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Math for Water Technology MTH 082 Lecture Chapter 8&15- Applied Math for Water Plant Operators

Flow From a Faucet, Flushing a Service Line, 90 th Percentile Lead and Copper Rule, Copper Sulfate Application Rates in Waterworks Operation. Math for Water Technology MTH 082 Lecture Chapter 8&15- Applied Math for Water Plant Operators

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Math for Water Technology MTH 082 Lecture Chapter 8&15- Applied Math for Water Plant Operators

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  1. Flow From a Faucet, Flushing a Service Line, 90th Percentile Lead and Copper Rule, Copper Sulfate Application Rates in Waterworks Operation Math for Water Technology MTH 082 Lecture Chapter 8&15- Applied Math for Water Plant Operators Water Sources and Storage (Chapter 8 pg 190-191) Laboratory Ch 15 (pg 404-407); Hydraulics Chapter 7 Flow Rate- Basic science Concepts and Applications (pg 311-319)

  2. Objectives Reading assignment: Chapter 8&15- Applied Math for Water Plant Operators Water Sources and Storage (Chapter 8 pg 190-191) Laboratory Ch 15 (pg 404-407); Hydraulics Chapter 7 Flow Rate- Basic science Concepts and Applications (pg 311-319) • Flow from a faucet • Flushing a service line • 90th percentileLeadandCopper • Copper Sulfate Application

  3. Conventional Treatment • Conventional Treatment– Where are these calculations applied in the big picture? 90th Percentile Lead and Copper Copper Sulfate Application Chlorination Ozone UV 5. Clear Well Disinfection 1. Coagulation 2. Flocculation 3. Sedimentation 4. Filtration Raw Water Fast Distribution Pretreatment Slow Rapid or flash Mixing Alum, polymer Clearwell Backwash pumps Faucet and Flushing Sludge Washwater

  4. Flow From a Faucet For small water systems: Record time it takes to fill a one gallon bucket Flow, gpm= Volume (gal) Time (min) Ideal flow rate is 0.5 gpm! 0.5 gallons

  5. The flow from a faucet filled up a 1 gallon container in 45 seconds. What was the gpm flow rate from the faucet? • 0.02 gpm • 0.75 gpm • 1.3 gpm • 2700 gpm sec= 45 seconds = 45 sec/60sec/min=0.75 min Flow, gpm= Volume (gal) Time (min) Flow, gpm = Vol (gal) 0.75 (min) Flow, gpm = 1 (gal) 0.75 (min) Flow, gpm= 1.3 gpm Given Formula Solve:

  6. The flow from a faucet filled up a 1 gallon container in 52 seconds. What was the gpm flow rate from the faucet? • 1.1 gpm • 0.02 gpm • 1.9 gpm • 3200 gpm sec= 52 seconds = 52 sec/60sec/min=0.87 min Flow, gpm= Volume (gal) Time (min) Flow, gpm = Vol (gal) 0.87 (min) Flow, gpm = 1 (gal) 0.87 (min) Flow, gpm= 1.1 gpm Given Formula Solve:

  7. The flow from a faucet filled up a 1 gallon container in 1 minute 12 seconds. What was the gpm flow rate from the faucet? • 1.2 gpm • 1.12 gpm • 0.9 gpm • 0.8 gpm sec= 12 seconds = 12 sec/60sec/min=0.2 min 1 minute + 0.2 min = 1.2 minutes Flow, gpm= Volume (gal) Time (min) Flow, gpm = Vol (gal) 1.2 (min) Flow, gpm = 1 (gal) 1.2 (min) Flow, gpm= 0.8 gpm Given Formula Solve:

  8. At a flow rate of 1.2 gpm, how long (minutes) should it take to fill a one gallon container? • 1.2 min • 1.8 min • 1.2 min • 0.8 min Flow = 1.2 gpm Flow, gpm= Volume (gal) Time (min) 1.2 gpm = 1 (gal) X (min) 1.2 gpm (x min) = 1 (gal) Min X = 1 gal 1.2 gpm X= 0.8 min Given Formula Solve:

