كيف تحسب سعة واحجام وقدرة معدات محطة صرف صناعى/Calculations Used in the Daily Operation of a Wastewater Treatment Facility

Calculations Used in the Daily Operation
of a Wastewater Treatment Facility

Several basic equations are used in calculations to determine things such as Weir Overflow Rate, Detention
Time, SRT, etc. The formulas for area and volume appear below, along with examples

AREA

Area Formula Example

Rectangle (Length) x (Width) (35' L) x (12' W) = 420 sq. ft
.
Circle (0.785) x (Dia. sq.) (0.785) x (55') x (55') = 2,375 sq. ft

VOLUME

Volume Formula Example

Rectangle L x W x H 80'L x 25'W x 10"H = 20,000 cu. ft

Cylinder (0.785) x (Dia. sq.) x (H) (0.785) x (55') x (55') x (25') = 59,366 cu. ft

Raw Influent Calculations
Plant Influent Loadings

Determines the influent loadings of pounds (BOD, TSS, etc.) to a wastewater treatment facility.

Plant Influent Loadings:

mg/l x mgd x 8.34 = lbs./day

Example (using TSS

Plant Conditions

TSS Influent = 135 mg/l

Influent flow = 1.5 mgd

To determine pounds into the plant, use the following calculation

mg/l x mgd x 8.34 = lbs/day

Example: 135 mg/l x 1.5 mgd x 8.34 = 1,688.85 lbs

This reflects the pounds of TSS into the plant on that particular day

Clarifier Capacity Calculations

Surface Loading Rate - RECTANGULAR Clarifier

Gallons of wastewater applied to each square foot of the clarifier area. This affects the settleable solids and
BOD removal efficiency in the sedimentation process

Surface Loading Rate (RECTANGULAR CLARIFIER

gpd

sq. ft. of clarifier

Example

Clarifier

Length = 75 feet

Width = 12 feet

Influent flow = 1.2 mgd (1,200,000 gpd
)
The square footage of the clarifier is L x W

75
x 12 = 900 sq. ft.
To determine the Surface Loading Rate of a rectangular clarifier, use

the following calculation

gpd

sq. ft. of clarifier

Example: 1,200,000 gpd = 1,333 gpd/sq. ft

900
sq. ft.
Clarifier Capacity Calculations

Surface Loading Rate - CIRCULAR Clarifier

Gallons of wastewater applied to each square foot of the clarifier area.

This affects the
settleable solids and
BOD removal efficiency in the se
dimentation process

Surface Loading Rate (C
IRCULAR CLARIFIER

gpd

sq. ft. of clarifier

Example

Clarifier

Diameter = 60 feet

Influent flow = 2,200,000 gpd

The square footage of the clarifier is .785 x Dia. sq

.785 x 60' x 60' = 2,826 sq. ft.
To determine the Surface Loading Rate of a circular clarifier, use the

following calculation

gpd

sq. ft. of clarifier

Example: 2,200,000 gpd = 778.48 gpd/sq. ft

2,826
sq. ft.
3

Clarifier Capacity Calculations

Detention Time

Period of time that a particle remains in a tank

Detention Time

tank volume x 24 hours

Influent flow

Example (rectangular clarifier

Clarifier

Length = 75 feet

Width = 30 feet

Depth = 10 feet

Influent flow = 2,850,000 gpd

The volume of the clarifier is L x W x H x 7.48

70
' x 25' x 10' x 7.48 = 130,900 gals.
To determine the Detention Time of a rectangular clarifier, use the

