C HAPTER 2 Basic Math Operations
2.9 DIMENSIONAL ANALYSIS
Dimensional analysis is a problem-solving method that uses the fact that one, without changing its value, can multiply any number or expression. It is a useful technique to check whether a problem is set up correctly. In using dimensional analysis to check a math setup, we work with the dimensions (units of measure) only — not with numbers.
An example of dimensional analysis common to everyday life is the unit pricing found in many hardware stores. A shopper can purchase a 1-lb box of nails for $0.98 in one store, whereas a warehouse store sells a 5-lb bag of the same nails for $3.50. The shopper will analyze this problem almost without thinking about it. The solution calls for reducing the problem to the price per pound.
The pound is selected as the unit common to both stores. A shopper will pay $0.70/lb for nails in the warehouse store or $0.98/lb in the local hardware store. Knowing the unit price, which is expressed in dollars per pound ($/lb), is implicit in the solution to this problem.
To use the dimensional analysis method, we must know how to perform three basic operations:
Note: Unit factors may be made from any two terms that describe the same or equivalent “amounts”
of the objects of interest. For example, we know that 1 in. = 2.54 cm.
1. Basic operation: to complete a division of units, always ensure that all units are written in the same format; it is best to express a horizontal fraction (such as gallons per square foot) as a vertical fraction. Horizontal to vertical:
The same procedures are applied in the following examples:
F M
amount of BOD amount of VSS
=
= ×
× 250 mg/L 7.2 L 1900 mg/L 1.6 L
= 0.59 1
= 0.59
gal/ft to gal ft
3
= 3
psi to lb in.2
ft /min becomes ft min
3 3
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48 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK
2. Basic operation: we must know how to divide by a fraction. For example,
In the preceding example, notice that the terms in the denominator were inverted before the frac-tions were multiplied. This is a standard rule that must be followed when dividing fracfrac-tions. An-other example is
3. Basic operation: we must know how to cancel or divide terms in the numerator and denominator of a fraction. After fractions have been rewritten in the vertical form and division by the fraction has been re-expressed as multiplication as shown earlier, the terms can be canceled (or divided) out.
Key point: For every term that is canceled in the numerator of a fraction, a similar term must be canceled in the denominator and vice versa, as shown below:
How are units that include exponents calculated?
When written with exponents (ft3, for example), a unit can be left as it is, or put in expanded form (ft)(ft)(ft), depending on other units in the calculation. However, it is necessary to ensure that square and cubic terms are expressed uniformly, as sq ft, cu ft, or as ft2, ft3. For dimensional analysis, the latter system is preferred.
For example, if we wish to convert 1400 ft3 volume to gallons and will use 7.48 gal/ft3 in the conversions, dimensional analysis can be used to determine whether we multiply or divide by 7.48.
To determine if the math setup is correct, only the dimensions are used.
First, try dividing the dimensions:
s/in. becomes s
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BASIC MATH OPERATIONS 49
Then, the numerator and denominator are multiplied to get
Thus, by dimensional analysis we determine that if we divide the two dimensions (ft3 and gal/ft3), the units of the answer are ft6/gal, not gal. Clearly, division is not the right way to go in making this conversion.
What would have happened if we had multiplied the dimensions instead of dividing?
Then, multiply the numerator and denominator to obtain
and cancel common terms to obtain
Obviously, by multiplying the two dimensions (ft3 and gal/ft3), the answer will be in gallons, which is what we want. Thus, because the math setup is correct, we would then multiply the numbers to obtain the number of gallons.
Now try another problem with exponents. We wish to obtain an answer in square feet. If we are given the two terms — 70 ft3/sec and 4.5 ft/sec — is the following math setup correct?
First, only the dimensions are used to determine if the math setup is correct. Multiplying the two dimensions yields:
Then, the terms in the numerators and denominators of the fraction are multiplied:
= ft
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50 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK
Obviously, the math setup is incorrect because the dimensions of the answer are not square feet.
Therefore, if we multiply the numbers as shown previously, the answer will be wrong.
Let us try division of the two dimensions instead:
Invert the denominator and multiply to get
= ft2
Because the dimensions of the answer are square feet, this math setup is correct. Therefore, by dividing the numbers as we did with units, the answer will also be correct:
Example 2.33
Problem:
We are given two terms — 5 m/sec and 7 m2 — and the answer to be obtained is in cubic meters per second (m3/sec). Is multiplying the two terms the correct math setup?
Solution:
Multiply the numerators and denominator of the fraction:
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BASIC MATH OPERATIONS 51
Because the dimensions of the answer are cubic meters per second (m3/sec), the math setup is correct. Therefore, multiply the numbers to get the correct answer:
Example 2.34
Problem:
Solve the following problem: Given that the flow rate in a water line is 2.3 ft3/sec, what is the flow rate expressed as gallons per minute?
Solution:
Set up the math problem:
Then, use dimensional analysis to check the math setup:
The math setup is correct as shown above. Therefore, this problem can be multiplied out to get the answer in correct units.
Example 2.35
Problem:
During an 8-h period, a water treatment plant treated 3.2 million gal of water. What is the plant total volume treated per day, assuming the same treatment rate?
Solution:
5 (m/sec) (7 m )2 = 35 m /sec3
(2.3 ft /sec) (7.48 gal/ft ) (60 sec/min)3 3
(ft /sec) (gal/ft ) (sec/min) ft sec
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52 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK
Example 2.36
Problem:
How many cubic feet per second (cfs; ft3/sec) are equal to 1 MGD?
Solution:
Example 2.37
Problem:
A 10-gal empty tank weighs 4.6 lb. What is the total weight of the tank filled with 6 gal of water?
Solution:
Example 2.38
Problem:
The depth of biosolids applied to the biosolids drying bed is 10 in. What is the depth in centimeters (2.54 cm = 1 in.)?
Solution:
1 MGD 10
1 day
= 6
10 gal 0.1337 ft /gal 1 day 86,400 se
6 3
= ×
× cc/day
133,700 ft 86,400 sec
= 3
= 1.547 cfs (ft /sec3 )
Weight of water = 6 gal × 8.34 lb/gal
= 50.04 lb
Total weight = 50.04 + 4.6 lb
54.6 lb
=
10 in. = 10 × 2.54 cm
= 25.4 cm
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BASIC MATH OPERATIONS 53