Water Meter Accuracy, Percent Solution Strength and Determining Chlorine Dosage Solution Strength in Waterworks Operation Math for Water Technology MTH 082 Lecture Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306- 311)
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Water Meter Accuracy, Percent Solution Strength and Determining Chlorine Dosage Solution Strength in Waterworks Operation Math for Water Technology MTH.
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Water Meter Accuracy, Percent Solution Strength and Determining
Chlorine Dosage Solution Strength in Waterworks Operation
Water Meter Accuracy, Percent Solution Strength and Determining
Chlorine Dosage Solution Strength in Waterworks Operation
Math for Water TechnologyMTH 082
Lecture Chapter 6- Applied Math for Water Plant Operators
Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators
Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)
Math for Water TechnologyMTH 082
Lecture Chapter 6- Applied Math for Water Plant Operators
Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators
Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)
ObjectivesObjectives
1. Water meter accuracy
2. Percent solution strength
3. Learn to calculate lbs formula for different specific gravities of basic chemical solutions (percent strength)
1. Water meter accuracy
2. Percent solution strength
3. Learn to calculate lbs formula for different specific gravities of basic chemical solutions (percent strength)
Reading assignment: Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)
Reading assignment: Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213) Chapter 12- Applied Math for Water Plant Operators Percent Strength of Chlorine Solution (pg 306-311)
Water Meter Accuracy Water Meter Accuracy Meter accuracy, %= (meter reading)* (100%)
Actual volume
IN % PROBLEMS UNITS MUST CANCEL!
(Gal and Gal, L and L)
Meter accuracy, %= (meter reading)* (100%)
Actual volume
IN % PROBLEMS UNITS MUST CANCEL!
(Gal and Gal, L and L)
•Given•Formula•Solve:
•Given•Formula•Solve:
If a water meter is being tested in the lab and reads 2,412 L. A volumetric tank
shows that the actual value in the tank is 2,445 L. What is the percent accuracy of
the meter? ?
If a water meter is being tested in the lab and reads 2,412 L. A volumetric tank
shows that the actual value in the tank is 2,445 L. What is the percent accuracy of
the meter? ?
101.
40%
98.6
5%
0.01
%1.
35%
0% 0%0%
100%
V1= 2,412 L, V2=2,445 L
Meter accuracy, %= (meter reading)* (100%) Actual volume
MA%= (2,412L)(100%)/2,445 LMA%= 98.65%
V1= 2,412 L, V2=2,445 L
Meter accuracy, %= (meter reading)* (100%) Actual volume
MA%= (2,412L)(100%)/2,445 LMA%= 98.65%
1. 101.4%
2. 98.65%
3. .010%
4. 1.35%
1. 101.4%
2. 98.65%
3. .010%
4. 1.35%
•Given•Formula•Solve:
•Given•Formula•Solve:
A water meter is being tested in the lab and it reads 500 ft3. A volumetric tank
shows that the actual value is 14,883 L. What is the percent accuracy of the
meter? ?
A water meter is being tested in the lab and it reads 500 ft3. A volumetric tank
shows that the actual value is 14,883 L. What is the percent accuracy of the
meter? ?
98.4
2%
101.
60%
96.6
6%
1.98
%
8%0%
77%
15%
V= 500 ft3 * 7.48 gal/1ft3 = 3740 gal,, V2=14,883 L and 1 Liter = 0.26 Gallons
Meter accuracy, %= (meter reading)* (100%) Actual volume
Meter accuracy, %= (3740, gal)* (100%) 3870 gal
MA%= 96.66%
V= 500 ft3 * 7.48 gal/1ft3 = 3740 gal,, V2=14,883 L and 1 Liter = 0.26 Gallons
Meter accuracy, %= (meter reading)* (100%) Actual volume
Meter accuracy, %= (3740, gal)* (100%) 3870 gal
MA%= 96.66%
1. 98.42%
2. 101.6%
3. 96.66%
4. 1.98%
1. 98.42%
2. 101.6%
3. 96.66%
4. 1.98%
•Given•Formula•Solve:
•Given•Formula•Solve:
If a meter had an accuracy of 99.1% and the reading on the meter indicated 1,500 L, what is the actual volume in liters?
If a meter had an accuracy of 99.1% and the reading on the meter indicated 1,500 L, what is the actual volume in liters?
148
6 L
15.
14 L
151
.4 L
151
3.6
L
21%
71%
0%7%
99.1% =.991 and 1,500 L 100% = 1.0
Meter accuracy, %= (meter reading)* (100%) Actual volume
Actual volume, L= (Meter reading)* (100%) (Meter Accuracy, %)
AV= (1,500)(1.0)/(0.991) = 1513.6 LAV= 1513.6 L
99.1% =.991 and 1,500 L 100% = 1.0
Meter accuracy, %= (meter reading)* (100%) Actual volume
Actual volume, L= (Meter reading)* (100%) (Meter Accuracy, %)
AV= (1,500)(1.0)/(0.991) = 1513.6 LAV= 1513.6 L
1. 1486 L
2. 15.14 L
3. 151.4 L
4. 1513.6 L
1. 1486 L
2. 15.14 L
3. 151.4 L
4. 1513.6 L
Percent Strength of a SolutionPercent Strength of a Solution% = Part *100 Whole
Amount of solute dissolved in solution!
