basic chemical calculations, determining chlorine dose in waterworks operation
DESCRIPTION
Basic Chemical Calculations, Determining Chlorine Dose in Waterworks Operation. Math for Water Technology MTH 082 Lecture 1 Handout on water chemistry, Basic Science Chemistry Ch 3,4,7 Chapter 3 and 12- Math for Water Technology Operators Lbs/day formula, Dose Demand Residual. - PowerPoint PPT PresentationTRANSCRIPT
Basic Chemical Calculations, Determining Chlorine Dose in
Waterworks Operation
Basic Chemical Calculations, Determining Chlorine Dose in
Waterworks Operation
Math for Water TechnologyMTH 082
Lecture 1Handout on water chemistry, Basic Science Chemistry Ch 3,4,7
Chapter 3 and 12- Math for Water Technology OperatorsLbs/day formula, Dose Demand Residual
Math for Water TechnologyMTH 082
Lecture 1Handout on water chemistry, Basic Science Chemistry Ch 3,4,7
Chapter 3 and 12- Math for Water Technology OperatorsLbs/day formula, Dose Demand Residual
Week 2-3 Objectives Week 2-3 Objectives
1. Review Temperature
2. Learn to calculate basic chemical solutions
3. Understand new formulas for liquid and solid chlorine application
1. Review Temperature
2. Learn to calculate basic chemical solutions
3. Understand new formulas for liquid and solid chlorine application
Reading assignment: Chapter 3 and 12- Math for Water Technology OperatorsChemistry Chapter 3,4 and Chapter 7 Chemistry Chemical Dosage Problems (Basic Science Concepts and Applications)
Reading assignment: Chapter 3 and 12- Math for Water Technology OperatorsChemistry Chapter 3,4 and Chapter 7 Chemistry Chemical Dosage Problems (Basic Science Concepts and Applications)
Temperature ConversionsTemperature Conversions
oF= (9 * oC) + 32 5
oF= (9 * oC) + 32 5
oC= 5 * (oF – 32) 9
oC= 5 * (oF – 32) 9
Convert 17oC to Fahrenheit Convert 17oC to Fahrenheit
Convert 451oF to degrees CelsiusConvert 451oF to degrees Celsius
oF= (9 *17)+32=62.6oF= 63oF 5
oF= (9 *17)+32=62.6oF= 63oF 5
Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32.
Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32.
Fahrenheit to Celsius1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5.
Fahrenheit to Celsius1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5.
oC= 5* (oF -32)=232.7oC= 233oC 9
oC= 5* (oF -32)=232.7oC= 233oC 9
•Given•Formula:•Solve:
•Given•Formula:•Solve:
Convert 88oF to oC? Convert 88oF to oC?
31
OC
67
OC
17
OC
100%
0%0%
88 oF oC= 5 * (oF – 32) 9oC= 5 * (88-32)/9
oC= 31
88 oF oC= 5 * (oF – 32) 9oC= 5 * (88-32)/9
oC= 31
oC= 5 * (oF – 32) 9oC= 5 * (88-32)/9
oC= 31
oC= 5 * (oF – 32) 9oC= 5 * (88-32)/9
oC= 31
1. 31 OC
2. 67 OC
3. 17 OC
1. 31 OC
2. 67 OC
3. 17 OC
•Given•Formula:•Solve:
•Given•Formula:•Solve:
Convert 16oF to oC? Convert 16oF to oC?
-23
OC
-9 O
C
26
OC
4% 0%
96%
16 oF oC= 5 * (oF – 32) 9oC= 5 * (16-32)/9
oC= -9
16 oF oC= 5 * (oF – 32) 9oC= 5 * (16-32)/9
oC= -9
oC= 5 * (oF – 32) 9oC= 5 * (16-32)/9
oC= -9
oC= 5 * (oF – 32) 9oC= 5 * (16-32)/9
oC= -9
1. -23 OC
2. -9 OC
3. 26 OC
1. -23 OC
2. -9 OC
3. 26 OC
•Given•Formula:•Solve:
•Given•Formula:•Solve:
Convert 35oC to oF? Convert 35oC to oF?
