FE Review for
Environmental
Engineering
Problems, problems, problems
Presented by L.R. Chevalier, Ph.D., P.E.
Department of Civil and Environmental Engineering
Southern Illinois University Carbondale
WATER TREATMENT
FE Review for Environmental Engineering
D
H
v
feeder pipe diameter
Problem
Strategy
Solution
Overflow Basin
For the dimensions given below, determine the average vertical velocity
(HLR) and hydraulic retention time.
D = 15 m
H = 3 m
v
= 0.28 m/s
Feeder pipe diameter 40 cm
Problem
Strategy
Solution
•
Determine Q from the feeder pipe
•
Determine hydraulic loading rate or vertical velocity
•
Calculate HRT
Problem
Strategy
Solution
A
r
cm
m
cm
m
Q
vA
m
s
m
m
s
m
day
p
2
2
2
2
2
3
3
20
1
100
0
126
0
28
0
126
0
0352
3040
.
.
.
.
IS THIS THE SAME Q OVERFLOWING THE BASIN?
1. Determine Q from the feeder pipe
Problem
Strategy
Solution
day
m
m
day
m
A
Q
v
m
m
m
r
A
b
2
.
17
177
3040
177
71
.
176
5
.
7
2
3
2
2
2
2
2. Determine the vertical velocity (HLR) in the basin
Problem
Strategy
Solution
V
A
h
m
m
m
V
Q
m
m
day
days
hrs
b
177
3
531
531
3040
0
175
4
19
2
3
3
3
.
.
.
3. Calculate HRT,
Problem
Strategy
Solution
Calculate the carbonate hardness (CH) and the
noncarbonate
hardness
(NCH) for two water samples listed below. Reported concentrations are as
mg/L CaCO
3
Example
Solution
Sample 1
Sample 2
Alkalinity
327
498
Total Hardness
498
327
CH
NCH
Example
Solution
Sample 1
Sample 2
Alkalinity
327
498
Total Hardness
498
327
CH
327
327
NCH
171
0
Compute the TH, CH, NCH of the following water at a
pH of 7.2.
CONSTITUENT
mg/L
as CaCO
3
Magnesium
107.7
Potassium
3.2
Phosphate
12.2
Calcium
296.3
Fluoride
0.8
Bicarbonate
136.5
Carbon dioxide
19.8
Problem
Strategy
Solution
•
Estimate alkalinity from bicarbonate and carbonate
•
Sum the multivalent metallic
cations
to get TH
•
Determine which is larger
•
Determine CH
•
Determine NCH
Problem
Strategy
Solution
Alkalinity = 136.5 mg/L as CaCO
3
TH = 404.0 mg/L as CaCO
3
CH = 136.5 mg/L as CaCO
3
NCH = 404.0

136.5 = 267.5 as CaCO
3
Problem
Strategy
Solution
Problem
Strategy
Solution
Consider a groundwater source that contains 2x10

4
moles of H
2
CO
3
(carbonic
acid). The rate of pumping from the aquifer is 1,000 m
3
/day. Determine the
amount of hydrated lime (Ca(OH)
2
) needed each day for neutralizing the
carbonic acid and the amount of calcium carbonate (CaCO
3
) produced as a
result. Report your answer in kg/day.
•
Determine the molecular weight of Ca(OH)
2
and CaCO
3
•
Review the governing chemical reaction to determine the
ratio between the moles of H
2
CO
3
, Ca(OH)
2
and CaCO
3
•
Use this ratio, the pumping rate and the MW to
determine the mass per day (in kg/day)
Problem
Strategy
Solution
O
H
s
CaCO
OH
Ca
CO
H
2
3
2
3
2
2
d
kg
D
g
kg
mole
g
C
m
L
d
m
B
L
moles
A
1000
1000
3
3
Problem
Strategy
Solution
•
Determine the molecular weight of Ca(OH)
2
and CaCO
3
Ca(OH)
2
= 74.1 g/mol
CaCO
3
= 100 g/mol
Problem
Strategy
Solution
•
Review the governing chemical reaction to determine the
ratio between the moles of H
2
CO
3
, Ca(OH)
2
and CaCO
3
O
H
s
CaCO
OH
Ca
CO
H
2
3
2
3
2
2
1 mole H
2
CO
3
: 1mole Ca(OH)
2
: 1 mole CaCO
3
2x10

