Volume
Basin, ft2 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the square feet (ft2) of the side of a rectangular water basin that is 10 ft tall and 25 feet long?
SECOND. By using the KEY WORDS you see this is a (Area) problem, that includes (length, width).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, ft2 = Length x Width or Ht (F)
THIRD. Follow the solution procedure below to solve the problem!
25 ft x 10 ft = 250 ft2
Basin ft3 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
If a trench is 400 ft long, 4.5 ft wide and 6 ft deep, how many cubic ft of dirt was removed?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (length width height, ft3).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, ft3 = Length, ft x Width, ft x Height, ft (F)
THIRD. Follow the solution procedure below to solve the problem!
400 ft/long x 4.5 ft/wide x 6 ft/deep = 10,800 ft3
Basin Gal - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A rectangular basin is 50 ft long, 25 ft wide and 20 ft deep. How many gallons of water will it hold?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (length width height, gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (F)
THIRD. Follow the solution procedure below to solve the problem!
50 ft/long x 25 ft/wide x 20 ft/deep x 7.48 gal/ft3 = 187,000 gal
Basin lbs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
There is a basin in your water treatment plant that is 75 ft long and is 35 ft wide and 15 ft deep how many pounds of water is in it?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (length width height, pounds).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (C&C)1
Lbs/Gal = Gallons x 8.34 lbs/gal (F)2
THIRD. Follow the solution procedure below to solve the problem!
75 ft/long x 35 ft/wide x 15 ft/deep x 7.48 gal/ft3 = 294,525 gal
294,525 gal x 8.34lbs/gal = 2,456,338.5 lbs
Basin MG - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A water treatment plant has a basin that is 60 ft long, 30 ft wide and 30 ft deep. How many million gallons (MG) will it hold?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (length width depth, million gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, Gal = Length, ft x Width, ft x Height, ft x 7.48 gal/ft3 (C&C)1
MG = Gallons divided by 1,000,000 (F)2
THIRD. Follow the solution procedure below to solve the problem!
60 ft/long x 30 ft/wide x 30 ft/deep x 7.48 gal/ft3 = 403,920 Gal
403,920 Gal / 1,000,000 = .40392 MG
Cylinder, ft2 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the square feet (ft2) of a 10 inch water main?
SECOND. By using the KEY WORDS you see this is a (Area) problem, that includes (diameter, ft2).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Cyl, Ft2 = 0.785 x (Dia,ft)2 (F)2
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
Cylinder ft3 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You just installed 1,000 ft of pipe that is 18 inches in diameter, how many cubic feet (ft3) of water will it hold?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (dia, ft3).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)1
Cyl, ft3 = 0.785 x (Dia/ft)2 x Hight, Depth or Length (F)2
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 x 1,000 ft = 1,766.25 ft3
Cylinder Gal - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A pipe is 2.5 ft in diameter and is 800 ft long. How many gallons of water is in the pipe?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (dia, gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (F)
THIRD. Follow the solution procedure below to solve the problem!
.785 x 2.5 dia/ft x 2.5 dia/ft x 800 ft/long x 7.48 gal/ft3 = 29,359 Gal
Cylinder Gal - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A reservoir is 25 ft tall and 40 ft in dia, the gauge at the bottom the tank says 6.50 PSI. how many gallons of water is in the reservoir?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Head, ft = PSI x 2.31 ft/PSI (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48gal/ft3 (F)2
THIRD. Follow the solution procedure below to solve the problem!
6.50 PSI x 2.31 ft/PSI = 15 ft tall
.785 x 40 dia/ft x 40 dia/ft 15 ft x 7.48 gal/ft3 = 140,923.20 gal
Cylinder lbs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
There is a reservoir in your water distribution sytem that is 35 ft dia and is 15 ft tall how many pounds of water is in it?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (dia, lbs).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48gal/ft3 (F)1
Lbs/Gal = Gallons x 8.34 lbs/gal (C&C)2
THIRD. Follow the solution procedure below to solve the problem!
