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In order to understand specific gravity, we need to have a better understanding of density and its relationship to mass. Mass is the amount of matter (atoms) in a given substance. Take saltwater for example. We know that one gallon of water weighs 8.34 gallons. However, if one pound (453 grams) of salt is added to create a saltwater solution, then that same gallon of water would weigh 9.34 pounds per gallon instead. It is the same volume (one gallon) but has a different weight due to the additional mass that was added from the salt. Hence the definition of density, mass per unit volume. The amount of “stuff” in a given volume. As discussed, the specific gravity of water is used as the reference point for comparisons. If something has a specific gravity less than water (<1), then the substance will float on water. Conversely, if a substance has a specific gravity greater than 1, it will sink in water. The table below lists common specific gravities and weights of substances used in the waterworks industry. On any State exam, you will be given the specific gravity or corresponding weight of the substance in the question.
Note that everything listed in the table after water is heavier, and before water is lighter. Also, it’s important to consider that sodium hypochlorite has a different specific gravity depending on the concentration of the solution. This is true of all chemical solutions. Since water is the reference, then a specific gravity (SG) of 1 and a weight of 8.34 lbs/gal are the reference numbers needed to calculate the SG and weight of other substances. We can use this information as a conversion factor to determine the SG or weight of other substances. The conversion can be expressed as follows:
PARTS-PER NOTATION Parts-per notation is used to describe very small quantities of chemical concentrations. Most of the time, chemical concentrations are expressed in percentages. However, it is important to understand the relationship between the concentration expressed as a percentage and the concentration expressed in parts-per notation.
Because the amounts are small, examples will help. One part per million is the equivalent of a drop of ink in 55 gallons of water. One part per billion is the equivalent of one drop of ink in 55 barrels of water. One part per billion is the equivalent of the width of a human hair within 68 miles.
In chemical dosage-related problems, concentrations are typically expressed in parts per million, ppm. Therefore, we are most interested in converting a percentage to ppm and ppm to a percentage. If you divide 1,000,000 (1 ppm) by 100 (100%), you get the following:
To convert between a percentage and ppm, multiply the percent solution by 10,000. Note that you do not convert the percentage to a decimal before multiplying. It has already been accounted for in the conversion.
Now let’s look at the differences between parts per million (ppm), parts per billion (ppb), and parts per trillion (ppt.) As water quality regulations become more stringent and laboratory analysis techniques get better and better, contaminants are being identified at smaller and smaller concentrations. Most water quality standards are expressed in ppm or milligrams per liter (mg/L), but many are expressed in ppb or micrograms per liter (ug/L), and a few are expressed in ppt or nanograms per liter (ng/L). Another way to express the amount of contaminant in water supplies is as follows.
The expression above says that 1 part of a small number (ppm) equals 1,000 parts of a smaller number (ppb) which equals 1,000,000 parts of an even smaller number (ppt). You can further simplify the difference between ppb and ppt as follows.
MIXING AND DILUTING SOLUTIONS The water industry uses a variety of different chemicals for various processes in water treatment. These chemicals can come in different forms including gas, solid, or liquid (typically a solution). A solution is a liquid often made up of water and one or more chemicals. The concentration of a solution is dependent on the amount of chemical that is diluted within the water and can be represented as a percentage. The percentage strength of a disinfectant solution represents the amount of active chemical available to inactivate pathogenic bacteria. For example, if you were to separate the two components of a 10% bleach solution, you would see that it is made up of 90% water and 10% sodium hypochlorite (active chemical).
Chemicals sometimes need to be diluted or mixed so that operators can safely work with them or to match the needs of a treatment process. The following formula can be used to calculate either the new concentration of a mixed solution, or the required volume needed to achieve a desired concentration.
C = Concentration represented as a percentage V = Volume of the solution in mL, gallons, etc. Example: If 700 mL of water is added to 250 mL of a 65% solution, what is the resulting solution’s diluted concentration strength?
Key Terms density – mass per unit of volume parts-per notation – a way of describing small quantities of chemical concentrates, often expressed as percentages solution – a liquid often made up of water and one or more chemicals specific gravity – the ratio of a solution's density compared to the density of water
UNIT 4 CHEMICAL DOSAGE ANALYSIS One of the most common and useful formulas in waterworks mathematics is used to calculate the amount of chemical needed to add to water. It is commonly known as the Pound Formula because it is a calculation for the weight of a chemical that is being added to water. Typically, this chemical is chlorine or a chlorine-related compound. However, it can be also used to calculate alum, ferric chloride, or any other type of chemical dosage.
You may want to use the Pie Wheel to solve pound formula problems. In the pie wheel, the horizontal line across the middle of the pie wheel represents a division sign. Anything above the line should be divided by anything below the line. Everything below the line, which are variables next to one another, are multiplied together. Anything in the bottom half of the pie wheel is multiplied together to get the answer in the top half of the wheel.
It's pretty easy to use the Pound Formula to calculate how many pounds of chlorine are needed to provide a certain dosage if we are using 100% concentration of any chemical. However, most chemicals used are not in pure 100% form. In many treatment plants and at treatment sites within distribution systems, the use of chlorine gas is in decline, unless the plant is of considerable size. In gas form, 100 lbs of gas chlorine is 100 lbs of available chlorine. The reduction in chlorine gas usage is primarily due to safety concerns and other forms of chlorine being less expensive. For example, groundwater wells are commonly disinfected with solid (calcium hypochlorite) or liquid (sodium hypochlorite) chlorine. In addition, other chemicals such as alum, ferric chloride, sodium hydroxide are used in varying concentration strengths at treatment plants in addition to chlorine.
When solving dosage problems with chemicals of different concentrations, you will still use the Pound Formula, but you need to adjust for the concentration strength of the chemical in the calculation.
If you are calculating the number of pounds needed, you divide by the decimal equivalent of the percent concentration. Since it is not at 100% concentration, you need more of the chemical. When you divide by a number less than one, you get a larger number, in this case, the lbs of chemical needed at the reduced concentration level.
It takes 417 lbs per day of 100% Alum to dose 5 MGD to 10 ppm. However, the Alum being used is only a 10% concentration. To adjust for the percent concentration, divide lbs per day by the percent concentration. Remember that to change a percent to a decimal, you need to move the decimal point over two places to the left.
