# Waste water treatment: Equalization filtration

Physical Treatment Processes
Process

Purpose

Equalization

Equalize flow entering wastewater treatment plants
Equalize a pumping rate into a storage tank before
gravity supply of water.

Bar Screens

Entrance of wastewater treatment plants and some
water treatment plants

Mixing

Rapid
Slow: flocculation

Settling

Type I: Grit removal
Type II: Low density flocs removal
Type III and IV: high density flocs removal

Filtration

Removal of low levels of suspended solids.

Desalination

To remove dissolved solids.

Equalization Tanks
Equalization tanks are used to dampen fluctuation in flow rates coming
into the WWTP or to pump a constant flow from a WTP

Flow, gallon per minute

1200

800

Water pumping out
of the plant

400

0
0

2

4

6

8

Flow, gallon per minute

Water Treatment Plant
1200
800

Water consumption

400
0
0

10 12 14 16 18 20 22 24

2

4

6

8

10 12 14 16 18 20 22 24
Time, hr

Time, hr

Flow, gallon per minute

1200
800

400

Wastewater generation

0
0

2

4

6

8 10 12 14 16 18 20 22 24
Time, hr

Flow, gallon per minute

Wastewater Treatment Plant
1200

800

Wastewater pumping into the
plant

400

0
0

2

4

6

8

10 12 14 16 18 20 22 24
Time, hr

In water treatment plants
ET

Raw
water

To
consumers

Water Treatment Processes
Pump

In wastewater treatment plants
PS

GC

SS

Cl

ET
BS

AT

Pump

Example
Consider the data shown for the
hourly consumption of water in a
typical city. Determine the volume
of the equalization tank using the

Time

Consumption
gpm

01-02
02-03
03-04
04-05
05-06
06-07
07-08
08-09
09-10
10-11
11-12
12-13
13-14
14-15
15-16
16-17
17-18
18-19
19-20
20-21
21-22
22-23
23-24
24-01

866
866
600
634
1000
1330
1830
2570
2500
2140
2080
2170
2130
2170
2330
2300
2740
3070
3330
2670
2000
1330
1170
933

Solution
1. Find the volume
consumed during each
hour (V=Q*t where t is 60
minutes in this example).
This is column 3.
2. Find the average of the
volume consumed
(=111.9 thous gal).
3. Subtract the average
from the volume
consumed (column 4).
4. Place the positives in
one column and the
negatives in another
(column 5 and 6).
5. Sum either the positive
or the negative to obtain
the capacity (Capacity=
485.4 thous gal).
purposes: Design
capacity= 1.2*485.4
thous gal

Time
01-02
02-03
03-04
04-05
05-06
06-07
07-08
08-09
09-10
10-11
11-12
12-13
13-14
14-15
15-16
16-17
17-18
18-19
19-20
20-21
21-22
22-23
23-24
24-01

Consumption

Volume
consumed

Average- volume
consumed

Positive

Negative

gpm

Thous gal

thous gal

thous gal

thous gal

866
866
600
634
1000
1330
1830
2570
2500
2140
2080
2170
2130
2170
2330
2300
2740
3070
3330
2670
2000
1330
1170
933
Average

52.0
52.0
36.0
38.0
60.0
79.8
109.8
154.2
150.0
128.4
124.8
130.2
127.8
130.2
139.8
138.0
164.4
184.2
199.8
160.2
120.0
79.8
70.2
56.0
111.9

59.9
59.9
75.9
73.9
51.9
32.1
2.1
42.338.116.512.918.315.918.327.926.152.572.387.948.38.132.1
41.7
55.9

59.9
59.9
75.9
73.9
51.9
32.1
2.1

Sum

485.4

42.338.116.512.918.315.918.327.926.152.572.387.948.38.132.1
41.7
55.9
485.4-

Tank Capacity-Graphical Method
Time

Consumption

Volume
consumed

Cummulative

gpm

thous gal

thous gal

The y-axis represents the
cumulative volume (column 4)

0
01-02
02-03
03-04
04-05
05-06
06-07
07-08
08-09
09-10
10-11
11-12
12-13
13-14
14-15
15-16
16-17
17-18
18-19
19-20
20-21
21-22
22-23
23-24
24-01

866
866
600
634
1000
1330
1830
2570
2500
2140
2080
2170
2130
2170
2330
2300
2740
3070
3330
2670
2000
1330
1170
933

52.0
52.0
36.0
38.0
60.0
79.8
109.8
154.2
150.0
128.4
124.8
130.2
127.8
130.2
139.8
138.0
164.4
184.2
199.8
160.2
120.0
79.8
70.2
56.0

52.0
103.9
139.9
178.0
238.0
317.8
427.6
581.8
731.8
860.2
985.0
1115.2
1243.0
1373.2
1513.0
1651.0
1815.4
1999.6
2199.4
2359.6
2479.6
2559.4
2629.6
2685.5

For design: The tank capacity is
increased by 20% over the one found
from the graph or the spreadsheet
method.

Filtration
Objective: to remove turbidity (suspended solids) from water

Turbidity of water
> 4NTU

Mixing

Flocculation

Sedimentation

Filter

Turbidity of water < 4NTU
Filter

Granular Media filter

Flow through filter = 2- 10 gpm/ft2

Dual Media filter

Anthracite (coal)
Specific gravity 1.4-1.6
Effective size 0.9-1.1 mm
Uniformity coefficient <1.7
Sand
Specific gravity 2.65
Effective size 0.45-0.55 mm
Uniformity coefficient <1.7
Coarse sand
Fine to coarse gravel
Underdrain

Example
A filter has a surface area of 15 ft by 30 ft and filters 2.5 mil gallon
every day. After 24-hr of filtering, the filter is backwashed at a rate of
15 gpm/sq ft for 12 min. Determine the average filtration rate per unit
. area and the quantity and percentage of wash water used every day

Solution
Q (2.5 ×106 gal / day )(day / 24 × 60 min)
Filtration rate =
=
= 3.9 gpm / ft 2
As
15 ft × 30 ft

Quantity of washwater used daily = q × As × t
= (15 gpm / ft 2 )(15 ft × 30 ft )(12 min) = 81,000 gal
81000 gal
Percent of wash water =
× 100 = 3.2%
6
2.5 × 10

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