4
mole H
2
CO
3
: 2x10

4
mole Ca(OH)
2
: 2x10

4
mole CaCO
3
Problem
Strategy
Solution
•
Use this ratio, the pumping rate and the MW to determine
the mass per day (in kg/day)
d
kg
D
g
kg
mole
g
C
m
L
d
m
B
L
moles
A
1000
1000
3
3
2
3
3
4
8
.
14
1000
1
.
74
1000
1000
10
2
OH
Ca
d
kg
g
kg
mole
g
m
L
d
m
L
moles
3
3
3
4
10
1000
50
1000
1000
10
2
CaCO
d
kg
g
kg
mole
g
m
L
d
m
L
moles
WASTEWATER TREATMENT
FE Review for Environmental Engineering
One wastewater treatment process, activated sludge, which will be discussed
later, requires
either a
detention time of 4 hrs, or
the ability to treat
approximately
20
gal/capita

day.
If a city has a population of 10,000, and an average flow of 1.2 MGD, what
approximate tank volume is required?
Example
Solution
The tank size can be estimated based on flow and typical detention times or on
population and the size per capita. SAME ANSWER EITHER WAY!
a) The typical detention time,
is 4 hours. Thus the tank volume is:
gal
hr
day
hr
Q
V
day
gal
000
,
200
24
1
4
10
2
.
1
6
Example
Solution
The tank size can be estimated based on flow and typical detention times or on
population and the size per capita.
b) Based on population requirements, the volume is:
gal
capita
V
capita
gal
000
,
200
000
,
10
20
Example
Solution
Estimate the area needed for bar racks given a city
population of 150,000. Clearly state all
assumptions.
Problem
Strategy
Solution
(Answers May Vary Depending On Assumptions)
•
Assume a peaking factor of 2.8 ( range 2

5)
•
Assume 150 gal/capita

day
•
Q
peak
= (2.8)(150 gal/capita

day)(150,000)
•
Q
peak
= 63.0 MGD = 238,140 m
3
/day
•
Limit approach velocity to 0.8 m/s (acceptable range 0.6

1.0
m/s)
•
A = Q/v
Problem
Strategy
Solution
•
0.8 m/s = 69120 m/day
•
A = (238140 m
3
/day) / (69120 m/day)
•
A = 3.45 m
2
=
3.5 m
2
•
Want to order 2 in case one is off line for maintenance or
repair
Problem
Strategy
Solution
Estimate the settling velocity of sand (density = 2650 kg/m
3
) with a mean diameter
of 0.21 mm. Assume the sand is approximately spherical. Using a safety factor of 1.4
to account for inlet and outlet losses, estimate the area required for a grit chamber
to remove sand if the flow rate is 0.25 m
3
/s.
The density and viscosity of water at
20
°
C
is 998 kg/m
3
and 1.01 x 10

3
N
∙
s/m
2
,
respectively.
Problem
Strategy
Solution
•
Review the governing equations
•
Note the given parameters
•
d = 0.21 mm
•
g =9.8 m
2
/s
•
r
p
= 2650 kg/m
3
•
At 20
°
C,
n
w
=
1.01 x 10

