General
Cost/Day - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A water treatment plant is treating 2.4 MGD with a dose of 15 ppm. At $0.93 per pound what is the cost a day, per pound of chlorine (CL2) gas?
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (cost per day).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Gas, lbs = Vol, MG x ppm or mg/l x 8.34 lbs/gal (C&C)1
Cost/day = lbs/day x Cost/lbs (F)2
THIRD. Follow the solution procedure below to solve the problem!
2.4 MGD x 15 ppm x 8.34 lbs/gal = 300.24 lbs
300.24 lbs x $0.93/lb = $279.22 lb/day
Paint, Cylinder - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You have been tasked to paint the outside perimeter and top of a cylindrical tank. The tank is 15 feet tall and 30 ft in diameter. How many gallons of paint will you need? Each gallon of paint can cover 250 square feet.
SECOND. By using the KEY WORDS you see this is a (Surface area) problem, that includes (square feet, gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
CIR,ft2 = Pie,3.14 x Dia/ft x Height (F)1
Top,ft2 = .785 x (Dia/ft)2 (F)2
Total,ft2 = CIR,ft2 + Top,ft2 (F)3
Paint,gal = Total ft2 (F)4
gal/ft2
THIRD. Follow the solution procedure below to solve the problem!
3.14 x 30 dia/ft x 15 ft = 1,413 CIR,ft2
.785 x 30 dia/ft x 30 dia/ft = 706.5 Top,ft2
1,413 CIR,ft2 + 706.5 Top,ft2 = 2,118.5 ft2
2,118.5 ft2 = 8.47 gal,
250 ft2/gal
round up to 9 gal/paint
Paint, Basin - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
All four sides of a basin in the water treatment plant needs to be painted, the dimesions are 10 ft tall 20 ft wide and 40 ft long. How many gallons of paint will you need? One gallon of paint covers 150 square feet.
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (square feet, gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Area,ft2,1 = Height x Width x 2 sides (C&C)1
Area,ft2,2 = Height x Length x 2 sides (C&C)2
Total,ft2 = ft2,1 + ft2,2 (C&C)3
Paint,gal = Total ft2 (F)4
gal/ft2
THIRD. Follow the solution procedure below to solve the problem!
10 ft x 20 ft x 2 = 400 ft2
10 ft x 40 ft x 2 = 800 ft2
400 ft2 + 800 ft2 = 1,200 ft2
1,200 ft2 = 8 gal/paint
150 ft2
Paint, Wall - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
There is a wall in front of your water agency in bad shape, you have been asked to paint bot sides of it, the wall is 100 ft long and 6 feet high. How many gallons of paint o you need? Each gallon spreads 200 square feet.
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (square feet, gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Area,ft2 = Height x Length x 2 sides (C&C)1
Paint, gal = Area, ft2 (F)2
gal/ft2
THIRD. Follow the solution procedure below to solve the problem!
6 ft x 100 ft x 2 = 1,200 ft2
1,200 ft2 = 6 gal/paint
200 ft2/gal
Gals/Day - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Solvang California has an estimated population of 5,920 people. each person uses about 65 gallons of water per day. How many total gallons per day is used?
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (gallons per day).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Gal/Day = Population x Gal/Capita/Day (F)
THIRD. Follow the solution procedure below to solve the problem!
5,920 pop x 65 gal/day = 384,800 Gals/Day
GPD - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
You are assigned to read a water meter once a month, on day 1 the meter reads 5300.5 ft3 after 30 days the meter reads 5480.0 ft3. What is the average GPD?
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (gallons per day).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Gallons = Ft3 x 7.48 gal/ft3 (C&C)1
GPD = Meter Read 2 – Meter Read 1 (F)2
Number of Days
THIRD. Follow the solution procedure below to solve the problem!
5300.5 ft3 x 7.48 gal/ft = 39,647.74 gal
5480.0 ft3 x 7.48 gal/ft = 40,990.40 gal
40,990.40 gal – 39,643.74 gal = 1,346.66 gal = 45.49 GPD
30 days 30 days
% Removal - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
What is the removal percent of a reclaimed water treatment plant if the influent turbidity is 25 ntu’s and the effluent Turbidity is .10 ntu’s
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (percent removal).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Percent Removal = In – Out x 100 (F)
In
THIRD. Follow the solution procedure below to solve the problem!
25 ntu’s - .10 ntu’s x 100 = 99.6 Percent Removal
25 ntu
Specific Capacity GPM/ft - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
A well yields 250 gpm and the drawdown is 20 ft. what is the specific capacity in GPM/ft?
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (specific capacity).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Specific Capacity, GPM/ft = Well Yield, GPM (F)
Drawdown, ft
THIRD. Follow the solution procedure below to solve the problem!
250 gpm = 12.5 gpm/ft
20 ft
Volume Gals - 1
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
Water is flowing into a basin at 30 GPM. How many gallons will be in the basin after 240 minutes?
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (gallons per minute).
The Formula (F) and Conversions & Calculations (C&C) shown below.
Volume,Gal = GPM x Time, Minutes (F)
THIRD. Follow the solution procedure below to solve the problem!
30 GPM x 240 min = 7,200 Gal
Volume Gals - 2
FIRST. Read the question CAREFULLY and look for KEY WORDS to help you decide what formula to use.
How long would it take to empty a tank that's 28' H and 28' radius, at rate 2 cubic feet per second (ft3/sec) through an 8" pipe.
SECOND. By using the KEY WORDS you see this is a (General) problem, that includes (Time, gallons).
The Formula (F) and Conversions & Calculations (C&C) are shown below.
Dia/ft = Radius x 2 (C&C)1
Cyl, Gal = 0.785 x (Dia/ft)2 x Ht, Dpth or Lgth x 7.48 gal/ft3 (C&C)2
Gal/sec = ft3/sec x 7.48 gal/ft3 (C&C)3
Gal/sec,Time = Cyl,Gal (F)4
Gal/sec
Gal/hr = gal/sec (C&C)5
60 sec/min x 60 min/hr
Minutes = decimal/hr x 60 min/hr (C&C)6
Seconds = decimal/min x 60 sec/min (C&C)7
THIRD. Follow the solution procedure below to solve the problem!
28 ft Radius x 2 = 56 Dia/ft
.785 x 56 Dia/ft x 56 Dia/ft x 28 ft x 7.48 gal/ft3 = 515,591 gal
2 CFS x 7.48 gal/ft3 = 14.96 gal/sec
515,591 gal = 34,465 sec
14.96 gal/sec
34,465 sec = 9.57 hr
60 sec/min x 60 min/hr
.57 hr x 60 min/hr = 34.2 min
.2 min x 60 sec/min = 12 sec