wastewater Calculations

BOD CALCULATION BOD bottle 300 ml Sample 30 ml The DO before was 6.0 mg/L and the DO after was 3.5 mg/L. What is the reportable BOD?

(6-3.5) =2.5 300 30 10 X 2.5 = 25

Calculate % Dry Solids Empty crucible = 110.3642 grams Crucible w/ wet sludge = 466.4642 grams Crucible w/ dry sludge = 136.8726 grams % Dry Solids=

(dry sludge - crucible empty weight)/ (wet weight - crucible empty weight) x 100 = % solids Dry sludge =(136.8726 - 110.3642) =26.5084 ÷ Wet Sludge (466.4642 - 110.3642)=356.1 .0744 x 100 = 7.44%

Calculate Volatile Solids Empty crucible = 110.3642 grams Crucible with wet sludge = 466.6742 grams Crucible with dry sludge = 136.8726 grams Crucible with ash = 122.4672 grams Volatile Solids content=

(dry weight - ash weight ) ÷ ( crucible weight- dry weight) x 100 (136.8726 - 122.4672)/ (110.3642- 136.8725) x 100= VS% % volatile solids = 14.4054/26.5083 x 100 % volatile solids = 54.6%

Covert 5.5 percent to milligrams

. covert percent to decimal 5.5/100=.055 Covert to Milligrams .055X1,000,000.= 55,000

HORSE POWER INCREASE NEED TO INCREASE SPEED If an existing centrifugal pump motor is replaced with a motor with 25 percent more speed, the motor horsepower required will increase by ?

1.25 x 1.25 x 1.25 = 1.95 = 95%

HORSE POWER INCREASE NEED TO INCREASE SPEED If you want to increase the speed of a pump by 35 percent what % increase in horse power would be required ?

1.35 x 1.35 x 1.35 = 2.46 = 246%

HORSE POWER INCREASE NEED TO INCREASE SPEED If you want to increase the speed of a pump with 2 HP motor by 35 percent what increase in horse power would be required ?

1.35 x 1.35 x 1.35 = 2.46 = 4.92 HP

WASTEWATER STABILIZATION VOLUME CALCULATION A wastewater stabilization pond has 130,680 sq. ft. It has a three to one slope on the embankment and has an average depth of 5 ft. Approximately how many people will this pond provide wastewater treatment for?

130,680 sq ft = 3 acres x 35 lbs / BOD/ acre = 105 lbs BOD =618 0.17 lbs BOD/ person 43,560 sq ft l acre 600 people

BOD Loading on trickling filter/ 1000ft3 What is the BOD loading in pounds per 1000 ft.3 if a trickling filter 80 ft. in diameter and 6 ft. deep with a flow of 1.5 MGD has an applied BOD of 140 mg/L?

3.14 × 40 ft × 40 ft × 6 ft = 30,144 sq. ft 1,751.4 lb. 30,144 /1,000 =58 lb./1000

Hydraulic Loading per/ acre on a trickling Filter A standard rate rock trickling filter is 120 ft. in diameter. deep. What is the hydraulic loading in MG per acre per day if 1.0 MGD is treated? Is this within the design loading for a standard rate filter? acre feet= 43,560

3.14 × 60 ft × 60 ft = 11,304 sq. ft. 1 MGD X (11,304 / 43560) = 3.85 MG /per Acre /day Standard rate 1-4 MGAD Yes, within design limits

Calculate Circumference what is the weir length of a 30 ft. diameter circular tank

3.14x d = circumference 3.14 x 30= 94.2

Weir Loading 2 clarifiers have a diameter of 36 ft. Each clarifier has a circular weir 1 ft. from the side wall. Find the weir loading at a peak flow of 4.0 MGD in gallons per day per linear foot of weir length

36 -2 ft. = 34 ft diameter 3.14 × 34 ft = 107 ft. linear weir length 2,000,000/107= 18,691 gallons per linear ft.

