# Shared Flashcard Set

## Details

Wastewater Math
Formula Flash cards
30
Mathematics
Professional
02/21/2013

## Cards Return to Set Details

Term
 Velocity, in ft/sec = Velocity is distance traveled over time.
Definition
 distance,  in feet  ÷  time, in seconds Example What is the velocity in feet /sec. if a stick travles a 240 foot channel in 120 seconds?   240 feet ÷ 120 seconds = 2.0 feet/sec.
Term
 Flow Rate, in (cubic feet per second),  ft3/ sec.  =
Definition
 (Velocity, ft/sec.)  (area, ft2)   Example   (2 ft/sec.)  (3 ft2) = 6 ft3 (cubic feet)/second
Term
 Change Cubic feet per second (ft3/sec )  to gallons per minute (gpm)
Definition
 Gallons per minute = (ft3/sec) (7.48 gal/ft3 ) (60 sec/min)   Example (3 ft3/sec) (7.48 gal./ft3) (60 sec/min) =   1346 gal/min or 1346 gpm
Term
 Detention Time =   This is the time, it theoretically takes a drop of water to travel from the inlet to the outlet of a tank or treatment or piping system.
Definition
 Volume, in gallons ÷ Flow, gallons per minute   Example   10000 gallon tank ÷ 500 gal/min =   20 min of detention time Note * The detention time can be in days, hours, minutes, and seconds. Your flow units must be the same as the unit of time your are using for your detention time. So if you are looking for hours then your flow should be in gal/hr.
Term
 Pounds Formula   This formula is used to find the pounds of chemical needed to maintain a partticular parts per million or miilgram per liter of a chemical of 100% purity at a fixed flow in million gallons per day or to find the pounds of solids in an areation basin, or the pounds being wasted in a WAS flow in MGD with a known WASss  The answer is always in pounds per day or pounds lbs/day  or lbs
Definition
 (Flow, in MGD or MG) (8.34 lbs/gal) (Concentration, in mg/L or ppm)   Example A plant has an Influent flow of .500MGD and an influent TSS of 200 mg/L. How many pounds of TSS come into the plant daily?   (0.500 MGD)  (8.34lbs/gal) (200 mg/L) = 834 lb. of TSS
Term
 Weir overflow rate, gal/day/foot of weir = [image]
Definition
 Total Flow, gal./day ÷ length of weir , in feet Example Your plant has an Influnet flow of 0.250MGD and a RAS flow of 50% of influent flow. The clarifier is 50 feet in diameter with a weir around the circumference. What is the weir overflow rate?   (0.250 MGD X 1,000,000) (3.14)  (50 ft)   250000 gpd  ÷ (3.14)  (50 ft)   250000gpd ÷ 157 ft   250000 gpd ÷ 157 ft 1592.4 gal./day/ft of weir
Term
 Solids Loading, lbs./day/ft2 =   This refers to the amount of Solids being sent from the Aeration basin to the Clarifier in the activated sludge process.
Definition
 Solids applied, in lbs./day ÷ Surface area, in ft2 EXAMPLE: What is the solids loading of a 50 ft. diameter clarifier if your MLSS is 2600 mg/L and you Inf. flow is .500MGD and you RAS flow is .250MGD ?                      Q    +    R (.500MGD + .250MGD) ( 8.34 lbs/gal)  (2600 mg/L) (3.14(∏))  (25 ft)  (25 ft(radius2))   16263 lbs./day 1962.5 ft2   8.3 lbs./day/ft2
Term
 Hydraulic loading,in gal./day/ft2 = This refers to the amount of flow goin into the Clarifier from the Aeration basin.
