Math Operations Review

Authored by: Frank R. Spellman

Handbook of Environmental Engineering

Print publication date:  September  2015
Online publication date:  September  2015

Print ISBN: 9781498708616
eBook ISBN: 9781498708623
Adobe ISBN:

10.1201/b18873-4

 

Abstract

Note everything that counts can be counted, and not everything that can be counted counts.

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Math Operations Review

Note everything that counts can be counted, and not everything that can be counted counts.

—Albert Einstein

Introduction

Most calculations required by environmental engineers require a knowledge of the basics, such as addition, subtraction, multiplication, division, and sequence of operations, among others. Although many of these operations are fundamental tools within each environmental practitioner’s toolbox, these tools must be used on a consistent basis to remain sharp. Environmental practitioners should master basic math definitions and the formation of problems, as daily operations require calculation of percentages, averages, simple ratios, geometric dimensions, threshold odor numbers, and force, pressure, and head, as well as the use of dimensional analysis and advanced math operations. With regard to advanced math operations, an in-depth knowledge of algebra, linear algebra, vectors, trigonometry, analytic geometry, differential calculus, integral calculus, and differential equations is required in certain environmental engineering design and analysis operations; however, we leave discussion of these higher operations to the math textbooks.

Basic Math Terminology and Definitions

The following basic definitions will aid in understanding the material that follows:

  • An integer, or an integral number, is a whole number; thus, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 are the first 12 positive integers.
  • A factor, or divisor, of a whole number is any other whole number that exactly divides it; thus, 2 and 5 are factors of 10.
  • A prime number in math is a number that has no factors except itself and 1; examples of prime numbers are 1, 3, 5, 7, and 11.
  • A composite number is a number that has factors other than itself and 1. Examples of composite numbers are 4, 6, 8, 9, and 12.
  • A common factor, or common divisor, of two or more numbers is a factor that will exactly divide each of them. If this factor is the largest factor possible, it is called the greatest common divisor. Thus, 3 is a common divisor of 9 and 27, but 9 is the greatest common divisor of 9 and 27.
  • A multiple of a given number is a number that is exactly divisible by the given number. If a number is exactly divisible by two or more other numbers, it is a common multiple of them. The least (smallest) such number is called the lowest common multiple. Thus, 36 and 72 are common multiples of 12, 9, and 4; however, 36 is the lowest common multiple.
  • An even number is a number exactly divisible by 2; thus, 2, 4, 6, 8, 10, and 12 are even integers.
  • An odd number is an integer that is not exactly divisible by 2; thus, 1,3, 5, 7, 9, and 11 are odd integers.
  • A product is the result of multiplying two or more numbers together; thus, 25 is the product of 5 × 5. Also, 4 and 5 are factors of 20.
  • A quotient is the result of dividing one number by another; for example, 5 is the quotient of 20 ÷ 4.
  • A dividend is a number to be divided; a divisor is a number that divides; for example, in 100 ÷ 20 = 5, 100 is the dividend, 20 is the divisor, and 5 is the quotient.
  • Area is the area of an object, measured in square units.
  • Base is a term used to identify the bottom leg of a triangle, measured in linear units.
  • Circumference is the distance around an object, measured in linear units. When determined for other than circles, it may be called the perimeter of the figure, object, or landscape.
  • Cubic units are measurements used to express volume (cubic feet, cubic meters, etc.).
  • Depth is the vertical distance from the bottom of the tank to the top. This is normally measured in terms of liquid depth and given in terms of sidewall depth (SWD), measured in linear units.
  • Diameter is the distance from one edge of a circle to the opposite edge passing through the center, measured in linear units.
  • Height is the vertical distance from the base or bottom of a unit to the top or surface.
  • Linear units are measurements used to express distances: feet, inches, meters, yards, etc.
  • Pi (π) is a number in calculations involving circles, spheres, or cones (π= 3.14).
  • Radius is the distance from the center of a circle to the edge, measured in linear units.
  • Sphere is a container shaped like a ball.
  • Square units are measurements used to express area, square feet, square meters, acres, etc.
  • Volume is the capacity of the unit (how much it will hold), measured in cubic units (cubic feet, cubic meters) or in liquid volume units (gallons, liters, million gallons).
  • Width is the distance from one side of the tank to the other, measured in linear units.

Key Words

  • Of means to multiply.
  • And means to add.
  • Per means to divide.
  • Less than means to subtract.

Sequence of Operations

Mathematical operations such as addition, subtraction, multiplication, and division are usually performed in a certain order or sequence. Typically, multiplication and division operations are done prior to addition and subtraction operations. In addition, mathematical operations are also generally performed from left to right using this hierarchy. The use of parentheses is also common to set apart operations that should be performed in a particular sequence. Consider the expression 2 + 3 × 4. You might say the solution is 20 but if you know the rules then you would answer (correctly) 24. The preceding expression may be rendered 2 + (3 × 4), but the brackets are unnecessary if you know the rules, as multiplication has precedence even without the parentheses.

Note: It is assumed that the reader has a fundamental knowledge of basic arithmetic and math operations; thus, the purpose of the following section is to provide a brief review of the mathematical concepts and applications frequently employed by environmental practitioners.

Sequence of Operations Rules

Rule 1

In a series of additions, the terms may be placed in any order and grouped in any way; thus, 4 + 3 = 7 and 3 + 4 = 7; (4 + 3) + (6 + 4) = 17, (6 + 3) + (4 + 4) = 17, and [6 + (3 + 4)] + 4 = 17.

Rule 2

In a series of subtractions, changing the order or the grouping of the terms may change the result; thus, 100 – 30 = 70, but 30 – 100 = –70, and (100 – 30) – 10 = 60, but 100 – (30 – 10) = 80.

Rule 3

When no grouping is given, the subtractions are performed in the order written, from left to right; thus, 100 – 30 – 15 – 4 = 51 (by steps, it would be 100 – 30 = 70, 70 – 15 = 55, 55 – 4 = 51).

Rule 4

In a series of multiplications, the factors may be placed in any order and in any grouping; thus, [(2 × 3) × 5] × 6 = 180 and 5 × [2 × (6 × 3)] = 180.

Rule 5

In a series of divisions, changing the order or the grouping may change the result; thus, 100 ÷ 10 = 10 but 10 ÷ 100 = 0.1, and (100 ÷ 10) ÷ 2 = 5 but 100 ÷ (10 ÷ 2) = 20. Again, if no grouping is indicated, the divisions are performed in the order written, from left to right; thus, 100 ÷ 10 ÷ 2 is understood to mean (100 ÷ 10) ÷ 2.

Rule 6

In a series of mixed mathematical operations, the convention is as follows: Whenever no grouping is given, multiplications and divisions are to be performed in the order written, then additions and subtractions in the order written.

Sequence of Operations Examples

In a series of additions, the terms may be placed in any order and grouped in any way:

4 + 6 = 10  and 6 + 4 = 10 ( 4 + 5 ) + ( 3 + 7 ) = 19 , ( 3 + 5 ) + ( 4 + 7 ) = 19 ,  and  [ 7 + ( 5 + 4 ) ] + 3 = 19

In a series of subtractions, changing the order or the grouping of the terms may change the result:

100 20 = 80 , but 20 100 = 80 ( 100 30 ) 20 = 50 , but 100 ( 30 20 ) = 90

When no grouping is given, the subtractions are performed in the order written, from left to right:

100 30 20 3 = 47

or by steps:

100 30 = 70 , 70 20 = 50 , 50 3 = 47

In a series of multiplications, the factors may be placed in any order and in any grouping:

[ ( 3 × 3 ) × 5 ] × 6 = 270  and 5 × [ 3 × ( 6 × 3 ) ] = 270

In a series of divisions, changing the order or the grouping may change the result:

100 ÷ 10 = 10 , but 10 ÷ 100 = 0.1 ( 100 ÷ 10 ) ÷ 2 = 5 , but 100 ÷ ( 10 ÷ 2 ) = 20

If no grouping is indicated, the divisions are performed in the order written—from left to right:

100 ÷ 5 ÷ 2  is understood to mean  (100 ÷ 5 ) ÷ 2

In a series of mixed mathematical operations, the rule of thumb is that, whenever no grouping is given, multiplications and divisions are to be performed in the order written, then additions and subtractions in the order written.

Consider the following classic example of sequence of operations (Stapel, 2012):

Problem: Simplify 4 – 3[4 – 2(6 – 3)] ÷ 2.

