Datasets:

izumi-lab
/

open-text-books

Dataset card Viewer Files Files and versions Community

Split (1)

train · 150k rows

text stringlengths 200 319k
Example: A well with pumps located 125 ft bgs pumps against a discharge head pressure of 85 psi to a tank located at an elevation 150 ft above the well. What is the level of water in the tank and what is the total head?
1 1 psi The problem indicates that the tank is located 150 feet above the well. Therefore, the height of the water in the tank is the difference between the elevation and the discharge head pressure in feet.
Practice Problems 10.2 A well is pumping water from an aquifer with a water table 55 feet below ground surface (bgs) to a tank 190 feet above the well. If the well flows 870 gpm, what is the required horsepower? (Assume the wire-to-water efficiency is 88%.) A booster pump station is pumping water from Zone 1 at an elevation of 1,537 ft above sea level to Zone 2 which is at 1,745 ft above sea level. The pump station is located at an elevation of 1,124 ft above sea level. The pump was recently tested and the efficiencies for the pump and motor were 76% and 88% respectively. The losses through the piping and appurtenances equate to a total of 15 ft. If the pump flows 1,500 gpm, what is the required motor horsepower? A well with pumps located 46 ft bgs pumps against a discharge head pressure of 140 psi to a tank located at an elevation 278 ft above the well. The well pumps at a rate of 1,260 gpm. What is the level of water in the tank and what is the required water horsepower? (Assume the wire-to-water efficiency is 70%.) A 430 hp booster pump is pulling water from a 50-foot tall tank that is 85 feet below the pump line. It is then pumping against a discharge head pressure of 150 psi. What is the flow rate in gpm? Assume the wire-to-water efficiency is 92% and the tank is full.
A well is pumping water from an aquifer with a water table 30 feet below ground surface (bgs) to a tank 150 feet above the well. If the well flows 1,000 gpm, what is the required horsepower? (Assume the wire-to-water efficiency is 68%.) A booster pump station is pumping water from Zone 1 at an elevation of 1,225 ft above sea level to Zone 2 which is at 1,445 ft above sea level. The pump station is located at an elevation of 1,175 ft above sea level. The pump was recently tested and the efficiencies for the pump and motor were 62% and 78% respectively. The losses through the piping and appurtenances equate to a total of 11 ft. If the pump flows 1,200 gpm, what is the required motor horsepower? A well with pumps located 75 ft bgs pumps against a discharge head pressure of 125 psi to a tank located at an elevation 253 ft above the well. The well pumps at a rate of 1,050 gpm. What is the level of water in the tank and what is the required water horsepower? Assume the wire-to-water efficiency is 55%. A 200 hp booster pump is pulling water from a 32-foot-tall tank that is 50 feet below the pump line. It is then pumping against a discharge head pressure of 112 psi. What is the flow rate in gpm? Assume the wire-to-water efficiency is 65% and the tank is full. CALCULATING POWER COSTS It is important for water managers to determine the potential costs in electricity for pumping water. Units used for measuring electrical usage are typically in kilowatt hours (kW-Hr). In order to convert horsepower to kilowatts of power, the following conversion factor is used.
Once you know the hp that is needed, you can determine the amount of kW-Hr needed. Then, costs can be determined depending on what the local electric company charges per kW-Hr. Water utilities will calculate estimated budgets for pumping costs since these are typically the largest operating costs. Example: A utility has 3 pumps that run at different flow rates and supply water to a 450,000 MG storage tank. Assume that only one pump runs per day. The TDH for the pumps is 73 ft. The utility needs to fill the tank daily and power costs are to be calculated at a rate of $0.088 per kW‐Hr. Complete the table below.
A utility has 3 pumps that run at different flow rates and supply water to an 800,000 gallon storage tank. Assume that only one pump runs per day. The TDH for the pumps is 130 ft. The utility needs to fill the tank daily and power costs are to be calculated at a rate of $0.10 per kW‐Hr. Complete the table below. A well draws water from an aquifer that has an average water level of 100 ft bgs and pumps to a tank 300 ft above it. Friction loss to the tank is approximately 28 psi. If the well pumps at a rate of 1,900 gpm and has a wire‐to‐water efficiency of 45%, how much will it cost to run this well 10 hours per day. Assume the electrical rate is $0.22 per kW‐ Hr. A utility manager is trying to determine which hp motor to purchase for a pump station. A 500 hp motor with a wire‐to‐water efficiency of 70% can pump 3,300 gpm. Similarly, a 300 hp motor with a wire‐to‐water efficiency of 80% can pump 2,500 gpm. With an electrical rate of $0.111 per kW‐Hr, how much would it cost to run each motor to achieve a daily flow of 1.5 MG? Which one is less expensive to run? Approximately 170 kW of power are needed to run a certain booster pump. If the booster has a wire‐to‐water efficiency of 81% and is pumping against 205 psi of head pressure, what is the corresponding flow in gpm?
