Filtration

PHEH 401 for SAN4

Asst Prof Shalasai Huangprasert

Dept.of Env H Sc, Public Health, Mahidol U.

1. Purpose

 

·        The fundamental system that removes particulate matter is filtration.

·        The most common filtration process employs a granular media of a certain size and depth.

·        The pre-treated water, typically pre-treated by chemical addition, coagulation and sedimentation passes through the filter bed where a majority of the particulates are removed in the top portion of the filter media as well as throughout the entire depth of the bed.

·        Under certain raw water conditions, adequate treatment of the raw water can be carried out in the filter alone, and the need for the ordinary flocculation and sedimentation processes may possibly be eliminated. This process is called direct filtration.

 

 

2. Considerations

 

·        Site Topography. The ideal plant site has a constant slope of 2-3% which allows the filter wash tank, filters and wash-waste holding tank to be easily situated and economically built since there is no need for excessive excavation.

·        Plant Size. Very large plants, > 200MGD, can have their total number of filters exceed 30. The total number of filters my be reduced if the filter size is increased, up to 200ft², or by using a high filtration rate such as 8 gpm/ft² or both.

·        Raw Water Quality and Type of Pretreatment. If the underground water contains high levels of iron or manganese, the best method for removing these compounds is by oxidation with chlorine or potassium permanganate, followed by pressure filtration.

 

 

In the soil (insoluble)		In the water (soluble)		In the water (when O2 available)
Fe+++	®	Fe++	®	Fe+++
Mn+4	anaerobic conditions such as in a ground water aquifer	Mn++	O2 when water reaches surface air	Mn+4 colloidal precipitate

 

Example:

Given: Well water. 3.2mg/l of Fe^{++  and} .8mg/l of Mn⁺⁺  at pH 7.8. 1mg/l of potassium permanganate will oxidize 1.06mg/l of iron and .52mg/l of manganese in accordance with the following equations:

3Fe⁺⁺ + MnO₄ ® 3Fe⁺⁺⁺ + MnO₂, pH>7.5

3Mn⁺⁺ + 2MnO₄ ®  5MnO₂, pH>9.5

Find: the potassium permanganate treatment requirements

KMnO₄ (for iron) = 3.2mg/ Fe^{++  x} 1KMnO₄ /1.06mg/l of Fe⁺⁺

KMnO₄ (for iron) = 3.02mg/l

KMnO₄ (for Mn) = 0.8mg/ Mn⁺⁺ x 1KMnO₄ /.52mg/l of Mn⁺⁺

KMnO₄ (for Mn) = 1.54mg/l

Total KMnO₄ requirement = Fe+Mn = 3.02mg/l + 1.54mg/l

Total KMnO₄ requirement = 4.56mg/l

·        Allowing for Future Filter Modifications or Additions. In many cases, water treatment plants “grow” by adding basins and filters in stages. The filter influent channel, effluent channel and backwash system must be designed to accommodate the new flow rate, based on the original plant specifications. In the future, GAC, granulated activated carbon, may be required to adsorb objectionable organic compounds. GAC beds require a 15-30 minute EBCT, empty bed contact time, necessitating a depth of 6-8’ which can not be provided by ordinary filters.

·        Filter Washing System. Four basic schemes: elevated wash tank, direct pump, self-backwash (Greenleaf) and continuous backwash. Elevated tank and direct pump are the traditional and therefore proven systems. With respect to the washing system, the basic alternatives are : backwash alone, ordinary air-scour wash, and a combination.

·        Filter Rate Control: There are 2 modes of filtration: constant rate and declining-rate. Both are capable of producing <1.0NTU effluents. There are 4 basic types of rate control systems: 1.constant rate with a flow meter and flow modulation valve;2. constant level filtration with equal flow splitting inlet weirs, water level sensor and flow modulator valve; 3.declining rate filtration with a submerged inlet; 4.constant rate filtration with equal flow spitting inlet weirs and a weir to control the inlet level.

·        Type of Filter Bed. Contingent upon the availability of bed material.  Choices include: sand, anthracite coal, GAC and garnet. Third world counties use sand and gravel because other types are too expensive. If the main purpose is the filtration of sewage as in a tertiary treatment process, a monosand or mono-anthracite coal is best. For slow sand filtration, the filtration rate is less than .2gpm/ft². Rapid sand filtration rates range from 2-5gpm/ft². High rate beds are from 5-10gpm/ft² and always consist of a reverse graded filter bed or a deep large grain monomedium bed.

