Water by the Numbers - Practical Math for Water Operators

In this Section

In the picture above Rudy is using tools to construct a new water pipeline. Math can also be used as a tool to help water system operators, like Rudy. They help him perform a variety of tasks and answer questions. They also help him meet the challenges of operating today’s water storage, treatment and distribution systems. System operators are becoming more involved with work such as:

These things are very important in making sure that enough good quality water is available to the public at all times. Like using a hammer to drive nails when building a wood structure, various math tools can be used to: analyze certain conditions, make accurate predictions and solve a number of practical problems that the operators come across.

Section 1. Basic Arithmetic

Arithmetic can be described as the study of math using numbers. Arithmetic can be used to add, subtract, multiply or divide numbers together arriving at a final figure. These four actions are often referred to as the basic arithmetic functions.

Before proceeding, be certain that you are comfortable with the arithmetic functions. These can be performed by hand or with an electronic calculator. A calculator will be required to effectively complete the training. It is important to review the owner’s calculator instruction manual to fully understand its features. Calculators can differ in the way they’re used and function.

Mathematical symbols are used are used in math. Many of these are very common. You probably know the symbols for addition + or multiplication x, for example. Take a minute to ensure you recognize the following relevant math symbols and their corresponding meaning.

1.1 Basic Arithmetic Symbols and Functions

The basic arithmetic symbols and their meanings are listed below:

Symbol What is it?
+ Adding Sign. Often referred to as the ‘plus’ sign.
- Subtracting Sign. Often referred to as the ‘minus’ sign.
x * ()() ∙ Multiplication Signs. Often referred to as the ‘times’ sign. (See note 1.)
÷ / — Division Signs. (See note 1.)
= Equal Sign.
≠ Not Equal to.
( ) Parenthesis.
[ ] Square Brackets.
% Percent Sign - Out of 100.
32 Where 2 is an Exponent. The number of times a number is multiplied to itself (in this case, twice, or 3 times 3).
√9 Where √ is a Square Root. Working backwards, the answer multiplied to itself gives the number under the √ (in this case, √9 = 3, and, working backwards, 3 x 3 = 9).
< Inequality sign. Less Than.
> Inequality sign. Greater Than.
π Pi
∴ Therefore

Note 1: Be aware that more than one symbol can be used to denote multiplication and division. Regardless of which symbol is used, the mathematical action is always the same.

Examples of the basic arithmetic functions are given below. Please ensure you are able to perform these calculations.

Addition

Subtraction

Multiplication

Division

Exponents

Note: The superscript numbers in the examples below are called exponents: they indicate the how many times the number under them is multiplied to itself.

Square Roots

Note: The square root of any value is a number that, when multiplied by itself, provides that value.

Section 2. Imperial and Metric

Meet Peter as he does his work for the municipal waterworks department.

Peter keeps busy traveling from site to site. He looks over the jobs before the crews arrive to dig up and install new mains and services or repair leaking pipelines.

Most places where people are living have water, sewer and natural gas piping systems. Most lines used to be overhead on poles. In recent times, new developments often have to install electrical, telephone and cable television lines underground. They have them underground rather than being strung overhead between power poles. When they are overhead they can be damaged during wind storms or motor vehicle accidents. Also they do not obstruct scenic views. Specialty services such as oil and fuel pipe lines can be found underground as well. These can often be found near residential areas and industrialized developments.

There are advantages to installing utility lines underground and out of sight. However, it is very important that the different types of lines are kept separate from each other. It is also important to know where the lines are located. This is very important for operators to know when they are working. Having the information protects work crews from the dangers of digging into electrically charged or pressurized underground utility lines.

Peter works as an M-Scope Operator. One of Peter’s main jobs is to locate all of the underground utility lines and place paint markings on the road surface. He does this so when the crews are digging up the road at a later date, they know where the pipes are.

Locating (finding) all underground pipelines near the project before excavation is a mandatory (must do) requirement in most places operators work. This is so workers are protected and lines are not damaged while working. Peter takes extra care to make sure the lines are marked as accurately as possible.

Sometimes the places where the underground lines are located are very congested. Peter has to be accurate when locating these underground lines. He uses many different techniques and a variety of tools that help him locate lines. One tool he uses is called an electronic pipe locator.

One of the pieces of information that helps Peter the most is the original construction drawing. These drawings come from the utility companies. They also have recorded measurements that Peter uses.

