# SOLUTION TO CASE STUDY #3: Municipal Watewater Treatment Facilities

## Problem Statement for Case 3.

The population increase in the state of Easy Living has brought about growing demands for a new supply of potable water, especially in the area of Freedom City (a residential community). It is anticipated that the city will have a population potential for approximately 50,000 new residents over the next 50 years. 1,000 visitors per day are also epected when the new shopping mall is completed within the next two years. The existing treatment facilites are limited and allow no future expansion. The purpose of any treatment plant is to provide water that is chemically and bacteriologically safe for human consumption and domestic use. A surface water reservior is located 8 miles north of Freedom City and is assumed to contain enough water supply at all times. Preliminary studies by the Pussy Cat water company indicate that the average use of water per capita is 300 gallons per day. This includes residential interior, business, schools, recreation and fire fighting use. It also indicates a peaking factor equal to 2. The company also studied the quality of the reservior water. Analysis of a typical sample of the surface reservior water gave the following information:
1. amount of dissolved oxygen = 5 mg/L
2. turbidity = 8 NTU
3. suspended solids = 30 mg/L
4. BOD5 = 8 mg/L
5. dissolved inroganic salts (mostly calcium) = 300 mg/L
6. 2000 fecal coliform per 100 mL
7. some debris and fish
8. no radioactive contaminants
The city has designated a space located 4 miles south of the reservior as a site for the construction of a new water treatment plant. It has an elevation 400 ft lower than the reservior. This elevation is also 200 ft higher than the ground surface of the city.

Design a suitable water treatment plant that solves the water shortage problem a nd meets the EPA standards. The following design criteria apply:
1. Minimum detention time in a sedimentation tank is 4 hours
2. Maximum overflow rate in a sedimentation tank is 30 m3/m2 a day.
3. Efficiency for sedimentation tank no less than 85%
4. Filtrate rate in a treatment filter is 3 L/m2 per second
5. Pumping station efficiency is no less than 80%
6. Flocculation average detention time no less than 30 minutes
Justify and design parameters used and give references for any assumed values.

## Introduction

Water treatment plants are designed to achieve two basic goals: "speed up the natural purification process that occur in streams, rivers and lakes and reduce toxic contaiminants that might otherwise interfere with the natural process."

In the design of a water treatment facility for the city of Freedom, environmental percussionsmust be adressed. The areas of most concern are the air, water and local ecology surrounding the plant. The following should be considered:
(1) Air
any gases or vapors resulting from the treatment be non-toxic,
any chemicals used in the treatment are properly stored and processed,
emergency backup system for contamination is installed.
(2) Water
water returning to the environment is non-toxic,
waste products are treated before discharging,
efficient distribution system,
prevention design for flood or overflow.
(3) Ecology
safety of fish and marine life,
preservation of non-aquatic animal wildlife,
maintenance of existing wildlife habitat,
reforestation of local vegetation,
environmental landscaping.
The building facilities will include besides the treatment sections, an administration section, a laboratory and workshop section, water storage facility and several parking lots. The site area should allow further expansion. The design should be flexible to permit any control adjustments required to compensate for any changes in the composition and behavior of the influent and its contents. Finally a proper on-site safety program that includes regular training in safety and emergency procedures should be established.

The major stages of the treatment process are:
1. Intake and Screening
2. Coagulation and Flocculation
3. Sedimintation
4. Filtration
5. Disinfection

## The Design Process

#### Intake and Screening

Because of the geography of the reservoir location, the treatment site and city elevation, the screening operation will take place along-side the water reservoir. The design will rely on the natural water pressure of the reservoir to force the water through the screening section. Placing the screening operation in close proximity to the reservoir allows quick and safe return of the retained fish and other aquatic life to their natural habitat. Effective screening should remove approximately 10% of the suspended solids and all the debris.

