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 / ft^{2}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 ft^{3}
Use four flocculation tanks each of volume 139,528/4 = 34,882 ft^{3}
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 = v_{p}
P  = Power = C A_{p} (Density) v_{p}^{3} / 2

 = 1.8 (0.2) (Crosssectional area of tank) (62.4) (2^{3}) / 2

 = 37.908 L^{2} * Watt 
Let
G * t  = 105 = (50 * 60) G

G  = 33.33 sec^{1}


G^{2}  = P / (V * viscosity) = 37.908 L^{2} / (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 A_{p}  = 0.2 * L^{2}

 = 134.88 ft2 
Using 20 paddles with one ft. spacing,
Area of paddle = 134.88/20 = 6.74 ft^{2}
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 ft^{3} 
3.14 * r^{2} * v = 11.625 = 3.14 * r^{2} * 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,
v_{small}  = (density_{s}  density)
* g * d^{2} / (18 * viscosity)

 = (66  62.4) (32.17) (0.00033)^{2} / 18 * 0.0007)

 = 0.00102 ft/sec. 
Assuming 50% of these particles is removed,
v_{average}  = 0.00102 / 0.5 = 0.00204 ft/sec

d_{average}  = [(18 * viscosity * v_{average})
/ (g * (density_{s}  density))]^{0.5}

 = 0.00047 ft

Re  = (0.00047 * 0.00204 * 62.4) / 0.0007 = 0.085 < 0.3

q_{o}  = 30 m^{3} / (m^{2} * day)
= 0.0011574 ft^{3} / (ft^{2} * sec)

Using 4 hours sedimentation time,
V  = volume  = Q*t  = 46.5 * 4 * 3600 = 669,600 ft^{3}

Hmax  = height  = v*t  = 0.00204 * 4 * 3600 = 29.38 ft

A    = 46.5 / qo = 46.5 / 0.0011574 = 40,184 ft^{2}

H    = 669,600 / 40,184 = 16.66 ft < 29.38 ft

Two tanks will be used , each of volume = 669,600 / 2 = 335,000 ft^{3}
approximately.
A  = 335,000 / 16.66 = 20,100 ft2

q_{o}  = 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.msec 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. ftsec
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 * v^{2} (1porosity)
/ d * porosity^{3} *
150 * (1porosity) / Re + 1.75)
Assuming L = 2 ft for anthracite and 1 ft. for sand
delta P_{sand} = 1.8
delta P_{anthracite} = 4.72
delta P_{gravel} = 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 ft^{2}
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 ft^{2}
at a rate 75 gpm or 0.167 ft^{3}/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
N_{t}/N_{o} = 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 ft^{3}
Use two 100,000 ft^{3} 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.