Some Equations and Applications for Drinking Water Operations Let’s Look at Math Some Equations and Applications for Drinking Water Operations Patricia Quinn and Russell Koenig

Patricia Quinn - Environmental Health and Water Quality Technology Student at Milwaukee Area Technical College, Certified Surface Water Treatment Operator, WAV Monitor, Member, WIAWWA Russell Koenig - Environmental Health and Water Quality Technology Student at Milwaukee Area Technical College, Certified Surface Water Treatment Operator, interned at Northshore Water Commission

Why Are We Here? Clean, potable, water is vital to life and health. Which makes water operators really important! (Hey, that’s us!)

Way more important in case of a zombie apocalypse…sorry doctors. Why Are We Here? Way more important in case of a zombie apocalypse…sorry doctors.

Why Are We Here? It’s also important to be able to “do the math” the old-fashioned way (in event of a worst-case-scenario computer/automated system crash…or a literal zombie apocalypse, of course).

Basic Equations 1 gal H2O = 8.34lbs The weight of water in pounds: The volume of water in gallons: 1 ft.3 H2O = 7.48gal

Basic Equations Density = Mass (g) Volume (mL) Surface area = (Length)(Width)

Basic Equations Velocity: V = Distance Time Flow: Q = AV (Area)(Velocity)

Basic Equations Volume of a rectangular basin (in gal): (Length)(Width)(Depth)(7.48 gal/ft.3) Volume of a circular basin (in gal): (0.785)(Diameter2)(Depth)(7.48 gal/ft.3)

lb/day = (Cmg/L )(QMGD)(8.34 lb/gal) Basic Equations Our Favorite, The Pounds Formula: lb/day = (Cmg/L )(QMGD)(8.34 lb/gal)

DT = (basin volume (gal))(24 hrs./day) Explain how to calculate theoretical hydraulic detention time in rectangular and circular basins: Formula for Detention Time: DT = (basin volume (gal))(24 hrs./day) Flow (in gal/day)

We need to first calculate the volume For example: Calculate the theoretical hydraulic detention time for a circular basin that is 40ft. in diameter,12ft. deep, with a flow of 2 MGD: We need to first calculate the volume (in gal) of the basin:

Volume of the Circular Basin Vgal= (0.785)(40ft.2)(12ft.)(7.48 gal/ft.3) V= 112,738.56 gal Note: for a rectangular basin, we would use (length)(width)(depth)(7.48 gal/ft.3)

Next, we simply “plug and chug” our information into the Detention Time Formula: DT = (112,738.56 gal)(24 hrs./day) 2,000,000 gal/day DT = 1.35 hrs., which is 81 min.

Solution: The theoretical hydraulic detention time for a circular basin that has a 40ft. diameter, is 12ft. deep and has a flow of 2 MGD, is 1 hour and 20 minutes.

Explain how to calculate backwash flow rate: Formula for backwash flow rate: Backwash Rate(gpm/ft.2) = Flow Rate (gpm) Filter Area (ft.2)

We need to first calculate the surface area of the filter: For example, calculate the backwash flow rate of a filter bed that is 30ft. in length, 10ft. in width, and has a flow rate of 3120 gpm: We need to first calculate the surface area of the filter: Area = (30ft.)(10ft.) = 300ft.2

Backwash Rate = 10.4 gpm/ft.2 300ft.2 Now, let’s “plug and chug” the information we know into the Backwash Flow Rate formula: Backwash Flow Rate(gpm/ft.2)= 3120 gpm 300ft.2 Backwash Rate = 10.4 gpm/ft.2

Solution: The backwash flow rate of a filter bed that is 30ft. in length, 10ft. in width, and has a flow rate of 3120 gpm, is 10.4 gpm/ft.2.

Formula for backwash percentage: Explain how to calculate the percent of total water production used for backwashing: Formula for backwash percentage: Backwash % = (Backwash Water (gal))(100%) Filtered Water (gal)

For example, calculate the percentage of total water production used for backwashing when a plant uses 74,000 gallons and produces 18.1 MGD: Backwash % = (74,000 gal)(100%) 18,100,000 gal Backwash % = 0.41

Solution: The percentage of total water production used for backwashing when a plant uses 74,000 gallons and produces 18.1 MGD, is 0.41%.

Where do zombies go swimming? THE DEAD SEA!

