Introduction to Water Distribution Pomona FairPlex July 2008 Day 8. Math. Student Handbook. CW Wulff Associates

Transcription

1 Introduction to Water Distribution Pomona FairPlex July 2008 Day 8 Math Student Handbook

2 Four Primary Functions 2

3 Unexecuted Division Function What does the following unexecuted fraction signify? 1 2 = Numerator Denominator Decimal equivalent = 0.5 Percent equivalent = 50% 3

4 Constants/Conversions 4

5 Inches/Feet Conversion Unless specifically instructed otherwise or because dimensional analysis calls for an exception, normally surface areas and volumes with be in units of feet # inches 12 inches / feet # 5

6 Inches/Feet Conversion Unless specifically instructed otherwise or because dimensional analysis calls for an exception, normally surface areas and volume with be in units of feet 24 inches 12 inches feet 2 feet 6

7 Inches/Feet Conversion Unless specifically instructed otherwise or because dimensional analysis calls for an exception, normally surface areas and volume with be in units of feet 6 inches 12 inches feet 7

8 Inches/Feet Conversion Unless specifically instructed otherwise or because dimensional analysis calls for an exception, normally surface areas and volume with be in units of feet 2 inches 12 inches feet 8

9 Inches/Feet Conversion Unless specifically instructed otherwise or because dimensional analysis calls for an exception, normally surface areas and volume with be in units of feet 33 inches 9

10 Gallon of water Common Constants 8.34 pounds One Gallon Fresh Water 8.60 pounds One Gallon Saline Water Why? 10

11 Common Constants Cubic Foot How many gallons? 7.48 gallons 11

12 Common Constants Cubic Foot How many pounds? 12

13 Common Constants Cubic Yard How many ft 3? 3 feet feet feet feet 27 ft 3 13

14 Common Constants Gallon 3.78 liters 14

15 Common Constants - Acre-foot Football Field 1 Foot Deep 1 ft 43,560 ft 2 Approximately 326,000 gals 15

16 Common Constants Mile Day 1 Mile = 5,280 feet Day = 1,440 minutes 16

17 Conversion of Gallons Conversion of gallons to MG 2,500, ,000 10,000 1,000 Divide by 1,000,000 gal/mg 2,500,000 gal 1,000,000 gal/mg 925,000 1,000,000 10,000 1,000,000 1,000 1,000, MG MG 0.01 MG MG 17

18 Specific Gravity 18

19 What is Specific Gravity (SG)? Specific Gravity is the relationship between a substance and water Wt. Substance (lbs.) Wt. H 2 0 (8.34 lbs.) 19

20 What is Specific Gravity (SG)? Specific Gravity is the relationship between a substance and water Wt. Substance (lbs.) Wt. H 2 0 (8.34 lbs.) 20

21 What is Specific Gravity (SG)? What is the specific gravity of alum if it weighs lbs/gal? Wt. Substance (lbs/gal) Wt. H 2 0 (8.34 lbs/gal) lbs/gal 8.34 lbs/gal =

22 What is Specific Gravity (SG)? What is the SG of NaOH if it weighs lbs/gal? Wt. Substance (lbs/gal) Wt. H 2 0 (8.34 lbs/gal) 22

23 What is Specific Gravity (SG)? What is the weight per gallon of NaOCl if SG is 1.1? Wt. H 2 0 (8.34 lbs/gal) * SG Wt. H 2 0 (8.34 lbs/gal) * 1.1 = 9.17 lbs/gal 23

24 What is Specific Gravity (SG)? What is the weight per gallon of garnet sand if SG is 4.4? Wt. H 2 0 (8.34 lbs/gal) * SG 24

25 Proper Mechanics 25

26 Proper Algebraic Mechanics Good a b * c 30 5 * 2 (10) = 3 No Good a b * c 30 5 * 2 = 12 Best a b * c 30 5 * 2 = 3 26

27 Scientific Notation Metric System Dimensional Analysis 27

28 Powers & Scientific Notation Powers Notations Shorthand used for expressing multiplication 28

29 Powers & Scientific Notation Powers Notations Shorthand used for expressing multiplication x 2 = x * x x 2 = 2x or x * = 4 * 4 = = 2 * 4 = 8 x 3 = x * x * x x 3 = 3x or x * = 4 * 4 * 4 = = 3 * 4 = 12 29

30 Powers & Scientific Notation Powers Notations Shorthand used for expressing multiplication of units ft 2 = ft * ft ft 2 = ft * 2 ft 3 = ft * ft * ft ft 3 = ft * 3 30

31 Metric System Metric System (based on 10-system) Basic metric prefixes: Kilo Hecto Deca - 10 Deci Centi Milli ppm Micro ppb Nano ppt 31

