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.

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.

Sequence of Operations Examples

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

$$\begin{array}{c}4+6=10\text{ and 6}+4=10\\ (4+5)+(3+7)=19,(3+5)+(4+7)=19,\text{ and }[7+(5+4\left)\right]+3=19\end{array}$$

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

$$\begin{array}{c}100-20=80,\text{but 20}-100=-80\\ (100-30)-20=50,\text{but}100-(30-20)=90\end{array}$$

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:

$$\left[\right(3\times 3)\times 5]\times 6=270\text{ and 5}\times [3\times (6\times 3\left)\right]=270$$

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

$$\begin{array}{c}100÷10=10,\text{but 10}÷100=0.1\\ (100÷10)÷2=5,\text{but 100}÷(10÷2)=20\end{array}$$

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

$$100÷5÷2\text{ is understood to mean}\text{ (100}÷5)÷2$$

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:

$$\begin{array}{l}4-3[4-2(6-3\left)\right]÷2\\ 4-3[4-2(3\left)\right]÷2\\ 4-3[4-6]÷2\\ 4-3[-2]÷2\\ 4+6÷2\\ 4+3=7\end{array}$$

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 $$\text{Percent (\%)}=\frac{{\displaystyle \text{Part}}}{{\displaystyle \text{Whole}}}$$

  
  For example, 22 percent (or 22%) means 22 out of 100, or 22/100. Dividing 22 by 100 results in the decimal 0.22:
  $$22\%=\frac{{\displaystyle 22}}{{\displaystyle 100}}=0.22$$

  
• 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?
  $$\begin{array}{l}\frac{{\displaystyle 26\text{ ft}}}{{\displaystyle 50\text{ ft}}}=0.52\text{ (decimal equivalent)}\\ 0.52\times 100=52\end{array}$$

  
  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:

$$\text{Percent}=\frac{{\displaystyle 325\text{ gal}}}{{\displaystyle 6500\text{ gal}}}\times 100=5\%$$

 

Problem: Convert 48% to decimal percent.

Solution:

$$\text{Decimal percent}=\frac{{\displaystyle \text{Percent}}}{{\displaystyle 100}}=\frac{{\displaystyle 48}}{{\displaystyle 100}}=0.48$$

 

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

Solution:

$$\text{Decimal percent}=\frac{{\displaystyle 5.8\%}}{{\displaystyle 100}}=0.058$$

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 $$\text{Quantity}=\text{Total}\times \text{Decimal percent}$$

 

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:

$$\text{Gallons}=\frac{{\displaystyle 5\%}}{{\displaystyle 100}}\times 2800\text{ gal}=140\text{ gal}$$

 

Problem: Convert 0.55 to percent.

Solution:

$$0.55=\frac{{\displaystyle 55}}{{\displaystyle 100}}=55\%$$

(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:

$$\frac{{\displaystyle 7}}{{\displaystyle 22}}=0.318=0.318\times 100=31.8\%$$

 

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:

$$\begin{array}{ccc}3\text{ mg/L}& =& \frac{{\displaystyle 3\text{ mg}}}{{\displaystyle 1\text{ L}\times 1,000,000\text{ mg}/\text{L}}}\times 100\%\hfill \\ & =& \frac{{\displaystyle 3}}{{\displaystyle 10,00}}\%=0.0003\%\hfill \end{array}$$

 

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

Solution:

$$1.4\%=\frac{{\displaystyle 1.4}}{{\displaystyle 100}}\times 1,000,000\text{ mg/L (the weight 1 L water to}{10}^6)=14,000\text{ mg/L}$$

 

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

Solution: Because 1 gal of water = 8.34 lb,

$$1\text{ ppm}=\frac{{\displaystyle 1\text{ gal}}}{{\displaystyle {10}^6\text{ gal}}}=\frac{{\displaystyle 1\text{ gal}\times 8.34\text{ lb}/\text{gal}}}{{\displaystyle 1,000,000\text{ gal}}}=8.34\text{ lb}/1,000,000\text{ gal}$$

 