Solution:

4 3 [ 4 2 ( 6 3 ) ] ÷ 2 4 3 [ 4 2 ( 3 ) ] ÷ 2 4 3 [ 4 6 ] ÷ 2 4 3 [ 2 ] ÷ 2 4 + 6 ÷ 2 4 + 3 = 7

Percent

The word “percent” means “by the hundred.” Percentage is usually designated by the symbol %; thus, 15% means 15 percent or 15/100 or 0.15. These equivalents may be written in the reverse order: 0.15 = 15/100 = 15%. In wastewater treatment, percent is frequently used to express plant performance and for control of biosolids treatment processes. When working with percent, the following key points are important:

  • Percents are another way of expressing a part of a whole.
  • Percent means “by the hundred,” so a percentage is the number out of 100. To determine percent, divide the quantity we wish to express as a percent by the total quantity, then multiply by 100:
    3.1 Percent (%) = Part Whole
    For example, 22 percent (or 22%) means 22 out of 100, or 22/100. Dividing 22 by 100 results in the decimal 0.22:
    22 % = 22 100 = 0.22
  • When using percentages in calculations (such as when calculating hypochlorite dosages and when the percent available chlorine must be considered), the percentage must be converted to an equivalent decimal number; this is accomplished by dividing the percentage by 100. For example, calcium hypochlorite (HTH) contains 65% available chlorine. What is the decimal equivalent of 65%? Because 65% means 65 per hundred, divide 65 by 100: 65/100, which is 0.65.
  • Decimals and fractions can be converted to percentages. The fraction is first converted to a decimal, then the decimal is multiplied by 100 to get the percentage. For example, if a 50-foot-high water tank has 26 feet of water in it, how full is the tank in terms of the percentage of its capacity?
    26  ft 50  ft = 0.52  (decimal equivalent) 0.52 × 100 = 52
    Thus, the tank is 52% full.

 

Problem: The plant operator removes 6500 gal of biosolids from the settling tank. The biosolids contain 325 gal of solids. What is the percent solids in the biosolids?

Solution:

Percent = 325  gal 6500  gal × 100 = 5 %

 

Problem: Convert 48% to decimal percent.

Solution:

Decimal percent = Percent 100 = 48 100 = 0.48

 

Problem: Biosolids contain 5.8% solids. What is the concentration of solids in decimal percent?

Solution:

Decimal percent = 5.8 % 100 = 0.058

Note: Unless otherwise noted, all calculations in the text using percent values require the percent to be converted to a decimal before use.

To determine what quantity a percent equals, first convert the percent to a decimal and then multiply by the total quantity:

3.2 Quantity = Total × Decimal percent

 

Problem: Biosolids drawn from the settling tank are 5% solids. If 2800 gal of biosolids are withdrawn, how many gallons of solids are removed?

Solution:

Gallons = 5 % 100 × 2800  gal = 140  gal

 

Problem: Convert 0.55 to percent.

Solution:

0.55 = 55 100 = 55 %

(To convert 0.55 to 55%, we multiply by 100, or simply move the decimal point two places to the right.)

 

Problem: Convert 7/22 to a decimal percent to a percent.

Solution:

7 22 = 0.318 = 0.318 × 100 = 31.8 %

 

Problem: What is the percentage of 3 ppm?

Note: Because 1 liter of water weighs 1 kg (1000 g = 1,000,000 mg), milligrams per liter is parts per million (ppm)

Solution: Because 3 parts per million (ppm) = 3 mg/L:

3  mg/L = 3  mg 1  L × 1 , 000 , 000  mg / L × 100 % = 3 10 , 00 % = 0.0003 %

 

Problem: How many mg/L is a 1.4% solution?

Solution:

1.4 % = 1.4 100 × 1 , 000 , 000  mg/L (the weight 1 L water to   10 6 ) = 14 , 000  mg/L

 

Problem: Calculate pounds per million gallons for 1 ppm (1 mg/L) of water.

Solution: Because 1 gal of water = 8.34 lb,

1  ppm = 1  gal 10 6  gal = 1  gal × 8.34  lb / gal 1 , 000 , 000  gal = 8.34  lb / 1 , 000 , 000  gal
 

Problem: How many pounds of activated carbon (AC) are needed with 42 lb of sand to make the mixture 26% AC?

Solution: Let x be the weight of AC; thus,

x 42 + x = 0.26 x = 0.26 ( 42 + x ) = 10.92 + 0.26 x x = 10.92 0.74 = 14.76  lb
 

Problem: A pipe is laid at a rise of 140 mm in 22 m. What is the grade?

Solution:

Grade = 140  mm 22  m × 100 % = 140  mm 22  m × 1000  mm × 100 % = 0.64 %

 

Problem: A motor is rated as 40 horsepower (hp). However, the output horsepower of the motor is only 26.5 hp. What is the efficiency of the motor?

Solution:

Efficiency = hp  output hp input × 100 % = 26.5  hp 40  hp × 100 % = 66 %

Significant Digits

When rounding numbers, the following key points are important:

  • Numbers are rounded to reduce the number of digits to the right of the decimal point. This is done for convenience, not for accuracy.
  • A number is rounded off by dropping one or more numbers from the right and adding zeroes if necessary to place the decimal point. If the last figure dropped is 5 or more, increase the last retained figure by 1. If the last digit dropped is less than 5, do not increase the last retained figure. If the digit 5 is dropped, round off preceding digit to the nearest even number.

 

Problem: Round off the following numbers to one decimal.

Solution:

34.73 = 34.7 34.77 = 34.8 34.75 = 3.48 34.45 = 34.4 34.35 = 34.4

 

RULE

Significant figures are those numbers that are known to be reliable. The position of the decimal point does not determine the number of significant figures.

 

Problem: Round off 10,546 to 4, 3, 2, and 1 significant figures.

Solution:

10,546 = 10,550 to 4 significant figures

10,546 = 10,500 to 3 significant figures

10,546 = 11,000 to 2 significant figures

10,547 = 10,000 to 1 significant figure

When determining significant figures, the following key points are important:

  1. The concept of significant figures is related to rounding.
  2. It can be used to determine where to round off.

Note: No answer can be more accurate than the least accurate piece of data used to calculate the answer.

 

Problem: How many significant figures are in a measurement of 1.35 in.?

Solution: Three significant figures: 1, 3, and 5.

 

Problem: How many significant figures are in a measurement of 0.000135?

Solution: Again, three significant figures: 1, 3, and 5. The three zeros are used only to place the decimal point.

 

Problem: How many significant figures are in a measurement of 103,500?

Solution: Four significant figures: 1, 0, 3, and 5. The remaining two zeros are used to place the decimal point.

 

Problem: How many significant figures are in 27,000.0?

Solution: There are six significant figures: 2, 7, 0, 0, 0, 0. In this case, the .0 in 27,000.0 means that the measurement is precise to 1/10 unit. The zeros indicate measured values and are not used solely to place the decimal point.

Powers and Exponents

When working with powers and exponents, the following key points are important:

  • Powers are used to identify area, as in square feet, and volume, as in cubic feet.
  • Powers can also be used to indicate that a number should be squared, cubed, etc. This later designation is the number of times a number must be multiplied times itself.
  • If all of the factors are alike, such as 4 × 4 × 4 × 4 = 256, the product is called a power. Thus, 256 is a power of 4, and 4 is the base of the power. A power is a product obtained by using a base a certain number of times as a factor.
  • Instead of writing 4 × 4 × 4 × 4, it is more convenient to use an exponent to indicate that the factor 4 is used as a factor four times. This exponent, a small number placed above and to the right of the base number, indicates how many times the base is to be used as a factor. Using this system of notation, the multiplication 4 × 4 × 4 × 4 is written as 44. The 4 is the exponent, showing that 4 is to be used as a factor 4 times.
  • These same consideration apply to letters (a,b,x,y, etc.) as well; for example:
    z 4 = z × z z 4 = z × z × z × z

Note: When a number or letter does not have an exponent, it is considered to have an exponent of one.

 

Problem: How is the term 23 written in expanded form?

Solution: The power (exponent) of 3 means that the base number (2) is multiplied by itself three times:

2 3 = 2 × 2 × 2

 

Problem: How is the term (3/8)2 written in expanded form?

Note: When parentheses are used, the exponent refers to the entire term within the parentheses.

Solution: In this example, (3/8)2 means:

( 3 / 8 ) 2 = ( 3 / 8 × 3 / 8 )

Note: When a negative exponent is used with a number or term, a number can be rewritten using a positive exponent:

6 3 = 1 / 6 3

Another example is

11 5 = 1 / 11 5
 

POWERS OF 1

POWERS OF 10

10 = 1

100 = 1

11 = 1

101 = 10

12 = 1

102 = 100

13 = 1

103 = 1000

14 = 1

104 = 10,000

 

Problem: How is the term 8–3 written in expanded form?

Solution:

8 3 = 1 8 3 = 1 8 × 8 × 8

Note: A number or letter written as, for example, 30 or X0 does not equal 3 × 1 or X × 1, but simply 1.