32 Math Review Dimensional Analysis Method of determining whether you have set up a math problem correctly by using proper units Unit cancellations Cost $/mo. = 90.1 lbs/day * 30 days/mo. * $350 ton 2000 lbs/ton $ 473/mo. 32

33 Temperature Math 33

34 Math Temperature Conversions 34

35 Option 1 Temperature Conversion Fahrenheit to Celsius C o = 5 * (F o - 32 o ) 9 Celsius to Fahrenheit F o = 9 * (C o ) + 32 o 5 35

36 Option 2 Temperature Conversions Step 1 = Add 40 Step 2 = Multiply by.555 or 1.8 Multiply by.555 if F o Multiply by 1.8 if C o C o F o Step 3 = Subtract 40 36

37 Temperature Conversion Convert 70 0 F to 0 C Step 1 = Add = 110 Step 2 = Multiply by *.555 = 60.5 Step 3 = Subtract = 21 0 C Convert 70 0 F to 0 C C o = 5 * ( o F - 32 o ) 9 C o = 5 * (70 o F - 32 o ) 9 C o = 21 0 C 37

38 Head/psi 38

39 Fluid Hydraulics Why don t t both outlets produce the same parabolic discharge? 39

40 40

41 41

42 42

43 Determining Pressure/Head Pressure Head 1 ft of water elevation =.433 psi 1 psi = 2.31 ft of water elevation PSI 2.31 * * 2.31 Ft of Head 43

44 Determining Pressure/Head Convert a pressure of 14 feet to psi * PSI Ft of Head 14 ft * psi/ft = 6 psi 44

45 Determining Pressure/Head A head of 250 feet of water is equivalent to what pressure in psi? * PSI Ft Head 250 feet of head * ft / psi = 108 psi 45

46 Determining Pressure/Head What is the pressure at ground level of a full tank 75 feet high? Answer: Known: 1 ft = psi * ft =? psi PSI Ft Head 75 ft psi = ft * psi/ft 75 ft * psi/ft? 32 psi 46

47 Determining Pressure/Head How many feet of head is equivalent to 100 psi? 1 psi = 2.31 ft/head 100 psi =? ft/head 100 psi * 2.31 ft/head/psi PSI * 2.31 Ft Head 231 ft/head 47

48 Determining Pressure/Head A pressure gauge reads 150 psi.. How many feet of head are exerted on the gauge? PSI 2.31 Ft Head 150 psi * 2.31ft/psi = 346 ft 48

49 Determining Pressure/Head If a system hydrant registers 75 psi,, what would be the system head? 1 psi = 2.31 ft/head 75 psi =? ft/head 75 psi * 2.31 ft/head/psi PSI * 2.31 Ft Head 173 ft/head 49

50 Determining Pressure/Head How high can a pump with a discharge capacity of 110 psi lift water? 110 psi 1 psi = 2.31 ft of lift 110 psi = 2.31 ft per psi * 110 psi 254 ft PSI * 2.31 Ft Head 50

51 Determining Pressure/Head A water tank is 150 feet in diameter and 36 feet in elevation with a spill level of 28 feet. If the tank is full, what is the pressure reading at the bottom of the tank? * foot = psi PSI = psi/feet * 28 feet Ft Head 12.1 psi 51

52 Determining Pressure/Head A water tank is 150 feet in diameter and 36 feet in elevation with a spill level of 28 feet. If the tank is full, what is the pressure reading from a gauge measuring 3 feet below the bottom in a vault? * psi/feet * (28+3) feet 13.4 psi PSI Ft Head 28 ft? 3 ft 150 ft 52

53 Determining Pressure/Head An elevated stand pipe has a diameter of 20 feet and a height of 50 feet. What is the reading on a pressure gauge located at the bottom of the stand pipe if the standpipe is 37% full? (assume 100% capacity at 50 feet of water elevation) 50 ft *.37(%) = 18.5 ft of head 50 ft 18.5 feet * psi/ft 20 ft? 37% of what? = 8 psi 53

54 A pump has a suction supply pressure reading of 35 psi and a discharge pressure reading of 125 psi. The pump nameplate indicates the pump will produce 600 gpm at 185 ft/tdh. What would you expect the pump to do? 35psi 125psi Pump more than 600 gpm Pump less than 600 gpm Pump the same 600 gpm Cannot determine 54

55 30 ft? feet 100 feet A tank measures 30 in diameter. A pressure gauge 100 below the bottom of the tank reads 50 psi. What is the water level in the tank? Step 1 1psi = 2.31ft of water elevation 50psi * 2.31ft/psi = ft total head 50 psi PSI * 2.31 Ft Head Step = 15.5ft in tank 55

