Introduction - Basic Mathematics for Water and Wastewater Operators - page 8

Because the dimensions of the answer are square feet, this math setup is correct;
therefore, by dividing the numbers as was done with units, the answer will also be
correct:

3

70 ft /sec
4.5 ft/sec

2

= 15.56 ft

■ Example 1.1
Problem: We are given two terms, 5 m/sec and 7 m 2 , and the answer to be obtained
should be in cubic meters per second (m 3 /sec). Is multiplying the two terms the cor-
rect math setup?

Solution:

m
sec

2 2

m/sec ×=× m m

Multiply the numerator and denominator of the fraction:

2 3

mm
sec

× m

sec

=

Because the dimensions of the answer are cubic meters per second (m 3 /sec), the
math setup is correct; therefore, multiply the numbers to get the correct answer:

5 m/sec × 7 m 2 = 35 m 3 /sec

■ Example 1.2
Problem: The flow rate in a water line is 2.3 ft 3 /sec. What is the flow rate expressed
as gallons per minute?

Solution: Set up the math problem and then use dimensional analysis to check the
math setup:

(2.3 ft 3 /sec) × (7.48 ga l /f t 3 ) × (60 sec/min)

Dimensional analysis can be used to check the math setup:

3

ft
sec

gal
ft

sec
mi

gal
min

3 3

ft /sec × gal/ft × sec/min = × × =

3

n

The math setup is correct as shown above; therefore, this problem can be multiplied
out to get the answer in correct units:

(2.3 ft 3 /sec) × (7.48 ga l /f t 3 ) × (60 sec/min) = 1032.24 gal/min