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If the volatile solids concentration in raw sludge is 66% and the volatile solids concentration in the digested sludge is 56%, what was the volatile solids reduction?

% VS Reduction	=	% Raw VS - % Digested VS	x 100%
				% Raw VS - (% Raw VS x  % Digested VS)
 
All % must be expressed as decimals.  To get decimals, divide % by 100
Raw = 66%  100 = 0.66
Digested  = 56%  100 = 0.56
% VS Reduction	= 	0.66  0.56	x  100%
				0.66  (0.66 X 0.56)
	=	0.1		x  100%
		0.66  0.37
 
	=	0.1		x  100%
		0.29
 
	=  0.34 x 100%
% VS Reduction	= 34%
 

Match the following:

1.  MCRT

A.  Final Effluent TSS

2.  Sludge Age

B.  Primary Effluent TSS
3.  SRT C.  WAS
  D.  WAS + Final Effluent TSS
  E.  WAS + Primary Effluent

1.  MCRT              D.  WAS + Final Effluent TSS

2.  Sludge Age    B.  Primary Effluent TSS

3.  SRT                 C.  WAS

 

What is the formula for the solids loading rate on a final clarifier in an activated sludge plant?

Solids Loading, lbs/day/ft2    Solids Applied, lbs/day

                                                          Surface Area, sq ft

 

mg/L x 8.34 x MGD is the formula for calculating

A. Chlorine demand

B. lbs

C. lbs/day

D. Volume of a cylindrical tank

The correct answer is C.  lbs/day

 

Your raw TSS is 230 mg/L and your primary effluent TSS is 175 mg/L. Your flow is 25 MGD. Calculate the lbs/day of TSS removed and % removal efficiency.

    TSS removed, mg/L = 230 mg/L - 175 mg/L = 55 mg/L

    TSS removed, lbs/day = 55 mg/L x 8.34 lbs/gal x 25 MGD= 11,467.5 lbs/day

    % Removal Efficiency    = [(230 mg/L - 175 mg/L) 230 mg/l] x 100

                                            = [55 230] x 100 = 0.24 x 100 = 24%

 

Click on the thumbnail below for the answers for the chart of BOD and TSS math.

 

Fill-In the Missing Information
Clarifiers 6
Diameter 130 ft
SWD 12 ft
Area 13,267 ft2
Volume 1.19 MG
Weir Diameter 125 ft
Weir Length 393 ft
Avg Day Flow 51.6 MGD
Max 30-day Avg 100.6 MGD
Max Day 153.5 MGD
Peak 195.5 MGD
Avg Day HDT  3.4 hrs
Max 30-day Avg HDT  1.8 hrs
Max Day HDT  1.2 hrs
Peak HDT  0.9 hrs
Avg Day SLR 650 gpd/ft2
Max 30-day Avg SLR 1,260 gpd/ft2
Max Day SLR 2,460 gpd/ft2

 

Click on the thumbnail below for the answers for the chart of pump station math for cylindrical tanks.

 

Click on the thumbnail below for the answers for the chart of pump station math for rectangular tanks.

 

Click on the thumbnail below for the answers for the chart of pipe math.

 

Match
1 The units for pressure are A Divided by
2 Head is expressed in B 2.31
3 Velocity is measured in C 0.434 psi
4 Units of flow D psi
5 Pressure times _______ = Head E fps
6 Head __________ 2.31 = Pressure F 1440
7 20 psi =  G 694 gpm
8 1 MGD =  H cfs
9 A head loss of 7 feet = I 3 psi
  J Times
  K ft
  L 46.2 ft
  M Subtracted by

1 = D

2 = K

3 = E

4= H

5 = B

6 = A

7 = L

8 = G

9 = I

 

The formula for the volume of a rectangular tank is

A.  (Length)(Width)

B.  (0.785)(Diameter2)

C.  (Length)(Width)(Height)

D.  (0.785)(Diameter2)(Height)

The correct answer is C.  (Length)(Width)(Height)

A is the formula for the area of a rectangle

B is the formula for the area of a circle

D is the formula for the volume of a cylinder

 

The formula for the volume of a cylindrical tank is

A.  (Length)(Width)

B.  (0.785)(Diameter2)

C.  (Length)(Width)(Height)

D.  (0.785)(Diameter2)(Height)

The correct answer is D.  (0.785)(Diameter2)(Height)