56 Determining Pressure/Head A pump is located 100 feet below a water storage tank. If the tank is 35 feet high, and the pump is capable of a discharge pressure of 50 psi,, is there sufficient pumping to fill the tank (assume 100% hydraulics)? Known: tank height - 35 feet tank head feet pump discharge (psi( psi) - 78 psi hydraulic eff % 56

57 35 feet 1 psi = 2.31 feet 100 feet 50 psi * 2.31 feet/psi feet 50 psi 57

58 Determining Pressure/Head An elevated tank has an 16 common inlet/outlet main. A pressure gauge located at the bottom of the main, 100 feet below the bottom, om, reads 78 psi.. If the tank is 150 feet in diameter and 95 feet high, how many feet full is the tank? 150 feet? 95 feet PSI * 2.31 Ft Head 1) 78 psi * 2.31 fthd/psi 180 fthd 2) 180 fthd 100 fthd feet 78 psi 80 fthd/tank 58

59 Determining Pressure/Head A pump at 400 ft. msl has a discharge capacity of 195 psi.. A tank at 900 ft. msl has a height of 32 ft. Does the pump have sufficient discharge capability to fill the tank? (Assume 100% efficiencies and no friction losses) 900 msl Known: 900 ft tank elevation + 32 ft additional tank height 932 ft total discharge head 400 ft msl ft pump elevation 532 ft differential (pumping head) 59

60 Pump psi capacity psi 195 psi * 2.31ft/psi PSI * 2.31 Ft Head 900 msl feet of total discharge lift 532 ft lift required Answer: 400 ft msl

61 What is the psi at a hydrant located at 290 feet msl if the water supply is delivered from a storage facility located at 430 feet of msl with a water depth of 20 feet? PSI * ft * psi/ft Ft Head ft * psi/ft = pressure 290 msl msl psi 61

62 Averages 62

63 Determining Averages If a water meter reads 2,538,000 (total gallons) on Monday, and 7 days later it reads 5,667,000 what is the daily average delivery in gallons? Average is obtained by determining the differential between Monday and the ending reading 7 days later and then dividing by the number of days Ending reading: 5,667,000 gallons Beginning reading: - 2,538,000 gallons Differential reading: 3,129,000 gallons 3,129,000 gallons 7 days = 447,000 gpd 63

64 Determining Averages If a meter registers 123,000 ft 3 on Monday, 235,000ft 3 on Tuesday, and 105,000 ft 3 each day on Wednesday, Thursday, and Friday, what is the average production for that 5-day 5 period? Average is determined by adding up the daily totals and dividing by the number of days. Monday 123,000 ft 3 Tuesday + 235,000 ft 3 Wednesday + 105,000 ft 3 Thursday + 105,000 ft 3 Friday + 105,000 ft 3 673,000ft 3 (5-day total) 64

65 Determining Averages Average production = total ft 3 divided by the total number of days (5). 673,000 ft 3 = 5 days 134,600 ft 3 /day 65

66 Determining Averages If a utility truck odometer reads 76,564 miles on Monday and two weeks later the reading is 78,675 what is the average daily miles driven? Average is obtained by determining the differential between Monday and two weeks later and then dividing by the number of days Ending reading: 78,675 miles Beginning reading: - 76,564 miles Differential reading: 2,111 miles 2,111 miles 14 days = 151 mpd 66

67 Determining Average Mileage A vehicle is driven based upon the following mileage: Monday 28.4 miles Tuesday 17.4 miles Wednesday 22.3 miles Thursday 9.3 miles Friday 33.4 miles Saturday miles Sunday 0 miles miles miles 7 days = 22.8 miles What is the average daily mileage driven? 67

68 Meter Madness 68

69 Determining Metered Production Meter questions will at times require you to be aware of the values of the numbers given 69

70 Determining Metered Production Another meter registers on the first of the month. The previous first of the month reading was If the far right number represents 1,000 gallons, what was the total water produced for the month? 56,734,523-56,489, ,650 * 1,000 After factoring with multiplier 244,650,000 gallons 70

71 Determining Metered Production A meter reading shows as the ending reading for a 30-day month. If the starting reading was , what was the daily average consumption (gpd)) if the meter reading has a 10x gallon factor? 675, ,421 42,142 * 10 After factoring with multiplier 421,420 gallons 30 days = 14,047 gpd 71

72 Determining Metered Production A master meter read in gallons x The beginning reading of the month is At the end of a 30 day month the reading is What was the average flow in acft/day? gal gal 3000 gal * 1000 factor gal 3,000,000 gal 30 days = 1,000,000 gal/day 326,000 gal/acft acft 3.07 acft/day 72

