with a pH of above 4 , the formation of hypochlorous acid
is most complete and leaves little free chlorine. Hypochlorous acid being a
weak acid dissociates poorly below a pH of 6. Thus any Cl2 or
OCl ions added to water will immediately form hypochlorous acid or hypochlorite
ions depending on the pH. Hypochlorous acid is a stronger disinfectant than
OCl- ions. Therefore it is water should be chlorinated at a pH 5—7
2.2 Reaction with other impurities
Reaction with H2S
and NH3
Hydrogen sulfide and
ammonia are two inorganic substance which may be present at the stage of
disinfection. Their presence can complicate the use of chlorine for
disinfection purposes. H2S and NH3 are reducing agents and give up their
electron easily. Chlorine reacts with these compound rapidly
Producing some
undesirable results
H2S +Cl2 Ã
S
+ H2O + 2Cl-
The chlorine required to
oxidize H2S to sulfur and water is 2.08 mg/liter of Cl2 for
1 mg/liter of H2S.
The complete oxidation of H2S to
sulphate is has shown below H2S + 4 Cl2
+ 4 H2O -------Ã H2 SO4 +
8HCl
Thus 8.32 mg/liter of
chlorine is required for complete oxidation of H2S.
When chlorine is
added to water containing ammonia it reacts rapidly to form chloroamines. This
means less chlorine is available for disinfecting purpose. As the ammonia
concentration increases the disinfecting power of chlorine falls of
rapidly.
2.3 Reaction with
organic materials
Chlorine will react with
any organic material present to form suspected carcinogenic product,
for Example trihalomethanes. This can be best avoided by limiting the
prechlorination and removing organic materials by other means prior to
chlorination.
2.4 Breakpoint
Chlorination
As explained earlier
chlorine reacts with various impurities in water .If you see the graph
below In the first stage
chlorine is destroyed by reducing agents. Then it reacts with organic matters
present in water. After reaction with organics it reacts with ammonia present
to produce chloroamines. The type of chloroamines formed depends on molar
ratio of chlorine added to ammonia nitrogen present and on the pH. At this
stage we get what is known as combined residual –a chlorine residual combined
with other substances and has less disinfecting strength. Further addition of
chlorine will destroy the chloroamines and some chloroorganics, Any
further addition of chlorine will give us –what is known as free
available residual chlorine. Free in the sense that it has not reacted with
anything and is available for disinfection. Free residual chlorine is the best
residual for disinfection.
Refer to the graph and
you will see that when reaction with ammonia is complete there is a dip in the
curve. The point where there is a dip is known as the break point.
Chlorination is always done beyond breakpoint to ensure that there is free residual
chlorine available.
2.5 Process Calculation
There is basically two
chlorination process calculation
calculating chlorine
dose mg/Liter
Estimating chlorine
demand
Chlorine dose mg/liter
= chlorine demand + chlorine residual
Chlorine mg/liter* M3/day
flow
Chlorine feed per day
=
---------------------------------------------
1000
Let us work out an
example
Chlorine dose
= 2.5mg/liter
Flow
= 4000M3/day
1 day
= 24 hours
Chlorine mg/liter* flowM3/day
Chlorine feed per
day in kgs =
-----------------------------------------------------------
1000
2.5*4000
=
---------------- = 10 Kgs/day
1000
Now calculate the Dry
calcium hypochlorite required for the above example
Chlorine
required
= 10 kgs/ day
Available
chlorine
= 65%
Of Calcium hypochlorite
Kg/day
Chlorine
10 kgs/day
Dry calcium hypochlorite
required =
---------------------------
= ---------------
%available
chlorine/100
0.65
=
15.38kgs
Similarly for sodium
hypochlorite which has available chlorine of 30%
Kgs of Cl2 10kgs
/ day
Dry sodium
hypochlorite =
------------------------------- =
--------------------
% available
chlorine/100
0 .30
=
33.33 kgs /day
Calculating the water required to make 5 % solution
Amount of water required
to dilute to 5 % solution
Chlorine
Kgs
Water
required(liters)
= --------------------
- Chlorine Kgs
% strength
15.38
=
-
15.38
5/100
=
292.2 liters ( Note sp. Gravity of water is 1 )
3.0 Filtration
Filtration is a process
that consists of passing a solid liquid mixture through a porous material
which retains the solid and allows the clear liquid to pass through . The
porous media can be any filtering media like sand, coal etc. The clear
liquid is called the filtrate
3.1 Types of filters
Filters are classified
in two ways
1. By mode of operation
2. By basis of media used
3.2 Classification by
operation
1. Gravity Filters
2. Pressure Filters
Filters are classified
based on the function of system, if the filter operates under the systems
gravity head, it is called GRAVITY Filter. Similarly if a filter
functions under the pressure of incoming water, it is called
PRESSURE FILTER.
3.3 Classification by
Media
Filters are also many
times classified by the type of media
used . Sand, anthracite,
garnet and some other material used for filtration. The filters
classified on the basis of media are
|
TYPE OF
FILTER MEDIA
|
|
Sand
filter
Sand
|
|
Dual media
filter
Sand & Anthracite
|
|
Multimedia
filter
Sand
,Anthracite & Garnet
|
|
Activated carbon
Filter
Activated carbon
|
|
Manganese zeolite
Filter
Greensand (MnO2)
|
3.4 Filtration Rates
It is a common
practice to state the working rate of filters in terms of
flow through a unit surface Area in a given time (Meter/hour) .
The
Filtration rate is taken as
follows
1.
Pressure sand
filters
10-----12 Meter/Hour
2.
Dual media
filter
12-----15 Meter/Hour
3.
Multimedia
filter
15-----20 Meter/Hour
3.5 Parameters required for designing a pressure
filter.
The following parameters
are required for designing a filter.
Total output in
meter cube/day (M3/day) or Flow in M3/Hr
Velocity in Meter/Hour
(M/Hr)
Backwash velocity in
M/Hr
Blower air velocity in
M/Hr
Backwash tank capacity
in M 3 or Backwash pump capacity in M3/Hr
|
Parameters
PSF
DMF
MMF
|
|
Velocity
M/Hr
10—12
15---18
18---20
|
|
|
|
BW Velocity M/Hr
36
30
24
|
|
(Without air scour)
|
|
BW Velocity
M/hr
24
24
20
|
|
(With air scour )
|
|
Blower velocity M/Hr
36
36
24
|
|
Rinse Velocity M/Hr
|
|
|
BW = Backwash
3.6 Calculating filter
Diameter
Step1 :-
Select flow as output in
Meter cube /day or In meter cube/hour
Step2 :-
Select type
of filter you wish to supply. Remember this will depend on
the turbidity or suspended solids in the raw water to
be treated
Step3:-
Select Velocity. Cleaner the
water , higher the velocity.
