Water Treatment Notes

Protect public health by providing chemically and biologically safe potable water.
Protect ground water and surface water quality.
Removal of solids in Water.

Removal of contaminants achieved by physical forces such as gravity and screening.
- E.g., Sedimentation and Filtration

Removal is achieved by chemical and biological reactions.
- E.g., coagulation, flocculation, precipitation, disinfection

Process selection is normally based upon the expected characteristics of the influent flow.
Includes screening and grit removal; however, preliminary treatment may also include other processes, each designed to remove a specific type of material that presents a potential problem for down-stream unit treatment process.
These processes include shredding, flow measurement, pre-aeration, chemical addition, and flow equalization.

Removes large solids such as rags, cans, rocks, branches, leaves, and roots from the flow before the flow moves on to downstream processes.
Large bulky objects can be removed by Bar Screens, which consist of vertically placed bars with openings generally on the order of $25mm$ ( $1$ in).
Screens have openings of approximately $6 mm$ ( $1/4$ in) and thus remove leaves, twigs, and fish.

Wastewater operators responsible for screenings disposal are typically required to keep a record of the amount of screenings removed from the flow. To keep and maintain accurate screening records, the volume of screenings withdrawn must be determined.

A total of $65$ gallons of screening is removed from the wastewater flow during a $24$ -hour period. What is the screening removal reported as cubic feet per day?
- Solution: $1 ft^3 = 7.48$ gal
- $Screenings removed = \frac{65 \frac{gal}{}}{7.48 \frac{gal}{cu. ft}} = 8.6898 cu. ft \approx 8.69 \frac{cu. ft}{day}$
During $1$ week, a total of $310$ gallons of screening was removed from waste water screens. What is the average removal in cu-ft/day?
- Solution: $Screenings removed = \frac{310 \frac{gal}{}}{7.48 \frac{gal}{cu. ft}} = 41.444 cu. ft$
- $Screenings removed = \frac{41.444 cu. ft}{7 days} = 5.921 \frac{cu. ft}{day}$

Detention Time is the time required for flow to pass through a basin or tank or the time required to fill a basin or tank at a given flow rate.

A screening Pit has a capacity of $500$ cu.ft. If an average of $3.4$ cu.ft. of screenings are removed daily from the wastewater flows, in how many days will the pit be full?
- Solution: $Screening pit fill time = \frac{500 cu. ft}{3.4 \frac{cu. ft}{day}} = 147.06 days$
A plant has been averaging a screening removal of $2$ cu.ft./MG. If the average daily flow is $1.8$ MGD, how many days will it take to fill the pit with an available capacity of $125$ cu.ft?
- Solution: $Screening pit fill time = \frac{125 cu. ft}{2 \frac{cu. ft}{MG} * 1.8 MGD} = \frac{125 cu. ft}{3.6 \frac{cu. ft}{day}} = 34.72 days$
A screening pit has a capacity of $12$ cu.yd available for screenings. If the plant removes an average of $2.4$ cu.ft. of screenings per day, in how many days will the pit be filled?
- Solution: $1 cu.yd = 27 cu.ft$
- $Screening pit fill time = \frac{12 cu. yd * 27 \frac{cu. ft}{cu. yd}}{2.4 \frac{cu. ft}{day}} = \frac{324 cu. ft}{2.4 \frac{cu. ft}{day}} = 135 days$

Objective is to remove inorganic solids (sand, gravel, clay, egg shells, metal fillings, etc.) that cause excessive mechanical wear.
Grit removal may be accomplished in grit chambers or by centrifugal separation of bio solids. Processes use gravity / velocity, aeration, or centrifugal force to separate the solids from the wastewater.

A treatment plant removes $10$ cu.ft. of grit in $1$ day. How many cu.ft of grit are removed per million gallons if the plant flow is $9$ MGD.
- Solution: $Grit removed = \frac{10 \frac{cu. ft}{day}}{9 MD} = 1.111 \frac{cu. ft}{MG}$
The monthly average grit removal is $2.5$ cu.ft/MG. If the monthly average flow is $2,500,000$ gpd, how many cubic yards must be available for grit disposal if the disposal pit is to have a $90$ -day capacity?
- Solution: $2.5 \frac{cu. ft}{MG} * 2.5 MGD = 6.25 \frac{cu. ft}{day}$
- $6.25 \frac{cu. ft}{day} * 90 days = 562.5 cu. ft$
- $Grit removed = \frac{562.5 cu. ft}{27} = 20.83 cu. yd$

The optimum velocity in sewers is approximately $2$ fps at peak flow, because the velocity normally prevents solids from settling from the lines; however, when the flow reaches the grit channel, the velocity should decrease to about $1$ fps to permit the heavy inorganic solids to settle.
This calculation can be used for a single channel or tank or for multiple channels or tanks with the same dimensions and equal flow.
If the flow through each unit of the unit dimensions is unequal, the velocity for each channel or tank must be computed individually.

A plant is currently using two grit channels. Each channel is $3$ ft wide and has a water depth of $1.3$ ft. What is the velocity when the influent flow rate is $4.0$ MGD.
- Solution: $Velocity = \frac{4.0 MGD * 1.55 \frac{cfs}{MGD}}{2 * 3 ft * 1.3 ft} = 0.795 fps$
A plant’s grit channel is designed to remove sand, which has a settling velocity of $0.080$ fps. The channel is currently operating at a depth of $2.3$ ft. How many seconds will it take for a sand particle to reach the channel bottom.
- Solution: $Settling time = \frac{2.3 ft}{0.080 fps} = 28.75 s$
The grit channel of a plant is designed to remove sand which has a settling velocity of $0.080$ fps. The channel is currently operating at a depth of $3$ ft. The calculated value of flow through the channel is $0.85$ fps. The channel is $36$ ft long. Is the channel long enough to remove the desired sand particle size.
- Solution: $Required channel length = \frac{3 ft * 0.85 fps}{0.080 fps} = 31.875 ft$