57
OF
51
OF
95
OF
35
OF
0% 0%
100%
0%
35 oC oF= (9 * oC) + 32 5oF= (9 * 35oC) + 32
5 oF= 95 oF
35 oC oF= (9 * oC) + 32 5oF= (9 * 35oC) + 32
5 oF= 95 oF
oF= (9 * oC) + 32 5
1. 57 OF
2. 51 OF
3. 95 OF
4. 35 OF
1. 57 OF
2. 51 OF
3. 95 OF
4. 35 OF
Solutions/ProblemsSolutions/Problems• Mole (mol): chemical mass unit, defined to be 6.022 x 1023 molecules,
atoms, or some other unit. A mole of a substance is a number of grams of that substance, where the number equals the substances molecular weight.
• Molarity (mol/kg, molal, or M) denotes the number of moles of a given substance per liter of solvent
Molarity (M)– Moles of solute Liters of solution
Molarity formula• Grams=(formula weight, grams/mole)(liters)(M moles/liter)
• M (moles/liter) = _________grams____________
(formula weight, grams/mole)(final volume, liters)
• Mole (mol): chemical mass unit, defined to be 6.022 x 1023 molecules, atoms, or some other unit. A mole of a substance is a number of grams of that substance, where the number equals the substances molecular weight.
• Molarity (mol/kg, molal, or M) denotes the number of moles of a given substance per liter of solvent
Molarity (M)– Moles of solute Liters of solution
Molarity formula• Grams=(formula weight, grams/mole)(liters)(M moles/liter)
• M (moles/liter) = _________grams____________
(formula weight, grams/mole)(final volume, liters)
Solutions/ProblemsSolutions/Problems• Normality: a measure of concentration: it is equal to the
number of gram equivalents of a solute per liter of solution. Depends on the valence or charge (old way)
Normality formula
• Grams=(equival. weight, grams/equival.)(liters)(N, equivalent./liter)
• N (equivl./liter) = __________Grams ___________
(equival. weight, grams/equival.)(final volume, liters)
• Normality: a measure of concentration: it is equal to the number of gram equivalents of a solute per liter of solution. Depends on the valence or charge (old way)
Normality formula
• Grams=(equival. weight, grams/equival.)(liters)(N, equivalent./liter)
• N (equivl./liter) = __________Grams ___________
(equival. weight, grams/equival.)(final volume, liters)
Formulas Formulas • Percent Strength by Weight: Weight of solute X 100 Weight of solution
Molarity formula• Grams=(formula weight, grams/mole)(liters)(M moles/liter)
Normality Formula• Grams=(equival. weight, grams/equival.)
(liters)(N, equivalent./liter)
• Percent Strength by Weight: Weight of solute X 100 Weight of solution
Molarity formula• Grams=(formula weight, grams/mole)(liters)(M moles/liter)
Normality Formula• Grams=(equival. weight, grams/equival.)
(liters)(N, equivalent./liter)
_________ is defined as the number of equivalents of solute dissolved in one liter of solution.
_________ is defined as the number of equivalents of solute dissolved in one liter of solution.
Norm
ality
Mola
rity
Alk
alin
ity
Aci
dity
65%
20%
5%10%
1. Normality
2. Molarity
3. Alkalinity
4. Acidity
1. Normality
2. Molarity
3. Alkalinity
4. Acidity
The three most commonly used coagulants in water treatment
are:
The three most commonly used coagulants in water treatment
are:
Alu
min
um h
ydro
...
Alu
min
um s
ulfa...
Lim
e, s
odiu
m h
...
Soda,
lim
e an
d...