Step4:-
Calculate
Filter Diameter
For example let us
consider a filter for a flow of 600 M 3 /Day
Then flow/hour is equal
to output/24 (600/24)
In this case it is
25 M 3 /Hr
We know
Flow = Area
X Velocity
Flow
Area =
----------------------
Velocity
Area = Õ D 2 /
4 ( D is diameter of filter and Õ is equal to 3.14)
Area
= 0 .785 D 2 (3.14
divided by 4 is equal to .785 )
Therefore
0 .785 D 2 = Flow / Velocity
D 2 =
Flow / .785 Velocity
D
= Ö Flow
/ .785 Velo
Let us consider a
pressure sand filter
|
|
Then D
= Ö 25 / ( .785* 10)
D
= 1.78 Meter ( say 1.8 Meter)
Keeping the same flow of
25 m 3 /Hr , The diameter for Dual media filter and
Multimedia
filter with a velocity of 15 M/Hr & 20 M/Hr
will be 1.4 meter &
1.2 Meter respectively
4.0 Introduction
Activated carbon is used
in water and waste water treatment for removal of organics,
dechlorination and also for removal of taste and odour . Activated carbon can
be used in either in granulated or powdered form. Powdered activated carbon is
introduced as a Chemical. Granulated activated carbon is generally used as a
filter media similar to Pressure filter. Activated carbon filter or ACF as it
is generally called is a Pressure vessel Similar in construction to
pressure sand filter containing the carbon media. Many raw is chlorinated and
such water has to be dechlorinated before it can be passed through
exchange or through a membrane. The most widely used method of
dechlorination is by activated carbon filter.
4.1 Design of Activated Carbon Filter
The designing of
activated carbon filter is similar to designing a Pressure sand filter.
In designing
activated carbon filter the following parameters are important
1.
Volumetric flow rate M3 / Hr
2.
Velocity
M / Hr
3. Bed
Volume
M3
4.
EBCT
Minutes (5---30)
EBCT means empty bed
contact time.
It is far simpler to
design ACF for industrial water treatment than for Industrial waste water
treatment. In water treatment ACF is generally used for dechlorination
and the diameter of vessel can be calculated by velocity . Velocity
is normally between10-20M3/Hr.
We know
Area = Flow /
Velocity
EBCT = Bed
Volume (Vb ) / Volumetric Flow( M3 / Hr)
Total bed volume =
Volume of media + void volume
Vb =
Area Of Vessel M 2 * Height (media +freeboard)
M
Therefore EBCT
= A*H / Q
Where A= Area in Sq.
mts, H= Height in meters
and Q= Volumetric
flow rate (M3 /Hr)
Now we can calculate
area either by the use of EBCT or by just using velocity
4.2 By Velocity
We Know
Velocity * Area = Flow
So
Area = Flow
/ Velocity
Area = Õ*D2 /4 where D =
Diameter
Area = 3.14/4*D2 = 0 .785 D2
------ Eq ( 1 )
Velocity of
ACF is normally between 10---20 M/Hr
We select a velocity of
15 M/Hr and flow of 10 M3 /Hr
Area = Flow / velocity
0 .785 D2 =
10 /15
D2 = 10/15*0.785
D
= 0 0.921 (say 1 M or 1000 mm )
This is the diameter of
the vessel. The height of Media is also 1 M & freeboard is 0.5 Meter
Volume of Media
= Area of vessel * height
In waste water treatment
and also for organic removal EBCT method is used for finding the area of vessel
and then the diameter.
|
Velocity
10
15
20
|
|
Flow
10
10
10
|
|
Height
01.5
01.5
01.5
|
|
Area
01.0
0.66
00.5
|
|
EBCT
09.0
05.94
04.5
|
|
Velocity
10
15
20
|
||
|
Flow
10
10
10
|
||
|
Height
1.75
1.75
1.75
|
||
|
Area
1
0.66
0.5
|
||
|
EBCT
10.5
6.93
5.25
|
||
|
Velocity
10
15
20
|
||
|
Flow
10
10
10
|
||
|
Height
2.0
2.0
2.0
|
||
|
Area
1
0.66
0.5
|
||
|
EBCT
12
7.92
6.0
|
||
There are many charts
which gives the required contact time required for removal of
particular contaminants by activated carbon . Once the contact time [d1]is known , the best Area can be selected based on the velocity
from the above table.
ACF designing is not as
simple as it looks especially in case of waste water treatment. As our aim is
to keep this design manual as simple as possible we are not going into details.
Designing by BDST& EBCT method. Anybody interested in
Designing by BDST&
EBCT method can consult the book “ Theory & practice of water and
wastewater treatment” by Ronald L.Droste (page 505-506)
5.0 Introduction
We have already
discussed the various methods available for removing iron from water. The
most effective means of removing iron and manganese from water is by Manganese zeolite .
5.1 Design criteria for Manganese zeolite
|
Characteristics
of manganese greensand
|
|
Color
Black
|
|
Density
1360Kg/M3
|
|
Effective
size
0.30- 0.35 mm
|
|
Uniformity
coefficient
1.6
|
|
Mesh
size
16—60
|
|
Attrition loss per annum
%
2—4 %
|
|
Raw water pH
|
|
Bed Depth
(minimum)
700mm
|
|
Freeboard
50% of bed depth
|
|
Service flow
rate
5 –12 M3/hr/M2
|
|
Backwash flow
rate
20—25
M3/hr/M2
|
KMnO4 is used either in
conjunction with chlorine or alone. KMnO4 dosage differs depending on whether
it is used alone or with chlorine.
5.2 Dosage of KMnO4.
With chlorine
1 mg/liter ofCl2 /
1ppm of Fe
KMnO4 mg/liter
=( 0.2mg/literKMnO4 for 1ppm of Fe)+ (2 mg/liter of
KMnO4 for
1ppm Of Mn ) + ( 5mg/liter of KMnO4
for1ppm of H2S
)
Without Chlorine
KMnO4 mg/liter
= (1.mg/literKMnO4 for 1ppm of Fe)+ (2 mg/liter of KMnO4
for 1ppm Of Mn )+ (
5mg/liter of KMnO4 for 1ppm ofH2S)
5.3 Continuous KMnO4 injection
Step 1
Decide your flow rate .
Let us consider 5 M3 / Hr and the concentration of Fe &
Mn is 1 PPM
Step2
Calculate the KMnO4
consumption1 mg KMnO4 for 1mg of Fe + 2 mg of KMnO4 for
1 mg of Mn = 3 mg KMnO4
Step3:-
Select filter velocity
from here
|
KMnO4
Consumption
|
Velocity
|
|
Mg/liter
|
M3/Hr/M2
|
|
0.5
|
12
|
|
1
|
10
|
|
2
|
08
|
|
3
|
06
|
|
4
|
05.5
|
|
5
|
05
|
Step 4
Selected filter velocity
–6 M/Hr
Step 5
Calculate filter
diameter
We know Area =
Flow/ Velocity
We also know that area = ÕD2/4 p=3.14
Therefore Õ/4 = 3.14 /4 =0.785 :pD2/4
=0.785 D2
= flow/velocity
D = SQ RT (
flow /0.785*velocity)
We know flow = 5 M3 /
hr , Velocity 6 M/ hr
Therefore
D = SQ RT ( 5/0.785*6 )
=
SQRT (1.06)
=
round filter diameter
Step 6
Calculate new filter
surface velocity & surface
area
Surface
Area = 0.785 D2
Velocity
= 5/ new surface area
= 5 / 0.785
= 6.36 M/Hr
Step7
Calculate
Volume of filter Media
Bed depth is the height
of media in the vessel. Bed
depth
Area * bed depth
= media volume
0.785 *
0.7 =
0.5495 M3
Weight of media = Volume
* density
Step8
Calculate operating time
till backwash
Length of filtered water
column = capacity / KMnO4 Consumption
= 400gm/M3 /
3gm/M3(Fe + Mn )
= 133.3 M
Time until
backwash
= 133.3/6.36
=
21 Hours
5.4 Discontinuous KMnO4 injection(intermittent regeneration)
Step
1
Decide your flow rate .
