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
FE Review for Environmental Engineering
WATER TREATMENT
Problem
Strategy
Solution
Overflow Basin
D
H
v
feeder pipe diameter
Problem
Strategy
Solution
For the dimensions given below, determine the average vertical velocity
(HLR) and hydraulic retention time.
D = 15 m
H=3m
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
1. Determine Q from the feeder pipe
2
2  1m 
2
Ap   r    20 cm 
.
m2
  0126
 100cm
3
3
m
m
m

Q  vA   0.28   0126
. m2   0.0352
 3040

s
s
day
IS THIS THE SAME Q OVERFLOWING THE BASIN?
2. Determine the vertical velocity (HLR) in the basin
Ab   r 2   7.5m 
2
 176.71 m 2  177m 2
m3
3040
Q
m
day
v 
 17.2
2
A 177m
day
3. Calculate HRT, 
V  Ab h  177 m2  3 m  531 m3
V
 
Q
531 m3
 
 0175
.
days  4.19 hrs.
3
m
3040
day
Example
Solution
Calculate the carbonate hardness (CH) and the noncarbonate hardness
(NCH) for two water samples listed below. Reported concentrations are as
mg/L CaCO3
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
Problem
Strategy
Solution
Compute the TH, CH, NCH of the following water at a
pH of 7.2.
CONSTITUENT
mg/L as CaCO3
Magnesium
Potassium
Phosphate
Calcium
Fluoride
Bicarbonate
Carbon dioxide
107.7
3.2
12.2
296.3
0.8
136.5
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 CaCO3
TH = 404.0 mg/L as CaCO3
CH = 136.5 mg/L as CaCO3
NCH = 404.0 - 136.5 = 267.5 as CaCO3
Problem
Strategy
Solution
Consider a groundwater source that contains 2x10-4 moles of H2CO3 (carbonic
acid). The rate of pumping from the aquifer is 1,000 m3/day. Determine the
amount of hydrated lime (Ca(OH)2) needed each day for neutralizing the
carbonic acid and the amount of calcium carbonate (CaCO3) produced as a
result. Report your answer in kg/day.
• Determine the molecular weight of Ca(OH)2 and CaCO3
• Review the governing chemical reaction to determine the
ratio between the moles of H2CO3, Ca(OH)2 and CaCO3
• Use this ratio, the pumping rate and the MW to
determine the mass per day (in kg/day)
H 2CO3  CaOH 2  CaCO3 s   2H 2O
3
g  kg 
kg
 moles  m  1000 L 


  D
A
 B

 C

3

L  d  m  mole  1000 g 
d

Problem
Strategy
Solution
• Determine the molecular weight of Ca(OH)2 and CaCO3
Ca(OH)2 = 74.1 g/mol
CaCO3 = 100 g/mol
• Review the governing chemical reaction to determine the
ratio between the moles of H2CO3, Ca(OH)2 and CaCO3
H 2CO3  CaOH 2  CaCO3 s   2H 2O
1 mole H2CO3: 1mole Ca(OH)2: 1 mole CaCO3
2x10-4 mole H2CO3: 2x10-4 mole Ca(OH)2: 2x10-4 mole CaCO3
• Use this ratio, the pumping rate and the MW to determine
the mass per day (in kg/day)
3
g  kg 
kg
 moles  m  1000 L 


  D
A
 B

 C

3

L  d  m  mole  1000 g 
d

m3  1000 L 
g  kg 
kg

 4 moles 


  14.8
CaOH 2
 2 10
1000 
 74.1

3
L 
d  m 
mole  1000 g 
d

m3  1000L 
g  kg 
kg

 4 moles 


  10 CaCO3
 2 10
1000 
 50

3
L 
d  m  mole  1000 g 
d

FE Review for Environmental Engineering
WASTEWATER TREATMENT
Example
Solution
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?
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:
V  Q
1day 

