THE FORMULA TO CONVERT

DEGREES CELSIUS (C^o)   TO  DEGREES FAHRENHEIT (F^o)

F^o= C^o x 1.8 + (32)

Example:

To convert 45^o Celsius to Fahrenheit 

F^o= 45 x 1.8 + (32)

F^o= 81 + 32

F^o = 113

distance,  in feet  ÷  time, in seconds

Example

What is the velocity in feet /sec. if a stick travles a 200 foot channel in 120 seconds?

200 feet ÷ 120 seconds = 1.67 feet/sec.

Flow Rate, in (cubic feet per second),

 ft.³/ sec.  =

Velocity, ft./sec.  X area, ft²

Example

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

Change

Cubic feet per second

(ft³/sec )

 to

gallons per minute

(gpm)

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

Detention Time, minutes =

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

The PIE Chart

Pounds =

Flow, in MGD or MG X 8.34 lb./gal X concentration, in mg/L or ppm

Example

A plant has a flow of .500MGD and and need to have a chlorine residual of 2.5 mg/L. How many pounds 100% chlorine  per day will be used to maintain this residual, assume no demand?

.500 MGD X 8.34lb./gal X 2.5 mg/L = 10.43 pounds of 100% chlorine.

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

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

Example:

Your filter has  flow of .300MGD and , the filter has a diameter of 100 feet. What is the Surface Loading of this clarifier?

.300MGD X 1000000 

3.14(∏) X 50 ft. X 50 ft.

300000gpd 

7850 ft²

38.2 gal./day/ft²

{(IN)- (OUT) ÷ (IN)} X 100%

EXAMPLE

What is the efficency of removal for Iron for a plant with an Influent Iron of 25 mg/L and an Effluent Iron of .5 mg/L?

{(25 mg/L - .5 mg/L ) ÷ 25 mg/L} X 100%

0.98 X 100%

98% Efficency

Change

Cubic feet per second

(ft³/sec )

 to

gallons per minute

(gpm) Quick Method

This is a quick, ball park estimate

this is not as accurate as the long method.

CFS X 449 = gpm

Examlpe:

6 ft³ (cfs) X 449= 2694 gpm

Change gpm (gallons per minute) to MGD (Million Gallons a Day)

This is needed if you are figuring the pounds formula and are given the flow in gallons per minute.

gpm X 1440 (minutes in a day)

1,000,000

Example:

300 gpm X 1440 

1,000,000

432000

1,000,000

.432 MGD 

Change gpm (gallons per minute) to MGD (Million Gallons a Day)

Quick Method

This is a quick, ballpark estimate, it is not as accurate as the long method.

gpm ÷ 700 = MGD

Example:

350 gpm ÷ 700= 0.500 MGD

This is useful if you are using the pounds formula and are given the flow in gallons per day. It is also the same formula you would use to change the gallons in a tank to MG (Million Gallons).

gpd ÷ 1,000,000 = MGD

gallons in tank ÷ 1,000,00= MG

or

gpd X .000001= MGD

gallons in tank X .000001= MG

Example:

125,000 gpd ÷ 1,000,000 = .125 MGD

125,000 gallons in tank ÷ 1,000,000 = .125 MG

125,000 gpd X .000001 = .125 MGD

125,000 gallons in tank X .000001=.125 MG

Change gpm (gallons per minute) to ft³/sec. (cfs, cubic feet per second)

Quick Method

This a quick method, it is as accurate as the long method.

gpm ÷ 449 = ft³/ sec.(cfs)

Example:

600 gpm ÷ 449 = 1.3 ft³/ sec.

This is the long method , it is more accurate then the quick method.

gpm ÷ 60 sec./min. ÷ 7.48 gal./ft³= ft³/sec.

600 gpm ÷ 60 sec./min. ÷ 7.48 gal./ft³ =

1.3 ft³/sec.

There are two methods for finding the volume of water in gallons in alength of pipe.

1. ∏ R² X length X 7.48 gal./ft³= gallons

all dimension are in feet or must be converted to feet in this method.

2. D(in inches)² X 0.0408 X Length (in feet) = gallons

this method uses both inches and feet. The diameter of the pipe is in inches, the length of the pipe is in feet.

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

Calculating LIME FEED in mg/L, based on Source Water

and Finished Water Constituents.

(A+B+C+D)×1.15 ÷ Lime Purity %, as decimal

A = CO₂ in Source Water, mg/L X 56/44

B = Bicarbonate Alkalinity removed, mg/L X 56/100

C = Hydroxide Alkalinity, in Effluent,mg/L X 56/100

D = Magnesium removed, mg/L X 56/24.3

1.15 is for 15% excess lime dose.

If you are using Hydrated Lime in stead of Quick Lime.

