tag:blogger.com,1999:blog-90064767581461993732018-03-06T17:44:18.624-08:00Brad Williams' Math Tutorial for Introductory-Level Water Treatment Facility Operator ExaminationsThis is the only one hundred percent free and one hundred percent complete math tutorial for introductory-level water treatment facility operator students that is available on the internet. To view all of the postings on this site, click on "older posts" at the bottom of each page. A companion site with several introductory-level math practice tests can be accessed at http://water-treatment-math-practice-tests.blogspot.com.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comBlogger22125tag:blogger.com,1999:blog-9006476758146199373.post-6644705075357051292009-06-07T18:12:00.003-07:002015-12-02T21:24:45.900-08:00Introduction to this SiteThis site contains a complete, thorough, and concise review of the mathematical information needed to pass the math portion of introductory-level water treatment operation exams. No information contained on this site is useless or unimportant. Anyone who learns all of the information on this site will be able to answer most or all of the mathematical questions on the mathematical portion of his or her water treatment certification exam. <br /> Please read each post carefully. As you finish studying each page, click on <em>older posts</em> at the bottom of the page to reach the posts on the next page. As you study this site, always remember this: Water treatment facility operator certification board members throughout the United States repeatedly say that almost no one fails an introductory-level water treatment facility operator certification exam if he or she performs well on the math portion of the exam. Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-82729048649612366152009-06-07T18:11:00.004-07:002010-02-28T12:20:09.802-08:00Lesson One: Converting Fractions & Percents to DecimalsFractions are always written with a top number that is separated from a bottom number by a line. This line simply means "divided by." To convert a fraction into a decimal, use your calculator to divide the bottom number into the top number.<br /><br />PULL OUT YOUR CALCULATOR AND CONVERT THESE FRACTIONS TO DECIMAL FORM.<br />3/4 (Answer: 3 divided by 4 = .75)<br />5/8 (Answer: 5 divided by 8 = .625)<br />1/2 (Answer: 1 divided by 2 = .50)<br />3/16 (Answer: 3 divided by 16 = .1875)<br />1/8 (Answer: 1 divided by 8 = .125)<br /><br />Percentages, when expressed with the % symbol, are converted to decimal form by dropping the % symbol and multiplying the remaining number by .01.<br /><br />Examples:<br />5% = 5 x .01 = .05<br />23% = 23 x .01 = .23<br />50% = 50 x .01 = .50<br />70% = 70 x .01 = .70Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-40430720488317547412009-06-07T18:11:00.003-07:002009-09-18T19:53:50.596-07:00Lesson Two: Calculation of Area<span style="font-size:0;"></span><span style="font-size:0;"></span>Calculation of area is the most basic and most important concept that must be understood in order to perform introductory-level water treatment math. Area calculations form the single building block upon which four of the six most important introductory-level math formulas are based. Area calulations are needed for such operations as calculating flow through filters, determining filter backwash rates, and calculating detention times in flocculation tanks and sedimentation basins.<br />On all introductory-level water treatment examinations in the United States, area is expressed in terms of inches and feet. The two key words, which indicate area and which indicate the scope of the area, are <em>square</em> and <em>cubic</em>. When area is expressed in inches, it will be designated as <em>square inches</em> or <em>cubic inches</em>. When area is expressed in feet, it will be designated as<em> square feet</em> or <em>cubic feet</em>.<br />A square measurement of area is always flat. To remember that the term <em>square</em> means <em>flat</em>, think of ordinary household carpeting, which is flat and which is usually measured in "square feet". A square foot is always 1 foot by 1 foot square. A square inch is always 1 inch by 1 inch square.<br />A cubic inch or cubic foot is always shaped like a perfect cube with the width, length, and height being exactly equal. A cubic inch is a cube that is 1 inch by 1 inch by 1 inch. A cubic foot is a cube that is 1 foot by 1 foot by 1 foot.<br />To obtain the square inches or square feet within a flat square or flat rectangular area, multiply the length of one side by the length of the other side. To obtain the cubic inches or cubic feet within a three-dimensional square or three-dimensional rectangular area, multiply the width of the area by the length of the area by the depth of the area. Below are several practice problems to illustrate this procedure.<br /><br />1. A mixed media filter is 10 feet wide by 15 feet long. How many square feet are in the surface area of the filter? Answer: 10 feet x 15 feet = 150 square feet.