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Wastewater Math
Formula Flash cards
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Mathematics
02/21/2013

Additional Mathematics Flashcards

 


 

Cards

Term
Velocity, in ft/sec =
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, ft2

 

Example

 

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

Term

Change

Cubic feet per second

(ft3/sec )

 to

gallons per minute

(gpm)

Definition

Gallons per minute = ft3/sec X 7.48 gal./ft3 X 60 sec./min.

 

Example

3 ft3/sec X 7.48 gal./ft3 X 60 sec/min =

 

1346.4 gpm or 1346.4 gal/min

Term

Detention Time =

 

 

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/ft2 =
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 ?

 

(.500MGD + .250MGD) X 8.34 lb./gal. X 2600 mg/L


3.14(∏) X 25 ft. X 25 ft.(radius2)

 

16263 lbs./day


1962.5 ft2

 

8.3 lbs./day/ft2

Term
Hydraulic loading,in gal./day/ft2 =
Definition

Flow rate, in gal./day


Surface area, in ft2

 

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 ft2

300 gal./day/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 ?

 

.350MGD X 8.34 lb./gal. X 125 mg/L


(3.14(∏) X 25 ft. X 25 ft. X 8 ft.) ÷ 1000 ft3

 

364.9 lbs./day


15.7 /1000 ft3

 

23.2 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 X 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 X 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 .350MGD and the INF. soluble CBOD5 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 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) 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 (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) X 8.34 lbs./gal X (Inf. CBOD5, 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 CBOD5 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.

 

(28147.5-25020)÷ 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. 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. (ft2) =
Definition

Total Flow, gal. / day ÷ surface area, ft2

 

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 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