American Water Collegeamericanwatercollege.org/Course_files/Water/PDF_Course_Handouts/... · Quantity = Area x Velocity Flow (ft 3/sec) ... (Surface Area, sq. ft.) Detention Time

American Water College

Water Treatment Math Book 1

First Edition 2006 American Water College Parkfield, CA 93451 Copyright © 2006 by American Water College


Table of Contents

Page Formulas and Conversion Factors i Lesson 1: Unit Conversions 1 Lesson 2: Working With Formulas 10 Lesson 3: Understanding Percentages 16 Lesson 4: Calculating Area 22 Lesson 5: Calculating Volume 30 Lesson 6: Weight/Volume Relationships 37 Lesson 7: Force-Pressure-Head 48 Lesson 8: Velocity and Flow Rate 55 Lesson 9: Pumps 59 Lesson 10: The Metric System 66 Lesson 11: Problem Solving 70 Lesson 12: Test Taking Strategies / Practice Test 74 Answer Key 87

i

Formulas and Conversion Factors

EQUIVALENTS 1 cubic foot of water weighs 62.3832 lb 1 gallon of water weighs 8. 34 lb 1 liter of water weighs 1,000 gm 1 mg/L = 1 part per million (ppm) 1 ug/L = 1 part per billion (ppb) 1 mile = 5,280 feet (ft) 1 yd = 3 feet 1 yd3 = 27ft3 1 acre (a) = 43,560 square feet (ft2) 1 acre foot = 325,829 gallons 1 cubic foot (ft3) = 7.48 gallons (gal) 1 gal = 3.785 liters (L) 1 L = 1,000 milliliters (ml) 1 pound (lb) = 454 grams (gm) 1 lb = 7,000 grains (gr) 1 grain per gallon (gpg) = 17.1 mg/L 1 gm = 1,000 milligrams (mg) 1 gm = 1,000, 000 micrograms (ug) SOLUTIONS Lb/Gal = (Solution %)_x 8.34 lb/gal x Specific Gravity

100

Lb Chemical = Specific Gravity x 8.34 lb/gal x Solution (gal) Specific Gravity = Chemical Wt. (lb/gal) 8.34 (lb/ gal) % of Chemical in Solution = (Dry Chemical, lb) x 100___ (Dry Wt. Chemical, lb) + (Water, lb) GPD = (Vol, MG) x (Conc., mg/L) x (8.34 lb/gal) (% Strength) x Chemical Wt. (lb/gal) GPD = (Feed, ml/ min. x 1,440 min/ day) (1,000 ml/L x 3.785 L/gal)


ii

Two – Normal Equations: a) C1 V1 = C2 V2 b) C1 V1 + C2 V2 = C3 V3 C = Concentration V = Volume Q = Flow PUMPING 1 horsepower (Hp) = 746 watts = 0. 746 kw = 3, 960 gal/ min/ ft Water Hp = (GPM) x (Total Head, ft) (3,960 gal/ min/ ft) Brake Hp = (GPM) x (Total Head, ft) (3,960) x (Pump % Efficiency) Motor Hp = (GPM ) x ( Total Head, ft) (3,960) x (Pump % Eff.) x (Motor % Eff.) “Wire to Water” Efficiency = (Motor, % Efficiency x Pump % Efficiency) Cost, $ = (Hp) x (0.746 Kw/Hp) x (Operating Hrs.) x (cents/ Kw- Hr) VOLUME Rectangular Basin Volume, gal = (Length, ft) x (Width, ft) x (Height, ft) x 7.48 gal/ft3 Cylinder Volume, gal = (0.785) x (Dia, ft)2 x (Height or Depth, in ft.) x 7.48 gal/ft3

Time, Hrs. = ( Volume, gallons ) (Pumping Rate, GPM, x 60 Min/ Hr) Supply, Hrs.= Storage Volume, Gals (Flow In, GPM – Flow Out, GPM) x 60 min/ hr PRESSURE PSI = ( Head, ft. ) PSI = Head, ft. x 0. 433 PSI/ ft. 2.31ft./ psi lb Force = (0. 785) x (D, ft.)2 x 144 in2 /ft2 PSI


iii

SCADA = 4 mA to 20 mA analog signal (live signal mA - 4 mA off set) x process unit and range (16 mA span) 4 mA = 0 20 mA = full–range FLOW -VELOCITY-AREA Q = A x V Quantity = Area x Velocity Flow (ft3/sec) = Area (ft2) x Velocity (ft/sec) CHLORINATION Dosage, mg/L = (Demand, mg/L) + (Residual, mg/L) (Gas) lb/ day = (Vol, MG) x (Dosage, mg/L) x (8.34 lb/ gal) HTH Solid (lb/day) = (Vol, MG) x (Dosage, mg/L) x (8.34lb/gal) ( % Strength ) Liquid (gal/day) = (Vol, MG) x (Dosage, mg/L) x (8. 34 lb/gal) (% Strength) x (Specific Gravity x 8.34) GENERAL ($) Cost /day = lb/day x ($) Cost/lb Removal, Percent = (In - Out) x 100 In Specific Capacity, GPM/ ft = Well Yield, GPM Drawdown, ft. Gals/Day = (Population) x (Gals/Capita/ Day) GPD = (Meter Read 2 - Meter Read 1) (Number of Days) Volume, Gals = GPM x Time, minutes