% Strength = Chem lbs * 100 Total Sol, Lbs
% Strength = Chem lbs * 100
Water lb + Chem, Lbs
% Strength = Chem lbs * 100
(Solution gal)(Water lb) + Chem, LbsNEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!
Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213)
Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213)
•Given•Formula
•Solve:
•Given•Formula
•Solve:
If a total of 8 ounces of dry polymer are added to 10 gallons of water, what is the
percent strength (by weight) of the polymer solution?
If a total of 8 ounces of dry polymer are added to 10 gallons of water, what is the
percent strength (by weight) of the polymer solution?
6% 20%
0.05
%0.
60%
8%
77%
15%
0%
8 ounces of polymer= 1lb = 0.5 lbs 0.0625 oz% = Part *100 Whole
% Strength = Chem lbs *100 Water lb + Chem, Lbs
% Strength = 0.5 lbs *100
(10 gal) (8.34 lb/gal) + 0.5 lbs
% Strength = 0.5 lbs * 100% 83.9 lbs Sol
8 ounces of polymer= 1lb = 0.5 lbs 0.0625 oz% = Part *100 Whole
% Strength = Chem lbs *100 Water lb + Chem, Lbs
% Strength = 0.5 lbs *100
(10 gal) (8.34 lb/gal) + 0.5 lbs
% Strength = 0.5 lbs * 100% 83.9 lbs Sol 1. 6%
2. 20%
3. 0.05%
4. 0.6 %
1. 6%
2. 20%
3. 0.05%
4. 0.6 %
•Given•Formula
•Solve:
•Given•Formula
•Solve:
If 100 grams of dry polymer are dissolved in 5 gallons of water, what is the percent
strength (by weight) of the polymer solution?
If 100 grams of dry polymer are dissolved in 5 gallons of water, what is the percent
strength (by weight) of the polymer solution?
0.52
%5.
20%
2.41
%
24.6
3%
100%
0%0%0%
100 grams= 0.0022 lb = 0.22 lbs of chemical grams% = Part *100 Whole
% Strength = Chem lbs *100 Water lb + Chem, Lbs
% Strength = 0.22 lbs *100
(5 gal) (8.34 lb/gal) + 0.22 lbs
% Strength = 0.22 lbs * 100% 41.92 lbs Sol
% strength =0.52%
100 grams= 0.0022 lb = 0.22 lbs of chemical grams% = Part *100 Whole
% Strength = Chem lbs *100 Water lb + Chem, Lbs
% Strength = 0.22 lbs *100
(5 gal) (8.34 lb/gal) + 0.22 lbs
% Strength = 0.22 lbs * 100% 41.92 lbs Sol
% strength =0.52%
1. 0.52%
2. 5.2%
3. 2.41%
4. 24.63%
1. 0.52%
2. 5.2%
3. 2.41%
4. 24.63%
•Given•Formula
•Solve:
•Given•Formula
•Solve:
How many lbs of dry weight polymer due you need to add to a 25 gallon tank to make a 1%
solution?
How many lbs of dry weight polymer due you need to add to a 25 gallon tank to make a 1%
Amount of solute dissolved in solution!% Strength of mixture = Chem lbs in Mix *100 Total lbs Sol in Mix
% Str of mix = Chem lbs in Sol #1 + Chem lbs in Sol #2 *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!1 gram= 0.0022 lbs 1lb= 454 grams1 ounce = 0.0625 pounds
1 gal = 8.34 lbs (if water)
Amount of solute dissolved in solution!% Strength of mixture = Chem lbs in Mix *100 Total lbs Sol in Mix
% Str of mix = Chem lbs in Sol #1 + Chem lbs in Sol #2 *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
NEED UNITS IN LBS FOR IT TO WORK (COVERSION NECESSARY)!!1 gram= 0.0022 lbs 1lb= 454 grams1 ounce = 0.0625 pounds
1 gal = 8.34 lbs (if water)
Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213)
Chapter 6- Applied Math for Water Plant Operators Percent Strength of Solution (pg 114-117) Chapter 9- Applied Math for Water Plant Operators Solution Mixtures (pg 208-213)
•Given•Formula
•Solve:
•Given•Formula
•Solve:
If 20 lbs of a 10% strength solution are mixed with 50 lbs of a 1%
solution, what is the percent strength of the new solution?
If 20 lbs of a 10% strength solution are mixed with 50 lbs of a 1%
solution, what is the percent strength of the new solution?
37%
3.70
%0.