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,

$$\begin{array}{l}\frac{{\displaystyle x}}{{\displaystyle 42+x}}=0.26\\ x=0.26(42+x)=10.92+0.26x\\ x=\frac{{\displaystyle 10.92}}{{\displaystyle 0.74}}=14.76\text{ lb}\end{array}$$

 

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

Solution:

$$\text{Grade}=\frac{{\displaystyle 140\text{ mm}}}{{\displaystyle 22\text{ m}}}\times 100\%=\frac{{\displaystyle 140\text{ mm}}}{{\displaystyle 22\text{ m}\times 1000\text{ mm}}}\times 100\%=0.64\%$$

 

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:

$$\text{Efficiency}=\frac{{\displaystyle \text{hp}\text{ output}}}{{\displaystyle \text{hp input}}}\times 100\%=\frac{{\displaystyle 26.5\text{ hp}}}{{\displaystyle 40\text{ hp}}}\times 100\%=66\%$$

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:

$$\begin{array}{ccc}34.73& =& 34.7\hfill \\ 34.77& =& 34.8\hfill \\ 34.75& =& 3.48\hfill \\ 34.45& =& 34.4\hfill \\ 34.35& =& 34.4\hfill \end{array}$$

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:
  $$\begin{array}{c}{z}^4=z\times z\\ z^4=z\times z\times z\times z\end{array}$$

  

 

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:

$$2^3=2\times 2\times 2$$

 

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:

$${(3/8)}^2=(3/8\times 3/8)$$

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

$${11}^{-5}={1/11}^5$$

 

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

Solution:

$$8^{-3}=\frac{{\displaystyle 1}}{{\displaystyle 8^3}}=\frac{{\displaystyle 1}}{{\displaystyle 8\times 8\times 8}}$$

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.

 

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:

$$\begin{array}{c}115\text{ gpm}+100\text{ gpm}+125\text{ gpm}+90\text{ gpm}=430\\ 430÷4=107.5\text{ gpm}\end{array}$$

 

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

$$\begin{array}{c}100,000+100,00+100,000+1,000,000=1,300,000\\ 1,300,000÷4=325,000\text{ gal}\end{array}$$

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:

$$\begin{array}{c}22+33+21+13=89\\ 89÷4=22.3\text{ mg/L}\end{array}$$

 

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:

$$\begin{array}{ccc}\text{Average SS}& =& \frac{{\displaystyle \text{Su of all measurements}}}{{\displaystyle \text{Number of measurements used}}}\\ & =& \frac{{\displaystyle 2206\text{ mg/L}\text{ SS}}}{{\displaystyle 7}}=315.1\text{ mg/L}\text{ SS}\end{array}$$

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:
  $$\frac{{\displaystyle \text{A}}}{{\displaystyle \text{B}}}=\frac{{\displaystyle \text{C}}}{{\displaystyle \text{D}}}$$

  
• 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):
  $$\text{A}\times \text{D}=\text{B}\times \text{C}\text{ (or,}\text{ AD}=\text{B}\text{C}\text{)}$$

  
• 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:
  $$\frac{{\displaystyle 1\text{ lb}\text{ alum}}}{{\displaystyle 500\text{ gal water}}}=\frac{{\displaystyle x\text{ lb}\text{ alum}}}{{\displaystyle 10,000\text{ gal water}}}$$

  
  Cross-multiplying,
  $$500\times x=1\times 10,000$$

  
  Transposing,
  $$\frac{{\displaystyle 1\times 10,000}}{{\displaystyle 500}}=20\text{ lb alum}$$

  
  To calculate proportion, suppose, for example, that 5 gal of fuel costs $5.40. What will 15 gal cost?
  $$\begin{array}{ccc}\hfill \frac{{\displaystyle \text{5 gal}}}{{\displaystyle \$5.40}}& =& \frac{{\displaystyle 15\text{ gal}}}{{\displaystyle \$y}}\hfill \\ \hfill 5\text{ gal}\times \$y& =& 15\text{ gal}\times \$5.40=81\hfill \\ \hfill \$y& =& \frac{{\displaystyle 81}}{{\displaystyle 5}}=\$16.20\hfill \end{array}$$

