343 C HAPTER 15 Water Treatment Process Calculations 15.1 INTRODUCTION Gupta (1997) points out that because of huge volume and flow conditions, the quality of natural water cannot be modified significantly within a body of water. Consequently, the quality control approach is directed to water withdrawn from a source for a specific use; the drawn water is treated prior to use. Typically, the overall treatment of water (for potable use) consists of physical and chemical methods of treatment — unlike wastewater treatment, where physical, chemical, and/or biological unit processes are used, depending on the desired quality of the effluent and operational limitations. The physical unit operations used in water treatment include: • Screening — used to remove large-sized floating and suspended debris. • Mixing — coagulant chemicals (alum, for example) are mixed with the water to make tiny particles stick together. • Flocculation — water mixed with coagulants is given low-level motion to allow particles to meet and floc together. • Sedimentation (settling) — water is detained for a time sufficient to allow flocculated particles to settle by gravity. • Filtration — fine particles that remain in the water after settling and some microorganisms present are filtered out by sending the water through a bed of sand and coal. Chemical unit processes used in treating raw water, depending on regulatory requirements and the need for additional chemical treatment, include disinfection, precipitation, adsorption, ion exchange, and gas transfer. A flow diagram of a conventional water treatment system is shown in Figure 15.1. Note : In the following sections (for water and wastewater treatment processes presented in Chapter 16), we present basic, often used daily operational calculations (operator’s math) along with engineering calculations (engineer’s math) used for solving more complex computations. This presentation method is in contrast to normal presentation methods used in many engineering texts. We deviate from the norm based on our practical real-world experience; we have found that environmental engineers tasked with managing water or wastewater treatment plants are responsible not only for computation of many complex math operations (engineering calculations) but also for overseeing proper plant operation (including math operations at the operator level). Obviously, engineers are well versed in basic math operations; however, they often need to refer to example plant operation calculations in a variety of texts. In this text, the format used, although unconven- tional, is designed to provide basic operations math as well as more complex engineering math in one ready reference. L1681_book.fm Page 343 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 344 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 15.2 WATER SOURCE AND STORAGE CALCULATIONS Approximately 40 million cubic miles of water cover or reside within the Earth. The oceans contain about 97% of all water on Earth; the other 3% is fresh water. Snow and ice on the surface of Earth contain about 2.25% of the water; usable ground water is approximately 0.3%, and surface fresh- water is less than 0.5%. In the U.S., for example, average rainfall is approximately 2.6 ft (a volume of 5900 km 3 ). Of this amount, approximately 71% evaporates (about 4200 km 3 ), and 29% goes to stream flow (about 1700 km 3 ). Uses of freshwater include manufacturing; food production; domestic and public needs; recre- ation; hydroelectric power production; and flood control. Stream flow withdrawn annually is about 7.5% (440 km 3 ). Irrigation and industry use almost half of this amount (3.4% or 200 km 3 per year). Municipalities use only about 0.6% (35 km 3 per year) of this amount. Historically, in the U.S., water usage is increasing (as might be expected). For example, in 1900, 40 billion gal of fresh water were used. In 1975, the total increased to 455 billion gal. Projected use for 2000 (the latest published data) was about 720 billion gal. The primary sources of fresh water include: •Water captured and stored rainfall in cisterns and water jars • Groundwater from springs, artesian wells, and drilled or dug wells • Surface water from lakes, rivers, and streams • Desalinized seawater or brackish groundwater • Reclaimed wastewater 15.2.1 Water Source Calculations Water source calculations covered in this subsection apply to wells and pond or lake storage capacity. Specific well calculations discussed include well drawdown; well yield; specific yield; well-casing disinfection; and deep-well turbine pump capacity. 