TRENCH CUBIC YARDS CALCULATION Trench is 4,100 feet long, four 4 wide and 6 feet deep. There is a 12-inch diameter pipe in the bottom of the trench. How many cubic yards of material were removed from the trench?

4100 x4 x 6=

FIND AMOUNT OF REDUCTION If 5,000 gallons of sludge containing 1.5% solids is dewatered to 6% solids, what would the new volume be in gallons?

5,000 X (1.5%/ 6%= .25) = 1,250 gal

Detention Time What is the detention time in hours of a tank 50 ft. long, 25 ft. wide and 10 ft. deep, with a flow of 1,500,000 gallons per day?

50 ft × 25 ft × 10 ft × 7.48 gal/ft3 = 93,500 gal 1,500,000/24= 62,500 GPH 93,500/65,500=1.5 hrs.

FIND DOSAGE 75 lbs. cl2 1.2 MG Chlorine is being fed at a rate of 75 lbs/day. Plant flow is 1,2000,000 gpd. The chlorine residual is found to be 1.2 mg/L. Calculate the chlorine demand.

75 lbs. /day 1.2 X 8.34 = 7.5 mg/l 7.5 - 1.2 = 6.3

Surface Loading on Clarifier Question: A treatment plant has two primary clarifiers. Each has a diameter of 36 ft. and a depth of 10 ft. If the design flow is 1.0 MGD, find the surface loading on each clarifier in gal./ft.2/day. Assume the flow is equally split between the two clarifiers.

Answer 1.0 MGD/ 2 clarifiers=.5 MGD × 1,000,000 gal/MG = gpd 0.5 MG × 1,000,000 gal/MG = 500,000 gpd 3.14 × 18 ft × 18 ft = 1017 ft2 gal/ ft2 of clarifier= gal/ft2/day 500,000 gal/1017 ft2 = 492 gal/ft2/day

lbs. BOD removal from Primary How many pounds of BOD5 are removed from the primary clarifier per day? Raw BOD entering primary clarifier: 280 mg/L BOD leaving primary clarifier: 180 mg/L Flow: 170,000 gpd

Answer 280 - 180 = 100 mg/L Removed 170,000 gal/ 1,000,000 gal/MG = 0.17 MG 0.17 × 8.34 lb/gal × 100 mg/L = 142 lb/day

Calculate treatment capacity based on population Wastewater treatment plant capacity needed (flow rate in MGD) treat the the discharge from 300,000 people? Also calculate this flow rate in gpm.

Answer: # people × 100 gal/person/day = gpd 300,000 people × 100 gal/person/day = 30,000,000 gpd 30,000,000 gpd/1,000,000 gal/MG = 30 MGD 30,000,000 gpd/1440= 20,833 gpm

BOD removal Efficiency The raw BOD of a wastewater treatment plant is 220 mg/L and the effluent BOD is 20 mg/L. Find the percent reduction of BOD.

Answer: (In - Out)/In * 100 (220-20) /200*100= 90.9% --------------------

CAPACITY OF TANKS primary clarifier is 50 ft. in diameter and 10 ft. deep. What is the capacity of the tank in gallons?

Answer: 3.14× R2 × D × 7.48 gal/ft3 = Gallons 3.14 × 25 ft × 25 ft × 10 ft × 7.48 gal/ft3 = 146,795 gal

Calculate SVI 30-minute settling test = 250 mL/L MLSS = 2,500 mg/L

Answer: SVI = [settled volume in 30 mins (mL/L) ÷ MLSS (mg/L)] × 1,000 = [250 mL/L ÷ 2,500 mg/L] × 1,000 = 100

Calculate MLSS lbs under aeration . Aeration basin volume = 90,000 gals MLSS = 1,800 mg/L

Answer: vol. (MG) × MLSS (mg/L) × 8.34 0.09 MG × 1,800 mg/L × 8.34 =1,351 lbs of MLSS

BOD Loading per Acre feet trickling filter An intermediate rate rock trickling filter 160 ft. in diameter 6 ft. deep Influent= 4.0 MGD. Influent BOD is 132 mg/L. What is the BOD loading in pounds per acre sq. foot per day? Is this within the design of organic loading for an intermediate rate filter? Acre feet= 43,560

BOD lbs. x 43,560 Cubic feet

Calculate the pounds of volatile solids pumped to the digester. 6,000 gallons raw sludge pumped to digester. TS= 5.5% %VS= 74.8%.