Definition
 Flow rate, in gal./day Surface area, in ft2   EXAMPLE What is the hydraulic loading of a 50 ft. diameter clarifier if your  Inf. flow is .500MGD and you RAS flow is .250MGD ?                                   Q     +    R  (500,000 gpd + 250,000 gpd)  (3.14(∏))  (25 ft)  (25 ft(radius2)) 750,000 ÷ 1962.5 =    382 gpd/ft2
Term
 Trickling Filter Organic Loading, in lbs. CBOD5 /day/1000 ft3 of Media
Definition
 CBOD5 applied, in lbs./day Volume of media, in 1000 ft3 units EXAMPLE What is the Organic loading on a trickling filter 50 ft. in diameter and 8 feet deep, with a P.E. CBOD5 of 125 mg/L, and a flow of .350MGD ?   0.350MGD X 8.34 lb./gal. X 125 mg/L (3.14) X 25 ft. X 25 ft. X 8 ft.) ÷ 1000 ft3   365 lbs./day 16 /1000 ft3   23 lbs./day/1000 ft3
Term
 Soluble CBOD5, in mg/l =   *Soluble CBOD5 is the amount of CBOD5 that is dissolved in the water and available for food for the microorganisms. It is used in calculating the Organic loading of RBC's
Definition
 (Total CBOD5  ,mg/L) - [(K)  (TSS, mg/L)] * (K is a constant that is 0.5 to 0.7 for most domestic wastewaters) EXAMPLE What is the soluble CBOD5 of a P.E. with the following make up, CBOD5 of 125 mg/L , a TSS of 150 mg/L and a K factor of  0.6 ? 125 mg/L - [(0.6) (150 mg/L)] 125 mg/L - 90 mg/l 35 mg/L of Soluble CBOD5
Term
 RBC (Rotating Biological Contactor) Organic Loading, in lbs. CBOD5 /day/ 1000 ft2 =
Definition
 Soluble CBOD5 applied, lbs./day Surface Area of Media, in 1000 ft2 units EXAMPLE An RBC has three units in parallel, each measures 10 feet in diameter and 15 feet long. The INF. Flow is 0.350 MGD and the INF. soluble CBOD5 is 45 mg/L. What is the Organinc Loading on this system?   (0.350MGD) (8.34 lbs/gal) (45 mg/L)  {(3.14) (10 ft) (15 ft) (3)}÷ 1000 ft2 131.4 lbs. ÷ 1.41 /1000 ft2 93.2 lbs/1000 ft2
Term
 SVI, in mL/gm= SLUDGE VOLUME INDEX THIS IS A CALCULATION THAT REPRESENTS THE TENDANCY OF ACTIVATED SLUDGE SOLIDS (AERATED SOLIDS) TO THICKEN AND BECOME CONCENTRATED DURING THE SEDIMENTATION PROCESS.
Definition
 (Settled Sludge Volume / Sample Volume(Settleometer result), mL/L ÷ MLSS concentration in mg/L) X (1000mg ÷  gram) EXAMPLE What is the SVI if the MLSS is 3500 mg/L and the Settleometer is 450 mL/L? (450 mL/L ÷ 3500 mg/L) (1000mg ÷  gram) (450 mL/L ÷ 3500 mg/L)  (1000mg ÷  gram) (0.129 mL/mg ) (1000 mg/gm) 129 mL/gm
Term
 SDI = *Sludge Density Index - used to calculate the settlability of Activated Sludge in a Secondary Clarifier. Similar to the SVI.
Definition
 100 SVI EXAMPLE The SDI of a sludge going to a secondary clarifier with a SVI of 129 mL/gm is? 100 ÷ 129 .78 SDI
Term
 Solids inventory, lbs. * used to calulate the ponud of TSS in an aeration tank, or in a clarifier, or a combination of both.
Definition
 Tank volume, in Million Gallons MG  X 8.34 lbs./gal. X MLSS mg/L, (for aeration), or TSS mg/L (for other tanks)   EXAMPLE What is the solids inventory of an aeration basin that is  0.750 MG with a MLSS of 4500 mg/L? (0.750 MG) (8.34 lbs/gal) (4500 mg/L) 28147.5 lbs This is just a basic pounds' formula.
Term
 Sludge Age ,  in days =   * used in the process control of an Activated Sludge plant and is controlled by wasting.