Let us consider 5 M3 / Hr and the concentration of Fe is 0.5ppm Mn
is 1.25 PPM
Step2
Calculate the KMnO4
consumption
0.5 mg KMnO4 for
mg of Fe + 2 .5mg of KMnO4 for 1.25 mg of Mn
= 3 mg /lKMnO4
Step3
Select filter velocity
from here
|
KMnO4 Consumption
|
Velocity
|
|
Mg/liter
|
M3/Hr/M2
|
|
0.5
|
12
|
|
2.0
|
10
|
|
3.0
|
08
|
|
4.0
|
07.5
|
|
5.0
|
7
|
|
10.0
|
6
|
Step4
Selected filter velocity
–8 M/Hr
Step 5
Calculate filter
diameter
We know
Area =
Flow/ Velocity
We also know
that area = pD2/4
p
= 3.14
Therefore p/4 =
3.14 /4 =0.785
pD2/4
= 0.785 D2 =
flow/velocity
D
= SQ RT (
flow /0.785*velocity)
We know flow = 5 M3 / hr
,Velocity 6 M/ hr
Therefore
D = SQ RT ( 5/0.785*8 )
=
SQRT (0.792)
=
0.892m= 0.9 round filter diameter
Step 6
Calculate new filter
surface velocity & surface area
Surface
Area = 0.785 D2
=
0.785* (.9)* (.9)
=
0.63585
Velocity
= 5 / new surface
area
=
5 / 0..63585
=
6.36 M/Hr
Step 7
Calculate Volume
of filter Media
Bed depth is the height
of media in the vessel. Bed depth
Area * bed
depth = media volume
0.0.63585 *
0.7 = 0.445095 M3
Weight of
media = Volume * density
Step 8
In the last design we calculated
the operating hours till backwash.
Here we calculate the
capacity / regeneration
445 liter media * 1.4 gm
/liter / 3.0 gm/liter = 207(M3/Regn.)
Hours of operation
= 207 /5 = 41.4 hours.
Softener designing
Step
1
Convert ions to CaCO3 equivalent.
Step
2
Calculate total Cations
and Total Anions
TC = TA
Total Cations as CaCO3 Total
Anions as CaCO3
|
Ca
Hardness
T Alkalinity
Mg
Cl
Na
SO4
--------------
---------------------------
Total Cation
Total anion
Step
3
Calculate Na/TC
Step
4:
Decide on a Hardness
leakage in the effluent and select the regeneration level Table
1.
Step 5
Most designers select a
space velocity of 16.6 M3/hr/M3 and a linear
velocity of 20 M3/hr/M2.
From this select the exchange capacity at the regeneration level selected Table
2.
Step
6
Bed depth can be 0.75 to
2M but for all practical purpose the bed depth is kept between 1 – 1.5
meter. Minimum bed depth is kept for the retention time.
Restriction of higher bed depth is due to the following reasons – increase in Dp due to increase in depth increase in height of
vessel taking void space into consideration.
Step
7
Calculation of
Theoretical Bed Depth.
Sp.Velocity X Resin
Volume
Resin Volume
-----------------------------------
=
-----------------
Linear Velocity M3/hr/M2
Bed
depth.
Linear Velocity
Bed depth
= ---------------------
Space velocity.
Linear velocity of
20M/hr/M2 and space velocity of 16.6 gives a bed depth of 1.2M.
Step
8
Calculate the Resin
Quantity (in liters) as follows
Load X OBR
Resin Quantity
= --------------------------------------------------------
Ex. Change capacity X Correction factors
Correction factor can be
selected Table 3, A , B
OBR = Output
between regeneration in M3
Load = Ionic
Load. (Hardness in case of Softener)
OBR = Flow X
Time.
Step
9
Resin Quantity and the
bed depth gives the are of vessel.
Resin Volume in M3
Area =
-------------------------
Bed depth in meters.
Step 10
Check linear
velocity. The value should be in the range of 10 – 25 M3/hr/M2.
A slightly lower value can be tolerated if the water is clear and slightly more
if slight increase in the hardness leakage in the effluent can be
tolerated.
Step
11
Find the diameter of
vessel
Step
12
Now we know the diameter
of vessel and Resin Quantity. Calculate Regenerant Quantity.
Regenerant Quantity =
Resin Qty(liters) X R/L in gm Nacl /liter.
Step
13
Calculate the Regenerant
flow Regenerant time. Regenerant flow should be carried out at a velocity
of 3 M3/hr/M3 of resin and the time should not be
less than 15mts. It should also not be more than 45 Mts.
Step
14
Find the rinse
volume required. 2.5 – 3 M3/M3 of resin.
It is also dependent on plant design and end of hardness desired in the
rinse. Normal tine for rinsing is between 30 – 50.
Step 15
Select unit up flow or
Down flow. Up flow unit is also called packed bed
column. In this unit row water passes from the bottom and treated water
is collected from the top. The down flow unit is in
which the raw water passes from the top and treated water collected from the
bottom. Merits and demerits of both has been described elsewhere.
WORKED OUT EXAMPLE
Step 1
A Water analysis shows
the following report
Hardness as CaCO3
= 500PPM
TDS
= 1000
Na
= 1000 – 500
= 500
Therefore Na / TC
= 50%
Step
2
R/L from resin
catalogue for hardness leakage less than 1ppm is 160 gm/liter.
Step 3
Ex. Capacity at 16.6 M3/hr/M3 and
R/L of 160 gm/liter is 70. (See table 2).
Step 4
Correction factor for
exchange capacity for NA/TC and hardness.
Actual Ex.
Capacity = Ex. capacity X 0.85 X 0.94
= 70 X 0.85 X 0.94
= 55.93 = 56.
(See Table 3 and
34)
Step
5
Treated water required
per day is 240 M3 (Per day = 24hrs)
Load X OBR
Resin Quantity
= -----------------------
Ex. Capacity
500 X
240
= ------------------
= 2142 liters = 2142/1000 = 2.142 cubic
Meters
56
240
Flow M3/hr
= --------
= 10 M3/hr
24
Step
6
Calculate the area of
vessel
Volume of Resin in
M3 2.142
Area =
---------------------------- = ---------
Bed
Depth
1.2
Area = 1.785
Flow
Linear velocity
= --------- = 5.6 M3/hr/M2 (it is very
less from assumed range)
Area
5.6 X 2 gives 11.2 M3/hr/M2 thus
two region per day gives the required velocity.
It is better to have 2
regeneration per day if flow of 10 M3/hr is required. Then OBR
for each regeneration
becomes 120 M3.
Load X OBR
500 X 120
Resin Qty (liters) =
--------------------- -------------------
Ex. Capacity
56
= 1071.42 liters =
1.071
1.071
Area
= -------- = .89
1.2
10
Linear Velocity
= ------ = 11.23 M3/hr/M2
0 .89
Step
7
Calculate the vessel
diameter
r2
= 0.89
0.89
r2
= -------
3.14
0.89
r
= -------
3.14
0.89
2r = d
= 2 X ------
= 1064 mm
3.14
Select 1000 mm diameter
vessel.
Step
8
Calculate the actual bed
depth for 1000mm Dia vessel.
Area =
r2 = 3.14 X 0.5 X 0.5 = .785
Volume 1.071
Bed depth =
------------ = -------- = 1.36
Area
.785
This is within the limit
of 1.0M to 1.5 M.
Step
9
Calculate Regenerant
flow time and flow .
At 3M3/hr/M3 the
Regenerant flow in M3/hr is resin Quantity X Velocity.
