Sunday, June 7, 2009
Lesson Seven: Feed Rate Terminology
A 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).
Lesson Eight: Chemical Dosage Terminology
Chemical 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.
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.
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.
Lesson Nine: Calculating Detention Time
Detention 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.
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.
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.
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?
Answer: 50,000 gallons divided by 50,000 gph = 1 hour
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?
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.
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?
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.
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?
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.
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?
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.)
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.
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.
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?
Answer: 50,000 gallons divided by 50,000 gph = 1 hour
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?
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.
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?
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.
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?
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.
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?
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.)
Lesson Ten: Calculating Filtration Rates
A 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.
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.
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?
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.
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?
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).
NOTE: Filtration rates are sometimes called Filter Loading Rates.
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.
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?
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.
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?
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).
NOTE: Filtration rates are sometimes called Filter Loading Rates.
Lesson Eleven: Calculating Backwash Percentage
Backwash 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.
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
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?
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.
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?
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%).
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
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?
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.
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?
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%).
Lesson Twelve: Pressure and PSI
Introductory-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.
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.
1.0 psi = 2.31 feet of water
.433 psi = 1 foot of water
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 equivalent
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.
1.0 psi = 2.31 feet of water
.433 psi = 1 foot of water
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 equivalent
Lesson Thirteen: Feed Rate/Flow Proportionality
Most 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.
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.
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.
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?
Answer: 840 ml/min x 2950 = 2,478,000 divided by 2350 = 1054 ml/min
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?
Answer: 980 ml/min x 3975 = 3,895,500 divided by 4730 = 824 ml/min
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?
Answer: 780 ml/min x 5.5 = 4290 divided by 4.4 = 1021 ml/min
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?
Answer: 1240 ml/min x 4.8 = 5952 divided by 5.7 = 1044 ml/min
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.
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.
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?
Answer: 840 ml/min x 2950 = 2,478,000 divided by 2350 = 1054 ml/min
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?
Answer: 980 ml/min x 3975 = 3,895,500 divided by 4730 = 824 ml/min
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?
Answer: 780 ml/min x 5.5 = 4290 divided by 4.4 = 1021 ml/min
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?
Answer: 1240 ml/min x 4.8 = 5952 divided by 5.7 = 1044 ml/min
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