It costs $103.61 in electricity to run a well for 10 hours a day. The well has a TDH of 167 psi and an overall efficiency of 82.3%. The cost per kW‐Hr is $0.156. What is the cost of the water per gallon?
A utility has 3 pumps that run at different flow rates and supply water to a 500,000- gallon storage tank. Assume that only one pump runs per day. The TDH for the pumps is 210 ft. The utility needs to fill the tank daily and power costs are to be calculated at a rate of $0.135 per kW‐Hr. Complete the table below. A well draws water from an aquifer that has an average water level of 150 ft bgs and pumps to a tank 225 ft above it. Friction loss to the tank is approximately 22 psi. If the well pumps at a rate of 2,300 gpm and has a wire‐to‐water efficiency of 62%, how much will it cost to run this well 14 hours per day. Assume the electrical rate is $0.13 per kW‐ Hr. A utility manager is trying to determine which hp motor to purchase for a pump station. A 400 hp motor with a wire‐to‐water efficiency of 65% can pump 3,000 gpm. Similarly, a 250 hp motor with a wire‐to‐water efficiency of 75% can pump 2,050 gpm. With an electrical rate of $0.155 per kW‐Hr, how much would it cost to run each motor to achieve a daily flow of 2 MG? Which one is less expensive to run? Approximately 224 kW of power are needed to run a certain booster pump. If the booster has a wire‐to‐water efficiency of 67.5% and is pumping against 135 psi of head pressure, what is the corresponding flow in gpm?
It costs $88.77 in electricity to run a well for 7 hours a day. The well has a TDH of 100 psi and an overall efficiency of 58.3%. The cost per kW‐Hr is $0.17. What is the cost of the water per gallon? UNIT 11 PER CAPITA WATER USE AND WATER USE EFFICIENCY Water use is often expressed as gallons per capita per day (GPCD). The term “per capita” means per person and the term “per day” means in a 24-hour period. Since the last drought, this term has been increasingly used to describe water use within communities, particularly for comparison of communities to one another.
There are many communities that are hubs of employment and tourism in Southern California that can distort GPCD. For example, in Anaheim, the total systemwide GPCD is 105 GPCD in March 2020, but the residential GPCD is only 58.45 Can you imagine why? Anaheim is home to Disneyland, so there are many tourists who come and stay overnight to visit Disneyland and use a lot of water along the way. Total GPCD does not necessarily measure residential GPCD (R- GPCD).
One way to understand residential GPCD is to consider the pattern in the map. Communities that are on the coasts have a low GPCD while communities that are inland have a much higher GPCD. This reflects that R-GPCD is related to both indoor and outdoor water use. The demands for water by plants inland communities are much greater than in coastal communities, so the R- GPCDs for inland communities are far greater than in coastal communities.
Example: If a small water system has one well that can pump 250 gallons per minute to serve a population of 1,575 people with an average GPCD of 100, how many hours must this pump run to meet the demand?
You can see how GPCD can be a useful calculation in reporting to the state and estimating how to meet demand of customers. Sometimes it is useful to focus on the residential GPCD or R- GPCD to measure conservation.
The biggest component is typically landscaping. In California, 60-70% of residential use is for outdoor landscaping, including front yards, backyards, pools, spas and any other water features. The state of California and local water agencies have encouraged residents to remove turf grass, particularly in their front yards where it is mostly “aesthetic” and replace it with lower-water use plants. Slowly this is decreasing outdoor residential water use.
Green Building Code) has resulted in considerable savings. The table below measures flow rates for various fixtures in units of gallons per minute (gpm) for showers and faucets, gallons per flush (gpf) for toilets and gallons per cubic foot (gpft2) for clothes washers. Example: How much water would a family of four save over a year from replacing two toilets from 1975 with two toilets purchased in 2020. Assume each person flushes each two times a day. (Typically, people use the bathroom 4-7 times per day, but not always at their house. Some of this water use is accounted for at schools or businesses and restaurants).
UNIT 12 BLENDING AND DILUTING Dilution is not the solution to pollution, but dilution can be used to reduce the level of a contaminant in drinking water supplies. Blending water sources of different water quality is common practice. However, when a water utility wants to blend sources of supply to lower a certain contaminant to acceptable levels, they must receive approval from the governing Health Department. A Blending Plan must be created that specifies what volumes of water from each source will be used and what the expected resulting water quality will be. In addition, a sampling strategy must be included in the plan. The Health Department may not allow blending for all contaminants. For example, the local health agency may not approve a blending plan for a contaminant that poses an acute health effect or is deemed to be too high of a risk to public health. An acceptable blending plan may be for reducing manganese in a source that has exceeded the California Secondary Maximum Contaminant Level (MCL) of 0.05 mg/L. Manganese causes black water problems for customers at levels over the secondary MCL. Additionally, an approved blending plan may involve a Primary MCL for nitrate. Nitrates above the MCL of 45 mg/L as NO3 can cause methemoglobinimia in infants under 6 months old. These are just two examples of blending plans. How are blended water quality results calculated? The blending of water supplies is nothing more than comparing ratios. For example, if 100 gallons of one source was mixed with 100 gallons of another source, the resulting water quality would be the average between the two sources. However, when you mix varying flows with varying water quality, the calculations become a little more complex. Using the diagram below will assist you in solving blending problems.