At only a 10% concentration, you will need 4,170 lbs of 10% Alum, significantly more, to get the equivalent of 417 lbs at full concentration. Therefore, to determine the amount of 10% Alum needed, you divide the total amount by 10% (or 0.1).
If the number of pounds is known, multiplying by the decimal equivalent of the percent concentration will calculate how much of that chemical is available in the total pounds of the substance. Multiplying by a number less than one yields a smaller number.
To solve for the dosage, you need to rearrange the terms in the equation. In this step, you can see that you are multiplying the lbs per day by the decimal equivalent of the percent concentration (15% is equal to 0.15)
Once you understand the concept behind the problem, it makes solving them easier. Think of it this way: it takes much more 10% ferric chloride in the coagulation process than 75% ferric chloride. The same is true if you are using calcium hypochlorite as opposed to gas chlorine, because gas is at a greater strength (100%) than calcium hypochlorite.
The last chemical dosage concept we need to look at is when the chemical being used is in the form of a liquid. Since the Pound Formula is measuring chemicals in pounds, the liquid chemical needs to be expressed as pounds. In Unit 3, we learned about specific gravity and how it affects the weight of a substance. You will need to use that information when presented with a pound formula question where the chemical used is a liquid.
You know that 220 lbs per day are being used and that it weighs 15.76 lbs per gallon. This allows you to calculate the gallons per day being used. Make sure to write your fractions in a way that allows for the lbs to cancel and the units in your answer to be gallons per day.
The reason drinking water is disinfected is to prevent pathogenic organisms from causing illness in the population drinking the water. These organisms are controlled by the amount of disinfectant, typically chlorine, added to the water. The amount of chlorine required to inactivate the various pathogens is called the chlorine demand. Once the demand is satisfied, the remaining chlorine in the water supply is termed the residual. In order to keep the water supply safe for consumption, there must be a residual after the dosage has been applied. Water operators may be required to measure the original dosage or the residual in a water supply.
Key Terms chlorine demand – the amount of chlorine required to inactivate the various pathogens dosage – the amount added to satisfy demand Pound Formula – a calculation for the weight of a chemical, such as chlorine or chlorine-related, that is added to water; can be used to calculate alum, ferric chloride or other chemical doses residual – remaining chlorine in the water supply once the demand is satisfied
A water utility produced 6,000 AF of water last year. The entire amount was dosed at an average rate of 0.6 ppm. If the chemical of choice was 35% HTH at a per pound cost of $2.43, what was the annual budget? Ferric chloride is used as the coagulant of choice at a 10.1 MGD rated capacity treatment plant. If the plant operated at the rated capacity for 60% of the year and operated at 30% of rated capacity for 40% of the year, how many pounds of the coagulant was needed to maintain a dosage of 65 mg/L?
A water utility produced 11,275 AF of water last year. The entire amount was dosed at an average rate of 1.5 ppm. If the chemical of choice was 65% HTH at a per pound cost of $1.85, what was the annual budget? Ferric chloride is used as the coagulant of choice at a 5.75 MGD rated capacity treatment plant. If the plant operated at the rated capacity for 75% of the year and operated at 60% of rated capacity for 25% of the year, how many pounds of the coagulant was needed to maintain a dosage of 45 mg/L?
UNIT 5 WEIR OVERFLOW RATE A weir is an overflow structure that is used to alter flow characteristics. Weirs are used in many different circumstances, including water treatment facilities and irrigation canals. The weir raises the water level and causes the water to flow over the weir structure at a constant flow rate known as the Weir Overflow Rate (WOR). WORs are expressed as the flow of water by the length of the weir, typically as MGD per foot (MGD/ft) or gpm per foot (gpm/ft).
Weirs can either be sharp-crested or broad-crested. Broad-crested weirs are flat-crested structures and are commonly used in dam spillways. Sharp-crested weirs (most common are “V” notch) allow the water to fall cleanly away from the weir and are typically found in water treatment plants. The same WOR formula can be used no matter which style of weir is used.
Calculating the length of the weir is required in order to calculate the WOR. Sometimes the weir can be a circular structure requiring the circumference to be calculated in order to find the actual length. Other times it is a linear structure, in which case the length would be known.
The units of the values provided in the problem statement, do not match. You have MGD for the daily flow and gpm/ft for the weir overflow rate. Therefore, in order to solve for length, either MGD must be converted to gpm or gpm/ft must be converted to MGD/ft. Both will provide you with the same final answer.
It doesn’t matter whether you calculate the weir length using MGD or gpm. In either case, the weir length is 57.43 ft. However, it is important that all of the units in the problem are the same – MGD to MGD/ft or gpm to gpm/ft.
Since the plant flow rate is provided in MGD and the WOR is provided in gpm per foot, you have the option of converting gpm to MGD or MGD to gpm in order to solve. Here is the conversion for gpm to MGD.
Key Terms weir – an overflow structure that is used to alter flow characteristics weir overflow rate (WOR) – the rate at which water flows over the weir structure; the flow of water by the length of the weir
An aqueduct is being reconstructed to widen the width across the top. The width across the bottom is 25 feet and the average water depth is 40 feet. The aqueduct must maintain a constant weir overflow rate of 15 gpm per foot with a daily flow of 0.88 MGD. What is the length of the weir? An engineering report determined that a minimum weir overflow rate of 25 gpm per foot and a maximum weir overflow rate of 30 gpm per foot were needed to meet the water quality objectives of a certain treatment plant. The existing weir is 120 feet long. What is the daily treatment flow range of the plant?