15.2.1.1 Well Drawdown Drawdown is the drop in the level of water in a well when water is being pumped (see Figure 15.2). Drawdown is usually measured in feet or meters. One of the most important reasons for measuring drawdown is to make sure that the source water is adequate and not being depleted. The data collected to calculate drawdown can indicate if the water supply is slowly declining. Early detection can give the system time to explore alternative sources, establish conservation measures, or obtain any special funding that may be needed to get a new water source. Figure 15.1 Conventional water treatment model. Addition of coagulant Water supply Mixing tank Flocculation basin Settling tank Sand filter To storage and distribution Screening Sludge processing Disinfection L1681_book.fm Page 344 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC WATER TREATMENT PROCESS CALCULATIONS 345 Well drawdown is the difference between the pumping water level and the static water level: (15.1) Example 15.1 Problem : The static water level for a well is 70 ft. If the pumping water level is 90 ft, what is the drawdown? Solution : Drawdown, ft, = Pumping Water Level, ft, – Static Water Level, ft Example 15.2 Problem : The static water level of a well is 122 ft. The pumping water level is determined using the sounding line. The air pressure applied to the sounding line is 4.0 psi, and the length of the sounding line is 180 ft. What is the drawdown? Solution : First, calculate the water depth in the sounding line and the pumping water level: Figure 15.2 Hydraulic characteristics of a well. Drawdown, ft Pumping Water Level, ft St= − aatic Water Level, ft = −90 ft 70 ft = 20 ft L1681_book.fm Page 345 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 346 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Then, calculate drawdown as usual: 15.2.1.2 Well Yield Well yield is the volume of water per unit of time produced from well pumping. Usually, well yield is measured in terms of gallons per minute (gpm) or gallons per hour (gph). Sometimes, large flows are measured in cubic feet per second (cfs). Well yield is determined using the following equation. (15.2) Example 15.3 Problem : Once the drawdown level of a well stabilized, operators determined that the well produced 400 gal during the 5-min test. What is the well yield in gpm 2 ? Solution : Example 15.4 Problem : During a 5-min test for well yield, a total of 780 gal are removed from the well. What is the well yield in gallons per minute? in gallons per hour? Water Depth in Sounding Line (4.0 psi) (2= 31 ft/psi) = 9.2 ft Pumping Water Level 180 ft 9.2 ft 170= − = 8 f t Drawdown, ft Pumping Water Level, ft St= − aatic Water Level, ft = −170.8 ft 122 ft = 48.8 ft Well Yield, gpm Gallons Produced Duration = oof Test, min Well Yield, gpm Gallons Produced Duration = oof Test, min 400 gallons 5 minutes = = 80 gpm L1681_book.fm Page 346 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC WATER TREATMENT PROCESS CALCULATIONS 347 Solution : Then convert gallons per minute flow to gallons per hour flow: 15.2.1.3 Specific Yield Specific yield is the discharge capacity of the well per foot of drawdown. The specific yield may range from 1-gpm/ft drawdown to more than 100-gpm/ft drawdown for a properly developed well. Specific yield is calculated using the equation: (15.3) Example 15.5 Problem : A well produces 260 gpm. If the drawdown for the well is 22 ft, what is the specific yield in gallons per minute per foot, and what is the specific yield in gallons per minute per foot of drawdown? Solution : Example 15.6 Problem : The yield for a particular well is 310 gpm. If the drawdown for this well is 30 ft, what is the specific yield in gallons per minute per foot of drawdown? Well Yield, gpm Gallons Removed Duration o = ffTest, min = 780 gallons 5 minutes = 156 gpm (156 gal/min) (60/hr) 9360 gph= Specific Yield, gpm/ft Well Yield, gpm Dra = wwdown, ft Specific Yield, gpm/ft Well Yield, gpm Dra = wwdown, ft = 260 gpm 22 ft = 11.8 gpm/ft L1681_book.fm Page 347 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 348 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution : 15.2.1.4 Well Casing Disinfection A new, cleaned, or repaired well normally contains contamination that may remain for weeks unless the well is thoroughly disinfected. This may be accomplished by ordinary bleach in a concentration of 100 ppm (parts per million) of chlorine. The amount of disinfectant required is determined by the amount of water in the well. The following equation is used to calculate the pounds of chlorine required for disinfection: (15.4) Example 15.7 Problem : A new well is to be disinfected with chlorine at a dosage of 50 mg/L. If the well casing diameter is 8 in. and the length of the water-filled casing is 110 ft, how many pounds of chlorine will be required? Solution : First, calculate the volume of the water-filled casing: Then, determine the pounds of chlorine required, using the milligrams-per-liter to pounds equation: 15.2.1.5 Deep-Well Turbine Pump Calculations Deep well turbine pumps are used for high-capacity deep wells. Usually consisting of more than one stage of centrifugal pumps, the pump is fastened to a pipe called the pump column; the pump is located in the water. The pump is driven from the surface through a shaft running inside the pump column, and the water is discharged from the pump up through the pump column to the Specific Yield, gpm/ft Well Yield, gpm Dra = wwdown, ft = 310 gpm 30 ft = 10.3 gpm/ft Chlorine, lb (Chlorine, mg/L) (Casing Vol= ,MG) (8.34 lb/gal) (0.785) (.67) (67) (110 ft) (7.48 gal/ft ) 3 == 290 gallons Chlorine, lb (chlorine, mg/L)(Volume, MG)= (8.34 lb/gal) (50mg/L) (0.000290 MG) (8.34 lb/gal) 0.1= 22lbChlorine L1681_book.fm Page 348 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC WATER TREATMENT PROCESS CALCULATIONS 349 surface. The pump may be driven by a vertical shaft, electric motor at the top of the well, or some other power source, usually through a right angle gear drive located at the top of the well. A modern version of the deep-well turbine pump is the submersible type pump in which the pump, along with a close-coupled electric motor built as a single unit, is located below water level in the well. The motor is built to operate submerged in water. 15.2.3 Vertical Turbine Pump Calculations The calculations pertaining to well pumps include head, horsepower, and efficiency calculations. Discharge head is measured to the pressure gauge located close to the pump discharge flange. The pressure (psi) can be converted to feet of head using the equation: (15.5) Total pumping head ( field head ) is a measure of the lift below the discharge head pumping water level (discharge head). Total pumping head is calculated as: (15.6) Example 15.8 Problem : The pressure gauge reading at a pump discharge head is 4.1 psi. What is this discharge head expressed in feet? Solution : Example 15.9 Problem : The static water level of a pump is 100 ft. The well drawdown is 26 ft. If the gauge reading at the pump discharge head is 3.7 psi, what is the total pumping head? Solution : (15.7) Five types of horsepower calculations are used for vertical turbine pumps; a general under- standing of these five horsepower types is important: Discharge Head, ft (press, psi) (2.31 ft/= ppsi) Pumping Head, ft Pumping Water Level, ft=++ DischargeHead, ft (4.1 psi) (2.31 ft/psi) 9.5 ft= Total Pumping Head, ft Pumping Water Leve= ll, ft Discharge Head, ft+ (100 ft 26 ft) (3.7 psi) (2.31 ft/ps=++ ii) =+126 ft 8.5 ft = 134.5 ft L1681_book.fm Page 349 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 350 