If two sources are to be blended, the water quality data for both sources is known. One of the sources with a poor- or high-water quality result for a certain constituent will need to be blended with a source that has good or low water quality data. Source A will be the high out of compliance data point and Source B will be the low in compliance data point. Source C is the desired blended result. Typically, this value is an acceptable level below an MCL. Once these values are established the ratios of the differences between these numbers can be calculated. For example, the ratio of C ‐ B to A ‐ B yields the quantity of Source A that is needed. Therefore, in the example below, the quantity of A needed is 37.5%. The same thing holds true for Source B. Simply take the ratio of the difference between the high (A) and desired (C) values and divide it by the difference between the high (A) and low (B) values. However, once you solve for the quantity of one source, simply subtract it from 100% to get the value for the other source. See the example below. It is expected that water quality results can and will fluctuate. It is always a good idea to take the highest result from recent sampling when calculating needed blend volumes to reduce the impacted water to acceptable levels. For example, if a well is being sampled for trichloroethylene (TCE) quarterly and the results are 6 ug/L, 7.8 ug/L, 5.9 ug/L, and 8.5 ug/L from a recent year of sampling, it would be prudent to use the 8.5 mg/L result when calculating blending requirements. It is also important to note that the local health authority should be consulted with respect to any blending plan. Example: A water utility would like to blend water source A and water source B. Water source A has 10 ppm of a particular contaminant and water source B has 2 ppm of the contaminant. The goal is to reduce the total contaminant level to 5 ppm. How much water from source A and source B need to be blended to produce this result?
Once the percentage of each source has been calculated the actual flows can be determined. Sometimes the total flow from both sources is known. In this case you would take that known flow rate and multiply it by the respected percentages of each source.
Practice Problems 12.1 A well has a nitrate level that exceeds the MCL of 63 mg/L. Over the last 4 sample results it has averaged 71 mg/L. A nearby well has a nitrate level of 40 mg/L. If both wells combined pump up to 1,575 gpm, how much flow is required from each well to achieve a nitrate level of 50 mg/L? A well (A) has shown quarterly arsenic levels above the MCL over the last year, of 16 ug/L, 22 ug/L, 20 ug/L and 10 ug/L. A utility wants to blend this well to a level of 6.0 ug/L with a well (B) that has a level of 2.1 ug/L. The total production needed from both of these wells is 4,100 gpm. How much can each well produce? A well with a PCE level of 12.4 ug/L is supplying approximately 65% of total water demand. It is being blended with a well that has a PCE level of 1.5 ug/L. Will this blended supply meet the MCL for PCE of 7.0 ug/L? Well A has a total dissolved solids (TDS) level of 625 mg/L. It is pumping 2,300 gpm, which is 50% of the total production from two wells. The other well (B) blends with well A to achieve a TDS level of 450 mg/L. What is the TDS level for Well B? Two wells need to achieve a daily flow of 2.1 MG and a total hardness level of 110 mg/L as calcium carbonate (CaCO3.) Well #1 has a total hardness level of 390 mg/L as CaCO3 and Well #2 has a level of 63 mg/L as CaCO3. What is the gpm that each well must pump? The State Health Department has requested a blending plan to lower levels of sulfate from a small water utility well. The well has a constant sulfate level of 480 mg/L. The utility needs to purchase the water to blend with the well. The purchased water has a sulfate level of 55 mg/L. They need to bring the sulfate levels down to 225 mg/L and supply a demand of 2.0 MGD. The purchased water costs $475/AF. How much will the purchased water cost for the entire year?