A 75-mile aqueduct is being reconstructed to widen the width across the top. The width across the bottom is 10 feet and the average water depth is 15 feet. The aqueduct must maintain a constant weir overflow rate of 25 gpm per foot with a daily flow of 0.63 MGD. What is the length of the weir? An engineering report determined that a minimum weir overflow rate of 15 gpm per foot and a maximum weir overflow rate of 20 gpm per foot were needed to meet the water quality objectives of a certain treatment plant. The existing weir is 80 feet long. What is the daily treatment flow range of the plant? An aqueduct has a weir that is 5 feet narrower than the distance across the aqueduct. Assuming a constant weir overflow rate of 28.75 gpm/ft, an average depth of 12 feet, a distance across the bottom of 8 feet, a length of 22 miles, and a daily flow of 0.85 MG, what is the capacity of the aqueduct in AF? UNIT 6 DETENTION TIME Detention Time is an important process that allows large particles to “settle out” from the flow of water through gravity, prior to filtration. It is the time it takes a particle to travel from one end of a sedimentation basin to the other end. Conventional filtration plants require large areas of land to construct sedimentation basins and employ the detention time process. Not all treatment plants have the available land and may decide that direct filtration is suitable. Therefore, in direct filtration plants the sedimentation process is eliminated. However, in direct filtration plants, the filters have shorter run times and require more frequent backwashing cycles to clean the filters. A term used that is interchangeable with detention time is contact time. Note that this should not be confused with CT, Concentration Time which will be discussed in Unit 7. Contact times represent how long a chemical (typically chlorine) is in contact with the water supply prior to delivery to customers. For example, contact time can be measured from the time a well is chlorinated until it reaches the first customer within a community or, it could be how long the water mixes in a storage tank before it reaches a customer. Calculating the Detention Time and Contact Time requires two elements, the volume of the structure holding the water (sedimentation basin, pipeline, and storage tank) and the flow rate of the water (gallons per minute, million gallons per day.) Since detention times and contact times are typically expressed in hours, it is important that the correct units are used. When solving Dt problems be sure to convert to the requested unit of time. As with all water math-related problems, there are other parameters that can be calculated within the problem. For example, if the detention time and volume are known, then the flow rate can be calculated. Or, if the flow rate and detention time are known, the volume can be calculated. Sometimes the flow rate and the desired detention time is known and the size of the vessel holding the water needs to be designed. In this example, the area or dimensions of the structure can be calculated. The pie wheel (below) shows a simple way of calculating the variables. If the variables are next to each other (Dt and Flow Rate) then multiply. If they are over each other (Volume and Dt or Volume and Flow Rate) then divide.
Units are extremely important when using this formula. There are three variables in this formula: Detention Time, Flow, and Volume. Each variable can be provided using a variety of units. You cannot calculate the solution unless all of the units align. If the units are similar (matching), then dividing volume by flow will yield a time (Detention Time). However, simply dividing a volume by a flow will not result in a time. For example, if you divide gallons by cubic feet per second there is no resulting answer. This is because “gallons” and “cubic feet” will not cancel each other.
In the first two examples, the terms can be divided. However, in the third example they cannot. Dt should be expressed as a unit of time (i.e., sec, min, hours). If you divide the first two examples (gal/gpm and cf/cfs), you will end up with minutes and seconds respectively. However, in the third example, gallons and million gallons cannot cancel each other out. Therefore, if you had 100,000 gallons as the volume and 1 MGD as the flow rate:
Sometimes this can be the simplest way to solve detention time problems. However, people can be confused when they get an answer such as 0.1 days. There are other ways to solve these problems. One way is to convert MGD to gpm. Using the above example, convert 1 MGD to gpm.
The problem asks for the detention time to be expressed in hours and minutes. Based on the calculation above, you know that you have 2 full hours and a portion of a third hour. To calculate the exact number of minutes, take the decimal amount and multiply by 60 min per hour.
Example: A water utility is designing a transmission pipeline collection system in order to achieve a chlorine contact time of 2 hours 10 mins once a 1,125 gpm well is chlorinated. How many feet of 30” diameter pipe are needed?
Rearranging the terms in the detention time equation to solve for volume results in volume equals detention time times flow. Substitute the minutes calculated and the flow rate into this equation in order to solve for the total volume in gallons.
Now that you know the total volume of the pipe in cubic feet, you can use the formula for volume of a pipe to determine the pipe length. Substitute the pipe diameter and the pipe volume into the equation and rearrange the terms to solve for length.
A water utility engineer is designing a sedimentation basin to treat 12 MGD and maintain a minimum detention time of 3 hours 30 minutes. The basin cannot be longer than 100 feet and wider than 65 feet. Under this scenario, how deep must the basin be? A water utility is designing a transmission pipeline collection system in order to achieve a chlorine contact time of 75 minutes once a 3,400 gpm well is chlorinated. How many feet of 36” diameter pipe are needed? The chlorine residual decay rate is 0.7 mg/L per 3/4 hour in a 4 MG water storage tank. If the storage tank needs to maintain a minimum chlorine residual of 6.5 mg/L what is the required dosage if the tank is filling at a rate of 900 gpm until the tank is full? A 44-foot-tall water storage tank is disinfected with chloramines through an onsite disinfection system. The average constant effluent from the tank is 680 gpm through a 20-inch diameter pipe. If the first customer that receives water from the tank is 4,972 feet from the tank, would the required 30-minute contact time be achieved?
A water utility engineer is designing a sedimentation basin to treat 10 MGD and maintain a minimum detention time of 2 hours 15 minutes. The basin cannot be longer than 80 feet and wider than 40 feet. Under this scenario, how deep must the basin be? Chlorine is injected into an 18” diameter pipe at a well site. The pipeline is 2,000 ft long before it reaches the first customer. Assuming a well flow rate of 1,700 gpm, what is the detention time (contact time) in minutes? A water utility is designing a transmission pipeline collection system in order to achieve a chlorine contact time of 40 minutes once a 2,250 gpm well is chlorinated. How many feet of 24” diameter pipe are needed? A fluoride tracer study is being conducted at a 15.5 MGD capacity water treatment plant. The contact time through the coagulation and flocculation process is 2.45 hours. If the sedimentation basin has a capacity of 500,000 gallons, what is the total detention time through the 3 processes? The chlorine residual decay rate is 0.2 mg/L per ½ hour in a 5 MG water storage tank. If the storage tank needs to maintain a minimum chlorine residual of 10.0 mg/L what is the required dosage if the tank is filling at a rate of 1,500 gpm until the tank is full? A drinking water well serves a community of 2,000 people. The customer closest to the well is 1,250 feet away. The above ground portion of the well piping is 12” diameter and 25 feet long. The below ground portion is 750 feet of 10” diameter and 475 feet of 8” diameter piping. What is the chlorine contact time in minutes from the well head to the first customer? Assume a constant flow rate of 3.90 cfs. A 32-foot-tall water storage tank is disinfected with chloramines through an onsite disinfection system. The average constant effluent from the tank is 550 gpm through a 16-inch diameter pipe. If the first customer that receives water from the tank is 3,220 feet from the tank, would the required 45-minute contact time be achieved?
