# FE Review for Environmental

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

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

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