<br /><br />2. A sedimentation basin is 12 feet wide by 60 feet long by 30 feet deep. How many cubic feet of space is in the basin? Answer: 12 feet x 60 feet x 30 feet = 21,600 cubic feet<br /><br />3. How many square inches are in a rectangle that is 3 feet by 5 inches? Answer: 36 inches x 5 inches = 180 square inches (Don't forget to convert the feet to inches.)<br /><br />In addition to calculating square and cubic areas in straight-sided figures, introductory-level water treatment students must be able to calculate the area within a circle. Since a circle is always flat, versus a sphere which is three-dimensional, the area within a circle will always be expressed in the units that are used to express flat area - namely, SQUARE inches and SQUARE feet. Introductory-level water treatment students do not need to know the formula for calculating cubic inches and cublic feet within sphere-shaped objects.<br />The very clever formula for calculating area within a circle was devised over two thousand years ago by one of the greatest mathematicians in the ancient world. The simplified version of this formula, which all water treatment schools in the United States teach, is "D2 x .785". D2 is the diameter of the circle squared. To square a number, just multiply the number times itself.<br />The diameter of a circle might best be described as the full width of the circle. The radius of a circle is one-half of the full diameter, much like the spoke on a bicycle wheel is one-half the full width of the circular rim. Sometimes, water treatment math problems only give the length of a circle's radius. To obtain the circle's diameter, just multiply the radius by two.<br />There are three primary applications on introductory-level water treatment exams that require the use of the formula to calculate area within a circle. The first application involves the calculation of area within a circular basin, such as a solids-contact basin. The second application involves the calculation of area within a structure that has a round opening, such as a tube, hose, or pipe. The third application involves the calculation of area within a circular bulk storage tank or a circular day tank.<br />Below are two practice problems involving the area within a circle. Study these examples carefully.<br /><br />1. A large circular basin is 32 feet across and 50 feet deep. How many cubic feet of water does the basin hold?<br />ANSWER: The first step is to find the area in square feet for the flat circle that forms the top surface of the basin. To do this, square the 32-foot diameter and multiply the answer by .785. (32 x 32 = 1024 x .785 = 803.84 square feet.) Next, multiply this answer by the depth of the basin to obtain the three-dimensional cubic area within the basin. 803.84 x 50 = 40,192 cubic feet.<br /><br />2. How many cubic inches of water does a 100-foot length of 8-inch pipe hold?<br />ANSWER: First, find the flat area that forms the surface opening of the 8-inch pipe. (8 x 8 = 64 x .785 = 50.24 square inches.) Next, because inches can only be multipled by inches, convert the 100-foot length to inches. (100 x 12 inches = 1200 inches.) Now, multiply the flat surface area by the three-dimensional length. (50.24 square inches x 1200 inches of depth = 60,288 cubic inches.) The 100-foot segment of 8-inch pipe in the question will hold 60,288 cubic inches of water.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-26468317879632089792009-06-07T18:10:00.004-07:002009-07-13T20:13:17.368-07:00Lesson Three: Learn These Two Constants NowLearn these two constants now. Repeat them in your mind several times a day. Learn to recite them with the same ease in which you recite your telephone number.<br /><br />1. There are 8.34 pounds of water in one gallon of water.<br /><br />2. There are 7.48 gallons of water in one cubic foot of water.<br /><br />If you have trouble remembering numbers, think of them as time. Think of 8.34 as 8:34 pm or 7.48 as 7:48 am.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-31982313854496457202009-06-07T18:10:00.003-07:002009-11-15T20:16:44.549-08:00Lesson Four: Million-Gallon NotationOn introductory-level water treatment exams, flow is expressed as "number of gallons within a given frame of time." The four primary expressions of flow on introductory-level exams are <em>gallons per minute </em>(gpm), <em>gallons per hour </em>(gph), <em>gallons per day </em>(gpd), and <em>million-gallons per day </em>(MGD) .<br />When flow is expressed as gpm, gph, or gpd, the number before the gpm, gph, or gpd is written without any type of abbreviated notation. As examples, one thousand gallons per minute is written as "1,000 gpm", one hundred thousand gallons per hour is written as "100,000 gph", and ten million gallons per day is written as "10,000,000 gpd".<br />When flow is expressed as MGD, the full six-digit million number is never written before the MDG. A flow of ten million gallons per day, correctly written as 10,000,000 gpd, is not correctly written as 10,000,000 MDG. Ten million gallons per day, written in the MGD form, is "10 MDG".