Example:

Given: The village of Salimati in east Africa decides that they need filtration. The population is 1567 people @ 15gpcd.

Find: Estimate the size of the filter beds

Filter area = 1567 cap x 15gal/cap.day x 1day/24hours x 1hour/60min / 2gpm/ft² 

Filter area = 81.6 ft²

·        Chemical Application Points. The most common chemicals added are: disinfectants, filter aids, adsorbents (for taste and odor control) and alkali chemicals for pH control. The three major concerns regarding chemical additions are: minimize the application points; certain chemicals have the potential to break through to the treated water; ensure the chemicals are adequately distributed in the water.

·        Miscellaneous Items. The wash troughs are an essential part of the filter design due to the predominant use of a high backwash rate. Gravity filters generally have an available headloss of 8-10’. The 4 essential factors in selecting the filter underdrain system are : uniform flow distribution of the backwash, durability, reliability and cost effectiveness.

 

3. Type and Selection Guide

 

A. Available Alternatives.

 

·        There are three basic types of granular medium filters: slow sand filters, rapid sand filters and high rate filters.  In addition there are pressure filters and proprietary filters.

·        Coagulation and sedimentation pretreatments are common to remove excessive amounts of suspended solids, color and some minerals.

·        A common design flaw is that the pretreatment does not match the filter requirements.

·        Bench scale or pilot studies are helpful to evaluate the treatability and seasonal variations involved with the raw water.

B. Selection Guide

 

·        Type of Filter. If the plant is small,<15MGD, proprietary systems may be considered. Proprietary systems have several advantages: “plug and ‘em’”  application, lower capital cost, easy operation and maintenance, acceptable effluent quality, shortened design and construction periods. The main disadvantage is a shortened life span of 20-30 years. Local plants such as Weymouth have been humming along for over 50 years and there is no end in sight. The most common types of filters plants >15MGD are gravity downflow and the dual media, sand and coal, filters.

·        Size and Number of Filters. The practical maximum size of an individual filter is 1500ft². A minimum of 4 filters preferably 6 should be provided for medium to large size plants. If a self-backwash type of filter is selected, the minimum number of filters per module must be 4 regardless of plant size. If a plant is designed with a small number of filters the filtration rate in the remaining filters substantially increases whenever one or two of the filters are place off-line due to washing or repair. The engineer must select an adequate number of filters and the proper size for each filter, so that the filtration rate of the remaining units will not be excessive when at least one is out of service. It is recommended that no more than 33% and preferably a maximum of 15-20% of hydraulic surcharge be allowed to flow to the remaining filters whenever one unit is off-line.

·        Filtration Rate and Terminal Headloss. Rates of over 10 gpm/ft² are impractical because of the high head losses as opposed to the quality of the filter water. When processing physically weak floc such as alum floc, the quality of the degrades if the filter rate rises above 4 gpm/ft²  a phenomenon exacerbated by cold weather and for rapid sand filters. Consequently, a designed filtration rate of 6 gpm/ft² is recommended for high rate filters and 3 gpm/ft² for rapid sand filters. The available head loss for filtration is generally designed to be 8-10’ for gravity filters but only 1’ for automatic backwash filters which are pre-designed by equipment manufacturers.

Example:

Given: A bed run at 5 gpm/ft² and 10 gpm/ft².

Find: The difference in headloss after 15 hours.

From F 3.2.7-3, p.210

headloss for 5 gpm/ft²: .8m

headloss for 10 gpm/ft²: 1.7m

Dheadloss = 1.7-.8

Dheadloss = .9m, over 100% difference for the same time period

·        Filter Flow Rate. Constant rate and declining rate. In recent years, the most popular scheme is the constant-rate filtration system since it provides better operational control, has proven performance and is overwhelmingly preferred by plant operators. Declining rate units lack positive filtration control.

·        Depth, Size and Composition of the Filter Bed. Three methods for sizing: pilot studies, existing data from similar facilities, depth:size ratios.  Ordinarily high rate filters have a porosity ratio of .45, a sphericity of .8 and a bed depth to effective media size ratio of 1000. Thus, the total surface area of the filter media grains per unit area is approximately 3000 ft² in area per square foot of bed area.  If the unit is dual media should have the same terminal velocities as determined by the following formula:

p. 212  = []^0.667

            Where d are the particle sizes and r are the densities.