Flow in M3/hr
= 3 X 1.071 = 3.21 M3/hr
Volume in M3
1.071
Time
= ------------------- = --------
Flow in M3/hr
3.21
= 0 .33 hrs. = 20Mts
i.e. time for injection
of salt solution should not be less than 20Mts.
Step
10
Calculate Rinse timing
:- Fast rinse in normally recommended between 12 – 16 M3/hr/M3 velocity.
Let us be conservative and use 12 M3/hr/M3 for
calculation purpose which is normally taken by all designers.
Fast Rinse
flow = 12 X Resin Qty = 12
X 1.071 = 12.85 M3/hr
Volume of water per M3 of Resin
Fast Rinse period
(minutes) = --------------------------------------------
Flow
Rinse volume is
considered to be 5M3/M3 of resin. (5 bed volumes)
In this case
= 5 X 1.071 = 5.355.
5.355
Final Rinse period =
------- = 0.41 hours
12.85
= 0.41 X 60 = 25,00 Minutes.
Selection of Option
Refer to the detail worked
out example. In working out the example, we have taken the following into
consideration.
Hardness
:- 500ppm as CaCO3
Na/TC
:- < 50%
Flow
:- 10M3/hr
Linear
velocity :-
10 – 25 M3/hr/M2
Space
Velocity
:- 5 - 50 M3/hr/M3
Bed
depth
:- 1 – 1.5 Meters
Output between
regeneration :- 240M3
By calculation, we found
that for 240M3 output of treated water the amount of resin
required is 2.142M3.
With this volume of
resin i.e. 2.142 M3 and bed depth of 1.2 Meters we got area of
vessel to be 1.785M2.
Now we know flow is 10M3/hr.
10
Then Linear
velocity = ------- = 5.6M3/hr/M3.
1.785
The linear velocity thus
calculated is too 1ow (Refer Point 4) and hence has to be increased,
By reducing area or
By increasing head of
the pump.
Option 1 By Reducing Area
Instead of having a
single regeneration per day we can have two regeneration per day. OBR now
for 12 hours is 120M3. Total output is 2 X 120 = 240M3 per
day. (24 hours) Now we calculate the resin quantity for 120 M3
which is 1.071 M3. Keeping the same bed depth we get an area
of .785 M2 which fits into all our assumptions.
Options 2
By Increasing
Head of Pump
Increase head of pump so
that 20M3/hr of raw water is passed through the ion exchange
vessel. The draw off is only 10M3/hr. Hence storage is
required for additional water being pumped. The operation is to be
regulated depending upon the requirement.
Dealkalizer means
removal of alkalinity . It is also referred to as partial demineralization
since in most of the process TDS is reduced by the amount of
alkalinity removed.
There are basically four
methods by which alkalinity can be removed using ion exchange resin
7.1 By using Weak acid cation (WAC
RESIN).
Generally used
when total alkalinity is greater than total hardness.. When
Total Alkalinity
> Total Hardness all hardness is temporary hardness.
We all know that
Weak acid cation removes Hardness associated with alkalinity
but does not
remove permanent Hardness .
Basic arrangement for
this process is
WAC resin in H+ Form ,
Degasser and Softener
7.2 Split stream process .
This system is usually
used for water of variable composition . The system consists of
Hydrogen cation unit
Softener
Degasser
In the Hydrogen Cation
unit Cations associated with alkalinity is converted to carbonic acid and
the CO2 is blown out in the dgassifier . Cl & SO4 present
will also get converted to their corresponding acids . This acidity is
neutralized by blending with softener water.
7.3 Starvation process .
This system is similar
to above but used with water of fixed composition.
7.4 Chloride anion exchange process
In some places it is
difficult to handle acids , for example hotels, hospitals etc . All
these places uses water
in considerable amount for utility purposes. In such places when water
has high bicarbonate alkalinity then this process is used. Bicarbonate ions are
replaced by chloride ions and there is no reduction of TDS in this
process .
Except for starvation
process we have given examples for designing of the
system for all the
three process.
Designing of Split
stream system.
Let us consider a water
analysis as shown below .
Cations as CaCO3 Anions as CaCO3
Ca 2+
--
150
HCO3 - ---- 130
Mg
2+
130
Cl - ---- 180
Na
+
100
SO4 2- --- 70
Total
Cation
380
Total Anion
380
Now let us consider the
quantity of water required . Say it is 200 M3 / day. This water is
required over a period of twenty hours. Then the service cycle is 200 /
20 = 10 M3/ Hr. Regeneration per day is 1.
Quality of water
required is alkalinity to be reduced to 25 ppm
%age of water to
be treated in hydrogen cation unit
= ( Raw
water Alkalinity - Alkalinity desired) /
(TDS - Cation leakage )
In this case
(130 – 25)
=
---------------------- X 100
(380 - 5)
=
28 %
Now we know that flow
rate is 10 M3/Hr through the Dealkalizer system
And 28 % of this has to
flow through Hydrogen cation unit and rest through
the softener .
28 % of 10
M3/ Hr
= 10* .28 = 2.8 M3/ Hr flow from H+ Unit
72 % of 10 M3/
Hr
= 10* ..72 = 7.2 M3/ Hr flow from softener
And the
total 10 M3/ Hr has to flow from the dgassifier unit .
Cation H + Resin
required
( total cation X Flow M3/Hr X time in Hours
=
----------------------------------------------------------
(Exchange capacity X deaeration factor 0.9)
Cation Na + Resin
required(softener)
( total Hardness X Flow M3/Hr X time in Hours
=
----------------------------------------------------------
(Exchange capacity X deaeration factor 0.9)
( 375)*(2.8)*20
Cation H + Resin
required =
-------------------------
= 416 liters
(56)*(0.9)
Cation Na + Resin
required (280)*( 7.2) *
20
(softener)
=
-------------------------- =
720 liters
(62) * (0.9)
Degasser
Area
= flow (M3 /Hr ) /
Velocity
=
10/ 50 = 0.2
Therefore Degasser
diameter = 0.2/.785
( Because Pd2/4 =area and P/4= 3.14/4=.785 and therefore d2 =
Area/.785 )
d2 =
Area/.785 =0.2/.785 = 0.255
d = sq. rt
(0.255) = 0.504 meters *1000= 504mm (say 500mm_)
Diameter of Cation
vessel
Bed depth normally
recommended by Resin manufacturers is between 700mm to
1800mm and normally
1000mm bed depth is the most commonly used. Linear
velocity is ( 10 -40) M3/Hr/M2 and
space velocity is between 10 - 45 M3/Hr/M3 .
linear Velocity =
flow/area
& space
velocity = Flow/ Volume
In this case Resin
Quantity Cation H+ Resin = 416 liters & Softener Resin
= 720 liters
Cation H+
Resin = 416/1000 & Softener Resin = 720 /1000
Cation H+
Resin = 0.416M3 & Softener Resin =
0.720M3
Bed depth =1000mm= 1 meter
,
therefore vessel
cross sectional area = Volume/Height(bed depth)
Area = Pd2/4 and P/4= 3.14/4=.785 and
Therefore d2 =
Area/.785
Cation Vessel diameter
= Ö (volume/height)/.785 = Ö 0.416/.785= 0.727metres = say 800mm
Softener Vessel diameter
= Ö (volume/height)/.785
=Ö 0..720/.785=
0.957metres = say 1000mm
If you increase
the bed depth of resin to 1500mm (1.5 M) you will
get the vessel Dia for Cation as 600mm and for softener
the diameter will be 800 mm.