73 Determining Meter Calibration A household meter registers 148 ft 3 during a bench test when in fact the flow was 150 ft 3. What is the meter s s accuracy deviation and does it fall within industry standards? % = metered amount 148 ft = 3 = actual amount 150 ft 3 Meter Accuracy = 98.7% Meter Deviation = 100 % % = 1.3% 73

74 Determining Meter Calibration If a bench test for a PD meter shows that 75 gallons flowed through the tested meter and it recorded 77 gallons, what would be the accuracy of the meter and does it meet AWWA standards? / yes / no 97.4 / yes 97.4 / no 77 gal recorded 75 gal actual 1.027* * 100 = % 74

75 Determining Unaccounted for Water What percent of unaccounted for water would a utility experience if the system meters read 4,578,755 gallons and a master meter showed 4,789,983 gallons? % = system meters master meter = 4,578,755 4,789,983 = % Unaccounted for water = 100 % % = 4.4% 75

76 Surface Areas 76

77 Surface Area 2-dimensional geometric shapes 77

78 Determining Surface Area Most common 2-dimensional 2 geometric shapes A ft2 = d 2 *.785 or π r 2 A ft2 or or = h * w = h * length = w * length 78

79 Cylinder 79

80 Cylinder or dπ 80

81 Cylinder or dπ 81

82 Cylinder Determining Surface Area or dπ A ft2 ft2 = 2 π r * length (height) A ft2 ft2 = d * π * length (height) A ft2 = π r 2 or d 2 (.785) 82

83 Determining Surface Area What is the outer surface area of a cylinder that measures 50 feet in diameter and 125 feet long? Area = d π * length or dπ Area = (50 ft) (3.14) * 125 ft Area = 19,625 ft 2 83

84 Determining Surface Area What is the surface area of that same cylinder s two ends (just one side)? Area = 2 [d 2 *.785] Area = 2[(502 ft) 2 *.785] Area = 3,925 ft 2 84

85 Determining Surface Area What is the outer surface area of the tank wall and the roof, and inner surface of the floor of a cylindrical water tank that measures 125 feet in diameter and is 42 feet high? Area = d π * length Area = (125 ft) (3.14) * 42 ft 16,485 ft 2 24,531 ft 2 Area = 16,485 ft 2 Area = 2[d2 2 *.785] Area = 2[(1252 ft) 2 *.785] = 24,531 ft 2 24,531 ft ,485 ft 2 41,016 ft 2 85

86 Determining Surface Area How many gallons of paint will need to be purchased (assume 1-gal. 1 cans) to completely paint a tank that has a combined surface area of 11,000 ft 2? Assume each gallon covers 150 ft 2. 11,0000 ft 2 No. of cans paint = ft 2 Paint Coverage (ft 2 ) ,000 ft ft 2 /can =

87 Determining Surface Area What is the cross sectional surface area of a 3 foot diameter pipe? 3 ft 3 ft Answer: Area = d 2 *.785 (3 ft ) 2 * ft 2 * ft 2 87

88 Determining Surface Area What is the cross sectional area of a pipe 33 inches in diameter? Answer: 33 inches =? Feet 12 inches = 1 foot 33 inches /12 inches/foot 33 inches 33 inches = 2.75 ft Area = d 2 *.785 (2.75 ft ) 2 * ft2 88

89 Volumes 89

90 Determining Volume Most common 3-dimensional geometric shapes Cylinder: Storage Tanks Cube: Basins Vol = d 2 *.785 * L or π r 2 * L Vol = h * w * L 90

91 Determining Volume A basin measures 200 feet long, 52 feet wide, and 12 feet high. What is the volume in cf & gallons? V = h * w * length V = 12 ft * 52 ft * 200 ft V = 124,800 ft 3 12 ft 200 ft 52 ft V gal = ft 3 * 7.48 gal / ft 3 V gal = 124,800 ft 3 * 7.48 gal / ft 3 933,500 gal 91

92 Determining Volume A basin measures 200 feet long, 52 feet wide, and 12 feet high. What is the volume in acre-feet feet? Acre-feet = gallons gallons/acre-foot 12 ft 200 ft Acre-feet = 933,500 gallons 325,851 gallons/acre-foot 52 ft 2.86 acre-feet 92

93 36 A 24 I.D. pipe is to be installed in a trench 1.5 miles long. If the trench is over dug by 1 foot on either side with an intended cover of 36,, how many yds 3 of spoil must be hauled off if the O.D. is 26? 26 Determining Volume 1.5 miles Vol pipe = d 2 *.785 * length 24 Vol pipe = (2.17 ft) 2 *.785 * 7,920 ft Vol pipe = 29,276 ft 3 1 ft 1 ft 93