Averages (Arithmetic Mean)

Whether we speak of harmonic mean, geometric mean, or arithmetic mean, each represents the “center,” or “middle,” of a set of numbers. They capture the intuitive notion of a “central tendency” that may be present in the data. In statistical analysis, an “average of data” is a number that indicates the middle of the distribution of data values.

An average is a way of representing several different measurements as a single number. Although averages can be useful in that they tell us “about” how much or how many, they can also be misleading, as we demonstrate below. You will find two kinds of averages in environmental engineering calculations: the arithmetic mean (or simply mean) and the median.

 

Problem: The operator of a waterworks or wastewater treatment plant takes a chlorine residual measurement every day; part of the operator’s log is shown below. Find the mean.

Monday

0.9 mg/L

Tuesday

1.0 mg/L

Wednesday

0.9 mg/L

Thursday

1.3 mg/L

Friday

1.1 mg/L

Saturday

1.4 mg/L

Sunday

1.2 mg/L

Solution: Add up the seven chlorine residual readings: 0.9 + 1.0 + 0.9 + 1.3 + 1.1 + 1.4 + 1.2. = 7.8. Next, divide by the number of measurements—in this case, 7:

7.8 ÷ 7 = 1.11

The mean chlorine residual for the week was 1.11 mg/L.

 

Definition

The mean (what we usually refer to as an average) is the total of values of a set of observations divided by the number of observations. We simply add up all of the individual measurements and divide by the total number of measurements we took.

 

Problem: A water system has four wells with the following capacities: 115 gpm (gallons per minute), 100 gpm, 125 gpm, and 90 gpm. What is the mean?

Solution:

115  gpm + 100  gpm + 125  gpm + 90  gpm = 430 430 ÷ 4 = 107.5  gpm

 

Problem: A water system has four storage tanks. Three of them have a capacity of 100,000 gal each, while the fourth has a capacity of 1 million gal. What is the mean capacity of the storage tanks?

Solution: The mean capacity of the storage tanks is

100 , 000 + 100 , 00 + 100 , 000 + 1 , 000 , 000 = 1 , 300 , 000 1 , 300 , 000 ÷ 4 = 325 , 000  gal

Notice that no tank in Example 2.24 has a capacity anywhere close to the mean.

 

Problem: Effluent biochemical oxygen demand (BOD) test results for the treatment plant during the month of August are shown below:

Test

1 22 mg/L

Test 2

33 mg/L

Test 3

21 mg/L

Test 4

13 mg/L

What is the average effluent BOD for the month of August?

Solution:

22 + 33 + 21 + 13 = 89 89 ÷ 4 = 22.3  mg/L

 

Problem: For the primary influent flow, the following composite-sampled solids concentrations were recorded for the week:

Monday

310 mg/L SS

Tuesday

322 mg/L SS

Wednesday

305 mg/L SS

Thursday

326 mg/L SS

Friday

313 mg/L SS

Saturday

310 mg/L SS

Sunday

320 mg/L SS

Total

2206 mg/L SS

What is the average SS?

Solution:

Average SS = Su of all measurements Number of measurements used = 2206  mg/L  SS 7 = 315.1  mg/L  SS

Ratio

A ratio is the established relationship between two numbers; it is simply one number divided by another number. For example, if someone says, “I’ll give you four to one the Redskins over the Cowboys in the Super Bowl,” what does that person mean? Four to one, or 4:1, is a ratio. If someone gives you four to one, it’s his or her $4 to your $1. As another more pertinent example, if an average of 3 cubic feet (ft3) of screenings are removed from each million gallons (MG) of wastewater treated, the ratio of screenings removed to treated wastewater is 3:1. Ratios are normally written using a colon (such as 2:1) or as a fraction (such as 2/1). When working with ratios, the following key points are important to remember.

  • One place where fractions are used in calculations is when ratios are used, such as calculating solutions.
  • A ratio is usually stated in the form A is to B as C is to D, which can be written as two fractions that are equal to each other:
    A B = C D
  • Cross-multiplying solves ratio problems; that is, we multiply the left numerator (A) by the right denominator (D) and say that the product is equal to the left denominator (B) times the right numerator (C):
    A × D = B × C  (or,  AD = B C )
  • If one of the four items is unknown, dividing the two known items that are multiplied together by the known item that is multiplied by the unknown solves the ratio. For example, if 2 lb of alum are needed to treat 500 gal of water, how many pounds of alum will we need to treat 10,000 gal? We can state this as a ratio: “2 lb of alum is to 500 gal of water as x lb of alum is to 10,000 gal of water.” This is set up in this manner:
    1  lb  alum 500  gal water = x  lb  alum 10 , 000  gal water
    Cross-multiplying,
    500 × x = 1 × 10 , 000
    Transposing,
    1 × 10 , 000 500 = 20  lb alum
    To calculate proportion, suppose, for example, that 5 gal of fuel costs $5.40. What will 15 gal cost?
    5 gal $ 5.40 = 15  gal $ y 5  gal × $ y = 15  gal × $ 5.40 = 81 $ y = 81 5 = $ 16.20
 

Problem: If a pump will fill a tank in 20 hr at 4 gpm, how long will it take a 10-gpm pump to fill the same tank?

Solution: First, analyze the problem. Here, the unknown is some number of hours. But, should the answer be larger or smaller than 20 hr? If a 4-gpm pump can fill the tank in 20 hr, a larger (10-gpm) pump should be able to complete the filling in less than 20 hr. Therefore, the answer should be less than 20 hours. Now set up the proportion:

x  hr 20  hr = 4  gpm 10  gpm x = ( 4 × 20 ) 10 = 8  hr

 

Problem: Solve for the unknown value x in the problem given below.

Solution:

36 180 = x 4450 4450 × 36 180 = x = 890

 

Problem: Solve for the unknown value x in the problem given below.

3.4 2 = 6 x

Solution:

3.4 × x = 2 × 6 x = 2 × 6 3.4 = 3.53

 

Problem: 1 lb of chlorine is dissolved in 65 gal of water. To maintain the same concentration, how many pounds of chlorine would have to be dissolved in 150 gal of water?

Solution:

1  lb 65  gal = x  lb 150  gal 65 × x = 1 × 150 x = 1 × 150 65 = 2.3  lb

 

Problem: It takes 5 workers 50 hr to complete a job. At the same rate, how many hours would it take 8 workers to complete the job?

Solution:

5 workers 8  workers = x  hr 50  hr x = 5 × 50 8 = 31.3  hr

 

Problem: If 1.6 L of activated sludge (biosolids) with volatile suspended solids (VSS) of 1900 mg/L are mixed with 7.2 L of raw domestic wastewater with BOD of 250 g/L, what is the food-to-micro-organism (F/M) ratio?

Solution:

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

Dimensional Analysis

Dimensional analysis is a problem-solving method that uses the fact that any number or expression can be multiplied by 1 without changing its value. It is a useful technique used to check if a problem is set up correctly. When 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 that is common to everyday life is the unit pricing found in many hardware stores. A shopper can purchase a 1-lb box of nails for 98¢ at a local hardware store, but a nearby warehouse store sells a 5-lb bag of the same nails for $3.50. The shopper will analyze this problem almost without thinking about it. The solution calls for reducing the problem to the price per pound. The pound is selected without much thought because it is the unit common to both stores. The shopper will pay 70¢ a pound for the nails at the warehouse store but 98¢ at the local hardware store. Implicit in the solution to this problem is knowing the unit price, which is expressed in dollars per pound ($/lb).

Note: Unit factors may be made from any two terms that describe the same or equivalent amounts of what we are interested in; for example, we know that 1 inch = 2.54 centimeters.

In order to use the dimensional analysis method, we must know how to perform three basic operations.

Basic Operation: Division of Units

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 gal/ft2) as a vertical fraction.

Horizontal to vertical

gal/ft 3  to  gal ft 3 psi to  lb in. 2

The same procedures are applied in the following examples.

ft 3 / min becomes  ft 2 min s/min  becomes   s min

Basic Operation: Divide by a Fraction

We must know how to divide by a fraction. For example,

( lb day ) ( min day ) becomes  lb day × day min

In the above, notice that the terms in the denominator were inverted before the fractions were multiplied. This is a standard rule that must be followed when dividing fractions.

Another example is

mm 2 ( mm 2 m 2 ) becomes mm 2 × m 2 mm 2

Basic Operation: Cancel or Divide Numerators and Denominators

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 above, then the terms can be canceled (or divided) out.

Note: 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:

 

Question: How do we calculate units that include exponents?

Answer: When written with exponents, such as ft3, a unit can be left as is or put in expanded form, (ft)(ft)(ft), depending on other units in the calculation. The point is that it is important to ensure that square and cubic terms are expressed uniformly (e.g., sq ft, ft2 cu ft, ft3). For dimensional analysis, the latter system is preferred.

