biofoundation-fereview.pptx

7/27/2019 BioFoundation-FEReview.pptx

1/34

FE Review for

EnvironmentalEngineeringProblems, problems, problems

Presented by L.R. Chevalier, Ph.D., P.E.

Department of Civil and Environmental Engineering

Southern Illinois University Carbondale


2/34

BIOLOGICAL FOUNDATIONS

FE Review for Environmental Engineering


3/34

Given the following data, calculate BOD5

Initial DO of sample: 9.0 mg/L

Volume of sample: 10 ml

Final DO of bottle after 5 days: 1.8 mg/L

Volume of BOD bottle: standard 300 ml

Problem Strategy Solution


4/34

Review and understand the terms of the governing equation

PDODO

V

VDODOBOD fi

b

s

fit



5/34

Standard Bottle: 300 ml

P = 10/300 =0.033

Lmg

P

DODOBOD

fi218

033.0

8.10.95



6/34

Time (days)

BOD(mg

/L)

BOD5

Typical Curve


7/34

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


8/34

Time (days)

BOD(mg

/L)

carbonaceous

nitrogenous

BOD5Lo


9/34

If the BOD3 of a waste is 75 mg/L and

k=0.345 day-1, what is the ultimate BOD?



10/34

For some of you there may be a confusion as to

which equation to use:

ktot eLL

kt

oteLBOD

1



11/34

Recall the equation for BODt

P

DODO

V

V

DODOBOD

fi

b

s

fit

The amount of DO measured will decrease over

time. Does BOD increase or decrease over time?



12/34

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

BODremaining,L

t

Oxygenconsumed,BODt Lo

OXYGEN DEMAND REMAINING

OXYGEN CONSUMED


13/34

LmgL

LeL

o

o

o

/116

645.0175

3345.0



14/34

Given: DOi = 9.0 mg/L

DO = 3.0 mg/L after 5 days

Dilution factor P = 0.030

Reaction 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?



15/34

Review and understand the equations needed for the solution

P

DODO

V

V

DODOBOD fi

b

sfit

ktot eLBOD

1

0

100

200

300

400

0 5 10 15 20 25

Time, days

BODremaining,Lt

Oxygenconsumed,

BODt

Lo

yt

Lt

BODt



16/34

a) What is the 5 day BOD?

Lmg

PDODOBOD

fi200

03.039

5



17/34

b) What is the ultimate BOD?

Lmg

eeBODL

kto300

1200

1 522.05



18/34

c) What is the remaining oxygen demand after 5

days?

300 - 200 = 100 mg/L



19/34

Determine the ThOD of a 400 mg/L solution of glucose C6H12O6



20/34

Balance the equation

Determine the MW of compound and O2

Calculate ThOD

oxygenchemicalmoles

oxygenmoles

chemicalMW

chemicalThOD mol

gLmg

Lmg 32

#

#



21/34

1. Balance the following equation

OHCOOOHC

OHCOOOHC

2226126

2226126

666

______



22/34

2. Determine the MW of glucose and O2

MW C6H12O6 = 12(6) + 12 + 16(6) = 180 g/mol

MW O2 = 2(16) = 32 g/mol

3. Calculate the ThOD

Lmg

molg

molg

Lmg

Lmg

oxygenecosglumoles

oxygenmolesThOD

7.426

321

6

180

400



23/34

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]



24/34

Balance two equations

Determine the MW of both compounds

Calculate ThOD for both, then add

oxygenchemicalmoles

oxygenmoles

chemicalMW

chemicalThOD mol

gLmg

Lmg 32

#

#



25/34

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



26/34

2. ThOD of EtOH is calculated as follows:

26.62

321

3

46

30

O

oxygenEtOHmoles

oxygenmolesThOD

Lmg

molg

molg

Lmg

Lmg



27/34

3. Calculate the ThOD for wastewater containing

40 mg/L sucrose [C6H12O6]

OHCOOOHC 2226126 666

MW Sucrose = 180 mg/L

27.42

321

6

180

40

O

oxygensucrosemoles

oxygenmoles

ThOD

Lmg

mol

g

molg

Lmg

L

mg



28/34

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



29/34

A chemical plant produces the amino acid

glycine [C2H5O2N]. The wastewater from the

facility contains approximately 25 mg/L ofthis acid. Calculate both the carbonaceous

and nitrogenous ThOD for the wastewater.

Example Solution


30/34

1. As in the previous example, write the

balance equation, but include NH3 as an end

product.

3222252 ???? NHOHCOONOHC

Example Solution


31/34

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

321

5.1

75

25

O

oxygenacidmoles

oxygenmolesThOD

Lmg

molg

molg

Lmg

Lmg

Example Solution


32/34

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


33/34

5. To determine the nitrogenous oxygen demand:

23.21

321

2

75

25

O

oxygenammoniamoles

oxygenmolesNOD

Lmg

molg

molg

Lmg

Lmg

Example Solution


34/34

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

biofoundation-fereview.pptx

Documents