iv

FILTRATION Filtration Rate (GPM/ sq. ft) = Filter Production (gallons per day) (Filter area sq. ft.) x (1,440 min/day) Filtration Rate (GPM) = (Filter Area, sq. ft.) x (GPM/sq. ft.) Loading Rate Loading Rate (GPM/sq. ft.) = (Flow Rate, GPM) (Filter Area, sq. ft.) Daily Filter Production (GPD) = (Filter Area, sq. ft.) x (GPM/sq. ft. x 1,440 min/day) Backwash Pumping Rate (GPM) = (Filter Area, sq. ft.) x (Backwash Rate, GPM/sq. ft.) Backwash Vol. (Gal) = (Filter Area, sq. ft.) x (Backwash Rate, gpm/sq. ft.) x (Time, min) Backwash Rate, GPM/ sq. ft. = (Backwash Volume, gallons) (Filter Area, sq. ft.) x (Time, min) Rate of Rise (inches per min.) = (backwash rate gpm/sq. ft.) x 12 inches /ft 7.48 gal/ cu. ft. CHEMICAL DOSE CALCULATIONS Note (% purity) and (% commercial purity) used in decimal form lb/day gas feed dry = MGD x 8. 34 lb/gal x (ppm or mg/L) lb/day = MGD x 8.34 lb/gal x (ppm or mg/L) % purity GPD = (MGD) x (8.34 lb/ gal) x (ppm or mg/L) (% purity) x lb/ gal GPD = MGD x 8. 34 lbs/ gal x (ppm or mg/L) (commercial purity %) x (ion purity %) x (lb/gal) ppm or mg/L = _____lb/day____ or gallons x % purity x lb/gal MGD x 8.34 lb/gal MG x 8.34 lb/gal


v

C • T CALCULATIONS C • t = (Chlorine Residual, mg/L) x (Time, minutes) Time, minutes = ________(C • t)________ (Chlorine Residual, mg/L) Chlorine Residual (mg/L) = ___( C • t )____ (Time, minutes) Inactivation Ratio = (Actual System C • t) (Table “E” C• t) C • t Calculated = T10 Value, minutes x Chlorine Residual, mg/L Log Removal = 1.0 - % Removal x Log key x (-1) 100 SEDIMENTATION Surface Loading Rate, (GPD/sq. ft.) = (Total Flow, GPD ) (Surface Area, sq. ft.) Detention Time = Volume Flow Flow = Volume Time Weir Overflow Rate, GPD/L.F. = (Flow, GPD) (Weir length, ft

Unit Conversions

1

Lesson 1 Unit Conversions Key Concepts

• Units are labels that distinguish one measurable quantity from other

measurable quantities • A given value divided by an equivalent value is equal to 1 • Any number multiplied by 1 remains the same

• By using equivalents, units can be changed to the desired units

• Same units above and below the division line cancel each other out

Example A: 1 acre = 1 43,560 ft2

Example B: 87,120 ft2 x 1 acre = 2 acres 43,560 ft2

Monorail Method of Converting Units

Step 1 – Draw a horizontal line with and a vertical line in the middle of it

Step 2 – Write down your known number with appropriate units in the upper left space

Step 3 – Choose an appropriate equivalent and write in the next space (make sure the value with the same units as your starting number is

below the line so that the units will cancel each other out)

Step 4 – Cross out units that appear above and below the division line Step 5 – Do the math by multiplying your beginning number by all numbers above the line and dividing by all numbers below the line

2

Example 1:

Convert 50 feet of head to psi.

Example 2:

Convert 100,000 ft2 to acres.

Example 3:

Convert 10 liters of water to pounds.

Example 4:

Convert 6 lb/day of polymer to mg/sec.

Unit Conversions

3

Convert between acres and square feet: 1 acre = 43,560 square feet

1. 150,000 square ft = _____________ acres

2. 23,000 square ft = _____________ acres

3. 500,000 square ft = _____________ acres

4. 5 acres = _____________ square feet

5. 3.5 acres = _____________ square feet

6. Find the acreage of a pond surface with an average width of 150 feet and a length of 200 feet.

7. Find the area in acres of a reservoir that is 1,000 feet long by 900 feet wide.

4

Convert the following volumes: 1 cubic foot of water = 7.48 gallons

8. A basin that measures 20 ft wide by 40 ft long by 10 ft deep = _____________ gallons



11. 250,000 gallons of water = _____________ cubic feet




Unit Conversions

5

Convert the following volume/weight relationships: 1 gallon of water = 8.34 pounds

15. 1,500 gallons of water = _____________ pounds

16. 120,000 gallons of water = _____________ pounds

17. 50 gallons of water = _____________ pounds

18. 30,000 pounds of water = _____________ gallons

19. 5,000 pounds of water = _____________ gallons Convert between English and Metric volumes: 1 gallon = 3.785 liters

20. 7 gallons = _____________ liters

21. 25 gallons = _____________ liters

6

22. 125 gallons = _____________ liters

23. 19 liters = _____________ gallons

24. 50 liters = _____________ gallons

Convert the following concentrations: 1 mg/L = 1 ppm and 1% = 10,000 ppm 25. 5 ppm = _____________ mg/L

26. 58 mg/L = _____________ ppm

27. 5% = _____________ mg/L

28. 12.5% = _____________ mg/L

Unit Conversions

7

29. 5,000 mg/L = _____________ %

30. 90,000 mg/L = _____________ % Convert between feet of Head and psi: 1 foot of water = .43 psi 1 psi = 2.31 feet of Head