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK • Motor horsepower refers to the horsepower supplied to the motor. The following equation is used to calculate motor horsepower: (15.8) • Total brake horsepower ( bhp ) refers to the horsepower output of the motor. The following equation is used to calculate total bhp: (15.9) • Field horsepower refers to the horsepower required at the top of the pump shaft. The following equation is used to calculate field horsepower: (15.10) • Bowl or laboratory horsepower refers to the horsepower at the entry to the pump bowls. The following equation is used to calculate bowl horsepower: (15.11) • Water horsepower refers to the horsepower at the pump discharge. The following equation is used to calculate water hp: (15.12) or the equivalent equation: Example 15.10 Problem : The pumping water level for a well pump is 150 ft, and the discharge pressure measured at the pump discharge centerline is 3.5 psi. If the flow rate from the pump is 700 gpm, what is the water horsepower? (Use Equation 15.12.) Solution : First, calculate the field head. The discharge head must be converted from psi to ft: Motorhp(input hp) Field bhp Motor Effici = eency 100 Total bhp Field bhp Thrust Bearing Loss=+ ,, h p Field bhp Bowl bhp Shaft Loss, hp=+ Bowl bhp (Lab bhp) = (Bowl Head, ft) (Capac iity, gpm) (3960) (Bowl Efficiency) 100 Water hp (Field Head, ft) (Capacity, gpm) 3 = 9960 Water hp (Field Head, ft) (Capacity, gpm) 3 = 33,000 ft-lb/min L1681_book.fm Page 350 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC WATER TREATMENT PROCESS CALCULATIONS 351 The water horsepower is therefore: The water horsepower can now be determined: Example 15.11 Problem : The pumping water level for a pump is 170 ft. The discharge pressure measured at the pump discharge head is 4.2 psi. If the pump flow rate is 800 gpm, what is the water horsepower? (Use Equation 15.12.) Solution : First, determine the field head by converting the discharge head from psi to ft: Now, calculate the field head: and then calculate the water horsepower: Example 15.12 Problem : A deep-well vertical turbine pump delivers 600 gpm. If the lab head is 185 ft and the bowl efficiency is 84%, what is the bowl horsepower? (Use Equation 15.11.) (3.5 psi) (2.31 ft/psi) 8.1 ft= 150 ft 8.1 ft 158.1 ft+= = +=150 ft 8.1 ft 158.1 ft 33,000 ft-lb/min = 28 whp (4.2 psi) (2.31 ft/psi) 9.7 ft= 170 ft 9.7 ft 179.7 ft+= whp (179.7 ft) (800 gpm) 3960 = = 36 whp L1681_book.fm Page 351 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 352 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution : Example 15.13 Problem : The bowl brake horsepower is 51.8 bhp. If the 1-in. diameter shaft is 170 ft long and is rotating at 960 rpm with a shaft fiction loss of 0.29 hp loss per 100 ft, what is the field bhp? Solution : Before you can calculate the field bhp, factor in the shaft loss: Now determine the field bhp: Example 15.14 Problem : The field horsepower for a deep-well turbine pump is 62 bhp. If the thrust bearing loss is 0.5 hp and the motor efficiency is 88%, what is the motor input horsepower? (Use Equation 15.8.) Bowl bhp (Bowl Head, ft) (Capacity, gpm) (3 = 9960) (Bowl Efficiency) 100 (185 ft) (600 gpm) (3960) (84.0) 100 = (185) (600) (3960) (84.0) = = 33.4 bowl bhp (0.29 hp loss) (170 ft) 0.5 hp loss= 100 Field bhp Bowl bhp Shaft Loss, hp=+ =+51.8 bhp 0.5 hp = 52.3 bhp L1681_book.fm Page 352 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC [...]... 5, 2004 10:51 AM 370 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Flow Rate, gpm = (0.785) (D 2 ) (Drop in Level, ft) (7.48 gal/ft 3 Duration of Test, min (15. 37) Example 15. 41 Problem: For a pumping rate calibration test conducted for a 1 5- min period, the liquid level in the 4-ft diameter solution tank is measured before and after the test If the level drops 0.5 ft during the 15min test, what is the... (0.5 mg/L) (15 MG) (8.34 lb/gal) 25 100 = 250 lb Copper Sulfate For calculating pounds of copper sulfate per acre-foot, use the following equation (assume the desired copper sulfate dosage is 0.9 lb/acre-ft): Copper Sulfate, lb = (0.9 lb Copper Sulfate) (acre-ft) 1 acre-ft (15. 