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Math should work for you, not intimidate you!

Math should work for you, not intimidate you. This guide will help you understand and use math in everyday situations. Learn important constants, conversion factors, and formulas for length, area, volume, and more. Discover tricks and techniques to solve math problems step by step. Take a deep breath, relax, and let math become your ally.

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Math should work for you, not intimidate you!

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  1. Math should work for you, not intimidate you! • Math is a means to tell us “how much” • Fundamentally it is a tool for you to use! • Math has power • Math can help you communicate and CONVINCE! • Use numbers to your advantage. • Math can actually be fun. (like a puzzle) • As in any discipline, Take a deep breath and don’t stress -- one step at a time. • A few tricks of the trade can help!

  2. Constants/ Conversion Factors Pay attention to Units • 1’ head = .434 psi • 1 psi = 2.31 ‘ head • 1 ft3 of volume = 7.48 gallons • 1 gallon water = 8.34 pounds • 1 gallon water = 3.785 Liters • 1 ton = 2000 lbs • 1 mile = 5280 ft • 1 acre = 43,560 ft2

  3. Constants/Conversion Factors • 1 hp = 746 watts or .746 kW • 1% = 10,000 mg/l • F = (C x 9/5) + 32 • C = (F – 32) x 5/9 • Pie = 3.14 • 1 kg = 2.2 lbs

  4. Length, Area and Volume – square/rectangular • Length: size of a line, how long is a pipeline or a road, etc. Measure in inches, feet, miles, cm, meters, km. • Area: how big is a surface, how big is a wall, or parking lot, etc. Measure in ft2 , acres, m2, hectares. • Volume: how much does it hold, as in a tank, a bottle, a truck bed • Area = length X width • Volume = area x depth (or length x width x depth) • What is the floor area of a tank 4 ft wide by 8 ft long? • If that tank is 6 feet deep what is its volume? • How many gallons is that? • If 25% full? • Volume of 18” long by 32” wide and 2’ deep tank holds how much water? • (hint: units can trip you up here)

  5. Volumes- Circular many problems can be seen as cylinders from circular clarifiers to pipelines. • Area = Pi x R2 • Volume = area x depth (or if laying down like a pipe V = A x length) • A 10 MGD facility has an 118’(diameter) clarifier with 13.5’ sidewall depth, how many gallons can it hold? (First figure volume in cubic feet then convert to gallons.) • How much wastewater is held in an 18 inch interceptor running for 1 and ½ miles? (note 2 different units are in play) Then in the real world, does the pipe run full? Diameter Radius

  6. Volumes • An aerobic digester is 50 ft in diameter, with a side water depth of 10 ft. If the desired air supply for this digester was determined to be 40 cfm/1000 - cu ft of digester capacity, what is the total cfm required for this digester? • (.785*50*50*10)/1000)= 19.6 1,000 cu ft • 40 / 1000 = x / 19.6 1000 cu ft • 40 * 19.6 =784 cfm Air

  7. Concentration and mass relationships • “Loading” formula: flow(mgd) X Concentration(mg/L) X 8.34 = pounds/day • (what is this “magic factor” 8.34?) basically it is the weight of water in English units! One gallon of water weighs 8.34 lbs. Thus, looking at units because we are dealing with dilute solutions, the approximation is used that 1 liter of solution weighs the same as one liter of water, thus, 1L weighs 1Kg = 1,000 grams, and a milligram is 1/1,000 gram, so when calculating the million in million gallons (or mgd) is cancelled by the mg/L (which by the way is essentially the same as a part per million) • So you have millions time parts per million then multiply by the unit weight of water (in English units) to get pounds! • What is the influent loading to a plant that treats 2.5 mgd if the BOD5 concentration is 210 mg/L? • How many pounds of solids are in an aeration basin holding 1.1 million gallons if the mixed liquor concentration is 3150 mg/L?

  8. Pounds Formula • A 10 MGD facility has an effluent TSS of 12 mg/l, how many lbs of TSS do they discharge daily? • A 1.4 MGD facility has an effluent BOD of 9.3 mg/l, how many lbs of BOD are discharged daily? • A sludge tank has 85,000 gallons of 2.5% solids in it, how many lbs of solids are in the sludge tank? • A 1.2 MGD facility doses CL2 at 7 mg/l and has a residual of 1.2 mg/l, how many lbs of cl2 were demanded?

  9. Pounds @ LAWPCA • Aeration tank 216’ long by 76’ wide with a depth of 15.95’ and holding a MLSS of 7520 mg/l hold how many lbs of MLSS? • A 118’ secondary clarifier with a depth of 13.50’ has a RAS concentration of 5760 mg/l, how many lbs of solids are in the clarifier?