  

 

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:

$$\begin{array}{c}\frac{{\displaystyle x\text{ hr}}}{{\displaystyle 20\text{ hr}}}=\frac{{\displaystyle 4\text{ gpm}}}{{\displaystyle 10\text{ gpm}}}\\ x=\frac{{\displaystyle (4\times 20)}}{{\displaystyle 10}}=8\text{ hr}\end{array}$$

 

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

Solution:

$$\begin{array}{ccc}\frac{{\displaystyle 36}}{{\displaystyle 180}}& =& \frac{{\displaystyle x}}{{\displaystyle 4450}}\\ \frac{{\displaystyle 4450\times 36}}{{\displaystyle 180}}& =& x=890\end{array}$$

 

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

$$\frac{{\displaystyle 3.4}}{{\displaystyle 2}}=\frac{{\displaystyle 6}}{{\displaystyle x}}$$

Solution:

$$\begin{array}{c}3.4\times x=2\times 6\\ x=\frac{{\displaystyle 2\times 6}}{{\displaystyle 3.4}}=3.53\end{array}$$

 

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:

$$\begin{array}{ccc}\hfill \frac{{\displaystyle 1\text{ lb}}}{{\displaystyle 65\text{ gal}}}& =& \frac{{\displaystyle x\text{ lb}}}{{\displaystyle 150\text{ gal}}}\hfill \\ \hfill 65\times x& =& 1\times 150\hfill \\ \hfill x& =& \frac{{\displaystyle 1\times 150}}{{\displaystyle 65}}=2.3\text{ lb}\hfill \end{array}$$

 

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:

$$\begin{array}{ccc}\frac{{\displaystyle \text{5 workers}}}{{\displaystyle 8\text{ workers}}}& =& \frac{{\displaystyle x\text{ hr}}}{{\displaystyle 50\text{ hr}}}\\ x=\frac{{\displaystyle 5\times 50}}{{\displaystyle 8}}& =& 31.3\text{ hr}\end{array}$$

 

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:

$$\frac{{\displaystyle F}}{{\displaystyle M}}=\frac{{\displaystyle \text{Amount of BOD}}}{{\displaystyle \text{Amount of VSS}}}=\frac{{\displaystyle 250\text{ mg/L}\times 7.2\text{ L}}}{{\displaystyle 1900\text{ mg}/\text{L}\times 1.6\text{ L}}}=\frac{{\displaystyle 0.59}}{{\displaystyle 1}}=0.59$$

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).

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

$$\begin{array}{c}{\text{gal/ft}}^3\text{ to }\frac{{\displaystyle \text{gal}}}{{\displaystyle {\text{ft}}^3}}\\ \text{psi to }\frac{{\displaystyle \text{lb}}}{{\displaystyle {\text{in.}}^2}}\end{array}$$

The same procedures are applied in the following examples.

$$\begin{array}{c}{\text{ft}}^3/\text{min becomes }\frac{{\displaystyle {\text{ft}}^2}}{{\displaystyle \text{min}}}\\ \text{s/min}\text{ becomes}\frac{{\displaystyle \text{s}}}{{\displaystyle \text{min}}}\end{array}$$

Basic Operation: Divide by a Fraction

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

$$\frac{\left(\frac{{\displaystyle \text{lb}}}{{\displaystyle \text{day}}}\right)}{\left(\frac{{\displaystyle \text{min}}}{{\displaystyle \text{day}}}\right)}\text{becomes }\frac{{\displaystyle \text{lb}}}{{\displaystyle \text{day}}}\times \frac{{\displaystyle \text{day}}}{{\displaystyle \text{min}}}$$

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

$$\frac{{\displaystyle {\text{mm}}^2}}{\left(\frac{{\displaystyle {\text{mm}}^2}}{{\displaystyle {\text{m}}^2}}\right)}{\text{becomes mm}}^2\times \frac{{\displaystyle {\text{m}}^2}}{{\displaystyle {\text{mm}}^2}}$$