 Daily amount of water used by residents = 50,000 x 300 = 15,000,000 gallons Daily amount of water used by visitors = 1000 x 300 x 2/24 = 25,000 gallons Maintenance, spillage and evaporation = 5,000 gallons/day Average daily consumption = 15,030,000 gallon Peak flow = 2 x 15,030,000 = 30,060,000 gallons/day (347.92 gallons/sec)

To avoid potential intake of surface debris and bottom sludge, the intake pipe entrance is located at 30 ft. below the surface. The pipe opening is covered with a grating composed of metal bars spaced 2 inches apart to prevent the influx of extremely large debris.
Inlet velocity should not exceed 3 ft/sec to prevent debris from being forced through the screens.
 v = 3 ft/sec Q = 347.92 x 8.33 / 62.4 = 46.5 ft3/sec A = Q / v = 15.5 ft2 = 3.14 D2 / 4 D = diameter of pipe = 3.44 ft

take D = 4.5 ft. Use concrete as material of the pipe due to its durability and low cost.

As recommended, head loss due to each screen should not exceed 0.3 ft.

Now, the equation of Bernoulli will be used to calculate the location of the screening facility. Taking the inlet velocity as zero (very large reservoir) and assuming atmospheric pressure and using Darcy-Weisbach expression for head loss, the equation reduces to:
30 = v2 / 2 g + Hb + 0.0006 L * v2
where
Hb = height of screen above intake level
L = the length of the pipe.

Solving the above equation iteratively gives a slight drop in elevation from reservoir surface. Taking into account the natural grade in the vicinity of the reservoir, location of screening would be 30.5 ft from water edge.
Two wedgewire rotary screens will be used, one with 1/2 inch mesh to remove large debris and aquatic life, the second with 1/8 inch mesh to remove suspended particulate matter. Based on recommended value of screen capacity of 10 to 30 gal / ft2-min, screen dimensions are 5 ft diameter and 10 ft long each.

Waste accumulates on the screening surface is scraped using a rubber flange and sprayer system. Debris is collected in a waste tank of dimensions 9 x 6 x 6 ft. Aquatic life will be returned to the reservoir. Periodic cleaning of tank is required. The filtered water enters a catch basin tank, from which it proceeds to the second screen. The basin tank has dimensions of 8 x 15 x 12 ft assuming 30 sec lag.

A backup screening system should be available and should be easily mounted in case of any rotary screening failure. Screening is housed on floor of dimensions 50 x 30 ft.

After screening, water is transported in concrete piping system of 4.5 ft diameter to the treatment facility using decreasing gradual elevation mechanism.

#### Coagulation and Flocculation

This operation is carried out to form larger particles to facilitate their settling and reduce turbidity. Chemicals such as metalic salts (coagulants) are added to water at the inlet section of the flocculator to help agglomerate the particles. Mixing takes place in this section to bring more collision between particles ,so flocs can be formed. The flocs then move to the detention section where wooden paddles regulate the motion of these particles. Chemical dosages can be determined experimentally or by searching existing data. In our case a mixture of Alum, cationic polymer and caustic soda is used. The dosage is as follows:
 Alum 2.4 to 3.0 mg/L Cationic polymer 0.5 to 2.0 mg/L Caustic soda 2.0 to 4.0 mg/L
These data were obtained from a typical treatment facility.

Storage and piping for chemicals are made of polyethylene to resist corrosion. Hydraulic jump is chosen for mixing because it requires minimum maintenance. Coagulant should be added at the point of maximum turbulence.

 Mixing Time = 3 minutes Detection Time t = V/Q

Assuming t = 50 min,
V = 50 * 60 * 46.5 = 139,528 ft3

Use four flocculation tanks each of volume 139,528/4 = 34,882 ft3

Water leaves screening system and enters the mixing section with a velocity = 3 ft/sec. Assuming head loss of 2 ft through the hydraulic jump mixing section, water enters the settling section with velocity = 1 ft/sec.

Velocity of paddle = 2 ft/sec = vp
 P = Power = C Ap (Density) vp3 / 2 = 1.8 (0.2) (Cross-sectional area of tank) (62.4) (23) / 2 = 37.908 L2 * Watt

Let
 G * t = 105 = (50 * 60) G G = 33.33 sec-1 G2 = P / (V * viscosity) = 37.908 L2 / (34882 * 0.00066)

Therefore,
 L = 25.97 ft V = L2W = 34,882 ft3 = tank volume W = 51.72 ft

The tank dimensions are 25.97 * 25.97 * 51.72
 Total Ap = 0.2 * L2 = 134.88 ft2

Using 20 paddles with one ft. spacing,
Area of paddle = 134.88/20 = 6.74 ft2
with dimensions 3.54 * 1.9 for the paddle.