A water treatment plant is in the process of redesigning their sedimentation basin. The plant treats 4.5 MGD with an average detention time of 1.85 hours. Portable storage tanks will be used when the basin is under construction. The portable storage tanks are 25 ft tall and 20 ft in diameter. How many tanks will be needed? FILTRATION RATES One of the most important processes in a Water Treatment Plant is filtration. It is the last barrier between the treatment process and the customer. Filters trap or remove particles from the water further reducing the cloudiness or turbidity. There are different shapes, sizes, and types of filters containing one bed or a combination of beds of sand, anthracite coal, or some other form of granular material. Slow sand filters are the oldest type of municipal water filtration and have filtration rates varying from 0.015 to 0.15 gallons per minute per square foot of filter bed area, depending on the gradation of filter medium and raw water quality. Rapid sand filters on the other hand can have filtration rates ranging from 2.0 to 10 gallons per minute per square foot of filter bed area. Typically, rapid sand filters will require more frequent backwash cycles to remove the trapped debris from the filters. Backwashing is the reversal of flow through the filters at a higher rate to remove clogged particles from the filters. Backwash run times can be anywhere from 5 – 20 minutes with rates ranging from 8 to 25 gallons per minute per square foot of filter bed area, depending on the quality of the pre-filtered water. Filtration and backwash rates are calculated by dividing the flow rate through the filter by the surface area of the filter bed. Typically, these rates are measured in gallons per minute per square foot of filter bed area.
Although filtration rates are commonly expressed as gpm/ft2 they are also expressed as the distance of fall (in inches) within the filter per unit of time (in minutes). This “fall” references the fact that filtration rates are reduced over time as particles are lodged into the filtration media during operation. Backwashing, a process of cleaning the media by reversing the flow through the filter, is then employed in order to try and recover some of the filtering capacity and prolong the operational life of the filter. During backwashing, the formula is expressed with the same units as “fall” but is described as “rise” in the filter instead. This lets you know how much filtering capacity is recovered through your backwashing cycle. Use the formulas below.
Example: What is the filtration rate through a 20’ by 20’ filter if the average flow through the treatment process is 2.5 MG? Express the filtration rate as in/min. First, convert 2.5 MGD to gpm. To do this divide 2.5 MGD by 1,440.
Key Terms contact time – detention time; do not confuse contact time with concentration time detention time – time that allows large particles to settle out from the flow of water through gravity filtration –the last barrier between the treatment process and the customer; filters trap or remove particles from the water further reducing the cloudiness or turbidity; conventional filtrate plants require large areas of land to construct sedimentation basins and employ detention time process; the direct filtration process eliminates sedimentation to have a shorter run time, use less land, and requires more frequent backwashing to clean the filters.
A water treatment plant processes a maximum of 7.50 MGD. The plant has 3 filters measuring 28 ft by 33 ft each. Assuming that each filter receives an equal amount of flow what is the filtration rate in gpm/ft2?
A filter needs to be backwashed when the fall rate exceeds 6.3 inches per minute. It was determined that this rate is reached after 4.7 MG flows through a 27 ft by 28 ft filter. How often does the filter need backwashing? Give your answer in the most logical time unit.
A water treatment plant processes a maximum of 18.65 MGD. The plant has 6 filters measuring 20 ft by 22 ft each. Assuming that each filter receives an equal amount of flow what is the filtration rate in gpm/ft2?
A filter needs to be backwashed when the fall rate exceeds 3.1 inches per minute. It was determined that this rate is reached after 2.3 MG flows through a 17 ft by 17 ft filter. How often does the filter need backwashing? Give your answer in the most logical time unit. UNIT 7 CT CALCULATIONS Concentration and Time are critical variables in water treatment. CT stands for Concentration and Time. As soon as a disinfectant is added to water, it begins the disinfection process. What is the concentration of the disinfectant and how long does it need to be in contact with the water? Well, it takes time to complete the disinfection process once chemical is added to water. In addition, there are other variables that can delay the disinfection process such as, pH, water temperature, turbidity, and the amount of pathogens in the water, among other things. Therefore, knowing the concentration of the disinfectant and the time the disinfectant has to do its “work” is very important in ensuring water is properly disinfected and safe for human consumption. There are many different types of bacteria in natural water sources that can cause sickness if not properly treated. The disinfection process kills and/or inactivates pathogenic (disease causing) bacteria to make it safe for human consumption. In order to follow the Surface Water Treatment Rule (SWTR), drinking water treatment plants must meet the following inactivation requirements:
Cryptosporidium will not be discussed in this class due to the complexities in the 2003 update to the SWTR known as the Long Term 2 Enhanced Surface Water Treatment Rule. Instead, we will focus on Giardia and viruses. Table 7.2 (below) depicts the log requirements that are required for both Giardia and viruses. Notice in the second column that Giardia must be disinfected to 3 Log (99.9%) and viruses to 4 Log (99.99%). Various treatment processes (shown in the first column) account for some of the inactivation or removal of pathogens from the raw water. Therefore, the SWTR provides “credits” toward the inactivation of Giardia and viruses (shown in the third column). Credits are subtracted from the inactivation requirements to determine the level of disinfection still required after water goes through treatment.
