Velocity, in ft/sec =

Velocity is distance traveled over time.

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.

Flow Rate, in (cubic feet per second),

 ft³/ sec.  =

(Velocity, ft/sec.)  (area, ft²)

Example

(2 ft/sec.)  (3 ft²) = 6 ft³ (cubic feet)/second

Change

Cubic feet per second

(ft³/sec )

 to

gallons per minute

(gpm)

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

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. 

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.

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 WAS_ss 

The answer is always in pounds per day or pounds

lbs/day  or lbs

(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

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

[image]

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

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.

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

Hydraulic loading,in gal./day/ft² =

This refers to the amount of flow goin into the Clarifier from the Aeration basin.

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

Trickling Filter Organic Loading,

in

lbs. CBOD₅ /day/1000 ft³ of Media

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

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

(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

RBC (Rotating Biological Contactor) Organic Loading,

in lbs. CBOD₅ /day/ 1000 ft² =

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. ÷ 1.413 1000 ft²

93.0 lbs/1000 ft²

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.

(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

SDI =

*Sludge Density Index - used to calculate the settlability of Activated Sludge in a Secondary Clarifier. Similar to the SVI.

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

Solids inventory, lbs.

* used to calulate the ponud of TSS in an aeration tank, or in a clarifier, or a combination of both.

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.

Sludge Age ,  in days =

* used in the process control of an Activated Sludge plant and is controlled by wasting.

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

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

(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

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.

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

WAS, lbs./day =

*used to figure out the amount of solids (in pounds) that need to be wasted to maintain the MCRT

{(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

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.

(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

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.

(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

Population loading, person / acre =

* used in calculating the population loading of wastewater ponds.

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

Pond Volume, acre feet, ac-ft =

*This is the volume in acre feet and not to be confused with volume in gallons.

(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

Pond Volume, gal

* This formula will calculate pond volume in gallons given the ac-ft of the pond.

(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./ ft^3.

14,988,124 gallons

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

Detention Time , days =

* Using ac-ft and ac-ft/day

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

(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

POND

Hydraulic Loading Rate, inches/day =

[(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

Dry solids, lbs. =

Used in sludge digestion to calculate the dry pounds of solids digested.

[(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

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

{(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