Basic Math Midterm Review
Math for Water Technology
MTH 082-2
RULES FOR MIDTERM
1. DRAW AND LABEL DIAGRAM
2. LIST FORMULAS
3 SOLVE EACH FORMULA INDIVIDUALLY
4. ISOLATE THE PARAMETERS NECESSARY
5. USE YOUR UNITS TO GUIDE YOU
6. SOLVE THE PROBLEM
Chlorine Concentrations
Hypochlorite or Available Purity = DIVIDE
1.Sodium hypochlorite
5 to 15% available chlorine
2. Calcium hypochlorite (HTH)
65-70% available chlorine
3. Chlorine gas
100% available chlorine or
100% active chlorine
% Dry Strength of Solution
1. Dry chlorine
% Chlorine Strength= Chlorine (lbs) X 100
Solution(lbs)
2. Dry chlorine
% Chlorine Strength= Chlorine (lbs)
X 100
Water, Lbs + Chlorine(lbs)
3. Dry chlorine
% Chlorine Strength= Hypo(lbs)(% available Cl)
100
X 100
Water, Lbs+ hypo(lbs) (% available Cl)
100
Dry Hypochlorite Feed Rate
1.Hypochlorite (lb/day)=lbs/day chlorine
% available chlorine
100
2. Hypochlorite (lb/day)= (mg/L chlorine)(MGD)(8.34)
%strength hypochl
100
3. Hypochlorite (lbs)= (mg/L chlorine)(MG tank Vol)(8.34)
%strength hypochl
100
% Liquid Strength of Solution
1. Liquid chlorine
Lbs 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
3. Liquid chlorine
(liq. Hypo gal)(8.34)(%Strength liq. Hyo)=(hypo Sol gal)(8.34)(%Strength liq. Hyp)
100
100
Lbs/day
When do you multiply?
Available to Active (multiply)
Twenty 20 pounds of 50% available chlorine equals
10 lbs pounds of 100% active chlorine?
20 lbs
20 lbs (0.50)= 10 lbs
10 lbs
Active is pure
Available is unpure
Lbs/day
When do you divide?
Active to Available
Forty 40 pounds of 100% active chlorine (chlorine
gas) has the same disinfection power as 61 lbs
pounds of 65% available HTH chlorine?
61 lbs
40 lbs
40 lbs = 61 lbs
(.65)
Active is pure
Available is unpure
Lbs/day
When do you multiply?
Available to Active (multiply)
HW #4 Problem 3
Twenty two 22 pounds of 65% available chlorine
equals 14.3 lbs pounds of 100% active chlorine!
22 lbs
22 lbs (0.65)= 14.3 lbs
14.3 lbs
Active is pure
Available is unpure
Lbs/day
When do you multiply?
Available to Available (multiply)
HW #3 Problem 3
How many pounds of available chlorine are in 2
pounds of 65% powdered dry (available) chlorine?
2 lbs (0.65)= 1.3 lbs
2 lbs of 65% ava
1.3 lbs is available
Available is unpure
Available is unpure
Lbs/day
When do you divide?
When do you multiply?
PURITY IS THE KEY!!!!!
When using hypochlorite, more lbs/day must be used
because it is not pure! Hypochlorite is “available” chlorine
If you have a certain amount of <100% chlorine (available)
and you want to know how much pure (active chlorine) that
is you multiply!!!! (Need less of a more pure substance)
If you have a certain # of lbs of 100% pure chlorine (active)
and what you need is a certain percentage (available) then
you divide (its not pure so you will need more of it!)
Chlorine in lbs DOSE in a tank!
1. Wells are disinfected
2. Tanks and reservoirs get chlorinated
3. Pipelines get chlorinated after repair or installation
Example 5: A storage tank needs to be disinfected with a
50 mg/L chlorine solution. If the tank holds 70,000
gallons, how many lbs of chlorine gas will be needed?
Step 1 Chlorine mg/L to lbs/day formula
( mg / L )( MG )( 8 . 34 lbs / gal )  lbs
Conc
Volume
Conversion
( 50 mg / L )( 0 . 07 MG )( 8 . 34 lbs / gal )  lbs
x  29 . 2 lbs
Hypochlorite DOSAGE
1. Calculate lbs/day chlorine
2. Calculate lbs/day hypochlorite needed by dividing the
lbs/day chlorine by the percent available chlorine
Example 6: A wastewater flow of 85,000 gpd requires a
chlorine dose of 25 mg/L. If sodium hypochlorite (15%
available chlorine) is used, how many lbs/day of sodium
hypochlorite are required? How many gal/day is this?
( mg / L )( MGD )( 8 . 34 lbs / gal )  lbs / day
Conc
Flow
Conversion
( 25 mg / L )( 0 . 85 MGD )( 8 . 34 lbs / gal )  177 lbs / day chlorine
Step 2 Calculate lbs/day sodium hypochlorite
(177 lbs / day Cl 2
15
100
Avail . Cl 2
(1180 lbs/d)/(8.34 lbs/gal)=
 1180 lbs / day hypochlori te
141 gal/day
sodium hypochlorite
Determining Cl Concentrations from
Hypochlorite dosage
1. 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) = Total lbs/d
= (280 lbs) = 430.77 lb/d CaOCL
% concentration
(.65)
Solutions with Specific Gravity
1. How many pounds of available chlorine are in 5
gallons of 12% hypochlorite solution if its specific
gravity is 1.19?