A well has a nitrate level that exceeds the MCL of 45 mg/L. Over the last 3 sample results it has averaged 52 mg/L. A nearby well has a nitrate level of 32 mg/L. If both wells combined pump up to 2,275 gpm, how much flow is required from each well to achieve a nitrate level of 40 mg/L? A well (A) has shown quarterly arsenic levels above the MCL over the last year, of 14 ug/L, 20 ug/L, 18 ug/L and 16 ug/L. A utility wants to blend this well to a level of 8.0 ug/L with a well (B) that has a level of 4.5 ug/L. The total production needed from both of these wells is 3,575 gpm. How much can each well produce? A well with a PCE level of 7.5 ug/L is supplying approximately 35% of total water demand. It is being blended with a well that has a PCE level of 3.25 ug/L. Will this blended supply meet the MCL for PCE of 5.0 ug/L? Well A has a total dissolved solids (TDS) level of 850 mg/L. It is pumping 1,500 gpm which is 40% of the total production from two wells. The other well (B) blends with well A to achieve a TDS level of 375 mg/L. What is the TDS level for Well B? Two wells need to achieve a daily flow of 3.24 MG and a total hardness level of 90 mg/L as calcium carbonate (CaCO3.) Well #1 has a total hardness level of 315 mg/L as CaCO3 and Well #2 has a level of 58 mg/L as CaCO3. What is the gpm that each well must pump? The State Health Department has requested a blending plan to lower levels of sulfate from a small water utility well. The well has a constant sulfate level of 525 mg/L. The utility needs to purchase the water to blend with the well. The purchased water has a sulfate level of 135 mg/L. They need to bring the sulfate levels down to 265 mg/L and supply a demand of 1.15 MGD. The purchased water costs $550/AF. How much will the purchased water cost for the entire year? UNIT 13 SCADA AND THE USE OF MA SCADA SCADA is the acronym for Supervisory Control and Data Acquisition. It is a computerized system allowing a water system to operate automatically. This does not mean that human beings are not involved. Distribution and treatment operators learn to use SCADA to help them in their work.
A SCADA system usually consists of three (3) basic components: field instrumentation, communications (telemetry), and some type of central control equipment. The field instrumentation will measure various parameters such as flow, chemical feed rates, chemical dosage levels, and tank levels. These instruments will then gather a series of signals and transmit them through some type of communication device(s) known as telemetry. The telemetry communication can be radio signals, telephone lines, or fiber optics. This information is sent to a central control computer typically located at an office or operations control center.
A common measurement used to analyze the various field parameters of a water system is the 4-20 milliamp (mA). A 4-20 mA signal is a point-to-point circuit which is used to transmit signals from instruments and sensors in the field to a controller. The 4 to 20 mA analog signal represents 0 to 100% of some process variable. For example, this 0 to 100% process variable can be a chlorine residual from 0.2 to 4.0 mg/L or a tank level of 0 to 40 feet. The 0% would represent the lowest allowed value of the process and 100% the highest. These mA signals are then sent through the SCADA system and processed into understandable values such as mg/L or feet, depending on the parameter being measured. When using this system to measure tank levels there are a couple of things to consider. First, assume the tank in the image below is 40 feet tall. Although the height of the tank is 40 ft, the water is never filled to that height. Why? Because the inside roof of the tank would be damaged. Therefore, all storage tanks have an “overflow” connected at the top of the tank. In this image you can see it on the top right side of the tank. The second thing to point out is that the “bottom” or zero level of the tank is never at the actual bottom of the tank. Why? Because you never want to run a tank empty. There is always a several foot distance from the actual
bottom to what is referred to as the “zero” level. In solving water related problems, the “overflow” (actual top level) and the “bottom” (actual location of the zero level) may be provided in the problem statement.
Example: A water utility has a 40 ft tall tank with a diameter of 30 ft as shown below. They are using the 4-20 mA signal to measure the level of water in the storage tank. What is the mA reading if the tank is half full (20 ft)? Since there is no reference in the problem statement to an overflow or where the zero level is located, the 4 mA signal would represent 0 ft and the 20 mA signal 40 ft. What this is saying is if your meter sends out a signal of 20 mA, then the corresponding level in feet would be 40. Likewise, if the signal was 4 mA, the corresponding level would be 0 ft.
If you initially thought 10 mA, that would be a logical guess. However, let’s think about this for a minute. Since the bottom or 0 ft is at 4 mA and the top or 40 ft is at 20 mA, the span or difference between 4 and 20 is only 16…not 20. This “span” is an important number when solving these problems.
span of 0 - 16, we are dealing with a span of 4 - 20. Therefore, half of 16 is 8, but the halfway distance between 4 and 20 is 12! Anyone who guessed 12 mA, give yourself a hand. Whatever read you have on your meter, you must subtract out the 4 mA offset.
Practice Problems 13.1 A 4-20 mA signal is being used to measure the water level in a water storage tank. The tank is 50 feet tall and the low level signal is set at 0 feet and the high level at 50 feet. What is the level in the tank with a 17 mA reading? A 48 ft tall water tank uses a 4-20 mA signal for calculating the water level. If the 4 mA level is set at 5 feet from the bottom and the 20 mA is set at 5 feet from the top, what is the level in the tank with a 9 mA reading?
A utility uses a 4-20 mA signal to determine the level in a well based on pressures. The set points are based on pressures in psi below ground surface (bgs). The 20 mA signal is set at 205 psi bgs and the 4 mA signal at 15 psi bgs. If the reading is 14 mA, what is the water level in feet? A water utility uses a 4-20 mA signal to determine groundwater elevations in a well. The set points are based on actual elevations above the mean sea level (MSL). The ground surface elevation at this well is 1,400 ft and this is where the 4 mA signal is set. The 20 mA signal is set at 740 ft. What is the elevation and the feet bgs with an 11 mA reading?
