Flow Velocity
Ft3/sec - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water in a water main is traveling at a velocity of 4 ft/sec and the pipe is 8 inches diameter. What is the flow in cubic feet per second (ft3/sec)?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
THIRD. Follow the solution procedure below to solve the problem!
8 in/ft / 12 in/ft = .667 ft
0.785 x .667 dia/ft x .667 dia/ft = .350 ft2
.350 ft2 x 4 ft/sec = 1.4 ft3/sec
Ft3/sec - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
An 8 inch water main’s velocity is 4 ft/sec. What is the flow a cubic feet per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (Flow, ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, Ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
THIRD. Follow the solution procedure below to solve the problem!
8 dia/in / 12 in/ft = .667 dia/ft
.785 x .667 dia/ft x .667 dia/ft = .350 ft2
.350 ft2 x 4 ft/sec = 1.4 ft3/sec
Ft3/sec - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing through a water main at 38 gallons per second, what is the flow at cubic feet per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Flow, Ft3/sec = Gal/sec (F)
7.48 gal/ft3
THIRD. Follow the solution procedure below to solve the problem!
38 gal/sec = 5 ft3/sec
7.48 gal/ft3
Ft3/sec - 4
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
2,693 gallons per minute of water is flowing through a main, what is the flow in Ft3/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Flow, Ft3/sec = Gal/min (F)
60 sec/min x 7.48 gal/ft3
THIRD. Follow the solution procedure below to solve the problem!
2,693 Gal/min = 2,693 Gal/min = 6 ft3/sec
60 sec/min x 7.48 gal/ft3 448.8 gal/sec
Ft3/sec - 5
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing through a 4” water main at 108,000 gallons per hour, what is the flow in ft3/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (Ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Flow, Ft3/sec = Gal/hr (F)
60 sec/min x 60 min/hr x 7.48 gal/ft3
THIRD. Follow the solution procedure below to solve the problem!
108,000 gal/hr = 108,000 gal/hr = 4 ft3/sec
60 sec/min x 60 min/hr x 7.48 gal/ft3 26,928 gal/hr
Ft3/sec - 6
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing out of a pipe at 5,816,448 gal/day. What is the flow in ft/3/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Flow, Ft3/sec = Gal/day (F)
86,400 sec/day x 7.48 gal/ft3
THIRD. Follow the solution procedure below to solve the problem!
5,816,448 gal/day = 5,816,448 gal/day = 9 ft3/sec
86,400 sec/day x 7.48 gal 646,272 gal/day
Ft3/sec - 7
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
The velocity in your canal is 4 ft/sec, and the measurements are 15 ft across the top, 10 across the base and 6 feet deep. What is the flow in ft3/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Trapezium, Ft2 = (Top + Base) x Height / Depth (C&C)1
2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)2
THIRD. Follow the solution procedure below to solve the problem!
(15 ft/top + 10 ft/base) x 6 ft = 150 = 75 ft2
2 2
75 ft2 x 4 ft/sec = 300 ft3/sec
Ft3/min - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A 10 inch water main is flowing at aa velocity of 6 ft/sec. What is the flow in cubic feet per minute?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/min).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
Ft3/min = Ft3/sec x 60 sec/min (C&C)4
THIRD. Follow the solution procedure below to solve the problem!
10 dia/in / 12 in/ft = .833 dia/ft
.785 x .833 dia/ft x .833 dia/ft = .545 ft2
.545 ft2 x 6 ft/sec = 3.27 ft3/sec
3.27 ft3/sec x 60 sec/min = 202.74 ft3/min
Ft3/hr - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A 6 inch water main is flowing at a velocity of 5 ft/sec. What is the flow in cubuc feet per hour?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/hr).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
ft3/hr = Ft3/sec x 60 sec/min x 60 min/hr (C&C)4
THIRD. Follow the solution procedure below to solve the problem!
6 dia/in / 12 in/ft = .5 dia/ft
.785 x .5 dia/ft x .5 dia/ft = .196 ft2
.196 ft2 x 5 ft/sec = .98 ft3/sec
.98 ft3/sec x 60 sec/min x 60 min/hr = 3,528 ft3/hr
Ft3/day - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A 12 inch ductile iron water main is flowing at a velocity of 8 ft/sec. What is the flow in cubic feet per day?