If raw water is treated using direct filtration, it would receive 1 Log credit. After water leaves the treatment plant, the disinfection requirement remaining for viruses would be 3 Log, as shown in the fourth column of Table 7.2 (complete calculation illustrated below the table). The remaining 3 Logs will need to be inactivated by disinfecting the water using the appropriate
Now let’s discuss CT units. When dealing with CT calculations, concentration will be expressed in mg/L and Time will be expressed in minutes. Therefore, CT is expressed as mg/L · min (all one unit). When solving CT problems, the concentration of the disinfectant is typically provided in the question. However, there may be times when the “Pound Formula” is needed to calculate the chemical concentration. In order to calculate the contact time of an applied chemical, the detention time (Dt) formula from Unit 6 will be needed. Time is defined as the moment the disinfectant is in contact with the water to the point where the Concentration is measured. These times are easily calculated through pipelines and reservoirs of known volumes but can be difficult to calculate through various treatment plant processes. To solve this issue, Tracer Studies (aka T10) are sometimes conducted. A tracer study can be accomplished by adding a unique tracer chemical to the raw water before it goes through the treatment plant and measuring how long before it is detected in the effluent of the plant. More specifically, T10 represents the time for 10% of an applied tracer mass to be detected through a treatment process or, the time that 90% of the water and pathogens are exposed to the disinfectant within a given treatment process. Some problems will require the calculation of the contact time using the Dt formula while others will provide T10 values.
Next, there are two crucial terms that are required in order to calculate whether water has been adequately disinfected, “actual CT” and “required CT.” Actual CT is the actual concentration of chemical through the treatment process and the actual time the disinfectant is in contact with the water. Required CT is found in the CT tables using the information provided in the problem. Once the concentration of chemical and the contact time are calculated, they can be multiplied together to determine the Actual CT.
Once the actual CT values have been calculated, the final step in the CT calculation process involves CT Tables. The U.S. Environmental Protection Agency (USEPA) as part of the SWTR, created a series of tables that list the type of disinfectant, the pH of the water, the concentration of the disinfectant, the contact time, and the pathogen in question. Using all this information, the required CT (mg/L · min) values can be found. For your reference, the CT Tables are provided at the end of this text. They can be confusing at first, but once you understand what information you need to look for, the CT values can be easily found. See the example below. Example: What is the required 1.0 log inactivation from disinfection (value after credits are applied) for the Inactivation for Giardia in 10 degrees Celsius water with a pH of 7.5 using a free chlorine dosage of 1 mg/L?
The supporting information (outlined in the boxes above) will help determine which exact column and row you need to use to find the answer. For this example, the chlorine concentration is 1 mg/L and the Log Inactivation is 1.0.
The CT Tables provide the required CT needed to inactivate either Giardia or viruses . The ratio of the actual CT (calculated portion of the problem) and the required CT (found in the CT tables) is then calculated to determine if the water has been properly disinfected. If the actual CT is equal to or greater than the required CT then the ratio is equal to or greater than 1.0 and CT is met. If the actual CT is less than the required CT then the ratio would be less than 1.0 and CT would not be met.
Finding the Correct CT Table Typical CT problems will provide the pH, the temperature, the pathogen of interest, the type of disinfectant, the dosage or a way to calculate the dosage, and the type of treatment in the problem statement. You can use this information to identify which CT table to use. Example: Given the following information, which CT table would you use? pH – 7.5 Temperature – 10°C Disinfectant – Free chlorine Dosage – 0.2 mg/L Pathogen – Giardia Treatment – Direct Filtration
Therefore, Table C-3 is the correct table to use for this data set. The title of the table tells you which CT value the table will provide. Table C-3 above is for Giardia, with free chlorine as the disinfectant, at a temperature of 10°C. Finding Required CT Now you need the other information in the problem statement. Specifically, the pH and the dosage concentration. There are seven (7) boxes in the table each with different pH values. On the far left of the table, you can find the varying disinfectant concentrations, starting with less than or equal to 0.4 mg/L going up to 3 mg/L. Example: Given the following information, what is the Required CT? pH – 7.5 Temperature – 10°C Disinfectant – Free chlorine Dosage – 0.2 mg/L Pathogen – Giardia Treatment – Direct Filtration
To find the required CT on the table, look for the box in the table that says pH = 7.5. Now look at the far left of the table and find the dosage of 0.2 mg/L as indicated in the problem statement. You’ll need to use the first row of the table. You now need the last bit of information, the treatment process. In this instance, it is Direct Filtration. Remember, Giardia has an inactivation requirement of 3 Log. Referring back to Table 7.1 you can identify the Log credit
Log required – 2 Log credit for direct filtration = 1 Log remaining Using the first row for disinfectant concentration and the second column from the 7.5 pH portion of the table, you should come up with a required CT of 42 mg/L · min (circled above). Calculating Actual CT There can be multiple locations where a disinfectant is added to the water during the treatment process. Sometimes the water is pre-chlorinated in the raw water pipeline leaving a storage reservoir prior to entering the treatment facility. Sometimes the water is disinfected before the coagulation flocculation process and many times the water is disinfected after filtration prior to delivery to customers. Every time chlorine is added to the water supply it counts towards the inactivation of pathogens. Each step of the way CT will need to be calculated. The example information below will help illustrate this concept. Example: Free chlorine is added at a concentration of 0.2 mg/L in a 12” diameter 5,000- foot-long pipeline leaving a storage reservoir prior to entering the treatment plant. The flow through the pipeline is 2 MGD. What is the time 0.2 mg/L of free chlorine is in contact with the water?
Since the ratio of actual to required CT is less than 1.0, then CT is not met. If a treatment plant does not meet CT it can either increase the detention time through the pipeline or plant or it can increase the dosage. In a situation where two different disinfection chemicals are used, the required CT values would be different, and you would not add the different disinfecting locations together. The next example illustrates this scenario. Example: A conventional water treatment plant receives water with a 0.4 mg/L free chlorine residual from 9,000 feet of 3-foot diameter pipe at a constant flow rate of 10 MGD. The water has a pH of 7.5 and a temperature of 10°C. Tracer studies have shown a contact time (T10) for the treatment plant to be 30 minutes. The
The first step in solving this problem is identifying the CT Tables to use to find the required CT values. This particular problem uses CT Tables C-3 and C-10. Remember to subtract out the 2.5 Log credit for conventional treatment.
The sum of the CT ratios equals 1.42 mg/L · min. Therefore, CT is met. You may have noticed that CT was achieved through the pipeline only and the chloramination through the plant is not needed. This is true. So, when solving one of these problems, once you meet the ratio of 1.0 or greater, CT is met, and you can stop solving the problem.
Practice Problems 7.1 What is the required CT inactivation in a conventional filtration plant for Giardia by free chlorine at 20°C with a pH of 8.0 and a chlorine concentration of 2.0 mg/L? (Look up value in the CT tables and remember to apply any credits.)