  10. Pounds Formula • A wastewater flow of 850,000 gpd requires a chlorine dosage of 25 mg/l. If hypochlorite (65% available chlorine)is to be used, how many lbs/day of hypochlorite are required ? • 850,000/1,000,000 = .85 • .85 * 25 * 8.34 = 177 lbs/day • 177 / .65 = 272 lbs/day

  11. Organic “loading” rate • Organic load (lb Bod/Day) Volume (ft3) • A trickling filter, 85 feet in diameter with a media depth of 5 feet, receives a flow of 1,200,000 gpd. If the BOD concentration of the primary effluent is 160 mg/L, what is the organic loading on the trickling filter in lbs BOD/day/1000 cu.ft.? 1.2 * 160 * 8.34 85 * 85 * .785 * 5/1000 = 56.5 lbs BOD/day/1000ft3

  12. Detention Time • Can also be referred to as sludge retention time (SRT) in certain applications Volume (gals) Flow (gpd)(1 day/24 hr) • Units may not always be in days • If you were to fill your 20’ wide and 4’ deep swimming pool with water at 20gpm how many minutes will it take you?

  13. Detention Time • A circular clarifier has a capacity of 120,000 gals. If the flow to the clarifier is 1,600,000 gpd what is the detention time? 120,000 / (1600000/24) = 1.8 hrs • Two 690,000 digesters receive 25,000 gallons of primary and 20,000 gallons of TWAS daily, what is the detention time in days?

  14. Mean Cell Residence Time (MCRT) Plant flow = 4.0 MGD, Final Eff. TSS 20mg/l, Aer. Vol. = 800,000 gal. and mlss=2650. • Suspended solids in Aeration (lb) SS wasted(lb/d) + SS lost(lb/d) • The goal at a particular activated sludge plant is to operate with a 4 day sludge age Given the following how many lbs of solids need to be wasted today to reach the sludge age? • 2650 mlss * .8 mg * 8.34 = 17681 lbs under Aeration • 17681 / 4 days target mcrt = 4420 lb/day • 4420 - (20 mg/l * 4.0 * 8.34) 667 lbs = 3753 lbs

  15. Food To Mass • Food is the BOD loading (primary effluent) to the reactor (aeration basin) • F:M ratio .8- 1.2 normal BOD loading (lb/d) Microorganism in reactor (lb) • 1.4 MGD facility with a primary effluent BOD of 57 mg/l flows to an aeration basin 150’ long, 70’ wide, and 15’ deep with a MLVSS of 6250 mg/l has a F:M ratio of?

  16. Food to Mass Ratio • The desired F/M ratio for an activated sludge plant is 0.3 lbs BOD/lb MLVSS. It has been determined that 5900 lbs/day BOD enter the aeration tank. The MLSS Volatile Solids content is 70 %. Based on the desired ratio,what is the desired lbs MLSS in the aeration tanks? • .3 lbs/day Bod/ 1lb MLVSS = 5900 lbs/day BOD / x lbs MLVSS • x = 5900 / .3 = 19,667 lbs MLVSS • 19,667/.7 = 28096 lbs MLSS

  17. Food to Mass Ratio • The desired F/M ratio at a particular activated sludge plant is 0.5 lbs COD/l mixed liquor volatile suspended solids. If the 3.6 MGD primary effluent flow has a COD of 165 mg/l how many lbs of MLVSS should be maintained in the aeration tanks? • F/M = COD Lbs/day / MLVSS LBs • .5 = (165 * 3.6 * 8.34) / (x) • x = 165 * 3.6 * 8.34 / .5 • x= 9908 Lbs MLVSS

  18. Velocity • What is the velocity in ft/sec in a 12" diameter force main carrying a flow of 1736 gpm ? • Pipe x-section area = πr2=.785ft2 • 1 ft3 = 7.48 gallons • 1 ft3/sec =450 gpm= 7.48x60 • 1736/450 = 3.86 ft3/sec • Velocity = flow rate/Area= (3.86 ft3/sec)/(.785 ft2) • Ans.= 4.9 ft/sec

  19. Slope • If a sewer line is 1.5 miles long and the elevation drop is 10 feet over this length, What is the percent grade? • Slope = Rise/Run * 100 • 1.5 x 5280 ft = 7920 ft • 1.5 mile = 7920 ft • 10/7920 = .13% grade

