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Water Treatment Math Problems and Charts
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Math Problems & Charts
What is the Langlier Index if the pH_{A}
is 9.2 and the pH_{S} is 8.6? Is this stable, depositing or corrosive?
Langlier Index = pH_{A}
– pH_{S }= 9.2 – 8.6 =
0.6
Depositing
If the P alkalinity
is 0 mg/L and the T alkalinity is 100 mg/L, calculate the
following:

Bicarbonate alkalinity

Carbonate alkalinity

Hydroxide alkalinity
Titration
Results 
Bicarbonate 
Carbonate 
Hydroxide 
P = 0 
T 
0 
0 
P = T 
0 
0 
T 
2P < T 
T – 2P 
2P 
0 
2P = T 
0 
T 
0 
2P > T 
0 
2T – 2P 
2P – T 
P alkalinity is 0
mg/L and the T alkalinity is 100 mg/L
Bicarbonate alkalinity =
T =
100 mg/L
Carbonate alkalinity =
0 mg/L
Hydroxide alkalinity =
0 mg/L
If the P alkalinity
is 35 mg/L and the T alkalinity is 60 mg/L, calculate the following:

Bicarbonate alkalinity

Carbonate alkalinity

Hydroxide alkalinity
P alkalinity is 35 mg/L and T
alkalinity is 60 mg/L →
2P > T
Titration
Results 
Bicarbonate 
Carbonate 
Hydroxide 
P = 0 
T 
0 
0 
P = T 
0 
0 
T 
2P < T 
T – 2P 
2P 
0 
2P = T 
0 
T 
0 
2P > T 
0 
2T – 2P 
2P – T 
Bicarbonate alkalinity =
0 mg/L
Carbonate alkalinity = 2T 
2P = (2)(60 mg/L)  (2)(35 mg/L) = 120 mg/L  70 mg/L =
50 mg/L
Hydroxide alkalinity = 2P  T
= (2)(35 mg/L)  60 mg/L = 70 mg/L  60 mg/L =
10 mg/L
120 mg/L of
Ca as Ca is equivalent to ______ mg/L of Ca as CaCO_{3}
Ca as CaCO_{3}
= (Ca as Ca)(2.5) = 120 mg/L x 2.5 =
300 mg/L Ca as
CaCO_{3}
If you
are removing 300 mg/L of hardness and your flow is 2 mgd, how many pounds of
hardness are you removing?
lbs/day =
(300 mg/L)(8.34 lbs/gal)(2
mgd) = 5004
lbs/day
You are adding 6 lbs/day of chlorine gas to a flow of 277 gpm and your total
chlorine residual is 1.5 mg/L. What is your chlorine demand in mg/L?
Flow, MGD = 277 gpm ÷ 694
gpm/mgd = 0.4 MGD
Chlorine Dose, mg/L = 6
lbs/day ÷ 8.34 lbs/gal ÷ 0.4 MGD = 1.8 mg/L
Chlorine demand, mg/L = 1.8
mg/L = 1.5 mg/L = 0.3 mg/L
Click on the
thumbnail below for the answers for the chart of pump station math
for cylindrical tanks.
Click on the
thumbnail below for the answers for the chart of pump station math
for rectangular tanks.
Click on the
thumbnail below for the answers for the chart of pipe math.
Match 
1 
The units for pressure are 
A 
Divided by 
2 
Head is expressed in 
B 
2.31 
3 
Velocity is measured in 
C 
0.434 psi 
4 
Units of flow 
D 
psi 
5 
Pressure times _______ = Head 
E 
fps 
6 
Head __________ 2.31 = Pressure 
F 
1440 
7 
20 psi = 
G 
694 gpm 
8 
1 MGD = 
H 
cfs 
9 
A head loss of 7 feet = 
I 
3 psi 


J 
Times 


K 
ft 


L 
46.2 ft 


M 
Subtracted by 
1 = D
2 = K
3 = E
4= H
5 = B
6 = A
7 = L
8 = G
9 = I
The formula for the volume of
a rectangular tank is
A. (Length)(Width)
B.
(0.785)(Diameter^{2})
C. (Length)(Width)(Height)
D.
(0.785)(Diameter^{2})(Height)
The correct answer is
C. (Length)(Width)(Height)
A is the formula for the area of
a rectangle
B is the formula for the area of
a circle
D
is the formula for the volume of a cylinder
The formula for the volume of
a cylindrical tank is
A. (Length)(Width)
B.
(0.785)(Diameter^{2})
C. (Length)(Width)(Height)
D.
(0.785)(Diameter^{2})(Height)
The correct answer is
D.
(0.785)(Diameter^{2})(Height)
A is the formula for the area of
a rectangle
B is the formula for the area of
a circle
C is the formula for the volume
of a rectangle
Given a tank measures 50
feet long, 25 feet wide, and 15 ft deep, calculate:
 Surface area in ft^{2}
 Crosssectional area in
square feet
 Volume in cf
 Capacity in gallons
The correct answers are:
Surface area in ft^{2}
= Length, ft x Width, ft = 50 ft x 25 ft =
1250 ft^{2}

Crosssectional area in square feet
Crosssectional area is ft^{2} = Width, ft x Depth, ft = 25 ft x
15 ft =
375 square feet
Volume in cf = Length, ft x Width, ft x
Depth, ft = 50 ft x 25 ft x 15 ft =
18,750 cf
Capacity in gallons = Vol in cf x 7.5
gal/cf = 18,750 cf x 7.5 gal/cf =
140,625 gal
For questions or comments,
call 5153131159
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