31. 100 feet of water = _____________ psi

32. 525 feet of water = _____________ psi

33. 375 feet of water = _____________ psi

34. 55 psi = _____________ feet of head

35. 125 psi = _____________ feet of head

8

36. 250 psi = _____________ feet of head Convert the following power relationships: 1 HP = .746 Kw = 550 ft-lb/sec = 33,000 ft-lb/min

37. 3 HP = _____________ Kw

38. 18 HP = _____________ Kw

39. 65 Kw = ____________ HP

40. 20 Kw = _____________ HP

41. 6,000 ft-lb/sec = _____________ HP

42. 17,000 ft-lb/sec = _____________ HP

Unit Conversions

9

43. 23,000 ft-lb/sec = _____________ Kw

44. 42,000 ft-lb/sec = _____________ Kw Perform the following complex conversions:

45. 8 lb/day = _____________ ml/sec

46. 4 MGD = _____________ CFS (cubic feet per second)

47. 10 MG = ______________ acre-feet

10

Lesson 2 Working With Formulas Key Concepts

• You can move factors from one side of the equal sign to the other side in a

diagonal manner • Unknown factor must be alone and on top • If there isn’t a division line, the factor is considered on top • When plugging numbers into a formula, they must be in the same units as the

formula

Example 1: Solve the formula for (x)

(x) (2) = (3) (6)

Example 2: Solve the formula for (a)

(3) (10) = (5) (a) Example 3: Solve for (x) 10 = 25 2 (x)

Working With Formulas

11

Example 4:

Solve the formula for (Dose, mg/L) lb/day = (Flow, MGD) x (Dose, mg/L) x (8.34) Example 5: Find the chemical consumption in lb/day. Plant flow is 2 CFS and chemical dose is 2.8 mg/L.

12

Rearrange the detention time formula: Detention Time = Tank Volume Flow

1. Solve for tank volume

2. Solve for flow Rearrange the Lbs/day formula lb/day = (Flow, MGD) x (Dose, mg/L) x (8.34)

3. Solve for flow

4. Solve for dose


13

Rearrange the velocity formula: Velocity = Distance Time 5. Solve for time

6. Solve for distance Rearrange the solids loading formula: Solids loading, lb/day/sq.ft. = Solids Applied, lb/day Surface Area, sq ft

7. Solve for solids applied

8. Solve for surface area

14

Rearrange the hydraulic loading formula: Hydraulic loading, gpd/sq ft = Flow, gpd Surface Area, sq. ft

9. Solve for flow

10. Solve for Surface area

11. Find the chemical consumption in lb/day if plant flow is 1.5 CFS and chemical dose is 2.5 mg/L

lb/day = (Flow, MGD) (Dose, mg/L) (8.34)


15

12. Find the detention time in minutes of a basin that has a volume of 15,000 ft3 and a flow of

350 gpm. Detention Time = Volume, gal Flow, gpm

13. Find the detention time in minutes of a basin that has a volume of 40,000 gallons and a flow of 350 gpm.

Detention Time = Volume, gal Flow, gpm

16

Lesson 3 Understanding Percentages Key Concepts

• To convert from a decimal or fraction to percent, multiply by 100 • To convert from percent to a decimal or fraction, divide by 100

Example 1:

Convert 0.33 to percent Example 2: Convert ¼ to percent Example 3: Convert 27% to a decimal Example 4: Convert 80% to a fraction

Understanding Percentages

17

Convert the following fractions to percent:

1. 1/4 = _____________ %

2. 2/3 = _____________ %

3. 5/8 = _____________ %

4. 7/13 = _____________ %

5. 9/16 = _____________ % Convert the following decimals to percent:

6. 0.23 = _____________ %

7. 0.256 = _____________ %

8. 0.683 = _____________ %

9. 0.3 = _____________ %

18

10. 0.05 = _____________ %

Convert the following percentages to fractions:

11. 33% = _____________

12. 17% = _____________

13. 75% = _____________

14. 80% = _____________

15. 90% = _____________ Convert the following decimals to fractions:

16. .35 = _____________

17. .70 = _____________


19

18. .08 = _____________

19. .15 = _____________

20. .45 = _____________ Convert the following fractions to decimals:

21. 2/5 = _____________

22. 4/8 = _____________

23. 3/9 = _____________

24. 6/18 = _____________

25. 25/100 = _____________

20

Convert the following percentages to decimals: 26. 88% = _____________

27. 75.3% = _____________

28. 27.9% = _____________

29. 8% = _____________

30. 0.3% = _____________ Review Problems

31. Convert 8 liters of water to pounds.

32. Convert 18 cubic feet to gallons.


21

33. Convert 2.5 MGD to CFS.

34. Solve for Dose: lb/day = (Flow, MGD) (Dose, mg/L) (8.34)

35. Solve for Flow: Detention Time = Tank Volume

Flow

22

Lesson 4 Calculating Area Key Concepts

• Area is a way of describing a two dimensional object • Units are usually ft2 or acres

• Rectangles A = L x W

• Triangles A = ½ x B x H • Circles A = π x r2 or A = 0.785 x D2 • Cylinders C = π x D and A = π x r2 • Spheres A = π x D2

Example 1: Find the surface area in acres of a pond measuring 100 ft by 350 ft. Example 2: Find the area of a triangle with a base that is 30 ft and a height of 40 ft. Example 3: Find the surface area of a clarifier with a 40 ft diameter.