A is the formula for the area of a rectangle

B is the formula for the area of a circle

C is the formula for the volume of a rectangle

 

Given a tank measures 50 feet long, 25 feet wide, and 15 ft deep, calculate:

  • Surface area in ft2
  • Cross-sectional area in square feet
  • Volume in cf
  • Capacity in gallons

The correct answers are:

  • Surface area in ft2

Surface area in ft2 = Length, ft x Width, ft = 50 ft x 25 ft = 1250 ft2

  • Cross-sectional area in square feet

Cross-sectional area in ft2 = Width, ft x Depth, ft = 25 ft x 15 ft = 375 square feet

  • Volume in cf

Volume in cf = Length, ft x Width, ft x Depth, ft = 50 ft x 25 ft x 15 ft = 18,750 cf

  • Capacity in gallons

Capacity in gallons = Vol in cf x 7.5 gal/cf = 18,750 cf x 7.5 gal/cf = 140,625 gal

 

The flow through an activated sludge plant in 15 mgd.  The BOD of the primary effluent is 110 mg/L and the BOD of the final effluent is 8 mg/L.  The rate of air flow to the aeration tank is 2,000 cfm.  The MLSS is the aeration tank is 2,500 mg/L.  The 30-minute settleable solids test is 225 mL/L.  The TSS of the primary effluent is 115 mg/L and the TSS of the final effluent is 12 mg/L.  The return sludge ratio is 20%.  60,000 gallons of 2% sludge are wasted each day.  The flow is split evenly between two rectangular aeration tanks.  All tanks are the same size.  Each tank is 175 feet long, 75 feet wide, and 12 feet deep.

The perimeter of one aeration tank is ________ feet.

175 ft + 75 ft + 175 ft +75 ft = 500 ft

 

The surface area of one aeration tank is ________ square feet.

175 ft x 75 ft = 13,125 sf

The volume of one aeration tank is ________ cubic feet.

175 ft x 75 ft x 12 ft = 157,500 cf

The volume of one aeration tank is ________ 1000-cubic feet.

157,500 cf 1000 = 157.5 1000-cf

 

The total volume of both aeration tanks together is ________ cubic feet.

157,500 cf x 2 = 315,000 cf

The total volume of both aeration tanks together is ________ 1000-cubic feet.

315,000 cf 1000 = 315 1000-cf

One aeration tank holds  ______ gallons.

157,500 cf x 7.48 gal/cf = 1,178,100 gal

One aeration tank holds  ______ million gallons.

1,178,100 gal 1,000,000 = 1.18 MG

The two aeration tanks together hold  ______ gallons.

1,178,100 gal x 2 = 2,356,200 gal

The two aeration tanks together hold is ______ million gallons.

1.18 MG x 2 = 2.36 MG

The return sludge flow is ________ mgd.

20%/100 = 0.2

0.2 x 15 mgd = 3 mgd

 

The applied flow to the aeration tanks is ________ mgd.

15 mgd + 3 mgd = 18 mgd

The applied flow to one aeration tank is ________ mgd.

18 mgd 2 = 9 mgd

The applied flow to one aeration tank is ________ gpd.

9 mgd 1,000,000 = 9,000,000 gpd

The detention time is ________ hrs.

(1,178,100 gal) x 24 hrs/day = 0.1309 days x 24 hrs/day = 3.14 hrs

9,000,000 gpd

 

The total organic load to both aeration tanks is ________  lbs BOD/day.

110 mg/L x 8.34 lbs/gal x 15 mgd = 13,761 lbs BOD/day

 

The organic load to one aeration tank is ________ lbs BOD/day.

6,881 lbs BOD/day 2 = 6,881 lbs BOD/day

The organic loading rate on one aeration tank is ________ lbs BOD/day/1000-cf.

3,440 lbs BOD/day 157.5 1000-cf = 21.8 lbs BOD/day/1000-cf

 

The MLSS in one aeration tank is ________  lbs MLSS.

2,500 mg/L x 8.34 lbs/gal x 1.18 mgd = 24,603 lbs MLSS

 

The total MLSS in both aeration tanks is ________  lbs MLSS.

2,500 mg/L x 8.34 lbs/gal x 2.36 mgd = 49,206 lbs MLSS

 

The F/M is ________.