Calculate the pounds of volatile solids pumped to the digester. 5.5 100 = 0.55 =.055 x 1,000000 =55,000 mg/l MGD X mg/l X 8.34= lbs. .006 X 55,000 X8.34= 2,752 lbs. 2,752 x .748= 2,059 lbs. VS

calculate pounds to waste to reach target mg/l MLSS . Aeration basin volume = 250,000 gals Current MLSS = 2,200 mg/L Desired MLSS = 2,000 mg/L

Current MLSS (lbs) = basin volume (MG) × MLSS (mg/L) × 8.34 = 0.250 MG × 2,200 mg/L × 8.34 =4,587 lbs Desired MLSS (lbs) = basin volume (MG) × MLSS (mg/L) × 8.34 = 0.250 MG × 2,000 mg/L × 8.34 = 4,170 lbs 4,587 lbs - 4,170 lbs = 417 lbs

CHLORINE DEMAND The chlorine demand of the final effluent is 1.7 mg/L. The chlorine residual is measured to be 0.6 mg/L. If the ·plant flow rate is 1.8 MGD at what rate (lbs/day) would you set the gas chlorinator?

Dosage =demand + Residual 1.8 X 2.3 X 8.34 = 34.5

Calculate lbs. F/M ratio Influent flow = 0.275 MGD Influent BOD = 230 mg/L Aeration basin volume = 0.432 MG Aeration basin MLSS = 1,750 mg/L

F/M ratio = influent BOD (lbs) ÷ MLSS under aeration (lbs) = 528 lbs ÷ 6,305 lbs = 0.08 F/M

FIND PISTON PUMP RATE A sludge pump with a piston of 10 inches in diameter and a stroke of 6 inches runs for 20 minutes at a rate of 50 strokes per minute. What would be the volume of sludge pumped during this time?

First Step: covert inches to feet

Calculate the flow through a grit chamber given the following information: the chamber is 2. 3" wide, wastewater flowing at a depth of 8 inches, velocity in the chamber is 1.3 ft/sec

Flow through the grit chamber: Area = 2.25 ft X 0.67 ft = 1 .5 ft2 Q = V X A = 1.3 ft/sec X 1.5 ft2 = 1 .95 ft3/sec (1.95 ft3/sec X 86,400 sec/day X 7.48 gal/ft3) (1,000,000 gal/MG)= 1.26MGD

Calculate Return Rate or Mass Balance Flow = 2.0 MGD MLSS: 4000 mg/L RSS: 9,000 mg/L

Formula Q xMLSS RSS-MLSS = R 2.0 MGD x 4000 mg/L 9000 - 4000 mg/L = 1.6 MGD

PUMP EFFICIENCY A centrifugal pump recirculates water to a trickling filter at a rate of 1,200 gpm. The head is 25 ft. The pump has an efficiency of 65%. Find the water horsepower and the brake horsepower.

GPM X head/ 3960=WHP 1200 X25/ 3960=7.6 WHP GPM X head /3960x PE= BHP (1200 X 25)/(3960 X.65) = 11.7 BHP

Calculate Gould Sludge Age Q: 2.0MGD ATV: 650,000 gallons MLSS: 4000 mg/L TSS inf { 190 mg/L

GSA is calculated by dividing the total pounds of MLSS in the aeration basin by the total pounds of TSS entering the aerator.