For example, to convert a volume of 1400 ft3 to gallons, we will use 7.48 gal/ft3 in the conversions. The question becomes do we multiply or divide by 7.48? In this instance, it is possible to use dimensional analysis to answer this question of whether we multiply or divide by 7.48.

To determine if the math setup is correct, only the dimensions are used. First, try dividing:

ft 3 gal/ft 3 = ft 3 ( gal ft 3 )

Multiply the numerator and denominator to obtain

ft 6 gal

So, by dimensional analysis, we have determined that if we divide the two dimensions (ft3 and gal/ft3) then the units of the answer are ft6/gal, not gal. It is clear that division is not the right approach to making this conversion.

What would have happened if we had multiplied the dimensions instead of dividing?

ft 3 × ( gal/ft 3 ) = ft 3 × ( gal ft 3 )

Multiply the numerator and denominator to obtain

ft 3 × gal ft 3

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:

( 1400  ft 3 ) × ( 7.48  gal/ft 3 ) = 10 , 472  gal

Now, let’s try another problem with exponents. We wish to obtain an answer in square feet. If we are given the two terms—70 ft3/s and 4.5 ft/s—is the following math setup correct?

( 70  ft 3 / s ) × ( 4.5  ft/s )

First, only the dimensions are used to determine if the math setup is correct. By multiplying the two dimensions, we get

( ft 2 / s ) × ( ft/s ) = ft 3 s × ft s

Multiply the terms in the numerators and denominators of the fraction:

ft 3 × ft s × s = ft 4 s 2

Obviously, the math setup is incorrect because the dimensions of the answer are not square feet; therefore, if we multiply the numbers as shown above, the answer will be wrong.

Let’s try division of the two dimensions instead:

( ft 3 / s ) = ( ft 3 s ) ( ft s )

Invert the denominator and multiply to get

 

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.

70  ft 3 / s 4.5  ft/ s = 15.56  ft 2
 

Problem: We are given two terms, 5 m/s and 7 m2, and the answer to be obtained should be in cubic meters per second (m3/s). Is multiplying the two terms the correct math setup?

Solution:

( m/s ) × ( m 2 ) = m s × m 2

Multiply the numerators and denominator of the fraction:

= m × m 2 s = m 3 s

Because the dimensions of the answer are cubic meters per second (m3/s), the math setup is correct; therefore, multiply the numbers to get the correct answer:

5  m/s × 7  m 2 = 35  m 3 / s

 

Problem: The flow rate in a water line is 2.3 ft3/s. 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 / s ) × ( 7.48  gal/ ft 3 ) × ( 60  s/min )

Dimensional analysis can be used 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:

( 2.3  ft 3 / s ) × ( 7.48  gal/ ft 3 ) × ( 60  s/min ) = 1032.24  gal/min

 

Problem: During an 8-hr period, a water treatment plant treated 3.2 million gallons of water. What is the plant total volume treated per day, assuming the same treatment rate?

Solution:

3.2  million gal 8  hr × 24  hr day = 3.2   × 24 8  MGD = 9.6  MGD

 

Problem: One million gallons per day equals how many cubic feet per second (cfs)?

Solution:

1  MGD = 10 6 1  day = 10 6  gal × 0.1337  ft 3 / gal 1  day × 86 , 400  s/day = 133 , 700 86 , 400 = 1.547  cfs

 

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:

Weight of water = 6  gal × 8.34  lb/gal = 50.04  lb Total weight = 50.04 + 4.6  lb = 54.6  lb
 

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:

10  in . = 10 × 2.54  cm = 25.4  cm

Threshold Odor Number

The environmental practitioner responsible for water supplies soon discovers that taste and odor are the most common customer complaints. Odor is typically measured and expressed in terms of a threshold odor number (TON), the ratio by which the sample has to be diluted with odor-free water for the odor to become virtually unnoticeable. In 1989, the USEPA issued a Secondary Maximum Contaminant Level (SMCL) of 3 TON for odor.

Note: Secondary Maximum Contaminant Levels are parameters not related to health.

When a dilution is used, a number can be devised in clarifying odor.

3.3 TON  (threshold odor number) = V T + V P V T

where

  • VT = Volume tested
  • VP = Volume of dilution with odor-free distilled water
  • For VP = 0, TON = 1 (lowest value possible)
  • ForVP = VT, TON = 2
  • For VP = 2VT, TON = 3

 

Problem: The first detectable odor is observed when a 50-mL sample is diluted to 200 mL with odor-free water. What is the TON of the water sample?

Solution:

TON = 200 V T = 200  mL 50  mL = 4

Geometrical Measurements

Environmental engineers involved in fisheries, water/wastewater treatment plants, and other operations dealing with tanks, basins, and ponds operations must know the area and volume of all tanks, basins, and ponds they deal with. For example, in water and wastewater treatment plant operations, the plant configuration usually consists of a series of tanks and channels. Proper design and operational control requires engineers and operators to perform several process control calculations. Many of these calculations include parameters such as the circumference or perimeter, area, or volume of the tank or channel as part of the information necessary to determine the result. Many process calculations require computation of surface areas. Moreover, in fisheries operations, exact measurements of area and volume are essential to calculate stocking rates and chemical applications. Stocking fish in a pond of uncertain area can result in poor production, disease, and possibly death. Chemical treatments can be ineffective if the volume or area is underestimated and can be potentially lethal if they are overestimated (Masser and Jensen, 1991). To aid in performing these calculations, the following definitions and relevant equations used to calculate areas and volumes for several geometric shapes are provided.

Definitions

Area—The area of an object, measured in square units.

Base—The term used to identity the bottom leg of a triangle, measured in linear units.

Circumference—The distance around an object, measured in linear units. When determined for other than circles, it may be called the

perimeter of the figure, object, or landscape.

Cubic units—Measurements used to express volume (cubic feet, cubic meters, etc.).

Depth—The vertical distance from the bottom the tank to the top. It is normally measured in terms of liquid depth and given in terms of

sidewall depth (SWD), measured in linear units.

Diameter—The distance, measured in linear units, from one edge of a circle to the opposite edge passing through the center.

Height—The vertical distance, measured in linear units, from one end of an object to the other.

Length—The distance, measured in linear units, from one end of an object to the other.

Linear units—Measurements used to express distance (e.g., feet, inches, meters, yards).

Pi (π)—A number in the calculations involving circles, spheres, or cones (π = 3.14).

Radius—The distance, measured in linear units, from the center of a circle to the edge.

Sphere—A container shaped like a ball.

Square units—Measurements used to express area (e.g., square feet, square meters, acres).

Volume—The capacity of a unit (how much it will hold), measured in cubic units (e.g., cubic feet, cubic meters) or in liquid volume units (e.g., gallons, liters, million gallons).

Width—The distance from one side of the tank to the other, measured in linear units.

Relevant Geometric Equations

Circumference C of a circle:

C = πd = 2πr

Perimeter P of a square with side a:

P = 4a

Perimeter P of a rectangle with sides a and b:

P = 2a + 2b

Perimeter P of a triangle with sides a, b, and c:

P = a + b + c

Area A of a circle with radius r (d = 2r):

A = πd2/4 = πr2

Area A of duct in square feet when d is in inches:

A = 0.005454d2

Area A of a triangle with base b and height h:

A = 0.5bh

Area A of a square with sides a:

A = a2

Area A of a rectangle with sides a and b:

A = ab

Area A of an ellipse with major axis a and minor axis b:

A = πab

Area A of a trapezoid with parallel sides a and b and height h:

A = 0.5(a + b)h

Area A of a duct in square feet when d is in inches:

A = πd2/576 = 0.005454d2

Volume V of a sphere with a radius r (d = 2r):

V = 1.33πr3 = 0.1667πd3

Volume V of a cube with sides a:

V = a3

Volume V of a rectangular solid (sides a and b and height c):

V = abc

Volume V of a cylinder with a radius r and height H:

V = πr2h = πd2h/4

Volume V of a pyramid:

V = 0.33

Geometrical Calculations

Perimeter and Circumference

On occasion, it may be necessary to determine the distance around grounds or landscapes. To measure the distance around property, buildings, and basin-like structures, either the perimeter or circumference must be determined. The perimeter is the distance around an object; a border or outer boundary. Circumference is the distance around a circle or circular object, such as a clarifier. Distance is a linear measurement that defines the distance (or length) along a line. Standard units of measurement such as inches, feet, yards, and miles and metric units such as centimeters, meters, and kilometers are used.

The perimeter (P) of a rectangle (a four-sided figure with four right angles) is obtained by adding the lengths (Li) of the four sides (see Figure 3.1):

3.4 Perimeter = L 1 + L 2 + L 3 + L 4

 

Problem: Find the perimeter of the rectangle shown in Figure 3.2.

Solution:

P = 35  ft + 8  ft + 35  ft + 8  ft = 86  ft

 

Problem: What is the perimeter of a rectangular field if its length is 100 ft and its width is 50 ft?