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:

$$\frac{{\displaystyle {\text{ft}}^3}}{{\displaystyle {\text{gal/ft}}^3}}=\frac{{\displaystyle {\text{ft}}^3}}{\left(\frac{{\displaystyle \text{gal}}}{{\displaystyle {\text{ft}}^3}}\right)}$$

Multiply the numerator and denominator to obtain

$$\frac{{\displaystyle {\text{ft}}^6}}{{\displaystyle \text{gal}}}$$

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?

$${\text{ft}}^3\times \left({\text{gal/ft}}^3\right)={\text{ft}}^3\times \left(\frac{{\displaystyle \text{gal}}}{{\displaystyle {\text{ft}}^3}}\right)$$

Multiply the numerator and denominator to obtain

$$\frac{{\displaystyle {\text{ft}}^3\times \text{gal}}}{{\displaystyle {\text{ft}}^3}}$$

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:

$$\left(1400{\text{ ft}}^3\right)\times \left(7.48{\text{ gal/ft}}^3\right)=10,472\text{ gal}$$

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?

$$(70{\text{ ft}}^3/\text{s})\times \left(4.5\text{ ft/s}\right)$$

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

$$({\text{ft}}^2/\text{s})\times \left(\text{ft/s}\right)=\frac{{\displaystyle {\text{ft}}^3}}{{\displaystyle \text{s}}}\times \frac{{\displaystyle \text{ft}}}{{\displaystyle \text{s}}}$$

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

$$\frac{{\displaystyle {\text{ft}}^3\times \text{ft}}}{{\displaystyle \text{s}\times \text{s}}}=\frac{{\displaystyle {\text{ft}}^4}}{{\displaystyle {\text{s}}^2}}$$

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:

$$({\text{ft}}^3/\text{s})=\frac{{\displaystyle \left(\frac{{\text{ft}}^3}{\text{s}}\right)}}{{\displaystyle \left(\frac{\text{ft}}{\text{s}}\right)}}$$

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.

$$\frac{{\displaystyle 70{\text{ ft}}^3/\text{s}}}{{\displaystyle 4.5\text{ ft/}\text{s}}}=15.56{\text{ ft}}^2$$

 

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:

$$\left(\text{m/s}\right)\times \left({\text{m}}^2\right)=\frac{{\displaystyle \text{m}}}{{\displaystyle \text{s}}}\times {\text{m}}^2$$

Multiply the numerators and denominator of the fraction:

$$=\frac{{\displaystyle \text{m}}\times {\text{m}}^2}{{\displaystyle \text{s}}}=\frac{{\displaystyle {\text{m}}^3}}{{\displaystyle \text{s}}}$$

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:

$$5\text{ m/s}\times 7{\text{ m}}^2=35{\text{ m}}^3/\text{s}$$

 

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:

$$(2.3{\text{ ft}}^3/\text{s})\times \left(\text{7.48}\text{ gal/}{\text{ft}}^3\right)\times \left(60\text{ s/min}\right)$$

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:

$$(2.3{\text{ ft}}^3/\text{s})\times \left(\text{7.48}\text{ gal/}{\text{ft}}^3\right)\times \left(60\text{ s/min}\right)=1032.24\text{ gal/min}$$

 

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:

$$\frac{{\displaystyle 3.2\text{ million gal}}}{{\displaystyle 8\text{ hr}}}\times \frac{{\displaystyle 24\text{ hr}}}{{\displaystyle \text{day}}}=\frac{{\displaystyle 3.2\times 24}}{{\displaystyle 8}}\text{ MGD}=9.6\text{ MGD}$$

 

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

Solution:

$$1\text{ MGD}=\frac{{\displaystyle {10}^6}}{{\displaystyle 1\text{ day}}}=\frac{{\displaystyle {10}^6\text{ gal}\times 0.1337{\text{ ft}}^3/\text{gal}}}{{\displaystyle 1\text{ day}\times 86,400\text{ s/day}}}=\frac{{\displaystyle 133,700}}{{\displaystyle 86,400}}=1.547\text{ cfs}$$

 

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:

$$\begin{array}{c}\text{Weight of water}=6\text{ gal}\times 8.34\text{ lb/gal}=50.04\text{ lb}\\ \text{Total weight = 50.04}+4.6\text{ lb}=54.6\text{ lb}\end{array}$$

 

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:

$$10\text{ in}\mathrm{.}=10\times 2.54\text{ cm}=25.4\text{ cm}$$

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.

