Term

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),
ft.^{3}/ sec. = 

Definition
Velocity, ft./sec. X area, ft^{2}
Example
2 ft./sec. X 3 ft^{2 }= 6 ft.^{3} (cubic feet)/second 


Term
Change
Cubic feet per second
(ft^{3}/sec )
to
gallons per minute
(gpm) 

Definition
Gallons per minute = ft^{3}/sec X 7.48 gal./ft^{3} X 60 sec./min.
Example
3 ft^{3}/sec X 7.48 gal./ft^{3} X 60 sec/min =
1346.4 gpm or 1346.4 gal/min 


Term

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
The PIE Chart
Pounds = 

Definition
Flow, in MGD or MG X 8.34 lb./gal X 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?
.500 MGD X 8.34lb./gal X 200 mg/L = 834 lb. of TSS 


Term
Weir overflow rate, gal/day/foot of weir = 

Definition
Total Flow, gal./day ÷ length of weir , in feet
Example
Your plant has an Influnet flow of .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?
(.250 MGD X 1000000) + [(.250 MGD X 1000000) X .50]
3.14 (∏) X 50 ft.
250000gpd + (250000gpd X .50) ÷ 3.14(∏) X 50 ft.
250000gpd + 125000gpd ÷ 157 ft.
375000gpd ÷ 157ft.
2388.5 or 2389 gal./day/ft. of weir 


Term
Solids Loading, lbs./day/ft^{2} = 

Definition
Solids applied, in lbs./day ÷ Surface area, in ft^{2}
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 ?
(.500MGD + .250MGD) X 8.34 lb./gal. X 2600 mg/L
3.14(∏) X 25 ft. X 25 ft.(radius^{2})
16263 lbs./day
1962.5 ft^{2}
8.3 lbs./day/ft^{2} 


Term
Hydraulic loading,in gal./day/ft^{2} = 

Definition
Flow rate, in gal./day
Surface area, in ft^{2}
EXAMPLE
What is the hydraulic loading on a pond 100' X 50', with a daily flow of 1.500MGD?
(1.500MGD X 1000000) ÷ (100 ft. X 50 ft.)
1500000 gpd ÷ (100 ft. X 50 ft.)
1500000 gpd ÷ 5000 ft^{2}
300 gal./day/ft^{2}



Term
Trickling Filter Organic Loading,
in
lbs. CBOD_{5} /day/1000 ft^{3} of Media 

Definition
CBOD_{5} applied, in lbs./day
Volume of media, in 1000 ft^{3} units
EXAMPLE
What is the Organic loading on a trickling filter 50 ft. in diameter and 8 feet deep, with a P.E. CBOD_{5} of 125 mg/L, and a flow of .350MGD ?
.350MGD X 8.34 lb./gal. X 125 mg/L
(3.14(∏) X 25 ft. X 25 ft. X 8 ft.) ÷ 1000 ft^{3}
364.9 lbs./day
15.7 /1000 ft^{3}
23.2 lbs./day/1000 ft^{3} 


Term
Soluble CBOD_{5}, in mg/l =
*Soluble CBOD_{5} is the amount of CBOD_{5} 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 CBOD_{5 },mg/L)  (K X TSS, mg/L)
* (K is a constant that is 0.5 to 0.7 for most domestic wastewaters)
EXAMPLE
What is the soluble CBOD_{5} of a P.E. with the following make up, CBOD_{5} of 125 mg/L , a TSS of 150 mg/L and a K factor of 0.6 ?
125 mg/L  (0.6 X 150 mg/L)
125 mg/L  90 mg/l
35 mg/L of Soluble CBOD_{5} 


Term
RBC (Rotating Biological Contactor) Organic Loading,
in lbs. CBOD_{5 }/day/ 1000 ft^{2} = 

Definition
Soluble CBOD_{5} applied, lbs./day
Surface Area of Media, in 1000 ft^{2 }units
EXAMPLE
An RBC has three units in parallel, each measures 10 feet in diameter and 15 feet long. The INF. Flow is .350MGD and the INF. soluble CBOD_{5} is 45 mg/L.
What is the Organinc Loading on this system?
.350MGD X 8.34 lbs./gal X 45 mg/L
{(3.14(∏) X 10 ft. X 15 ft.) X 3}÷ 1000 ft^{2}
131.4 lbs. ÷ 1.41 /1000 ft^{2}
93.2 lbs/1000 ft^{2} 


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) X (1000mg ÷ gram)
(450 mL/L ÷ 3500 mg/L) X (1000mg ÷ gram)
.129 mL/mg X 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 .750MG with a MLSS of 4500 mg/L?
.750MG X 8.34 lbs./gal. X 4500 mg/L
28147.5 lbs. 