Assume velocity of inflow to settling section = 1 ft/sec

Using steel pipes to inlet and exit from tank:

Q = 46.5

Using 4 pipes,

 Q = 46.5 / 4 = 11.625 ft3 3.14 * r2 * v = 11.625 = 3.14 * r2 * 1 r = 1.92 and d = 3.84 ft.

Tanks are made of concrete and lined with polyethylene because of its chemical resistance.

#### Sedimentation

Suspended solids originally large and those formed by flocculation are given enough time to settle by gravity. Sedimentation is responsible for the removal of the majority of the suspended solids (over 80%). The technique used here is that of a horizontal flow sedimentation. Quiescent conditions should be established in the sedimentation tank through slow and smooth flow of water. A large detention time is required, i.e. a long and narrow tank is used. Sedimentation tanks have four zones: (1) inlet zone where very little settling is observed and the flow should be evenly distributed throughout the tank, (2) settling zone where particles start to settle as a result of quiescent conditions, (3) sludge storage zone where most of the settling takes place. The sludge at the bottom should be removed preriodically to allow for more settling and prevent particles from moving back upward and disturb the flow, and (4) outlet zone where partially clear water exits the tank and moves to the filtration step.

To facilitate the design, four particle sizes ranging from 0.004 inch to 0.2 inch, all with the same density of 66 lb/ft3 and each size contributes the same amount to the total suspended solid concentration. An automated sludge removal will be constructed as part of the tank. This is a conveyor type chai n pulling transverse blades which sweep the sludge toward a sump.

 Total suspended solids concentration = 30 mg/L = 46.5 * 28.316 * 0.03 = 39.50 g/sec = 0.087 lb/sec

Assuming the smallest particles as 0.004 inch or 0.1 mm, particles with diameter s larger than 0.1 mm will be removed completely.

Using Stokes Equation for the smallest particles,
 vsmall = (densitys - density) * g * d2 / (18 * viscosity) = (66 - 62.4) (32.17) (0.00033)2 / 18 * 0.0007) = 0.00102 ft/sec.

Assuming 50% of these particles is removed,
 vaverage = 0.00102 / 0.5 = 0.00204 ft/sec daverage = [(18 * viscosity * vaverage) / (g * (densitys - density))]0.5 = 0.00047 ft Re = (0.00047 * 0.00204 * 62.4) / 0.0007 = 0.085 < 0.3 qo = 30 m3 / (m2 * day) = 0.0011574 ft3 / (ft2 * sec)

Using 4 hours sedimentation time,
 V = volume = Q*t = 46.5 * 4 * 3600 = 669,600 ft3 Hmax = height = v*t = 0.00204 * 4 * 3600 = 29.38 ft A = 46.5 / qo = 46.5 / 0.0011574 = 40,184 ft2 H = 669,600 / 40,184 = 16.66 ft < 29.38 ft

Two tanks will be used , each of volume = 669,600 / 2 = 335,000 ft3 approximately.
 A = 335,000 / 16.66 = 20,100 ft2 qo = 46.5 / (2 * 20,100) = 0.0011567 ft3/ft2

Assuming 300 ft length,
 w = 20,100 / 300 = 67 ft Stream exit velocity = 300 / (4 * 3600) = 0.0208 ft/sec
Water exits the sedimentation tank and continues on to filtration.

#### Filtration

The remaining suspended solids (10 to 15 %) are removed by filteration. The resulting effluent is clear and colorless.
Filteration rate is constant at 3 lit / sq.m-sec and the head loss should not exceed 9.8 ft. Eight filter, each with filteration capacity of 4. million gallons per day are used. Because of the high filteration rate, the rapid sand filteration process is used. A layer of anthracite occupies the upper part of filter, sand and anthracite are chosen because of their particle size, porosity and specific gravity. The particles that escapes the lage and light anthracite layer will be trapped in the find sand layer. An 8 inches gravel layer is placed between the sand layer and the underdrain system to prevent sand from slipping into the underdrain system.The design is based on the peak conditions to ensure smooth operation at any time. The underdrain system consists of central manifold with laterals, 7 inches diameter steel pipes on each side containing a number of orifices (radius = 0.25 inches). A back washing system is to be used to remove any clogging and prevent any dramatic increase in the headloss or shearing stress on the cake. A recommended rate of washing is 15 to 20. gpm / sq.ft. The backwash valve has to be opened slowly to avoid upsetting the filter layers. The differences between the specific gravities of the layers materials allow the filter media to expand and fall without intermixing. The low specific gravity suspended particles are disposed through a system of troughs. A control system is used to control the constant flow rate operation.