Solution:

P = ( 2 × length ) + ( 2 × width ) = ( 2 × 100  ft ) + ( 2 × 50  ft) = 200  ft + 100  ft = 300  ft

Perimeter.

Figure 3.1   Perimeter.

Perimeter of a rectangle for

Figure 3.2   Perimeter of a rectangle for Example 3.40.

Diameter of circle.

Figure 3.3   Diameter of circle.

 

Problem: What is the perimeter of a square with 8-in. sides?

Solution:

P = ( 2 × length ) + ( 2 × width ) = ( 2 × 8  in . ) + ( 2 × 8  in. ) = 16  in. + 16  in . = 32  in.

The circumference is the length of the outer border of a circle. The circumference is found by multiplying pi (π) times the diameter (D) (a straight line passing through the center of a circle, or the distance across the circle; see Figure 3.3):

3.5 C = π × D

where

  • C = Circumference.
  • π = pi = 3.1416.
  • D = Diameter.

Use this calculation if, for example, the circumference of a circular tank must be determined.

 

Problem: Find the circumference of a circle that has a diameter of 25 feet (π = 3.14)

Solution:

C = π × 25  ft C = 3.14 × 25  ft = 78.5  ft

 

Problem: A circular chemical holding tank has a diameter of 18 m. What is the circumference of this tank?

Solution:

C = π × 18  m C = 3.14 × 25  m = 56.52  m

 

Problem: An influent pipe inlet opening has a diameter of 6 ft. What is the circumference of the inlet opening in inches?

Solution:

C = π × 6  ft C = 3.14 × 6  ft = 18.84  ft

Area

For area measurements in water/wastewater operations, three basic shapes are particularly important—namely, circles, rectangles, and triangles. Area is the amount of surface an object contains or the amount of material it takes to cover the surface. The area on top of a chemical tank is called the surface area. The area of the end of a ventilation duct is called the cross-sectional area (the area at right angles to the length of ducting). Area is usually expressed in square units, such as square inches (in.2) or square feet (ft2). Land may also be expressed in terms of square miles (sections) or acres (43,560 ft2) or in the metric system as hectares. In fisheries operations, pond stocking rates, limiting rates, and other important management decisions are based on surface area (Masser and Jensen, 1991).

If contractor’s measurements or country field offices of the U.S Department of Agricultural Soil Conservation Service do not have records on basin, lake, or pond measurements, then surveying basins, tanks, lagoons, and ponds using a transit is the most accurate way to determine area. Less accurate but acceptable methods of measuring basin or pond area are chaining and pacing. Inaccuracies in these methods come from mismeasurements and measurement over uneven or sloping terrain. Measurements made on flat or level areas are the most accurate.

Chaining uses a surveyor’s chain or tape of known length. Stakes are placed at each end of the tape. The stakes are used to set or locate the starting point for each progressive measurement and to maintain an exact count of the number of times the tape was moved. Sight down the stakes to keep the measurement in a straight line. The number of times the tape is moved multiplied by the length of the tape equals the total distance.

Pacing uses the average distance of a person’s pace or stride. To determine your pace length, measure a 100-foot distance and pace it, counting the number of strides. Pace in a comfortable and natural manner. Repeat the procedure several times and get an average distance for your stride. It is good practice to always pace a distance more than once and average the number of paces (Masser and Jennings, 1991). The formula for calculating distances from pacing is

Distance  (ft) = Total number of paces × Length of average pace

A rectangle is a two-dimensional box. The area of a rectangle is found by multiplying the length (L) times width (W) (see Figure 3.4).

3.6 Area = L × W
Rectangle.

Figure 3.4   Rectangle.

Area of a rectangle for

Figure 3.5   Area of a rectangle for Example 3.46.

 

Problem: Find the area of the rectangle shown in Figure 3.5.

Solution:

Area = L × W = 14  ft × 6  ft = 84  ft 2

To find the area of a circle, we need to introduce a new term, the radius, which is represented by r. The circle shown in Figure 3.6 has a radius of 6 ft. The radius is any straight line that radiates from the center of the circle to some point on the circumference. By definition, all radii (plural of radius) of the same circle are equal. The surface area of a circle is determined by multiplying π times the radius squared:

3.7 A = π × r 2

where

  • A = Area.
  • π = pi = 3.14.
  • r = Radius of circle = one half of the diameter.
 

Problem: What is the area of the circle shown in Figure 3.6?

Solution:

Area of cicle = π × r 2 = π × 6 2 = 3.14 × 36 = 113  ft 2

Area of a circle for

Figure 3.6   Area of a circle for Example 3.47.

Pond area calculation.

Figure 3.7   Pond area calculation.

If we are assigned to paint a water storage tank, we must know the surface area of the walls of the tank to determine how much paint is required. In this case, we need to know the area of a circular or cylindrical tank. To determine the surface area of the tank, we need to visualize the cylindrical walls as a rectangle wrapped around a circular base. The area of a rectangle is found by multiplying the length by the width; in the case of a cylinder, the width of the rectangle is the height of the wall, and the length of the rectangle is the distance around the circle (circumference).

Thus, the area (A) of the side wall of a circular tank is found by multiplying the circumference of the base (C = π × D) times the height of the wall (H):

3.8 A = π × D × H A = π × 20  ft × 25  ft = 3.14 × 20  ft × 25  ft = 1570  ft 2

To determine the amount of paint needed, remember to add the surface area of the top of the tank, which is 314 ft2. Thus, the amount of paint needed must cover 1570 ft2 + 314 ft2 = 1884 ft2. If the tank floor should be painted, add another 314 ft2.

Many ponds are watershed ponds that have been built by damming valleys. These ponds are irregular in shape. If no good records exist on the pond, then a reasonable estimate can be made by chaining or pacing off the pond margins and using the following procedures to calculate area:

  1. Draw the general shape of the pond on graph paper.
  2. Draw a rectangle over the pond shape that would approximate the area of the pond if some water was eliminated and placed onto an equal amount of land. This will give you a rectangle on which to base the calculation of area (see Figure 3.7).
  3. Mark the corners of the rectangle (from the drawing) on the ground around the pond and chain or pace its length and width. For example, a length of 375 paces and a width of 130 paces and a pace length of 2.68 (for example) would be equal to 1005 ft (375 paces × 2.68 ft/pace) by 348.4 ft.
  4. Multiply the length times width to get the approximate pond area. For example, 1005 ft × 348.4 ft = 350,142 ft2 or 8.04 acres (350,142 ÷ 43,500).

Volume

Volume is the amount of space occupied by or contained in an object (see Figure 3.8). It is expressed in cubic units, such as cubic inches (in.3), cubic feet (ft3), or acre-feet (1 acre-foot = 43,560 ft3). The volume (V) of a rectangular object is obtained by multiplying the length times the width times the depth or height:

3.9 V = L × W × H

Volume.

Figure 3.8   Volume.

where

  • V = Volume.
  • L = Length.
  • W = Width.
  • H (or D) = Height (or depth).
 

Problem: A unit rectangular process basin has a length of 15 ft, width of 7 ft, and depth of 9 ft. What is the volume of the basin?

Solution:

V = L × W × D = 15  ft × 7  ft × 9  ft = 945  ft 3

For environmental practitioners involved with fisheries and water/wastewater operators, representative surface areas are most often rectangles, triangles, circles, or a combination of these. Practical volume formulas used in fisheries and water/wastewater calculations are given in Table 3.2.

When determining the volume of round pipe and round surface areas, the following examples are helpful.

 

Problem: Find the volume of a 3-in. round pipe that is 300 ft long.

Solution:

  1. Change the diameter (D) of the duct from inches to feet by dividing by 12:
    D = 3 ÷ 12 = 0.25  ft

    Table 3.2   Volume Formulas

    Sphere volume=(π/6) × (Diameter)3
    Cone volume=1/3 × (Volume of a cylinder)
    Rectangular tank volume=(Area of rectangle) × (D or H)
    =(L × W) × (D or H)
    Cylinder volume=(Area of cylinder) × (D or H)
    =πr2 × (D or H)
  2. Find the radius (r) by dividing the diameter by 2:
    r = 0.25  ft ÷ 2 = 0.125
  3. Find the volume (V):
    V = L × π × r 2 V = 300  ft × 3.14 × 0.0156 = 14.72  ft 2

 

Problem: Find the volume of a smokestack that is 24 in. in diameter (entire length) and 96 in. tall.

Solution: First find the radius of the stack. The radius is one half the diameter, so 24 in. ÷ 2 = 12 in. Now find the volume:

V = H × π × r 2 V = 96  in. × π × ( 12  in. ) 2 V = 96  in . × ( 144  in. 2 ) = 43 , 407  ft 3

To determine the volume of a cone and sphere, we use the following equations and examples.