6,881 lbs BOD 24,603 lbs MLSS = 0.28

 

The total amount of sludge wasted daily is ________  lbs WAS/day.

20% 100 = 0.02

60,000 gal x 8.34 lbs/gal x 0.02 = 10,008 lbs/day

 

The amount of sludge wasted from each aeration tank is ________ lbs WAS/day.

10,008 lbs WAS/day 2 = 5,004 lbs/day

 

The SRT is ________ days.

24,603 lbs MLSS 5,004 lbs/day = 4.92 days

 

The effluent TSS is ________  lbs TSS/day.

12 mg/L x 8.34 lbs/gal x 7.5 mgd = 751 lbs/day

 

The MCRT is ________ days.

             24,603 lbs MLSS             = 24,603 lbs MLSS/5,755 lbs/day = 4.28 days

   (5,004 lbs/day + 751 lbs/day)

 

The influent TSS is ________ lbs TSS/day.

115 mg/L x 8.34 lbs/gal x 7.5 mgd = 7,193 lbs/day

The sludge age is ________ days.

24,603 lbs MLSS 7,193 lbs/day = 3.42 days

 

The SVI is ________.

            225 mL/L x 10,000     =      2,250,000         =  900

                                        2,500                          2,500

 

The flow through an activated sludge plant is 15 mgd.  The BOD of the primary effluent is 110 mg/L and the BOD of the final effluent is 8 mg/L.  The MLSS is the aeration tank is 2,500 mg/L.  The 30-minute settleable solids test is 225 mL/L.  The TSS of the primary effluent is 115 mg/L and the TSS of the final effluent is 12 mg/L.  The return sludge ratio is 20%.  60,000 gallons of 2% sludge are wasted each day.  The flow is split evenly between two equal-sized rectangular aeration tanks and final clarifiers.  Each aeration tank is 175 feet long, 75 feet wide, and 12 feet deep.  Each final clarifier has a 75 feet diameter, a SWD of 8 feet, and they are 12 feet deep in the center.

The circumference of a final clarifier is ________ feet.

3.14 x 75 ft = 235.5 ft

The surface area of a final clarifier is ________ square feet.

0.785 x 75 ft x 75 ft = 4,415.625 sf

The volume of a final clarifier is ________ cubic feet.

Cylinder:  0.785 x 75 ft x 75 ft x 8 ft = 35,325 cf

Cone:   (0.785 x 75 ft x 75 ft x 4 ft) 3 = 5,888 cf

Tank:    35,325 cf + 5,888 cf = 41,213 cf

A final clarifier holds  ______ gallons.

41,213 cf x 7.48 gal/cf = 308,273 gal

The detention time of a final clarifier is _______ minutes.

(308,273 gal 9,000,000 gpd) x 1440 min/day = 49 min

The detention time of a final clarifier is _______ hours.

(308,273 gal 9,000,000 gpd) x 24 hrs/day = 0.82 hrs

 

49 min 60 min/hr = 0.82 hrs

The surface settling rate of a final clarifier is _______ gpd/square foot

9,000,000 gpd/4415.625 sf = 2,038 gpd/sf

The weir overflow rate of a final clarifier is ________ gpd/ft

9,000,000 gpd/235.5 ft = 38,216 gpd/ft

The solids loading rate of a final clarifier is ________ lbs/day/sq ft

2,500 mg/L x 8.34 x 9 mgd =  187,650 lbs/day

(187,650 lbs/day)/4415.625 sf = 42.5 lbs/day/sq ft

The BOD removed is ______ mg/L.

110 mg/L 8 mg/L = 102 mg/L

The BOD removed is ______ lbs/day.

102 mg/L x 8.34 lbs/gal x 15 mgd = 12,760 lbs/day

The BOD discharged by the plant is ______ lbs/day.

8 mg/L x 8.34 lbs/gal x 15 mgd = 1,000.8 lbs/day

The TSS removed is ______ mg/L.

115 mg/L 12 mg/L = 103 mg/L

The TSS removed is ______ lbs/day.

103 mg/L x 8.34 lbs/gal x 15 mgd = 12,885 lbs/day

The TSS discharged by the plant is ______ lbs/day.

12 mg/L x 8.34 lbs/gal x 15 mgd = 1,501.2 lbs/day

The BOD removal efficiency is _____ %.