<br />Before beginning the practice portion of this post, it is important to understand the difference between MGD and MG. Both expressions are written with the same type of abbreviated notation, but MGD contains a timeframe (per day), which denotes flow. The term "MG", or <em>million-gallons,</em> has no timeframe attached to it. The term "MG" can express flow, under the right conditions, but it can also be used to denote capacity or quantity, such as a 2.5 MG clearwell or a 1.25 MG elevated tank.<br />For introductory-level water treatment operators, writing one million gallons per day as 1 MGD is easy. Writing ten and one-half million gallons per day as 10.5 MGD is also easy. But many introductory-level water treatment operators start to experience a problem when they have to write less than one million gallons per day in the MGD form.<br />To convert gallons into million-gallon notation, multiply the gallons by .000001 and add the abbreviation MGD or MG, depending on the context in which the million-gallon notation is used. The figure .000001 is a decimal point that is followed by five zeros and a 1. Below are several examples to illustrate how easily gallons can be converted into the million-gallon form.<br /><br />PULL OUT YOUR CALCULATOR AND TRY EACH OF THESE PROBLEMS FOR YOURSELF. YOU MUST KNOW THIS INFORMATION IN ORDER TO ANSWER SEVERAL QUESTIONS ON YOUR EXAM. THESE EXAMPLES ARE PROBABLY MORE COMPLICATED THAN THE QUESTIONS ON THE EXAM, BUT IF YOU CAN PERFORM THESE PROBLEMS, YOU'LL HAVE NO DIFFICULTY AT ALL WITH ANY OF THE MILLION-GALLON CONVERSIONS ON YOUR EXAM.<br /><br />3,250,000 gallons x .000001 = 3.25 MG<br /><br />45,000 gallons x .000001 = .045 MG<br /><br />999,999 gallons x .000001 = .999999 MG<br />(This is actually one gallon away from being 1 MG.)<br /><br />1 gallon x .000001 = .000001 MG<br />(This is 999,999 gallons away from being 1 MG.)<br /><br />655,000 gallons x .000001 = .655 MG<br /><br />351,200 gallons x .000001 = .3512 MG<br /><br />17,600 gallons x .000001 = .0176 MG<br /><br />18,379,000 gallons x .000001 = 18.379 MG<br /><br />To check and verify that your calculation is correct, multiply your MG figure by 1,000,000. You should get the original figure with which you started. You will need to use this procedure also if you are asked to convert MG or MGD into gallons.<br /><br />3.25 MG x 1,000,000 = 3,250,000 gallons<br /><br />.045 MG x 1,000,000 = 45,000 gallons<br /><br />.999999 MG x 1,000,000 = 999,999 gallons<br /><br />.000001 MG x 1,000,000 = 1 gallon<br /><br />.655 MG x 1,000,000 = 655,000 gallons<br /><br />.3512 MG x 1,000,000 = 351,200 gallons<br /><br />.0176 MG x 1,000,000 = 17,600 gallons<br /><br />18.379 MG x 1,000,000 = 18,379,000 gallons<br /><br />VERY IMPORTANT NOTE: Whenever a test question or math problem requires an answer that involves a large number of gallons, be sure to ascertain if the answer should be in gallons or in million-gallons. Frequently, introductory-level math questions will ask "how many million gallons" are involved in a certain situation. But test questions can also ask "how many gallons" are involved in a situation. Always assess if questions involving large numbers of gallons are asking for the answer to be in gallons or in million gallons.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-48877517349471915132009-06-07T18:10:00.001-07:002009-07-18T18:44:48.934-07:00Lesson Five: Rounding Off NumbersVery often, the multiple-choice answers on the math portion of water treatment facility operator certification exams are supplied in a "rounded off" form. For this reason, the answers on certification tests may be very close, but not identical, to the answers that appear on your calculator screen. When selecting a multiple-choice answer on your certification exam, choose the multiple-choice answer that is closest to the figure on your calculator screen.<br /><br />EXAMPLES OF ROUNDED-OFF NUMBERS:<br /><br />310,020 gallons might be written as 310,000 gallons.<br /><br />1206 cubic feet might be written as 1200 cubic feet.<br /><br />65.34 mg/l might be written as 65 mg/l.<br /><br />8.247 MGD might be written as 8.2 or 8.25 MGD.<br /><br />Note: Math questions and answers on introductory-level water treatment operator certification exams are not intended to be particularly tricky. There will generally be enough difference in the multiple-choice answers on these exams, so there will be no doubt as to which answer is correct.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-35393033253154310692009-06-07T18:09:00.000-07:002009-09-20T13:39:45.875-07:00Lesson Six: Averaging a Set of NumbersTo average a set of numbers, add the numbers together and divide the total by the number of numbers that were added together. Below are a couple of practical examples to illustrate this procedure.<br /><br /><br />1. A water treatment facility feeds an alum dosage of 45 mg/l on Sunday, Monday, and Tuesday and feeds an alum dosage of 75 mg/l on Wednesday, Thursday, Friday, and Saturday. What was the water treatment facility's average daily alum dosage for the week?<br /><br />Answer: 45 + 45 + 45 + 75 + 75 + 75 + 75 = 435 divided by 7 = 62 mg/l (rounded off)<br /><br />2. A water treatment plant has six filters. At the end of the day, the filter hours on these filters are 14.5, 59.0, 71.5, 2.0, 88.5, and 33.0 hours. What is the filter-hour average for these filters?