Sedimentation
Surface Loading Rate - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
The wastewater treatment plant is flowing at 6 MGD. The area of the tank is 100 ft long by 30 ft wide. What is the surface loading rate?
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (surface loading rate).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Gal/Day = MGD x 1,000,000 (C&C)1
Area, Ft2 = Length x Width (C&C)2
Surface Loading Rate, GPD/ft2 = Total flow, GPD (F)3
Surface Area, ft2
THIRD. Follow the solution procedure below to solve the problem!
6 MGD x 1,000,000 = 6,000,000 gal/day
100 ft x 30 ft = 3,000 ft2
6,000,000 gal/day = 2,000 GPD/ft2
3,000 ft2
Surface Loading Rate - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A sedmination basin measures 200 ft long by 50 ft wide by 10 feet deep. Flow throught the basin is 3,125 gallons per minute. What is the surface loading rate in GPD/ft2?
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (surface loading rate).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Gal/Day = Gal/min x 1,440 min/day (C&C)1
Area, Ft2 = Length x Width (C&C)2
Surface Loading Rate, GPD/ft2 = Total flow, GPD (F)3
Surface Area, ft2
THIRD. Follow the solution procedure below to solve the problem!
3,125 gal/min x 1,440 min/day = 4,500,000 GPD
200 ft x 50 ft = 10,000 ft2
4,500,000 GPD = 450 GPD/ft2
10,000 ft2
Detention Time - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the detention time of 950,000 gal/hr of water flowing into a settling basin that holds 150,000 gallons.
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (detention time).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Detention time = Volume (F)1
Flow
Minutes = decimal/hr x 60 min/hr (C&C)2
Seconds = decimal/min x 60 sec/min (C&C)3
THIRD. Follow the solution procedure below to solve the problem!
.785 x 50 dia/ft x 50 dia/ft x 10 ft x 7.48 gal/ft3 = 146,795 gal
950,000 gal/hr = 6.33 hrs
150,000 gal
.33 hrs x 60 min/hr = 19.8 min
.8 min x 60 sec/min = 48 sec
6hr 19 min 48 sec
Detention Time Hrs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A Basin measures 20 ft in Diameter and is 30 ft tall. What is the Detention time in hours, if there is a daily flow of 3.5 MGD?
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (detention time, hrs).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Cylinder, ft3 = 0.785 x (Dia/ft)2 x Hight, Depth or Length (C&C)1
Gal/Day = MGD x 1,000,000 (C&C)2
Detention Time Hours = Volume, ft3 x 7.48 gal/ft3 x 24 hr/day (F)3
Gal/day
Minutes = decimal/hr x 60 min/hr (C&C)4
THIRD. Follow the solution procedure below to solve the problem!
.785 x 20 dia/ft x 20 dia/ft x 30 ft/tall = 9,420 ft3
3.5 MGD x 1,000,000 = 3,500,000 Gal/day
9,420 ft3 x 7.48 gal/ft3 x 24 hr/day = .48 hrs
3,500,000 Gal/day
.
.48 hr x 60 min/hr = 29 min
Detention Time Hrs - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the Detention time in hours of a sedimentation basin that is 10 ft tall, 45 ft long, and 25 ft wide. With a daily flow of 2 MGD.
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (detention time, hrs).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Cube, ft3 = Length, ft x Width, ft x Height, ft (C&C)1
Gal/Day = MGD x 1,000,000 (C&C)2
Detention Time Hours = Volume, ft3 x 7.48 gal/ft3 x 24 hr/day (F)3
Gal/day
THIRD. Follow the solution procedure below to solve the problem!
10 ft/tall x 25 ft/wide x 45 ft/long = 11,250 ft3
2 MGD x 1,000,000 = 2,000,000 Gal/day
11,250 ft3 x 7.48 gal/ft3 x 24 hr/day = 1.0 hrs
2,000,000 Gal/day
Flow Rate - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
350 Gallons of water has been flowing through a sedimentation tank for 8 hours. What is the flow rate in gallons per hour?
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (flow rate).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Flow Rate = Volume (F)
Time
THIRD. Follow the solution procedure below to solve the problem!
350 Gallons = 43.75 Gal/hr
8 Hours
Flow Rate - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the flow rate of the clarifier that is flowing 4 MGD for 24 hours?
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (flow rate).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Gal/Day = MGD x 1,000,000 (C&C)1
Flow Rate = Volume (F)1
Time
THIRD. Follow the solution procedure below to solve the problem!
4 MGD x 1,000,000 = 4,000,000 GPD
4,000,000 GPD = 166,667 Gal/hr
24 Hrs/day
Flow Rate - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the flow rate in minutes for a settling tank that is flowing 3,000 gallons in 8 hours?
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (flow rate).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Minutes = Hours x 60 min/hr (C&C)1
Flow Rate = Volume (F)
Time
THIRD. Follow the solution procedure below to solve the problem!
8 hr x 60 min/hr = 480 min
3,000 gal = 6.25 gal/min
480 min
Weir Overflow Rate - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the weir over flow rate in gallons per minute, per foot through a clarifier if it flows 80,000 gallons per day and the weir length is 45 feet.
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (wir overflow length).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Weir Overflow Rate, GPD/ft = Flow, GPD (F)1
Weir Length, ft
Gal/min = Gal/day (C&C)2
1,440 min/day
THIRD. Follow the solution procedure below to solve the problem!
80,000 GPD = 1,777.78 GPD/ft
45 ft
1,777.78 GPD/ft = 1.23 gpm/ft
1,440 min/day
Weir Overflow Rate - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the weir over flow rate in gallons per day, per foot if the plant is flowing 5 MGD and the weir length is 100 feet.
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (weir overflow rate).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Gal/Day = MGD x 1,000,000 (C&C)1
Weir Overflow Rate, GPD/ft. = Flow, GPD (F)2
Weir Length, ft
THIRD. Follow the solution procedure below to solve the problem!
5 MGD x 1,000,000 = 5,000,000 GPD
5,000,000 GPD = 50,000 GPD/ft
100 ft
Weir Overflow Rate - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the weir over flow rate in MGD/ft, per foot through a water treatment plant if it flows 9,500,000 Gal/day and the weir length is 250 feet.
SECOND. By using the KEY WORDS you see this is a (Sedimentation) problem, that includes (weir overflow length).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Weir Overflow Rate, GPD/ft = Flow, GPD (F)1
Weir Length, ft
MGD = Gallons divided by 1,000,000 (C&C)2
THIRD. Follow the solution procedure below to solve the problem!
9,500,000 gal/day = 38,000 gal/ft
250 ft
38,000 gal/ft / 1,000,000 = .038 MGD/ft
Weir Overflow Rate - 4
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the weir overflow length if the flow is 175,000 GPD and the weir overflow rate is 1,591 gpd/ft?
SECOND. By using the KEY WORDS you see this is a () problem, that includes ().
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Weir length, ft = Flow, GPD (C&C)
overflow rate, GPD/ft
THIRD. Follow the solution procedure below to solve the problem!
175,000 GPD = 110 ft
1,591 GPD/ft