A conventional water treatment plant is fed from a reservoir 1.5 miles away through a 7-foot diameter pipe. Disinfection is provided from the supply reservoir to the plant influent at a free chlorine residual of 0.6 mg/L. The daily flow is a constant 40 MGD. And the water is 10°C and has a pH of 8.5. The treatment plant maintains a chloramines residual of 2.0 mg/L. Tracer studies have shown the contact time (T10) for the treatment plant at the rated capacity of 40 MGD to be 22 minutes. Does this plant meet compliance for CT inactivation for Giardia? A conventional water treatment plant is fed from a reservoir 4 miles away through a 4- foot diameter pipe. Disinfection is provided from the supply reservoir to the plant influent at a free chlorine residual of 0.1 mg/L. The daily flow is a constant 25 MGD. The water is 10°C and has a pH of 7.0. The treatment plant maintains a chloramines residual of 1.5 mg/L. Tracer studies have shown the contact time (T10) for the treatment plant at the rated capacity of 25 MGD to be 55 minutes. Does this plant meet compliance for CT inactivation for viruses? A direct filtration water treatment plant is fed from a reservoir 0.5 miles away through a 3-foot diameter pipe. Disinfection is provided from the supply reservoir to the plant influent at a free chlorine residual of 0.6 mg/L. The daily flow is a constant 15 MGD. The water is 15°C and has a pH of 7.0. The treatment plant maintains a chloramines residual of 0.4 mg/L. Tracer studies have shown the contact time (T10) for the treatment plant at the rated capacity of 15 MGD to be 30 minutes. Does this plant meet compliance for CT inactivation for Giardia? A direct filtration plant is operated at a designed flow of 20 MGD with a contact time of 35 minutes. A free chlorine dose of 1.2 mg/L is maintained through the plant. Upon leaving the plant, the effluent is chloraminated (and maintained to the distribution system) to a dose of 0.4 mg/L through a pipeline with a contact time of 12 minutes into a 650,000-gallon reservoir. The pH of the water is 8.5 and has a temperature of 15°C. Does this treatment process meet compliance for CT inactivation for viruses? Does a water utility meet CT for viruses by disinfection if only the free chlorine concentration is 0.5 ppm through 200 ft of 24” diameter pipe at a flow rate of 730 gpm? Assume the water is 15°C and has a pH of 8.0. Exercise 7.1 What is the required CT inactivation in a conventional filtration plant for Giardia by free chlorine at 20°C with a pH of 9.0 and a chlorine concentration of 1.0 mg/L? (Look up value in the CT tables and remember to apply any credits.)
A conventional water treatment plant is fed from a reservoir 3 miles away through a 5- foot diameter pipe. Disinfection is provided from the supply reservoir to the plant influent at a free chlorine residual of 0.3 mg/L. The daily flow is a constant 50 MGD. And the water is 10°C and has a pH of 8.0. The treatment plant maintains a chloramines residual of 1.0 mg/L. Tracer studies have shown the contact time (T10) for the treatment plant at the rated capacity of 50 MGD to be 30 minutes. Does this plant meet compliance for CT inactivation for Giardia? A conventional water treatment plant is fed from a reservoir 2 miles away through a 6- foot diameter pipe. Disinfection is provided from the supply reservoir to the plant influent at a free chlorine residual of 0.2 mg/L. The daily flow is a constant 55 MGD. The water is 10°C and has a pH of 7.5. The treatment plant maintains a chloramines residual of 1.0 mg/L. Tracer studies have shown the contact time (T10) for the treatment plant at the rated capacity of 55 MGD to be 40 minutes. Does this plant meet compliance for CT inactivation for viruses? A direct filtration water treatment plant is fed from a reservoir 2.5 miles away through a 4-foot diameter pipe. Disinfection is provided from the supply reservoir to the plant influent at a free chlorine residual of 0.4 mg/L. The daily flow is a constant 30 MGD. The water is 15°C and has a pH of 8.5. The treatment plant maintains a chloramines residual of 0.75 mg/L. Tracer studies have shown the contact time (T10) for the treatment plant at the rated capacity of 30 MGD to be 20 minutes. Does this plant meet compliance for CT inactivation for Giardia? A direct filtration plant is operated at a designed flow of 10 MGD with a contact time of 15 minutes. A free chlorine dose of 0.5 mg/L is maintained through the plant. Upon leaving the plant, the effluent is chloraminated (and maintained to the distribution system) to a dose of 1.0 mg/L through a pipeline with a contact time of 10 minutes into a 500,000-gallon reservoir. The pH of the water is 8.0 and has a temperature of 20°C. Does this treatment process meet compliance for CT inactivation for viruses? Does a water utility meet CT for viruses by disinfection if only the free chlorine concentration is 2.60 ppm through 150 ft of 48” diameter pipe at a flow rate of 1,200 gpm? Assume the water is 15°C and has a pH of 7.0. UNIT 8 PRESSURE Pressure is the amount of force that is “pushing” on a specific unit area. What does this mean? When you turn on your water faucet or shower you feel the water flowing out, but why is it flowing out? Water flows through pipes and out of faucets because it is under pressure. It could be that a pump is turned on in which case the pump and motor are providing the pressure. More commonly, the pressure is being provided by water being stored at a higher elevation. This is why you see water tanks on top of hills.
Pressures are usually expressed as pounds per square inch (psi), but they can be expressed as pounds per square foot or pounds per square yard as well. The key is that the force is expressed per unit area. Typically, water operators will measure pressures with gauges and express the unit answer as psig. The “g” is this case represents gauge. However, it is also common to express pressure in feet. Feet represent the height of the water in relation to the location that the pressure is being measured. There are two commonly used factors to convert from feet to psi and vice versa. For every foot in elevation change there is a 0.433 change in psi. Conversely, for every psi change there is a 2.31 foot in elevation change.
From this example, it is clear that the pressure exerted on the bottom of the tank is the same for either tank, but what about the force? Is the force exerted on the bottom of both tanks the same too?
Example: Looking at these tanks again, assume that the diameter of the narrower tank is 10 feet and the diameter of the wider tank is 40 feet. Assume both tanks are filled with water. What is the force exerted on the bottom of each tank?
