# Math Basics I Area, Volume, Circumference, Converting Inches to Feet, Converting Flow, Converting Percent/Decimal Point, Decimal Point, Detention Time.

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Math Basics I Area, Volume, Circumference, Converting Inches to Feet, Converting Flow, Converting Percent/Decimal Point, Decimal Point, Detention Time

Volume for Rectangular Tank, cuft www.wastewater101.net2  Next step will be to find the VOLUME, in cubic feet  The formula to find the Volume, cuft is: Volume, cuft = Length, ft X Width, ft X Height, ft (notice you are multiplying 3 feet lengths together and the finished product is in cubic feet or feet cubed)  When dealing with Volume include the depth or liquid level to the surface area

Circular container calculations www.wastewater101.net  Finding the volume in cubic feet of a circular tank, basin or container will utilize 3.14 (pi or π ), the radius in feet and the depth or height.  The formula is: Volume, cuft = 3.14 X r 2 - OR - Volume, cuft = 3.14 x R, ft X R, ft X H, ft (R 2 isn’t radius times 2 it is radius times radius) (notice you are multiplying 3 feet lengths together and when you are finished the product is in cubic feet or feet cubed – 3.14 is considered a “constant” not a number)  When dealing with Volume you include the depth (height) or liquid level 3

Wastewater Math Basics - II Basic Level – Converting Cubic Feet, Gallons and Pounds, Population Equivalency, Ponds, Chlorine www.wastewater101.net 4

cfs cfm gps gpm lbs/sec lbs/min cfd lbs/day 60 sec/min 1440 min/day 7.48 gal/cuft 8.34 lbs/gal gpd 7.48 gal/cuft 8.34 lbs/gal 60 sec/min 1440 min/day 5

Working Problems (Review) www.wastewater101.net  1. I have 1360 lbs of solids in the effluent of my plant. How many gallons of solids is that equal to?  Answer: (1360 lbs) / (8.34 lbs/gallon) = 163 gallons  2. I am using chlorine in the plant for disinfection and I add 15 pounds per day, how many lbs per hour is that? Answer: (15 lbs/day) / (24 hours/day) = 0.625 lbs/hour  3. Convert 4678 cuft/day to cuft/sec. Answer: (4678 cuft/day) / (1440 min/day) / (60 sec/min) = 0.054 cuft/sec 6

Your Turn www.wastewater101.net7  Find the Flow, MGD Velocity – 1.9 ft/sec Width – 2 ft 7 inches Depth – 11 inches  Answer: 1.9 ft/sec X 2.58 ft X 0.92 ft = 4.5 cuft/sec 4.5 cuft/sec X 7.48 gal/cuft = 33.7 gal/sec 33.7 gal/sec X 60 sec/min X 1440 min/day = 2911680 gpd 2.9 MGD = (7”) / (12”/ft) = 0.58 ft + 2 ft = 2.58 ft = (11”) / (12”/ft) = 0.92 feet

Wastewater Math Basics - III Basic Level – Celsius and Fahrenheit, Davidson Pie Chart, Loadings, Sludge Age www.wastewater101.net 8

How it Works Anything in the bottom half of the circle is multiplied together to get the answer on the top of the circle (lbs/day). If the top half of the circle is given and the question you are performing is asking for either the flow or the concentration you must divide the top half of the circle by sum of numbers multiplied in the bottom half. www.wastewater101.net9 TIP – Cover up the part you are trying to solve and either divide or multiply accordingly.

Water/Wastewater Math Basics - IV Basic Level = Average, Median, Mode, mg/L and ppm, Understanding Word Problems, Efficiency, BOD Information, Pressure, Force, Horsepower, Samples www.wastewater101.net 10

Problems www.wastewater101.net11  The influent TSS is 135mg/L to a primary clarifier, the primary effluent TSS is 45mg/L and the secondary TSS effluent is 2mg/L. What is the primary removal efficiency in %?  Known –  (Infl – Eff) X 100% (Infl)  Primary Influent TSS – 135 mg/L  Primary Effluent TSS – 45 mg/L  Secondary Effluent TSS – 2 mg/L  Unknown – Primary % Removal Plug in the correct information and do the math: (135 mg/L – 45 mg/L) X 100% 135 mg/L = 90 mg/L X 100% 135 mg/L = 0.67 or 67% Effl 45 mg/L Primary Infl 135 mg/L Sec Effl 2 mg/L

Problem www.wastewater101.net12  Motor Horsepower (HP) = (Flow, gpm)(Total Head, ft) (3960 gal/min/ft)(Pump Eff/100)(Motor Eff/100)  Find the Motor Horsepower for a pump discharging 4.0 MGD against a Total Head of 14 feet. Assume the pump is 70% efficient and the motor is 90% efficient.  Answer: 4.0 MGD = 4,000,000 gpd 4,000,000 gpd / 1440 min/day = 694 gpm 70% = 0.7 90% = 0.9 (694 gpm)(14 ft) =3.9 HP (3960)(0.7)(0.9)  Where Did 3,960 Come From? One HP is 550 foot-pounds per second. Multiply this by 60 seconds in a minute and we have 33,000 foot pounds per minute, or horsepower-minutes. Next, a gallon of water (at sea level and 70 F°) weighs 8.333 pounds. Divide the 33,000 ft.-pounds by 8.333 pounds per gallon and we have 3,960. The “3,960” is a horsepower expressed in pump terminology. So the units that “you don’t see” for 3960 is gal/min/ft  (Pump Eff/100) & (Motor Eff/100) Changes the % to decimal

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