Definition
 Solidis under aeration, lbs. solids added, lbs./day   EXAMPLE What is the sludge age in days of an activated sludge plant with the following data, aeration tank volume of 0.500MG and an MLSS of 3200 mg/L, an influent flow of 0.350MGD and an influent TSS of 165 mg/L ?   (0.500MG) (8.34 lbs/gal) (3200 mg/L) (0.350MGD) (8.34 lbs/gal) (165 mg/L)   13344 lbs. ÷  481.6 lbs./day   27.7 days
Term
 FOOD / MICROORGANISM RATIO =   * used to calculate the ration between the incomming food (CBOD5) and the microorganisms in the aeration basin (MLVSS). **MLVSS is the organincs in the aeration basin and is a measure of the microorganism mass. It is sometimes expressed as a percentage of the MLSS
Definition
 (Inf. Flow MGD) (8.34 lbs/gal) (Inf. CBOD5, mg/L)  (Aeration Tank Volume, in MG) ( 8.34 lbs/gal) (MLVSS, mg/L)   EXAMPLE Given an Ilfluent Flow of 0.450MGD, an Ifluent CBOD5 of 225 mg/L,an Aeration Tank with a volume of 0.350MG, and a MLVSS of 1975 mg/L. What is the F/M ratio? (0.450MGD) (8.34 lbs/gal) (225 mg/L) (0.350MG) (8.34 lbs/gal)  (1975 mg/L)   844.4 lbs.  ÷ 5765 lbs. F/M Ratio is 0.15
Term
 Mean Cell Retention Time (MCRT) =   *The amount of time a microorganism is in the system working on breaking down the organic matter. This is controled by wasting. **It uses a solids inventory that may or may not take into account the level and concentration of the clarifier blanket.
Definition
 Solids Inventory, lbs. Eff. Solids, lbs. + WAS Solids, lbs. EXAMPLE What is the MCRT of a plant with the following: Inf. Flow of 0.600MGD Aeration Volume of 1.000MG with an MLSS of 5000 mg/L WAS TSS is 12500 mg/L and a flow of .025MG the Eff. TSS is 0.9 mg/L (1.000MG) (8.34 lbs/gal) (5000 mg/L) (0.600MGD) (8.34 lbs/gal) (0.9 mg/L) + (0.025MG) (8.34 lbs/gal) (12500 mg/L)   41700 lbs. ÷ (4.5 lbs.) + (2606.25 lbs.) 41700 ÷ 2610.75 15.9 or 16 days
Term
 WAS, lbs./day =   *used to figure out the amount of solids (in pounds) that need to be wasted to maintain the MCRT
Definition
 {(Solids inventory, lbs.) ÷ MCRT, days} - (Solids lost in Effluent, lbs./day) EXAMPLE A plant with a 0.500MG aeration tank and an MLSS of 4500 mg/L , an INF. Flow of 0.350MGD and an EFF. TSS of 1.2 mg/L, and a MCRT of 8 days. How many pounds need to be wasted per day? {( 0.500MG X 8.34 lbs./gal. X 4500 mg/L)÷ (8 days)} - ( 0.350MGD X 8.34 lbs./gal. X 1.2 mg/L) 2345.6 lbs. - 3.5 lbs. 2342.1 lbs./day to be wasted
Term
 Change in WAS flow rate, MGD =   *Calculates the amount the WAS flow need to be adjusted in MGD, a positive number means an increase if the flow, a negative number indicates a decrease in the flow rate is required.
Definition
 (Current Solids Inventory, lbs.) - (Desired Solids Inventory, lbs.) WAS, mg/L X 8.34 lbs./gal. EXAMPLE A plant has a 0.750MG aeration basin and an MLSS of 4500 mg/L, and a WAS TSS of 11500 mg/L. How much would the WAS flow need to be changed to get to an MLSS of 4000 mg/L? (0.750MG X 8.34 lbs./gal. X 4500 mg/L)-(0.750MG X 8.34 lbs./gal. X 4000 mg/L) 11500 mg/L X 8.34 lbs./gal.   (28147.5-25020)÷ 95910 3127.5 ÷ 95910 .0326 or .033MG increase
Term
 Return Sludge Rate, MGD=   *used to calculate the flow rate of RAS based on the results from the Settlometer in mL/L and to maintain that rate of settling.
Definition
 (Settleable Soilds,mL) X (Inf Flow, MGD) (1000 mL) - (Settleable solids, mL)   Example Your plant has an influent flow of 1.500 MGD and you settleometer result is 400 mL/L. What should your RAS flow rate be ? (400 mL x 1.500 MGD) ÷ (1000 mL - 400 mL) 600 ÷ 600 1.000 MGD
Term
 Population loading, person / acre =   * used in calculating the population loading of wastewater ponds.
Definition
 Population Served, persons ÷ Pond Area, acres   Example A town of 2500 has wastewater pond 1000 feet long and 2000 feet wide, what is the population loading on this pond?   2500 ÷ [( 1000 ft. X 2000 ft.)÷ 43560 ft2/acre] 2500 ÷ 45.9 54.5 or 55 persons/acre
Term
 Pond Volume, acre feet, ac-ft =   *This is the volume in acre feet and not to be confused with volume in gallons.