Volume of Cone

Volume of cone Diameter Diameter Height

3.10 Volume  of cone = π 12 × Diameter × Diameter × Height

Note that

π 12 = 3.14 12 = 0.262

Note: The diameter used in the formula is the diameter of the base of the cone.

 

Problem: The bottom section of a circular settling tank has the shape of a cone. How many cubic feet of water are contained in this section of the tank if the tank has a diameter of 120 ft and the cone portion of the unit has a depth of 6 ft?

Solution:

Volume   ( ft 3 ) = 0.262 × 120  ft × 120  ft × 6  ft = 22 , 637  ft 3

Volume of Sphere

3.11 Volume of sphere = π 6 × Diameter × Diameter × Diameter

Note that

π 6 = 3.14 6 = 0.524
Circular or cylindrical water tank.

Figure 3.9   Circular or cylindrical water tank.

 

Problem: What is the volume (ft3) of a spherical gas storage container with a diameter of 60 ft?

Solution:

Volume  ( ft 3 ) = 0.524 × 60  ft × 60  ft × 60  ft = 113 , 184  ft 3

Circular process and various water and chemical storage tanks are commonly found in water/ wastewater treatment. A circular tank consists of a circular floor surface with a cylinder rising above it (see Figure 3.9). The volume of a circular tank is calculated by multiplying the surface area times the height of the tank walls.

 

Problem: If a tank is 20 feet in diameter and 25 feet deep, how many gallons of water will it hold?

Hint: In this type of problem, calculate the surface area first, multiply by the height, and then convert to gallons.

Solution:

r = D ÷ 2 = 20  ft ÷ 2 = 10  ft A = π × r 2 = π × 10  ft × 10  ft = 314  ft 2 V = A × H = 314  ft 2 × 25  ft = 7850  ft 3 × 7.48  gal/ft 3 = 58 , 718  gal

Force, Pressure, and Head Calculations

Before we review calculations involving force, pressure, and head, we must first define these terms:

  • Force—The push exerted by water on any confined surface. Force can be expressed in pounds, tons, grams, or kilograms.
  • Pressure—The force per unit area. The most common way of expressing pressure is in pounds per square inch (psi).
  • Head—The vertical distance or height of water above a reference point. Head is usually expressed in feet. In the case of water, head and pressure are related.

One cubic foot of water weighs 62.4 lb.

Figure 3.10   One cubic foot of water weighs 62.4 lb.

Force and Pressure

Figure 3.10 helps to illustrate these terms. A cubical container measuring 1 foot on each side can hold 1 cubic foot of water. A basic fact of science states that 1 cubic foot of water weighs 62.4 lb and contains 7.48 gal. The force acting on the bottom of the container is 62.4 lb/ft2. The area of the bottom in square inches is

1  ft 2 = 12  in . × 12  in. = 144  in. 2

Therefore, the pressure in pounds per square inch (psi) is

62.4  lb/ft 2 1  ft 2 = 62.4  lb/ft 2 144  in. 2 / ft 2 = 0.433  lb/in. 2 ( psi )

If we use the bottom of the container as our reference point, the head would be 1 foot. From this, we can see that 1 foot of head is equal to 0.433 psi—an important parameter to remember. Figure 3.11 illustrates some other important relationships between pressure and head.

Note: Force acts in a particular direction. Water in a tank exerts force down on the bottom and out of the sides. Pressure, however, acts in all directions. A marble at a water depth of 1 foot would have 0.433 psi of pressure acting inward on all sides.

Relationship between pressure and head.

Figure 3.11   Relationship between pressure and head.

Using the preceding information, we can develop Equations 3.12 and 3.13 for calculating pressure and head:

3.12 Pressure  (psi) = 0.433 × Head ( ft )

3.13 Head  (ft) = 2.31 × Pressure  (psi)

Head

Head is the vertical distance the water must be lifted from the supply tank or unit process to the discharge. The total head includes the vertical distance the liquid must be lifted (static head), the loss to friction (friction head), and the energy required to maintain the desired velocity (velocity head):

3.14 Total head = Static head + Friction  head + Velocity head

Static Head

Static head is the actual vertical distance the liquid must be lifted.

3.15 Static Head = Discharge elevation Supply elevation
 

Problem: The supply tank is located at elevation 108 ft. The discharge point is at elevation 205 ft. What is the static head in feet?

Solution:

Static Head  (ft) = 205  ft 108  ft = 97  ft

Friction Head

Friction head is the equivalent distance of the energy that must be supplied to overcome friction. Engineering references include tables showing the equivalent vertical distance for various sizes and types of pipes, fittings, and valves. The total friction head is the sum of the equivalent vertical distances for each component:

3.16 Friction head (ft) = Energy losses due to friction

Velocity Head

Velocity head is the equivalent distance of the energy consumed in achieving and maintaining the desired velocity in the system:

3.17 Velocity head (ft) = Energy losses to maintain velocity

Total Dynamic Head (Total System Head)

3.18 Total head = Static head + Friction head + Velocity head

Pressure and Head

The pressure exerted by water/wastewater is directly proportional to its depth or head in the pipe, tank, or channel. If the pressure is known, the equivalent head can be calculated:

3.19 Head (ft) = Pressure (psi) × 2.31  ft/psi

 

Problem: The pressure gauge on the discharge line from the influent pump reads 75.3 psi. What is the equivalent head in feet?

Solution:

Head (ft) = 75.3 × 2.31  ft/psi = 173.9  ft

Head and Pressure

If the head is known, the equivalent pressure can be calculated by

3.20 Pressure (psi) = Head  (ft) 2.31  ft/psi
 

Problem: A tank is 15 ft deep. What is the pressure in psi at the bottom of the tank when it is filled with wastewater?

Solution:

Pressure (psi) = 15  ft 2.31  ft/psi = 6.49  psi

Before we look at a few example problems dealing with force, pressure, and head, it is important to review the key points related to force, pressure, and head:

  1. By definition, water weighs 62.4 lb/ft3.
  2. The surface of any one side of a 1-ft3 cube contains 144 in.2 (12 in. × 12 in. = 144 in.2); therefore, the cube contains 144 columns of water that are 1 ft tall and 1 inch square.
  3. The weight of each of these pieces can be determined by dividing the weight of the water in the cube by the number of square inches:
    Weight = 62.4  lb 144  in. 2 = 0.433  lb/in. 2  or 0.433  psi
  4. Because this is the weight of one column of water 1 ft tall, the true expression would be 0.433 pounds per square inch per foot of head, or 0.433 psi/ft.

Note: 1 foot of head = 0.433 psi.

In addition to remembering the important parameter that 1 ft of head = 0.433 psi, it is important to understand the relationship between pressure and feet of head—in other words, how many feet of head 1 psi represents. This is determined by dividing 1 ft by 0.433 psi:

Pressure (psi) = Head  (ft) 2.31  ft/ psi

If a pressure gauge reads 12 psi, the height of the water necessary to represent this pressure would be 12 psi × 2.31 ft/psi = 27.7 feet.

Note: Both of the above conversions are commonly used in water/wastewater treatment calculations; however, the most accurate conversion is 1 ft = 0.433 psi. This is the conversion we use throughout this text.

 

Problem: Convert 40 psi to feet head.

Solution:

40  psi 1 × ft 0.433  psi = 92.4  ft

 

Problem: Convert 40 ft to psi.

Solution:

40 ft 1 × 0.433  psi 1  ft = 17.32  psi

As the above examples demonstrate, when attempting to convert psi to feet we divide by 0.433, and when attempting to convert feet to psi we multiply by 0.433. The above process can be most helpful in clearing up confusion about whether to multiply or divide; however, there is another approach that may be easier for many operators to use. Notice that the relationship between psi and feet is almost 2 to 1. It takes slightly more than 2 feet to make 1 psi; therefore, in a problem where the data are provided in pressure and the result should be in feet, the answer will be at least twice as large as the starting number. For example, if the pressure were 25 psi, we intuitively know that the head is over 50 feet, so we must divide by 0.433 to obtain the correct answer.

 

Problem: Convert a pressure of 45 psi to feet of head.

Solution:

45 psi 1 × 1  ft 0.433  psi = 104  ft

 

Problem: Convert 15 psi to feet.

Solution:

15 psi 1 × 1  ft 0.433  psi = 34.6  ft

 

Problem: Between the top of a reservoir and the watering point, the elevation is 125 feet. What will the static pressure be at the watering point?

Solution:

125 psi 1 × 1  ft 0.433  psi = 288.7  ft

 

Problem: Find the pressure (psi) in a tank 12 ft deep at a point 5 ft below the water surface.

Solution:

Pressure  (psi) = 0.433 × 5  ft = 2.17  psi

 

Problem: A pressure gauge at the bottom of a tank reads 12.2 psi. How deep is the water in the tank?