Exercise 13.1 A 4-20 mA signal is being used to measure the water level in a water storage tank. The tank is 32 feet tall and the low level signal is set at 0 feet and the high level at 32 feet. What is the level in the tank with a 15 mA reading? A 75 ft tall water tank uses a 4-20 mA signal for calculating the water level. If the 4 mA level is set at 4 feet from the bottom and the 20 mA is set at 4 feet from the top, what is the level in the tank with a 10 mA reading?
A utility uses a 4-20 mA signal to determine the level in a well based on pressures. The set points are based on pressures in psi below ground surface (bgs). The 20 mA signal is set at 182 psi bgs and the 4 mA signal at 10 psi bgs. If the reading is 9 mA, what is the water level in feet? A water utility uses a 4-20 mA signal to determine groundwater elevations in a well. The set points are based on actual elevations above the mean sea level (MSL). The ground surface elevation at this well is 1,180 ft and this is where the 4 mA signal is set. The 20 mA signal is set at 930 ft. What is the elevation and the feet bgs with a18 mA reading?
UNIT 14 WATER UTILITY MANAGEMENT Every water utility has a management staff that directs, plans, organizes, coordinates, and communicates the direction of the organization. One important function of utility managers is financial planning. Managers are responsible for preparing budgets, working on water rate structures, and calculating efficiencies within the organization. Budgets How much money does a utility need to perform the routine, preventative, and corrective action maintenance items? How much money is needed to operate the utility? How much needs to be spent on Capital Improvement Projects? How much needs to be saved for emergencies? Does the utility have any debt to pay off? How much are salaries and benefits? These are some of the main items that managers look at when determining budgets. Many times, budgets are not only prepared for the upcoming year. Frequently, utilities will look 5, 10, even 20 years into the future for budgetary analysis. Let’s define some of these budget items. Operations and Maintenance (O&M) These two items typically go hand and hand. There are certain costs that the utility must cover and must properly budget for to keep the water flowing. Chemical costs for treating water, repairs on vehicles and mechanical equipment, power costs to pump water, leak repairs, and labor are just a few of the items that fall under this budgetary classification. Some are known, such as labor (salaries), as long as overtime isn’t too large. Others are predictable, such as power and chemicals. Based on historical water production, power and chemicals can be predicted within a reasonable amount of accuracy. Others, like water main breaks can be estimated based on history, but other factors come into play such as age, material, location, and pressures. Regardless of the predictability of O&M costs, managers must come up with an accurate budget number and then make sure that number is covered with revenue. Capital Improvement Projects (CIP) In addition to the reoccurring O&M costs, utilities need to plan and budget for future growth and the replacement of old infrastructure, such as pipelines and storage structures. Depending on the age of the utility and the expected future growth, CIP investment can be quite extensive. Typically, utilities can recover the costs of new infrastructure from the developers that are planning to build within the utilities service area. However, as infrastructure ages, it eventually needs to be replaced. The timing and funding of these replacements is an important part of a manager’s responsibility.
mean extra funds to pay for security during a time of crisis or overtime for staff for an unplanned outage due to an earthquake. It is hard to anticipate the precise nature of a crisis, but often having a contingency fund for such emergencies is useful. Debt More times than not, utilities will take on large amounts of debt to cover major capital improvement projects that expand their systems for future water users. Financing a project with debt allows current and future water users to share in the costs rather than saddling only current users with the cost of the project by paying all costs at the time of the project. Additionally, if a utility were to cover the cost of replacing major infrastructure projects through rates, the water rate could be too high for many people to pay. With a proper debt structure, the utility can spread out the costs over many years to help keep rates lower and have the right people pay for the right project Revenues and Rates For water utilities to pay for all their expenses (i.e., pumping, chemicals, material, salaries, benefits) they need to collect enough money. This is known as Revenue Requirements. A utility must identify all revenue requirements and then identify the means for collecting this revenue. Utilities can have different revenue sources such as property taxes, rents, leases. However, most water utility revenues are collected through the sale of water. The cost of water is determined through a rate study. A rate study is a report that lists the revenue requirements and then calculates how much the rate of water needs to be to collect these requirements. Water rates can be set in a variety of different structures (flat rate, single quantity rate, tiered rate), but regardless of the structure, the utility must sell water at the calculated rate to recover the needed revenue. Efficiencies As part of the budgetary process, managers need to identify when certain pieces of equipment will fail. Calculating the return on investment and identifying when the cost of maintenance exceeds the cost to replace the asset is crucial. An example of this is looking at the efficiencies of pumps and motors. Over time the efficiency decreases and the cost to operate and maintain the pump and motor increases. Another example is with pipelines. As pipes age more and more leaks occur. At some point in time the cost to repair leaks becomes greater than the cost to replace the pipe.
Example: Using the above table, calculate the percentage of the annual budget for each O&M Item listed in the table below. (If you need to review how to calculate a percentage, see Unit 3 in the Water 130 textbook.) Per the table, the total annual budget for the small water utility is $6,286,030. The total annual cost for water production is $3,950,400. The question is asking what percentage of $6,286,030 is $3,950,400.