Definition
 (Pond area ,in acres) X (Depth, in feet)   Example What is the volume in acre feet (ac-ft) of a pond 250 ft. long and 300 ft. wide and 6 ft. deep?   [(250 ft. X 300 ft.) ÷ 43560 ft2 / acre] X 6 ft. 1.7 acres X 6 ft. 10.2 ac-ft
Term
 Pond Volume, gal   * This formula will calculate pond volume in gallons given the ac-ft of the pond.
Definition
 (Volume, ac-ft) X (43560 ft2/acre) X (7.48 gal./ft3)   Example What is the volume in gallons of a 46 ac-ft pond?   46 ac-ft X 43560 ft2/acre x 7.48 gal./ ft3.   14,988,124 gallons
Term
 Pond Flow, ac-ft/day =
Definition
 Flow, gal/day ÷ [(7.48 gal./ft3) x (43560 ft2/acre)]   Example What is the flow in ac-ft /day if the Inffluent flow is 2.000MGD ? 2.000MGD X 1000000 =   2000000 gpd ÷ (7.48 gal./ft3 X 43560 ft2/acre) 2000000 gpd ÷ 32582.8 gal/ ac-ft 61.4 ac-ft/day
Term
 Detention Time , days =   * Using ac-ft and ac-ft/day
Definition
 Volume, ac-ft ÷ Flow, ac-ft / day   Example What is the detention time in days of a pond that is 300 feet long 400 feet wide and 8 feet deep and gets a flow of 2.500 MGD? [(300 ft X 400 ft X 8 ft) ÷ 43560 ft2/ acre] ÷ [(2.500 MGD X 1000000) ÷ (7.48 gal./ft2 X 43560 ft2/acre)] 22.04 ac-ft  ÷ 7.7 ac-ft / day 2.9 days
Term
 Pond, Organic Loading, lbs./day/acre =
Definition
 (Influent CBOD5 , lbs./day ) ÷ (Pond Area , acres )   Example What is the organic loading of a 40 acre pond with a ifluent flow of 0.750 MGD and an influent CBOD5 of 250 mg/L ? (0.750 MGD X 8.34 lbs./gal X 250 mg/L) ÷ 40 acres 1563.75 lbs/day ÷ 40 acres 39.1 lbs./day/acre
Term
 POND Hydraulic Loading Rate, inches/day =
Definition
 [(Flow, ac-ft/day) ÷ (Pond Area, acres)] X 12 in./ ft   Example What is the hydraulic loading in inches/day for a 55 acre pond with flow of 10 ac-ft/day?   (10 ac-ft/day ÷ 55 acres ) X 12 in. / ft 2.2 inches / day
Term
 Dry solids, lbs. =   Used in sludge digestion to calculate the dry pounds of solids digested.
Definition
 [(Sludge volume, gal) (Sludge solids conc. % )  (8.34 lbs/gal)] ÷ 100 % EXAMPLE What is the weight  in pounds of 0.025MGD of a 16% solids sludge? [(25000 gal ) (16 %) (8.34 lbs/gal) ] ÷ 100 % 3336000 ÷ 100 33360 lbs. Dry weight an Alternative way and to check (0.025 MG) (8.34 lbs/gal) (160,000mg/L)  = 33360 lbs. Dry weight (REMEBER 1% = 10,000 mg/L) so 16% =(16 X 10,000 mg/L)  = 160,000 mg/L
Term
 Surface Loading, gal./ day/ sq. ft. (ft2) =
Definition
 Total Flow, gal. / day ÷ surface area, ft2   Example: Your palnt has an Influent flow of 0.300MGD and a RAS flow of 50% of the influent flow, the clarifier has a diameter of 100 feet. What is the Surface Loading of this clarifier? (0.300MGD) (1000000) + (0.300MGD) (1000000) (0.50) (3.14) (50 ft) (50 ft)   450000gpd  7850 ft2 57.3 gal./day/ft2
Term
 Efficency, %  =
Definition
 {(IN)- (OUT) ÷ (IN)} X 100% EXAMPLE What is the efficency of removal for CBOD5 for a plant with an Influent CBOD5 of 250 mg/L and an Effluent CBOD5 of 3.5 mg/L? {(250 mg/L - 3.5 mg/L ) ÷ 250 mg/L} X 100% 0.986 X 100% 98.6% Efficency
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