SECOND. By using the KEY WORDS you see this is a (flow) problem, that includes (Ft3/day).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
ft3/day = Ft3/sec x 86,400 sec/day (C&C)4
THIRD. Follow the solution procedure below to solve the problem!
12 dia/in / 12 in/ft = 1 dia/ft
.785 x 1 dia/ft x 1 dia/ft = .785 ft2
.785 ft2 x 8 ft/sec = 6.28 ft3/sec
6.28 ft3/sec x 86,400 sec/day = 542,594 ft3/day
Gal/sec - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is moving through a 4 inch water main at a velocity of 7 ft/sec, what is the flow in gallons per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gallons per second).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
Gal/sec = ft3/sec x 7.48 gal/ft3 (C&C)4
THIRD. Follow the solution procedure below to solve the problem!
4 in/dia / 12 in/ft = .333 dia/ft
.785 x .333 dia/ft x .333 dia/ft = .087 ft2
.087 ft2 x 7 ft/sec = .609 ft3/sec
.609 ft3/sec x 7.48 gal/ft3 = 4.55 gal/sec
Gal/sec - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing into a tank at a flow of 5.5 ft3/sec, what is the flow in gallons per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gal/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Gal/sec = ft3/sec x 7.48 gal/ft3 (F)
THIRD. Follow the solution procedure below to solve the problem!
5.5 ft3/sec x 7.48 gal/ft3 = 41.14 gal/sec
Gal/min - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing at a velocity of 10 ft/sec through an 18 inch CMLC water main. What is the flow in gallons per minute?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gal/min).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
Ft3/min = Ft3/sec x 60 sec/min (C&C)4
Gal/min = ft3/min x 7.48 gal/ft3 (C&C)5
THIRD. Follow the solution procedure below to solve the problem!
18 dia/in / 12 in/ft = 1.5 dia/ft
.785 x 1.5 dia/ft x 1.5 dia/ft = 1.413 ft2
1.413 ft2 x 10 ft/sec = 14.13 ft3/sec
14.13 ft3/sec x 60 sec/min = 847.80 ft3/min
847.80 ft3/min x 7.48 gal/ft3 = 6,341.5 gal/min
Gal/min - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing through a main at 7 ft3/sec. What is the flow in Gal/min?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gal/min).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Ft3/min = ft3/sec x 60 sec/min (C&C)1
Gal/min = ft3/min x 7.48 gal/ft3 (F)2
THIRD. Follow the solution procedure below to solve the problem!
7 ft3/sec x 60 sec/min = 420 ft3/min
420 ft3/min x 7.48 gal/ft3 = 3,141.6 gal/min
Gal/hr - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is moving through a 24 inch water main a a velocity of 8 ft/sec. What is the flow in gallons after 8 hours.
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gallons, hours).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
Ft3/hr = Ft3/sec x 60 sec/min x 60 min/hr x 8 hr (C&C)4
Gal/hr = ft3/hr x 7.48 gal/ft3 (C&C)5
THIRD. Follow the solution procedure below to solve the problem!
24 dia/in / 12 in/ft = 2 dia/ft
.785 x 2 dia/ft x 2 dia/ft = 3.14 ft2
3.14 ft2 x 8 ft/sec = 25.12 ft3/sec
25.12 ft3/sec x 60 sec/min x 60 min/hr x 8hr = 723,456 ft3/8hr
723,456 ft3/8hr x 7.48 gal/ft3 = 5,411,450.88 gal/8hr
Gal/hr - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
3.72 ft3/sec is flowing into a lake, what is the flow in gallons per hour?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft3/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Ft3/hr = ft3/sec x 60 sec/min x 60 min/hr (F)1
Gal/hr = ft3/hr x 7.48 gal/ft3 (C&C)2
THIRD. Follow the solution procedure below to solve the problem!
3.72 ft3/sec x 60 sec/min x 60 min/day = 13,392 ft3/hr
13,392 ft3/hr x 7.48 gal/ft3 = 100,172.16 gal/hr
Gal/day - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Flow through an 8 inch water main is traveling at 5 ft/sec. What is the flow in gallons per day?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gallons per day).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
Ft3/day = Ft3/sec x 86,400 sec/day (C&C)4
Gal/day = ft3/day x 7.48 gal/ft3 (C&C)5
THIRD. Follow the solution procedure below to solve the problem!