Lb=(% concentration) X (Specific gravity) X (lb/gal) X (gal)
Lb= (.12) X (1.19) X (8.34 lb/gal) X (5 gal)
Lb=5.95 lb
Calcium hypochlorite, gal= (lb CaOCl2 @ 100%)/(8.34 lb/gal)(CaOCl2 %)
Calcium hypochlorite, lbs= (gal CaOCl2)(8.34 lb/gal)(62.5%)/(100%)
Hypochlorite Solutions
is disinfected with 62.5% calcium hypochlorite solution,
If 3.03 gal of CaOCl2 were used, what must have
been the dosage of the calcium hypochlorite?
Assume the calcium hypochlorite weighs 8.34
lb/gal. Get answer in (lbs) to solve for dosage in
lbs formula!
Calcium hypochlorite, lbs= (gal CaOCl2)(8.34 lb/gal)( % Available)
(100%)
Calcium hypochlorite, lbs= (3.03 gal CaOCl2)(8.34 lb/gal)(.625)
Calcium hypochlorite, lbs= 15.79 lbs
Solve lbs formula
Number of lb=(MG)(Dosage)(8.34 lb/gal)
Dosage mg/L CaOCl2=15.79 lbs CaOCl2/(0.038MG)(8.34 lb/gal)
Calcium Hypochlorite Solutions
Midterm 3. storage tank requires disinfection with a
65% calcium hypochlorite solution. How many
gallons of hypochlorite solution are required if the
desired dose is 50.0 mg/L. Assume the calcium
hypochlorite solution weighs 10.21 lb/gal.?
Calcium hypochlorite, gal= (lb CaOCl2 @ 100%)
(10.21 lb/gal)(CaOCl2 % available)
Calcium hypochlorite, gal= (51.7 lb CaOCl2)
(10.21 lb/gal)(.65)
= 7.79 gal
Specific Gravity, LBS, Gallons,
Solution Strength
Lbs  (% chlorine
sol . strength )( full strength chlorine )( 8 . 34 lbs / gal )( specific g ravity )( gal )
( lbs )
Gallons 
(% chlorine
% chlorine
sol . strength ) (full strenght
chlorine)
( 8 . 34 lb / gal )( Specific Gravity )
( lbs )
Sol . Strength 
( 8 . 34 lb / gal )( full strength chlorine )( Specific Gravity )( gallons )
( lbs )
Specific Gravity 
(% chlorine
sol . strength )( full strength
chlorine)
( 8 . 34 lb / gal )( gallons )
Problem #1
1. Determine the
lbs / day  ( mg / L )( MGD )( 8 . 34 lbs / gal )
chlorinator setting lbs/day  (1.7 mg/L)(3.2 MGD)(8.34 lbs/gal)
(lbs/day) needed to lbs/day  45 lbs/day Cl
treat a flow of 3.2
MGD with a chlorine
dose of 1.7 mg/L (5
pts).
Problem #2
2. The chorine demand
was calculated to be
11 lb/day, and the
residual was
measured at 0.5
mg/L. What was the
chlorine dose (lbs)
that day, if the flow
was 1.2 MGD?
lbs / day  ( mg / L )( MGD )( 8 . 34 lbs / gal )
lbs/day  (0.5 mg/L)(1.2
MGD)(8.34
lbs/gal)
lbs/day  5 lbs/day Cl
1 . Cl 2 Dose ( lbs / day )  Cl 2 Demand ( lbs / day )  Cl 2 Re sidual ( lbs / day )
Cl 2 Dose ( lbs / day )  5 lbs / day  11 lbs / day
Cl 2 Dose ( lbs / day )  16 lbs / day
Problem #3
3. How many gallons of
15% liquid chorine
(s.g. = 1.2) would be
required to dose a
circular tank to a 50
mg/L concentration?
The tank diameter is
25 ft. and the water
depth in the tank is
5 ft.
•Volume= 0.785(diameter2)(depth)
•Volume= 0.785(25ft)2(5 ft)
•Volume=2453ft3
•Conversion= 2453 ft3 (7.48 gal/1ft3)
•Conversion=18349 gallons
•Conversion=.018349 MG
( chlorine weight lbs)  dosage(mg/
L) (volume
of container
(MG)) ( 8 . 34 lbs / gal )
( chlorine weight lbs)  ( 50 mg/L) (.018349 MG) ( 8 . 34 lbs / gal )
( chlorine weight lbs)  7 . 6 lbs
( lbs )
Gallons 
(% chlorine
Gallons 
sol . strength )( 8 . 34 lb / gal )( Specific Gravity )
7 . 6 lbs chlorine
 5 . 06 gallons
(. 15 )( 8 . 34 lb / gal )( 1 . 2 )
? Gallons
1.2 SG
15% OCl
25 ft
5 ft
Problem #4
Lbs  (% chlorine
sol . strength )( 8 . 34 lbs / gal )( specific g ravity )( gal )
( chlorine weight lbs)  (. 12 ) (8 . 34 lbs / gal )( 1 . 19 ) (30gal)
( chlorine weight lbs)  35 . 72 lbs of pure Cl 2 in the 12% portion of the hypochlori
te solution
4. A barrel contains 30
(%concentration)(X lb)= total lbs required
X lb= 35.72 lbs/.65
gallons of 12%
X=54.96 lbs of 65% HTH
liquid hypochlorite
The lbs/day formula calculates the pure
concentration of chlorine within a % solution of
at a specific gravity
chlorine. Thus in the 12% hypochlorite solution
of 1.19. How many there are 35 lbs of pure Cl (A). Imagine the same
size barrel of a higher concentration of OCl- as
pounds of dry HTH 65% HTH (B). Which solution has more lbs of
pure Cl ? Well, the 65% it’s a more concentrated
(65% active
solution so there will be more pure chlorine in it.
hypochlorite) would
1.19 SG
65% OCl as HTH
be needed to
12% OCl= 35 lbs pure Cl2
55 lbs pure Cl2
replace the chlorine
in the barrel?
A
B
2