 1.2 10 day  4hr 

24
hr


 200,000 gal
6 gal
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:
V  20
gal
capita
10,000capita 
 200,000 gal
Problem
Strategy
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
Qpeak = (2.8)(150 gal/capita-day)(150,000)
Qpeak = 63.0 MGD = 238,140 m3/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 m3/day) / (69120 m/day)
A = 3.45 m2 = 3.5 m2
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/m3) 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 m3/s.
The density and viscosity of water at 20°C is 998 kg/m3 and 1.01 x 10-3 N·s/m2,
respectively.
• Review the governing equations
• Note the given parameters
•
•
•
•
•
•
d = 0.21 mm
g =9.8 m2/s
rp = 2650 kg/m3
At 20°C, nw = 1.01 x 10-3 N·s/m2 rw= 998 kg/m3
Q = 0.25 m3/s
SF = 1.4
r  r d 2g
vp


p
w

18
Q
SF 
A
OFR
The Stokes’ settling velocity can thus be calculated:
vp 

r
p
 r w d 2 g
18
2650
kg
m
 0.039 ms
3

4

2
 998 m3 2.1  10 m 9.8 sm2
kg

18 1.01  10 3
kg
m s


Knowing the overflow rate, we can calculate the area required for the grit
chamber. Note, the safety factor 1.4
m3
s
m
s
0.25
Q
SF  
A
OFR
0.039
1.4
 8.97 m 2
So the area of the grit chamber must be 9 m2 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.
diameter
depth
Problem
Strategy
Solution
• Determine the Design Data needed for your solution
• surface over flow rate of 1,000 gal/ft2/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)
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/ft2/day, we can compute the area of each clarifier
Q  vA
6 gal
10

10
Q
day
A 
v 1000 gal ft 2 day
 10,000 ft 2
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.
A
d
d 2
4
4A



4 10,000 ft 2


 112.8 ft  115 ft
2


 115 ft
A
4
 10,387 ft 2
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
V  Q
3




1
day
1
ft
6 gal


 10  10 day 2 hr 
 24 hr  7.48 gal 
3
 111,408 ft


Clarifiers should be as shallow as possible, but not less that 7 ft deep. The approximate
depth, h, can now be calculated.
V  Ah
3
V 111,408 ft
h 
 10.73 ft  11 ft
2
A 10,387 ft
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 BOD5 and 30 mg/L SS.
Assuming that the BOD5 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.
Existing Plant Effluent Characteristics
Flow = 0.150 m3/s
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 BOD5 = 30 mg/L
BOD5 = 84 mg/L
Use the following “typical values”
Parameter
Value
Ks
100.00 mg BOD5/L
m
2.5 d-1
kd
0.05 d-1
Y
0.5 mg VSS/mg BOD5 removed
Problem
Strategy
Solution
(Q + Qr)
X, S
Q,So
(Q-Qw), S, Xe
Qr ,Xr ,S
S
K s 1  k d c 
 c  m  kd   1
Qw ,Xw ,S
X
 cY  S o  S 
 1  k d  c 
V  Q
Problem
Strategy
Solution
A portion of the SS is BOD5. Therefore, an estimate of the allowable soluble
BOD5 in the effluent can be made using the 63% assumption.
S = 30 - (0.63)(30) = 11.1 mg/L
Note: S is soluble BOD5
Problem
Strategy
Solution
The mean cell residence time can be estimated using
K s 1  k d c 
S
 c  m  kd   1
 
 
100 mg BOD 5  1  0.05 day 1 

c
L 
mg

11.1

L
 c 2.5 day 1  0.05 day 1  1

c = 5.0 days

Problem
Strategy
Solution
To solve for the hydraulic residence time:
 cY  S o  S 
X 
 1  k d  c 
5.0 days 0.5 84.0 mg L  11.1mg L 


2000 mg 

1
L
 1  0.05 day 5.0 day 
 = 0.073 d or 1.8 h
The volume of the aeration tank is then estimated using:

m
V  Q  0150
.
3
s 18
3600
. h

s
h
 972 m3  970 m3
....end of example