Peter is responsible for transposing (taking from one place to another) the recorded information to the workplace. He utilizes several techniques (procedures or skills used in a specific task) to do this. Determining distances by using a ruler to scale measurements (a system of measurement based on a series of marks laid down at regular intervals and representing numerical values) from construction blueprints (a set of drawings used as a plan or guide) is sometimes necessary. This requires a good understanding of measurements and scales.

At work, Peter often has to convert (change) measurements from the Imperial (or British) System, often used in older drawings, to Metric units, used in our industry today. It is very important to be able to convert units from Imperial to Metric. Peter has to be able to compare information found on different drawings. For instance he has to calculate distances as accurately as possible so he knows how much pipe he will need to do the work.

2.1 Using Imperial and Metric

The following explains the Imperial and Metric Systems. The information and charts describe the procedure used to convert units of measure from one system to the other.

There are two measurement systems that are commonly used in the industry. They are the Imperial system and the International System of Units, generally referred to as the metric system.

The Imperial system uses inches, feet, yards, and miles for linear measurements. Weight is measured in pounds and ounces. Liquids are measured using pints and gallons. Imperial measurement is often used on job sites in the construction industry to order and measure building materials. For example, paint is ordered in gallons, pipe is ordered according to its diameter in inches, and loads are measured in pounds. Blueprints on many job sites use feet and inches. Therefore, it is important to be able to use these units of measurement and to be able to convert between units of measurement within the Imperial system. For example, you need to be able to convert between inches, feet, and yards.

The Imperial System uses many different units of measurement, for example, inches, feet, yards, bushel, ounce, pint, gallon, cup, teaspoon, and so on. Unfortunately, the names of units do not repeat and the number of units is not consistent – each one needs to be learned. For example, there are 12 inches in a foot. But there are 16 ounces in a pound. In technical training, abbreviations for feet and inches are: ft. and in. However, they are not usually followed by periods (for example: ft and in).

Some common units of measurement in the Imperial System are:
Linear Measure
12 in = 1 ft
36 in = 1 yd
3 ft = 1 yd
1,760 yd = 1 mile
5,280 ft = 1 mile		 Area Measure
144 in2 = 1 ft2
9 ft2 = 1 yd2
640 acres = 1 mi2
43 560 ft2 = 1 acre		 Volume Measure for Solids
27 ft3 = 1 yd3
1,728 in3 = 1 ft3
Volume Measure for Fluids
1 quart (qt) = 2 pints (pt)
1 Imperial gallon (gal) = 4 qt
Volume Measure Equivalents
1 Imperial gallon = 231 in3

The International System of Units (SI) or the metric system is common in most countries including Canada. Operators attending technical training primarily use the metric system. The metric system uses millimetres, centimetres, metres, and kilometres for linear measurements. Weight is measured in grams, for example, milligrams and kilograms.

Liquids are measured using litres. It is important to be comfortable using the metric system and to be able to convert between units of measurement within the metric system. For example, you need to be able to convert between millimetres and metres.

The International System of Units (SI) is often referred to as the metric system within the construction industry. It sets the standards for the metric system and provides exact measurements that have the same meaning everywhere in the world.

The metric system is based on 10. Calculations within the metric system are done with numbers such as 10, 100, and 1 000. Converting units within the metric system means moving the decimal point a specific number of places to the right or to the left.

Prefixes in the metric system are the same for measuring length, volume and weight, for example, kilometre, kilolitre, and kilogram. The prefixes are:

Tera T 1012 1 000 000 000 000
Giga G 109 1 000 000 000
Mega M 106 1 000 000
kilo k 103 1 000
hecto h 102 100
deka da 101 10
no prefix Base Unit 100 1
deci d 10-1 0.1
centi c 10-2 0.01
milli m 10-3 0.001
micro µ (mu) 10-6 0.000 001
nano n 10-9 0.000 000 001
pico p 10-12 0.000 000 000 001

Common metric operations:

Description Abbreviation Magnitude
Units of Length
metre m standard unit of length
centimetre cm 1/100 of a metre
millimetre mm 1/1000 of a metre
kilometre km 1000 metres
Units of Weight
gram g standard unit of weight
milligram mg 1/1000 of a gram
kilogram kg 1000 grams
Units of Area
square metre m2 or cu m standard unit of volume
cubic millimetres mm3 or cu mm 1/1 000 000 000 of a cubic metre
litres L 1/1000 of a cubic metre
megalitres ML 1 000 000 litres
Units of Flow
litres/second L/sec standard unit of flow
cubic metres/second m3/sec 1000 litres/second
litres/minute L/min 60 x L/sec = L/min
megalitres/day ML/day 86,400 x L/sec = L/day
Units of Pressure
pascal Pa standard unit of pressure
kilopascal kPa 1000 pascals

When you’re ready, click here to tryout the Imperial Quiz.
When you’re ready, click here to tryout the Metric Quiz.
When you’re ready, click here to tryout the Imperial to Metric Conversion Quiz.