3
N
∙
s/m
2
r
w
= 998 kg/m
3
•
Q = 0.25 m
3
/s
•
SF = 1.4
Problem
Strategy
Solution
SF
OFR
Q
A
g
d
v
w
p
p
r
r
18
2
The Stokes’ settling velocity can thus be calculated:
s
m
s
m
kg
s
m
m
kg
m
kg
w
p
p
m
g
d
v
039
.
0
10
01
.
1
18
8
.
9
10
1
.
2
998
2650
18
3
2
4
2
2
3
3
r
r
Problem
Strategy
Solution
Knowing the overflow rate, we can calculate the area required for the grit
chamber. Note, the safety factor 1.4
2
97
.
8
4
.
1
039
.
0
25
.
0
3
m
SF
OFR
Q
A
s
m
s
m
So the area of the grit chamber must be 9 m
2
to remove 0.21mm grit.
Problem
Strategy
Solution
Sizing
a Primary Clarifier for WWT
Use the typical design values to estimate the size for two
circular clarifiers
used
to treat wastewater at a design flow of 20 MGD. Each clarifier is to treat half the
flow
.
Report the diameter and depth.
Problem
Strategy
Solution
diameter
depth
•
Determine the Design Data needed for your solution
•
surface over flow rate of 1,000 gal/ft
2
/day
•
average detention time,
, of 2.0 hr
•
Each clarifier should receive half the flow, Q/2 = 10 MGD
•
Estimate the area (A=Q/v)
•
Estimate the diameter (d) assuming a circular clarifier
•
Clarifiers diameters are generally available in multiples of 5 ft in
the US, or in multiples of 2 m outside the US
•
Estimate the volume using the detention time (V=
Q
•
Estimate the depth of the tank (V/A = d)
Problem
Strategy
Solution
Each clarifier should receive half the flow, or 10 MGD. Using the typical design value for
the surface over flow rate of 1,000 gal/ft
2
/day, we can compute the area of each clarifier
2
6
000
,
10
1000
10
10
2
ft
v
Q
A
vA
Q
day
ft
gal
day
gal
Problem
Strategy
Solution
From this area, we can now calculate the diameter of the clarifier. Clarifiers
are generally available in multiples of 5 ft in the US, or in multiples of 2 m
outside the US.
2
2
2
2
387
,
10
4
115
115
8
.
112
000
,
10
4
4
4
ft
ft
A
ft
ft
ft
A
d
d
A
Problem
Strategy
Solution
3
3
6
408
,
111
48
.
7
1
24
1
2
10
10
ft
gal
ft
hr
day
hr
Q
V
day
gal
To determine volume, we will need a detention time. Again, we will use a design
value. In this case, consider the typical design value for the average detention
time,
, of 2.0 hr
Problem
Strategy
Solution
Clarifiers should be as shallow as possible, but not less that 7 ft deep. The approximate
depth,
h
, can now be calculated.
ft
ft
ft
ft
A
V
h
Ah
V
11
73
.
10
387
,
10
408
,
111
2
3
Problem
Strategy
Solution
A township has been directed to upgrade its primary WWTP to a secondary
plant that can meet an effluent standard of 30 mg/L BOD
5
and 30 mg/L SS.
Assuming that the BOD
5
of the SS may be estimated as equal to 63% of the SS
concentration, estimate the required volume of the aeration tank. The
following data are available from the existing primary plant.
Problem
Strategy
Solution
Existing Plant Effluent Characteristics
Assume that the secondary clarifier can produce an
effluent with only 30 mg/L SS
Assume MLVSS = 2000 mg/L
Want to meet an effluent standard
SS = 30 mg/L BOD
5
= 30 mg/L
BOD
5
= 84 mg/L
Flow = 0.150 m
3
/s
Problem
Strategy
Solution
Parameter
Value
K
s
100.00 mg BOD
5
/L
m
2.5 d

1
k
d
0.05 d

1
Y
0.5 mg VSS/mg BOD
5
removed
Use the following “typical values”
Problem
Strategy
Solution
Q
r
,
X
r
,
S
Q,S
o
(Q + Q
r
)
X, S
(Q

Q
w
), S, X
e
Q
w
,
X
w
,
S
1
1
d
m
c
c
d
s
k
k
K
S
X
Y
S
S
k
c
o
d
c
1
Q
V
A portion of the SS is BOD
5
. Therefore, an estimate of the allowable soluble
BOD
5
in the effluent can be made using the 63% assumption.
S = 30

(0.63)(30) = 11.1 mg/L
Note: S is soluble BOD
5
Problem
Strategy
Solution
1
1
d
m
c
c
d
s
k
k
K
S
1
05
.
0
5
.
2
05
.
0
1
100
1
.
11
1
1
1
5
day
day
day
L
BOD
mg
L
mg
c
c
The mean cell residence time can be estimated using
c
= 5.0 days
Problem
Strategy
Solution
X
Y
S
S
k
c
o
d
c
1
day
day
L
mg
L
mg
days
L
mg
0
.
5
05
.
0
1
1
.
11
0
.
84
5
.
0
0
.
5
2000
1
To solve for the hydraulic residence time:
= 0.073 d or 1.8 h
Problem
Strategy
Solution
V
Q
m
s
s
h
h
m
m
0
150
3600
1
8
972
970
3
3
3
.
.
....end of example
The volume of the aeration tank is then estimated using:
Problem
Strategy
Solution
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