ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 15 potx

First, calculate the volume of the water-filled casing: Then, determine the pounds of chlorine required, using the milligrams-per-liter to pounds equation: 15.2.1.5 Deep-Well Turbine Pump

The physical unit operations used in water treatment include:

• Screening — used to remove large-sized floating and suspended debris.

• Mixing — coagulant chemicals (alum, for example) are mixed with the water to make tiny particles stick together.

• Flocculation — water mixed with coagulants is given low-level motion to allow particles to meet and floc together.

• Sedimentation (settling) — water is detained for a time sufficient to allow flocculated particles to settle by gravity.

• Filtration — fine particles that remain in the water after settling and some microorganisms present are filtered out by sending the water through a bed of sand and coal.

Chemical unit processes used in treating raw water, depending on regulatory requirements and theneed for additional chemical treatment, include disinfection, precipitation, adsorption, ion exchange,and gas transfer A flow diagram of a conventional water treatment system is shown in Figure 15.1

Note: In the following sections (for water and wastewater treatment processes presented in Chapter

16 ), we present basic, often used daily operational calculations (operator’s math) along with engineering calculations (engineer’s math) used for solving more complex computations This presentation method is in contrast to normal presentation methods used in many engineering texts.

We deviate from the norm based on our practical real-world experience; we have found that environmental engineers tasked with managing water or wastewater treatment plants are responsible not only for computation of many complex math operations (engineering calculations) but also for overseeing proper plant operation (including math operations at the operator level) Obviously, engineers are well versed in basic math operations; however, they often need to refer to example plant operation calculations in a variety of texts In this text, the format used, although unconven- tional, is designed to provide basic operations math as well as more complex engineering math in one ready reference.

344 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

15.2 WATER SOURCE AND STORAGE CALCULATIONS

Approximately 40 million cubic miles of water cover or reside within the Earth The oceans containabout 97% of all water on Earth; the other 3% is fresh water Snow and ice on the surface of Earthcontain about 2.25% of the water; usable ground water is approximately 0.3%, and surface fresh-water is less than 0.5% In the U.S., for example, average rainfall is approximately 2.6 ft (a volume

of 5900 km3) Of this amount, approximately 71% evaporates (about 4200 km3), and 29% goes tostream flow (about 1700 km3)

Uses of freshwater include manufacturing; food production; domestic and public needs; ation; hydroelectric power production; and flood control Stream flow withdrawn annually is about7.5% (440 km3) Irrigation and industry use almost half of this amount (3.4% or 200 km3 per year).Municipalities use only about 0.6% (35 km3 per year) of this amount

recre-Historically, in the U.S., water usage is increasing (as might be expected) For example, in

1900, 40 billion gal of fresh water were used In 1975, the total increased to 455 billion gal.Projected use for 2000 (the latest published data) was about 720 billion gal

The primary sources of fresh water include:

• Water captured and stored rainfall in cisterns and water jars

• Groundwater from springs, artesian wells, and drilled or dug wells

• Surface water from lakes, rivers, and streams

• Desalinized seawater or brackish groundwater

• Reclaimed wastewater

15.2.1 Water Source Calculations

Water source calculations covered in this subsection apply to wells and pond or lake storage capacity.Specific well calculations discussed include well drawdown; well yield; specific yield; well-casingdisinfection; and deep-well turbine pump capacity

15.2.1.1 Well Drawdown

Drawdown is the drop in the level of water in a well when water is being pumped (see Figure15.2) Drawdown is usually measured in feet or meters One of the most important reasons formeasuring drawdown is to make sure that the source water is adequate and not being depleted Thedata collected to calculate drawdown can indicate if the water supply is slowly declining Earlydetection can give the system time to explore alternative sources, establish conservation measures,

or obtain any special funding that may be needed to get a new water source

Figure 15.1 Conventional water treatment model.

Addition of coagulant

Water

supply

Mixing tank

Flocculation basin

Settling tank

Sand filter

To storage and distribution Screening

Sludge processing

Disinfection

WATER TREATMENT PROCESS CALCULATIONS 345

Well drawdown is the difference between the pumping water level and the static water level:

First, calculate the water depth in the sounding line and the pumping water level:

Figure 15.2 Hydraulic characteristics of a well.