Clearly, the force exerted on the bottom of the larger, 40-foot diameter tank, is significantly greater than the force exerted on the 10-foot diameter tank. Let’s look at this problem a different way. The formula for force is:
Typically, pressure would be calculated in psi, but this requires that the area be expressed in square inches in order to calculate force. Since the tank diameters are given in feet, you can convert pressure from psi, pounds per square inch, to pounds per square foot.
As you can see, the outcome is the same. The force being exerted on the bottom of the larger tank is significantly greater. The calculated value of the force is slightly different from the previous calculation due to rounding in the standard conversion factors.
A home sits at an elevation of 900 ft above sea level. The base of a water tank that serves the home sits at an elevation of 1,281 ft above sea level. The tank is 20 feet tall and ¾ full. What is the pressure in psi at the home? Two houses are served by a nearby water storage tank. House A is 108 ft above House B which sits at 432 ft above sea level. The base of the tank sits at 705 ft above sea level. The low water level in the tank is at 2.0 ft. At the low level, will House A meet the minimum pressure requirements of 60 psi? House A sits at an elevation of 1,300 ft. Another house (B) needs to be built 125 ft below House A. At what elevation should the tank be built in order to give House B the maximum pressure of 210 psi?
A home sits at an elevation of 1,301 ft above sea level. The base of a water tank that serves the home sits at an elevation of 1,475 ft above sea level. The tank is 35 feet tall and ¾ full. What is the pressure in psi at the home? Two houses are served by a nearby water storage tank. House A is 55 ft above House B which sits at 725 ft above sea level. The base of the tank sits at 855 ft above sea level. The low water level in the tank is at 7.5 ft. At the low level, will House A meet the minimum pressure requirements of 35 psi?
HEAD LOSS As water travels through objects including pipes, valves, and angle points, or goes up hill, there are losses in pressure (or head) due to the friction. These losses are called “friction” or head loss. There are standard tables listing head loss factors (also termed C factor) for pipes of differing age and material, different types of valves and angle points. However, in this text we will focus on the theory more than the actual values. Example: If water is traveling through 10,000 feet of pipe that has head loss of 3 feet, passes through 4 valves that have head loss of 1 foot for each valve, and passes through 2 angle points that have head loss of 0.5 feet each, calculate the total head loss.
In distribution systems, water is pumped from lower elevations to higher elevations in order to supply customers with water in different areas called zones. Water is also pumped out of the ground using groundwater wells and from treatment plants that typically treat water from surface water sources. The water is then sent throughout the distribution system to supply customers in different pressure zones. As water makes its way through the distribution system head loss is realized (as mentioned in the previous paragraph) and pumps must overcome the head loss from elevation changes, interior pipe conditions, valves, and sudden angles. Knowing the head losses will help determine what size pump is needed in a specific pressure zone. However, there are other forces acting on the pump that either help or hinder its ability to pump water. The diagrams below help illustrate the differences between suction lift and suction head. Suction lift requires more work by the pump to move the water from point A to point B because it has to lift water from a lower elevation. Suction head provides some help (head pressure) to the pump in order to get water from point A to point B.
Friction will need to be added at the first measuring point because the pump will need to work harder to push water up to the tank. However, if water is traveling to a lower elevation like a home, then friction will need to be subtracted because the home will experience less pressure than it would have if there was no friction in the supply pipe. Key Terms head loss – loss in pressure or head in a water system pressure – the amount of force “pushing” on a specific unit area suction lift – (Tank A Elevation + Pump Elevation) + (Pump Elevation + Tank B Elevation) = Total Head suction head – (Tank B Elevation - Pump Elevation) - (Tank A Elevation – Pump Elevation) = Total Head Practice Problems 8.2 A well pumps directly to a 60-foot tall water tank that sits 500 feet above the elevation of the well. If the total head loss in the piping up to the tank is 14 feet, what is the total pressure in psi on the discharge side of the well?
A well located at 200 feet above sea level has a below ground surface water depth of 50 ft and pumps to a water tank at an elevation of 840 ft above sea level. The water main from the well to the tank has a total head loss of 6 psi. What is the TH in feet? A housing tract is located at an approximate average elevation of 5,000 ft above sea level and is served from a storage tank that is at 5,460 ft. The average head loss from the tank to the housing tract is 20.3 psi. What is the minimum water level in the tank to maintain a minimum pressure 30 psi? A water utility has two different pressure zones (1 and 2.) The zone 1 Tank is 15 ft tall and sits at an elevation of 625 ft and the zone 2 Tank is 50 feet tall and sits at 1,300 ft. The booster pump from zone 1 to 2 sits at an elevation of 700 ft. The head loss is 11 psi. Tank 1 is full, and Tank 2 needs to be 1/2 full. What is the TH?
A well pumps directly to a 25-foot tall water tank that sits 200 feet above the elevation of the well. If the total head loss in the piping up to the tank is 5 feet, what is the total pressure in psi on the discharge side of the well?
A well located at 750 feet above sea level has a below ground surface water depth of 38 ft and pumps to a water tank at an elevation of 1,030 ft above sea level. The water main from the well to the tank has a total head loss of 11 psi. What is the TH in feet? A housing tract is located at an approximate average elevation of 2,225 ft above sea level and is served from a storage tank that is at 2,330 ft. The average head loss from the tank to the housing tract is 15.5 psi. What is the minimum water level in the tank to maintain a minimum pressure 40 psi? A water utility has two different pressure zones (1 and 2.) The zone 1 Tank is 30 ft tall and sits at an elevation of 850 ft and the zone 2 Tank is 40 feet tall and sits at 1,061 ft. The booster pump from zone 1 to 2 sits at an elevation of 925 ft. The head loss is 19 psi. Tank 1 is half full and Tank 2 needs to be ¾ full. What is the TH? UNIT 9 WELL YIELD, SPECIFIC CAPACITY, AND DRAWDOWN Many people in rural areas rely on their own well water as their primary and only source of water supply. Water agencies also rely on well water, in some cases, as their primary and only supply of water. While the diagram below is of a single-family household well, the key parts are the same: well casing, well screen, and a submersible pump. The well casing is a tube that maintains the opening in the ground. The well screen is attached to the bottom of the casing and decreases the amount of sand that enters the well. The pump brings the water to the surface.
pump tests are performed to determine if the underlying aquifer can supply enough water. Continuous pumping for an extended period is usually performed and the yield is calculated based on the amount of water extracted. Well yields are typically measured in the field with a flow meter. Drawdown In order to understand the term drawdown, you must also understand static water level and pumping water level. The static water level is defined as the distance between the ground surface and the water level when the well is not operating. The pumping level is defined as the distance between the ground surface and the water level when a well is pumping. Therefore, the pumping water level is always deeper than the static water level. The difference between these two levels is the drawdown. Depending on the aquifer, static water levels can be 20 feet below ground surface (bgs) or several hundred feet bgs.
