Waste water treatment: Equalization filtration

Physical Treatment ProcessesEqualization Equalize flow entering wastewater treatment plants Equalize a pumping rate into a storage tank before gravity supply of water.. Bar Screens Entra

Physical Treatment Processes

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

Slow: flocculation

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

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

Equalization Tanks

0 400 800 1200

0 2 4 6 8 10 12 14 16 18 20 22 24

Time, hr

0

400

800

1200

0 2 4 6 8 10 12 14 16 18 20 22 24

Time, hr

Water Treatment Plant

Water pumping out

of the plant

Water consumption

0

400

800

1200

0 2 4 6 8 10 12 14 16 18 20 22 24

Time, hr

0 400 800 1200

0 2 4 6 8 10 12 14 16 18 20 22 24

Time, hr

Wastewater Treatment Plant

Wastewater generation Wastewater pumping into the plant

In wastewater treatment plants

BS

ET GC

PS AT SS

Cl Pump

Raw

water

ET

Water Treatment Processes

Pump

To consumers

In water treatment plants

Time Consumption

gpm

Tank Capacity-Spreadsheet Method

Example

Consider the data shown for the

hourly consumption of water in a

typical city Determine the volume

of the equalization tank using the

spreadsheet method

Time Consumption

Volume consumed

Average- volume consumed Positive Negative gpm Thous gal thous gal thous gal thous gal 01-02 866 52.0 59.9 59.9

02-03 866 52.0 59.9 59.9 03-04 600 36.0 75.9 75.9 04-05 634 38.0 73.9 73.9 05-06 1000 60.0 51.9 51.9 06-07 1330 79.8 32.1 32.1 07-08 1830 109.8 2.1 2.1 08-09 2570 154.2 42.3- 42.3 -09-10 2500 150.0 38.1- 38.1 -10-11 2140 128.4 16.5- 16.5 -11-12 2080 124.8 12.9- 12.9 -12-13 2170 130.2 18.3- 18.3 -13-14 2130 127.8 15.9- 15.9 -14-15 2170 130.2 18.3- 18.3 -15-16 2330 139.8 27.9- 27.9 -16-17 2300 138.0 26.1- 26.1 -17-18 2740 164.4 52.5- 52.5 -18-19 3070 184.2 72.3- 72.3 -19-20 3330 199.8 87.9- 87.9 -20-21 2670 160.2 48.3- 48.3 -21-22 2000 120.0 8.1- 8.1 -22-23 1330 79.8 32.1 32.1

23-24 1170 70.2 41.7 41.7 24-01 933 56.0 55.9 55.9

Average 111.9

Sum 485.4 485.4

-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)

6 Add 20% for design

purposes: Design

capacity= 1.2*485.4

thous gal

Tank Capacity-Graphical Method

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

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

Time Consumption consumedVolume Cummulative

gpm thous gal thous gal

0 01-02 866 52.0 52.0

02-03 866 52.0 103.9

03-04 600 36.0 139.9

04-05 634 38.0 178.0

05-06 1000 60.0 238.0

06-07 1330 79.8 317.8

07-08 1830 109.8 427.6

08-09 2570 154.2 581.8

09-10 2500 150.0 731.8

10-11 2140 128.4 860.2

11-12 2080 124.8 985.0

12-13 2170 130.2 1115.2

13-14 2130 127.8 1243.0

14-15 2170 130.2 1373.2

15-16 2330 139.8 1513.0

16-17 2300 138.0 1651.0

17-18 2740 164.4 1815.4

18-19 3070 184.2 1999.6

19-20 3330 199.8 2199.4

20-21 2670 160.2 2359.6

21-22 2000 120.0 2479.6

22-23 1330 79.8 2559.4

23-24 1170 70.2 2629.6

24-01 933 56.0 2685.5

Turbidity of water < 4NTU

Filter

Objective: to remove turbidity (suspended solids) from water

Filter

Turbidity of water

> 4NTU

Flow through filter = 2- 10 gpm/ft2 Granular Media filter

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

6

/ 9

.

3 30

15

min) 60

24 / )(

/ 10

5 2

(

ft

gpm ft

ft

day day

gal A

Q rate

Filtration

s

=

×

×

×

=

=

gal ft

ft ft

gpm

t A q

daily used

washwater of

000 ,

81 min)

12 )(

30 15

)(

/ 15

=

×

×

=

% 2 3

100 10

5 2

81000

×

water wash

of Percent

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

Example

Solution