following calculation

tank volume x 24 hours

Influent flow

Example: 130,900 gals x 24 = 1.10 hours

2,850,000

Weir Overflow Rate

gpd flow

Total feet of weir

Example

Plant Conditions

Weir length = 100 feet

Flow rate = 1,500,000 gpd

To determine the Weir Overflow Rate, use the following calculation

gpd flow

Total feet of weir

Example: 1,500,000 gpd = 15,000 gpd/ft

100
feet of weir

Activated Sludge Calculations
Sludge Age
Average time a particle of suspended solids remains in the activated sludge system. Sludge age is used to
maintain the proper amount of activated sludge in the aeration system.
To determine sludge age, you must calculate both the daily amount of suspended solids and the total amount
of solids in the aerator.
Sludge Age:
Total lbs. of MLSS in aerator
Daily lbs. of MLSS in aerator
Example:
Suspended solids of primary effluent = 110 mg/l
Influent flow = 2.5 mgd
Aeration tank volume = .5 mg
MLSS concentration = 2,200 mg/l
The plant influent loading = mg/l x mgd x 8.34
Example:
110 x 2.5 x 8.34 = 2,293.5 lbs. daily of SS (lbs of MLSS in aerator)
2,200 x .5 x 8.34 = 9,174 lbs. total of MLSS in aerator
With these calculations, you can determine Slude Age:
Total lbs. of MLSS in aerator = Days Sludge Age
Daily lbs. of MLSS in aerator
Example:
9,174 lbs. total of MLSS in aerator = 4 Days Sludge Age
2,293.5 lbs. daily of MLSS in aerator
Example:
Influent suspended solids = 250 mg/l
Influent flow = 2.5 mgd
Aeration tank volume = .5 mg
MLSS concentration = 2,200 mg/l
The plant influent loading = mg/l x mgd x 8.34
Example:
250 x 2.5 x 8.34 = 5,213 lbs. daily of SS (lbs of MLSS in aerator)
2,200 x .5 x 8.34 = 9,174 lbs. total of MLSS in aerator
With these calculations, you can determine Slude Age:
Total lbs. of MLSS in aerator = Days Sludge Age
Daily lbs. of MLSS in aerator
Example:
9,174 lbs. total of MLSS in aerator = 1.76 Days Sludge Age
5,213 lbs. daily of MLSS in aerator
5
Sludge Age - Plants with Primary Clarifier
Sludge Age - Plants without Primary Clarifier
Using these figures, calculate the amount
of suspended solids leaving the system:
Example:
(7,250 mg/l waste sludge) x (.08 waste sludge flow) x (8.34 lbs.)
+ (20 mg/l SS final effluent) x (4.3 MGD influent flow) x (8.34 lbs.)
= 4,837.2 lbs. SS + 717.2 lbs. SS = 5,554.4 lbs. SS leaving system
For this example, the volume of the aeration tank and clarifier equals
1.7 mg.
Pounds of Aeration Tank MLSS =
Aeration Tank mg/l MLSS x mg x 8.34
Example:
2,500 mg/l MLSS x 1.7 mg volume x 8.34 lbs. = 35,445 lbs. MLSS
Solids Retention Time (SRT):
SRT days = lbs./day of SS in system
lbs./day of SS leaving system
Example:
35,445 lbs. MLSS = 6.38 SRT
5,554.4 lbs. SS leaving system
Activated Sludge Calculations
Solids Retention Time (SRT)
Average time a given unit of cell mass stays in the activated sludge system. SRT is based on solids leaving the
Activated Sludge process, and includes solids in the clarifiers. Also called MCRT (Mean Cell Residence Time).
Solids Retention Time (SRT):
SRT days = lbs./day of SS in system
lbs./day of SS leaving system
Factors in determining SRT:
1. Solids leaving the system (wasted)
2. Solids leaving the system (SS in the final effluent)
3. Solids in the system (aeration basins and clarifiers)
Example:
Tank volumes: Aeration volume = 1.6 mg
Final Clarifer volume = .10 mg
SS Concentration: Final effluent = 20 mg/l
Waste sludge SS = 7,250 mg/l
Aeration Tank MLSS = 2,500 mg/l
Flows: Influent flow = 4.3 mgd
Waste sludge flow = 0.08 mgd (24-hour period)
6
Solids to be Wasted
Solids (measured in pounds) to be wasted over a set period of time (typically daily).
Solids to be Wasted:
Current inventory
- Desired inventory to achieve peak performance
= Pounds to be wasted
After calculating the MLSS Concentration (see above), use the
following formula to determine the pounds of solids to be wasted.
Example:
Peak performance = 22,000 pounds of MLSS in aeration system
Current inventory = 25,000 pounds of MLSS in aeration system
25,000 pounds (current inventory)
- 22,000 pounds (desired inventory)
= 3,000 pounds to be wasted
7
MLSS Concentration in Pounds
Pounds of MLSS in the aeration system.
MLSS Concentration:
MLSS (mg/l) x volume of aeration tank x 8.34
Example:
MLSS in aeration tank = 2,500 mg/l
Aeration system volume = 1.4 mg
MLSS (mg/l) x volume of aeration tank x 8.34
2,500 mg/l x 1.4 x 8.34 = 29,190 lbs. of MLSS
Activated Sludge Calculations
F/M Ratio
Ratio of food (biochemical oxygen demand) entering the aeration tank to microorganisms in the tank.
F/M Ratio:
lbs./BOD
lbs. MLSS
Example:
lbs. of BOD entering plant daily = 4,340 lbs.
lbs. of MLSS in aerator = 8,201 lbs.
lbs. BOD or 4,340 lbs. of BOD entering plant daily = .53 F/M
lbs. MLSS 8,201 lbs. of MLSS in aerator
Activated Sludge Calculations
Waste Sludge Pumping Rate
Rate at which to pump the sludge to be wasted from the system. Normally expressed in gallons per minute (gpm).
Example:
Plant Conditions:
lbs. of MLSS to be wasted = 1,400 lbs.
MLSS concentration = 5,900 mg/l
lbs. to be wasted = flow rate (mgd)
MLSS concentration x 8.34
Example:
1,400 lbs. to be wasted = .0284 mgd wasting rate
5,900 mg/l x 8.34
Convert million gallons per day to gallons per day:
mgd wasting rate x 1,000,000 = gpd wasting rate
Example: .0284 x 1,000,000 = 28,400 gpd
Convert gallons per day to gallons per minute (over 8 hours):
gpd wasting rate
480 minutes (8 hours)
Example: 28,400 gpd = 59.166 gpm
480 minutes