<br /><br />Answer: 14.5 + 59.0 + 71.5 + 2.0 + 88.5 + 33.0 = 268.5 divided by 6 = 44.75 hoursBradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-36480311352746284622009-06-07T18:08:00.001-07:002009-09-18T20:41:34.986-07:00Lesson Seven: Feed Rate TerminologyA feed rate is the rate at which a liquid volume or a gram weight of treatment chemical is injected into the water being treated during the course of a specific timeframe, which is usually one minute. When feed rates are expressed in terms of liquid volume, the rate of injection is generally written in milliliters-per-minute (ml/min). When feed rates are expressed in terms of gram weight, the rate of injection is generally written in grams-per-minute (g/min) or in milligrams-per-minute (mg/min).Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-53529914351394799392009-06-07T18:07:00.004-07:002009-09-19T08:38:20.817-07:00Lesson Eight: Chemical Dosage TerminologyChemical dosages for water treatment chemicals are expressed as milligrams-per-liter (mg/l). Milligrams-per-liter are often refered to as parts-per-million (ppm). One mg/l is exactly equal to one ppm.<br />Chemical dosages are actually a ratio of the weight of the chemical to the weight of the water. In the metric system, one liter of water weighs exactly 1000 grams. In the metric system, one milligram weighs exactly one-thousandth of one gram. Since there are one thousand milligrams in one gram and there are one thousand grams in one liter of water, there are one million milligrams in a liter of water. If a chemical compound is present in a liter of water at a weigh of one milligram, it is present at one ppm.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-9142932804335938492009-06-07T18:07:00.003-07:002009-10-18T07:25:08.766-07:00Lesson Nine: Calculating Detention TimeDetention time is the amount of time that water remains in a basin as the water travels from the entrance point to the exit point of the basin. All detention time calcuations on introductory-level water treatment examinations will involve water that is passing through flocculation basins or sedimentation basins. The answers to detention time problems will be expressed in minutes, hours, or days.<br />An interesting and easily understandable way to view detention time is to view it as the time needed to completely fill a basin if the basin is completely empty. Two numerical figures are always needed to calculate detention time. The first figure is the total number of gallons that the basin holds. The second figure is the flow of water through the basin, expressed as gpm, gph, or gpd.<br />To calculate detention time, simply divide the number of gallons that the basin holds by the flow through the basin in either gpm, gph, or gpd. Below are several good examples of detention time problems.<br /><br />1. If a basin holds 50,000 gallons and the flow through the basin is 50,000 gph, what is the detention time through the basin as expressed in hours?<br /><br />Answer: 50,000 gallons divided by 50,000 gph = 1 hour<br /><br />2. If a basin holds 75,000 gallons and the flow through the basin is 2 MGD, what is the detention time through the basin expressed in hours?<br /><br />Answer: The first step is to convert 2 MDG to gph. To do this, divide 2,000,000 by 24 to obtain 83,333 gph. Next, divide 75,000 by 83,333. The answer is .9 hours or 9/10 of one hour. Thus, water is detained in this basin for .9 hours.<br /><br />3. If a basin holds 75,000 gallons and the flow through the basin is .85 MGD, how many hours does it take for water to flow through the basin?<br /><br />Answer: First, divide 850,000 by 24 to obtain 35,417 gph. Next, divide 75,000 by 35,417. The answer is 2.1 hours.<br /><br />4. A water treatment plant has a flocculation basin that holds 240,000 gallons. This plant is located in a state that requires a minimum flocculation time of thirty minutes. A water treatment plant operator wishes to increase the plant flow from 9.2 MGD to 10.8 MGD. At a flow of 10.8 MGD, what will be the detention time through the flocculation basins in minutes?<br /><br />Answer: Divide 10,800,000 by 1440 (minutes in a day) to obtain a flow of 7500 gpm. Next, divide the 240,000 gallons that the flocculation basin holds by the flow of 7500 gpm to obtain a detention time of 32 minutes. Yes, this plant operator can legally increase the plant flow to 10.8 MGD.<br /><br />5. A sedimentation basin is 150 feet wide, 120 feet long, and 45 feet deep. What is the basin detention time in minutes if the plant flow is 24.7 MDG?<br /><br />Answer: Convert the area of the basin into gallons. To do this, multiply 150 x 120 x 45 to obtain 810,000 cubic feet. Then, multiply 810,000 by 7.48 gallons per cubic foot to obtain a basin volume of 6,058,800 gallons. Next, divide 24,700,000 gallons by 1440 minutes in a day to obtain a flow of 17,153 gpm. Lastly, divide 6,058,800 by 17,153 to obtain a detention time of 353 minutes. (If the problem had asked for detention time in hours, 353 minutes divided by 60 equals 5.88 hours.)Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com1tag:blogger.com,1999:blog-9006476758146199373.post-52786737703033109862009-06-07T18:07:00.001-07:002009-10-18T08:36:32.591-07:00Lesson Ten: Calculating Filtration RatesA filtration rate is the number of gallons that passes through one square foot of filter surface in one minute. Filtration rates are always expressed in "gpm per square-foot." Because of regulations, which limit maximum filtration rates, the answers to filtration rate problems will always be limited to single-digit numbers. The typical range of answers for filtration rate problems is 1.5 gpm/ft2 to 8 gpm/ft2, with the 2 gpm/ft2 to 4 gpm/ft2 range being the most probable range.<br />To calculate filtration rates, divide the gpm that passes through the filter by the square footage of the filter's surface. Always remember this, if a problem provides the the gpm of the entire plant, divide the plant gpm by the number of filters to obtain the gpm for the individual filters. You must use only the gpm for the individual filter to calculate the filtration rate for that filter. Below are two representative filtration rate practice problems.<br /><br />1. A rapid-sand filter is 15 feet wide by 20 feet long. The flow through the filter is 36,000 gph. What is the filtration rate of the filter?<br /><br />Answer: First, multiply 15 x 20 to obtain a filter surface area of 300 square-feet. Next, divide 36,000 by 60 to obtain a filter flow of 600 gpm. Lastly, divide 600 by 300 to obtain a filtration rate of 2 gpm/ft2.<br /><br />2. A water treatment plant has eight 12-feet by 16-feet mixed media filters and has a plant flow of 6.6 MGD. What is the plant's filtration rate?<br /><br />Answer: First, multiply 12 x 16 to obtain 192 square-feet as the square footage of one filter. Next, divide 6,600,000 by 1440 to obtain 4583 gpm as the flow of the entire plant. Then divide 4583 gpm by eight to obtain 573 gpm as the individual filter flow. Finally, divide 573 gpm by 192 to obtain a filtration rate of 2.98 gpm/ft2 ( 3 gpm/ft2, if rounded off).<br /><br />NOTE: Filtration rates are sometimes called Filter Loading Rates.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-67629969472374265482009-06-07T18:06:00.002-07:002009-09-26T16:36:20.242-07:00Lesson Eleven: Calculating Backwash PercentageBackwash percentage, which is often called "percent backwash", is the ratio of backwash water used by a plant versus the total water produced by the plant, as expressed in a percentage form. All backwash percentage problems will provide both the backwash water used and the total plant production. Generally speaking, backwash percentage math problems are easily calculable from the basic information provided within the problem. The answers to most backwash percentage problems will fall somewhere into the .5 percent to 8 percent range.<br />To calculate backwash percentage, divide the numbers of backwash gallons used by the total finished production of the plant. (A ten-digit calculator should be used.) Then, multiply this tiny number, which usually ranges from .005 to .080, by 100<br /><br />1. A water treatment plant used 28,923 gallons of backwash water while producing 3,175,000 gallons of finished water. What was the plant's backwash percentage?<br /><br />Answer: Divide 28,923 by 3,175,000 to obtain a number that is approximately .0091. Then, multiply .0091 by 100 to obtain a backwash percentage of .9%. This plant used slighly under 1% of its water to backwash filters.<br /><br />2. In the wintertime, a water treatment plant experienced shorter filter runs because of increased turbidity break-through on its filters. This plant used 86,755 gallons of backwash water, while producing 1,650,000 gallons of finished water. What is this plant's backwash percentage?<br /><br />Answer: Divide 86,755 by 1,650,000 to obtain a number that is approximately .0526. Then, multiply .0526 by 100 to obtain a backwash percentage of 5.26% (rounded to 5.3% or just 5%).Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-26102737684818508632009-06-07T18:06:00.001-07:002009-09-20T16:08:38.383-07:00Lesson Twelve: Pressure and PSIIntroductory-level water treatment students need to know only one simple fact pertaining to pressure and psi (pounds per square-inch). This fact, which is generally found on most mathematical constants sheets, is that 1.0 psi = 2.31 feet of water. This means that every 2.31 feet of water elevation in a tank exerts 1 psi of pressure onto the walls of any pipe or tube that the water in the tank drains downward into.<br />Many constants sheets also state that 1.0 foot of water = .433 psi. This constant is more easily understandable if it is reversed to state .433 psi = 1 foot of water. By reversing this constant, it matches the form of the 1.0 psi = 2.31 constant.