Calculating Area

23

Example 4: Find the surface area of a round tank with a diameter of 30 ft. and a height of 15 ft. (do not include the top of the tank)

Example 5: Find the surface area of a sphere that is 20 feet in diameter.

24

Find the area of the following rectangles: A = L x W 1 Acre = 43, 560 square feet

1. A wall that is 5 feet tall and 20 feet long = _____________ square feet

2. A wall that is 72 inches tall and 30 feet long = _____________ square feet

3. A parking lot that measures 300 feet by 150 feet = _____________ square feet

4. A facility that measures 600 ft. by 2,500 ft = _____________ acres

5. A reservoir that is 500 ft by 500 ft = _____________ acres

Calculating Area

25

Find the area of the following triangles: A = ½ B x H

6. A triangle with a base of 10 feet and a height of 12 feet = _____________ ft2





26

Find the area of the following circles: A = π x r2 (π = 3.14)

11. The surface area of a round clarifier that measures 50 feet across = _____________sq ft

12. The top of a circular storage tank with a diameter of 30 feet = _____________ sq ft

13. The cross sectional area of a pipe that is 3 feet in diameter = _____________ sq ft

14. The cross sectional area of a pipe that has a 24 inch radius = _____________ sq ft

15. The area of a circle that measures 60 feet across = _____________ sq ft

Calculating Area

27

Find the surface area of the following cylinders: Circumference = π x D Area = π x D x H

16. A round tank that is 20 feet tall and 15 feet across = ______________ sq ft





28

Find the surface area of the following spheres: A = π x D2

21. A methane storage sphere that is 15 feet wide has a surface area of ____________ sq ft

22. A propane storage sphere that is 25 feet wide has a surface area of ____________ sq ft


24. A sphere that is 12 feet wide has a surface area of ____________ sq ft


Calculating Area

29

Review

26. Solve for Flow: lb/day = (Flow, MGD) (Dose, mg/L) (8.34)

27. Solve for MLSS: Aerator Solids lb = (Tank Volume, MG) (MLSS, mg/L) (8.34)

30

Lesson 5 Calculating Volume Key Concepts

• Volume is used for describing 3 dimensions • Units are usually ft3, gallons, or acre-ft

• Cubes Vrectangle = L x W x H

• Cylinders Vcylinder = π x r2 x H

• Cones Vcone = 1/3 x π x r2 x H

• Spheres Vsphere = 1/6 x π x D3

Example 1: Find the volume of a basin that is 15 feet deep by 60 feet long by 30 feet wide. Example 2: Find the volume of a round tank that is 12 feet tall with a diameter of 10 feet.

Calculating Volume

31

Example 3: Find the volume of a cone that has a base that is 12 feet across and is 15 feet tall. Example 4: Find the volume of a sphere that measures 13 feet from side to side.

32

Find the volume of the following rectangular basins and flow channels:

Vrectangle = L x W x H

1. 20 feet tall by 60 feet long by 30 feet wide = _____________ cubic feet

2. 17 feet deep by 20 feet wide by 33 feet long = _____________ cubic feet

3. 2 feet wide by 18 inches deep by 6 feet long = _____________ cubic feet



Calculating Volume

33

Find the volume of the following cylindrical objects: Vcylinder = π x r2 x H

6. A 12 foot pipe with a diameter of 24 inches = _____________ cubic feet

7. A tank that is 15 feet tall with a diameter of 40 feet = _____________ cubic feet

8. A 17 feet deep clarifier that has a sludge collector arm that is 13 feet long = _________ ft3

9. A 48 inch pipe that is 50 feet long = ____________ cubic feet

10. A 36 inch pipe that is 1 mile long = _____________ cubic feet (1 mile = 5,280 feet)

11. How many gallons are in the pipe described in #5 above? _____________ gallons

34

Find the volume of the following cones: Vcone = 1/3 x π x r2 x H

12. A cone with a base that is 18 inches and is 36 inches tall = _____________ cubic feet

13. A cone with a base that is 2 feet across and is 48 inches tall = _____________ cubic feet

14. A cone with a base that is 15 feet across and is 25 feet tall = _____________ cubic feet

15. A cone with a base that is 7 feet across and is 18 feet tall = _____________ cubic feet

16. A cone with a base that is 48 inches across and is 5 feet tall = _____________ cubic feet

Calculating Volume

35

Find the volume of the following spheres: Vsphere = 1/6 x π x D3

17. A sphere with a diameter of 12 feet = _____________ cubic feet

18. A sphere with a radius of 12 feet = _____________ cubic feet

19. A sphere with a radius of 15 feet = _____________ cubic feet

20. A sphere with a diameter of 7 feet = _____________ cubic feet

36

Review Problems

21. Convert 500 lb/day to mg/sec

22. Convert 30 acre-ft to MG

23. Convert 504,345 square feet to acres

24. Convert 135 feet of head to psi

25. Convert 835,000 gallons to cubic feet

Weight/Volume Relationships

37

Lesson 6 Weight/Volume Relationships Key Concepts

• 1 gallon of water weighs 8.34 lbs • 1 cubic foot of water contains 7.48 gallons • 1 cubic foot of water weighs 62.4 lbs