Determine the amount of a concentrated solution required to achieve a diluted concentration in a larger volume: Formula: C1V1 = C2V2

For example, how many mLs of formazine at 4000 NTU are needed to make 1 liter at 20 NTU concentration? Known Variables: C1 = 4000 NTU V1 = unknown (mL) C2 = 20 NTU V2 = 1 liter (1000mL)

We can now solve for V1 by inserting the known variables into our formula: (4000 NTU)( V1) = (20 NTU)(1000mL) (4000)(V1) = (20,000) V1 = (20,000 / 4000) V1 = 5 mL

Solution: It would take 5 mL of formazine at 4000 NTU concentration to make 1 liter at 20 NTU concentration.

I encountered a problem: By the numbers: (4000 NTU)( V1) = (20 NTU)(1000mL) (4000)(V1) = (20,000) V1 = (20,000 / 4000) V1 = 5 mL

Maybe I missed a step? Bench Top Verification Hach Model 2100AN 16.5 NTU? An additional batch was prepared. Again 16.5 NTU was the result. Maybe I missed a step?

What went wrong? Was the primary standard solution mislabeled?

What went wrong? Assuming the concentration to be unknown: C1 = ? V1 = 5 mL C2 = 16.5 NTU V2 = 1000 mL

The calculation in reverse: (C1)(5 mL) = (16.5 NTU)(1000mL) (C1)(5 mL) = (16,500) C1 = (16,500 / 5 mL) C1 = 3300 NTU

How many mLs are really needed? (3300 NTU)( V1) = (20 NTU)(1000mL) (3300)(V1) = (20,000) V1 = (20,000 / 3300) V1 = 6.06 mL

Only a Couple More, We Promise! Calculate the pounds of Alum (Al2(SO4)3) required per day if the dosage is 7.2 mg/L, the flow is 1.2 MGD, and the alum used is 42% solution:

Pound Formula lb/day = (Cmg/L)(MGD)(8.34 lb/gal) % Available lb/day = (7.2mg/L)(1.2 MGD)(8.34 lb/gal) 0.42 lb/day = 171.57 (or about 172)

Solution: 172 lbs of Alum are required per day if the dosage is 7.2 mg/L, the flow is 1.2 MGD and the Alum used is a 42% solution.

Calculate Hardness via Titration Lab Results The Formula: (Total amt Titrant UsedmL)(Molarity)(1000mg/g)(Molar Wt CaCO3) Sample Size mL

Example Results Sample Size: 50mL Titrant Used: 0.02M Standardized EDTA solution Initial Burret Reading: 0.0mL Endpoint Burret Reading: 3.7mL Total Amount of Titrant Used: 3.7mL

Plug and Chug! (3.7mL)(0.02M)(1000mg/g)(100.1g/mol CaCO3) 50mL = 148.1 mg/L CaCO3

Convert to Grains Per Gallon 1 gpg = 17.1 mg/L (or ppm) CaCO3 Therefor, 148.1 mg/L = 8.7 gpg 17.1 mg/L/gpg According to a table on hardness by the WQA (Water Quality Association), a range of 7.0-10.5gpg or 120-180mg/L CaCO3 indicates hard water.

In Conclusion: Math for water operators may seem daunting at first, but with increased familiarity and some practice, we see that a lot of the formulas and concepts are easily applied.

In Conclusion: But of course, we thankfully no longer have to rely upon hand-written calculations to determine every setting, dosage and concentration. In SCADA we trust!

In Conclusion: Hopefully the zombie apocalypse never actually occurs…but if it does and we know this math, we can rule the water supply! Muahaha!

Thank You! Thank you so much for your time and attention! We are honored to be here! Thanks to Bill Fowler, President of MWAA, for inviting us to present!

Thank You! A HUGE, special thank you to Kathy Bates- the Environmental Health and Water Quality Technology instructor at Milwaukee Area Technical College and our mentor!

Are You an MATC Alumnus? We are celebrating the Environmental Health and Water Quality Technology program’s 50th anniversary on Tuesday, March 20th 2018! We would be honored if you joined us for our big party!

Please Join Us! Contact Kathy Bates batesks@matc.edu

One Last Equation: How many adult beverages do Water Operators require after listening to an entire presentation of spoken mathematics? Solution: At least as many as the students who painstakingly and lovingly prepared this presentation will be having to celebrate!