Drawdown, ft = Pumping Water Level, ft − aatic Water Level, ftSt

= 90 ft − 70 ft

= 20 ft

346 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

Then, calculate drawdown as usual:

15.2.1.2 Well Yield

Well yieldis the volume of water per unit of time produced from well pumping Usually, well yield

is measured in terms of gallons per minute (gpm) or gallons per hour (gph) Sometimes, large flowsare measured in cubic feet per second (cfs) Well yield is determined using the following equation

Drawdown, ft = Pumping Water Level, ft − aatic Water Level, ftSt

WATER TREATMENT PROCESS CALCULATIONS 347

Specific Yield, gpm/ft Well Yield, gpm

Dra

=wwdown, ft

= 260 gpm

22 ft

= 11.8 gpm/ft

348 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

15.2.1.4 Well Casing Disinfection

A new, cleaned, or repaired well normally contains contamination that may remain for weeks unlessthe well is thoroughly disinfected This may be accomplished by ordinary bleach in a concentration

of 100 ppm (parts per million) of chlorine The amount of disinfectant required is determined bythe amount of water in the well The following equation is used to calculate the pounds of chlorinerequired for disinfection:

(15.4)

Example 15.7

A new well is to be disinfected with chlorine at a dosage of 50 mg/L If the well casing diameter

is 8 in and the length of the water-filled casing is 110 ft, how many pounds of chlorine will berequired?

First, calculate the volume of the water-filled casing:

Then, determine the pounds of chlorine required, using the milligrams-per-liter to pounds equation:

15.2.1.5 Deep-Well Turbine Pump Calculations

Deep well turbine pumps are used for high-capacity deep wells Usually consisting of more thanone stage of centrifugal pumps, the pump is fastened to a pipe called the pump column; the pump

is located in the water The pump is driven from the surface through a shaft running inside thepump column, and the water is discharged from the pump up through the pump column to the

Specific Yield, gpm/ft Well Yield, gpm

Dra

=wwdown, ft

WATER TREATMENT PROCESS CALCULATIONS 349

surface The pump may be driven by a vertical shaft, electric motor at the top of the well, or someother power source, usually through a right angle gear drive located at the top of the well A modernversion of the deep-well turbine pump is the submersible type pump in which the pump, alongwith a close-coupled electric motor built as a single unit, is located below water level in the well.The motor is built to operate submerged in water

15.2.3 Vertical Turbine Pump Calculations

The calculations pertaining to well pumps include head, horsepower, and efficiency calculations

The pressure (psi) can be converted to feet of head using the equation:

(15.5)

water level (discharge head) Total pumping head is calculated as:

under-Discharge Head, ft = (press, psi) (2.31 ft/ppsi)

Pumping Head, ft = Pumping Water Level, ft ++ Discharge Head, ft

350 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

• Motor horsepowerrefers to the horsepower supplied to the motor The following equation is used

to calculate motor horsepower:

(15.8)

• Total brake horsepower (bhp) refers to the horsepower output of the motor The following equation

is used to calculate total bhp:

First, calculate the field head The discharge head must be converted from psi to ft:

Motor hp (input hp) Field bhp

Motor Effici

=

eency100

Total bhp = Field bhp + Thrust Bearing Loss,, hp

Field bhp = Bowl bhp + Shaft Loss, hp

Bowl bhp (Lab bhp) = (Bowl Head, ft) (Capaciity, gpm)

Water hp (Field Head, ft) (Capacity, gpm)

3

=

33,000 ft-lb/min

WATER TREATMENT PROCESS CALCULATIONS 351

The water horsepower is therefore:

The water horsepower can now be determined:

Example 15.11

The pumping water level for a pump is 170 ft The discharge pressure measured at the pumpdischarge head is 4.2 psi If the pump flow rate is 800 gpm, what is the water horsepower? (UseEquation 15.12.)

First, determine the field head by converting the discharge head from psi to ft:

Now, calculate the field head:

and then calculate the water horsepower:

352 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

Example 15.13

The bowl brake horsepower is 51.8 bhp If the 1-in diameter shaft is 170 ft long and is rotating

at 960 rpm with a shaft fiction loss of 0.29 hp loss per 100 ft, what is the field bhp?