<br /><br />1.0 psi = 2.31 feet of water<br />.433 psi = 1 foot of water<br /><br /><br />The vertical elevation of water from one point to another point is called "head." Sometimes mathematical problems will use the term "head" instead of the term "water elevation." For introductory-level math purposes, the two terms can be considered as equivalentBradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-6105945359621990482009-06-07T18:05:00.006-07:002012-11-13T21:15:31.263-08:00Lesson Thirteen: Feed Rate/Flow ProportionalityMost introductory-level water treatment facility operator exams do not contain problems on feed rate/flow proportionality. Nevertheless, all introductory-level water treatment facility operators need to know this information to effectively operate a water treatment facility. <br />Feed rate/flow proportionality is the mathematical procedure by which a water treatment facility operator increases or decreases chemical feed rates whenever the plant's flow increases or decreases. If a water treatment facility operator is unable to accurately perform this procedure, he or she runs the risk of encountering serious water-quality problems. <br />To increase or decrease a chemical feed rate based on an increase or decrease in plant flow, multiply the existing feed rate by the new flow, and then divide this answer by the old flow. Below are several examples to fully illustrate this procedure. The first two examples are based on gpm flow rates. The last two examples are based on MGD flow rates. <br /><br />1. A water treatment facility is feeding 840 ml/min of alum for a raw flow rate of 2350 gpm. What alum feed rate is needed if the raw flow rate is increased to 2950 gpm? <br /><br />Answer: 840 ml/min x 2950 = 2,478,000 divided by 2350 = 1054 ml/min <br /><br />2. A water treatment facility is feeding 980 ml/min of sodium hydroxide for a raw flow rate of 4730 gpm. What sodium hydroxide feed rate is needed if the raw flow rate is decreased to 3975 gpm? <br /><br />Answer: 980 ml/min x 3975 = 3,895,500 divided by 4730 = 824 ml/min <br /><br />3. A water treatment facility is feeding 780 ml/min of alum for a raw flow rate of 4.2 MGD. What alum feed rate is needed if the raw flow rate is increased to 5.5 MGD? <br /><br />Answer: 780 ml/min x 5.5 = 4290 divided by 4.4 = 1021 ml/min <br /><br />4. A water treatment facility is feeding 1240 ml/min of alum for a raw flow rate of 5.7 MGD. What alum feed rate is needed if the raw flow rate is decreased to 4.8 MGD? <br /><br />Answer: 1240 ml/min x 4.8 = 5952 divided by 5.7 = 1044 ml/min Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-73669008291936979642009-06-07T18:05:00.003-07:002009-09-21T10:49:42.201-07:00Page SpacerSpacer.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-23203475680990743512009-06-07T18:05:00.001-07:002009-09-21T10:51:39.360-07:00Page SpacerSpacer.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-40523382910364789722009-06-07T18:03:00.003-07:002009-09-22T03:55:25.571-07:00Lesson Fourteen: Calculating Basin Overflow RatesBasin overflow rates are calculated in exactly the same way that filtration rates are calculated. To easily visualize this procedure, think of a long sedimentation basin that is feeding its settled water into a filter. The filtration rate of the filter is the number of gallons per square foot of filter surface area that flow downward through the filter per minute. The basin overflow rate is the number of gallons per square foot of basin surface area that flow upward and out of the basin per minute, as if the basin level were slowly rising up and overflowing.<br />Filters have much less surface area than the much larger sedimentation basins. Yet, the same number of gallons is flowing out of the sedimentation basin that is flowing into and through the filter. If two gallons per minute is flowing through each square foot of filter surface, only a fraction of that two gallons is flowing up and out of each square foot of basin surface.<br />Below is a set of problems. Both problems assume that the flow is 600 gpm. One problem calculates filtration rate by dividing the flow by the number of square feet on the filter surface. The other problem calculates basin overflow rate by dividing the flow by the number of square feet on the basin surface.<br /><br />1. A filter is 10 feet by 15 feet. The flow through the filter is 600 gpm. What is the filtration rate?<br /><br />Answer: First, multiply 10 x 15 to obtain a filter surface area of 150 square feet. Next, divide 600 by 150 to obtain a filtration rate of 4 gallons per square foot per minute.<br /><br />2. A basin is 10 feet wide by 90 feet long. The flow through the basin is 600 gpm. What is the basin overflow rate?<br /><br />Answer: First, multiply 10 x 90 to obtain a basin surface area of 900 square feet. Next, divide 600 by 900 to obtain a basin overflow rate of .67 gallons per square foot per minute.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-50320453335330489232009-06-07T18:03:00.001-07:002009-10-18T20:14:43.419-07:00Lesson Fifteen: Chemical Dosages and PoundageThe formula for the calculation of chemical dosages and chemical poundage is the single most important formula that introductory-level water treatment facility operator students must learn to perform well on the math portion of introductory-level water treatment facility operator exams. In all likelihood, four to five questions on any introductory-level water treatment facility operator exam will involve questions on the calculation of chemical dosages and chemical poundage.