  9. Flushing a Line Flushing is considered adequate when the water in the service line has been replaced twice! Volume= 0.785 (D)(D)(L) Conversion= (7.48 gal/ft3) Flow time, min= Pipe Volume (gal) (2) Flow Rate, gpm Flow time, min = 0.785 (D)(D)(L) (7.48 gal/ft3) (2) Flow Rate, gpm Don’t forget its twice the line! 2 X

  10. How long (minutes) will it take to flush a 50 ft length of ¾ inch diameter service line if the flow through the line is 0.5 gpm? • 1.4 min • 2.11 min • 4.2 min • 1.8 min D= ¾ inch (1ft/12 inch) =0.06 ft; L= 50 ft; Rate=0.5 gpm A=0.785 (D)(D)(L)(7.48 gal/ft3) Flow time, min= Pipe Volume (gal) (2) Flow Rate, gpm D= 0.75 in/12in/1ft=0.06 ft Flow= 0.785 (0.06 ft)(0.06 ft)(50 ft)(7.48 gal/ft3)(2) 0.5 gpm Flow = 4.2 min Given Formula Solve:

  11. How long (minutes) will it take to flush a 40 ft length of ¾ inch diameter service line if the flow through the line is 0.7 gpm? • 7.0 min • 2.4 min • 1.2 min • 40.2 min D= ¾ inch, (1ft/12 inch) =0.06 ft;L= 40 ft; Rate=0.7 gpm A=0.785 (D)(D)(L)(7.48 gal/ft3) Flow time, min= Pipe Volume (gal) (2) Flow Rate, gpm D= 0.75 in/12in/1ft=0.06 ft Flow= 0.785 (0.06 ft)(0.06 ft)(40 ft)(7.48 gal/ft3)(2) 0.7 gpm Flow = 2.4 min Given Formula Solve:

  12. At a flow rate of 0.5 gpm, how long (in minutes and seconds) will it take to flush a 65 ft length ¾ inch diameter service? • 2 min, 74 sec • 5 min, 5 seconds • 91 min, 60 sec • 5 min, 30 seconds D= ¾ inch, (1ft/12 inch) =0.06 ft;L= 65ft; Rate=0.5 gpm A=0.785 (D)(D)(L)(7.48 gal/ft3) Flow time, min= Pipe Volume (gal) (2) Flow Rate, gpm D= 0.75 in/12in/1ft=0.06 ft Flow= 0.785 (0.06 ft)(0.06 ft)(65 ft)(7.48 gal/ft3)(2) 0.5 gpm Flow = 5.5 min or 5 min and 30 sec Given Formula Solve:

  13. Lead 0.015 mg/L is the Action Limit MCLG = 0 mg/L Health Effect Delays in physical or mental development Children could show slight deficits in attention span and learning abilities Source Corrosion of household plumbing systems Erosion of natural deposits Household dust containing lead from lead-based paint

  14. What are the MCLG and Action level for lead in drinking water? • 15 ppb and 15 ppb • 0 and 15 ppb • 15 ppb and 50 ppb • 10 ppb and 50 ppb

  15. The Lead and Copper Rule addresses which of the following items: • Requires water suppliers to use only lead pipes for water distribution • Establishes requirements for lead pipe replacement • Determines action levels for lead and copper in drinking water • Establishes a treatment technique including corrosion control • 2, 3 and 4

  16. A system will be required to replace service lines made of lead (Pb) after what series of events - check all that apply. • System fails to meet the lead action level in tap samples after installing corrosion control. • System fails to meet the lead action level in tap samples after source water treatment. • Systems that fail to monitor for lead on a monthly basis. • Systems that have budgeted for service line replacement. • 1 and 2 Only

  17. Which of the following are potential problems associated with lead exposure in children? • Brain damage and lower intelligence • Behavior and learning problems • Aches or pains in stomach • All of the above

  18. A component of the 1986 SDWA amendments was to “GET THE LEAD OUT”? • True • False

  19. What are some of the health effects of short term exposure to lead? • Causes excessively brittle bones • Delays in physical and mental development in babies and children • Causes kidney disease • There are no short term health effects

  20. According to the Lead and Copper Rule, the action for the 90th percentile lead level is: 1. 0.005 mg/l 2. 0.015 mg/l 3. 0.030 mg/l 4. 0.050 mg/l