Before you can calculate the field bhp, factor in the shaft loss:

Now determine the field bhp:

Example 15.14

The field horsepower for a deep-well turbine pump is 62 bhp If the thrust bearing loss is 0.5

hp and the motor efficiency is 88%, what is the motor input horsepower? (Use Equation 15.8.)

Bowl bhp (Bowl Head, ft) (Capacity, gpm)

(3

=9960) (Bowl Efficiency)100

(185 ft) (600 gpm)(3960) (84.0)100

=

(185) (600)(3960) (84.0)

WATER TREATMENT PROCESS CALCULATIONS 353

When we speak of the efficiencyof any machine, we speak primarily of a comparison of what

is put out by the machine (energy output) compared to its input (energy input) Horsepowerefficiency, for example, is a comparison of horsepower output of the unit or system with horsepowerinput to that unit or system — the unit’s efficiency For vertical turbine pumps, four efficiencytypes are considered:

×100

354 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

fluctu-• Increasing operating convenience

• Leveling pumping requirements (to keep pumps from running 24 h a day)

• Decreasing power costs

• Providing water during power source or pump failure

• Providing large quantities of water to meet fire demands

• Providing surge relief (to reduce the surge associated with stopping and starting pumps)

• Increasing detention time (to provide chlorine contact time and satisfy the desired CT (contact

time) value requirements)

• Blending water sources

Field Efficiency, % = (Field Head, ft) (Cappacity, gpm)

(3960) (Total bhp) ×100

(180 ft) (850 gpm)(3960) (61.3 bhp)

= 74.9%

WATER TREATMENT PROCESS CALCULATIONS 355

15.3.1 Water Storage Calculations

The storage capacity, in gallons, of a reservoir, pond, or small lake can be estimated using Equation

15.16:

(15.16)

Example 15.17

A pond has an average length of 250 ft, an average width of 110 ft, and an estimated average

depth of 15 ft What is the estimated volume of the pond in gallons?

Example 15.18

A small lake has an average length of 300 ft and an average width of 95 ft If the maximum

depth of the lake is 22 ft, what is the estimated gallons volume of the lake?

Note: For small ponds and lakes, the average depth is generally about 0.4 times the greatest depth.

Therefore, to estimate the average depth, measure the greatest depth and then multiply that number

by 0.4.

First, estimate the average depth of the lake:

Then determine the lake volume:

Still Water Body, gal = (Av Length, ft) (Avv Width, ft) (Av Depth, ft) (7.48 gal/ft3)

Vol, gal = (Av Length, ft) (Av Width, ft)(Av Depth, ft) (7.48 gal/ft )3

356 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

15.3.2 Copper Sulfate Dosing

Algal control is perhaps the most common in situ treatment of lakes, ponds, and reservoirs; this is

usually accomplished by the application of copper sulfate — the copper ions in the water kill the

algae Copper sulfate application methods and dosages vary depending on the specific surface water

body being treated The desired copper sulfate dosage may be expressed in milligrams per liter of

copper, pounds of copper sulfate per acre-foot, or pounds of copper sulfate per acre

For a dose expressed as milligrams per liter of copper, the following equation is used to calculate

pounds of copper sulfate required:

(15.17)

Example 15.19

For algal control in a small pond, a dosage of 0.5-mg/L copper is desired The pond has a

volume of 15 MG How many pounds of copper sulfate are required? (Copper sulfate contains

25% available copper.)

For calculating pounds of copper sulfate per acre-foot, use the following equation (assume the

desired copper sulfate dosage is 0.9 lb/acre-ft):

(15.18)

Example 15.20

A pond has a volume of 35 acre-ft If the desired copper sulfate dose is 0.9 lb/acre-ft, how

many pounds of copper sulfate are required?