2.2 Converting Units

The table below is used to demonstrate a quick method of converting from one unit to another within the metric system. When you convert from one unit to another, the way the measurement is expressed changes. In other words, the number and the prefix change, but the length, volume or weight of the object does not change.

thousands hundreds tens Base units tenths hundredths thousandths
kilometre (km) hectometre (hm) decametre (dam) metre (m) decimetre (dm) centimetre (cm) millimetre (mm)
kilolitre (kL) hectolitre (hL) decalitre (daL) litre (L) decilitre (dL) centilitre (cL) millilitre (mL)
kilogam (kg) hectogram (hg) decagram (dag) gram (g) decigram (dg) centigram (cg) milligram (mg)

Moving right to left 1 000 mm = 1 m

Move the decimal one place to the left for every cell you move in the table. You are moving from smaller units to bigger units.

Moving left to right 1 m = 1 000 mm

Move the decimal one place to the right for every cell you move in the table. You are moving from larger units to smaller units.

Example 1

A brick wall measures 450 000 mm in length. Convert 450 000 mm to metres. You are moving from small units to large units.

Steps:

Example 2

A piece of pipe measures 79.5 cm in length. Convert 79.5 cm to millimetres. You are moving from large units to small units.

Steps:

When you’re ready, click here to tryout the Metric Conversion Quiz.

2.3 Measuring Metric

Most workers in the water industry in Canada use the metric system. Millimetres are commonly used for measurements but can be awkward to calculate legths, areas and volumes because of the number of digits. Workers often convert millimetres to metres to make calculations and measure lengths.

Example

Length = 12,000 mm
Height = 3,000 mm

Area = L x W
= 12,000 mm x 3,000 mm
= 36,000 mm2

or

Area = L x W
= 12 m x 3 m
= 36 m2

It would be very easy to miss one of the zeros or to add one zero too many when working in mm. Converting milimetres to metres makes the calculation less awkward. The illustrations below represent segments of the metric ruler on the Numeracy Rules Pocket Guide.

When you’re ready, click here to tryout the Measuring Metric Quiz.

2.4 Ratios

Measurements come in a variety of units from 2 different systems. Converting units from one to the other can be achieved a number of ways. We have already looked at converting metric to imperial. Another way is to use ratios. Math techniques and ratios will be demonstrated in this section.

Ratios are an effective method when tables and calculators are not available. Ratios can be used to convert units within the same system. They can also be used to convert units between systems. Calculators and tables often only provide conversions between systems. Simple conversions between common units will be used to demonstrate this process using ratios.

The key is to set up 2 ratios that are in direct proportion with each other.

The first step is to set up a ratio between the equal amounts of the two units being converted. This means that when converting inches to feet, the first ratio is set up using 12" = 1ft. When converting millimetres to metres, the first ratio is set up using 1000 mm = 1 m.

More examples of this would be when converting acres to hectares the first ratio would use 2.47 acres = 1 hectare and converting litres to cubic metres would use 1000 L = 1 cu m etc.

The next step is to set up a second ratio between the amount being converted and the unknown amount. In setting up the second ratio, it is most important that both ratios are arranged in direct proportion to each other. This means ensuring that the same units in each ratio are in the same position, either above or below the line (i.e., the numerator and denominator).

Setting up these two ratios will be demonstrated in the following example.

Once these two ratios are set up, the last step in the process is to perform a simple calculation which will also be explained and demonstrated in the following example.

Example: How many inches are in 3 feet?

Problem: Convert 3 feet to the equivalent number of inches.

Solution:

The first ratio comes from the relationship between the units, 12 inches = 1 foot.

The second ratio comes from the amount being converted and the unknown quantity: 3 ft is the amount being converted and the equivalent amount of inches is the unknown.

Remembering to keep the same units above and below the line, the following direct proportions can be set up as ratios:

The first ratio is .  The second ratio is .

Notice the ratios are set up so both inch measurements are above the line and both foot measurements are below the line.

Once the ratios have been set up as equal proportions, the answer is then simply calculated by multiplying the two numbers on the diagonal together (circled below) and dividing by the third number. In this case, the 12 and the 3 are on the diagonal, and the third number is 1.