18) Example 15. 20 Problem: A pond has a volume of 35 acre-ft If the desired copper sulfate dose is 0.9 lb/acre-ft, how many... October 5, 2004 10:51 AM 376 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 15. 5.4 Mean Flow Velocity The measure of average velocity of the water as it travels through a rectangular sedimentation basin is known as mean flow velocity and is calculated using Equation 15. 52: Q (Flow), ft 3 /min = A (Cross-Sectional Area), ft 2 × V (Vol.) ft/min (15. 52) (Q = A × V) Example 15. 50 Problem: A sedimentation... feeder settings, and detention time 15. 4.4.1 Chamber and Basin Volume Calculations To determine the volume of a square or rectangular chamber or basin, use Equation 15. 22 or 15. 23: © 2005 by CRC Press LLC L1681_book.fm Page 360 Tuesday, October 5, 2004 10:51 AM 360 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Volume, ft 3 = (length, ft) (width, ft) (depth, ft) (15. 22) Volume, gal = (length, ft) (width,... 366 Tuesday, October 5, 2004 10:51 AM 366 15. 4.5.1 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Determining Percent Strength of Liquid Solutions When using liquid chemicals to make up solutions (liquid polymer, for example), a different calculation is required: Liq Poly., lb = Liq Poly (% Strength) Poly Sol (% Strength) = Poly Sol., lb 100 100 (15. 31) Example 15. 35 Problem: A 12% liquid polymer is used... CRC Press LLC (15. 14) L1681_book.fm Page 354 Tuesday, October 5, 2004 10:51 AM 354 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution: Field Efficiency, % = = (Field Head, ft) (Capacity, gpm) × 100 (3960) (Total bhp) (180 ft) (850 gpm) × 100 (3960) (61.3 bhp) = 63% Overall efficiency is a comparison of the horsepower output of the system with that entering the system Equation 15. 15 is used to calculate... AM 374 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK • Basic detention time equation: Volume of Tank, gal Flow Rate, gph (15. 48) (Length, ft) (width, ft) (Depth, ft) (7.48 gal/ft 3 ) Flow Rate, gph (15. 49) Detention Time, h = • Rectangular sedimentation basin equation: Detention Time, h = • Circular basin equation Detention Time, h = (0.785) (D 2 ) (Depth, ft) (7.48 gal/ft 3 ) Flow Rate, gph (15. 50)... mL/gal) 1440 min/day (15. 28) L1681_book.fm Page 364 Tuesday, October 5, 2004 10:51 AM 364 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Example 15. 31 Problem: The desired solution feed rate was calculated at 9 gpd What is this feed rate expressed as milliliters per minute? Solution: mL/min = = (gpd) (3785 mL/gal) 1440 min/day (9 gpd) (3785 mL/gal) 1440 min/day = 24 mL/min Feed Rate Example 15. 32 Problem:... 5, 2004 10:51 AM WATER TREATMENT PROCESS CALCULATIONS 15. 5.1.1 373 Calculating Tank Volume For rectangular sedimentation basins, we use Equation 15. 46: Volume, gal = (length, ft) (width, ft) (depth, ft) (7.48 gal/ft 3 ) (15. 46) For circular clarifiers, we use Equation 15. 47: Volume, gal = (0.785) (D2) (depth, ft) (7.48 gal/ft 3 ) (15. 47) Example 15. 44 Problem: A sedimentation basin is 25 ft wide, 80... October 5, 2004 10:51 AM 368 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Then, calculate the pounds per day feed rate: Chemical Feed Rate, lb/day = (0.06 lb/min) (1440 min/day) = 86.4 lb/day Feed Rate Example 15. 38 Problem: Calculate the actual chemical feed rate in pounds per day, if a container placed under the chemical feeder collects a total of 1.6 lb during a 20-min period Solution: First, calculate . Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 360 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK (15. 22) (15. 23) Example 15. 23 Problem: A flash mix chamber is 4 ft 2 with water to a. LLC 354 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution : Overall efficiency is a comparison of the horsepower output of the system with that entering the system. Equation 15. 15 is used. 2005 by CRC Press LLC 352 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution : Example 15. 13 Problem : The bowl brake horsepower is 51.8 bhp. If the 1-in. diameter shaft is 170