94 36 A 24 I.D. pipe is to be installed in a trench 1.5 miles long. If the trench is over dug by 1 foot on either side with an intended cover of 36,, how many yds 3 of spoil must be hauled off if the O.D. is 26? ft 1 ft Determining Volume 1.5 miles Vol 3 Yd HauledOff = Vol 3 Yd HauledOff = Vol 3 Yd HauledOff = 1,084 yds 3 ft 3 27 ft 3 yd 3 29,276 ft 3 27 ft 3 yd 3 94

95 A basin measures 200 feet long, 52 feet wide, and 12 feet high. If the basin contains 70,000 cf of water, what is the depth? 200 ft Vol = h * w * length 70,000 cf = h ft * 50 ft * 200 ft Invert 70,000 cf ft = h ft 50 ft * 200 ft 7 ft Invert 12 ft 70,000 cf 50 ft? 95

96 Determining Volume A 60,000 cubic foot tank holds how many gallons of water? Answer: 1 cubic foot = 7.48 gallons 60,000 cubic feet =? Gallons 60,000 cf Gallons = cubic feet * 7.48 gal/ft 3 = 60,000 ft 3 * 7.48 gal/ft 3 449,000 gallons 96

97 Determining Volume What is the volume of a pipe 2 feet in diameter and 5,280 feet long? Volume= d 2 *.785 * length Volume= (2 ft ) 2 *.785 * 5,280 ft Volume= 16,600 ft3 97

98 A tank measures 150 feet in diameter. It has a water level of 32 feet. What is the volume in gallons? V g = d 2 *.785 * height * 7.48 gal/ft 3 V g = (150 ft ) 2 *.785 * 32 ft * 7.48 gal/ft 3 V g = 4,227,696 gal Determining Volume 32 ft 150 ft 98

99 Dosage Math 99

100 Davidson Templates Pounds (Lbs) MG 8.34 ppm Pounds (Lbs) MG 8.34 ppm 100

101 Determining Pounds of Chemical (100 %) How many pounds of chemical are needed to treat 1 MG at a concentration of 1 mg/l? Lbs MG * 8.34 * ppm

102 Determining Pounds of Chemical (100 %) How many pounds of chlorine are needed to disinfect a tank holding 500,000 gallons of water to a concentration of 50 ppm? Lbs/Cl 2 = MG * 8.34 * ppm 500,000 gal 1,000,000 gal/mg = 0.5 * 8.34 *

103 Determining Pounds of Chemical (100 %) A A pipe has a flow of 2,500 gpm.. A 0.75 mg/l residual is desired (assume no demand). How many pounds of chlorine per day are required? Establish total 24-hour flow: 2,500 gpm * 1440 min/day = 3,600,000 gal/day = 3.6 mgd Lbs Lbs = MGD * 8.34 * mg/l 3.6 * 8.34 * 0.75 MG 8.34 ppm 22.5 lbs/cl 2 103

104 Determining ppm/mg/l What is the resulting concentration if 110 pounds of CuSO 4 is fed into a tank containing 5 million gallons? pounds mg/l (ppm) = MG * 8.34 Lbs * 8.34 MG 8.34 ppm 2.6 ppm 104

105 Determining Pounds of Chemical (100 %) What is the dosage in mg/l if a chemical metering pump has fed 3,000 lbs of chemical into a pipeline that is flowing at a rate of 12,000,000 gallons per day? Lbs (MGD) * (8.34) * (mg/l) 3,000 lb 12 MGD * 8.34 = 30 mg/l Lbs MG 8.34 ppm 105

106 Determining Pounds of Chemical (100 %) How many lbs of alum are used if a chemical metering pump is set to feed 45 mg/l into a flow rate of 32 MGD? 32 MGD * 8.34 * 45 mg/l 12,010 lbs/day Lbs MG 8.34 ppm 106

107 Determining Pounds of Chemical (100 %) How many lbs of chlorine are used in a day if a well is producing 1800 gpm and the residual at the discharge of the well shows 1.5 mg/l chlorine (assume 0 demand)? 1800 gpm * 1440min/day = 2,592,000 gal/day 2,592,000 gal/day 1,000,000 gal.mg = MG Lbs Lbs = (MGD) * (8.34) * (mg/l) 2.59 MGD * 8.34 * 1.5 mg/l MG 8.34 ppm 32.4 lbs/day 107

108 Determining Pounds of Chemical (100 %) A water has a chlorine demand of 1.1 mg/l. The desired residual is 1.2 mg/l How many pounds of chlorine will be required daily to chlorinate a flow of 2500 gpm? Dosage = 1.1mg/L (demand) mg/l(target) = 2.3 mg/l 2500 gpm * 1440 min/day = 3,600,000 gpd 3,000,000 gal/day 1,000,000 gal.mg = 3.0 MG Lbs Lbs = (MGD) * (8.34) * (mg/l) 3.0 MGD * 8.34 * 2.3 mg/l MG 8.34 ppm 57.5 lbs/day 108