Solution:

Head (ft) = 2.31 × 12.2  psi = 28.2  ft

 

Problem: What is the pressure (static pressure) 4 miles beneath the ocean surface?

Solution: Change miles to feet, then to psi:

5280  ft/mile × 4 = 21 , 120  ft 21 , 120  ft ÷ 2.31  ft/psi = 9143  psi

 

Problem: A 150-ft-diameter cylindrical tank contains 2.0 MG water. What is the water depth? At what pressure would a gauge at the bottom read in psi?

Solution:

  1. Change MG to cubic feet:
    2 , 000 , 000  gal ÷ 7.48 = 267 , 380  ft 3
  2. Using volume, solve for depth:
    Vloume = 0.785 × D 2 × Depth 267 , 380  ft 3 = 0.785 × ( 150 ) 2 × Depth Depth = 15.1  ft

 

Problem: The pressure in a pipe is 70 psi. What is the pressure in feet of water? What is the pressure in psf?

Solution:

  1. Convert pressure to feet of water:
    70  psi × 2.31  ft/psi = 161.7  ft of water
  2. Convert psi to psf:
    70  psi × 144  in. 2 / ft 2 = 10 , 080  psf

 

Problem: The pressure in a pipeline is 6476 psf. What is the head on the pipe?

Solution:

Head on pipe = Feet of pressure Pressure = Weight × height 6476  psf = 62.4  lb/ft 3 × height Height = 104  ft

Review of Advanced Algebra Key Terms and Concepts

Advanced algebraic operations (linear, linear differential, and ordinary differential equations) have in recent years become an essential part of the mathematical background required by environmental engineers, among others. It is not the intent here to provide complete coverage of the topics (environmental practitioners are normally well grounded in these critical foundational areas), but it is important to review the key terms and relevant concepts. Key definitions include the following:

  • Algebraic multiplicity of an eigenvalue—The algebraic multiplicity of eigenvalue c of matrix
  • A is the number of times the factor (tc) occurs in the characteristic polynomial of A. Basis for a subspace—A basis for subspace W is a set of vectors {v1, …, vk} in W such that
    1. {v1, …, vk} is linearly independent, and
    2. {v1, …, vk} spans W.
  • Characteristic polynomial of a matrix—The characteristic polynomial of n×n matrix A is the polynomial in t given by the formula det(AtI).
  • Column space of a matrix—The subspace spanned by the columns of the matrix considered as a set of vectors (also see row space).
  • Consistent linear system—A system of linear equations is consistent if it has at least one solution.
  • Defective matrix—Matrix A is defective if A has an eigenvalue whose geometric multiplicity is less than its algebraic multiplicity.
  • Diagonalizable matrix—A matrix is diagonalizable if it is similar to a diagonal matrix.
  • Dimension of a subspace—The dimension of subspace W is the number of vectors in any basis of W. (If W is the subspace {0}, then we say that its dimension is 0.)
  • Echelon form of a matrix—A matrix is in row echelon form if
    1. All rows that consist entirely of zeros are grouped together at the bottom of the matrix, and
    2. The first (counting left to right) nonzero entry in each nonzero row appears in a column to the right of the first nonzero entry in the preceding row (if there is a preceding row).
  • Eigenspace of a matrix—The eigenspace associated with the eigenvalue c of matrix A is the null space of A – cl.
  • Eigenvalue of a matrix—An eigenvalue of matrix A is scalar c such that Ax = cx holds for some nonzero vector x.
  • Eigenvector of a matrix—An eigenvector of square matrix A is a nonzero vector x such that Ax = cx holds for some scalar c.
  • Elementary matrix—A matrix that is obtained by performing an elementary row operation on an identity matrix.
  • Equivalent linear systems—Two systems of linear equations in n unknowns are equivalent if they have the same set of solutions.
  • Geometric multiplicity of an eigenvalue—The geometric multiplicity of eigenvalue c of matrix A is the dimension of the eigenspace of c.
  • Homogeneous linear system—A system of linear equations Ax = b is homogeneous if b = 0.
  • Inconsistent linear system—A system of linear equations is inconsistent if it has no solutions.
  • Inverse of a matrix—Matrix B is an inverse for matrix A if AB = BA = I.
  • Invertible matrix—A matrix is invertible if it has no inverse.
  • Least-squares solution of a linear system—A least-squares solution to a system of linear equations Ax = b is a vector x that minimizes the length of the vector Axb.
  • Linear combination of vectors—Vector v is a linear combination of the vectors v1, …, vk if there exist scalars a1, …, ak such that v = a1v1 + … + akvk.
  • Linear dependence relation for a set of vectors—A linear dependence relation for the set of vectors {v1, …, vk} is an equation of the form a1v1 + … + akvk = 0, where the scalars a1, …, ak are zero.
  • Linearly dependent set of vectors—The set of vectors {v1, …, vk} is linearly dependent if the equation a1v1 + … + akvk = 0 has a solution where not all the scalars a1, …, ak are zero (i.e., if {v1, …, vk} satisfies a linear dependence relation).
  • Linearly independent set of vectors—The set of vectors {v1, …, vk} is linearly independent if the only solution to the equation a1v1 + … + akvk = 0 is the solution where all the scalars a1, …, ak are zero (i.e., if {v1, …, vk} does not satisfy any linear dependence relation). Linear transformation—A linear transformation from V to W is a function T from V to W such that
    1. T(u + v) = T(u) + T(v) for all vectors u and v in V.
    2. T(av) = aT(v) for all vectors v in V and all scalars a.
  • Nonsingular matrix—Square matrix A is nonsingular if the only solution to the equation Ax = 0 is x = 0.
  • Null space of a linear transformation—The null space of linear transformation T is the set of vectors v in its domain such that T(v) = 0.
  • Null space of a matrix—The null space of m×n matrix A is the set of all vectors x in Rn such that Ax = 0.
  • Nullity of a linear transformation—The nullity of linear transformation T is the dimension of its null space.
  • Nullity of a matrix—The dimension of its null space.
  • Orthogonal complement of a subspace—The orthogonal complement of subspace S of Rn is the set of all vectors v in Rn such that v is orthogonal to every vector in S.
  • Orthogonal set of vectors—A set of vectors in Rn is orthogonal if the dot product of any two of them is 0.
  • Orthogonal linear transformation—Linear transformation T from V to W is orthogonal if T(v) has the same length as v for all vectors v in V.
  • Orthogonal matrix—Matrix A is orthogonal if A is invertible and its inverse equals its transpose; that is, A–1 = AT.
  • Orthonormal set of vectors—A set of vectors in Rn is orthonormal if it is an orthogonal set and each vector has length 1.
  • Range of a linear transformation—The range of linear transformation T is the set of all vectors T(v), where v is any vector in its domain.
  • Rank of a linear transformation—The rank of a linear transformation (and hence of any matrix regarded as a linear transformation) is the dimension of its range. Note that a theorem tells us that the two definitions of rank of a matrix are equivalent.
  • Rank of a matrix—The rank of matrix A is the number of nonzero rows in the reduced row echelon form of A; that is, the dimension of the row space of A.
  • Reduced row echelon form of a matrix—A matrix is in reduced row echelon form if
    1. The matrix is in row echelon form.
    2. The first nonzero entry in each nonzero row is the number 1.
    3. The first nonzero entry in each nonzero row is the only nonzero entry in its column.
  • Row equivalent matrices—Two matrices are row equivalent if one can be obtained from the other by a sequence of elementary row operations.
  • Row operations—The elementary row operations performed on a matrix are
    1. Interchange two rows.
    2. Multiply a row by a nonzero scalar.
    3. Add a constant multiple of one row to another.
  • Row space of a matrix—The subspace spanned by the rows of the matrix considered as a set of vectors.
  • Similar matrices—Matrices A and B are similar if there is a square invertible matrix S such that S–1AS = B.
  • Singular matrix—Square matrix A is singular if the equation Ax = 0 has a nonzero solution for x.
  • Span of a set of vectors—The span of the set of vectors {v1, …, vk} is the subspace V consisting of all linear combinations of v1, …, vk. One also says that the subspace V is spanned by the set of vectors {v1, …, vk} and that this set of vectors spans V.
  • Subspace—A subset W of Rn is a subspace of Rn if
    1. The zero vector is in W.
    2. x + y is in W whenever x and y are in W.
    3. ax is in W whenever x is in W and a is any scalar.
  • Symmetric matrix—Matrix A is symmetric if it equals its transpose; that is, A = AT.

Quadratic Equations

a x 2 + b x + c = 0 x = b ± b 2 4 a c 2 a

When studying a discipline that does not include mathematics, one thing is certain: The discipline under study has little or nothing to do with environmental practice.