Cone of Depression The triangular shape that results as a difference between the static water level and pumping water level is called the cone of depression. The bigger the well capacity, the bigger the cone of depression.
Specific Capacity Specific Capacity is helpful in assessing the overall performance of a well and the transmissivity, horizontal flow ability, of the aquifer. The specific capacity is used in determining the pump design in order to get the maximum yield from a well. It is also helpful in identifying problems with a well, pump, or aquifer. The specific capacity is defined as the well yield divided by the drawdown, expressed as gallons per minute per foot of drawdown.
Key Terms cone of depression – the triangular shape that results as a difference between static water level and pumping water level specific capacity – the well yield divided by the drawdown well yield – the amount of water a certain well can produce over a period of time
When a well was first constructed it was pumping 1,237 gpm. The efficiency of the well has dropped 42%. In addition, the drawdown has decreased by 22%. If the original drawdown was 66 ft what is the current specific capacity? A well pumped 468 AF over a one-year period averaging 14 hours of operation per day. For half the year the static water level was 41 ft bgs and half the year 30 ft bgs. The pumping level averaged 72 ft bgs for half the year and 83 ft bgs the other half. What was the average specific capacity for the year?
When a well was first constructed it was pumping 1,750 gpm. The efficiency of the well has dropped 35%. In addition, the drawdown has decreased by 15%. If the original drawdown was 42 ft what is the current specific capacity? A well pumped 538 AF over a one-year period averaging 10 hours of operation per day. For half the year the static water level was 25 ft bgs and half the year 42 ft bgs. The pumping level averaged 55 ft bgs for half the year and 68 ft bgs the other half. What was the average specific capacity for the year?
UNIT 10 HORSEPOWER AND EFFICIENCY We discussed the theory of pressure in both feet (head pressure) and psi (pounds per square inch.) In this unit, we will look at the “power” requirements to move water with pumps and motors. How does water get to the customer’s home? Water pressure is typically provided to customers because of differences in elevation with above ground tanks, reservoirs, elevated storage tanks. In the Santa Clarita, tanks are clearly visible on the hill tops surrounding the valley floor, which creates water pressure to the customer’s home
Most water utilities have maps that show how pressure is broken into different zones with mechanisms for moving water between points within each zone. Moving water throughout the zones within the overall system is something that is carefully considered by distribution operators during daily work as well as during the design of the system itself or new portions of the system.
How does the water move to the storage tanks? This is where the concept of horsepower comes in. Historically, the definition of horsepower was the ability of a horse to perform heavy tasks such as turning a mill wheel or drawing a load. It wasn’t until James Watt (1736‐ 1819) invented the first efficient steam engine that horsepower was used as a standard to which the power of an engine could be meaningfully compared. Watt's standard of comparing “work” to horsepower (hp) is commonly used for rating engines, turbines, electric motors, and water‐ power devices.
The above equation is used to calculate the power needed to move a certain flow of water a certain height. The constant, 3,960, is the result of converting the 33,000 ft-lb/min with the weight of water flow. For example, instead of using gallons per minute, pounds per minute would be needed because 33,000 is in foot-pounds. Water horsepower is the theoretical power needed to move water. In order to actually perform the work a pump and motor are needed. However, neither the pump nor the motor is 100% efficient. There are friction losses with each. If the pump and the motor were both 100% efficient, then the resulting answer would be 100% x 100% or 1.0 x 1.0 = 1. Hence, the actual horsepower would be the water horsepower and the equation is not affected. However, this is never the case. Typically, there are inefficiencies with both components. This efficiency is termed the wire-to-water efficiency.
The horsepower required by the pump (brake horsepower) can be calculated, but the actual horsepower needed looks at the efficiencies of both the pump and the motor. The formula below shows brake horsepower and motor horsepower, which includes the combined pump and motor inefficiencies.
As with all water-related math problems, it is important for the numbers being used to be in the correct units. For example, the flow needs to be in gallons per minute (gpm) and the total head in feet (ft). These will not always be the units provided in the questions. The example below demonstrates this concept. Example: What is the horsepower of a well that pumps 2.16 million gallons per day (MGD) against a head pressure of 100 pounds per square inch (psi)? Assume that the pump has an efficiency of 65% and the motor 85%.
What is the motor horsepower needed to pump 4,643 AF of water over a year with an average daily pumping operation of 6 hours? Assume the pump is pumping against 70 psi and has a pump efficiency of 90% and a motor efficiency of 75%.
What is the motor horsepower needed to pump 2,420 AF of water over a year with an average daily pumping operation of 12 hours? Assume the pump is pumping against 95 psi and has a pump efficiency of 70% and a motor efficiency of 80%. HEAD LOSS AND HORSEPOWER As discussed in Unit 8, suction pressure can either be expressed as “lift” or “head.” In other words, the location of the water on the suction side of the pump can either help or hinder the pump.
The diagram on the left (suction lift) requires work from the pump to bring the water up to the pump and then additional work to bring the water to the reservoir above the pump. The diagram on the right (suction head) receives “help” from the tank on the suction side and the pump only has to lift water the height difference between the two tanks. When calculating horsepower, the total head pressure (suction lift + discharge head) or (discharge head – suction head) needs to be calculated. Example: A booster pump station is pumping water from Zone 1 at an elevation of 2,500 ft above sea level to Zone 2 which is at 3,127 ft above sea level. The pump station is located at an elevation of 1,824 ft above sea level. The losses through the piping and appurtenances equate to a total of 31 ft. Is this an example of Suction Lift or Suction Head? What is the total head?