Math Basics I Area Volume Circumference Converting Inches
Math Basics I Area, Volume, Circumference, Converting Inches to Feet, Converting Flow, Converting Percent/Decimal Point, Detention Time
Volume for Rectangular Tank, cuft � 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 2 www. wastewater 101. net
Circular container calculations � 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 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 www. wastewater 101. net
Wastewater Math Basics - II Basic Level – Converting Cubic Feet, Gallons and Pounds, Population Equivalency, Ponds, Chlorine 4 www. wastewater 101. net
cfs cfd 60 sec/min gpm 1440 min/day 8. 34 lbs/gal lbs/sec 7. 48 gal/cuft gps 1440 min/day cfm 60 sec/min lbs/min gpd 8. 34 lbs/gal 1440 min/day lbs/day 5 www. wastewater 101. net
Working Problems (Review) � 1. I have 1360 lbs of solids in the effluent of my plant. How many gallons of solids is that equal to? e you tic care o N � Answer: t ’ e n m o i d ut t (1360 lbs) / (8. 34 lbs/gallon) = 163 gallons abo � 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: You used both quantity and time (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 Time 6 www. wastewater 101. net
Problem �A float travels 400 ft in a channel in 1 min 28 seconds. What is the estimated velocity in ft/sec? Formula: Distance Traveled, ft = Velocity, ft/sec Time, sec Example: 1 min 28 seconds = 60 sec/min + 28 sec = 88 sec 400 ft = 4. 5 ft/sec 88 sec 7 www. wastewater 101. net
Wastewater Math Basics - III Basic Level – Celsius and Fahrenheit, Davidson Pie Chart, Loadings, Sludge Age 8 www. wastewater 101. net
How it Works Lbs/day TIP – Cover up the part you are trying to solve and either divide or multiply accordingly. Flow, MGD 8. 34 lbs/gal Conc, mg/L 9 www. wastewater 101. net 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.
Water/Wastewater Math Basics - IV Basic Level = Average, Median, Mode, mg/L and ppm, Understanding Word Problems, Efficiency, BOD Information, Pressure, Force, Horsepower, Samples 10 www. wastewater 101. net
Problems � The influent TSS is 135 mg/L to a primary clarifier, the primary effluent TSS is 45 mg/L and the secondary TSS effluent is 2 mg/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 11 � Secondary Effluent TSS – 2 Infl 135 in the Effl 45 Primar Sec Effl Plug correct mg/L y mg/L information and do the 2 mg/L andmath: when in doubt draw it out! The information given is more The deals than you need. with (135 question mg/L – 45 mg/L) X the To find the primaryonly. removal primary clarifier So 100% you only need theefficiency information about the 135 mg/L what comes into the secondary clarifier is not primaries and what leaves the 90 primaries. Use your Known = mg/Lneeded. X 100% and Unknown 135 mg/Lcolumns answer was 0. 66666 we or You. The can multiply 0. 67 bybut 100 rounded up 0. 67 because the 6 is = 0. 67 or to 67% you can shuffle the decimal (2) greater than 5 and we went out to the places to the right and add the % 100 th place after the decimal because that is the most significant number www. wastewater 101. net sign
Problem � � Motor Horsepower (HP) = 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. (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. 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 Answer: Convert flow from MGD to 4. 0 MGD = 4, 000 gpd GPD by using the shuffle Convert gpm 4, 000 gpdgpd / 1440 to min/day = 694 gpm Convert 70% = 0. 7 the % Efficiencies to Decimals again just Shuffle – 90% = 0. 9 in. Plug Formula that inft) =3. 9 numbers andit (694 gpm)(14 HP is where showed (Pump Eff/100) and (3960)(0. 7)(0. 9) work formula (Motor Eff/100) 12 Where Did 3, 960 Come From? � (Pump Eff/100) & (Motor Eff/100) Changes the % to decimal www. wastewater 101. net
- Slides: 12