  20. Population Equivalent • Population = (BOD lbs/day) / Lbs Bod / day / person • A 100,000 gpd wastewater flow has a BOD content of 280 mg/l. • Using an average of 0.2 lbs/day BOD/person, what is the population equivalent of this flow. • (280 mg/l * 0.1 * 8.34) / .2 Lbs BOD/day / Person • Ans: 1,167

  21. Population Equivalent • What is the population equivalent of an industry that contributes 4165 lbs of B.O.D to a plant? (1 P.E is equivalent to .17 lb BOD/day.) • P.E= 4165 lb Bod / day =24500 .17

  22. Horsepower • 1 hp = 746 watts • If you are running a 5 hp pump continuous and paying $0.10/kw for electricity how much does it cost to run the pump per day? • If you are running a 200 hp blower for an average of 8 hours per day and paying $0.09/kw for electricity how much does it cost to run the blower per month (assume 30 days per month)?

  23. Horsepower • The pressure in a horizontal water pipe is measured between two points 200 ft apart. The total pressure drop between points is 15 psi and the flow rate is 90 gpm. What is the horsepower used in pumping flow between these two points? • Given 1 psi = 2.31 ft and 1 hp = 33000 ft-lb/min • 15psi * 2.31=34.65 ft • 90 gpm * 8.34 = 750.6 lb/min • 34.65 ft * 750.6lb/min = 26,008 • 26008 ft-lb/min / 33000 ft-lb/min = .79 hp

  24. BOD - Biochemical Oxygen Demand Initial DO (mg/l) – Final DO (mg/l) Sample used (ml)/Total sample (ml) Sample amount = 300 ml Influent, Primary, Effluent Seeded vs. unseeded Must deplete 2.0 mg/l over 5 days and have a day 5 result greater than 1 mg/l Example: Day 1: 8.90 mg/l Day 5: 3.40 mg/l Sample amount 10 ml 8.9 – 3.4 10 / 300 = 183.3 mg/l BOD

  25. Solids Concentration Weight (mg) Volume (L) =mg/l 1% = 10,000 mg/l • If you pour 2 liters of liquid through a filter and then weigh the filter to get a weight of 100 mg, what is the concentration of that liquid? • If you pour 100 ml of liquid through a filter and then weigh the filter to get a weight of 50 mg, what is the concentration of that liquid?

  26. Sludge Volume Index 30 min reading (ml/l) x 1000 ml/l Aeration system suspended solids (mg/l) 30 min reading: 200 ml/l MLSS: 5800 mg/l 800 x 1000 / 8700 = 92 80 to 120 good range

  27. Removal Efficiency (In – out) x 100 in Influent BOD: 200 mg/l Effluent BOD: 12 mg/l Percent removed? $10 in wallet Spend $3.25 on coffee What Percent of the money in wallet did you spend/remove? • LAWPCA has an 85% removal requirement for lbs of BOD And TSS, are they being met below? April Flow 13.2 MGD April INF TSS avg conc = 212 mg/l April EFF TSS avg conc = 27.3 mg/l April INF BOD avg conc = 196 mg/l April EFF BOD avg conc = 18.5 mg/l

  28. Dosage • Use pounds formula • Flow x concentration x 8.34 • DOSAGE – demand = residual • Your 1.7 mgd facility has a cl2 demand of 4 mg/l and would like a cl2 residual of .8 mg/l, how many lbs of CL2 must be dosed daily?

  29. Hydraulic Loading Rate or Surface loading rate Flow (gpd) Area (ft2) What is the hydraulic loading rate of a primary clarifier 215 feet long and 55 feet wide receiving a flow of 12.3 mgd? • What is the surface loading rate of a 110 foot diameter secondary clarifier receiving a flow of 9.2 mgd?

  30. Weir overflow rate Flow (gpd) Weir length (ft) • What is the overflow rate on a 70’ wide primary clarifier receiving 1.7 MGD? • A 50’ circular clarifier receives a flow of 3.7 mgd, what is the weir overflow rate of the clarifier?

  31. Weir overflow rate • A secondary Clarifier is 60 feet in diameter and it treats a flow of 2.5 mgd. It has a two sided,3 ft wide peripheral weir trough. The trough is set in 3 ft from the side wall. What is the weir overflow rate? • Outer weir diameter = 60 ft - 6ft = 54 • Length = π*D = 3.14 * 54 ft = 170 ft • Inner Weir Diameter = 60 ft - 12 = 48 ft • Length = π*D = 3.14 * 48 ft = 151 ft • Over flow rate = 2,500,000 / ( 170+151) 321ft = 7788 gpd/ft

  32. LAWPCA Tank Profiles

  33. LAWPCA Tank Profiles

  34. LAWPCA Tank Profiles

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