       

Practice Question

Problem: Convert 3.4 hectares to the equivalent number of acres.

Solution:

Step 1: set up a ratio between the equal amounts of the two units being converted.

 

Step 2: set up a ratio between the amount being converted and the unknown quantity.

 

Step 3: Set up equal proportions and multiply the two numbers on the diagonal together (circled below), then divide by the third number to arrive at the answer.

 

Answer: There are 8.2 acres in 3.4 hectares of land.

As previously mentioned, the above example and practice question deal with simplistic unit conversions; however, the technique of setting up ratios to convert units works for all different types of measuring units. The key is knowing the association between the units being converted to set up the first ratio.

Section 3. Math Rules and Principles

3.1 Order of Calculations

It is important that the arithmetic functions are performed in the right order. Unlike reading English, equations are not always solved by moving from left to right. The following acronym (a word made up from the initials of several words) is often used to describe the correct order the work is to be performed to solve the problem correctly:

BEDMAS is an acronym. It defines the correct order to perform the work. BEDMAS stands for:

If an equation involves a division line, all work is to be completed separately above the line and below the line before performing the final division. 

To better illustrate the importance of following the correct order when solving an arithmetic problem, notice that in the following example, different answers are arrived at when solving the problem in a different order:

Example:

 3.4 + (6 ÷ 2) x 42 = ?

          Wrong Order, left to right: 3.4 + 6 = 9.4
9.4 ÷ 2 = 4.7
4.7 x 4 = 18.8
18.82 = 353.44
 

         Correct Order, applying BEDMAS:

6 ÷ 2 = 3 solve brackets
     simplify 3.4 + 3 x 42
42 = 16 solve exponent
     simplify 3.4 + 3 x 16
3 x 16 = 48 solve multiplication
     simplify 3.4 + 48
3.4 + 48 = 51.4 solve addition

Solving the problem in the wrong order results in an answer of 353.44 rather than arriving at the correct answer of 51.4 when the proper order is followed.

Mandeep works in the maintenance shop of the waterworks department. He and his work group perform a variety of tasks. They analyze and repair suspect water meters. They answer customer inquiries about water usage. They also provide temporary hook ups during interruptions in the normal supply.

Conserving water is becoming more and more important. More cities are moving toward a user-based fee (charging users for the amount of water they use) for the supply of water. In the past they were charged a fixed, annual charge. Water users are now, often billed for the amount of water they use.

If there are big fluctuations (changes from high to low amounts used in an unpredictable way) in quarterly meter readings, Mani and his crew investigates. He and his co-workers visit the households or industries in question. They investigate what appears to be unusually high or low water consumption (amounts used). Sometimes they find the high consumption is caused by leaks in the pipes. It is very important that the municipal water meters are working properly. They must accurately register the quantities of water being used.

Mani is dispatched (sent out) to investigate these types of problems. He must trouble-shoot each situation to find out what is causing the problem. It may be that he finds the private piping is in satisfactory condition. He also finds that other plumbing considerations appear to be working properly. Mani may then suspect the meter to be faulty. If possible, he and his partner will then remove the meter, temporarily. He would bring it back to the shop for further testing. At the shop, the suspect meter is fitted on a test bench. The accuracy is tested. Before flowing water through the meter, Mani must be certain of the volume of water (how much) being used. This is easily achieved by installing a second test meter on the test bench in line with the suspect meter. The second meter will accurately record the amount of water flowing through both meters on the bench. It will used to check against the suspect meter.

Mani opens the valve and runs water through both meters for a period of time. He then records and compares the volume of water registered on both meters. The accuracy of the suspect meter is arrived at by:

Using actual figures:

After water flows through the meters for approximately 15 minutes, the suspect meter registers 325.3 litres. The test meter installed just downstream of the suspect meter shows a reading of 383.7 litres. What is the % accuracy of the suspect meter?

383.7 litres - 325.3 litres = 58.4 litres (the difference)

58.4 litres  ÷ 383.7 litres = 0.152

0.152 x 100 = 15.2 % difference

Often, with older worn meters, the meter can under-register. In this example the suspect meter is under-registering by 15.2%. It can therefore be said that the suspect meter is registering 84.8 % accurate. It should be noted that domestic meters are expected to be 98.5% accurate, or within 1.5% of the actual amount of water being consumed. The suspect meter Mani has been testing should be rebuilt. It might also salvaged as scrap metal.

