Calculations
In order to correcly solve a problem, all units of measure
must be converted to the proper units required for the formula. Examples
of some units the operator will be using are grains per gallon, grams, ounces,
pounds, cubic feet, square feet, liters, gallons, parts per million and
pounds per square inch.
Units of time are important in water calculations. To convert
gallons per minute into gallons per hour, multiply by 60. For gallons
per day when gallons per minute is known, multiply gallons per minute by
1440. This is the number of minutes in a day (24 x 60).
*Here are some formulas and figures that you will need in water
calculations.
Area and Volume of Square or Rectangular
Figures:
Example 1) Area
Example 2) Area of Triangles:
The area of a triangle is 1/2 of a square or rectangle.
Hence, the formula for the area is half the base (b) times the altitude or
height (a).
A = 1/2b
x a
A = 1/2b x a
A = 3' x
6'
A = 5' x 6'
A = 18 sq. ft.
A = 30 sq. ft.
Example 3) Area and Volume of Circular Figures:
Concerning circles, several terms are used. The distance
across a circle through the center is the diameter (D). Half the diameter
is called the radius (r). A number pi which is called a constant is
3.1416. Another term is "squared". When a number is squared,
it is multiplied by itself. The area of a circle equals the
radius (r), which is half the diameter, squared times pi.
Example 4) Find the area of a circle with a 20 foot diameter:
A = pi (r)^{2}
A = 3.14 x 10^{2}
A = 3.14 x 10 x 10
A = 314.0 sq. ft.
*The volume of a circular tank is the area of the floor surface
multiplied by the depth. This volume equals pi times the radius
squared times the depth, or the area of the bottom times the
depth. The formula for the volume is V = pi
(r)^{2} x depth (remember pi
(r)^{2} is the area of the circle).
Example 5) Volume
Volume equals length times width times depth.
V = L x W x D
V = 10' x 6' x 5'
V = 300 cubic feet
*Unless great accuracy is desired, the operator should
consider 7.5 gallons as being equal to 1 cubic foot. Actually 1 cubic
foot of water is 7.481 gallons. Thus, the volume of the tank in
Example 5 (6' wide, 10' long, 5' deep) is 2250 gallons. (10'
x 6' x 5' x 7.5 = 2250 gallons
or 300 cubic feet x 7.5 gallons = 2250 gallons).
Example 6)
Find the volume of a tank 20 feet in diameter and 20 feet
deep. The area of the tank bottom is 3.14 x 10^{2}
or 3.14 x 10 x 10 = 314 square feet. Next,
multiply the area by the depth and obtain the volume: 314 sq. ft.
x 20 ft. = 6280 cu. ft.
To find the number of gallons the tank will hold, multiply the
volume in cubic feet by
the number of gallons in one cubic foot: 6280 cu. ft. x
7.5 gal. = 47,100 gallons.
Circumference of Circles
The distance around a circle is called the circumference.
The circumference, C, may be determined by measuring the diameter and multiplying
by pi.
Example 7) Find the circumference of a circle 20 feet in diameter.
C = pi(D)
C = 3.14(20)
C = 62.8 ft
.
Area of Tanks or Cylinders
In order to find the area of the walls of a tank or cylinder,
the circumference is multiplied by the height. Such a calculation would
be useful in deciding how much paint would be necessary to coat a tank.
The area of the walls of a cylinder is called the "lateral area", A.
If a tank is 20 feet in diameter and 20 feet deep, its circumference is
3.14 x 20 = 62.8 feet. The area of the wall, the lateral
area, is 62.8' x 20' or 1256 sq. ft. Thus the formula for
lateral area is:
A = pi(D)(H)
A = pi x 20 x 20
A = 3.14 x 20 x 20
A = 1256 sq. ft.
Lateral Area (A) = pi(D)(H)
The area of the bottom of the cylinder is pi(r)^{2}, or 3.14(10)^{2}
= 314.6 sq. ft. The area of the top of the cylinder would also be
314.6 sq. ft. so total area of the inside of the cylinder would be 1256
+ 314.6 = 1885.2 sq. ft. If the inside and outside were to be painted,
the total area would be 1885.2 x 2 = 3770.4 sq. ft.
If one gallon of paint will cover 250 sq. ft., to paint the enclosed
cylinder inside and out would require 3770.4/250 = 15.08, or 16 gallons.
Suppose the tank has a conical top which must be painted.
The area of a cone is found by multiplying the circumference of its base
by half of the slant height. The top overhangs the tank 2 feet
all around making its diameter 24 feet, rather than 20 feet, as the tank
has. If measured it would be found that the slant height would be about
14.4 feet. Thus the area is:
A = 2 pi(r) x 1/2 slant height
A = 2 x 3.14 x 12' x 7.2'
= 542 sq. ft.
For the inside or outside of the cone.
If the tank has a rounded bottom, its approximate area can be found
by using the formula for half a sphere (ball). Even if the bottom is
not quite round, this formula should give a good approximation of the area.
Multiply the radius of the tank by the circumference. (Remember, circumference
= 2pi(r)).
Area = r x 2 (pi(r))
A = 10 x 2 x 3.14 x 10'
A = 628 sq. ft.
The total area to be painted is the sum of these areas (the walls,
top and bottom):
1256 + 542 + 628 = 2,456 sq. ft.
If the tank is to be painted inside and out, the area must be doubled.
The approximate volume of this bottom section, if it is half of
a sphere (ball) is 1/2 D^{3}
By adding this figure to the volume of the cylinder, which was figured
in Example 6, the total volume of the tank will be 47,000 + 15,000, or approximately
62,000 gallons of water.
Detention Time
Detention time, sometimes called the retention period, is the length
of time water is retained in a vessel or basin. It may also be considered
the period from the time the water enters a settling basin until it flows
out the other end. To calculate the detention period of a basin, the
volume of the basin must be first obtained. Using a basin 20' wide,
60' long and 10' deep the volume would be:
V = L x W x D
V =
60 x 20 x 10
V =
12,000 cu. ft.
Gallons
= V x 7.5 gal. per cu. ft.
Gallons
= 12,000 x 7.5 = 90,000 gallons
Assume that the plant filters 250 gpm. Therefore, 90,000/250
= 360 minutes, or 6 hours of detention time. Stated another way, the
detention time is the length of time theoretically required for the coagulated
water to flow through the basin.
If chlorine were added to the water as it entered the basin, then
the chlorine contact time would be six hours.
Example
Problems for Conversion Calculations
Example
Problems for Measurement Conversions