In both cases the linear
and space velocity is within the range . This calculation has been shown in
more detail in Vessel & tank designing.
Chemicals Required For
Regenerating Resin
Acid & Salts are
required for regeneration of Resin . How much Acid & salt is
required is
Calculated as shown
100% Chemical required
in Kgs = ( Quantity of resin in liters * Regeneration
level in grams /liter)/
1000
Dealkalization by Weak
Acid Cation
Let us consider a raw
water with the following analysis .
Cation as CaCO3
Anion as CaCO3
Ca
--
282
HCO3 -- - 265
Mg
--
8
Cl ---- 30
Na
--
22
SO4 ----- 17
Total
Cation
312
Total Anion
312
We all know that
Weak acid cation resin removes hardness associated with
alkalinity but is
not efficient to remove permanent hardness. Therefore this
system has a softener
to remove permanent hardness . The system consists of
Weak acid cation
unit
Degasser unit
Softener unit
Designing the unit
Permanent hardness
= total hardness - total alkalinity
=
290 - 265
= 25 ppm
Total quantity of water
required = 200 M3/ day
Service cycle
-- 20 Hrs.
Flow from WAC
unit = 200/20 = 10 M3/Hr
( Ionic load)*( M3/Hr)*Service cycle
Quantity Of
WAC Resin = (
------------------------------------------------------
(
Exchange Capacity) * 0.9
Where 0.9 is a
deaeration factor .
Capacity of weak acid
cation from graph (table ) is 100 gm/litre for exhaust
cycle of 20 hours
(
265)*( 10)*(20)
Quantity Of
WAC Resin =
-----------------------------------
100* 0.9
=
588.8 litres (say 590 litres)
Total hardness after WAC
Unit = Permanent hardness +
leakage)
Permanent
hardness + leakage
Softener resin
required =
----------- --- ----------------------------
Exchange capacity * 0.9
=
30* 10*20/62*0.9
=
107 litres
The diameters of units
are calculated as shown for split stream system
Dealkalization by chloride anion exchange
Strong base
anion absorb acids(cation Effluent ) as mn demineralization
and is used in the second stage of Demineralization. Strong base anion can also
split salts and this property of resin makes suitable for dealkalization.In
commercial establishment like hotels , hospitals & other similar
building & small factories softener were employed for providing feed
water to prevent scaling. Softener did not remove alkalinity and hence
sodium bicarbonate got converted to caustic soda and carbon dioxide and when
the steam condensed ,CO2 got dissolved in the condensate
,lowering the pH and creating return line corrosion. Later cation resin
in H + form was used but these created problem of acid handling . This
problem of acid handling was solved by utilizing SBA resin for Dealkalization.
A type 2 resin is
generally used for Dealkalization. The anion resin can be regenerated in
two ways
By using only common
salt
By using salt and
caustic
Normally the first
method of regeneration is used for Potable water and second method when Water
has to be used for boiler.
Let us design a system
with the following example . Consider the water analysis
shown below.
Cation as
CaCO3
Anion as CaCO3
Ca
….
200
HCO3……. 200
Mg
…
100
Cl
…….
150
Na….
100
SO4 ……
50
-------------
------------
Total
Cation 400
Total
Anion
400
Total
Anion
400
Alkalinity
200
% Alkalinity/ Total
anion
(200/400)*100=50%
|
Design
parameters
|
SBA
|
SOFTENER
|
|
Hourly Flow rate M3/Hr
|
5
|
5
|
|
Service cycle in hours
|
10
|
10
|
|
Exchange Capacity gm
CaCO3/L(Regeneration only with NaCl)
|
15
|
56
|
|
Exchange Capacity gmCaCO3/L(Regeneration
only with NaCl+NaOH)
|
20
|
56
|
|
Regenerant level
NaCl gm/L(case1)
|
56
|
160
|
|
Regenerant level
NaCl gm/L(case 2)
|
48
|
160
|
|
Regenerant level
NaOH gm/L(case 2)
|
48
|
CASE 1Salt regeneration
The water has to be
softened before dealkalization .
The percentage
alkalinity before dealkalization (200/400)*100= 50%
From the capacity table
given below using 50 % alkalinity and 400 TDS the capacity for SBA Resin is 15
gmCaCO3/L and For softener is 56 gm CaCO3/L . Correction factor
has been taken for both .
Load (alkalinity ) * Flow* Service cycle
Anion Resin
Quantity = ……………………………………………………..
Exchange capacity
=
( 200)
* (5)* 10/ 15 = 667 litres
Load (Hardness ) * Flow* Service cycle
Softener Resin
Quantity = ……………………………………………………..
Exchange capacity
=
( 300) * (5)* 10/ 56 = 268 litres
Once we Know the volume
of resin required it is very easy to calculate the vessel
diameter.
CASE 2
Regeneration with NaCl + NaOH
The method of
calculation will remain same as in case 1 but with different exchange capacity
Load (alkalinity ) * Flow* Service cycle
Anion Resin
Quantity = ---------------------------------------------
Exchange capacity
=
( 200) * (5)* 10/20 = 500 Litres
Load (Hardness ) * Flow* Service cycle
Softener Resin
Quantity = ---------------------------------------------------------
Exchange
capacity
=
( 300) * (5)* 10/ 56 = 268 litres
In conservative
designing the amount of resin obtained is divided by 0.9 which is known as
deaerating factor
Leakage will be 10 %
of the influent alkalinity 20 ppm CO3 or HCO3
as CaCO3
The calculation
for determining alkalinity when caustic soda is used for regeneration
Total carbonate
alkalinity = carbonate alkalinity + bicarbonate alkalinity
Carbonate
alkalinity
= 2* p alkalinity
Bicarbonate
Alkalinity
= M. Alkalinity - 2p Alkalinity
A water having the
following chemical analysis is treated to produce demineralized water.
|
Cations
as CaCO3
|
Anions
as CaCO3
|
||||
|
Ca
|
85
|
HCO3
|
190
|
||
|
Mg
|
25
|
CO3
|
0
|
||
|
Na
|
140
|
OH
|
0
|
||
|
K
|
30
|
Cl
|
60
|
||
|
Fe
|
SO4
|
30
|
|||
|
NO3
|
5
|
||||
|
Total
cation
|
280
|
Total
anion
|
280
|
||
|
Total Hardness as CaCO3(
Ca+Mg)
|
110
|
SiO2
|
10
|
||
|
%N a/TC
|
0.5
|
CO2
|
5
|
||
|
T.AlkalinityasCaCO3(HCO3+CO3+OH)
|
190.0
|
||||
|
EMA as CaCO3 (Cl+SO4+NO3)
|
95.0
|
||||
|
%alkalinity/TC
|
67.857
|
||||
|
%weak acids/T.A
|
5.357
|
||||
|
%Cl
|
21.428
|
||||
|
%SiO2/TA
|
1.785
|
||||
|
DESIGN
PARAMETERS
|
A
|
B
|
|||
|
Hourly
flow rate M3/Hr
|
1
|
15.8
|
|||
|
Service
cycle Hrs.
|
12
|
24
|
|||
|
Hourly
flow rate gallon/min
|
4.4
|
69.44
|
|||
Treated water Quality
Complete removal of
silica and CO2 not required
Conductivity
< 50 micro mhos
pH will not
remain constant and will vary during the service cycle between 6-10
Designing the system
|
Parameters
|
SAC
|
SBA
|
|
Type of resin
|
H+ resin
|
Type 2 Resin
|
|
Capacity of resin gmCaCO3/L
|
60
|
48.0
|
|
Regeneration Level Gm/L
|
100
|
96
|
|
Sodium leakage % of Total Cation
|
6
|
|
|
Total rinse M3/M3of
Resin
|
4.69
|
2.95
|
|
Displacement rinse BV
|
1.50
|
2.00
|
|
Final rinse M3/M3 of
resin
|
16.02
|
8.01
|
|
Backwash rate M3/Hr/M2
|
9.76
|
4.88
|
|
Regeneration rate
|
4.88
|
4.01
|
|
Bed depth M
|
1.0
|
1.0
|
Step By Step Calculation
In Ion exchange system
designing starts from designing the last unit first. Therefore we will design
the Anion unit first.