36%
5%0%
59%
1. Aluminum hydroxide, lime and sodium hydroxide
2. Aluminum sulfate, ferric chloride, and ferrous sulfate
3. Lime, sodium hydroxide, and chlorine
4. Soda, lime and chlorine
1. Aluminum hydroxide, lime and sodium hydroxide
2. Aluminum sulfate, ferric chloride, and ferrous sulfate
3. Lime, sodium hydroxide, and chlorine
4. Soda, lime and chlorine
A chemical commonly used for coagulation in water treatment
is:
A chemical commonly used for coagulation in water treatment
is:
Chlo
rine
Soda
ash
Alu
m
Coppe
r sul
fate
5%0%
86%
10%
1. Chlorine
2. Soda ash
3. Alum
4. Copper sulfate
1. Chlorine
2. Soda ash
3. Alum
4. Copper sulfate
The chemical symbol for the most common coagulant used in
water treatment, aluminum sulfate (alum), is:
The chemical symbol for the most common coagulant used in
water treatment, aluminum sulfate (alum), is:
Al2
(OH)6
Fe2
(SO4)
3
NH3(
OH)7
Al2
(SO
4)3
0%
100%
0%0%
1. Al2(OH)6
2. Fe2(SO4)3
3. NH3(OH)7
4. Al2(SO4)3
1. Al2(OH)6
2. Fe2(SO4)3
3. NH3(OH)7
4. Al2(SO4)3
Molecular Weights Molecular Weights • Step One: Determine how many atoms of each
different element are in the formula.
• Step Two: Look up the atomic weight of each element in a periodic table.
• Step Three: Multiply step one times step two for each element.
• Step Four: Add the results of step three together and round off as necessary.
• Step One: Determine how many atoms of each different element are in the formula.
• Step Two: Look up the atomic weight of each element in a periodic table.
• Step Three: Multiply step one times step two for each element.
• Step Four: Add the results of step three together and round off as necessary.
Solutions/ProblemsSolutions/Problems
Lime calcium oxide (CaO):
1 atom of calcium= 40 grams1 atom of oxygen= 16 grams Formula weight = 56 grams in 1 mole
So a .10 mole solution would contain how many grams?
(0.10)(56 grams)= 5.6 grams
Lime calcium oxide (CaO):
1 atom of calcium= 40 grams1 atom of oxygen= 16 grams Formula weight = 56 grams in 1 mole
So a .10 mole solution would contain how many grams?
(0.10)(56 grams)= 5.6 grams
•Given•Formula•Solve:
•Given•Formula•Solve:
Determine the molar mass of ALUM
chemical formula Al2(SO4)3?
Determine the molar mass of ALUM
chemical formula Al2(SO4)3?
75
g
198
.2 g
166
g
342
.14
g
0%
100%
0%0%
2 Al, 3 S, 12 O FIND Al=26.98 g, S=32.06 g, O = 16 g
MM= Al 2(26.98 g) + S 3(32.06g)+ O12(16g)MM= 2Al + 3S+ 12O
MM= 53.96 g + 96.18 g + 192 g
MM Al2(SO4)3 =342.14 g
2 Al, 3 S, 12 O FIND Al=26.98 g, S=32.06 g, O = 16 g
MM= Al 2(26.98 g) + S 3(32.06g)+ O12(16g)MM= 2Al + 3S+ 12O
MM= 53.96 g + 96.18 g + 192 g
MM Al2(SO4)3 =342.14 g
1. 75 g
2. 198.2 g
3. 166 g
4. 342.14 g
1. 75 g
2. 198.2 g
3. 166 g
4. 342.14 g
•Given•Formula•Solve:
•Given•Formula•Solve:
Determine the molar mass of sodium hexametaphosphate chemical formula
(NaPO3)6 ?
Determine the molar mass of sodium hexametaphosphate chemical formula
(NaPO3)6 ?
70
g
611
.28
g
700
.38
g
50.