.785 x 35 dia/ft x 35 dia/ft x 15 ft/tall x 7.48 gal/ft3 = 107,894.33 gal
107,894.33 gal x 8.34 lbs/gal = 899,839.71 lbs
Cylinder MG - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A water main is 2 ft in diameter and 4,250 ft long, how may million gallons (MG) does it hold?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (dia, MG).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48gal/ft3 (F)1
MG = Gallons divided by 1,000,000 (C&C)2
THIRD. Follow the solution procedure below to solve the problem!
.785 x 2 dia/ft x 2 dia/ft x 4,250 ft/long x 7.48 gal’ft3 = 99,820.6 gal
99,820.6 gal / 1,000,000 = .0998 MG
yd3 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You are digging a trench for a new pipe installation, the trench is 500 ft long 3 ft wide and 5 feet deep. How many cubic yards (yd3) of dirt is removed.
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (yd3).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, Ft3 = Length, ft x Width, ft x Height, ft (CWC)1
yd3 = ft3 divided by 27 ft3/yd (Prime)2
THIRD. Follow the solution procedure below to solve the problem!
500 ft x 3 ft x 5 ft = 7,500 ft3
7,500 ft3 / 27 ft3/yd = 278 yd3
yd3 - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You are installing an 12” water main 1 mile long. The trench is 3 feet wide and 6 feet deep. How much dirt do you have to put back in cubic yards (yd3) ?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (yd3).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Ft,Length = Mile,Length x 5,280 ft/mile (C&C)1
Basin, ft3 = Length, ft x Width, ft x Height, ft (C&C)2
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)3
Cyl, ft3 = 0.785 x (Dia/ft)2 x Hight, Depth or Length (C&C)4
Total, ft3 = Trench/ft3 – Pipe/ft3 (C&C)5
yd3 = Total, ft3 divided by 27 ft3/yd (F)6
THIRD. Follow the solution procedure below to solve the problem!
1 mile x 5,280 ft/mile = 5,280 ft
5,280 ft x 3 ft x 6 ft = 95,040 ft3
12 dia/in / 12 in/ft = 1 dia/ft
.785 x 1 dia/ft x 1 dia/ft x 5,280 ft = 4,144.8 ft3
95,040 Trench/ft3 – 4,144.8 pipe/ft3 = 90,895 ft3
90,895.2 ft3 / 27 ft3/yd = 3,367 yd3
yd3 - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You are installing an 18” water main 1,500 ft long. The trench is 4 feet wide and 6 feet deep. How many dump trucks will you need to backfill the trench in cubic yards (yd3) ? Each dump truck holds 25 yd3.
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (cubic yards, yd3).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Basin, ft3 = Length, ft x Width, ft x Height, ft (C&C)1
Dia/Ft = Dia/Inches divided by 12 In/Ft (C&C)2
Cyl, ft3 = 0.785 x (Dia/ft)2 x Hight, Depth or Length (C&C)3
Total, ft3 = Trench/ft3 – Pipe/ft3 (C&C)4
yd3 = Total, ft3 divided by 27 ft3/yd (C&C)5
yd3/Truck = yd3 divided by 25 yd3/Truck (F)6
THIRD. Follow the solution procedure below to solve the problem!
1,500 ft x 4 ft x 6 ft = 36,000 ft3
18 dia/in / 12 in/ft = 1.5 dia/ft
.785 x 1.5 dia/ft x 1.5 dia/ft x 1,500 ft = 2,649.38 ft3
36,000 trench/ft3 – 2,649.38 pipe/ft3 = 33,350.62 ft3
33,350.62 ft3 / 27 ft3/yd = 1,235.21 yd3
1,235.21 yd3 / 25 yd3/truck = 49.40, round up to 50 dump trucks
Trapezium, ft2 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the square feet (ft2) of a trapezium canal that is 15 ft on top, 5 ft base and is 10 ft tall.
SECOND. By using the KEY WORDS you see this is a (Area) problem, that includes (trapezium, ft2).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Trapezium, Ft2 = (Top + Base) x Height / Depth (F)
2
THIRD. Follow the solution procedure below to solve the problem!
(15 ft + 5 ft) x 10 ft = 200 ft = 100 ft2
2 2
Trapezium ft3 - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A trapezium canal is 35 feet across the top, 20 feet across the bottom and the water depth is at 8 feet. If the canal is 3,000 ft long what is the volume of water in cubic feet?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (ft3).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Trapezium, Ft2 = (Top + Base) x Height / Depth (F)1
2
Ft3 = Ft2 x Length (C&C)2
THIRD. Follow the solution procedure below to solve the problem!