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

3.3 $$\text{TON}\text{ (threshold odor number)}=\frac{{\displaystyle V_T+V_P}}{{\displaystyle V_T}}$$

where

• V_T = Volume tested
• V_P = Volume of dilution with odor-free distilled water
• For V_P = 0, TON = 1 (lowest value possible)
• ForV_P = V_T, TON = 2
• For V_P = 2V_T, 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:

$$\text{TON}=\frac{{\displaystyle 200}}{{\displaystyle V_T}}=\frac{{\displaystyle 200\text{ mL}}}{{\displaystyle 50\text{ mL}}}=4$$

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.

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 (L_i) of the four sides (see Figure 3.1):

3.4 $$\text{Perimeter}=L_1+L_2+L_3+L_4$$

 

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

Solution:

$$P=35\text{ ft}+8\text{ ft}+35\text{ ft}+8\text{ ft}=86\text{ ft}$$

 

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

Solution:

$$P=(2\times \text{length})+(2\times \text{width})=(2\times 100\text{ ft})+(2\times 50\text{ ft)}=200\text{ ft}+100\text{ ft}=300\text{ ft}$$

Figure 3.1   Perimeter.

Figure 3.2   Perimeter of a rectangle for Example 3.40.

Figure 3.3   Diameter of circle.

 

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

Solution:

$$\begin{array}{c}P=(2\times \text{length})+(2\times \text{width})\\ =(2\times 8\text{ in}\mathrm{.})+(2\times 8\text{ in.})=16\text{ in.}+16\text{ in}\mathrm{.}=32\text{ in.}\end{array}$$

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 $$C=\pi \times D$$

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:

$$\begin{array}{c}C=\pi \times 25\text{ ft}\\ C=3.14\times 25\text{ ft}=78.5\text{ ft}\end{array}$$

 

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

Solution:

$$\begin{array}{c}C=\pi \times 18\text{ m}\\ C=3.14\times 25\text{ m}=56.52\text{ m}\end{array}$$

 

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

Solution:

$$\begin{array}{c}C=\pi \times 6\text{ ft}\\ C=3.14\times 6\text{ ft}=18.84\text{ ft}\end{array}$$

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

$$\text{Distance}\text{ (ft)}=\text{Total number of paces}\times \text{Length of average pace}$$

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 $$\text{Area}=L\times W$$

Figure 3.4   Rectangle.

Figure 3.5   Area of a rectangle for Example 3.46.

 

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

Solution:

$$\text{Area}=L\times W=14\text{ ft}\times 6\text{ ft}=84{\text{ ft}}^2$$

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 $$A=\pi \times r^2$$

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:

$$\text{Area of cicle}=\pi \times r^2=\pi \times 6^2=3.14\times 36=113{\text{ ft}}^2$$

Figure 3.6   Area of a circle for Example 3.47.

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 $$\begin{array}{c}A=\pi \times D\times H\\ A=\pi \times 20\text{ ft}\times 25\text{ ft}=3.14\times 20\text{ ft}\times 25\text{ ft}=1570{\text{ ft}}^2\end{array}$$

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 $$V=L\times W\times H$$

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:

$$V=L\times W\times D=15\text{ ft}\times 7\text{ ft}\times 9\text{ ft}=945{\text{ ft}}^3$$

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:
   $$D=3÷12=0.25\text{ ft}$$

   

   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:
   $$r=0.25\text{ ft}÷2=0.125$$

   
3. Find the volume (V):
   $$\begin{array}{c}V=L\times \pi \times r^2\\ V=300\text{ ft}\times 3.14\times 0.0156=14.72{\text{ ft}}^2\end{array}$$

   

 

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:

$$\begin{array}{c}V=H\times \pi \times r^2\\ V=96\text{ in.}\times \pi \times {\left(12\text{ in.}\right)}^2\\ V=96\text{ in}\mathrm{.}\times \left(144{\text{ in.}}^2\right)=43,407{\text{ ft}}^3\end{array}$$