• 1 gallon is 3.785 liters

• 1 Kg is equal to 2.2 lb

• 1 pound is equal to 454 grams

Example 1: Convert 1,000 gallons of water to pounds Example 2: Convert 20,000 ft3 of water to gallons

38

Example 3: Convert 7,250 ft3 of water to pounds Example 4: Convert 4,500 gallons to liters Example 5: Convert 2,300 lb to Kg


39

Convert the following volumes of water to lb: 1 gallon = 8.34 lb

1. 300 gallons = _____________ lb

2. 25 gallons = _____________ lb

3. 3 gallons = _____________ lb

4. 230,000 gallons = _____________ lb

5. 50,000 gallons = _____________ lb

40

Convert the following weights of water to gallons: 1 gallon = 8.34 lb

6. 30,000 lb = _____________ gallons

7. 50 lb = _____________ gallons

8. 125 lb = _____________ gallons

9. 350 lb = _____________ gallons

10. 11 lb = _____________ gallons


41

Convert the following dimensional volumes to gallons: 1 cubic foot = 7.48 gallons

11. 5 cubic feet = _____________ gallons

12. 16 cubic feet = _____________ gallons

13. 8,500 cubic feet = _____________ gallons

14. A basin that is 20 ft. deep by 60 ft. wide by 80 ft. long = _____________ gallons

15. A round tank that is 15 ft. tall and 25 feet in diameter = _____________ gallons

42

Convert cubic feet of water to pounds: 1 cubic foot = 7.48 gallons 1 gallon = 8.34 lb

16. 65,000 cubic feet = __________ lb

17. 43,000 cubic feet = __________ lb

18. 5,577 cubic feet = __________ lb

19. 373,616 cubic feet = __________ lb

20. 935,500 cubic feet = __________ lb


43

Convert from gallons to liters: 1 gallon = 3.785 liters

21. 3 gallons = _____________ liters

22. 20 gallons = _____________ liters

23. 15 gallons = _____________ liters

24. 33 gallons = _____________ liters

25. 75 gallons = _____________ liters

44

Convert from liters to gallons: 1 gallon = 3.785 liters

26. 10 liters = _____________ gallons

27. 25 liters = _____________ gallons

28. 38 liters = _____________ gallons

29. 78 liters = _____________ gallons

30. 125 liters = _____________ gallons


45

Convert between lb and Kg 1 Kg = 2.2 lb

31. 5 lb = _____________ Kg

32. 20 lb = _____________ Kg

33. 27 lb = _____________ Kg

34. 175 lb = _____________ Kg

35. 500 lb = _____________ Kg

46

36. 35 Kg = _____________ lb

37. 135 Kg = _____________ lb

38. 650 Kg = _____________ lb

39. 5,000 Kg = _____________ lb

40. 750 Kg = _____________ lb


47

Review Problems

41. How many gallons of water can a basin measuring 30 feet wide by 50 feet long by 10 feet deep hold?

42. How many gallons can a tank with a diameter of 26 feet and a height of 12 feet hold?

43. How many acre-ft is 3.9 million gallons?

48

Lesson 7 Force – Pressure – Head Key Concepts

• Force – The push exerted by water on any surface being used to confine it (Usually expressed in units of pounds, tons, grams, or kilograms)

• Pressure – The force per unit area (Usually expressed in units of psi)

• Head – The vertical distance in feet from the water surface to a reference point

below the surface (Expressed in units of feet) • 1 foot of head is equal to 0.433 psi • 1 psi is equal to 2.31 feet of head

• 1 foot of head is equal to 62.4 lb/ft2

Example 1: How much force would be exerted on the bottom of a 1foot cube if it was full of water? (give answer in psi) Example 2:

Find the upward force on the bottom of a partially underground tank that has a bottom surface area of 400 ft2. Ground water level is 5 feet above the bottom of the tank

Example 3: Convert 45 feet of head to psi

Force-Pressure-Head

49

Find the upward force on the following underground tanks: F = 62.4 x A x H

1. The bottom of the tank is 30 feet in diameter and groundwater level is 10 feet above the bottom of the tank. How much force is being exerted on the bottom of the tank?

2. A basin is 15 feet wide by 30 feet long. The groundwater level is 7 feet above the bottom of the basin. How much force is being exerted on the bottom of the basin?

50

3. A tank is 24 feet in diameter and groundwater level is 8 feet above the bottom of the tank.

How much force is being exerted on the bottom of the tank?

4. Groundwater level is 13 feet above the bottom of a tank whose diameter is 20 feet. How much force is being exerted on the bottom of the tank?

Force-Pressure-Head

51

5. If the bottom of a tank with a 44 foot diameter is 12 feet below ground, and the groundwater level is 8 feet below grade, how much force is being exerted on the bottom of the tank?

6. If the bottom of a tank with a 34 foot diameter is 15 feet below ground, and the groundwater level is 10 feet below grade, how much force is being exerted on the bottom of the tank?