Typical WWTP Operational Parameters

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كيف تحسب سعة واحجام وقدرة معدات محطة صرف صناعى/Calculations Used in the Daily Operation of a Wastewater Treatment Facility Wwtp10

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Metal Salt Addition for Phosphorus Removal
Using metal salt chemical data, theoretically calculate the amount of chemical salt solution to add in
gallons per day to remove phosphorus.
* Values that need to be entered are shown in the gray boxes.
Step 1. Determine the amount of influent phosphorus to remove.
Influent flow in MGD Influent P (mg/l) lbs/gallon lbs of P to remove
2 x 8 x 8.34 = 133
Step 2. Determine the pounds of metal salt in a gallon of solution knowing the specific gravity.
Specific gravity lbs/gal lbs metal salt/gal
1.4 x 8.34 = 11.7
Chemical Specific Gravity
Alum 1.33
Ferric Chloride 1.441
Ferrous Chloride 1.23
Ferrous Sulfate 1.185
Step 3. Determine the pounds of actual metal in a gallon of metal salt solution with certain percentage
metal content (provided by chemical supplier).
lbs metal salt/gal % metal lbs metal/gal metal salt solution
11.7 x 12.5 = 1.5
Step 4. Look up removal ratio for the metal salt being used.
Chemical Removal Ratio
Alum 0.87 to1
Ferric Chloride 1.8 to 1
Ferrous Chloride 2.7 to 1
Ferrous Sulfate 2.7 to 1
Step 5. Determine the pounds of metal needed to remove the incoming pounds of phosphorus.
Removal ratio Influent lbs of P lbs of metal to add
1.8 x 133 = 240
Step 6. Determine the gallons of metal salt solution with a certain percentage metal content to thus add.
lbs of metal to add lbs of metal/gal gal/day of metal salt solution to add
240 ÷ 1.5 = 165
This estimate may be different than
the value obtained from a calculator
due to rounding differences.

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