number of units= 30.06 / 4 = 7.5 units
use 8 units each of capacity = 4 millon gallons per day
3 lit / sq.m -sec = 4.43 gallons / sq. ft-sec
filteration area per unit = 4 million / 24 * 60 * 4.43 = 630. sq.ft
with dimensions of 42 x 15 ft

total headloss = sum of headlosses over the three layers
delta P = { L / g ( fluid density * v2 (1-porosity) / d * porosity3 * 150 * (1-porosity) / Re + 1.75)
Assuming L = 2 ft for anthracite and 1 ft. for sand
delta Psand = 1.8
delta Panthracite = 4.72
delta Pgravel = 1.22
total pressure head loss = 7.73 ft
where
 sp gravity sand = 2.65, d = 0.002 ft. sp gravity anthracite = 1.45, d = 0.005 ft. sp gravity gravel = 4.5, d = 0.003 ft. superficial velocity v = 0.5 ft/sec.

Area of orifice / area of filter bed = 0.0001 - 0.2
N (3.14) (0.02 ft)2 / 630 = 0.0216
N = number of orifices = 10,000
Area of laterals / area of orifices = 2:4:1
take ratio = 3.14:1 = n (3.14) (3.5/12)2 / 10,000 (3.14) (.022)
Number of laterals n =160
Area of manifold / area of laterals = 1.5 - 3:1
a / 160 (3.14) (3.5/12)2 = 3
a = 128 ft2 with dimensions 40 * 3.2 ft
use space between orifices same as space between laterals. Each lateral has about 63 orifices. Tanks above the troughs are used for supplying the backwashing water. The filtrate now goes to the disinfection unit.

#### Disinfection

Disinfection is done using chlorine. Chlorine is effective in killing harmful organisms. It also controls odor.
Chlorine is stored in cylinders, each of one ton capacity in a climate control room with sufficient ventilation and away from the sun.
Chlorination facility is housed in a separate building. Safety procedures must be used in storage transporting and processing chlorine. Piping should be PVC schedule 80 to resist corrosion. It is released as a solution into the water through diffusers constructed on the pipes carrying the gas solution along the bottom third of the basin.
Chlorine monitors and feed control system are installed to ensure 1.0 mg residual chlorine. A plug flow is preferred to help retain the water in the basin for the required detention time. Contact time between the gas and the water depends on the residual concentration of the chlorine, temperature, and pH of the water.

Recommneded dosage for 1.0 mg/L chlorine residual is 6 mg/L
Rate of Chlorine supply / day

 = 6 (30.06) (10)6 (3.76) / (454 * 1000) = 1500 lbs/day
using a 2 inch diameter injection pipe
A = 3.14 (1) / 144 = 0.0218 ft2
at a rate 75 gpm or 0.167 ft3/sec
velocity of chlorine solution = 0.167/0.0218 = 7.66 ft/sec

Assuming a pH = 7 and average temperature of 35 F for winter,
to achieve 2 coliform / 100 mL

Nt/No = 2 / 1000 = (1 + 0.23 t)-3

t = 39.13
For a residual of 1. mg/L
Contact time = 39.13 / 1 = 39.13 minutes
Use detention time of 60 minutes.

For summer with average temperature of water=60F, use detention time 30 minutes.

Volume of detention basin:

V = 46.5 (60) (60) = 167,400 ft3
Use two 100,000 ft3 basins with L/W > 20
Dimensions of basin are
 h = 16 ft W = 15 ft L = 416 ft
velocity of water traveling the basin
v = 46.5 / (16 * 15) = 0.193 ft/sec

Baffles are used to regulate the flow along the basin. Four baffles, 16 x 15 x 1/3 ft for each basin.

The disinfected water now goes to the distribution system.