52

Find the water pressure in psi for the following water columns: 1 ft = .43 psi 7. 19 ft of water = _____________ psi

8. 29 ft of water = _____________ psi

9. 123 ft of water = _____________ psi

10. 75 ft of water = _____________ psi

11. 53 ft of water = _____________ psi

Force-Pressure-Head

53

12. 45 psi = ______________ ft of head

13. 55 psi = ______________ ft of head

14. 10 psi = ______________ ft of head

15. 25 psi = ______________ ft of head

16. 100 psi = ______________ ft of head

54

Review Problems

17. How many pounds does 325 gallons of water weigh?

18. How many liters are in 55 gallons?

19. How many acres is 653,400 square feet?

20. Convert 0.25 to apercentage.

21. Convert 75% to a decimal.

Velocity and Flow Rate

55

Lesson 8 Velocity and Flow Rate Key Concepts

• Velocity – The speed at which a particle or fluid is moving. Units are distance per unit of time. (mph, ft/sec)

Velocity = Distance Time

• Flow Rate – The rate at which a volume of fluid is moving. Units are volume per unit of time. (MGD, CFS, gpm)

Q = Velocity x Area

Example 1: Calculate the velocity in ft/sec if a particle of water travels 30 ft in 1 minute Example 2: Calculate the flow rate in CFS if a particle of water travels through a 12 inch pipe at 3 ft/sec Example 3: Calculate the flow rate in gpm of water traveling at 10 ft/sec in a 24 inch pipe

56

Calculate water velocity in feet per second: Velocity = Distance Time

1. A particle of water travels 60 feet in 15 seconds = _____________ feet per second 2. A particle of water travels 30 feet in 5 seconds = _____________ feet per second 3. A particle of water travels 120 feet in 60 seconds = ___________ feet per second 4. A particle of water travels 200 feet in 1 minute = ___________ feet per second 5. A particle of water travels 1 mile in 2 minutes = _____________ feet per second

Velocity and Flow Rate

57

Find the flow rate through the following pipes in cubic feet per second (CFS) Q = Velocity x Area

6. Flowing at 20 feet per second through a 36 inch pipe = _______________ CFS



9. Flowing at 200 feet per minute through a 36 inch pipe = _______________ CFS

10. Flowing at 150 feet per minute through a 24 inch pipe = _______________ CFS

58

Calculate the flow rate through the following pipes in gallons per minute (gpm) Remember that 1 cubic foot = 7.48 gallons

11. Flowing at 20 feet per second through a 36 inch pipe = _____________ gpm



14. Flowing at 200 feet per minute through a 36 inch pipe = _____________ gpm

15. Flowing at 150 feet per minute through a 24 inch pipe = _____________ gpm

Pumps

59

Lesson 9 Pumps Key Concepts

• Pressure – 1 foot of water height = 0.43 psi gage pressure • Head – The vertical distance between start and finish water level in feet

• Work – Work can be expressed as lifting a weight a certain vertical distance

(Units are usually ft-lbs)

Work = Weight, lb x Height, ft

• Power – Power is the rate of doing work (Units are usually ft-lb/min)

Power = Work, ft-lb Time, min.

• Horse Power – A unit of power defined as 33,000 ft-lb/min.

HP = Power, ft-lb/min 33,0000 ft-lb/min/HP

• Water Horse Power – The amount of power required to lift water

Water HP = (Flow, gpm) x (Head, ft) 3960 gpm-ft/HP

• Break Horse Power – Takes into account that pumps are not 100% efficient

Break HP = (Flow, gpm) x (Head, ft) 3960 gpm –ft/HP x Ep

• Motor Horse Power – Takes both pump and motor efficiency into account

Motor HP = (Flow, gpm) x (Head, ft) 3960 gpm-ft/HP x Ep x Em

• 1 Horse Power is equal to 0.746 Kw

60

Example 1: Calculate the Motor horse power required to pump water at a flow rate of 300 gpm against 25 ft of head. Pump efficiency is 65% and motor efficiency is 90%. Example 2:

Calculate the daily electrical cost to operate the pump in example 1 if the cost of electricity is $0.12 per Kw-hr.

Pumps

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Solve the following for Water Horse Power Water HP = (Flow, gpm) x (Head, ft.) 3,960 gpm-ft/HP

1. A flow of 300 gpm is pumped against a head of 70 feet. What is the Water HP?





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Solve for Brake Horsepower: Break HP = (Flow, gpm) x (Head, ft.) (3,960 gpm-ft/HP) x (Ep)

6. A flow of 200 gpm is pumped against a head of 25 feet by a pump that is 75% efficient. What is the Break HP?





Pumps

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Solve for Motor Horsepower:

Motor HP = (Flow, gpm) x (Head, ft) (3,960 gpm-ft/HP) x (Ep) x (Em)

11. A flow of 225 gpm is pumped against a head of 75 feet. The pump is 70% efficient and the motor is 80% efficient. What is the Motor HP?





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Calculate the electrical cost for the following: 1 HP = .746 Kw

16. Energy cost is $0.11 per Kw-hr. Pump HP is 17. What is the daily cost to operate this pump?

17. A pump is to pump 300 gpm against a head of 30 feet. Electricity cost $0.10 per Kw-hr. What will be the daily cost to operate this pump?

Pumps

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18. A 65% efficient pump is pumping 200 gpm against a head of 100 ft. If electricity cost $0.13 per Kw-hr, what will be the annual cost to operate this pump? Assume continuous operation.

19. A pump that is 70% efficient is pumping against a head of 160 feet. If electricity cost $0.12 per Kw-hr, what is the annual cost to operate this pump. Assume continuous

operation.

20. A 75% efficient pump is pumping 500 gpm against a head of 70 feet. If electricity cost $0.14 per Kw-hr, what is the annual cost to operate this pump? Assume continuous

operation.