109 Determining Pounds of Chemical (~ %) If a chlorine dose of 110 pounds per day of chlorine is necessary to treat a particular water, how many pounds of 65% HTH are needed? pounds of chemical decimal equivalent (%) 110 lbs

110 Determining Pounds of Chemical (~ %) If a chlorine dose of 110 pounds per day of chlorine is necessary to treat a particular water, how many pounds of 5% NaOCl solution is needed? pounds of chemical decimal equivalent (%) 110 lbs ,

111 Determining Pounds of Chemical (~ %) If a chlorine dose of 110 pounds per day of chlorine is necessary to treat a particular water, how many pounds of 0.8% NaOCl solution is needed? pounds of chemical decimal equivalent (%) 110 lbs ,

112 Determining Pounds of Chemical (~ %) A pipe has a flow of 2,500 gpm.. A 0.75 mg/l residual is desired (assume no demand). How many pounds of 65% HTH per day are required? Establish total 24-hour flow: 2,500 gpm * 1440 min/day = 3,600,000 gal/day Pounds chlorine: = 3.6 mgd MGD * 8.34 * mg/l 3.6 * 8.34 * lbs..65 lbs. Chlorine = 34.6 lbs 65%HTH 112

113 Determining Pounds of Chemical (~ %) If 51 pounds of 65% HTH is used in a day and the flow produced is i 1,400,000, what is the resulting concentration in ppm? Lbs (MGD) * (8.34) 51 lbs/hth (.65) (1.4 MG) * (8.34) 2.8 ppm Lbs MG 8.34 ppm 113

114 Determining Gallons of Chemical (~ %) How many gallons of 5% NaOCl are required to disinfect a tank 100 feet in diameter and 30 feet high with a dosage of 10 ppm (assume no chlorine demand and specific gravity of 1.1)? Volume gal = d 2 *.785 * height * 7.48 gal/cf (100 ft) 2 *.785 * 30 ft * 7.48 gal/cf 1,761,540 gallons Volume MG =

115 Determining Gallons of Chemical (~ %) How many gallons of 5% NaOCl are required to disinfect a tank 100 feet in diameter and 30 feet high with a dosage of 10 ppm (assume no chlorine demand and specific gravity of 1.1)? lbs. Cl 2 = MG * 8.34 * ppm 1.76 * 8.34 * ,940 lbs 2,940 lbs 9.17 lbs/gal 321 gallons 115

116 Determining Volume What is the volume of a basin in gallons if 75 pounds of 100% chemical is applied and a residual is measured as 3.2 ppm? pounds MG = 8.34 * ppm Lbs * 3.2 MG 8.34 ppm 2.81 MG 2,810,000 gallons 116

117 Determining ppm/mg/l A A vehicle with a filled 18 gallon tank of gasoline is driven to Pomona. After arriving, the driver is able to read his exacto gas gauge and sees that his tank now holds 7 gallons. What was the fuel consumed on the trip? Amount of gasoline started with (full tank) Amount of gasoline remaining Amount of gasoline consumed 18 gallons - 7 gallons 11 gallons 117

118 Chlorine (Demand-Free Free-Total) Full tank of gas Gas consumed Remaining gas in tank 118

119 Math Review What is the chlorine demand of a system if 3.0 mg/l of chemical is applied and after an appropriate amount of time, a residual test shows a concentration of 0.6 ppm. 3.0 ppm Demand ppm? 0.6 ppm 119

120 Math Review What is the chlorine demand of a system if 3.0 mg/l of chemical is applied and after an appropriate amount of time, a residual test shows a concentration of 0.6 ppm. Original chlorine application = 3.0 ppm Residual reading = ppm Total Chlorine Free Chlorine Demand 2.4 ppm Demand 120

121 Math Review If a water has a chlorine demand of 1.3 ppm,, how many mg/l of gaseous chlorine must be dosed to achieve a residual of 0.8 ppm? Total dosage ppm? 1.3 ppm 0.8 ppm 121

122 Math Review If a water has a chlorine demand of 1.3 ppm,, how many mg/l of gaseous chlorine must be dosed to achieve a residual of 0.8 ppm? Chlorine demand = 1.3 ppm Residual desired = ppm Demand Chlorine Free Chlorine Total chlorine 2.1 ppm Total Chlorine 122

123 Determining ppm/mg/l If 250,000 gallons are treated with 10 pounds of chlorine gas, what is the resulting concentration in ppm? Known: amount of chlorine applied (10 lbs) volume treated (250,000 gallons) Lbs. Total ppm = lbs MG * 8.34 MG 8.34 ppm.25 MG 10 * (total ppm) 123