<br />The formula for calculating chemical dosages and chemical poundage is "lbs. = MG x mg/l x 8.34." The term "lbs." in this formula is the total bulk poundage of a chemical used in a specific timeframe, which is usually a 24-hour period. The term "MG" in this formula does not necessarily mean MGD, but for purposes of introductory-level exams, it is used as MGD. The term "mg/l" in this formula can be used interchangeably with ppm, which may allow some students to better understand its context.<br />Practically everyone who is studying this website has completed one week of water treatment training school. In the math classes at that school, handout sheets containing a pie chart of the chemical dosage and poundage formula were distributed. This chart may prove helpful in performing chemical dosage and poundage calculations.<br />Nearly all chemical dosage and poundage problems involve solving the problem for dosage or poundage. Very few of these problems involve a calculation in which the MG part is the answer. Thus, the two most common versions of the dosage and poundage formula are "lbs. = MG x ppm x 8.34" and "ppm = lbs. / (MG x 8.34)". The third version, for those rarer times when it is needed, is "MG = lbs. / ppm x 8.34".<br /><br />1. A water treatment plant uses 1245 pounds of alum to treat 2.65 MG of water. What is the plant's alum dosage?<br /><br />Step one: ppm = 1245 divided by (2.65 x 8.34). Step two: ppm = 1245 divided by 22.1 = 56.3 ppm<br /><br />2. A water treatment plant uses a caustic dosage of 14.8 ppm to treat 12.7 MG of water. How many pounds of caustic did the plant use?<br /><br />Step one: lbs. = 12.7 x 14.8 x 8.34. Step two: lbs. = (187.96) x 8.34 or 12.7 x (123.432) = 1568<br />lbs.<br /><br />3. A water treatment plant has only 320 lbs. of coagulant-aid polymer left in its inventory. If the plant feeds this polymer at a dosage of 1.5 ppm, how many gallons of water can the plant treat before running out of polymer?<br /><br />Step one: MG = 320 divided by (1.5 x 8.34). Step two: MG = 320 divided by 12.51 = 25.6 MG<br /><br />A SPECIAL VARIATION OF THIS FORMULA: A special variation of this formula is used to calculate the available poundage of a treatment chemical when only a portion of the total chemical poundage of that chemical is available for actual treatment usage. For most introductory-level water treatment math purposes, the chemicals to which this refers will be HTH (calcium hypochlorite) and bleach (sodium hypochlorite).<br />Whenever chlorine gas is used to disinfect water, 100 percent of the chlorine in the gas is available for treatment purposes. Whenever calcium hypochlorite is used to disinfect water, only about 65 to 70 percent of the total calcium hypochlorite poundage is converted to usable chlorine poundage. Whenever commercial-grade sodium hypochlorite is used to disinfect water, the available chlorine poundage is only about 12 to 15 percent of the total sodium hypochlorite poundage. With laundry-grade sodium hypochlorite bleach, the available chlorine poundage is only 4 to 6 percent of the total sodium hypochlorite poundage.<br />A typical test question involving this variation of the poundage formula might read, "How many pounds of 12 percent pure sodium hypochlorite will be needed to treat eight million gallons of water with a chlorine dosage of 6 mg/l?" On introductory-level exams, these types of questions will almost always start with "How many pounds", since poundage is the only variable that is calculated with introductory-level questions of this type. Also, the word <em>pure</em>, which is not really a very descriptive term, is commonly used on test questions to refer to the "available treatment chemical" involved in the question.<br />To solve these special types of problems, first calculate the total chemical poundage as if the percentage factor didn't even exist. Then, divide the total poundage answer by the percentage in a decimal form. The final poundage answer will always be greater than the original poundage answer. Below are two examples to fully illustrate this procedure.<br /><br />1. How many pounds of 70% pure HTH will be needed to disinfect a newly installed 1.5 MG water storage tank with a 50 mg/l dosage of chlorine? (In this question, the term " 70% pure " denotes that the total HTH poundage yields 70 percent of itself to available chlorine poundage.)<br /><br />Step one: lbs. = 1.5 x 50 x 8.34 = 625.5 pounds. Step two: 625.5 divided by .70 = 894 pounds.<br /><br />2. How many pounds of sodium hypochlorite with 12.5% available chlorine will be needed to treat 10,000,000 gallons of water with a chlorine dosage of 5.5 ppm?<br /><br />Step one: lbs. = 10 x 5.5 x 8.34 = 458.7 pounds. Step two: 375.3 divided by .125 = 3669.6 pounds.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-48726518311702356012009-06-07T18:02:00.006-07:002010-02-28T14:47:15.410-08:00Test Preparation and Study TipsEven if you have been studying your test material before water school, review and categorize all of your study notes right after water school, when your school lessons are freshest on your mind.<br /><br />If you find yourself getting tense or stressed out while studying, take an immediate break.