Advanced Waterworks Mathematics, 2019a

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Well B: A - C = A - B 22 ug/L - 6 ug/L 16 ug/L = = 0.804020 22 ug/L - 2.1 ug/L 19.9 ug/L Well A: 4,100 gpm 0.20 = 820 gpm Well B: 4,100 gpm 0.80 = 3,280 gpm 3. 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: 12.4 ug/L Well B: 1.5 ug/L Desired result: ? ug/L Well A: ? ug/L - 1.5 ug/L = 0.65 12.4 ug/L - 1.5 ug/L ? ug/L - 1.5 ug/L = 0.65 10.9 ug/L ? ug/L - 1.5 ug/L = (0.65)(10.9 ug/L) ? ug/L = 7.085 ug/L + 1.5 ug/L = 8.585 ug/L = 8.6 ug/L NO! 4. 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? Well B Well A – 625 mg/L Well B – ? mg/L Desired result – 450 mg/L A - C = A - B 625 mg/L - 450 mg/L 175 mg/L = = 0.50 625 mg/L - ? mg/L 625 mg/L - ? mg/L 175 mg/L= 0.50(625 mg/L - ? mg/L) 175 mg/L = 312.5 mg/L - (0.50)(? mg/L) 175 mg/L - 312.5 mg/L = - (0.50)(? mg/L) 356 | A dvanced Waterworks Mathematics
-137.5 mg/L ? mg/L = = 275 mg/L - (0.50) OR Well A C - B = A - B 450 mg/L - ? mg/L = 0.5 625 mg/L - ? mg/L 450 mg/L - ? mg/L = 0.5(625 mg/L - ? mg/L) 450 mg/L - ? mg/L = 312.5 mg/L - (0.5)(? mg/L) 450 mg/L - 312.5 mg/L = ? mg/L - (0.5)(? mg/L) 0.5(? mg/L) = 137.5 mg/L 137.5 mg/L ? mg/L = = 275 mg/L 0.5 5. 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? Well 1 Well 1 – 390 mg/L Well 2 – 63 mg/L Desired result – 110 mg/L C - B = A - B 110 mg/L - 63 mg/L 47 mg/L = = 0.1437 390 mg/L - 63 mg/L 327 mg/L Well 2 A - C = A - B 390 mg/L - 110 mg/L 280 mg/L = = 0.8562 390 mg/L - 63 mg/L 327 mg/L 2,100,000 gal 1 day day 1,440 min = 1,458.3333 gpm = 1,458 gpm Well 1: 357 | A dvanced Waterworks Mathematics
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1 | A dvanced Waterworks Mathematic
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Acknowledgements College of the Can
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Unit 10 201 10.1 Horsepower and Eff
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Multiple units may need to be conve
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Practice Problems 1.1a Show how eac
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Word problems may involve unit conv
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6. A pipe flows at a rate of 6.3 cf
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6. A well flows at a rate of 1,250
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Well 2: Total Flow in MGD 1,000 gal
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Practice Problems 1.2 1. A water ut
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3. A water utility has 12,300 servi
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The shape of a tank is often a cyli
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Circles A circle is a round figure
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Example: What is the area of an 813
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Trapezoids Trapezoids are four-side
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Figure 2.9 9 Circumference of a Cir
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The formula for the area of a spher
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Example: What is the entire interio
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Practice Problems 2.1 1. What is th
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8. What is the entire interior surf
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5. A standpipe needs to be painted.
