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
MATHEMATICAL/PHYSICAL FOUNDATIONS
Problem
Strategy
Solution
Complete the following chart:
Total Solids (TS)
______ mg/L
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
VDS
FDS
VSS
FSS
70 mg/L
(_____%)
270 mg/L
(_____%)
180 mg/L
(_____%)
180 mg/L
(_____%)
Problem
Strategy
Solution
• Review the definitions of
•
•
•
•
•
•
•
TS
TDS
TSS
VDS
FDS
VSS
FSS
• Assume a 1 liter sample
• Divide 700 mg by the percentage shown or calculated
Problem
Strategy
Solution
Total Solids (TS)
700mg/L
TDS
TSS
340mg/L (49%)
360mg/L (51%)
VDS
FDS
VSS
FSS
70 mg/L (10%)
270 mg/L (39%)
180 mg/L (26%)
180 mg/L (26%)
Complete a flow chart using the following information
COM POUND
C O N C E N T R A T IO N
(m g/L )
D IS S O L V E S ?
V O L A T IL IZ E S O R
BU R N S A T 550 C
S o d iu m C h lo rid e
45
Y es
No
C alciu m su lfate
30
Y es
No
C la y
100
No
No
C o p p er ch lo rid e
10
Y es
No
A cetic a cid
20
Y es
Y es
C o ffee gro u n d s
25
No
Y es
Problem
Strategy
Solution
Total Solids (TS)
______ mg/L
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
Problem
Strategy
Solution
• Review definitions
• Fixed mean inorganic – it does not burn
• Volatile means organic – it does burn
Problem
Strategy
Solution
Sodium chloride
45 mg/L
Dissolves
Doesn’t volatilize
45
Total Solids (TS)
______ mg/L
45
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
45
Problem
Strategy
Solution
Calcium sulfate
30 mg/L
Dissolves
Doesn’t volatizes
45
30
Total Solids (TS)
______ mg/L
45
30
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
45
30
Problem
Strategy
Solution
Clay
100 mg/L
Doesn’t dissolve
Doesn’t volatizes
45
30
100
Total Solids (TS)
______ mg/L
45
30
100
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
45
30
100
Problem
Strategy
Solution
Copper chloride
10 mg/L
Dissolves
Doesn’t volatizes
45
30
10
100
Total Solids (TS)
______ mg/L
45
30
10
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
100
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
45
30
10
100
Problem
Strategy
Acetic acid
20 mg/L
Dissolves
Volatizes
Solution
45 20
30
10
100
Total Solids (TS)
______ mg/L
45 20
30
10
20
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
100
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
45
30
10
100
Problem
Strategy
Coffee grounds
25 mg/L
Doesn’t dissolves
Volatizes
Solution
45 20
30
10
100
25
Total Solids (TS)
______ mg/L
45 20
30
10
TDS
TSS
_____mg/L
(_____%)
_____mg/L
(_____%)
100
25
VDS
FDS
VSS
FSS
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
____ mg/L
(_____%)
20
45
30
10
25
100
Problem
Strategy
Solution
Total Solids (TS)
230mg/L
TDS
TSS
105mg/L (46%)
125 mg/L (54%)
VDS
FDS
VSS
FSS
20 mg/L (9%)
85 mg/L (37%)
25 mg/L (11%)
100 mg/L (43%)
Problem
Strategy
Solution
• Water flows into a heated tank at a rate of 150 gal/min.
• Evaporation losses are estimated to be 2000 lb/hr.
• Assuming the tank volume to be constant, what is the flow
rate out of the tank?
Problem
Strategy
Solution
• Draw a schematic (control volume)
• Convert to like units (Weight of water 8.34 lb/gal)
• Mass in = Mass out
• concept of density (Volume in = Volume out)
150
gpm
2000 lb/hr
?
2000
lb

1 gal

1hr
 4 gpm
hr 8 .34 lb 60 m in
dM
 0
dt
150  4  x  0
F low rate out  146 gpm
Example
Solution
Consider the following report from three supplies
into a reservoir. Is it correct?
Source
A
B
C
Flow
140 gpm
5 gpm
5 gpm
Quality
essentially clean
500 ppm toluene
500 ppm benzene
Total
150 gpm
1000 ppm
Example
Solution
Source
A
B
C
Flow
140 gpm
5 gpm
5 gpm
Quality
essentially clean
500 ppm toluene
500 ppm benzene
Total
150 gpm
1000 ppm
Do you see a
problem here?
Example
Solution
Important Rule: We can add mass (mass
balance) but not concentrations
Source
A
B
C
Flow
140 gpm
5 gpm
5 gpm
Quality
essentially clean
500 ppm toluene
500 ppm benzene
Total
150 gpm
1000 ppm
X
Example
Solution
1. What is the total volume of water per day?
140 gpm + 5 gpm + 5 gpm = 150 gpm
Converting to liters/day (L/d)
(150 gpm)(3.785 L/gal)(60 min/hr)(24 hr/day)
= 817560 L/d
Example
Solution
2. What is the mass from source B?
(5 gpm)(3.785 L/gal)(500 mg/L)(60 min/hr)(24 hr/d)
= 1.36 x 107 mg/d
Converting to ppm per day in total water
1.36 x 107 mg/ 817560 L = 16.67 mg/L = 16.67 ppm
Example
Solution
3. What is the mass from source C?
(5 gpm)(3.785 L/gal)(500 mg/L)(60 min/hr)(24 hr/d)
= 1.36 x 107 mg/d
Converting to ppm per day in total water
1.36 x 107 mg/ 817560 L = 16.67 mg/L = 16.67 ppm
Example
Solution
4. Therefore, we have 16.67 ppm benzene, and
16.67 ppm toluene! Not 1000 ppm!
5. Can we add these concentrations?
Simple Model Of Stream Pollution Based On Mass
Balance
Industrial Complex
Qu
Cu
Qe
Ce
Qd
Cd
Problem
Strategy
Solution
• A factory for copper and brass plating is dumping its
wastewater effluent into a near-by stream.
• Local regulations limit the copper concentration in the stream
to 0.005 mg/L.
• Upstream flow in stream, 0.5 m3/s. Concentration of copper in
upstream flow is below detection limits
• Effluent flow from plating factory 0.1 m3/s
• Determine the maximum concentration allowable in the
effluent from the factory’s wastewater.
Problem
•
•
•
•
Strategy
Solution
Draw a control volume diagram
Determine Qtotal = Qstream + Qeffleuent
Convert concentrations to mass (mass flux)
Use mass balance to determine the allowable concentration
(based on mass) of effluent
Problem
Strategy
Solution
Industrial Complex
Qu= 0.5 m3/s
Cu = 0 mg/L
Qe =0.1 m3/s
Ce = ?
Qd = ?
Cd = 0.005 mg/L
dm
dt
 Qu C u  Q eC e  Q d C d  0
Ce 
Qd C d  QuC u
Qe
0 .6m s  0 .005 mg L   0 .5 m s  0 mg L 
3

