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?
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 |
What is the formula for the solids loading rate on a final clarifier in an activated sludge plant?
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
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.
Click on the thumbnail below for a chart of BOD and TSS math and have some fun!!!
| Fill-In the Missing Information | |
| Clarifiers | 6 |
| Diameter | 130 ft |
| SWD | 12 ft |
| Area | |
| Volume | |
| Weir Diameter | 125 ft |
| Weir Length | |
| 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 | |
| Max 30-day Avg HDT | |
| Max Day HDT | |
| Peak HDT | |
| Avg Day SLR | |
| Max 30-day Avg SLR | |
| Max Day SLR |
Click on the thumbnail below for a chart of pump station math for cylindrical tanks and have some fun!!!
Click on the thumbnail below for a chart of pump station math for rectangular tanks and have some fun!!!
Click on the thumbnail below for a chart of pipe math and have some fun!!!
| 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 |
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 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)
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 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 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.
- The surface area of one aeration tank is ________ square feet.
- The volume of one aeration tank is ________ cubic feet.
- The volume of one aeration tank is ________ 1000-cubic feet.
- The total volume of both aeration tanks together is ________ cubic feet.
- The total volume of both aeration tanks together is ________ 1000-cubic feet.
- One aeration tank holds ________ gallons.
- One aeration tank holds ________ million gallons.
- The two aeration tanks together hold ________ gallons.
- The two aeration tanks together hold ________ million gallons.
- The return sludge flow is ________ mgd.
- The applied flow to the aeration tanks is ________ mgd.
- The applied flow to one aeration tank is ________ mgd.
- The applied flow to one aeration tank is ________ gpd.
- The detention time is ________ hours.
- The total organic load to both aeration tanks is ________ lbs BOD/day.
- The organic load to one aeration tank is ________ lbs BOD/day.
- The organic loading rate on one aeration tank is ________ lbs BOD/day/1000-cf.
- The MLSS in one aeration tank is ________ lbs MLSS.
- The MLSS in both aeration tanks is ________ lbs MLSS.
- The F/M is ________.
- The total amount of sludge wasted daily is ________ lbs WAS/day.
- The amount of sludge wasted from each aeration tank is ________ lbs WAS/day.
- The SRT is ________ days.
- The effluent TSS is ________ lbs TSS/day.
- The MCRT is ________ days.
- The influent TSS is ________ lbs TSS/day.
- The sludge age is ________ days.
- The SVI is ________
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.
- The surface area of a final clarifier is ________ square feet.
- The volume of a final clarifier is ________ cubic feet.
- A final clarifier holds ______ gallons.
- The detention time of a final clarifier is _______ minutes.
- The detention time of a final clarifier is _______ hours.
- The surface settling rate of a final clarifier is _______ gpd/square foot
- The weir overflow rate of a final clarifier is ________ gpd/ft
- The solids loading rate of a final clarifier is ________ lbs/day/sq ft
- The BOD removed is ______ mg/L.
- The BOD removed is ______ lbs/day.
- The BOD discharged by the plant is ______ lbs/day.
- The TSS removed is ______ mg/L.
- The TSS removed is ______ lbs/day.
- The TSS discharged by the plant is ______ lbs/day.
- The BOD removal efficiency is _____ %.
- The TSS removal efficiency is _____ %.
The surface area of a tank 120 feet in diameter and 14 feet deep is ________ ft2.
The surface area of a tank 90 feet in diameter and 9 feet deep is _____ ft2.
The cross-sectional area of a tank 100 ft long, 25 ft wide, and 12 ft deep is _____ ft2.
The surface area of a tank 100 feet long, 25 feet wide, and 12 feet deep is _____ ft2.
A lot that measures 1,200 feet by 2,500 feet contains ______ acres.
The strength of a chemical solution made with 6 pounds of chemical and 55 gallons of water is _____ %.
200 gallons of a 2.5% solution contains _____ pounds 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 solution contains _____ pound(s) 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.
A drum contains 55 gallons of a chemical solution with a specific gravity of 1.12. The dry chemical weighs _______ pounds.
A dosage of 7 mg/L into a flow of 250,000 gpd uses ________ lbs/day of chemical.
Twenty-five pounds of chemical per day are being added to a flow of 500 gpm is _______ mg/L.
The maximum rate for a chemical feeder is 640 lbs/day. The feed rate is _____ pounds/minute.
The solution in a 15 foot by 35 foot by 55 foot tank with a specific gravity of 1.05 weighs _____ pounds.
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 gallons of a 0.15% chemical solution contains _____ pound(s) 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.
A feed pump that lowers a 15 foot diameter chemical solution tank 8 inches in 20 minutes delivers _______ gpd.
A pump that lowers a 15 foot diameter chemical solution tank 8 inches in 20 minutes delivers _______ gpm.
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.
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.
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.
To dose a flow of 0.6 mgd at 1.8 mg/L, you need ____ 40 pound bags of the chemical.
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?
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.
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.
Calculate the molecular weight of H2CO3. The atomic weight of calcium is 40, oxygen is 16, carbon is 12, and hydrogen is 1.
Calculate the molecular weight of H2O. The atomic weight of calcium is 40, oxygen is 16, carbon is 12, and hydrogen is 1.
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.
Go the the next page to see the answers.
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/dayYour 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.
175 ft + 75 ft + 175 ft +75 ft = 500 ft
175 ft x 75 ft = 13,125 sf
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
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
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 2,500The 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 feetThe 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 feetThe 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 feetThe 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 feetA 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 acresThe 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 lbsTwo 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 chemicalTwo 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 solutionA 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 lbsA 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/dayTwenty-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/LThe 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/minThe 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 lbsA 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 lbs55 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 chemicalA 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 solutionA 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 solutionA 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 solutionA 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 chemicalIf 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 chemicalYou 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 bagsTo 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 bagYou 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 bagsThe 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 ft3The 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 ft3Calculate 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) = 62Calculate 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) = 18Calculate 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