Practice Question

Problem: How accurate is the suspect meter? Is it over or under registering?

Suspect meter reading: 4,367.3 litres
Test meter reading: 4,212.7 litres

Solution:

In this case the suspect meter is over registering by 3.7% and described as being 103.7% accurate.

The testing procedure we just described only works where the meter can be removed from the building easily. When a meter cannot be removed Mani and his partner have to use a different approach.

The first thing Mani checks is that no water is running within the facility. This includes all faucets, faulty toilet flush valves, automatic sprinklers, outside taps or any other fixture within the house and outside yard. Leak detection equipment can pick up the sound of running water. This will verify that water has stopped flowing through the meter and into the building.

Mani records the first reading on the meter register. He and his partner then open a water valve. They fill the pails with water. The volume of water in each pail can be calculated. It can then be compared to the amount the meter registered on the meter. The % difference is calculated the same way. This time the calculated amount of water in the pails is used to determine the actual volume (amount) consumed.

To calculate the volume of water in the pail, Mani first must measure the diameter of the pail and height of the water.

If the pail being used has a diameter of 400 mm and was filled to 432 mm, he calculates the amount of water in each pail by multiplying the area of the base by the height of the water in the pail, for each pail of water filled over a period of time.

To first calculate the area of the base we recognize that the base of the pail is circular. The area of a circle is calculated as follows:

where A is area of the base, d is the diameter and π = 3.14

As mentioned, by multiplying the area of the base by the height of the water in the pail, the amount of water can be calculated as follows:

(area of the base) x h (height of the water in the pail)

By inserting the actual measurements of the pail into the formula,

diameter of the pail = 400 mm or 0.4 m
height of the water = 432 mm or 0.432 m

The amount of water in the pail is calculated as follows:

         

If Mani fills 43 pails of water over a period of time, the total amount of water that actually passes through the meter is calculated by multiplying the amount of water in one pail by 43 pails:

Total amount of water = 0.05 m3 per pail x 43 pails

Total amount of water = 2.15 m3

Once Mani completes this calculation, it could be that he must convert units to compare the actual volume he’s calculated to the amount registered on the meter. For example, if the meter is recording in litres and Mani has calculated the volume in cubic meters (as done so above), then he must convert the two amounts to the same units in order to compare the numbers.

Because there are 1000 litres per cubic meter, cubic meters can be multiplied by 1000 to convert the equivalent amount to litres and on the other hand, litres can be divided by 1000 to convert the figure to cubic meters.

2.15 m3 x 1000 litres/cu metre = 2,150 litres

If the meter being tested registers 2,003 litres, the percent difference is

By performing these calculations, Mani is certain that the metre is reading 93% accurate and is under-registering by 7%, demonstrating that the metre should be replaced.

Practice Question

In order to determine the accuracy of a metre, the crew fills 23 cylindrical drums all of the same size with water. The water meter registers this consumption as being 19,941 litres. The base of the drums has a diameter of 450 mm and is filled with water to exactly 550 mm deep in each case. How much water in total was contained in the drums and how accurate is the water meter?

Problem: What is the volume of water contained in the drums? How accurate is the water meter?

Solution: Converting the dimensions of the water in the drum from millimetres to metres, the volume of each drum is calculated as follows:

Volume of a cylinder = πr2h
= 3.14 x 0.2252  x 0.55
= 3.14 x 0.051 x 0.55
= 0.088 m3 = 880 L

880 L of water per drum x 23 drums = 20,240 L of water in total were contained.

The percent difference between the actual amount and the meter reading is:

           

It can be concluded that the meter in this example is reading 98.5% accurate and under-registering by 1.5%. This is within the acceptable range of accuracy as stipulated by industry standards. The meter is therefore performing satisfactorily and requires no additional work to be performed.

Supplying pure, good tasting water requires a high standard of care. Water systems have to be properly operated and maintained. In Canada, the federal government sets standards to which drinking water must conform. Controlling the quality of the water we drink is a combined effort. It depends on a variety of agencies. These include provincial governments, local health authorities, and analytical laboratories. It also includes the regions and municipalities that treat, test and distribute the water to our homes and businesses.

Giuseppe works in the disinfection section of the regional waterworks department. His work group is responsible for the disinfection of newly installed pipelines. Occasionally they also clean the reservoirs to ensure the stored water is of good quality. They need to be sure it is ready to be used during periods of high demand. Those living in the area rely on Giuseppe and his crew for a clean, fresh and abundant supply of water at all times.