Clearly, the majority of the budget is allocated to water production and staffing. Fixed costs are costs that are the same from year to year. Variable costs are costs that change from year to year. List the fixed costs versus variable costs and give an explanation justifying your response. Some might seem fixed, but there are ways to look at them as a variable cost. Others might seem like a variable cost, but in reality, there is limited control of the cost and these would actually be considered fixed.
Fixed and Variable Costs Although the cost of water is “fixed”, sometimes water utilities can control the amount that is purchased versus the amount that is pumped from wells. Buying water from another entity can be quite costly. However, more information would be needed about the utility to understand their production flexibilities. Staffing and benefits would also be considered a “fixed” cost but staffing reductions or adjustments in benefits could also occur. There are certain fixed vehicle expenses, such as oil changes, tune ups, and tires. There are also some unknown maintenance issues such as a bad battery or a faulty water pump. These examples can be looked at as either fixed or variable costs. The idea is not to “pigeonhole” these expenses as fixed or variable. Instead, you want to be able to accurately estimate these and other expenses in a budget. It is extremely important that utility managers have a general understanding of the concepts associated with utility management as well as the mathematical computations necessary to support the budgetary decisions being made.
Practice Problems 14.1 A utility vehicle costs on average $730 per year for maintenance. A replacement vehicle would cost $32,000. The utility has a vehicle policy that states all vehicles with 155,000 miles or more shall be replaced. The policy also states that once maintenance costs exceed 50% of the cost of a replacement vehicle, the vehicle shall be replaced. This particular vehicle averages 22,000 miles per year.
A pump that has been in operation for 15 years pumps a constant 450 gpm through 65 feet of dynamic head. The pump uses 6,537 kW‐Hr of electricity per month at a cost of $0.095 per kW‐Hr. The old pump efficiency has dropped to 50%. Assuming a new pump that operates at 90% efficiency is available for $10,270, how long would it take to pay for replacing the old pump? A utility has annual operating expenses of $4.7 million and a need for $2.1 million in capital improvements. The current water rate is $1.30 per CCF. Last year the utility sold 7270 AF of water and did not meet their capital budget need. How much does the utility need to raise rates in order to cover both the operational and capital requirements? (Round your answer to the nearest cent.)
A 300 hp well operates 6 hours a day and flows 1,700 gpm. The electricity cost is $0.118 per kW‐Hr. The well is also dosed with a 55% calcium hypochlorite tablet chlorinator to a dosage of 1.65 ppm. The tablets cost $1.20 per pound. The labor burden associated with the well maintenance is $60 per day. What is the total operating expense for this well in one year?
A water treatment manager has been asked to prepare a cost comparison between gas chlorine and a chlorine generation system using salt. Gas chlorine is $3.40 per pound and salt is $0.50 per pound. It takes approximately 4 pounds of salt to create 1 gallon of 1.75% chlorine with a specific gravity of 1.20. Assuming that the plant is dosing 12.5 MGD to a dosage of 2.75, what would be the annual cost of each? Which one is more cost effective? Exercise 14.1 A utility vehicle costs on average $1,250 per year for maintenance. A replacement vehicle would cost $35,000. The utility has a vehicle policy that states all vehicles with 150,000 miles or more shall be replaced. The policy also states that once maintenance costs exceed 60% of the cost of a replacement vehicle, the vehicle shall be replaced. This vehicle averages 18,500 miles per year.
A pump that has been in operation for 25 years pumps a constant 600 gpm through 47 feet of dynamic head. The pump uses 6,071 kW‐Hr of electricity per month at a cost of $0.085 per kW‐Hr. The old pump efficiency has dropped to 63%. Assuming a new pump that operates at 86% efficiency is available for $9,370, how long would it take to pay for replacing the old pump? A utility has annual operating expenses of $3.4 million and a need for $1.2 million in capital improvements. The current water rate is $1.55 per CCF. Last year the utility sold 6550 AF of water and did not meet their capital budget need. How much does the utility need to raise rates in order to cover both the operational and capital requirements? (Round your answer to the nearest cent.)
A 250 hp well operates 9 hours a day and flows 2,050 gpm. The electricity cost is $0.135 per kW‐Hr. The well is also dosed with a 65% calcium hypochlorite tablet chlorinator to a dosage of 1.25 ppm. The tablets cost $1.85 per pound. The labor burden associated with the well maintenance is $75 per day. What is the total operating expense for this well in one year?
A water treatment manager has been asked to prepare a cost comparison between gas chlorine and a chlorine generation system using salt. Gas chlorine is $2.35 per pound and salt is $0.38 per pound. It takes approximately 5 pounds of salt to create 1 gallon of 0.8% chlorine with a specific gravity of 1.15. Assuming that the plant is dosing 15 MGD to a dosage of 2.25, what would be the annual cost of each? Which one is more cost effective?