(35 ft/top + 20 ft/base) x 8 ft = 440 ft = 220 ft2
2 2
220 ft2 x 3,000 ft = 660,000 Ft3
Trapezium MG - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
trapezium canal is 40 feet across the top, 25 feet across the bottom and the water depth is at 9 feet. If the canal is 3 miles long what is the volume of water in million gallons (MG) ?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Trapezium, Ft2 = (Top + Base) x Height / Depth (F)1
2
Ft,Length = Mile,Length x 5,280 ft/mile (C&C)2
Gallons = Ft2 x Ft/Length x 7.48 gal/ft3 (C&C)3
MG = Gallons divided by 1,000,000 (C&C)4
THIRD. Follow the solution procedure below to solve the problem!
(40 ft/top + 25 ft/base) x 9 = 585 ft = 292.5 ft2
2 2
3 miles x 5,280 ft/mile = 15,840 ft
292.5 ft2 x 15,840 ft x 7.48 gal/ft3 = 34,620,792 gal
34,620,792 gal / 1,000,000 = 34.62 MG
Gal % full - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You have a reservoir in your water distriution system that can hold 6 MG of water. How many gallons does it hold when it is 25% full?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (%full).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Decimal = % divided by 100 (C&C)1
Gal/MG = Gal/full x Decimal (F)2
THIRD. Follow the solution procedure below to solve the problem!
25% / 100 = .25
6 MG x .25 = 1.5 MG
Gal % full - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A reservoir in your water distriution system is 95 dia/ft and 45 ft tall. How many gallons does it hold when it is 33% full?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (%full).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)1
Decimal = % divided by 100 (C&C)2
Gallons = Gal/full x Decimal (F)3
THIRD. Follow the solution procedure below to solve the problem!
.785 x 95 dia/ft x 95 dia/ft x 45 ft x 7.48 gal/ft3 = 2,384,684.78 gal
33% / 100 = .33
2,384,684.78 gal/full x .33 = 786,945.72 gal
Gal % full - 3
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A reservoir in your water distriution system is 53 dia/ft and 30.5 ft tall. How many gallons does it hold when it is 50% full?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (%full).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)1
Decimal = % divided by 100 (C&C)2
Gallons = Gal/full x Decimal (F)3
THIRD. Follow the solution procedure below to solve the problem!
.785 x 53 dia/ft x 53 dia/ft x 30.5 ft x 7.48 gal/ft3 = 503,063.53 gal
50% / 100 = .50
503,063.53 gal/full x .50 = 251,531.77 gal
Gal % full - 4
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You have a reservoir in your water distriution system that can hold 10,000,000 gallons of water. How many gallons does it hold when it is 75% full?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (%full).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Decimal = % divided by 100 (C&C)1
Gallons = Gal/full x Decimal (F)2
THIRD. Follow the solution procedure below to solve the problem!
75% / 100 = .75
10,000,000 gal/full x .75 = 7,500,000 gal
Supply Hrs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A 450,000 gal water tank supply's a small town. If the water is flowing out at 250 gpm and the water is being pumped in at 550 gpm, how long will it take to fill the tank, in Hrs?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (supply, hrs).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Supply, Hrs = Storage volume, gallons (F)
Flow in, GPM – Flow out, GPM x 60 min/hr
THIRD. Follow the solution procedure below to solve the problem!
450,000 gal = 450,000 gal = 25 hrs
550 gpm – 250 gpm x 60 min/hr 18,000 gal/hr
Time Hrs - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
How many hours will it take to fill a 3,000 gallon tank if the is water flowing into it at 20 gpm?
SECOND. By using the KEY WORDS you see this is a (Volume) problem, that includes (time, hrs).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Time, Hrs = Volume, gallons (F)
Pumping rate, GPM x 60 min/hr
THIRD. Follow the solution procedure below to solve the problem!
3,000 gal = 3,000 gal = 2.5 hrs
20 gpm x 60 min/hr 1,200 gal/hr