124 Determining ppm/mg/l The chlorine demand of a water is 3 mg/l. If 250,000 gallons are treated with 10 pounds of chlorine gas, what is the residual? Known: amount of chlorine applied (10 lbs) volume treated (250,000 gallons) chlorine demand (3.0 ppm) Free Chlorine Combined Chlorine (Demand) Free + Combined = Total Chlorine 124

125 Determining ppm/mg/l The chlorine demand of a water is 3 mg/l. If 250,000 gallons are treated with 10 pounds of chlorine gas, what is the residual? Known: amount of chlorine applied (10 lbs) volume treated (250,000 gallons) chlorine demand (3.0 ppm) Total ppm = lbs MG * ,000 gal * ,000,000 gal/mg 4.8 (total ppm) 4.8 ppm total chlorine -3.0 ppm chlorine demand 1.8 ppm chlorine residual 125

126 Detention Time/Flow/Volume 126

127 Water Distribution Math Volume/Detention Time/Flow Q Time = Volume Time = Volume Q Volume Q T Volume = Q * Time 127

128 Math Review What flow (cfs( cfs) ) rate is utilized to fill a tank that contains 975 cubic feet (cf( cf) ) if the tank fills in exactly 600 seconds? Volume Q = Volume Time Q = 975 cf 600 sec = 1.63 cfs Q T 128

129 Math Review What flow (gpm( gpm) ) rate is utilized to fill a tank that contains 1,000,000 gallons if the tank fills in exactly 1,440 minutes? Volume Q = Volume Q Time = 1,000,000 gal 1,440 min Q T = 694 gpm 129

130 Math Review What flow (gpm( gpm) ) rate is utilized to fill a tank that contains 4,227,696 gallons if the tank fills in exactly 12 hours? Volume Q = Volume Q Time = 4,227,696 gal 12 hrs Q T Q gal/min = 4,227,696 gal 12 hrs * 60 min/hr = 5,870 gpm 130

131 T Detention = Volume Q gpm Math Review What is the detention time in hours to fill a 75 foot diameter tank, 35 feet high using a pump rated at 500 gpm? Volume Volume = (d ft ) 2 *.785 * h ft * 7.48 gal/ gal/cf Volume = (75 ft ) 2 *.785 * 35 ft * 7.48 gal/ Volume = 1,156,011 gallons gal/cf Q T T Detention/hr = 1,156,011 gals 500 gpm * 60 min/hr T Detention/hr =

132 Math Review What is the detention time in hours to fill a tank that holds 2,500,000 gallons if a pump produces 4,500 gpm? Q = Volume Time T = Volume Q T Hours Hours = 2,500,000 gal 4,500 gpm * 60 min/hr 9.26 hours 132

133 Detention Time A storage reservoir holds 5 MG. A well pumps into the reservoir at 2500 gph.. What is the detention time, in hours, in the reservoir? Time Which formula? T = Volume Q = Volume Q = 5,000,000 gals 2,500 gph Q = A * Velocity Q = Volume Time = 2,000 hrs or 133

134 Detention Time An elevated storage tank is 30 in diameter with an operating water level of 15.. Flow into the tank is 500 gpm. What is the detention time in the tank in hours? Volume = Time d 2 *.785 * h * 7.48 gal/cf = Volume Q (30 ft ) (30 ft ) *.785 * 15 ft *7.48gal/ft 3 = 79,269 gal Time = 79,269 gallons 500 gpm * 60 min / hr 2.64 hrs 134

135 Detention Time What is the detention time (in minutes) in a 24 inch pipeline that is 5000 feet long flowing at a rate of 3.3 ft 3 /s Time Volume = d 2 *.785 * h * 7.48 gal/cf = Volume Q (2 ft ) (2 ft ) *.785 * 5,000 ft *7.48gal/ft 3 = 117,436 gal Time = 117,436 gallons 3.3 cfs * 60 sec / min * 7.48 gal/cf 79.3 min 135

136 Detention Time A chlorine dosage change is made at noon on a 36 conduit 25 miles upstream of an analyzer. What time will the dosage change be detected if the flow is 36 cfs? Time Volume = d 2 *.785 * h * 7.48 gal/cf (3 ft ) (3 ft ) *.785 * (25 miles *5,280 ft ) = 932,580 cf = Volume Q Time = 932,580 cf 36 cfs * 60 sec / min * 60 min / hour 7.2 hours 0.2 hours * 60 min / hr = 12 min 7 hours & 12 min 136

137 Detention Time A chlorine dosage change is made at noon on a 36 conduit 25 miles upstream of an analyzer. What time will the dosage change be detected if the flow is 36 cfs? Time = Volume Q Dosage Change noticed = 12:00 pm + Detention Time Change noticed Dosage Change noticed = 12:00 pm + 7 hours & 12 min Change noticed Dosage Change noticed = 7:12 pm Change noticed 137