It is most important that after new sections of pipeline are built, they are clean before being put into service. To do this the water lines are thoroughly flushed with water. This is done at a high enough velocity to scour any construction debris from within the pipeline. Often referred to as the “first flush”, Giuseppe sends out his crew with the required equipment to do this work. In order to be effective, scouring velocities are recommended to be maintained at no less than 1.5 metres per second.

Giuseppe begins by connecting fire hoses from a fire hydrant to the new section of pipe. The quantity of water being used is captured by a flow meter installed at the discharge port of the hydrant.

Once this apparatus is in place, Giuseppe slowly opens the hydrant. He shuts off valve starting the flow of water through the line. During this procedure one of the operators must be sure to open the valve so enough water is flowing so that scouring velocities are reached.

He does this by calculating the exact amount of water flow required through the given sized pipe to maintain a velocity of at least 1.5 m/sec. Once he knows how much water is required, Giuseppe adjusts the hydrant valve until the desired rate of flow is displayed on the meter attached to the hydrant port.

In the section on Basic Hydraulics we learnt about force and pressure. Here there must be enough water pressure going through the pipes to make sure they are scoured (thoroughly cleaned).

Flows are calculated by multiplying the area of the pipe by the velocity of the water traveling through the pipe:

Flow = cross sectional area of the pipe x velocity of water.

Q = A∙v where Q is flow, A is area of the pipe and v is velocity of the water. Acknowledging that the shape of a pipe is circular, again the area of a circle is calculated:

A is area, d is diameter, π is 3.14

For any given velocity that is desired the amount of water (flow) required to achieve that velocity, can be calculated by applying the above formula.

By first calculating the area and then multiplying by the desired velocity, the required flow through the pipe can be determined.

For this particular application, Giuseppe is responsible to achieve velocities of 1.5 m/sec by flushing enough water through a 250 mm diameter pipe.

Desired scouring velocity: 1.5 m/sec

Pipe diameter: 250 mm or 0.25 m

Solution:

Calculate area of pipe:

Again, for circles area is calculated where A is area, d is diameter, π is 3.14

       

Calculate flow rate:

Q = A∙v
Q = 0.049 m2 x 1.5 m/sec
Q = 0.074 m3/sec or 74 L/sec

Giuseppe must ensure the hydrant is flowing at 74 L/sec if he is to achieve a velocity of 1.5 m/sec to effectively scour the newly installed pipeline.

Practice Question:

A 240 m length of newly installed 150 mm diameter water main must be flushed. What is the rate of flow to maintain a desired scouring velocity of 2 m/sec?

Problem: What is the flow rate given a desired scouring velocity of 2.0 m/sec and a pipe diameter of 150mm or 0.15 m?

Solution:

Calculate area of pipe:

       

Calculate flow rate:

Q = A∙v
Q = 0.018 m2 x 2 m/sec
Q = 0.036 m3/sec or 36 L/sec

The hydrant must flow at 36 L/sec to achieve a scouring velocity of 2 m/sec.

Once all of the noticeable debris is flushed from the pipe, the next and most important task Giuseppe must perform is to disinfect the inside of the line. This is done before it is put into service. The line is filled with a concentrated chemical solution (usually chlorine, supplied in the form of sodium hypochlorite or more commonly known as bleach). It is left for a period of 24 hours. In this case, municipal engineering standards call for a concentration of 25 mg/l to be applied.

Giuseppe knows the central stores at work carry bleach in 1-litre jugs. The bleach is highly concentrated with 5% chemical solution or 50 g of chemical per litre (50,000 mg/L). His job is to bring enough bleach to the job site to fill the section of pipe being disinfected with a quantity of solution diluted to 25 mg of chemical per litre of water (mg/L). To do this, he must first calculate how much solution is required to fill the pipe (i.e., the volume of the pipe).

The pipeline is 250 mm diameter and 67 m long:

The total amount of solution Giuseppe needs to fill the pipe is 3.29 m3 or 3,290 litres. If he is required to dose the pipe at a concentration of 25 mg chemical/litre, then 25mg/L x 3,290 litres = 82,250 mg of chemical is needed in total to adequately disinfect the pipeline.

Giusepe knows that the bleach kept in central stores is 5% concentration and contains 50 g or 50,000 mg of chemical per litre.

The amount of bleach required can be calculated by dividing the total amount of chemical required (82,250 mg) by the amount of chemical in 1 litre of bleach (50,000 mg).

82,250 mg of chemical desired ÷ 50,000 mg/litre of bleach = 1.65 litres of bleach.