Copper Sulfate, lb = Copper (mg/L) (Volume,, MG) (8.34 lb/gal)

% Available Copper100

Copper Sulfate, lb = Copper (mg/L) (Volume,, MG) (8.34 lb/gal)

% Available Copper100

(0.5 mg/L) (15 MG) (8.34 lb/gal)

25100

=

= 250 lb Copper Sulfate

Copper Sulfate, lb = (0.9 lb Copper Sulfate)) (acre-ft)

1 acre-ft

WATER TREATMENT PROCESS CALCULATIONS 357

Then solve for x:

The desired copper sulfate dosage may also be expressed in terms of pounds of copper sulfate

per acre The following equation is used to determine pounds of copper sulfate (assume a desired

dose of 5.2 lb of copper sulfate per acre):

(15.19)

Example 15.21

Problem:

A small lake has a surface area of 6.0 acres If the desired copper sulfate dose is 5.2 lb/acre,

how many pounds of copper sulfate are required?

Solution:

15.4 COAGULATION, MIXING, AND FLOCCULATION

15.4.1 Coagulation

After screening and the other pretreatment processes, the next unit process in a conventional water

treatment system is a mixer, in which the first chemicals are added in what is known as coagulation

The exception to this situation occurs in small systems using groundwater, when chlorine or other

taste and odor control measures are introduced at the intake and are the extent of treatment

Copper Sulfate, lb = (0.9 lb Copper Sulfate)) (acre-ft)

The term coagulation refers to the series of chemical and mechanical operations by which

coagulants are applied and made effective These operations comprise two distinct phases: (1) rapidmixing to disperse coagulant chemicals by violent agitation into the water being treated; and (2)flocculation to agglomerate small particles into well-defined floc by gentle agitation for a muchlonger time The coagulant must be added to the raw water and perfectly distributed into the liquid;such uniformity of chemical treatment is reached through rapid agitation or mixing

Coagulation results from adding salts of iron or aluminum to the water (Coagulation is thereaction between one of these salts and water.) Common coagulants (salts) include:

• Alum — aluminum sulfate

For complete mixing:

(15.20)

For plug flow:

(15.21)

where

t = detention time of the basin, minute

V = volume of basin, cubic meters or cubic feet

Q = flow rate, cubic meters per second or cfs

K = rate constant

C i= influent reactant concentration, milligrams per liter

C e= effluent reactant concentration, milligrams per liter

L = length of rectangular basin, meters or feet

v = horizontal velocity of flow, meters per second or feet per second

i e

Step 1 Find C e:

Step 2 Calculate t for complete mixing (Equation 15.20):

Step 3 Calculate t for plug flow using the following formula:

15.4.3 Flocculation

Flocculation follows coagulation in the conventional water treatment process; it is the physicalprocess of slowly mixing the coagulated water to increase the probability of particle collision.Through experience, we see that effective mixing reduces the required amount of chemicals andgreatly improves the sedimentation process, which results in longer filter runs and higher qualityfinished water Flocculation’s goal is to form a uniform, feather-like material similar to snowflakes

— a dense, tenacious floc that entraps the fine, suspended, and colloidal particles and carries themdown rapidly in the settling basin To increase the speed of floc formation and the strength andweight of the floc, polymers are often added

15.4.4 Coagulation and Flocculation General Calculations

In the proper operation of the coagulation and flocculation unit processes, calculations are performed

to determine chamber or basin volume, chemical feed calibration, chemical feeder settings, anddetention time

15.4.4.1 Chamber and Basin Volume Calculations

To determine the volume of a square or rectangular chamber or basin, use Equation 15.22 or 15.23:

1440 min1d

1440

90 In

404

i e

Then calculate basin volume:

Volume, ft3 = (length, ft) (width, ft) (deptth, ft)

Volume, gal = (length, ft) (width, ft) (deptth, ft) (7.48 gal/ft )3

Volume, gal = (length, ft) (width, ft) (deptth, ft) (7.48 gal/ft )3

15.4.4.2 Detention Time

Because coagulation reactions are rapid, detention time for flash mixers is measured in seconds,whereas the detention time for flocculation basins is generally between 5 and 30 min The equationused to calculate detention time is

Then calculate detention time:

15.4.4.3 Determining Dry Chemical Feeder Setting (Pounds per Day)

Adding (dosing) chemicals to the water flow calls for a measured amount of chemical The amount

of chemical required depends on such factors as the type of chemical used, the reason for dosing,and the flow rate being treated To convert from milligrams per liter to pounds per day, use thefollowing equation:

(15.25)

Example 15.28

Problem:

Jar tests indicate that the best alum dose for a water sample is 8 mg/L If the flow to be treated

is 2,100,000 gpd, what should the pounds per day setting be on the dry alum feeder?