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Then, calculate the volume of the t
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2 1 2 Trapezoidal Prism = H L 8.5
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Exercise 2.2 1. What is the volume
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2.3 AREA AND VOLUME WORD PROBLEMS W
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Figure 2.20 19 Step 4: Put the info
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Step 3: Make a sketch to clarify th
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Figure 2.22 21 When looking at a pr
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Calculate the volume of the cylinde
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4. A water tank truck delivered 30
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Exercise 2.3 1. A water utility ope
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6. A private contractor needs water
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2.4 FLOW RATE Flow rate is the meas
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Key Terms circumference - the dist
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4. What is the diameter of a pipe t
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4. What is the diameter of a pipe t
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Substance Specific Gravity Weight C
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5. The specific gravity of 25% Alum
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To convert between a percentage and
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And now we are moving to the row fo
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Practice Problems 3.2 1. An 87.5% c
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Exercise 3.2 Solve the following pr
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3.3 MIXING AND DILUTING SOLUTIONS T
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Total Volume of the new solution is
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3. A chlorine storage tank that is
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3. A chlorine storage tank that is
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The following charts or Pie Wheels
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At only a 10% concentration, you wi
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Now the question is asking how many
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Key Terms chlorine demand - the
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3. How many pounds of 24.5% sodium
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7. A water treatment operator adjus
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11. An operator added 422 gallons o
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3. How many pounds of 12.5% sodium
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7. A water treatment operator adjus
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11. An operator added 275 gallons o
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Weirs can either be sharp-crested o
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Weir Overflow Rate (gpm/ft) = ? gpm
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Key Terms weir - an overflow stru
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5. A treatment plant processes 8.4
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5. A treatment plant processes 15 M
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UNIT 6 6.1 DETENTION TIME Detention
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D = t Volume Flow gallons million
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Rearranging the terms in the detent
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3. A water utility engineer is desi
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7. A 70-foot diameter, 35-foot-deep
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3. A water utility engineer is desi
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7. The chlorine residual decay rate
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11. A circular clarifier processes
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Although filtration rates are commo
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Key Terms contact time - detenti
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4. A water treatment plant processe
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Exercise 6.2 1. A water treatment p
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7. An Engineer is designing a circu
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Pin It! Misconception Alert Sometim
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Next, there are two crucial terms t
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Finding the Correct CT Table Typica
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and resulting disinfection inactiva
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plant maintains a chloraminated res
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5. A conventional water treatment p
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7. A direct filtration plant is ope
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4. A conventional water treatment p
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6. A direct filtration water treatm
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UNIT 8 8.1 PRESSURE Pressure is the
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From this example, it is clear that
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5. Two houses are served by a nearb
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8.2 HEAD LOSS As water travels thro
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Friction will need to be added at t
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3. A well located at 200 feet above
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UNIT 9 9.1 WELL YIELD, SPECIFIC CAP
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Cone of Depression The triangular s
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Practice Problems 9.1 1. A well has
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7. A well has a specific capacity o
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4. A well has an hour meter attache
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UNIT 10 10.1 HORSEPOWER AND EFFICIE
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The above equation is used to calcu
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Practice Problems 10.1 1. What is t
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Exercise 10.1 Calculate the require
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10.2 HEAD LOSS AND HORSEPOWER As di
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Practice Problems 10.2 1. A well is
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Exercise 10.2 Calculate the followi
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10.3 CALCULATING POWER COSTS It is
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93.25 kW $ 0.088 8.33 hr $ 68.35
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Practice Problems 10.3 1. A well fl
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4. A well draws water from an aquif
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7. Complete the table below based o
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Exercise 10.3 Solve the following p
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4. A well draws water from an aquif
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7. Complete the table below based o
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UNIT 11 11.1 PER CAPITA WATER USE A
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That’s a lot of gallons. This is
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Green Building Code) has resulted i
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Practice Problems 11.1 1. What is t
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7. A water district has a goal of 1
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4. How much water would a family of
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UNIT 12 12.1 BLENDING AND DILUTING
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To determine the percentage of Sour
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Practice Problems 12.1 1. A well ha
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5. Two wells need to achieve a dail
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3. A well with a PCE level of 7.5 u
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UNIT 13 13.1 SCADA AND THE USE OF M
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ottom to what is referred to as the
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mA (reading) - mA (offset) = percen
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4. A water tank is 52 ft tall and h
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8. A chemical injection system is m
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4. A water tank is 45 ft tall and h
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8. A chemical injection system is m
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mean extra funds to pay for securit
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Per the table, the total annual bud
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Pumps and Motors: $72,000 x% $6,2
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Variable Costs Reason 273 | A dvanc
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Practice Problems 14.1 1. A utility
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3. A utility has annual operating e
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7. A small water company has a tota
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2. A pump that has been in operatio
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5. A 250 hp well operates 9 hours a
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PRACTICE PROBLEM SOLUTIONS Practice
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3.1 cf 3,600 sec 10 hr 7.48 gal 30
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1,346,400 gal 1 AF 365 day day 3
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Practice Problems 2.3 1. A water ut
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3 22 ft 217,800.13 ft = depth ft
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Q (cfs) 3.18 cfs A (ft ) = = = 1.17
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To convert ppb to ppm, you divide p
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Page 315 and 316: Weir Length (ft) = 4,791.67 gpm 41
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Page 327 and 328: 4. A conventional water treatment p
Page 329 and 330: Now look at Table C-7 Inactivation
Page 331 and 332: 500 mg/L min CT is required. Locati
Page 333 and 334: Location and Type of Disinfection T
Page 335 and 336: Practice Problems 8.1 1. What is th
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Page 343 and 344: Practice Problems 10.1 1. What is t
Page 345 and 346: 3. A well with pumps located 46 ft
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Page 349 and 350: electrical rate of $0.111 per kW-Hr
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Page 359 and 360: Practice Problems 13.1 1. A 4-20 mA
Page 361 and 362: 7. A water utility uses a 4-20 mA s
Page 363 and 364: NEW PUMP: (450 gpm)(65 ft) Motor hp
Page 365 and 366: Total operational cost per year. $2
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Advanced Waterworks Mathematics, 2019a

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