fe review for environmental engineering
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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. biological Foundations. - PowerPoint PPT PresentationTRANSCRIPT
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FE Review for Environmental EngineeringProblems, problems, problemsPresented by L.R. Chevalier, Ph.D., P.E.Department of Civil and Environmental EngineeringSouthern Illinois University Carbondale
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BIOLOGICAL FOUNDATIONSFE Review for Environmental Engineering
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Given the following data, calculate BOD5
Initial DO of sample: 9.0 mg/LVolume of sample: 10 mlFinal DO of bottle after 5 days: 1.8 mg/LVolume of BOD bottle: standard 300 ml
Problem Strategy Solution
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• Review and understand the terms of the governing equation
PDODO
VVDODO
BOD fi
b
s
fit
Problem Strategy Solution
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Standard Bottle: 300 ml
P = 10/300 =0.033
Lmg
PDODO
BOD fi 218033.0
8.10.95
Problem Strategy Solution
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Time (days)
BOD
(mg/
L)
BOD5
Typical Curve
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Further Discussion on BOD• Typical values• domestic sewage 250 mg/L• industrial waste as high as 30,000 mg/L• untreated dairy waste 20,000 mg/L
• After 5 days, BOD curve may turn sharply upward• demand of oxygen by microorganisms that decompose
nitrogeneous organic compounds into stable nitrate
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Time (days)
BOD
(mg/
L)
carbonaceous
nitrogenous
BOD5
Lo
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If the BOD3 of a waste is 75 mg/L and k=0.345 day-1, what is the ultimate BOD?
Problem Strategy Solution
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For some of you there may be a confusion as to which equation to use:
ktot eLL kt
ot eLBOD 1
Problem Strategy Solution
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Recall the equation for BODt
PDODO
VVDODO
BOD fi
b
s
fit
The amount of DO measured will decrease over time. Does BOD increase or decrease over time?
Problem Strategy Solution
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ktot eLBOD 1
ktot eLL
OXYGEN CONSUMED OXYGEN DEMAND REMAINING
Want to use the equation that shows an increase with time!
0
100
200
300
400
0 5 10 15 20 25
Time, days
BO
D r
emai
ning
, Lt
Oxy
gen
cons
umed
, BO
D t Lo
OXYGEN DEMAND REMAINING
OXYGEN CONSUMED
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LmgLLeL
o
o
o
/116645.0
175 3345.0
Problem Strategy Solution
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Given: DOi = 9.0 mg/LDO = 3.0 mg/L after 5 daysDilution factor P = 0.030Reaction rate, k = 0.22 day-1
a) What is the 5-day BOD?b) What is the ultimate BOD?c) What is the remaining oxygen demand after 5 days?
Problem Strategy Solution
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Review and understand the equations needed for the solution
PDODO
VVDODO
BOD fi
b
s
fit
ktot eLBOD 1
0
100
200
300
400
0 5 10 15 20 25
Time, days
BO
D r
emai
ning
, Lt
Oxy
gen
cons
umed
, BO
Dt
Lo
yt
Lt
BODt
Problem Strategy Solution
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a) What is the 5 day BOD?
Lmg
PDODO
BOD fi 20003.0
395
Problem Strategy Solution
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b) What is the ultimate BOD?
Lmg
eeBOD
L kto 3001
2001 522.0
5
Problem Strategy Solution
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c) What is the remaining oxygen demand after 5 days?
300 - 200 = 100 mg/L
Problem Strategy Solution
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Determine the ThOD of a 400 mg/L solution of glucose C6H12O6
Problem Strategy Solution
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• Balance the equation• Determine the MW of compound and O2
• Calculate ThOD
oxygenchemicalmolesoxygenmoles
chemicalMWchemicalThOD mol
gLmg
Lmg 32
##
Problem Strategy Solution
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1. Balance the following equation
OHCOOOHC
OHCOOOHC
2226126
2226126
666
______
Problem Strategy Solution
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2. Determine the MW of glucose and O2
MW C6H12O6 = 12(6) + 12 + 16(6) = 180 g/molMW O2 = 2(16) = 32 g/mol
3. Calculate the ThOD
Lmg
molg
molg
Lmg
Lmg oxygen
ecosglumolesoxygenmolesThOD
7.426
3216
180400
Problem Strategy Solution
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Ethanol, or ethyl, alcohol is used in beverages, as a gasoline additive, and in other industrial applications. Because small amounts of ethanol and sugar are used in the biological process to produce methanol, both of these compounds inevitable end up in the waste water of methanol plants.
Calculate the ThOD demand for waste water containing 30 mg/L ethanol [CH3CH2OH] and 40 mg/L sucrose [C6H12O6]
Problem Strategy Solution
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• Balance two equations• Determine the MW of both compounds• Calculate ThOD for both, then add
oxygenchemicalmolesoxygenmoles
chemicalMWchemicalThOD mol
gLmg
Lmg 32
##
Problem Strategy Solution
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1. Write the balanced equation for the oxidation of ethanol (often written EtOH) to the end products of CO2 and H2O.
OHCOOOHCHCH 22223 323
MW EtOH = 46 g/mol
Problem Strategy Solution
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2. ThOD of EtOH is calculated as follows:
26.62
3213
4630
O
oxygenEtOHmolesoxygenmolesThOD
Lmg
molg
molg
Lmg
Lmg
Problem Strategy Solution
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3. Calculate the ThOD for wastewater containing 40 mg/L sucrose [C6H12O6]
OHCOOOHC 2226126 666
MW Sucrose = 180 mg/L
27.42
3216
18040
O
oxygensucrosemolesoxygenmolesThOD
Lmg
molg
molgL
mg
Lmg
Problem Strategy Solution
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4. To calculate ThOD for waste water containing both 30 mg/L ethanol [CH3CH2OH] and 40 mg/L sucrose [C6H12O6], you can add the ThOD of the individual compounds.
ThOD tot = 62.6 mg/L O2 + 42.7 mg/L O2
= 105.3 mg/L O2... end of example
Problem Strategy Solution
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A chemical plant produces the amino acid glycine [C2H5O2N]. The wastewater from the facility contains approximately 25 mg/L of this acid. Calculate both the carbonaceous and nitrogenous ThOD for the wastewater.
Example Solution
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1. As in the previous example, write the balance equation, but include NH3 as an end product.
3222252 ???? NHOHCOONOHC
Example Solution
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322221
252 21 NHOHCOONOHC
2. Balanced equation:
3. The molecular weight of the acid is 75 g/mol. The amount of oxygen required to oxidize the carbonaceous portion is:
216
3215.1
7525
O
oxygenacidmolesoxygenmolesThOD
Lmg
molg
molg
Lmg
Lmg
Example Solution
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4. One mole of ammonia is produced for each mole of acid oxidized. The equation for oxidation of the ammonia is:
NH O NO H O H3 2 3 22
ammonia nitrate
Example Solution
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5. To determine the nitrogenous oxygen demand:
23.21
3212
7525
O
oxygenammoniamolesoxygenmolesNOD
Lmg
molg
molg
Lmg
Lmg
Example Solution
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6. The amount of oxygen required to oxidize the acid is the sum of both the carbonaceous and the nitrogenous oxygen demands.
ThOD = 16 + 21.33 = 37.33 mg/L O2
.....end of example
Example Solution