04%
0.07
%
0% 0%
15%
85%
Solution #1= 20 lbs of 10% Solution #2 = 50 lbs of 1%
% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
% Str of mix = 20 lbs (10/100) + 50 lbs (1/100) *100 20 lbs Sol #1 + 50 lbs Sol #2 in Mix
% Str of mix = 2 lbs + 0.50 lbs *100
70 lbs Sol
% Str of mix = 3.7%
Solution #1= 20 lbs of 10% Solution #2 = 50 lbs of 1%
% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
% Str of mix = 20 lbs (10/100) + 50 lbs (1/100) *100 20 lbs Sol #1 + 50 lbs Sol #2 in Mix
% Str of mix = 2 lbs + 0.50 lbs *100
70 lbs Sol
% Str of mix = 3.7%
1. 37%
2. 3.7 %
3. 0.036 %
4. .07%
1. 37%
2. 3.7 %
3. 0.036 %
4. .07%
•Given•Formula
•Solve:
•Given•Formula
•Solve:
If 5 gal of an 8% strength solution are mixed with 40 gal of a 0.5% solution, what is the
percent strength of the new solution? (Assume the 8% solution weighs 9.5 lbs/gal and the 0.5% solution weighs 8.34 lbs/gal)
If 5 gal of an 8% strength solution are mixed with 40 gal of a 0.5% solution, what is the
percent strength of the new solution? (Assume the 8% solution weighs 9.5 lbs/gal and the 0.5% solution weighs 8.34 lbs/gal)
0.04
%14
%1.
40%
1.70
%
9%0%
91%
0%
Solution #1= 5 gal of 8% Solution #2 = 40 gal of 0.5%
% Str of mix = Chem lbs in Sol #1 (% str/100) + Chem lbs in Sol#2 (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix
How many gallons of a 9% liquid polymer should be mixed with water to produce 50 gallons of a 0.5% polymer solution? The
density of the polymer liquid is 10.34 lb/gal, assume the density of the polymer solution is
8.34 lb/gal?
How many gallons of a 9% liquid polymer should be mixed with water to produce 50 gallons of a 0.5% polymer solution? The
density of the polymer liquid is 10.34 lb/gal, assume the density of the polymer solution is
8.34 lb/gal?
1 g
al
1.9
5 gal
2.2
gal
22
gal
9%18%
73%
0%
9% Liquid polymer ?gallons of liquid polymer =50 gal of 0.5% polymer solution---- Density of polymer 10.4 lb/gal Density of polymer solution= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions
9% Liquid polymer ?gallons of liquid polymer =50 gal of 0.5% polymer solution---- Density of polymer 10.4 lb/gal Density of polymer solution= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions
A 10% liquid polymer will be used in making up a solution. How many gallons of liquid polymer should be added to the water to make up 40 gallons of 0.35% polymer solution. The liquid polymer has a specific gravity of 1.1. Assume the polymer has a specific
gravity of 1.0?
A 10% liquid polymer will be used in making up a solution. How many gallons of liquid polymer should be added to the water to make up 40 gallons of 0.35% polymer solution. The liquid polymer has a specific gravity of 1.1. Assume the polymer has a specific
gravity of 1.0?
1.1
gal
0.9
17 g
al
1.3
gal
1.0
6 gal
0%
10%
80%
10%
10% Liquid polymer ?gallons of liquid polymer =40 gal of 0.35% polymer solution---- liquid polymer specific gravity= 1.1, polymer solution specific gravity =11.1 (8.34 lb/gal) = 9.17 lb/gal polymer solution 1(8.34 lb/gal)= 8.34 lb/galLbs of polymer in liquid polymer = lbs of polymer in polymer solutions(Liquid Polymer gal)(D)(%/100) = (Polymer sol, lbs)(D)(% str pol sol/100)
If 12 gallons of a 10% strength solution are added to 48 gallons of 0.6% strength solution, what is the
percent strength of the solution mixture? (Assume the 10% strength solution weighs 10.12 lbs/gal and
the 0.6% solution weighs 8.7 lbs /gal)?
If 12 gallons of a 10% strength solution are added to 48 gallons of 0.6% strength solution, what is the
percent strength of the solution mixture? (Assume the 10% strength solution weighs 10.12 lbs/gal and
the 0.6% solution weighs 8.7 lbs /gal)?
1% 25%
36.7
0%
2.70
%
0%
64%
9%
27%
Sol#1=12 gallons of a 10% solution Sol #2= 48 gal of 0.6%solution---- D=10.12 lbs/gal D2=8.7 lbs/gal
% Str of mix = (Chem lbs in Sol #1)(D) (% str/100) + (Chem lbs in Sol#2)(D) (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix % Str of mix = (12 gal)(10.12 lb/gal)(10/100) + (48 gal)(8.7 lb/gal)(0.6/100) * 100 (12 gal)(10.12 lb/gal) + (48 gal)(8.7 lb/gal)
Sol#1=12 gallons of a 10% solution Sol #2= 48 gal of 0.6%solution---- D=10.12 lbs/gal D2=8.7 lbs/gal
% Str of mix = (Chem lbs in Sol #1)(D) (% str/100) + (Chem lbs in Sol#2)(D) (% str/100) *100 Total lbs Sol #1 in Mix + Total lbs Sol #2 in Mix % Str of mix = (12 gal)(10.12 lb/gal)(10/100) + (48 gal)(8.7 lb/gal)(0.6/100) * 100 (12 gal)(10.12 lb/gal) + (48 gal)(8.7 lb/gal)