For this job, Giuseppe acquires 2 (two) 1-litre jugs of 5% bleach from central stores.

Practice Question

How much 10% bleach is required in order to properly disinfect a newly-installed large transmission pipeline 3 m in diameter that is 325 m long at a dosage concentration of 50 mg of chemical/litre.

Problem: How much 10% bleach is required to disinfect the above pipeline?

Solution: First calculate how much water is in the pipe:

Pipe length: 325 m

Pipe diameter: 2 m

Required Formula:

The total amount of solution required to fill the pipe is 1,020.5 m3.

The required concentration is 50 mg of chemical/litre or 50 g/cubic metre: i.e., 50 mg/L x 1000 L/m3 x 1 g/1000 mg = 50 g/m3

If a concentration of 50 g/m3 is required to effectively disinfect the pipeline and the pipe holds 1,020.5 m3 of solution then:

50 g/m3 x 1,020.5 m3 = 51,025 g of chemical is needed in total to adequately disinfect the pipeline.

If 10% bleach contains 100 g of chemical per litre, then the total amount of bleach required in order to provide a desired amount of chemical (51,025 g) is calculated by dividing the desired amount of chemical required by the amount of chemical in 1 litre of bleach:

51,025 g of chemical desired ÷ 100 g/L of bleach = 510 L or 0.51 m3 of bleach.

For large dosages, as is the case in this example, bleach is supplied in 20 L cardboard containers called carboys.

510 L of bleach ÷ 20 L/carboy = 25.5 or 26 carboys are required

This example is based on an actual project pictured below.

Reservoirs are drained for periodic (regular) inspections and preventative maintenance. They are also drained to do repairs to the steel or concrete structures. After being drained, and before being put back into service, they need to be disinfected. Giuseppe is involved with disinfecting water reservoirs.

Giuseppe and his crew first thoroughly scrub and wash down the inner walls of the reservoir. Once this is done, the crew sprays the walls with a chemical solution. The solution is similar to what is used to disinfect new water mains.

The degree of chemical concentration depends on the amount of time Giuseppe needs to let the chemical stand on the walls. He must leave it for enough time to get rid of any bacteria before he puts the reservoir back into service. The American Waterworks Association (AWWA) Standards indicate the degree of concentration of the solution for a given contact time.

Sometimes an alternate water supply is not available while the reservoir is taken out of service. This work must then be completed in a short time period. It would normally be done at night during periods of low demand.

Giuseppe knows that the shorter the time period, the more concentrated the chemical solution must be to achieve the desired results. He is often required to dilute industrial bleach concentrated at 50 g (or 50,000 mg) of chemical per litre to amounts between 100 mg to 300 mg per litre. Giuseppe does this by calculating just the right amount of water to mix into the bleach to achieve the desired concentration as follows:

In order to dilute the bleach from 50,000 mg/L to 250 mg/L for example, he calculates the amount of water to add. He divides the bleach concentration by the desired concentration of the diluted solution. This calculation determines how many times more concentrated the bleach is than the end product.

50,000 mg/L ÷ 250 mg/L = 200 times more concentrated.

In this case, Giuseppe adds about 200 times more water to the amount of bleach to achieve the desired results.

By adding 199 litres of water to 1 litre of bleach (199 L of water + 1 L of bleach = 200 L total solution) he ends up with a final solution which is exactly 50,000 mg of chemical per 200 litres of water, which is equivalent to 250 mg/L (50,000 mg/200 L = 250mg/l)

Perhaps Giuseppe doesn’t need 200 litres of solution to complete the job. If he doesn’t, he can lower the amount of bleach and water used proportionately (having the correct amount in relationship of size or quantity):

If he needs only half of the amount, he would use half the amount of bleach (0.5 litres) and half the amount of water (99.5 litres). This way the same proportion is maintained. He then has a concentration of 25,000 mg of chemical per100 litres of water. This is equivalent to 250 mg/L (25,000 mg/100L = 250 mg/L).

The same approach would apply if he only needs one quarter of the starting amount, or any other fraction of the starting amount. The key point here is using the same proportions. Giuseppe may use lesser or greater amounts of bleach and water to mix his solution. However, he does so in the same proportion. Then the concentration remains the same.

Practice Question

The crew has only 16 hours to disinfect a reservoir. They must apply a concentration of 300 mg/L of chemical to the walls of the reservoir.

How much water must be added to 0.25 litres of 5% bleach to obtain the desired concentration of solution?