Detention Time, sec Volume of Tank, gal

Fl

=o

Chem added, lb/day = (Chemical, mg/L) (Floww, MGD) (8.34 lb/gal)

Lb/day = (Chemical, mg/L) (Flow, MGD) (8.34lb/gal)

15.4.4.4 Determining Chemical Solution Feeder Setting (Gallons per Day)

When solution concentration is expressed as pounds of chemical per gallon of solution, therequired feed rate can be determined using the following equations:

Solution:

(15.26)Then, convert the pounds per day dry chemical to gallons per day solution:

(15.27)

Example 15.30

Problem:

Jar tests indicate that the best alum dose for a water sample is 7 mg/L The amount to be treated

is 1.52 MGD Determine the gallons-per-day setting for the alum solution feeder if the liquid alumcontains 5.36 lb of alum per gallon of solution

Solution:

First calculate the pounds per day of dry alum required, using the milligrams-per-liter to per-day equation:

pounds-Then, calculate the gallons per day solution required:

15.4.4.5 Determining Chemical Solution Feeder Setting (Milliliters per Minute)

Some solution chemical feeders dispense chemical as milliliters per minute (mL/min) To calculatethe milliliters per minute solution required, use the following procedure:

(15.28)

Chem., lb/day = (Chemical, mg/L) (Flow, MGD)) (8.34 lb/gal)

Solution, gpd Chemical, lb/day

The desired solution feed rate has been calculated at 25 gpd What is this feed rate expressed

as milliliters per minute?

Solution:

Sometimes we need to know milliliters per minute solution feed rate when we do not knowthe gallons per day solution feed rate In such cases, calculate the gallons per day solution feedrate first, using the following equation:

(15.29)

15.4.5 Determining Percent of Solutions

The strength of a solution is a measure of the amount of chemical solute dissolved in the solution

We use the following equation to determine percent strength of solution:

= 65.7 mL/min Feed Rate

gpd = (Chemical, mg/L) (Flow, MGD) (8.34 lb//gal)

Chemical, lb/Solution, gal

% Strength Chemical, lb

Water, lb Chem

=

+ iical, lb × 100

ll) + 0.625 lb × 100

0.625 lb Chemical125.7 lb Solution

=

+ 0.198 lb × 100

= 4%

15.4.5.1 Determining Percent Strength of Liquid Solutions

When using liquid chemicals to make up solutions (liquid polymer, for example), a differentcalculation is required:

15.4.5.2 Determining Percent Strength of Mixed Solutions

The percent strength of solution mixture is determined using the following equation:

(Liq Poly., lb) (Liq Poly % Strength)

100 = Poly Sol., lb Poly Sol (% Strength)

lb Solution 1

(Sol 2, + +

llb)(% Strength, Sol 2)

lb Solution 2

15.4.6 Dry Chemical Feeder Calibration

Occasionally we need to perform a calibration calculation to compare the actual chemical feed ratewith the feed rate indicated by the instrumentation To calculate the actual feed rate for a drychemical feeder, place a container under the feeder; weigh the container when it is empty and thenweigh it again after a specified length of time (30 min, for example)

The actual chemical feed rate can be calculated using the following equation:

lb Solution 1

(Sol 2,++

Chemical Feed Rate, lb/min = Chemical Applieed, lb

Length of Application, min

Feed Rate, lb/day = (Feed Rate, lb/min) (14440 min)

day

Chemical Feed Rate, lb/min = Chemical Applieed, lb

Length of Application, min

2 lb

30 min

=0.06 lb/min Feed Rate

=

Then, calculate the pounds per day feed rate:

First, calculate the pounds per minute feed rate:

Then, calculate the pounds per day feed rate:

15.4.6.1 Solution Chemical Feeder Calibration

As with other calibration calculations, the actual solution chemical feed rate is determined and thencompared with the feed rate indicated by the instrumentation To calculate the actual solutionchemical feed rate, first express the solution feed rate in million gallons per day (MGD) Once themillion gallons per day solution flow rate has been calculated, use the milligrams per liter equation

to determine chemical dosage in pounds per day

If solution feed is expressed as milliliters per minute, first convert milliliters per minute flowrate to gallons per day flow rate

(15.35)

Then, calculate chemical dosage, pounds per day

(15.36)

Chemical Feed Rate, lb/day = (0.06 lb/min)((1440 min/day)

= 86.4 lb/day Feed Rate

Chemical Feed Rate, lb/min = Chemical Applieed, lb

Length of Application, min

1.6 lb

20 min

=

= 0.08 lb/min Feed Rate

Chemical Feed Rate, lb/day = (0.08 lb/min)((1440 min/day)

= 115 lb/day Feed Rate

Example 15.39

Problem:

A calibration test is conducted for a solution chemical feeder During a 5-min test, the pumpdelivered 940 mg/L of the 1.20% polymer solution (Assume the polymer solution weighs 8.34lb/gal.) What is the polymer dosage rate in pounds per day?

Solution:

The flow rate must be expressed as million gallons per day; therefore, the milliliters per minutesolution flow rate must first be converted to gallons per day and then million gallons per day Themilliliters per minute flow rate are calculated as:

Next, convert the milliliters per minute flow rate to gallons per day flow rate:

Then, calculate the pounds per day polymer feed rate:

Example 15.40

Problem:

During a calibration test conducted for a solution chemical feeder, over a 24-h period, thesolution feeder delivers a total of 100 gal of solution The polymer solution is a 1.2% solution.What is the pounds-per-day feed rate? (Assume the polymer solution weighs 8.34 lb/gal.)

Solution:

The solution feed rate is 100 gal/day or 100 gpd Expressed as million gallons per day, this is0.000100 MGD Use the milligrams-per-liter to pounds-per-day equation to calculate actual feedrate, pounds per day:

The actual pumping rates can be determined by calculating the volume pumped during aspecified time frame For example, if 60 gal are pumped during a 10-min test, the average pumpingrate during the test is 6 gpm

Actual volume pumped is indicated by drop in tank level Using the following equation, wecan determine the flow rate in gallons per minute:

940 mL

5 min = 188 mL/min

(188 mL/min) (1440 min/day)

3785 mL/gal = 72gpd flow rate

(12,000 mg/L) (0.000072 MGD) (8.34 lb/day) == 7.2 lb/day Polymer

lb/day Chemical = (Chemical, mg/L) (Flow, MGGD) (8.34 lb/day)

= (12,000 mg/L) (0.000100 MGD) (8.34 lb/day))

= 10 lb/day Polymer

15.4.7 Determining Chemical Usage

One of the primary functions performed by water operators is the recording of data The per-day or gallons-per-day chemical use is part of these data From them, the average daily use ofchemicals and solutions can be determined This information is important in forecasting expectedchemical use, comparing it with chemicals in inventory, and determining when additional chemicalswill be required

pounds-To determine average chemical use, use Equation 15.38 (pounds per day) or Equation 15.39(gallons per day):

(15.38)or

(15.39)

Then we can calculate days’ supply in inventory:

(15.40)or

Days Supply in Inventory = Total Chemicaliin Inventory, lb

Average Use, lb/day

Days Supply in Inventory = Total Chemicaliin Inventory, gal

Average Use, gpd

15.4.7.1 Paddle Flocculator Calculations

The gentle mixing required for flocculation is accomplished by a variety of devices Probably themost common device in use is the basin equipped with mechanically driven paddles Paddleflocculators have individual compartments for each set of paddles The useful power input imparted

by a paddle to the water depends on the drag force and the relative velocity of the water withrespect to the paddle (Droste, 1997)

Day of week Amount of chemical (lb/day)

=

Days Supply in Inventory = Total Chemicaliin Inventory, lb

Average Use, lb/day

= 2800 lb in Inventory

77 lb/day Average Usee

= 36.4 days Supply in Inventory

For paddle flocculator design and operation, environmental engineers are mainly interested indetermining the velocity of a paddle at a set distance, the drag force of the paddle on the water,and the power input imparted to the water by the paddle.