3
0 . 1m
 0 . 03
3
s
mg
L
....end of example
SIMPLE PHOSPHOROUS MODEL
Want to estimate the amount of phosphorous control
needed to prevent eutrophication due to the overproduction of algae
Simple Phosphorous Model
Assumptions:
•
•
•
•
Completely mixed lake
Steady state
Constant settling rate
Phosphorous is the controlling nutrient
Schematic Of System
Waste water treatment plant
Q = 0.2 m3/s P=5.0 mg/L
Stream
Q = 15.0 m3/s P=0.01
mg/L
Settling rate
vs = 10 m/yr
Surface area of
lake 80 x 106 m2
Problem
Strategy
Solution
• Estimate P
• What rate of phosphorous removal at the wastewater
treatment plant would be required to keep the concentration
of phosphorous in the lake at an acceptable level of 0.01
mg/L?
Problem
Strategy
Solution
• Evaluate all inputs and outputs to the control volume
• Qin = Qout
• QCin = QCout
Problem
Strategy
Sources
QwwtPwwt
QstreamPstream
Using mass balance approach:
Rate of addition of P = Rate of removal of P
Solution
Problem
Strategy
Solution
Outflow rate, QtPlake
Area, A
Concentration, Plake
Settling rate, AvsPlake
Using mass balance approach:
Rate of addition of P = Rate of removal of P
Problem
Strategy
Solution
Rate of addition of P = Rate of removal of P
S = QTPlake + vsAPlake
where:
S = rate of addition of phosphorus from all sources (g/s)
P = concentration of phosphorus (g/m3)
QT = stream outflow rate (m3/s)
vs = the phosphorus settling rate (m/s)
A = surface area of the lake (m2)
which results in a steady-state concentration of
Plake 
S
QT  v s A
Of note, vs is empirically derived and difficult to predict with any confidence.
Suggest a settling rate of 3-30 m/year.
Problem
Strategy
Solution
1. Determine the mass loading from both sources
Phosphorous loading from incoming stream:
Ss = (15.0 m3/s)(0.01 mg/L)(g/1000mg)/(1000 L/m3) = 0.15 g/s
From the wastewater treatment plant:
Sw = (0.2 m3/s)(5.0 mg/L)(1 g/m3)/(mg/L) = 1 g/s
For a total loading of
S = 0.15 g/s + 1.0 g/s = 1.15 g/s
2. Determine the volume (mass) of water entering over time:
Neglecting evaporation
QT = 15 m3/s + 0.2 m3/s = 15.2 m3/s
3. Estimate the settling rate:
 10 m   yr   d   hr 
vs  




 yr   365 d   24 hr   3600 s 
 3 . 17  10
7 m
s
4. Apply model
PLake 
S
QT  v s A
1 . 15

15 . 2 m
 3 . 17  10
3
s
 0 . 028
g
 0 . 028
mg
m
g
s
7 m
s
80  10
6
m
2

3
L
This is above the 0.01 mg/L suggested for acceptable
concentration. We cannot reduce background levels in
the stream. Therefore, we need to determine the
reduction at the plant. To start with, solve for S with a
known value of P = 0.01 mg/L
S  Plake Q t  v s A 



g
6
2
 15 . 2 m 3  3 . 17  10  7 m
  0 . 01
80  10 m
3 
s
s
m 

 0 . 41

g
s
The amount that the wastewater treatment plant could contribute would be:
Sw = 0.41g/s – 0.15 g/s = 0.26 g/s
Since Sw is now at 1.0 g/s, there is a need for 74% phosphorous removal
Summary of Problem
• We basically did a mass balance for
• the water supply
• the contaminant
Q1
where M4 is the mass
in both the lake and
the outgoing stream
M1/T
M4/T=M1/T+M2/T-M3/T
Q3=Q1+Q2
M2/T
Q2
M3/T
(settling)