 [(110 mg/L 8 mg/L)/110] x 100% = 0.927 x 100% = 93%

The TSS removal efficiency is _____ %.

 [(115 mg/L 12 mg/L)/115 mg/L] x 100% =0.8956 x 100% = 90%

 

The surface area of a tank 120 feet in diameter and 14 feet deep is ________ square feet.

120 ft x 120 ft x 0.785 = 11,304 square feet

or

60 ft x 60 ft x 3.14 = 11,304 square feet

 

The surface area of a tank 90 feet in diameter and 9 feet deep is _____ square feet.

90 ft x 90 ft x 0.785 = 6,358.5 square feet

or

45 ft x 45 ft x 3.14 = 6,358.5 square feet

 

The cross-sectional area of a tank 100 ft long, 25 ft wide, and 12 ft deep is _____ square feet.

25 ft x 12 ft = 300 square feet

 

The surface area of a tank 100 feet long, 25 feet wide, and 12 feet deep is _____ square feet.

100 ft x 25 ft = 2,500 square feet

 

A lot that measures 1,200 feet by 2,500 feet contains ______ acres.

(1,200 ft x 2,500 ft) 43,560 square feet/acre = 68.87 acres

 

The strength of a chemical solution made with 6 pounds of chemical and 55 gallons of water is _____ %.

[6 lbs (55 gal x 8.34 lbs/gal)] x 100% = 1.31%

 

200 gallons of a 2.5% solution contains _____ pounds of chemical.

200 gal x 8.34 lbs/gal x (2.5% 100%) = 41.7 lbs

 

Two solutions are being combined.  250 gallons of a solution containing 1.5 lbs of chemical per gallon of solution is being added to 1,500 gallons containing 4.8 lbs of chemical per gallon of solution.  The solution contains _____ pound(s) of chemical.

        250 gal x 1.5 lbs of chemical/gal of solution = 375 lbs of chemical

       1,500 gal x 4.8 lbs of chemical/gal of solution = 7,200 lbs of chemical

        375 lbs + 7,200 lbs = 7,575 lbs of chemical

 

Two solutions are being combined. 250 gallons of a solution containing 1.5 lbs of chemical per gallon of solution is being added to 1,500 gallons containing 4.8 lbs of chemical per gallon of solution.  The weight of chemical in the new solution is _____ lbs/gal.

        250 gal + 1,500 gal = 1,750 gal of solution

        7,575 lbs 1,750 gal = 4.33 lbs of chemical/gallon of solution

 

A drum contains 55 gallons of a chemical solution with a specific gravity of 1.12. The dry chemical weighs _______ pounds.

        55 gal x 8.34 lbs/gal x 1.12 = 514 lbs

 

A dosage of 7 mg/L into a flow of 250,000 gpd uses ________ lbs/day of chemical.

        250,000 gpd (1 mgd/1,000,000 gpd) = 0.25 mgd

        0.25 mgd x 7 mg/L x 8.34 lbs/gal = 14.6 lbs/day

 

Twenty-five pounds of chemical per day are being added to a flow of 500 gpm is _______ mg/L.

        500 gpm x 1440 min/day = 720,000 gpd

        720,000 gpd (1 mgd/1,000,000 gpd) = 0.72 mgd

        25 lbs/day (0.72 mgd x 8.34 lbs/gal) = 4.16 mg/L

 

The maximum rate for a chemical feeder is 640 lbs/day. The feed rate is _____ pounds/minute.

        640 lbs/day 1440 min/day = 0.444 lbs/min

 

The solution in a 15 foot by 35 foot by 55 foot tank with a specific gravity of 1.05 weighs _____ pounds.

        15 ft x 35 ft x 55 ft x 7.48 gals/cubic foot x 8.34 lbs/gal x 1.05 = 1,891,381 lbs

 

 A drum contains 55 gallons of a solution with a specific gravity of 1.25.  The empty drum weighs 25 pounds.  The weight of the filled drum is _____ pounds.

55 gal x 8.34 lbs/gal x 1.25 = 573.4 lbs

573.4 lbs + 25 lbs = 598.4 lbs

 

55 gallons of a 0.15% chemical solution contains _____ pound(s) of dry chemical.