8 in/ft / 12 in/ft = .667 dia/ft
.785 x .667 dia/ft x .667 dia/ft = .350 ft2
.350 (ft2) x 5 (ft/sec) = 1.75 (ft3/sec)
1.75 ft3/sec x 86,400 sec/day = 151,200 ft3/day
151,200 ft3/day x 7.48 gal/ft3 = 1,130,967 gal/day
Gal/day - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
The velocity in a 30 inch water main is 4 ft/sec. What is the flow in gallons per day?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gallons per day).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Flow, ft3/sec = Area, ft2 x Velocity, ft/sec (F)3
Ft3/day = Ft3/sec x 86,400 sec/day (C&C)4
Gal/day = ft3/day x 7.48 gal/ft3 (C&C)5
THIRD. Follow the solution procedure below to solve the problem!
30 dia/in / 12 in/ft = 2.5 dia/ft
.785 x 2.5 dia/ft x 2.5 dia/ft = 4.91 ft2
4.91 ft2 x 4 ft/sec = 19.64 ft3/sec
19.64 ft3/sec x 86,400 sec/day = 1,696,896 ft3/day
1,696,896 ft3/day x 7.48 gal/ft3 = 12,692,782.1 gal/day
Gal/day - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing at 10 ft3/sec in a 12 inch water main. What is the flow in gallons for 2 days?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (gallons in 2 days).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Hr/2day= 2 days x 24 hr/day (C&C)1
Ft3/hr = ft3/sec x 60 sec/min x 60 min/hr x hr/day (F)2
Gal/day = ft3/day x 7.48 gal/ft3 (C&C)3
THIRD. Follow the solution procedure below to solve the problem!
2 days x 24 hr/day = 48 hr/2days
10 ft3/sec x 60 sec/min x 60 min/day x 48 hr/2days =
1,728,000 ft3/day x 7.48 gal/day = 12,925,440 gal/2days
Ft/sec - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
9 MGD of water is flowing through a main that is 2 feet in diameter. What is the Velocity in ft/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (velocity, ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
MGD x 1.55 ft3/sec/MGD = ft3/sec = ft/sec (F)
0.785 x Dia, ft2 ft2
THIRD. Follow the solution procedure below to solve the problem!
9 MGD x 1.55 cu ft/sec/MGD = 13.5ft3/sec = 4.30 ft/sec
0.785 x 2 dia/ft x 2 dia/ft 3.14 ft2
Ft/sec - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing through a 4” water main at 3 cubic feet per second, what is the velocity in ft/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft/sec = Ft3/sec (F)3
Ft2
THIRD. Follow the solution procedure below to solve the problem!
4 dia/in / 12 in/ft = .333 dia/ft
.785 x .333 dia/ft x .333 dia/ft = .087 ft2
3 ft3/sec = 3.48 ft/sec
.087 ft2
Ft/sec - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing through a 6 inch main at 100 cubic feet per minute, what is the velovity in feet per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft3/sec = Ft3/min (C&C)3
60 sec/min
Ft/sec = Ft3/sec (F)4
Ft2
THIRD. Follow the solution procedure below to solve the problem!
6 dia/in / 12 in/ft = .5 dia/ft
.785 x .5 dia/ft x .5 dia/ft = .196 ft2
100 ft3/min = 1.67 ft3/sec
60 sec/min
1.67 ft3/sec = 8.5 ft/sec
.196 ft2
Ft/sec - 4
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
105,000 ft3 has been flowing through an 8 inch main for 12 hours. What is the velocity in ft/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (Ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft3/sec = Ft3/12hr (C&C)3
60 sec/min x 60 min/hr x 12 hr
Ft/sec = Ft3/sec (F)4
Ft2
THIRD. Follow the solution procedure below to solve the problem!