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 .500MG and an MLSS of 3200 mg/L, an influent flow of .350MGD and an influent TSS of 165 mg/L ?
.500MG X 8.34 lbs./gal. X 3200 mg/L
.350MG X 8.34 lbs./gal. X 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 (CBOD_{5}) 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) X 8.34 lbs./gal X (Inf. CBOD_{5}, mg/L)
(Aeration Tank Volume, in MG) X 8.34 lbs./gal. X (MLVSS, mg/L)
EXAMPLE
Given an Ilfluent Flow of .450MGD, an Ifluent CBOD_{5} of 225 mg/L,an Aeration Tank with a volume of .350MG, and a MLVSS of 1975 mg/L.
What is the F/M ratio?
.450MGD X 8.34 lbs./gal. X 225 mg/L
.350MG X 8.34 lbs./gal. X 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 .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 X 8.34 lbs./gal. X 5000 mg/L
(.600MGD x 8.34 lbs./gal X 0.9 mg/L) + (.025MG X 8.34 lbs./gal X 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.
(28147.525020)÷ 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 ft^{2}/acre]
2500 ÷ 45.9
54.5 or 55 persons/acre 


Term
Pond Volume, acre feet, acft =
*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 (acft) of a pond 250 ft. long and 300 ft. wide and 6 ft. deep?
[(250 ft. X 300 ft.) ÷ 43560 ft^{2} / acre] X 6 ft.
1.7 acres X 6 ft.
10.2 acft 


Term
Pond Volume, gal
* This formula will calculate pond volume in gallons given the acft of the pond. 

Definition
(Volume, acft) X (43560 ft^{2}/acre) X (7.48 gal./ft^{3})
Example
What is the volume in gallons of a 46 acft pond?
46 acft X 43560 ft^{2}/acre x 7.48 gal./ ft^{3.}
14,988,124 gallons 


Term

Definition
Flow, gal/day ÷ [(7.48 gal./ft^{3}) x (43560 ft^{2}/acre)]
Example
What is the flow in acft /day if the Inffluent flow is 2.000MGD ?
2.000MGD X 1000000 =
2000000 gpd ÷ (7.48 gal./ft^{3 }X 43560 ft^{2}/acre)
2000000 gpd ÷ 32582.8 gal/ acft
61.4 acft/day 


Term
Detention Time , days =
* Using acft and acft/day 

Definition
Volume, acft ÷ Flow, acft / 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 ft^{2}/ acre] ÷ [(2.500 MGD X 1000000) ÷ (7.48 gal./ft^{2} X 43560 ft^{2}/acre)]
22.04 acft ÷ 7.7 acft / day
2.9 days 


Term
Pond, Organic Loading, lbs./day/acre = 

Definition
(Influent CBOD_{5} , 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 CBOD_{5} 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, acft/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 acft/day?
(10 acft/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. X (Sludge solids conc. % ) X (8.34 lbs./gal/)] ÷ 100 %
EXAMPLE
What is the weight in pounds of .025MGD of a 16% solids sludge?
[(25000 gal X 16 % X8.34 lbs./gal. ) ] ÷ 100 %
3336000 ÷ 100
33360 lbs. Dry weight
an Alternative way and to check
.025 MG X 8.34 lbs./gal. X 160,000 mg/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. (ft^{2}) = 

Definition
Total Flow, gal. / day ÷ surface area, ft^{2}
Example:
Your palnt has an Influent flow of .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?
.300MGD X 1000000 + (.300MGD X 1000000 X .50)
3.14(∏) X 50 ft. X 50 ft.
450000gpd
7850 ft^{2}
57.3 gal./day/ft^{2} 


Term

Definition
{(IN) (OUT) ÷ (IN)} X 100%
EXAMPLE
What is the efficency of removal for CBOD_{5} for a plant with an Influent CBOD_{5} of 250 mg/L and an Effluent CBOD_{5} of 3.5 mg/L?
{(250 mg/L  3.5 mg/L ) ÷ 250 mg/L} X 100%
0.986 X 100%
98.6% Efficency 