  21. Copper 1.3 mg/L is the MCL Health Effect Copper is an essential nutrient, required by the body in very small amounts. Short periods of exposure can cause gastrointestinal disturbance, including nausea and vomiting. Long periods of exposure liver or kidney damage Source Corrosion of copper plumbing Erosion of natural deposits Copper mining and smelting operations

  22. 90th percentile Lead and Copper The action level for lead is 0.015 mg/L and 1.3 mg/L copper? Given the following data for lead concentrations find the 90th percentile and determine if the utility has to proceed with a corrosion control program: 0.023 mg/L 0.018 mg/L 0.004 mg/L 0.019 mg/L 0.013 mg/L 0.006 mg/L 0.007 mg/L 0.026 mg/L 0.010 mg/L 0.006 mg/L

  23. 90th percentile Lead and Copper Line em up from largest at the top to smallest at the bottom. The action level for lead is 0.015 mg/L 0.026 mg/L 0.023 mg/L 0.019 mg/L 0.018 mg/L 0.013 mg/L 0.010 mg/L 0.007 mg/L 0.006 mg/L 0.006 mg/L 0.004 mg/L 90th percentile is 0.023 mg/L So it exceeds the Action Limit of 0.015 mg/L Corrosion control needed!!

  24. 90th percentile Lead and Copper The action level for lead is 0.015 mg/L and 1.3 mg/L copper? Given the following data for copper concentrations find the 90th percentile and determine if the utility has to proceed with a corrosion control program: 1.023 mg/L 1.018 mg/L 0.92 mg/L 0.87 mg/L 0.85 mg/L 0.83 mg/L 0.82 mg/L 0.80 mg/L 0.79 mg/L 0.78 mg/L

  25. 90th percentile Lead and Copper Line em up from largest at the top to smallest at the bottom. The action level for copper is 1.3 mg/L 1.023 mg/L 1.018 mg/L 0.92 mg/L 0.87 mg/L 0.85 mg/L 0.83 mg/L 0.82 mg/L 0.80 mg/L 0.79 mg/L 0.78 mg/L 90th percentile is 1.018 mg/L So it is less then the Action Limit of 1.3 mg/L No corrosion control needed!

  26. Copper Sulfate Application Copper Sulfate pentahydrate (spray or drag a bag): CuSO4 * 5H2O For algae control in a small lake, a dosage of 0.5 mg/L copper is desired. Copper Sulfate, lbs= (mg/L Cu)(vol, MG)(8.34 lb/gal) % available Cu 100 Desired Dose= Actual Dose 5.4 lbs CuSO4/1 ac = x lbs CUSO4/ Actual area, ac

  27. For algae control in a small lake, a dosage of 0.5 mg/L copper is desired. The lake has a volume of 20 MG. How many lbs of copper sulfate pentahydrate will be required? (Copper sulfate pentahydrate contains 25% available copper?) • 83.4 lbs • 40 lbs • 334 lbs • 42 lbs Given Formula Solve: Mg/L= 0.5; V=20 MG, 25% or 25/100 or .25% Cu Copper Sulfate, lbs= (mg/L Cu)(vol, MG)(8.34 lb/gal) % available Cu 100 Copper Sulfate, lbs= (0.5 mg/L Cu)(20 MG)(8.34 lb/gal) 25 % available Cu 100 Copper Sulfate, lbs= 334 lbs CUSO4

  28. Lake Oswego has a surface area of 415 acres. If the desired copper sulfate dose is 5.4 lbs/ac, how many lbs of copper sulfate are required? • 560.25 lbs • .013 lbs • 2241 lbs • 2 lbs Given Formula Solve: Desired 5.4 lbs/acre, Area 415 acres, Actual? Desired Dose= Actual Dose 5.4 lbs CuSO4 = x lbs CUSO4 1 ac Actual area, ac 5.4 lbs CuSO4 = x lbs CUSO4 1 ac 415 ac 2241 lbs CuSO4

  29. Today’s objective: Flow From a Faucet, Flushing a Service Line, 90th Percentile Lead and Copper Rule, Copper Sulfate Application Rates in Waterworks Operation has been met? • Strongly Agree • Agree • Neutral • Disagree • Strongly Disagree

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