Example:

Given: The size of the sand is .50mm and its density is 2.60. The density of the anthracite is 1.60. The fluid is water at density 1.

Find: What size coal particle is required to ensure the same settling velocity

 = []^0.667

 =[ ]^0.667

d₂ = .96mm

 

 

·        Filter Wash Systems and Wash Troughs. Two basic types of wash systems: fluidized bed backwash with or without surface wash and secondly, air-scour with partial fluidized backwash. Fluidized bed backwash with or without surface wash is used extensively in this country and Japan. The backwash troughs are arranged 6-8’ apart and 2.5-4’ above the filter. The second type is used in Europe.  According to the “Ten State Standards”, the distance for the top of the trough to the surface of a rapid sand filter bed should be the equivalent of the rate of backwash expressed in inches per minute. The troughs should be spaced so that the floc does not travel more than 3 ft horizontally before reaching a trough. The “Ten State Standards” criteria are OK when designing rapid sand filters but are not adequate for high rate filters with multi media beds. Therefore, for multimedia filter beds, the top of the troughs should be located 3.5-4ft above the surface of the filter bed and should be spaced 6-10 ft apart. The troughs should accommodate a flow rate for both backwash and surface wash in addition to 4-6” free fall into the main collection gullet at the lower end. The required height for a trough with level inverts and a rectangular cross section is:

H₀ = [ ]^0.667 English, H₀ = [ ]^0.667 metric units p.216

            A freeboard of at least 3 inches would be added to h₀.

Example:

Given: A 2’ wide wash trough that is carrying 3.5MGD.

Find: How high is it?

H₀ = [ ]^0.667 = [ ]^0.667

H₀ = 1.057ft + 3” Freeboard

H₀ = 1.307ft = 15.684 inches, pick the next larger even size made by the manufacturer

·        Filter Underdrains. Based on whether or not air-scour is chosen as the filter washing system. The strainer system is the predominant type of air-scour wash filter underdrain because it is effective and dependable.

·        Auxiliary Scouring of the Filter Wash. The 2 basic types of surface wash systems are : fixed grid and rotating arms. Auxiliary scouring is absolutely essential during filter backwash whenever coagulants are used in the pretreatment process, especially if polymer is used as a filter aid. Backwash alone cannot maintain the filter beds in a reasonably clean condition.  Regular surface wash systems are capable of adequately cleaning rapid sand and dual media filters.

·        Appurtenant Systems. The major appurtenances are: filter backwash tank, pumps, auxiliary scouring system, wash flow control system and wash-waste handling system. The backwash waster tank should hold at least 2 backwashes. The lowest water level in the tank should be about 40ft above the top of the wash-water troughs located in the filters. The fixed grid surface wash can take water from this tank, but the rotating arms require 80psi and require booster pumps to achieve this pressure. The source of the backwash water may be: the main distribution line, the water pumped from a clearwell, the effluent from the rest of the on-line filters. The elevated tank is the most common source of backwash.

 

4. Hydraulics of Filtration

 

            The hydraulics of filtration may be based on the Carmen-Kozeny equations. Other equations such as the Rose equation are available.

Consider the Darcy-Weisbach equation:

h_l = f

in which:

h_l = frictional headloss, ft

f = dimensionless friction factor

L = conduit length, ft

f = diameter of conduit, ft

v = mean velocity, fps

g = acceleration due to gravity, 32.2 ft²/s

            Since the flow path through the sand is not a conduit or pipe, but irregular, f=4R in which R equals the hydraulic radius. The length of the conduit, L, is more aptly described by D, the depth of the bed. Making these 2 substitutions:

 

h_l = f

            If there are n particles each with V_p , the total volume of the particles is N V_p . If the porosity, e, the total bed volume is:

V_total(total bed volume) =

            The total channel volume is the void space or the void ratio times the total volume:

eV_total = e

 

            If S_p is the surface area i.e. the wetted surface, the total surface area equals nS_p. Then the hydraulic radius, R, is the total bed volume divided by the wetted surface:

R = bed volume/wetted surface = e / n S_p

 = volume of sphere = f³

 = surface area of a sphere = pf²

R =   f³ /  pf²

R = f/6 for spherical particles

For non-spherical particles

R = kf/6 

where:

k=1 for spheres

k=.73 for crushed coal and angular sand

k=.82 for rounded sand

k=.75 for average sand

           

            The approach velocity, v_A = Q/A,

where:

Q=flow through the filter, cfs

A=filter surface area, ft²

Then the velocity through the pores, which is the velocity in the equation is:

v = v_A/e

Note: If e=100%, no sand, v = v_A/e, v = v_A/1, v = v_A

If e=50%, 1/2 sand, v = v_A/.5, v = v_A/.5, v = 2v_A

If e=1%, 99% sand, v = v_A/.01, v = 100v_A

Substituting into the original equation, h_l = f

h_l =

if f’=f/8 and rearranging:

h_l = f’ for uniform sand

 

 

f’=150(1-e)/N_r + 1.75 where

N_r=Reynolds number = kfv_A/n

where n=kinematic viscosity, ft²/s, p.604

For mixed beds:

where:

x=weight fraction of the particles between adjacent sieve sizes, apply as a fraction

d=geometric mean size of sieve openings, ft

h_l = f’ , for mixed beds

For stratified beds:

 =

h_l = , for stratified beds

Example:

Given: Rapid sand filter. D=20”. e=.42, T=55°F. filtration rate = 2gpm/ft². Spherical particles.

Sieve Analysis	% of sand retained given as a %, apply as a decimal	Geometric mean size, f (ftx10-3)
14-20	1.05	3.28
20-28	6.65	2.29
28-32	15.70	1.77
32-35	18.84	1.51
35-42	18.98	1.25
42-48	17.72	1.05
48-60	14.24	.88
60-65	5.15	.75
65-100	1.66	.59
	å=100%

 

Find: headloss using the Carmen-Kozeny equation

 

filtration rate = 2gpm/ft² x ft³/7.48gal x 1min/60s

v_A = 4.46 x 10^-3 fps

T=55°F=12.8°C

n = 1.304 x 10^-5 ft²/s p.604, interpolate

Sample calculations:

N_r=Reynolds number = kfv_A/n

N_r(f=3.28) = kfv_A/n = (1)(3.28x10^-3 ft)(4.46 x 10^-3 fps) / 1.304 x 10^-5 ft²/s

N_r(f=3.28) = 1.12

N_r(f=2.29) = kfv_A/n = (1)(2.29x10^-3 ft)(4.46 x 10^-3 fps) / 1.304 x 10^-5 ft²/s

N_r(f=2.29) = .783

etc.

f’=150(1-e)/N_r + 1.75

f’(f=3.28) = 150(1-.42)/1.12 + 1.75

f’(f=3.28) = 77.6

f’(f=2.29) = 150(1-.42)/.783 + 1.75

f’(f=2.29) = 114

etc.

 (f=3.28) = 77.6(.0105)/.00328

 (f=3.28) = 248

 (f=2.29) = 114(.0665)/.00229

 (f=2.29) = 3310

etc.

 

 

 

Nr Reynolds number  kfvA/n	f’ 150(1-e)/Nr + 1.75	x	Geometric mean size, f (ftx10-3)
1.12	78	.0105	3.28	248
.78	114	.0665	2.29	3310
.60	146	.1570	1.77	12,900
.51	169	.1884	1.51	21,100
.43	207	.1898	1.25	31,400
.36	245	.1772	1.05	41,300
.30	293	.1424	.88	47,500
.26	343	.0515	.75	23,600
.20	437	.0166	.59	12,300
		å=1.00		å=193,000

 

h_l = , for stratified beds

h_l =  x 193,000

 

h_l = 1.56ft

5. Operation and Maintenance

 

·        Regular Filter Performance. Three indicators are: turbidity of the effluent, <.5NTU; time of filter run, an average of 2 days for dual media filters; volume of backwash water to the volume of filtered water, <3%, 2% is very good, >5% is poor.

Example:

Given: Weymouth Q=600MGD

Find: What is the volume of backwash water, MGD, assuming that the backwash operation is very good.

2% is very good

Q_backwash = Q x .02 = 600 MGD x .02

Q_backwash = 12 MGD

Note: This amount although only 2% may equal the entire capacity of a plant in Iowa.