Step 1.
Calculating regeneration dilution requirement
4%
NaOH contains 41.76 gms NaOH / litre
50%
NaOH contains 763.68 gms NaOH / litre
4%
NaOH applied = 96 / 41.76 =
2.30 BV( or M3/Hr/M3 of Resin) -- -(eq 1) 50%
NaOH applied = 96 / 763.8 = 0.125 BV
( or M3/Hr/M3 of Resin) ---(eq 2)
Product dilution
water = eq1- eq2 = 2.30- 0.125 = 2.17 B
Step 2 Calculate Rinse requirement
Displacement Rinse
= 2.0 BV
Fast
rinse
= Total rinse - Displacement Rinse
= 2.95 – 2.0
= 0.95
Step 3 Calculating Resin Quantity
Actual Resin
Capacity = 48 gm CaCO3/L
Water used for
regeneration = 2.95 +2.30 = 5.25 BV
= capacity of anion
exchanger actually lost
See water analysis
T.Alk. + Cl + SO4 = 190+ 60 +30 =280 gm as CaCO3
Capacity lost due to
regeneration = 5.25*280/1000 = 1 .47
Therefore effective
resin capacity = 48.09 - 1.47= 46.62
Volume of Resin =
(Ionic load * M3/Hr * Service cycle) / Exchange Capacity
= (total
Anion ) * 1* 12 / 47.62
= (280)* 1*
12 / 46.62
= 280
*1*12 / 46.62
=
72 litres
Step
4 Check if single regeneration per day is
possible or not
Flow
rate
= 1 M3/Hr
Resin
Volume = .072 M3
Service
rate = 1
/ .072 = 13.68 M3/Hr/ M3 of resin
Which is within the
limit and hence 1 regeneration per day is
possible.
Step 5
Calculating vessel diameter
Bed
Depth = 1 M
Bed cross sectional
area = volume of resin / bed depth
= 0.72 / 1 = 0.72 M2
Area = 0.785 D2 Because
(p/ 4=3.14/4=0.785)
Therefore
D =
sq.rt ( area /0.785)
=
0.302 M
=
302 mm (say 300mm)
Step 6 Calculating Water Requirements
Backwash Flow (M3/Hr)
= Velocity * Area = 4.88* 0.072 = 0.
0.35136 M3/Hr
Backwash volume
(M3) = Flow * Time = 0.
0.35136* (15/60) = 0. 0.08784 M3
Because backwash time is
15 minutes
Regeneration
Dilution water
= Product Dilution water * Resin Volume
= 2.17* .072
=0.157
4% regenerant applied
4% NaOH
applied = (
-------------------------------------- ) *60
Regenerant flow rate
=
( 2.30/4.01)*60 = 34.397 minutes.
Displacement rinse
volume = Displacement rinse * Resin volume
= 2.0*
.072 = .144 M3
Displacement rinse
time = bed volume/ regenerant flow rate
=
(2/4.01)*60 = 30 minutes
Time For Displacement
& Regeneration = 34.39 + 30 = 65 minutes
Fast Rinse Volume
= Fast rinse BV * Resin Volume = 0.95* .072 =.068M3 m
Fast
Rinse time = Fast rinse BV/ Fast
Rinse Rate
= (.95/ 8.02)*60 =
7 minutes
|
Operation
time
water
used
Volume
|
|
Min
type
M3
|
|
Backwash
15
Cation
0.176
|
|
Regeneration
34 Cation
0.157
|
|
Displacement
30
Cation
0.108
|
|
Fast
Rinse
07
Cation
0. 100
|
|
|
|
Operation
time
water
used
Volume
|
|
Min
type
M3
|
|
Backwash
15
Raw
0.146
|
|
Regeneration
30
Raw
0.157
|
|
Displacement
18
Raw
0.090
|
|
Fast
Rinse
11
Raw
0.068
|
|
|
If you see in excel we have worked out
for the same water analysis a combination of SAC/WBA /DG (Check Alkalinity load
has not been considered) . The reason is here we are considering a flow rate of
1 M3/Hr.
Calculation For
Cation
Step 1.
Calculating regeneration dilution requirement
4%
HCl contains 40.72 gms
HCl/litre
30%
HCl contains 479.2 gms
HCl/litre
4% HCl
applied = 100 / 40.72
=2.45 BV( or M3/Hr/M3 of Resin)
---- (eq 1)
30% HCl
applied = 100 / 479.2 = 0.208 BV ( or M3/Hr/M3 of
Resin) ---- (eq 2)
Product dilution
water = eq1- eq2 = 2.45- 0.208 = 2.24 BV
Step 2 Calculate Rinse requirement
Displacement Rinse
= 1.5 BV
Fast
rinse
= Total rinse - Displacement Rinse
=
4.69 – 1.5 = 3.19
Step 3 calculating Resin Quantity
Actual Resin
Capacity
= 60gm CaCO3
Water used for
regeneration = 4.69 +2.45 = 7.14 BV
= capacity of
cation exchanger actually lost
See water analysis
Ca+Mg+Na+K = 85+25+140+30 =280 gm as CaCO3
Capacity lost due to
regeneration = 7.14*280/1000 = 1.99
Therefore effective
resin capacity = 60 - 1.99= 58.0
Volume of Resin =
(Ionic load * M3/Hr * Service cycle) / Exchange Capacity
= (
(total Cation)+1*12+ / 58)+(total cation)*(.5)/58
=
(280)* 1* 12 / 58 + (280)*0.5/58
=
58 + 2.4 =60.4 (say 60 litres)
Step 4 Check if single regeneration
per day is possible or not
Flow rate
=
1 M3/Hr
Resin
Volume = .060 M3
Service
rate = 1
/ .060 = 16.66. M3/Hr/ M3 of resin
Which is within the
limit and hence 1 regeneration per day is possible.
Step 5
Calculating vessel diameter
Bed
Depth = 1 M
Bed cross sectional
area = volume of resin / bed depth
= 0.060 / 1 = 0.060 M2
Area = 0.785 D2 Because
(p/ 4=3.14/4=0.785)
Therefore
D =
sq.rt ( area /0.785)
=
0.276 M
=
276 mm (say 300mm)
Step 6 Calculating Water Requirements
Backwash Flow (M3/Hr)
= Velocity * Area = 9.76* 0.060 = 0. 0.5856
M3/Hr
Backwash volume
(M3) = Flow *
Time = 0. 0.5856* (15/60) = 0.1464M3
Because backwash time is
15 minutes
Regeneration
Dilution
water = Product Dilution water * Resin Volume
= 2.17* .072
=0.157
4% regenerant applied
4% HCl applied
= (
-------------------------------------- ) *60
Regenerant flow rate
=
( 2.45/4.88)*60 = 30.12 minutes.