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 $$\text{Volume}\text{ of cone}=\frac{{\displaystyle \pi }}{{\displaystyle 12}}\times \text{Diameter}\times \text{Diameter}\times \text{Height}$$

Note that

$$\frac{{\displaystyle \pi }}{{\displaystyle 12}}=\frac{{\displaystyle 3.14}}{{\displaystyle 12}}=0.262$$

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:

$$\text{Volume}\left({\text{ft}}^3\right)=0.262\times 120\text{ ft}\times 120\text{ ft}\times 6\text{ ft}=22,637{\text{ ft}}^3$$

Volume of Sphere

3.11 $$\text{Volume of sphere}=\frac{{\displaystyle \pi }}{{\displaystyle 6}}\times \text{Diameter}\times \text{Diameter}\times \text{Diameter}$$

Note that

$$\frac{{\displaystyle \pi }}{{\displaystyle 6}}=\frac{{\displaystyle 3.14}}{{\displaystyle 6}}=0.524$$

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:

$$\text{Volume }\left({\text{ft}}^3\right)=0.524\times 60\text{ ft}\times 60\text{ ft}\times 60\text{ ft}=113,184{\text{ ft}}^3$$

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:

$$\begin{array}{c}r=D÷2=20\text{ ft}÷2=10\text{ ft}\\ A=\pi \times r^2=\pi \times 10\text{ ft}\times 10\text{ ft}=314{\text{ ft}}^2\\ V=A\times H=314{\text{ ft}}^2\times 25\text{ ft}=7850{\text{ ft}}^3\times 7.48{\text{ gal/ft}}^3=58,718\text{ gal}\end{array}$$

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.

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

$$1{\text{ ft}}^2=12\text{ in}\mathrm{.}\times 12\text{ in.}=144{\text{ in.}}^2$$

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

$$\frac{{\displaystyle 62.4{\text{ lb/ft}}^2}}{{\displaystyle 1{\text{ ft}}^2}}=\frac{{\displaystyle 62.4{\text{ lb/ft}}^2}}{{\displaystyle 144{\text{ in.}}^2/{\text{ft}}^2}}=0.433{\text{ lb/in.}}^2\left(\text{psi}\right)$$

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.

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 $$\text{Pressure}\text{ (psi)}=0.433\times \text{Head}\left(\text{ft}\right)$$

3.13 $$\text{Head}\text{ (ft)}=2.31\times \text{Pressure}\text{ (psi)}$$

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 $$\text{Total head}=\text{Static head}+\text{Friction}\text{ head}+\text{Velocity head}$$

Static Head

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

3.15 $$\text{Static Head}=\text{Discharge elevation}-\text{Supply elevation}$$

 

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:

$$\text{Static Head}\text{ (ft)}=205\text{ ft}-108\text{ ft}=97\text{ ft}$$

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 $$\text{Friction head (ft)}=\text{Energy losses due to friction}$$

Velocity Head

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

3.17 $$\text{Velocity head (ft)}=\text{Energy losses to maintain velocity}$$

Total Dynamic Head (Total System Head)

3.18 $$\text{Total head}=\text{Static head}+\text{Friction head}+\text{Velocity head}$$

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 $$\text{Head (ft)}=\text{Pressure (psi)}\times 2.31\text{ ft/psi}$$

 

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

Solution:

$$\text{Head (ft)}=75.3\times 2.31\text{ ft/psi}=173.9\text{ ft}$$

Head and Pressure

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

3.20 $$\text{Pressure (psi)}=\frac{{\displaystyle \text{Head}\text{ (ft)}}}{{\displaystyle 2.31\text{ ft/psi}}}$$

 

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:

$$\text{Pressure (psi)}=\frac{{\displaystyle 15\text{ ft}}}{{\displaystyle 2.31\text{ ft/psi}}}=6.49\text{ psi}$$

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:
   $$\text{Weight}=\frac{{\displaystyle 62.4\text{ lb}}}{{\displaystyle 144{\text{ in.}}^2}}=0.433{\text{ lb/in.}}^2\text{ or 0.433}\text{ psi}$$

   
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:

$$\text{Pressure (psi)}=\frac{{\displaystyle \text{Head}\text{ (ft)}}}{{\displaystyle 2.31\text{ ft/}\text{psi}}}$$

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:

$$\frac{{\displaystyle 40\text{ psi}}}{{\displaystyle 1}}\times \frac{{\displaystyle \text{ft}}}{{\displaystyle 0.433\text{ psi}}}=92.4\text{ ft}$$

 

Problem: Convert 40 ft to psi.