·        Pretreatment Optimization. The target turbidity is the LA is £.5 NTU. Two methods to optimize filter performance: coagulant type and amount: mixing level and time. A combination of alum ~2-4mg/l and cationic polymer, ~.5mg/l is used because they gives good results and is cost effective. Whenever the raw water has a pH >7.5, and an alkalinity >50mg/l, ferric iron salts should be evaluated as the primary coagulant.

 

Example:

Given: The above plant at alum of 3mg/l

Find: How much alum do they use per day.

 

 

 = X mg/l x 8.34  x Q(MGD)

 = 3 mg/l x 8.34  x 600(MGD)

 = 15,012 lb/day = 7.51 tons/day

Note: QADF was used, report the answer in units that mean something

·        Evaluation and Adjustment of the Filter Washing Procedure. Three methods to evaluate the effectiveness of the washing procedure: visually inspect the bed before and after washing; measuring the turbidity of the backwash water at 1 minute intervals: core sample the bed before and after the wash. A filter bed is clean but also ripened i.e. the media grains of a ripe filter are coated with the proper amount of floc or polymer. The operators look for mud balls, cracks and worms and debris on the surface. Overwashing may be detrimental; the wash may be terminated when the turbidity of the wash waste ranges from 10-15NTU which usually occurs after 5-6 minutes of non air-scour washing.

·        Mud Deposition Profile. A mud profile is bed depth vs. floc deposition p.208 measured as turbidity. A filter bed is properly conditioned if it is in a ripened stage: each grain is coated with a thin film of coagulant hydroxide or polymer. A bed that is too clean usually exhibits a distinct turbidity breakthrough at the beginning of each filter cycle lasting anywhere for 30-60 minutes. A turbidity of > 300NTU is indicative of a mud ball problem. When mud balls are found in the filter bed, they are measured by % volume. If the percentage is < .1%, the filter bed is clean; 1-5% indicates a bad condition. If the volume is > 5%, the bed must be replaced with new media.

·        Air-Binding. Air-binding is when large amounts of air bubbles accumulate in the filter bed. Air-binding may be alleviated by initiating a filter wash whenever the filter headloss reaches 4-5ft a practice which prevents the creation of negative pressure and the concomitant negative pressure in any part of the filter bed.

6. Design Criteria

See information starting on p.225 including T3.2.7-1, p.234

7. Example

 

Given: A 75MGD water treatment plant using a conventional process.

Find:

1.) number of filters

2.) filter bed, filtration rate, filter washing scheme

3.) size of filters

4.) filter arrangement

5.) filter media

6.)backwash rate, underdrain system, headloss through each orifice(<1ft)

 

1.) number of filters

N=1.2Q^.5 p.230

N = 1.2(75)^.5

N = 10.39, say 10. Leave room for future expansion.

 

2.) filter bed, filtration rate, filter washing scheme

Filter bed-standard dual media. Standard choice, no reason for choosing an alternative.

Filtration rate-From 5-10 gpm/ft², use 6 gpm/ft² p.210

Filter washing scheme- Elevated tank since the site has a nice hill

 

3.) size of filters

Filter area =  Q/Rate = 75MGD x 694.4gpm/MGD / 6 gpm/ft²

Filter area =  8680ft²

Individual filter bed area = Filter area / number of filters =  8680ft² /10

Individual filter bed area = 868ft²

Each filter will be divided into two filter cells and a central gullet. See note p.230.

Each filter cell = 868ft²/2

Each filter cell = 434ft²

From p.210

W=10-20ft,

L:W::2:1 to 4:1, 3:1 average

Area of filter cell 250-1000ft², 600ft² average

D=12-20ft, 17ft average

A = LW= 434ft²

L = 3W, midrange ratio

W(3W)=434ft²

W= 12.0ft

L =36.0ft

D = 17 ft (average p.210)

A actual = LW = 12x36

A actual = 432 ft²

Check the actual filtration rate, based on the actual dimensions:

filtration rate, actual = 75MGDx 694.4gpm/MGD / (432 ft² x 10filters x 2cells/filter)

filtration rate, actual = 6.03 gpm/ft² OK

 

4.) filter arrangement, see pps. 242-243

 

5.) filter media

The specifications for the filter media are as follows:

T3.2.7-1 p.234-235

Sand, bottom layer:

effective size: .45-.65, say .55

UC: 1.4-1.7, say 1.5

Depth: 1ft

Anthracite coal, top layer:

effective size: .9-1.4, say 1.2

UC: 1.4-1.7, say 1.6

Depth: 1.5ft

SG³ 4.0-4.1

 

6.)backwash rate, underdrain system, headloss through each orifice

The backwash flow rate = 18-22gpm/ft², say 20gpm/ft² p.213

Qbackwash = 20gpm/ft² x 868ft²

Qbackwash = 17,360 gpm

 

The underdrain system of choice is precast concrete laterals with a triangular cross section See F3.2.7.9, p.219. Precast concrete laterals are 14.5ft long, placed at 12” (1’) intervals and both side of the sloped walls have orifices that .5” in diameter and 3” OC.

Each filter cell is 36’ long.

Each length can accommodate 36’/1’ interval = 36 laterals

Each filter consists of 2 filter cells = 36 laterals x 2 cells/filter = 72 laterals total

 

Maximum backwash rate per filter = 17,360 gpm / 72 laterals total

Maximum backwash rate per filter = 241 gpm =.537cfs

The cross sectional area of the lateral , from manf = (.615x.6)/2 =.185ft²

v at the entrance to the lateral = Q/A = .537cfs / .185ft²

v = 2.90fps < 4.5fps, OK

 

The total number of orifices per lateral:

14.5’(length of lateral)/ 3” OC = 58 x 2 sides/lateral = 116 orifices per laterals

Qorifice = .537cfs / 116

Qorifice = .00463cfs

Aorifice = D² = .785(.5/12)²

Aorifice = .00136ft²

vbackwash through each orifice = Q/A = .00463cfs / .00136ft²

vbackwash through each orifice = 3.40fps

 

h_l = Kv²/2g = 2.4(3.40)²/(2x32.2)

h_l = .431ft < 1ft, OK

 

 

HOMEWORK No.7, Filtration

Read Chapter 3 pp. 194-292

Problems:

 

7A. Given: Well water. 6.1 mg/l of Fe and .9 mg/l of Mn at pH 7.8. 1mg/l of potassium permanganate will oxidize 1.06mg/l of iron and .52mg/l of manganese in accordance with the following equations:

3Fe⁺⁺ + MnO₄ ® 3Fe⁺⁺⁺ + MnO₂, pH>7.5

3Mn⁺⁺ + 2MnO₄ ®  5MnO₂, pH>9.5

Find: the potassium permanganate treatment requirements

 

7B. Given: Des Moines, Iowa feel  that they need filtration. The population is 230,000 @ 150gpcd. Assume they will use a high rate system.

Find: Estimate the area of the filter beds.

 

7C. Given: The above problem.

Find: Estimate the number of filters

 

7D. Given: A bed run at 5 gpm/ft² has a normal filter run of  30 hours.

Find: What would be the filter run if the rate were 20 gpm/ft²

 

7E.Given: The size of the sand is .50mm and its density is 2.60. The density of the anthracite is 1.60. The fluid is water at density 1. UC=1.7

Find: What size coal particle is required to ensure the same settling velocity, use F3.2.4-7, p 213, NOT the equation.

 

7F. Given: A .5m wide wash trough that is carrying .25m³/s.

Find: How high is it?

 

7G. Given: The same problem as the lecture notes. Rapid sand filter. D=20”. e=.42, T=55°F. filtration rate = 2gpm/ft². Spherical particles.

Rose equation:

h_l = , for stratified beds, where

C_D = 24/N_r

Sieve Analysis	% of sand retained given as a %, apply as a decimal	Geometric mean size, f (ftx10-3)
14-20	1.05	3.28
20-28	6.65	2.29
28-32	15.70	1.77
32-35	18.84	1.51
35-42	18.98	1.25
42-48	17.72	1.05
48-60	14.24	.88
60-65	5.15	.75
65-100	1.66	.59
	å=100%

Find: headloss using the Rose equation.

 

 

7H. Given: A 130 MGD water treatment plant using a conventional process.

Find:

1.) number of filters

2.) filter bed, filtration rate, filter washing scheme

3.) size of filters

4.) filter arrangement

5.) filter media

6.)backwash rate, underdrain system, headloss through each orifice(<1ft)

 

7I. Given: Design the filtration system for your project.

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