<br /><br />Don't worry about your test. The math test questions on certification exams are usually easier than most of the math practice test questions.<br /><br />Many people have better mental retention if they briefly review their material for about five minutes just before they go to bed each night. Do this in addition to your regular studying, which will certainly require more than five minutes a day.<br /><br />Bring a ten-digit calculator for your exam. Use the calculator your plan to use on the exam for all of your studying and pre-exam practice.<br /><br />Get a good night's sleep the night before the exam. Arrive early enough at your exam site so that your don't have to rush in at the last minute.<br /><br />IMPORTANT: Immediately after your test proctor instructs you to begin taking the exam, write your math formulas onto the scratch paper that you are given or onto the exam copy if it is used as scratch paper. This is a perfectly acceptable practice, and you will have your formulas written down so that you can proceed to take the exam with absolute confidence.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com0tag:blogger.com,1999:blog-9006476758146199373.post-21972414579908624172009-06-07T18:02:00.005-07:002011-02-05T15:52:05.057-08:00E-Mail Address for This SiteIf you used this site and found it to be beneficial, please send me a short e-mail. Any comments or constructive suggestions about this site would be most appreciated.<br /><br />My e-mail address for this site is <a href="mailto:bradleycharleswilliams2@gmail.com">bradleycharleswilliams2@gmail.com</a> .<br /><br />Thank you, and best wishes on your upcoming exam.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-76082282937154700882009-06-07T18:02:00.003-07:002010-02-24T17:08:07.599-08:00Math Practice Test WebsiteI have established a new website, which will contain nothing but math practice tests for introductory-level water treatment facility operator certification exams. I am personally writing all of the test questions on this site, so the questions will be completely original and will not be the same questions that are found on standard water school practice test handout sheets.<br /><br />The site address is <a href="http://water-treatment-math-practice-tests.blogspot.com/">http://water-treatment-math-practice-tests.blogspot.com/</a><br /><br />(Click on this link to go directly to the site.)<br />**************************************Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.comtag:blogger.com,1999:blog-9006476758146199373.post-92049307083394718022009-06-07T18:02:00.001-07:002011-02-05T16:05:43.344-08:00Comment on the Water Treatment ProfessionMany water treatment operators who view and study this site are new to the water treatment profession. I strongly encourage these operators to consider the advantages of remaining in the water treatment profession and making the water treatment profession a career that will last until retirement. <br /><br />The most essential commodity on earth is water. Water is needed for drinking, cooking, bathing, washing dishes, washing clothing, flushing toilets, preparing infant formula, and a host of other purposes that are vital to human existence. A clean, fresh supply of public drinking water is essential to the survival and growth of every modern community throughout the world.<br /><br />Employment in the drinking water profession offers a level of job security that is nearly unequalled in any other profession. Food is also an essential commodity, but food brands come and go, and food manufacturing plants routinely close down. Public water systems, on the other hand, are legal monopolies in the areas they serve. This means no other water production utility can come in and sell water to your customers, unless given permission to do so by the owner of your system. <br /><br />Electrical utilities are also monopolies in many of the areas they serve. Most electrical utility employees don't need the levels of licensing and certification that drinking water operators need. This makes electrical utility employees easier to replace than water treatment operators. <br /><br />The medical profession offers nearly equal job security to the water treatment profession. Of course, working as a licensed professional in the medical profession often entails being around sick and dying people. This is the type of profession many of us respect, but we do not wish to enter it. <br /><br />Water treatment plants and water utilities almost never go out of business. This is true during even the worst of economic times. Municipalities and water utilities may institute hiring freezes, and workloads may increase on water treatment operators, but water treatment plants and water utilities don't generally go out of business.<br /><br />I strongly encourage all new water treatment operators to consider the job security that the profession offers. Excellent job security, along with fairly respectable retirement packages, make the water treatment profession a good career choice for anyone seeking stable, long-term employment.Bradley C. Williamshttp://www.blogger.com/profile/10208871753298952462noreply@blogger.com