Solution:

$$40\frac{{\displaystyle \text{ft}}}{{\displaystyle 1}}\times \frac{{\displaystyle 0.433\text{ psi}}}{{\displaystyle 1\text{ ft}}}=17.32\text{ psi}$$

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:

$$45\frac{{\displaystyle \text{psi}}}{{\displaystyle 1}}\times \frac{{\displaystyle 1\text{ ft}}}{{\displaystyle 0.433\text{ psi}}}=104\text{ ft}$$

 

Problem: Convert 15 psi to feet.

Solution:

$$15\frac{{\displaystyle \text{psi}}}{{\displaystyle 1}}\times \frac{{\displaystyle 1\text{ ft}}}{{\displaystyle 0.433\text{ psi}}}=34.6\text{ ft}$$

 

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:

$$125\frac{{\displaystyle \text{psi}}}{{\displaystyle 1}}\times \frac{{\displaystyle 1\text{ ft}}}{{\displaystyle 0.433\text{ psi}}}=288.7\text{ ft}$$

 

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

Solution:

$$\text{Pressure}\text{ (psi)}=0.433\times 5\text{ ft}=2.17\text{ psi}$$

 

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

Solution:

$$\text{Head (ft)}=2.31\times 12.2\text{ psi}=28.2\text{ ft}$$

 

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

Solution: Change miles to feet, then to psi:

$$\begin{array}{c}5280\text{ ft/mile}\times 4=21,120\text{ ft}\\ 21,120\text{ ft}÷2.31\text{ ft/psi}=9143\text{ psi}\end{array}$$

 

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:
   $$2,000,000\text{ gal}÷7.48=267,380{\text{ ft}}^3$$

   
2. Using volume, solve for depth:
   $$\begin{array}{c}\text{Vloume}=0.785\times D^2\times \text{Depth}\\ 267,380{\text{ ft}}^3=0.785\times {\left(150\right)}^2\times \text{Depth}\\ \text{Depth}=15.1\text{ ft}\end{array}$$

   

 

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:
   $$70\text{ psi}\times 2.31\text{ ft/psi}=161.7\text{ ft of water}$$

   
2. Convert psi to psf:
   $$70\text{ psi}\times 144{\text{ in.}}^2/{\text{ft}}^2=10,080\text{ psf}$$

   

 

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

Solution:

$$\begin{array}{c}\text{Head on pipe}=\text{Feet of pressure}\\ \text{Pressure}=\text{Weight}\times \text{height}\\ 6476\text{ psf}=62.4{\text{ lb/ft}}^3\times \text{height}\\ \text{Height}=104\text{ ft}\end{array}$$

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 (t – c) occurs in the characteristic polynomial of A. Basis for a subspace—A basis for subspace W is a set of vectors {v₁, …, v_k} in W such that
  1. {v₁, …, v_k} is linearly independent, and
  2. {v₁, …, v_k} 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(A – tI).
• 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 Ax – b.
• Linear combination of vectors—Vector v is a linear combination of the vectors v₁, …, v_k if there exist scalars a₁, …, a_k such that v = a₁v₁ + … + a_kv_k.
• Linear dependence relation for a set of vectors—A linear dependence relation for the set of vectors {v₁, …, v_k} is an equation of the form a₁v₁ + … + a_kv_k = 0, where the scalars a₁, …, a_k are zero.
• Linearly dependent set of vectors—The set of vectors {v₁, …, v_k} is linearly dependent if the equation a₁v1 + … + a_kv_k = 0 has a solution where not all the scalars a₁, …, a_k are zero (i.e., if {v₁, …, v_k} satisfies a linear dependence relation).
• Linearly independent set of vectors—The set of vectors {v₁, …, v_k} is linearly independent if the only solution to the equation a₁v₁ + … + a_kv_k = 0 is the solution where all the scalars a₁, …, a_k are zero (i.e., if {v₁, …, v_k} 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 = A^T.
• 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, …, v_k} is the subspace V consisting of all linear combinations of v₁, …, v_k. One also says that the subspace V is spanned by the set of vectors {v₁, …, v_k} 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 = A^T.