Displacement rinse
volume = Displacement rinse * Resin volume
= 1.5*
.060= . 0.09M3
Displacement rinse
time = bed volume/ regenerant
flow rate
=
(1.5/4.88)*60 = 18.44 minutes
Time For Displacement
& Regeneration = 30 + 19 = 49 minutes
Fast Rinse Volume
= Fast rinse BV * Resin Volume = 3.19* .060 =.068M3 m
Fast
Rinse time = Fast rinse
BV/ Fast Rinse Rate
=
( 3.19/ 16.01)*60 = 11 minutes
Designing a demineralizing system with SBA & WBA
Get a raw water
analysis
Turbidity
10
Color
2
Free Cl2
2
Organics
2
pH
7.2
Cation as CaCO3
Anion as CaCO3
Ca
150
HCO3 225
Mg
75
Cl 175
Na
230
SO4 65
Fe
0
NO3
0
SiO2
10
---------------------------
---------------------------------
Total
Cation
465
Total
Anion
465
Design treated water analysis
Conductivity
< 5 m mhos
pH
7 +- 0.2
SiO2
< 0.03ppm
Ion Exchange
scheme
MB, SBA, WBA, DG, SAC
Pretreatment
DMF , ACF
Regeneration
CO
FLOW
Design parameters
Minimum service cycle is
8 hours
Minimum Bed depth
is 1 meter
Maximum bed depth is
dictated by pressure drop but we have limited it up to 1.5 M
Linear Velocity is
between 15 to 45 M2/Hr/M2 .
Volumetric velocity
(space velocity) is between 8 to 40 M3/Hr/M3 .
Sufficient backwash
velocity to have 50% bed expansion.(see Resin Chart ).
Chemical injection
-- 4—6 % HCl and 4—7
% NaOH .
Chemical injection
rate 2 to 5 M3/HR/M3 For Anion & 4
to 10 M3/Hr/M3
Chemical displacement
rate is same as chemical injection rate.
Displacement volume 1.5
to 2 bed volume Fast rinse (final rinse
) Flow rate is the greatest of the three value Service
flow or 12 M3/Hr/M3 of
resin or 15 m3/Hr/M2
Fast rinse volume 5 to
10 bed volume.
Other parameters
required for designing Volumetric flow
rate M3/Hr Regenerant level gm /l
or Kg/M3
Operating capacity
|
Parameters
|
WAC
|
SAC
|
WBA
|
SBA
(type1)
|
SBA
(type2)
|
MB
|
|
Regeneration
level for Cation
|
110
|
80
|
80
|
|||
|
Regeneration
level for Anion
|
55
|
80
|
80
|
80
|
||
|
Exchange
Capacity for Cation
|
110
|
60
|
|
25
|
32
|
40
|
|
Exchange
Capacity for Anion
|
|
50
|
25
|
32
|
20
|
|
The above
capacities and regeneration level is based on the following ratio
Alkalinity/ TC, Na/TC ,
SO4/EMA , SiO2/TA , TH/Alkalinity
Pretreatment
See pressure
filters and activated carbon filters.
Ion exchange system
It is always better to
start from the last unit first but in this case we are starting
from SBA i.e. Strong base Anion . We are designing the mixed bed
separately because nowadays it is quite common to give mixed bed as a standalone
unit.
Calculation Methodology
Calculate the resin
requirement by the following formula
Flow X Service cycle X ionic load
Anion Resin quantity in
litres = ----------------------------------------------------
Capacity
25 X 24 X 38.5
In this
example
= -------------------------------
24
=
962.5 litres
Area =
.785 D2 Where D is the diameter
Linear Velocity
= Flow / Area
Area
= Resin Volume in sq. meters / Bed depth
Bed depth = 1 to 1.5
Area = 962.5*.001/
1 ( 1 Meter
is the bed depth)
Area = .962 M2
Linear Velocity
= 25 /.96 = 26.6
M/Hr ( within the limits )
Area
= pD2 / 4
or p/4 * D2
=
.785 D2
D2 =
Area/.785
D
= SqRt (Area/.785 )
D =
1.10 M or 1100 mm
Calculation for WBA resin
25 X 24
X (Total Anion – T alkalinity)
Resin
Quantity
=-------------------------------------------------------
Ex capacity
= 25
X 24 X (465-–225)
57.6
= 600 X
240 / 57.6
= 2500 litres
Keeping a fixed bed
depth of 1 M we get an area of
= 2.5 M2
Linear velocity =
25/2.5= 10 M2/Hr/M2
This is slightly on the
lower side , hence we will increase bed depth and reduce cross sectional Area.
We
will use a bed depth of 1.2
Area =
2.5/1.2 = 2.0
Linear velocity =
25/2 =12.5
Let us keep a bed depth
of 1.5 M
Then Area =
2.5/1.5 =1.7
Linear Velocity = 25/1.7
=14.7
Slightly less but ok
D2 =
sort( 1.7/.785)
D = 1.5 M =1500 mm
Thus it can be seen
that both velocities are function of flow , service cycle and bed depth.
Flow is dependentent on requirement and bed depth is restricted due to pressure
drop, hence it is by reducing or increasing the service cycle that one can get
the optimum linear and space velocities. In the above example if we have
to increase linear velocity to double it’s value, then we to reduce
service cycle duration by half. The advantage is less resin, smaller vessel
size but number of regeneration per day will increase
Any way without
changing any values in the above calculated values for SBA & WBA , we
similarly calculate Resin quantity for SAC resin
Flow X Service cycle X ionic load
Cation Resin quantity in
liters =
-------------------------------------------------------------
Capacity
=
25 X 24 X 455/60
=
4550 litres
=
4.5 M3
Space velocity = flow/
volume = 25/ 4.5 =5.55
Linear velocity will be
slightly more depending on the bed depth . If bed depth is 1M , the linear
velocity will be same as space velocity. That means service
cycle should be Half or 16 hours . But we are keeping the service cycle
same as for other two units i.e. is 24 hours and bed depth as 1.2 meter.
Then area = volume of
resin / bed depth = 4.5/1.2 =
3.75M2
Therefore diameter =
sqrt ( 3.75/.785) = 2.185
= 2.2 M (say)
Here the resin
quantity we have calculated is for net output and not for gross output . The
addition resin required for meeting the anion regeneration water can be
calculated separately . The other way is to calculate the entire
regeneration requirement of anion and then cation treated water quantity
required for the same is added to the net output.
Gross output = Net
output require + water for regeneration of succeeding ion exchangers.
After we have done our
preliminary design , now we will work out the details .
We start with backwash
flow for down flow units . For up flow packed bed units there is no backwash
stage.
Backwash flow =
Area X backwash velocity
Strong Anion
backwash = .948M2 X
6 M/Hr = 5.70 M3/Hr
Similarly
for Weak Anion backwash = 1.766 M2
X 6 M/Hr = 10.6 M3/Hr
Vessel volume
= Area * shell height
= .948
* (2) =
1.896 M3 for
SBA and
= 1.766 *
(2.2) = 3.8852 M3
for WBA
Backwash
time = Vessel volume/ Backwash flow
=
1.896/5.70 = 0.33 hours = .33*60 = 19.8
Minutes For SBA
=
3.885/ 10.6 = 0.36 hours =21 Minutes
Regenerant
Chemical Amount = Resin
(litres) X Regenerant Dose gm/litre
The minimum dilute flow
rate of regeneration flow is 2..00 BV/Hr for NaOH and 3-4 BV/Hr for
Cation. Regeneration strength for both is 4-5 % .