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Lesson 10 The Metric System Key Concepts

Prefix Symbol Meaning micro µ 0.000001 milli m 0.001 centi c 0.01 deci d 0.1 “unit” 1 Deka da 10 Hecto h 100 Kilo K 1,000 Mega M 1,000,000 Common Units of Measure

Weight Volume Length mg (milligram) ml (milliliter) cm (centimeter) g (gram) L (Liter) m (meter) Kg (Kilogram) Kl (Kiloliters) Km (Kilometer) Temperature oC (celsius) Example 1: Convert 2,345 g to Kg

Example 2: Convert 4,356 ml to L Example 3: Convert 7,856 m to Km

Metric System

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Convert between the following weight units: 1 g = 1,000 mg 1 Kg = 1,000 g 1. 7,235 g = _____________ Kg

2. 2.8 g = _____________ mg

3. 2.6 Kg = _____________ mg

4. 0.75 Kg = _____________ g

5. 0.35 g = _____________ mg

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Convert between the following volume units: 1 L = 1,000 ml

6. 7,858 ml = _____________ L

7. 3.85 L = _____________ ml

8. 5,285 ml = _____________ L

9. 2 L = _____________ ml

10. 7.2 L = ____________ ml

Metric System

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Convert between the following distance units: 1 m = 100 cm 1 Km = 1,000 m

11. 250 cm = _____________ m

12. 750 cm = _____________ m

13. 2 m = _____________ cm

14. 12 Km = _____________ m

15. 700 m = _____________ Km

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Lesson 11 Problem Solving 5 Step Approach: 1. Identify the problem and list what is known and unknown 2. Select an appropriate formula to solve the problem 3. Arrange the formula to solve for the unknown 4. Plug the known values into the formula and solve for the unknown value 5. Convert answer to the appropriate units Example 1 The chlorination system is feeding the CCB at a dose of 3 mg/L. Find the plant flow in gpm if the chlorine scale shows that 500 pounds were used in the last 12 hours. Step 1 – Identify the problem and list what is known and what is unknown Known Unknown Step 2 – Select the appropriate formula Step 3 – Arrange formula to solve for unknown Step 4 – Plug the known values in and solve for unknown Step 5 – Convert answer to the appropriate units

Problem Solving

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Example 2 Jar testing shows that the optimum polymer dose is 12 mg/L. Find the proper setting on the chemical feeder in ml/min when plant flow is 4.7 MGD. The liquid polymer delivered to the plant contains 642.3 milligrams of polymer per milliliter of liquid solution. Step 1 – Identify the problem and list what is known and what is unknown Known Unknown Step 2 – Select the appropriate formula Step 3 – Arrange formula to solve for unknown Step 4 – Plug the known values in and solve for unknown Step 5 – Convert answer to the appropriate units

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Example 3 Jar testing shows that the optimum polymer dose is 12 mg/L. Find the proper setting on the chemical feeder in ml/min when plant flow is 4.7 MGD. The liquid polymer delivered to the plant contains 5.36 lb of polymer for each gallon of liquid solution. Step 1 – Identify the problem and list what is known and what is unknown Known Unknown Step 2 – Select the appropriate formula Step 3 – Arrange formula to solve for unknown Step 4 – Plug the known values in and solve for unknown Step 5 – Convert answer to the appropriate units

Problem Solving

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Example 4 Find the chemical feed rate in pounds of polymer per day from a chemical feed pump. The polymer solution is 1.5%. Assume a specific gravity of 1.0. During a test run, the chemical feed pump delivered 800 ml of polymer solution during a 5 minute period. Step 1 – Identify the problem and list what is known and what is unknown Known Unknown Step 2 – Select the appropriate formula Step 3 – Arrange formula to solve for unknown Step 4 – Plug the known values in and solve for unknown Step 5 – Convert answer to the appropriate units

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Lesson 12 Test Taking Strategies / Practice Test Before the Test

I. Establish a Study Program A. Begin well in advance of the test B. Establish the material you intend to cover C. Create a plan and systematically work your plan

II. Memorize common formulas A. Use flash cards to memorize formulas B. Learn when and how the formula is used

III. Memorize common equivalents

A. Use flash cards to memorize equivalents B. Learn when and how the equivalent is used

IV. Study concepts you aren’t sure about A. Don’t waste time studying concepts you already know B. Take time to get answers to concepts you are unsure of

V. Ensure proper rest the night before A. Don’t change your normal routine the night before the test B. Don’t try to cram the night before

During the Test

I. Arrive prepared A. Bring a properly functioning, basic calculator you are familiar with B. Have a good supply of pencils C. Proper ID D. Test admittance card

II. Review the test before beginning A. Look through the test once it has begun B. Work the easy math problems first C. Answer multiple choice questions next D. Work essay problems if applicable E. Finish with difficult math problems

III. Answer all questions A. Eliminate obvious incorrect answers B. Make your best guess

Practice Test

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1. Estimate the detention time for a rectangular basin that has a flow of 2.4 MDG.