138 Flow/Area/Velocity 138

139 Flow Math The continuity equation can be used to find flow rate (volume over time, ft 3 /sec), velocity (feet per second, fps), and area of pipe carrying the water, (ft 2 ). Quantity: rate of flow ft 3 /sec Area: square feet, ft 2 Velocity: speed fps 139

140 Flow Math Q = A * Velocity A = Q Velocity Velocity = Q A A Q Vel 140

141 Determining Flow (Q) An 8-inch 8 main needs to be flushed. If the required velocity through the main is 2.5 ft/sec what is the required flow rate in cfs? Q cfs = A [cross sectional area] * Velocity Q cfs = [(0.667 ft ) 2 Q cfs = cfs *.785] * 2.5 ft/sec 141

142 Determining Flow (Q) An 8-inch 8 main needs to be flushed. If the required velocity through the main is 2.5 ft/sec what is the required flow rate in gpm? Q cfs = A [cross sectional area] * Velocity Q cfs = [(0.667 ft ) 2 Q cfs = cfs Q gpm = Q gpm = 392 gpm *.785] * 2.5 ft/sec cf/sec * 60 sec/min * 7.48 gal/cf 142

143 Telemetry Math PV 4 MV 4 143

144 Telemetry Math

145 Telemetry Math 4-20 ma question You are using a 4 to 20 ma transducer to measure a standpipe that has a maximum sensed height of 120 feet. Your indicator is out of service so you have hooked up your digital volt meter, which reads 16 ma.. What is the level in the reservoir in feet? Answer:

146 Telemetry Math 4-20 ma question You are using a 4 to 20 ma transducer to measure a standpipe that has a maximum sensed height of 120 feet. Your indicator is out of service so you have hooked up your digital volt meter, which reads 16 ma.. What is the level in the reservoir in feet? Answer: PV 4 MV 4 16 ma 4 20 ma 4 12 ma 16 ma =.75 (or 75 %).75 * 120 ft = 90 ft 146

147 Telemetry Math A ma transducer mounted on a distribution system pipeline measures the line pressure. If the maximum sensed pressure is 250 psi what is the pipeline pressure if the transducer is sending a signal of 13 ma to the recorder? PV 4 MV 4 13 ma 4 20 ma 4 9 ma 16 ma =.56 (or 56 %).56 * 2500 psi = 141 psi 147

148 Telemetry Math A standpipe measures 26 ft tall and has a diameter of 15 ft. A ma transducer, located at the bottom of the standpipe measures pressure and has a maximum sensed pressure of 11 psi.. What is the elevation in the standpipe if the ma signal coming from the transducer is 15 ma? PV 4 MV 4 15 ma 4 20 ma 4 11 ma 16 ma =.68 (or 68 %) 7.56 psi.68 * 11 psi = 7.56 psi psi * 2.31 ft/psi = 17.5 ft 148

149 Quiz No A tank has a capacity of 250,000 gallons. If it takes a pump 180 minutes to fill the tank, what is the flow of the pump in gpm? 589 gpm 889 gpm 1,389 gpm 4,389 gpm 180 minutes to fill 250,000 gals Q gpm = Volume Time (min) Q gpm = 250,000 gal 180 min Volume Q T Q gpm = 1,

150 Quiz No What is the flow of a pump in gpm that is used to fill a tank holding 180,000 gallons if it takes the pump 60 minutes to fill? 3,000 gpm 401 gpm 180,000 gpm 1.8 gpm Q gpm = Volume Time Q = Volume Time Volume Q T Q gpm = 180,000 gallons 60 minutes 150

151 Quiz No A pipeline has three sections that include 2,000 feet of 18-inch pipe, 3,500 feet of 12-inch pipe, and 5,050 feet of 6-inch pipe. What is the total volume in cubic feet? 7,272 21,272 54, ,395 Volume= d 2 *.785 * length Pipe No. 1 Pipe No.2 Vol pipe1 = (1.5 ft) 2 *.785 * 2,000 ft 3,533 cf Vol pipe2 = (1.0 ft) 2 *.785 * 3,500 ft 2,748 cf Pipe No. 3 Vol pipe3 = (0.5 ft) 2 *.785 * 5,050 ft cf 7,272 cf 151

152 Quiz No A basin measures 200 feet long, 52 feet wide, and 12 feet deep. How many million gallons does the basin hold? , , , ft 200 ft Vol gal Vol gal = h * w * length * 7.48 gal/cf = 12 ft * 52 ft * 200 ft * 7.48 gal/cf 52 ft. 152

153 End of Day 8 Bill Wulff cwulff@bak.rr.com 153