Because of slip (factor k), the velocity of the water will be less than the velocity of the paddle.

If baffles are placed along the walls in a direction perpendicular to the water movement, the value

of k decreases because the baffles obstruct the movement of the water (Droste, 1997) The frictional dissipation of energy depends on the relative velocity, v The relative velocity can be determined

where N = rate of revolution of the shaft (rpm).

To determine the drag force of the paddle on the water, use Equation 15.44:

15.5.1 Tank Volume Calculations

The two common tank shapes of sedimentation tanks are rectangular and cylindrical The equationsfor calculating the volume for each type of tank are shown next

v = vp− vt = vp − kvp = v (1p − k)

260

vp = πNr

FD = 1/2pC AvD 2

dP=dF vD = 1 2/ pC v dAD 3

15.5.1.1 Calculating Tank Volume

For rectangular sedimentation basins, we use Equation 15.46:

(15.46)For circular clarifiers, we use Equation 15.47:

(15.47)

Example 15.44

Problem:

A sedimentation basin is 25 ft wide, 80 ft long, and contains water to a depth of 14 ft What

is the volume of water in the basin in gallons?

Volume, gal = (length, ft) (width, ft) (deptth, ft) (7.48 gal/ft )3

Volume, gal = (0.785) (D2) (depth, ft) (7.488 gal/ft )3

Volume, gal = (length, ft) (width, ft) (deptth, ft) (7.48 gal/ft )3

(80 ft) (25 ft) (14 ft) (7.48 gal/ft )3

=209,440 gal

=

Volume, gal = (length, ft) (width, ft) (deptth, ft) (7.48 gal/ft )3

140,000 gal = (75 ft) (24 ft) (x ft) (7.48ggal/ft )3

x ft 140,000(75)(24)(7.48)

=

x ft = 10.4 ft

• Basic detention time equation:

A sedimentation tank has a volume of 137,000 gal If the flow to the tank is 121,000 gph, what

is the detention time in the tank, in hours?

Solution:

Example 15.47

Problem:

A sedimentation basin is 60 ft long, 22 ft wide, and has water to a depth of 10 ft If the flow

to the basin is 1,500,000 gpd, what is the sedimentation basin detention time in hours?

15.5.3 Surface Overflow Rate

Surface loading rate — similar to hydraulic loading rate (flow per unit area) — is used to determineloading on sedimentation basins and circular clarifiers Hydraulic loading rate, however, measuresthe total water entering the process, whereas surface overflow rate measures only the water over-flowing the process (plant flow only)

Note: Surface overflow rate calculations do not include recirculated flows Other terms used

synonymously with surface overflow rate are surface loading rate and surface settling rate.

Surface overflow rate is determined using the following equation:

= 0.6 gpm/ft2

15.5.4 Mean Flow Velocity

The measure of average velocity of the water as it travels through a rectangular sedimentation basin

is known as mean flow velocity and is calculated using Equation 15.52:

Because velocity is desired in feet per minute, the flow rate in the Q = AV equation must be

expressed in cubic feet per minute (cfm):

Then the Q = AV equation can be used to calculate velocity:

Because velocity is desired in feet per minute, the flow rate in the Q = AV equation must be

expressed in cubic feet per minute (cfm):

Q (Flow), ft /min3 = A (Cross-Sectional Area)), ft2 ×V (Vol.) ft/min

(Q = A × V)

900,000 gpd(1440 min/day) (7.48 gal/ft )3 = 44 cfm8

The Q = AV equation can be used to calculate velocity:

15.5.5 Weir Loading Rate (Weir Overflow Rate)

Weir loading rate (weir overflow rate) is the amount of water leaving the settling tank per linearfoot of weir The result of this calculation can be compared with design Normally, weir overflowrates of 10,000 to 20,000 gal/day/ft are used in the design of a settling tank Typically, weir-loadingrate is a measure of the gallons per minute (gpm) flow over each foot of weir Weir loading rate

is determined using the following equation:

(15.53)

Example 15.52

Problem:

A rectangular sedimentation basin has a total of 115 ft of weir What is the weir loading rate

in gallons per minute per foot when the flow is 1,110,000 gpd?