1 1.547 cfs A water utility operator needs to report the total amount of water drained from two separate basins. The 18” pipe in Basin A drained water at a rate of 12.4 cfs for four hours each day. The 12” pipe in Basin B drained water at a rate of 3.1 cfs for 16 hours each day. What is the total amount of water drained in million gallons in 30 days?
sec 1 hr 1 day 1 cf 1 Total = Basin A + Basin B = 40.1 MG + 25.0 MG = 65.1 MG A water utility operator drove a total of 22,841 miles in one year. What were the average miles driven per day? Assume that the vehicle operated 6 days per week. 22,841 miles  1 year  1 week  73.2 miles 1 year 52 weeks 6 days day Water travels 52 miles per day through an aqueduct. What is the velocity of the water in feet per second?
Practice Problems 1.2 A water utility manager has been asked to prepare an end of year report for the utility’s board of directors. The utility has four groundwater wells and two connections to a surface water treatment plant. Complete the table below.
2  2    A 35-foot diameter spherical tank needs to be painted. If one gallon of paint will cover 400 sf, how many gallons of paint will be required to put two coats of paint on the exterior of the tank?
There are two ways to approach this problem since the aqueduct is only half full. First, we’ll calculate the volume of the entire aqueduct and then divide it by 2 to get the volume when it is half full.
Practice Problems 2.3 A water utility operator needs to determine the cost of painting an above ground storage tank. The tank is 50 feet tall and has a diameter of 28 feet. One gallon of paint can cover 200 sf and costs $36.27 per gallon. What is the total cost to paint the storage tank?
A water tank truck delivered 30 loads of water to a construction site. The water tank on the truck is shaped like a pill. Each end has a 10-foot diameter and the center section is 15 feet long. If the water costs $352 an AF, how much did the construction site pay for the water?
A maintenance crew is replacing an 18” meter at a well. The specifications state that there needs to be 6.5 times the pipe diameter in feet of straight pipe before the meter and 4 times the pipe diameter in feet of straight pipe after the meter. How many feet of 18” pipe are needed?
A 1,200-foot section of a trapezoidal shaped aqueduct needs to be drained for maintenance. The aqueduct contains 5 AF of water, is 8 feet wide at the bottom, and is 14 feet wide at the water line. What is the water depth?
A water utility has installed 900 feet of 28” diameter pipe. They want to wrap a corrosion resistant sleeve around the pipe and fill the pipe to pressure test it. How many gallons of water will the pipe hold and how many square feet of corrosion resistant sleeve are required to cover the whole pipe?
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?
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?
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 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.
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?
Now look at Table C-3 Inactivation of Giardia Cysts by Free Chlorine at 10°C. In the pH 8.5 section look down the 0.5 Log Inactivation column until it intersects with the 0.6 mg/L row. 31 mg/L min CT is required. Treatment Plant: From Table 7.2 Treatment Credits and Log Inactivation Requirements you can see that: 3 Log (Required) - 2.5 Log (Credit) = 0.5 Log (Remaining)
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 .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?
Now look at Table C-4 Inactivation of Giardia Cysts by Free Chlorine at 15°C. In the pH 7.0 section look down the 1.0 Log Inactivation column until it intersects with the 0.6 mg/L row. 24 mg/L min CT is required. Treatment Plant: From Table 7.2 Treatment Credits and Log Inactivation Requirements you can see that: 3 Log (Required) - 2.0 Log (Credit) = 1.0 Log (Remaining)
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.
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?
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 5,160 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?
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?
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%.
Practice Problems 10.2 A well is pumping water from an aquifer with a water table 55 feet below ground surface (bgs) to a tank 190 feet above the well. If the well flows 870 gpm, what is the required horsepower? (Assume the wire-to-water efficiency is 88%.)
A booster pump station is pumping water from Zone 1 at an elevation of 1,537 ft above sea level to Zone 2 which is at 1,745 ft above sea level. The pump station is located at an elevation of 1,124 ft above sea level. The pump was recently tested and the efficiencies for the pump and motor were 76% and 88% respectively. The losses through the piping and appurtenances equate to a total of 15 ft. If the pump flows 1,500 gpm, what is the required motor horsepower?
A well with pumps located 46 ft bgs pumps against a discharge head pressure of 140 psi to a tank located at an elevation 278 ft above the well. The well pumps at a rate of 1,260 gpm. What is the level of water in the tank and what is the required water horsepower? (Assume the wire-to-water efficiency is 70%.)
A 430 hp booster pump is pulling water from a 50-foot tall tank that is 85 feet below the pump line. It is then pumping against a discharge head pressure of 150 psi. What is the flow rate in gpm? Assume the wire-to-water efficiency is 92% and the tank is full.