20 g
0% 0%0%
100%
6 NA, 6 P, 18 O FIND Na=22.98 g, P=30.97 g, O = 16 g
MM= Na 6( 22.98g) + P 6(30.9g)+ O18(16g)MM= 6Na + 6P+ 18O
MM= 137.88 g + 185.4 g + 288 g
MM (NaPO3)6 =611.28 g
6 NA, 6 P, 18 O FIND Na=22.98 g, P=30.97 g, O = 16 g
MM= Na 6( 22.98g) + P 6(30.9g)+ O18(16g)MM= 6Na + 6P+ 18O
MM= 137.88 g + 185.4 g + 288 g
MM (NaPO3)6 =611.28 g
1. 70 g
2. 611.28 g
3. 700.38 g
4. 50.20 g
1. 70 g
2. 611.28 g
3. 700.38 g
4. 50.20 g
M1V1=M2V2M1V1=M2V2
M1V1 = M2V2
1 is starting (concentrated conditions)
2 is ending (dilute conditions)
M1V1 = M2V2
1 is starting (concentrated conditions)
2 is ending (dilute conditions)
•Given•Formula•Solve:
•Given•Formula•Solve:
If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L?
If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L?
0% 0%
100%
0%
M1 = 3 mol/L or 3 M, V1 = 1 L, V2 = 6 L
M1V1 = M2V2, M1V1/V2 = M2
M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M
M1 = 3 mol/L or 3 M, V1 = 1 L, V2 = 6 L
M1V1 = M2V2, M1V1/V2 = M2
M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M
V1 = 1 LM1 = 3 M
M1V1 = 3 mol
V2 = 6 LM2 = 0.5 M
M2V2 = 3 mol
1. 19 M
2. 2 M
3. 0.5 M
4. 20 M
1. 19 M
2. 2 M
3. 0.5 M
4. 20 M
•Given•Formula•Solve:
•Given•Formula•Solve:
What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl?
What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl?
6 L
24
L 1
2 L
1 L
25%
0%0%
75%
M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 LM1V1 = M2V2, M1V1/M2 = V2
V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L
M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 LM1V1 = M2V2, M1V1/M2 = V2
V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L
1. 6 L
2. 24 L
3. 12 L
4. 1 L
1. 6 L
2. 24 L
3. 12 L
4. 1 L
•Given•Formula•Solve:
•Given•Formula•Solve:
How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?
How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?
437
.5 m
L
31.
25 m
L
200
0 m
L
765
mL
0% 0%10%
90%
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mLV1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
V1 = 0.03125 L = 31.25 mL
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mLV1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
V1 = 0.03125 L = 31.25 mL
1. 437.5 mL
2. 31.25 mL
3. 2000 mL
4. 765 mL
1. 437.5 mL
2. 31.25 mL
3. 2000 mL
4. 765 mL
Chlorine ConcentrationsChlorine Concentrations
1.Sodium hypochlorite 5 to 15% available chlorine
2. Calcium hypochlorite 65-70% available chlorine
3. Chlorine gas100% available chlorine
1.Sodium hypochlorite 5 to 15% available chlorine
2. Calcium hypochlorite 65-70% available chlorine
3. Chlorine gas100% available chlorine
Determining Cl Concentrations from Hypochlorite dosageDetermining Cl Concentrations from Hypochlorite dosage
1. Disinfection requires 280 lb/day chlorine. If CaOCl (65% available Cl) is used how many lbs day are required1. Disinfection requires 280 lb/day chlorine. If CaOCl (65% available Cl) is used how many lbs day are required
(%concentration)(X lb/day)= total lbs/day required
Rearrange:(X lb /day) = (280 lbs) = 430.77 lb/d CaOCL
(.65)
(%concentration)(X lb/day)= total lbs/day required
Rearrange:(X lb /day) = (280 lbs) = 430.77 lb/d CaOCL
(.65)
Hypochlorite Solution Feed RateHypochlorite Solution Feed Rate
1. Actual Dose=Solution Feeder dose. (mg/L chlorine)(MGD)(8.34)= (mg/L chlorine)(MGD)(8.34)
1. Actual Dose=Solution Feeder dose. (mg/L chlorine)(MGD)(8.34)= (mg/L chlorine)(MGD)(8.34)
% Dry Strength of Solution% Dry Strength of Solution1. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100
Solution(lbs)
1. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100
Solution(lbs)
2. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100
Water, Lbs + Chlorine(lbs)
2. Dry chlorine % Chlorine Strength= Chlorine (lbs) X 100
Water, Lbs + Chlorine(lbs)
3. Dry chlorine % Chlorine Strength= Hypo(lbs)(% available Cl) X
100 100 Water, Lbs+ hypo(lbs) (% available Cl)
100
3. Dry chlorine % Chlorine Strength= Hypo(lbs)(% available Cl) X
100 100 Water, Lbs+ hypo(lbs) (% available Cl)
100
% Liquid Strength of Solution% Liquid Strength of Solution1. Liquid chlorineLbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution1. Liquid chlorineLbs Cl in liq. hypo= Lbs of chlorine in hypochlorite solution
2. Liquid chlorine (liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp)
100 100
2. Liquid chlorine (liq. Hypo lbs)(%Strength liq. Hyo) = (hypo Sol lbs) (% Strength liq. hyp)
100 100
3. Liquid chlorine(liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp)
100 100
3. Liquid chlorine(liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp)
100 100
•Given•Formula•Solve:
•Given•Formula•Solve:
A chlorinator setting of 20 lbs of chlorine per 24 hrs results in a residual of 0.4 mg/L. The chlorinator setting is 25 lb per 24 hrs. The chlorine residual
increased to 0.5 mg/L at this new dosage rate. The average flow being treated is 1.6 mgd. On the basis of
this data is the water being chlorinated beyond breakpoint?
A chlorinator setting of 20 lbs of chlorine per 24 hrs results in a residual of 0.4 mg/L. The chlorinator setting is 25 lb per 24 hrs. The chlorine residual
increased to 0.5 mg/L at this new dosage rate. The average flow being treated is 1.6 mgd. On the basis of
this data is the water being chlorinated beyond breakpoint?
yes n
o
33%
67%
Residual 1= 0.4 mg/L, residual 2=0.5 mg/L, 1.6 mgd, new 5 lb/dayLbs/d incr= Dose( flow)(8.34 lb/g)Act increase in residual=New residual-Old residualDose= Lb/day increase/flow 8.34 lb/g 5 lb/day/1.6 mgd(8.34 lb/d)=Residual = 0.37 mg/L
Actual increase in residual was 0.5mg/L-0.4mg/L=0.1 mg/L
Expected was 0.37 but the actual was 0.1. Not being met!
Residual 1= 0.4 mg/L, residual 2=0.5 mg/L, 1.6 mgd, new 5 lb/dayLbs/d incr= Dose( flow)(8.34 lb/g)Act increase in residual=New residual-Old residualDose= Lb/day increase/flow 8.34 lb/g 5 lb/day/1.6 mgd(8.34 lb/d)=Residual = 0.37 mg/L
Actual increase in residual was 0.5mg/L-0.4mg/L=0.1 mg/L
Expected was 0.37 but the actual was 0.1. Not being met!1. yes
2. no
1. yes
2. no
Specific Gravity, LBS, Gallons, Solution Strength
Specific Gravity, LBS, Gallons, Solution Strength
))(/34.8( chlorine)strength )( . chlorine (%
)(
))( )( )(/34.8(
)( . chlorine %
) )(/34.8(chlorine)strenght (full) . chlorine (%
)(
))(g )(/34.8)( )( . chlorine (%
gallonsgallbfullstrengthsol
lbsGravitySpecific
gallonsGravitySpecificchlorinestrengthfullgallb
lbsStrengthSol
GravitySpecificgallbstrengthsol
lbsGallons
galravityspecificgallbschlorinestrengthfullstrengthsolLbs
Team ScoresTeam Scores
Today’s objective: Review basic chemistry and solution making as it pertains to the waterworks industry
Calculate the chemical dosage using the standard “pounds formula” has been met?
Today’s objective: Review basic chemistry and solution making as it pertains to the waterworks industry
Calculate the chemical dosage using the standard “pounds formula” has been met?
Stro
ngly A
gree
Agre
e
Dis
agre
e
Stro
ngly D
isag
ree
15%
0%0%
85%
1. Strongly Agree
2. Agree
3. Disagree
4. Strongly Disagree
1. Strongly Agree
2. Agree
3. Disagree
4. Strongly Disagree