Quadratic Equations and Environmental Practice

A logical question at this point might be why is the quadratic equation important in environmental practice? The logical answer is that the quadratic equation is used in environmental practice to find solutions to problems primarily dealing with length and time determinations. Stated differently: The quadratic equation is a tool, an important tool that belongs in every environmental practitioner’s toolbox.

To the student of mathematics, this explanation might seem somewhat strange. Math students know, for example, that there will be two solutions to a quadratic equation. In environmental disciplines such as environmental engineering, many times only one solution is meaningful. For example, if we are dealing with a length, a negative solution to the equation may be mathematically possible but is not the solution we would use. Negative time, obviously, would also pose the same problem.

So what is the point? The point is that we often need to find a solution to certain mathematical problems. In environmental problems involving the determination of length and time using quadratic equations, we will end up with two answers. In some instances, a positive answer and a negative answer may result. One of these answers is usable; thus, we would use it. Real engineering is about modeling situations that occur naturally and using the model to understand what is happening or maybe to predict what will happen in future. The quadratic equation is often used in modeling because it is a beautifully simple curve (Bourne, 2013).

Key Terms

  • a is the coefficient of x2.
  • b is the coefficient of x.
  • c is a number in the quadratic equation (not a coefficient of any x term).
  • Simple equations are equations in which the unknown appears only in the first degree.
  • Pure quadratic equations are equations in which the unknown appears only in the second degree.
  • Affected quadratic equations are equations containing the first and second degree of an unknown.

Quadratic Equations: Theory and Application

The equation 6x = 12 is a form of equation familiar to most of us. In this equation the unknown appears only in the first degree, so it is a simple equation or linear equation. Those experienced in mathematics know that not all equations reduce to this form. For instance, when an equation has been reduced, the result may be an equation in which the square of the unknown equals some number, as in x2 = 5. In this equation, the unknown appears only in the second degree, so it is a pure quadratic equation. In some cases, when an equation is simplified and reduced, the resulting equation contains the square and first power of the unknown, which equal some number, such as x2 – 5x = 24. An equation containing the first and second degree of an unknown is an affected quadratic equation.

Quadratic equations, and certain other forms, can be solved with the aid of factoring. The procedure for solving a quadratic equation by factoring is as follows:

  1. Collect all terms on the left and simplify to the form ax2 + bx + c = 0.
  2. Factor the quadratic expression.
  3. Set each factor equal to zero.
  4. Solve the resulting linear equations.
  5. Check the solution in the original equation.

 

Problem: Solve x2x – 12 = 0.

Solution:

  1. Factor the quadratic expression.
    ( x 4 ) ( x + 3 ) = 0
  2. Set each factor equal to zero.
    x 4 = 0 x + 3 = 0
  3. Solve the resulting linear equations.
    x = 4 x = 3

Thus, the roots are x = 4 and x = –3.

  1. Check the solution in the original equation.
    ( 4 ) 2 4 12 = 0 ( 3 ) 2 12 = 0 0 = 0 0 = 0

Many times factoring is either too time consuming or not possible. The formula shown below is called the quadratic formula. It expresses the quadratic equation in terms of its coefficients. The quadratic formula allows us to quickly solve for x with no factoring.

3.21 x = b ± b 2 4 a c 2 a

To use the quadratic equation, just substitute the appropriate coefficients into the equation and solve.

Derivation of the Quadratic Equation Formula

The equation ax2 + bx +c = 0, where a, b, and c are any numbers, positive or negative, represents any quadratic equation with one unknown. When this general equation is solved, the solution can be used to determine the unknown value in any quadratic equation. The solution follows.

 

Problem: Solve ax2 + bx + c = 0 for x.

Solution:

  1. Subtract c from both members:
    a x 2 + b x = c
  2. Divide both members by a:
    x 2 + b a ( x ) = c a
  3. Add (b/2a)2 to both sides:
    x 2 + b a ( x ) + ( b 2 a ) 2 = c a + ( b 2 a ) 2
  4. Complete the square:
    ( x + b 2 a ) 2 = c a + ( b 2 a ) 2
  5. Take the square root of both members:
    x + b 2 a = ± c a + ( b 2 a ) 2
  6. Subtract b/2a from both members:
    x = b 2 a ± c a + ( b 2 a ) 2
    Thus, the following quadratic formula is obtained:
    x = b ± b 2 4 a c 2 a

Using the Quadratic Equation

 

Problem: After conducting a study and deriving an equation representing time, we arrive at the following equation:

x 2 = 5 x + 6 = 0

Solution: All like terms have been combined and the equation is set to equal zero. Use the quadratic formula to solve the problem:

x = b ± b 2 4 a c 2 a

From our equation, a = 1 (the coefficient of x2), b = –5 (the coefficient of x), and c = 6 (the constant or third term). Substituting these coefficients in the quadratic formula:

x = ( 5 ) ± ( 5 ) 2 4 ( 1 ) ( 6 ) 2 ( 1 ) x = 5 ± 25 4 2 x = 5 ± 1 2 x = 3 , 2

Note: The roots may not always be rational (integers), but the procedure is the same.

Trigonometric Ratios

sin  A = a / c cos  A = b / c tan   A = a / b

We owe a lot to the Indians, who taught us how to count, without which no worthwhile scientific discovery could have been made.

—Albert Einstein

Trigonometry is the branch of mathematics that is used to compute unknown angles and sides of triangles. The word trigonometry is derived from the Greek words for triangle and measurement. Trigonometry is based on the principles of geometry. Many problems require the use of geometry and trigonometry.

Smith and Peterson (2007)

Table 3.4   Definition of Trigonometric Ratios

Sine of angle A

Measure of leg opposite angle  A Measure of hypotenuse

sin A = a/c

Cosine of angle A

Measure of leg adjacent angle  A Measure of hypotenuse

cos A = b/c

Tangent of angle A

Measure of leg opposit angle  A Measure of leg  adjacent to angle  A

tan A = a/b

Trigonometric Functions and the Environmental Practitioner

Typically, environmental practitioners are called upon to make calculations involving the use of various trigonometric functions. Consider slings, for example; they are commonly used with cranes, derricks, and hoists to lift a load and move it to the desired location. For the environmental professional responsible for safety and health, knowledge of the properties and limitations of the sling, the type and condition of material being lifted, the weight and shape of the object being lifted, the angle of the sling to the load being lifted, and the environment in which the lift is to be made are all important considerations to be evaluated before the safe transfer of material can take place. Later, we put many of the following principles to work in determining sling load and working load on a ramp (inclined plane)—that is, to solve force-type problems. For now, we discuss the basic trigonometric functions used to make such calculations.

Trigonometric Ratios or Functions

In trigonometry, all computations are based on certain ratios (i.e., trigonometric functions). The trigonometric ratios or functions are sine, cosine, tangent, cotangent, secant, and cosecant. It is important to understand the definition of the ratios given in Table 3.4 and defined in terms of the lines shown in Figure 3.12.

Note: In a right triangle, the side opposite the right angle is the longest side. This side is called the hypotenuse. The other two sides are the legs.

 

Problem: Find the sine, cosine, and tangent of angle Y in Figure 3.13.

Solution:

sin  Y = Opposite leg Hypotenuse = 9 15 = 0.60 cos  Y = Adjacent leg Hypotenuse = 12 15 = 0.80 tan  Y = Opposite leg Adjacent leg = 9 12 = 0.75

Right triangle.

Figure 3.12   Right triangle.

Illustration for

Figure 3.13   Illustration for Examples 3.71 and 3.72.

Illustration for

Figure 3.14   Illustration for Example 3.73.

 

Problem: Using Figure 3.13, find the measure of angle x to the nearest degree.

Solution:

sin   x = Opposite leg Hypotenuse = 12 15 = 0.8

Use a scientific calculator to find the angle measure with a sine of 0.8.

Enter: 0.8 [2nd] or [INV]

Result: 53.13010235

So, the measure of angle x = 53°.

 

Problem: For the triangle shown in Figure 3.14, find sin C, cos C, and tan C.

Solution:

sin   C = 2 / 5 ; cos  C = 4 / 5 ; tan   C = 3 / 4

References And Recommended Reading

Bourne, M. (2013). Quadratic Equations. Interactive Mathematics, http://www.intmath.com/quadratic-equations/quadratic-equations-intro.php.
Masser, M.P. and Jensen, J.W. (1991). Calculating Area and Volume of Ponds and Tanks, USDA Grant 89-38500-4516. Washington, DC: Southern Regional Aquaculture Center, U.S. Department of Agriculture.
McKeague, M. and Charles, P. (1998). Algebra with Trigonometry for College Students. Philadelphia, PA: Saunders.
Smith, R.D. and Peterson, J.C. (2009). Mathematics for Machine Technology, 6th ed. Clifton Park, NY: Delmar.
Stapel, E. (2013). The Order of Operations: More Examples. Purplemath, http://www.purplemath.com/modules/orderops2.htm.
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