Problem: What is the amount of water to be added to 0.25 litres of 5% bleach to end up with a concentration of 300 mg/L mixture?

Solution: The amount of water to add to dilute 0.25 L of bleach to a concentration of 300 mg/L is as follows:

First calculate how many times more concentrated the bleach is than the end product.

50,000 mg/L ÷ 300 mg/L = 167 times more concentrated, therefore 166 litres of water must be added to 1 litre of bleach to end up with a final solution concentrated to 300 mg.

50,000 mg of chemical /167 L of water = 300 mg/L

If only 0.25 L of bleach is being used, which is one quarter of the amount of bleach used above, then only one quarter of the amount of water used above would be required to arrive at the same concentration.

167 L ÷ 4 = 41.75 L of total solution, therefore, adding 41.50 litres of water to 0.25 litres of bleach = 41.75 litres total solution.

0.25 L of 5% bleach provides 12,500 mg of chemical/41.75 L of solution = 300mg/L concentrated solution

Gord manages the Construction Section of the Waterworks Branch. He is the Superintendent. He oversees (watches over) the daily operations of this area. He is responsible for completing the various projects identified in the annual pipe rehabilitation (treatment) program.

     Today Gord has to review a set of construction drawings. He has to provide a cost estimate to the design department for
     supplying water to a new condominium development.

     He starts by gathering actual labour and equipment cost data (figures) of recently completed jobs. He gets these from a work
     order management system. Gord starts by looking at the hourly rates for labour and equipment. He also looks at material and
     other related work costs.

     Gord’s construction crew consists of 8 employees. They make an average of $37/hr per person including fringe benefits

      Labour costs for the crew per hour: 8 employees x $37/hr = $296/crew-hour

      Gord plans to perform this job using a backhoe and a skid steer loader. He will also use 2 tandem axel dump trucks to haul away
     the excavated material.

 Equipment costs: backhoe $32/hr
skid steer loader $19/hr
tandem axel dump trucks $23/hr each

The project calls for 200 mm diameter pipe at $13 per metre and imported gravel as backfill. He purchases this at $150 m3 per truckload (a total of 10 m3) delivered.

Gord visits the site. He also has past experience on similar projects. He estimates the crew can install 40 m of pipe per day. They will use approximately 45 to 50 m3 of gravel to backfill the trench.

Gord refers to the design drawings. He carefully measures the length of the water main. He notes any special details. These might be installation of a service line to the property, a new hydrant or the construction of a valve chamber at the end of the line. These pieces will add to the project costs. He estimates these as lump sums separately from the water mainline portion of the work.

Gord notes that a total of 123 metres of water main is to be installed and at a rate of 40 m of pipe per day. The total time to install the main would be:

123 m ÷ 40 m of pipe per day = 3.1 days to install the pipe, making it 3.5 days to completion including site clean up time.

He also determines that at 3.1 days to finish the water main x 50 m3 of gravel used per day = 155 m3 of gravel will be required to backfill the trench.

He then estimates the lump sum costs to install:

He continues his estimate by combining the costs of the same unit together. He applies those figures to the quantities required to complete the work:

Labour and Equipment
8 person crew $296/hr
backhoe $32/hr
skid steer loader $19/hr
dump trucks $23/hr each x 2 = $46/hr

Labour and equipment unit cost = $393/hr

3.5 days is required to complete the project
3.5 days x 8 hrs/day x $393/hr = $11,004
or approximately $11,000 for the labour and equipment components.

Backfill Material

backfill gravel used $50 m3/day
delivered at $150/truckload (10 m3)

Gravel unit cost

$15/m3

155 m3 of gravel is required in total is required to backfill the trench.
155 m3 of gravel x $15/m3 = $2,325
or approximately $2,500 total for backfill material.

Other equipment needed for the work:

Gord finally adds the cost components together. He adds a small amount for unexpected extras. He arrives at the final estimate for the project:

Labour and Equipment $11,000
Backfill Material $2,500
Hydrant $5,500
Service Pipe $3,300
Valve Chamber $6,500
Subtotal $28,800
10% contingency $2,880
Total Project Cost $31,680

We have now reviewed a few examples of the type of work typically performed in the waterworks industry by operators.

We hope you have developed an appreciation for the variety of math applications that can be very helpful in fulfilling our job duties and obligations.

Applying math in a practical way helps operators and supervisors complete work assignments. They help us work in a more professional and efficient manner. It is much like selecting just the right tool from our tool-belt or the back of the service truck to help to get the job done to the best of our ability.