You must substitue 74 for 56 in A,B,C,Dabove.

Lime Demand, mg/L

Lime demand, mg/L = 

(2.27 x CO2) + (Total Alkalinity) +(4.12 x Mg) x 0.56

Soda Ash required, mg/L =

Non-carbonate hardness, mg/L as CaCO₃ X 106/100

Non-ccabonate hardness,mg/L = (Total hardness removed, mg/L)- (Carbonate hardness removed, mg/L)

Note= if Alkalinity< Total hardness,

Alkalinity as CaCO₃ = Carbonate Hardness as CaCO₃.

Total CO₂ required in mg/L = (Excess lime dose in mg/L) ( 44 /74 ) + (Mg²⁺ in finished, mg/L) (44/24.3)

If using Quick Lime substitue 56 instead of 74 which is used for hydrated lime.

Non-Carbonate hardness, mg/L as CaCO₃ =

Total Hardness, mg/L - Total Alkalinity, mg/L

Raw water Non-Carbonate hardness, mg/l - Finished wate Non-Carbonate hardness = Non-Carbonate hardness removed, mg/L

Exchange Capacity, grains

The capacity of an Ion Exchange Media to exchange ions to the point of exhaustion for a volume of media, in cubic feet, of a given removal capacity, in grains/cu.ft. 

1 grain/ gallon = 17.1 mg/L

So

mg/L ÷ 17.1 = grains/gallon

EX. (171 mg/L ÷ 17.1 = 10 grains/gallon)

Alkalinty Constituents

A table to help define the type of Alkalinity present based on the values of Phenolphthalein Alkalinity =P

and Total Alkalinity = T

Alkalinity as CaCO₃

  Titration Result      Bicarbonate      Carbonate      Hydroxide

       P=0                     T               0                0

       P<½T                T-2P            2P              0

       P=½T                  0               2P              0

       P>½T                  0             2T-2P         2P-T

       P=T                     0                0               T

   P= Phenolphthalein Alkalinity

   T= Total Alkalinity

Lime Demad, lbs/MG

pounds of lime needed per Million gallons treated.

Lime demand, lbs / MG =

(Lime demand, mg/L) (1 MG) (4.67 lb/ MG / mg/L) (excess lime, .15 ) ÷ Calcium oxide purity (%)

CHLORINE DEMAND, mg/L =

CHLORINE DOSE, mg/L - CHLORINE RESIDUAL, mg/L

Example

A plant doses a 5 mg/L and carries a 3.5 mg/L residual leaving the plant. What is the chlorine demand?

5 mg/L - 3.5 mg/L = 1.5 mg/L Chlorine Demand

Chlorine Dosage, mg/L =

Chlorine Demand, mg/L + Chlorine Residual, mg/L

Example:

A plant has a chlorine demand of 2.5 mg/L and must carry a residual of 3.5 mg/L leaving the plant. What must the chlorine dosage be?

2.5 mg/L + 3.5 mg/L = 6.0 mg/L Chlorine Dosage

Chlorine Residual, mg/L =

Chlorine Dosage, mg/L - Chlorine Demand, mg/L

Example:

Find the chloine residual for plant that doses at 5.5 mg/L, with a chlorine demand of 2.0 mg/L.

5.5 mg/L - 2.0 mg/L = 3.5 mg/L Chlorine Residual

(ION EXCHANGE)

Hardness, grains/gallon =

(Hardness, mg/L)(1 grains/gallon)

17.1 mg/L

Example:

 A water has a hardness of 110 mg/L converts to a grains/gallon of ___?

110 mg/L ÷ 17.1 = 6.4 grains/gallon

*1 grain/gallon = 17.1 mg/L

(ION EXCHANGE)

Exchange Capacity, grains =

(Media Volume, ft³)(Removal Capacity, grains/ft³)

Example:

What is the exchange capacity of an ion exchange unit with 15 ft³ of media with a removal capacity of 1000 grains/ft³?

15 ft³× 1000 grains/ft³ =15,000 grains Exchange Capacity

(ION EXCHANGE)

Water Treated, gallons =

Exchange Capacity, grains

Hardness Removed, grains/gallon

Example:

How much water will an ion exchange unit with an Exchange capacity of 15,000 grains treat, if the hardness to be removed is 15 grains/gallon?

15,000 grains ÷ 15 grains/gallon = 1000 gallons

(ION EXCHANGE)

Operating Time, hr =

Water treated, gals. ÷ (Avg. Daily Flow, gpm)(60 min/hr)

Example:

How long will an Ion Exchange unit run before regeneration, that can treat 10000 gallons, if the average daily flow is 5 gpm?

10000 gals. ÷ ( 5 gpm )(60 min/hr)

10000 ÷ 300 gph = 33.33 hrs