771 gpm

115 ft

=

= 6.7 gpm/ft

15.5.6 Percent Settled Biosolids

The percent settled biosolids test (a.k.a “volume over volume” test, or V/V test) is conducted bycollecting a 100-mL slurry sample from the solids contact unit and allowing it to settle for 10 min.After 10 min, the volume of settled biosolids at the bottom of the 100-mL graduated cylinder ismeasured and recorded The equation used to calculate percent settled biosolids is

= 2465 gpm

283 ft

= 8.7 gpm/ft

% Settled Biosolids = Settled Biosolids Vollume, mL

Total Sample Volume, mL × 100

% Settled Biosolids = Settled Biosolids Vollume, mL

Total Sample Volume, mL × 100

22 mL

100 mL 100

= 19% Settled Biosolids

Example 15.55

Problem:

A 100-mL sample of slurry from a solids contact unit is placed in a graduated cylinder After

10 min, a total of 21 mL of biosolids settles to the bottom of the cylinder What is the percentsettled biosolids of the sample?

Solution:

15.5.7 Determining Lime Dosage (Milligrams per Liter)

During the alum dosage process, lime is sometimes added to provide adequate alkalinity (HCO3)

in the solids contact clarification process for the coagulation and precipitation of the solids.Determining the required lime dose in milligrams per liter uses three steps

In Step 1, the total alkalinity required is calculated Total alkalinity required to react with the

alum to be added and provide proper precipitation is determined using the following equation:

First, calculate the alkalinity that will react with 45 mg/L of alum:

% Settled Biosolids = Settled Biosolids Vollume, mL

Total Sample Volume, mL × 100

Next, calculate the total alkalinity required:

Example 15.57

Problem:

Jar tests indicate that 36 mg/L of alum is optimum for particular raw water If a residual mg/L alkalinity must be present to promote complete precipitation of the alum added, what is thetotal alkalinity required in milligrams per liter?

30-Solution:

First, calculate the alkalinity that will react with 36 mg/L of alum:

Then, calculate the total alkalinity required:

In Step 2, we make a comparison between required alkalinity and alkalinity already in the raw

water to determine how many milligrams per liter of alkalinity should be added to the water Theequation used to make this calculation is:

Total Alk Required, mg/L = Alk to React w/AAlum, mg/L + Residual Alk, mg/L

In Step 3, after determining the amount of alkalinity to be added to the water, we determine

how much lime (the source of alkalinity) needs to be added We accomplish this by using the ratioshown in Example 15.59

Example 15.59

Problem:

If 16-mg/L alkalinity must be added to a raw water, how many milligrams-per-liter lime will

be required to provide this amount of alkalinity? (1 mg/L alum reacts with 0.45 mg/L alkalinityand 1 mg/L alum reacts with 0.35 mg/L lime.)

Given the following data, calculate the lime dose required, in milligrams per liter:

• Alum dose required (determined by jar tests) — 52 mg/L

• Residual alkalinity required for precipitation — 30 mg/L

• 1 mg/L alum reacts with 0.35 mg/L lime

Alk to be Added, mg/L = Tot Alk Req., mg//L-Alk Present in the Water, mg/L

x= 12.4 mg/L Lime

• 1 mg/L alum reacts with 0.45 mg/L alkalinity

• Raw water alkalinity — 36 mg/L

Solution:

To calculate the total alkalinity required, we must first calculate the alkalinity that will reactwith 52 mg/L of alum:

The total alkalinity requirement can now be determined:

Next, calculate how much alkalinity must be added to the water:

Finally, calculate the lime required to provide this additional alkalinity:

Total Alk Required, mg/L = Alk to React w//Alum, mg/L + Residual Alk, mg/L

= 23.4 mg/L + 30 mg/L

= 53.4 mg/L Total Alkalinity Required

Alk to be Added, mg/L = Tot Alk Req., mg//L Alk Present, mg/L–

x = 13.5 mg/L Lime