Post on 14-Dec-2015
FE Review for Environmental EngineeringProblems, problems, problemsPresented by L.R. Chevalier, Ph.D., P.E.Department of Civil and Environmental EngineeringSouthern Illinois University Carbondale
MATHEMATICAL/PHYSICAL FOUNDATIONS
FE Review for Environmental Engineering
Complete the following chart:
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS70 mg/L (_____
%)
FDS270 mg/L (_____%)
TSS_____mg/L (_____%)
VSS180 mg/L (_____%)
FSS180 mg/L (_____%)
• 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
Problem Strategy Solution
Total Solids (TS)700mg/L
TDS340mg/L (49%)
VDS70 mg/L (10%)
FDS270 mg/L (39%)
TSS360mg/L (51%)
VSS180 mg/L (26%)
FSS180 mg/L (26%)
Complete a flow chart using the following information
COMPOUND CONCENTRATION(mg/L)
DISSOLVES? VOLATILIZES ORBURNS AT 550 C
Sodium Chloride 45 Yes No
Calcium sulfate 30 Yes No
Clay 100 No No
Copper chloride 10 Yes No
Acetic acid 20 Yes Yes
Coffee grounds 25 No Yes
Problem Strategy Solution
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
• Review definitions• Fixed mean inorganic – it does not burn• Volatile means organic – it does burn
Problem Strategy Solution
45
45
Sodium chloride 45 mg/L Dissolves Doesn’t volatilize
45
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
4530
4530
Calcium sulfate 30 mg/L Dissolves Doesn’t volatizes
4530
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
4530
4530
4530
Clay 100 mg/L Doesn’t dissolve Doesn’t volatizes 100
100
100
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
453010
453010
453010
100
100
100
Copper chloride 10 mg/L Dissolves Doesn’t volatizes
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
45 203010
45 203010
453010
100
100
100
Acetic acid 20 mg/L Dissolves Volatizes
20
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
45 203010
45 203010
453010
10025
10025
100
Coffee grounds 25 mg/L Doesn’t dissolves Volatizes
2025
Problem Strategy Solution
Total Solids (TS)______ mg/L
TDS_____mg/L (_____%)
VDS____ mg/L (_____%)
FDS____ mg/L (_____%)
TSS_____mg/L (_____%)
VSS____ mg/L (_____%)
FSS____ mg/L (_____%)
Problem Strategy Solution
Total Solids (TS)230mg/L
TDS105mg/L (46%)
VDS20 mg/L (9%)
FDS85 mg/L (37%)
TSS125 mg/L (54%)
VSS25 mg/L (11%)
FSS100 mg/L (43%)
• 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)
150gpm
2000 lb/hr
?
Problem Strategy Solution
20001
8 34
1
604
0
150 4 0
146
lb
hr
gal
lb
hrgpm
dM
dtx
Flow rate out gpm
. min
Problem Strategy Solution
Consider the following report from three supplies into a reservoir. Is it correct?
Source Flow Quality
A 140 gpm essentially cleanB 5 gpm 500 ppm tolueneC 5 gpm 500 ppm benzene
Total 150 gpm 1000 ppm
Example Solution
Source Flow Quality
A 140 gpm essentially cleanB 5 gpm 500 ppm tolueneC 5 gpm 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 Flow Quality
A 140 gpm essentially cleanB 5 gpm 500 ppm tolueneC 5 gpm 500 ppm benzene
Total 150 gpm 1000 ppmX
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?
Example Solution
Simple Model Of Stream Pollution Based On Mass Balance
Qu
Cu
Qd
Cd
Qe
Ce
Industrial Complex
• 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
Qd = ?Cd = 0.005 mg/L
Qe =0.1 m3/sCe = ?
Industrial Complex
Qu= 0.5 m3/sCu = 0 mg/L
Problem Strategy Solution
Lmg
sm
Lmg
sm
Lmg
sm
Q
CQCQC
CQCQCQdt
dm
e
uudde
ddeeuu
03.0
1.0
05.0005.06.0
0
3
33
....end of example
Problem Strategy Solution
SIMPLE PHOSPHOROUS MODEL
Want to estimate the amount of phosphorous control needed to prevent eutrophication due to the over-
production of algae
Simple Phosphorous Model
• Completely mixed lake• Steady state• Constant settling rate• Phosphorous is the controlling nutrient
Assumptions:
Stream Q = 15.0 m3/s P=0.01 mg/L
Waste water treatment plant Q = 0.2 m3/s P=5.0 mg/L
Surface area oflake 80 x 106 m2
Settling ratevs = 10 m/yr
Schematic Of System
• 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 Solution
Using mass balance approach:Rate of addition of P = Rate of removal of P
Sources
Problem Strategy Solution
QwwtPwwt
QstreamPstream
Settling rate, AvsPlake
Area, A
Outflow rate, QtPlake
Concentration, Plake
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)
Problem Strategy Solution
AvQ
SP
sTlake
which results in a steady-state concentration of
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
Problem Strategy Solution
sm
s shr
hrd
dyr
yrm
v
71017.3
36002436510
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:
Problem Strategy Solution
Lmgmg
m
AvQ
SP
sm
sm
sg
sTLake
028.0
028.0
10801017.32.15
15.1
3
2673
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
4. Apply model
Problem Strategy Solution
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
Problem Strategy Solution
sg
msm
sm
mg
AvQPS stlake
41.0
10801017.32.1501.0 2673
3
Summary of Problem
• We basically did a mass balance for• the water supply• the contaminant
Q1
Q2
Q3=Q1+Q2
M1/T
M2/T
M4/T=M1/T+M2/T-M3/T
M3/T(settling)
where M4 is the mass in both the lake and the outgoing stream