A utility has 3 pumps that run at different flow rates and supply water to an 800,000 gallon storage tank. Assume that only one pump runs per day. The TDH for the pumps is 130 ft. The utility needs to fill the tank daily and power costs are to be calculated at a rate of $0.10 per kW‐Hr. Complete the table below.
A well draws water from an aquifer that has an average water level of 100 ft bgs and pumps to a tank 300 ft above it. Friction loss to the tank is approximately 28 psi. If the well pumps at a rate of 1,900 gpm and has a wire‐to‐water efficiency of 45% how much will it cost to run this well 10 hours per day. Assume the electrical rate is $0.22 per kW‐ Hr.
A utility manager is trying to determine which hp motor to purchase for a pump station. A 500 hp motor with a wire‐to‐water efficiency of 70% can pump 3,300 gpm. Similarly, a 300 hp motor with a wire‐to‐water efficiency of 80% can pump 2,500 gpm. With an
Approximately 170 kW of power are needed to run a certain booster pump. If the booster has a wire‐to‐water efficiency of 81% and is pumping against 205 psi of head pressure, what is the corresponding flow in gpm?
8. It costs $103.61 in electricity to run a well for 10 hours a day. The well has a TDH of 167 psi and an overall efficiency of 82.3%. The cost per kW‐Hr is $0.156. What is the cost of the water per gallon?
Practice Problems 12.1 A well has a nitrate level that exceeds the MCL of 63 mg/L. Over the last 4 sample results it has averaged 71 mg/L. A nearby well has a nitrate level of 40 mg/L. If both wells combined pump up to 1,575 gpm, how much flow is required from each well to achieve a nitrate level of 50 mg/L? Well A – 71 mg/L Well B – 40 mg/L Desired result “C” – 50 mg/L To determine the percentage of Source A required for the blend, the desired result minus the low result is divided by the high result minus the low result. C - B = A - B
71 mg/L - 40 mg/L 31 mg/L This says that 32% of Well A is needed to achieve the desired blended result. To determine the percentage of Source B required for the blend, the high result minus the desired result is divided by the high result minus the low result. A - C = A - B
A well (A) has shown quarterly arsenic levels above the MCL over the last year, of 16 ug/L, 22 ug/L, 20 ug/L and 10 ug/L. A utility wants to blend this well to a level of 6.0 ug/L with a well (B) that has a level of 2.1 ug/L. The total production needed from both of these wells is 4,100 gpm. How much can each well produce?
A well with a PCE level of 12.4 ug/L is supplying approximately 65% of total water demand. It is being blended with a well that has a PCE level of 1.5 ug/L. Will this blended supply meet the MCL for PCE of 7.0 ug/L?
Well A has a total dissolved solids (TDS) level of 625 mg/L. It is pumping 2,300 gpm, which is 50% of the total production from two wells. The other well (B) blends with well A to achieve a TDS level of 450 mg/L. What is the TDS level for Well B?
Two wells need to achieve a daily flow of 2.1 MG and a total hardness level of 110 mg/L as calcium carbonate (CaCO3.) Well #1 has a total hardness level of 390 mg/L as CaCO3 and Well #2 has a level of 63 mg/L as CaCO3. What is the gpm that each well must pump?
The State Health Department has requested a blending plan to lower levels of sulfate from a small water utility well. The well has a constant sulfate level of 480 mg/L. The utility needs to purchase the water to blend with the well. The purchased water has a sulfate level of 55 mg/L. They need to bring the sulfate levels down to 225 mg/L and supply a demand of 2.0 MGD. The purchased water costs $475/AF. How much will the purchased water cost for the entire year?
Practice Problems 13.1 A 4-20 mA signal is being used to measure the water level in a water storage tank. The tank is 50 feet tall and the low-level signal is set at 0 feet and the high level at 50 feet. What is the level in the tank with a 17 mA reading?
A 48 ft tall water tank uses a 4-20 mA signal for calculating the water level. If the 4 mA level is set at 5 feet from the bottom and the 20 mA is set at 5 feet from the top, what is the level in the tank with a 9 mA reading?
A utility uses a 4-20 mA signal to determine the level in a well based on pressures. The set points are based on pressures in psi below ground surface (bgs). The 20 mA signal is set at 205 psi bgs and the 4 mA signal at 15 psi bgs. If the reading is 14 mA, what is the water level in feet?
A water utility uses a 4-20 mA signal to determine groundwater elevations in a well. The set points are based on actual elevations above the mean sea level (MSL). The ground surface elevation at this well is 1,400 ft and this is where the 4 mA signal is set. The 20 mA signal is set at 740 ft. What is the elevation and the feet bgs with an 11 mA reading?
Practice Problems 14.1 A utility vehicle costs on average $730 per year for maintenance. A replacement vehicle would cost $32,000. The utility has a vehicle policy that states all vehicles with 155,000 miles or more shall be replaced. The policy also states that once maintenance costs exceed 50% of the cost of a replacement vehicle, the vehicle shall be replaced. This particular vehicle averages 22,000 miles per year.