Quadratic Equations

$$\begin{array}{c}a{x}^2+bx+c=0\\ x=\frac{-{\displaystyle b}\pm \sqrt{b^2-4ac}}{{\displaystyle 2a}}\end{array}$$

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: 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 x² – x – 12 = 0.

Solution:

1. Factor the quadratic expression.
   $$(x-4)(x+3)=0$$

   
2. Set each factor equal to zero.
   $$\begin{array}{cc}x-4=0& x+3=0\end{array}$$

   
3. Solve the resulting linear equations.
   $$\begin{array}{cc}x=4& x=-3\end{array}$$

   

Thus, the roots are x = 4 and x = –3.

1. Check the solution in the original equation.
   $$\begin{array}{ccccc}\hfill {\begin{array}{c}\left(4\right)\end{array}}^2-4-12& =& 0& \hfill {\begin{array}{c}(-3)\end{array}}^2-12& \begin{array}{cc}\begin{array}{c}=\end{array}& 0\end{array}\\ \hfill 0& =& 0& \hfill 0& \begin{array}{cc}\begin{array}{c}=\end{array}& 0\end{array}\end{array}$$

   

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 $$x=\frac{{\displaystyle -b\pm \sqrt{b^2-4ac}}}{{\displaystyle 2a}}$$

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

Using the Quadratic Equation

 

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

$$x^2=5x+6=0$$

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

$$x=\frac{{\displaystyle -b\pm \sqrt{b^2-4ac}}}{{\displaystyle 2a}}$$

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:

$$\begin{array}{ccc}\hfill x& =& \frac{{\displaystyle -(-5)\pm \sqrt{{(-5)}^2-4\left(1\right)\left(6\right)}}}{{\displaystyle 2\left(1\right)}}\hfill \\ \hfill x& =& \frac{{\displaystyle 5\pm \sqrt{25-4}}}{{\displaystyle 2}}\hfill \\ \hfill x& =& \frac{{\displaystyle 5\pm 1}}{{\displaystyle 2}}\hfill \\ \hfill x& =& 3,2\hfill \end{array}$$

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

Trigonometric Ratios

$$\begin{array}{ccc}\text{sin }A=a/c& \text{cos }A=b/c& \text{tan}A=a/b\end{array}$$

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:

$$\begin{array}{ccc}\text{sin }Y& =\frac{{\displaystyle \text{Opposite leg}}}{{\displaystyle \text{Hypotenuse}}}=\frac{{\displaystyle 9}}{{\displaystyle 15}}& =0.60\\ \text{cos }Y& =\frac{{\displaystyle \text{Adjacent leg}}}{{\displaystyle \text{Hypotenuse}}}=\frac{{\displaystyle 12}}{{\displaystyle 15}}& =0.80\\ \text{tan }Y& =\frac{{\displaystyle \text{Opposite leg}}}{{\displaystyle \text{Adjacent leg}}}=\frac{{\displaystyle 9}}{{\displaystyle 12}}& =0.75\end{array}$$

Figure 3.12   Right triangle.

Figure 3.13   Illustration for Examples 3.71 and 3.72.

Figure 3.14   Illustration for Example 3.73.

 

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

Solution:

$$\text{sin}x=\frac{{\displaystyle \text{Opposite leg}}}{{\displaystyle \text{Hypotenuse}}}=\frac{{\displaystyle 12}}{{\displaystyle 15}}=0.8$$

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:

$$\sin C=2/5;\text{cos }C=4/5;\text{tan}C=3/4$$