55 gal x 8.34 lbs/gal x (0.15 100) = 0.688 lbs of dry chemical

 

A feed pump lowers a 15 foot diameter chemical solution tank containing a 4.5% solution 8 inches in 20 minutes.  The pump delivers ______ pounds/day of solution.

15 ft x 15 ft x 0.785 x (8 in 12 in/ft) x 7.48 gal/ft3 = 881 gal of solution

881 gal 20 min = 44 gpm of solution

44 gpm x 1440 min/day = 63,415 gpd of solution

63,415 gpd x 8.34 lbs/gal = 528, 885 lbs/day of solution

 

A feed pump that lowers a 15 foot diameter chemical solution tank 8 inches in 20 minutes delivers _______ gpd.

15 ft x 15 ft x 0.785 x (8 in 12 in/ft) x 7.48 gal/ft3 = 881 gal of solution

881 gal 20 min = 44 gpm of solution

44 gpm x 1440 min/day = 63,415 gpd of solution

 

A pump that lowers a 15 foot diameter chemical solution tank 8 inches in 20 minutes delivers _______ gpm.

15 ft x 15 ft x 0.785 x (8 in 12 in/ft) x 7.48 gal/ft3 = 881 gal of solution

881 gal 20 min = 44 gpm of solution

 

A slurry was prepared by adding two 40 pound bags of a chemical to a 55 gallon drum and adding enough water to fill the drum.  The slurry contains _____ pound(s) of chemical.

40 lbs x 2 = 80 lbs of chemical

 

If a bag of polymer weighs 60 pounds and contains 4.5% of a certain chemical by weight, the weight of that chemical is ______ pounds.

60 lbs x 4.5% 100% = 2.7 lbs of chemical

 

You need to control an algae bloom in your 20,000,000 gallon raw water reservoir.  The herbicide directions recommend applying 25 pounds/acre.  If the average depth of the reservoir is 15 feet and each bag contains 40 pounds of herbicide, you need ____ bags.

 20,000,000 gal 7.48 gal/ft3 = 2,673,797 ft3

 2,673,797 ft3 15 ft = 178,253 ft2

 178,253 ft2 43,560 ft/acre = 4.09 acres

 4.09 acres x 25 lbs/acre = 102.3 lbs

 102.3 lbs 40 lbs/bag = 2.56 bags

 Need 3 bags

 

To dose a flow of 0.6 mgd at 1.8 mg/L, you need ____ 40 pound bags of the chemical.

 0.6 mgd x 1.8 mg/L x 8.34 lbs/gal = 9 lbs/day

 Need 1 bag

 

You try to maintain a 120 day supply of all chemicals.  If a bag contains 40 pounds of chemical, how many bags of a chemical are needed to dose a flow of 0.6 mgd at 1.8 mg/L?

0.6 mgd x 1.8 mg/L x 8.34 lbs/gal = 9 lbs/day

9 lbs/day x 120 days = 1,080 lbs

1,080 lbs 40 lbs/bag = 27 bags

 

The hopper for your chemical feeder is conical in shape with a reserve tank on top.  The hopper has a diameter of 12 feet and a depth of 8 feet.  The reserve tank is 20 feet high.  The total volume of the hopper and the reserve tank is _______ ft3.

Hopper:  (12 ft x 12 ft x 0.785 x 8 ft) 3 = 301 ft3

Reserve:  12 ft x 12 ft x 0.785 x 20 ft = 2,260 ft3

Total:  301 ft3 + 2,260 ft3 = 2,561 ft3

 

The hopper for your chemical feeder is conical in shape.  If it has a diameter of 12 feet and a depth of 8 feet, the volume of the hopper is _______ ft3.

(12 ft x 12 ft x 0.785 x 8 ft) 3 = 301 ft3

 

Calculate the molecular weight of H2CO3.  The atomic weight of calcium is 40, oxygen is 16, carbon is 12, and hydrogen is 1.

(2 x 1) + (1 x 12) + (3 x 16) = 62

 

Calculate the molecular weight of H2O.  The atomic weight of calcium is 40, oxygen is 16, carbon is 12, and hydrogen is 1.

(2 x 10) + (1 x 16) = 18

 

Calculate the molecular weight of Ca(HCO3)2.  The atomic weight of calcium is 40, oxygen is 16, carbon is 12, and hydrogen is 1.

(1 x 40) + (2 x 1) + (2 x 12) + (6 x 16) = 162

 

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