Gallons to be pumped based on Settleable Solids Test The flow into a primary clarifier averages 2.6 MGD. A settleable solids test is performed using an Imhoff cone on the clarifier inlet and outlet. The inlet test shows 15 mL/L settleable solids and the outlet shows 3 mL/L. How many gallons of sludge should be pumped from the bottom of the clarifier each day?

In - Out = Settled 15 mL/L - 3 mL/L = 12 mL/L mL settled × 1,000 × MGD = Gallons to be removed 12 mL/L × 1,000 × 2.6 MGD = 31,200 gal

CALCULATE MCRT Q: 2.0MGD ATV: 650,000 gallons MLSS: 3500 mg/L CV: 180,000 gallons CSS: 500 mg/L W: 20,000 gpd WSS: 9,000 mg/L ESS: 6 mg/L

MCRT is used to estimate how long the living cells are kept in the plant. MCRT is calculated by dividing the total pounds of MLSS in the aeration basin by the pounds of MLSS wasted daily or lost in the effluent.

Calculate GSA ATV Aeration tank Volume= 300,000 AB MLSS =6,000 Influent TSS= 150mg/l Influent Q= 1.5 MGD

MG aeration * 8.34 lb/gal * mg/L MLSS/ MGD * 8.34 lb/gal * mg/L TSS Inf = GSA 0.3 MG * 8.34 lb/gal * 6,000 mg/L 1.5 MGD * 8.34 lb/gal * 150 mg/L = 8 days GSA

LBS. of BOD per /1000 ft.3 trickling Filter What is the BOD loading in pounds per 1000 ft.3 if a trickling filter 80 ft. in diameter and 6 ft. deep with a flow of 1.5 MGD has an applied BOD of 140 mg/L?

MGD × 8.34 lb/gal × mg/L = lb 1.5 × 8.34 lb/gal × 140 = 1,751.4 lb 3.14 × R2 × D = Volume 3.14 × 40 ft × 40 ft × 6 ft = 1,751.4 lb / (30,144 ft3/1000) = 58 lb/1,000 ft3

Calculate OUR ( 10 Minute Test) DO, 0 min: 5.0 mg/L DO, 10 min: 1.0 mg/L

OUR is a measure of the activity of microorganisms in the activated sludge system. A lower than normal value would indicate low activity, while a higher than normal value would indicate increased biological activity. To determine OUR, record the decrease in DO (mg/L) over a 10-minute period.

Calculate Respiration Rate (RR) OUR: 24 mg/L/hour MLVSS: 2500 mg/L

RR relates the OUR to a specific amount of sludge at a definite time. To calculate RR, multiply OUR by 1000 and then divide by the milligrams per liter of MLVSS. OUR: 24 mg/L/hour MLVSS: 2500 mg/L 24 X 1000 2500

Calculate Gallons to Remove from Digester to maintain balance 150,000 gallons per month added to the digester. Raw sludge content = 96% moisture 4% Solids, 30% ash After digestion = 93% moisture 7% solids and 50% ash. gallons of digested sludge to be withdrawn to maintain balance?

Raw sludge= 96% moisture 4% Solids, 30% ash After digestion =93% moisture 7% solids and 50% ash. (4% /7% =.57)x (30%/50% =.6) x150,000 =51,300 ( % solids before/ after)x (% ash before/after) x gallons

Calculate the Sludge Density Index SVI 87.5

SDI is the reciprocal of the SVI multiplied by 100, or simply 100 + SVI. It indicates the weight in grams per 100 milliliters of mixed liquor solids after settling for 30 minutes.