8 dia/in / 12 in/ft = .667 dia/ft
.785 x .667 dia/ft x .667 dia/ft = .350 ft2
105,000 ft3/12hr = 105,000 ft3/12hr = 2.43 ft3/sec
60 sec/min x 60 min/hr x 12 hr 43,200 sec/12hr
2.43 ft3/sec = 6.9 ft/sec
.350 ft2
Ft/sec - 5
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
470,880 ft3 a day is flowing through a 10 inch outlet, what is the velocity in feet per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft3/sec = Ft3/day (C&C)3
86,400 sec/day
Ft/sec = Ft3/sec (F)4
Ft2
THIRD. Follow the solution procedure below to solve the problem!
10 dia/in / 12 in/ft = .833 dia/ft
.785 x .833 dia/ft x .833 dia/ft = .545 ft2
470,880 ft3/day = 5.45 ft3/sec
86,400 sec/day
5.45 ft3/sec = 10 ft/sec
.545 ft2
Ft/sec - 6
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A basin is being filled at 5 gallons per second from a 4 inch outlet, what is the velocity in ft/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft3/sec = Gal/sec (C&C)3
7.48 gal/ft3
Ft/sec = Ft3/sec (F)
Ft2
THIRD. Follow the solution procedure below to solve the problem!
4 dia/in / 12 in/ft = .333 dia/ft
.785 x .333 dia/ft x .333 dia/ft = .087 ft2
5 gal/sec = 1.6 ft3/sec
7.48 gal/ft3
.67 ft3/sec = 7.7 ft/sec
.087 ft2
Ft/sec - 7
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
600 gal/min is flowing through a 6” water main,what is the velocity in feet per second?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft3/sec = Gal/min (C&C)3
60 sec/min x 7.48 gal/ft3
Ft/sec = Ft3/sec (F)4
Ft2
THIRD. Follow the solution procedure below to solve the problem!
6 dia/in / 12 in/ft = .5 dia/ft
.785 x .5 dia/ft x .5 dia/ft = .196 ft2
600 gal/min = 1.3 ft3/sec
60 sec/min x 7.48gal/ft3
1.3 ft3/sec = 6.6 ft/sec
.196 ft2
Ft/sec - 8
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A water main has flowed 370,000 gallons for 8 hours through an 8 inch water main, what is the velocity in ft/sec?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Area, Ft2 = 0.785 x (Dia/ft)2 (C&C)2
Ft3/sec = Gal/8hr (C&C)3
60 sec/min x 60 min/r x 8 hr x 7.48 gal/ft3
Ft/sec = Ft3/sec (F)4
Ft2
THIRD. Follow the solution procedure below to solve the problem!
8 dia/in / 12 in/ft = .667 dia/ft
.785 x .667 dia/ft x .667 dia/ft = .350 ft2
370,000 gal/8hr = 370,000 gal/8hr = 1.7 ft3/sec
60 sec x 60 min x 8 hr x 7.48 gal/ft3 215,424 gal/8hr
1.7 ft3/sec = 4.8 ft/sec
.350 ft2
Ft/sec - 9
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
3 MGD of water is flowign through a 12 inch water main, what is the velocity in ft/sec
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Ft/sec = MGD x 1.55 ft3/sec/MGD = ft3/sec (F)2
0.785 x Dia, ft2 ft2
THIRD. Follow the solution procedure below to solve the problem!
12 dia/in / 12 in/ft = 1 dia/ft
3 MGD x 1.55 ft3/sec/MGD = 4.65 ft3/sec = 5.9 ft/sec
.785 x1 dia/ft x 1 dia/ft .785 ft2
Ft/sec - 10
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A water district is flowing 175 MGD through a trapezium canal that has a 30 foot top, 20 foot base, and the water is 10 feet deep. What is the velocity in feet per second (ft/sec) ?
SECOND. By using the KEY WORDS you see this is a (Flow) problem, that includes (Ft/sec).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Ft3/sec = MGD x 1.55 ft3/sec/MGD (C&C)1
Trapezium, Ft2 = (Top + Base) x Ht / Depth (C&C)2
2
Ft/sec = Ft3/sec (F)3
Ft2
THIRD. Follow the solution procedure below to solve the problem!
175 MGD x 1.55 ft3/sec/MGD = 127.1 ft3/sec
(30 ft/top + 20 ft/base) x 10 ft = 500 ft = 250 ft2
2 2
271.25 ft3/sec = 1.1 ft/sec
250 ft2