The basin is 30 feet long, 17 feet wide, and 9 feet deep. a. 15 min b. 20 min c. 25 min d. 30 min

2. Calculate the correct setting for a dry alum feeder in pounds per day if jar testing

shows the optimum dose is 12 mg/L and plant flow is 5.4 MGD a. 500 lb/day b. 450 lb/day c. 540 lb/day d. 480 lb/day

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3. If plant flow averages 3.2 MGD, calculate how many pounds of polymer will be required to feed the plant at a dose of 2.1 mg/L for 14 days.

a. 785 lb b. 625 lb c. 750 lb d. 830 lb

4. If 30 pounds of cationic polymer were consumed to treat 2.7 million gallons of water in a 24 hour period, what was the polymer dose in mg/L?

a. 0.9 mg/L b. 1.1 mg/L c. 1.3 mg/L d. 1.5 mg/L

Practice Test

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5. How much liquid polymer that is a 7% solution should be added to a 200 gallon

mix tank to produce a 0.5% solution? a. 12.5 gal b. 13.2 gal c. 14.3 gal d. 15.3 gal

6. A liquid alum solution used to treat water contains 643 mg of alum for each ml of solution. Jar testing shows that the optimum dose is 11 mg/L. Find the setting in ml/min for the alum feed pump if plant flow is 3 MGD.

a. 125 ml/min b. 135 ml/min c. 145 ml/min d. 155 ml/min

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7. Calculate the volume of water in gallons of a tank with a diameter of 25 feet that has 12 feet of water in it.

a. 41,039 gal b. 42,039 gal c. 43,039 gal d. 44,039 gal

8. How much does a 55 gallon drum weigh if it is full of water? (ignore the weight of the drum)

a. 479 lb b. 469 lb c. 459 lb d. 449 lb

Practice Test

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9. Calculate the surface area of a circular clarifier with a 15 foot radius.

a. 607 ft2 b. 707 ft2 c. 807 ft2 d. 997 ft2

10. How much copper sulfate is required to produce a concentration of 0.6 mg/L of copper in a 15 million gallon reservoir? (copper sulfate is 25% copper)

a. 300 lb b. 400 lb c. 350 lb d. 75 lb

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11. Calculate the surface loading rate for a plant flowing 1.8 MGD and a

sedimentation basin that measures 70 feet long by 30 feet wide. a. 12 gpm/ft2 b. 6 gpm/ft2 c. 0.6 gpm/ft2 d. 0.12 gpm/ft2

12. Find the weir loading rate in gpm/ft if plant flow is 1.2 MGD and the combined weir length is 30 feet.

a. 28.7 gpm/ft b. 27.8 gpm/ft c. 23.4 gpm/ft d. 24.3 gpm/ft

Practice Test

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13. Calculate the filtration rate in gpm/ft2 for a treatment plant flowing 1.4 MGD with

540 ft2 of filter surface area. a. 1.4 gpm/ft2 b. 5.0 gpm/ft2 c. 4.2 gpm/ft2 d. 1.8 gpm/ft2

14. If there is 6 feet of head on a gravity filter, what is the pressure in psi on the filter surface?

a. 2.6 psi b. 1.6 psi c. 3.6 psi d. 4.6 psi

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15. Calculate the upward force in lb on the bottom of an empty tank due to 4 feet of ground water above the tank bottom. The tank bottom has a surface area of 250 ft2.

a. 45,400lb b. 54,500 lb c. 62,400 lb d. 74,800 lb

16. What percent of finished water is used for backwashing if 25,000 gallons is used for each 1.5 million gallons produced?

a. 2.8 % b. 1.7% c. 3.4% d. 5.6%

Practice Test

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17. Calculate the flow rate in gpm of water flowing through an 18 inch pipe at 3 feet

per second. a. 2,475 gpm b. 2,380 gpm c. 1,270 gpm d. 950 gpm

18. Estimate the flow velocity in a channel in which a float travels 30 feet in 15 seconds.

a. 5 ft/sec b. 4 ft/sec c. 3 ft/sec d. 2 ft/sec

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19. If a chlorine cylinder weighs 1400 lb at 12:00 p.m. and later weighs 1380 lb at

8:00 p.m., find the dose in pounds per day. a. 40 lb/day b. 50 lb/day c. 60 lb/day d. 70 lb/day

20. What is the chlorine demand of water being treated with a dose of 2.5 mg/L chlorine and has a residual of 1.6 mg/L after 30 minutes?

a. 4.1 mg/L b. 1.1 mg/L c. 0.9 mg/L d. 1.8 mg/L

Practice Test

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21. What is the wire to water efficiency of a pump if water HP is 8 and the power

input is 11 HP? a. 66% b. 71 % c. 73% d. 78%

22. Find the Motor HP required if the water HP required is 40 HP and the pump efficiency is 75% and the motor efficiency is 88%.

a. 59.8 HP b. 60.6 HP c. 61.4 HP d. 64.0 HP

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23. A temperature of 80oF is equivalent to how many degrees Celsius?

a. 27oC b. 28oC c. 29oC d. 30oC

24. A temperature of 12oC is the same as how many degrees Fahrenheit? a. 44oF b. 54oF c. 64oF d. 34oF

25. How long should a 60 feet long sample with a pipe ¾ inch diameter be flushed with a flow of 0.5 gpm if you need to flush two line volumes before sampling?

a. 2.5 min b. 3.25 min c. 5.5 min d. 4.5 min

Answer Key

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Lesson 1 Lesson 2 Lesson 3 Lesson 4 Lesson 5 Lesson 6 Lesson 7 Lesson 8 Lesson 9 Lesson 10 Lesson 11 Lesson 12

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