Optimal contact time is 60 Minutes .
Dilute Caustic
Rate = 2 BV/Hr for anion
The regeneration
of WBA & SBA Is thoroughfare and the alkali used
for SBA should be added to that calculated for WBA
Regenerant Chemical in
Kgs for WBA = 57.6 *2546 = 146600gms = 146.6 kgs
Regenerant Chemical in
Kgs for SBA = 80* 960= 76800gms = 76.8 kgs
Total Dilute
chemical
= 146.6 +76.8 =223.4 kgs
=
223400 gms / 41.6gms/litre =5370 litres =5.37 M3
=
(5.37/5)*60= 64 minutes
Displacement
Rinse 1 to 2 Bed volume is used
Displacement time
= 1 X Resin volume in M3 / Dilution
rate
=
( 3.5/5)*60 = 42 minutes
Note ;- We have
been conservative in this calculation . We have done the same calculation
in excel where you can change user values to slightly higher values.
Final Rinse or fast
rinse is calculated the greatest of the following figures .
Service flow
12 BV/ Hr
15 M3/Hr/M2
In our design we have
used the 15 M3/Hr/M2 throughout .
Weak Base Anion
fast rinse = 15 *1.766 = 26 M3/Hr
Strong base Anion
Rinse = 15* .948 =
14.2 M3/Hr
Volume of rinse water
required = 9 to 10 BV
WBA Rinse
Time = ( 10* 2.56 / 26)*60
= (25.6/26.4 )*60 = 58 Min
SBA Rinse
time = ( 10*.96 /14.2
)*60 = (9.6/14.2)*60 = 40 Min
Once you have
calculated the time you have to calculate the waste water . Water used for
anion is all cation water and hence the additional resin required
for this is calculated.
Regeneration flow and
time is calculated similarly for Cation unit . We leave this has an exercise
for You. We have done the complete calculation in Excel You can check
your answer there .
Mixed bed unit is a unit
in which cation and anion resin are in one unit. Mixed bed is generally
employed as a polishing unit. In earlier days it was generally employed with
other ion exchange units but today it is used alone for RO treated
water. Mixed bed unit is always a down flow unit and
free board is always taken as 100% .The cation and
anion load depends on the treated water used . If RO treated water is
used as feed Water for MB then the ionic load should be taken as 50
. For Ion Exchange treated water the load can be less than 35ppm.
Here we are taking 35
ppm ionic load both for Cation & Anion.
Volumetric flow
considered is 25M3/Hr & Service
time is 24 hours.
Minimum Bed depth
considered is 1 Meter & Maximum 1.5 Meter
Linear Velocity range
is 15 –45M3/Hr/M2
Space Velocity
range is 10—40 M3/Hr/M3
Step1
Finding Resin Quantity
Resin Quantity in
Litres = ( Ionic Load * Flow * Time) / Exchange capacity
Cation Resin
= (35*25*24) / (44 *.9)
=
530 litres = 530/1000= .530 M 3 .
Anion Resin
= (35*25*24) / (24 *.9)
=
972 litres = 972/1000= .972 M 3
Step 2
:- Finding Vessel Diameter by either by bed depth or by
linear velocity .
Area
= Resin volume / Bed
Depth =
(.53+.972)/1.2
= 1..25 M2
Area
= Flow/ Velocity =
25/25
= 1 M2
Diameter
= sqrt (area/.785)
For case
1 Diameter = 1.26 =
1400mm
A
For case
2 Diameter = 1.12 =
1100mm
B
Which one to select
?
A
Or B
Either of them is
correct but we have to select the most appropriate one required for our
need.
What will be the shell height
For case
1) Shell
height = Bed height + 100% free board
= 1.2
+ 100% * (1.2) = 2.4
For case
2) Bed
depth =
volume / area
=
1.5/1 =1.5
Now shell
height = Bed height + 100% free
board
=
1.5 + 100% * (1.5) = 3.0
Though diameter is
slightly larger in case1 it will preferable because shell height is
less & Linear & space velocity is within the limit.
Acid
required = Regenerant level *
Resin quantity
= 80*
.53 0 = 42 .4 Kgs
Caustic
required = Regenerant level *
Resin quantity
= 80*
.972 = 77.76 Kgs
Step 3 :-
Calculating Regeneration Schedule
Back washing
Flow = 8 M/hr *
1.25 = 10 M3/Hr
Back washing
time = 15 Minutes
Back washing
Volume = 10
*(15/60) = 2.5 M3
4% HCl
contains 40.7 gm/l HCl &
4% NaOH contains
41.6 gm/l NaOH
4%
Acid:
42400/40.6 =1044/1000= 1.044 M3
4%Caustic
77760/41.6 = 1869/1000 = 1.869 M3
Assuming 30 minute
contact time
Dilution (injection rate
) for NaOH = 1.869/30 = 1869 / 30 = 62.3 litres/min
Dilution
rate For
Acid
= 1.044/30 = 1044/30 = 34.8 litres/ min
Acid & Caustic
displacement rate
Voids =
40%*.972* 1 = 0.388 M3 (Chemical
is displaced for 1 bed Volume)
150 mm space above the
bed = .15 *1.25 =
0.187 M3
Total
= 0.388 +0.187 = 0.575M3 .
Keeping the same
dilution rate as that for caustic
Displacement Rinse for
Caustic = 613/62.5 = 9 Min
Similarly for
Acid = 40%
*.530 *1
=
.212 M 3
150 mm space above the
bed = .15 *1.5
=
.187 M 3.
Total
=0.212+0.187 = 0 .399 M 3.
Keeping the same rate as
that for acid injection
Displacement Rinse for
Acid = 399/34.8 = 12.0 Min
Note : Here
we have assumed the contact time should be minimum 30 minutes .
Step
5:- Drain Down
Draining is done to the
top of bed and takes about 15 minutes .
Step 6 :-
Air Mix
Air mix is done by
using 180M3/Hr per M 2
with enough pressure to overcome the height of water and that of resin.
Step 7 :-
Refill
The vessel is filled
with water
Step 8 :- Final
Rinse
Calculating the total
volume of water required per rinsing 4 Bed Volume
Flow rate required is
15M3/Hr/M2 of area.
15M3/Hr/M2 *
1.25 M2 =
18.75 M3 / Hr
Volume of water required
is = 4 * 1.5 = 6 M3 .
Time for
rinsing
= 6 / 18.75 =
.32 hours = .32*60 = 19.2 Minutes..
Note ;- For acid &
caustic injection we started our calculation with minimum contact
time of 30 Minutes . We could have found out the time for
injection by assuming injection rate
For caustic at 3.5 BV/Hr
and 4 BV/Hr For acid .
Dilute caustic rate =
3.5*.972 =3.402 M3/Hr
Time for injecting 4 %
Caustic = (1.869 / 3.40) *60= 32.9 Min ( Ignoring other
factor)
Similarly for acid
injection
Dilute acid rate
= 4*.530 =2.12 M3/Hr
Time for injecting 4%
Acid = (1.044/2.12 )*60 = 28.30 minutes
[d1] The residence time (contact time)
in packed bed reactors is often characterized by empty bed contact time
(EBCT) which is simply the depth of media divided by superficial velocity or
the total bed volume divided by the volumetric flow rate (v/Q)