Gallons to waste to reach Target Sludge Age 150 lbs/day WAS concentration =5000mg/l Calculate gallons of sludge to waste per day to waste 150 lbs/day

Sludge to waste (gpd) = (lbs) ÷ [WAS (mg/L) × 8.34]) × 1,000,000 = [ 150 lbs ÷ (5,000 × 8.34) ] × 1,000,000 = 3,597 gpd

Calculate how far grit will travel in channel A grit removal channel is 3 feet wide and has a wastewater flow depth of 18 inches. How far down the channel would an averaged size grit particle travel if the flow through the channel is 1. 7 MGD? (Hint: average sized grit particle settles at a rate of 0.075 ft/sec) a. 2.6 ft b. 5.1 ft c. 11.6 ft d. 20 ft e. 75 ft

Step 1 Calculate Time for grit to fall in channel = Water Depth / descend velocity = seconds 1.5 ft x .075= 20 seconds Step 2 calculate cubic feet of total flow Total gallons /gallons in a cubic ft = cu.ft 1,700.000 gal /7.48= 227,272 cu.ft. Step 3 calculate cu. ft / per sec cubic ft. / 86400= cu.ft /per sec 227,272 / 86400= 2.63 cu. ft/per sec Step 4 Calculate Area width x depth= sq. ft. (Area) = 3ft x1.5ft= 4.5 sq.ft Step 5 Calculate Velocity cu. ft per sec/ area = ft. per sec ( velocity) 2.63/4.5= 0.58 ft/sec Step 6 Calculate distance grit traveled Time to fall x velocity= distance 20 sec x .58 ft = 11.6 ft.

Calculate Mass Balance Q: 2.0MGD MLSS: 4000 mg/L RSS: 9,000 mg/L

The return flow (R) is monitored to maintain the mass balance in the clarifier. The return flow in MGD is calculated by multiplying the influent flow (Q), in MGD, times the mg/L of MLSS, and then dividing that sum by the mg/L of the return activated suspended solids (RSS) less the mg/L MLSS (RSS-MLSS).

Calculate Sludge yeild W: 0.02 MGD WSS: 9,000 mg/L Q: 2.0 MGD BOD infl{ 180 mg/L

Y is calculated by dividing pounds of waste suspended solids (WSS) by pounds of incoming food (BOD inf).

Digest Volume Reduction Calculation 35,000 gallons/day waste activated sludge pumped to a flotation thickener at a concentration of 0.4%. If this sludge is concentrated to 4.2% before it is sent to the digester, how much digester volume would be saved?

[(.4/4.2)x 35,000= 3,333 ] 35,000- 3,333 / 7.48 = 4,233.6

Lbs. of solids 15,000 gallons of sludge with 4% solids is run through a press to produce a 25% cake. How many metric dry tons are disposed of? [Conversion factor for metric tones is 2200 lbs]

[Gallons x 8.34 x Solids Content] / [2,200 [actual is 2,204.6]] 15,000 gal x 8.34 x 0.04 = 5,004 lbs 5,004 lbs/(2,200 lbs/mt) = 2.27 metric tons

Calculate gallons to waste needed each day to waste 417 lbs. lbs to waste=417 WAS Concentration=4,000

[sludge to waste (lbs) ÷ (WAS conc. (mg/L) × 8.34)] × 1,000,000 [417 lbs ÷ (4,000 mg/L × 8.34)] × 1,000,000 = 12,500 gals

Lbs. to waste each day to reach target sludge age Calculating Desire MLSS under aeration =1351 lbs Desired Sludge age= 9 days lbs. to waste to achieve 9 days

answer: Sludge to waste (lbs/day) = MLSS under aeration (lbs) ÷ desired sludge age (days) = 1,351 lbs ÷ 9 days = 150 lbs/day

100% to 65 % equivalent calculation How many pounds of 100% chlorine would be equivalent to 11 lb of 65% calcium hypochlorite?

lb. hypochlorite × decimal % = lb 11 lb × 0.65 = 7.15 lb 100% Cl2

Dosage Calculation lbs. to mg/l If 70 lb of gas chlorine is put into 3 000,000, the dosage is _________ mg/L.

lbs/ (MGD X 8.34)= mg/L 70 (3 X 8.34) =2.8 mg/L