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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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

• Review and understand the terms of the governing equation

PDODO

VVDODO

BOD fi

b

s

fit

Problem Strategy Solution

Standard Bottle: 300 ml

P = 10/300 =0.033

Lmg

PDODO

BOD fi 218033.0

8.10.95

Problem Strategy Solution

Time (days)

BOD

(mg/

L)

BOD5

Typical Curve

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

Time (days)

BOD

(mg/

L)

carbonaceous

nitrogenous

BOD5

Lo

If the BOD3 of a waste is 75 mg/L and k=0.345 day-1, what is the ultimate BOD?

Problem Strategy Solution

For some of you there may be a confusion as to which equation to use:

ktot eLL kt

ot eLBOD 1

Problem Strategy Solution

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

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

LmgLLeL

o

o

o

/116645.0

175 3345.0

Problem Strategy Solution

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

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

a) What is the 5 day BOD?

Lmg

PDODO

BOD fi 20003.0

395

Problem Strategy Solution

b) What is the ultimate BOD?

Lmg

eeBOD

L kto 3001

2001 522.0

5

Problem Strategy Solution

c) What is the remaining oxygen demand after 5 days?

300 - 200 = 100 mg/L

Problem Strategy Solution

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

Problem Strategy Solution

• Balance the equation• Determine the MW of compound and O2

• Calculate ThOD

oxygenchemicalmolesoxygenmoles

chemicalMWchemicalThOD mol

gLmg

Lmg 32

##

Problem Strategy Solution

1. Balance the following equation

OHCOOOHC

OHCOOOHC

2226126

2226126

666

______

Problem Strategy Solution

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

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

• Balance two equations• Determine the MW of both compounds• Calculate ThOD for both, then add

oxygenchemicalmolesoxygenmoles

chemicalMWchemicalThOD mol

gLmg

Lmg 32

##

Problem Strategy Solution

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

2. ThOD of EtOH is calculated as follows:

26.62

3213

4630

O

oxygenEtOHmolesoxygenmolesThOD

Lmg

molg

molg

Lmg

Lmg

Problem Strategy Solution

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

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

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

1. As in the previous example, write the balance equation, but include NH3 as an end product.

3222252 ???? NHOHCOONOHC

Example Solution

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

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

5. To determine the nitrogenous oxygen demand:

23.21

3212

7525

O

oxygenammoniamolesoxygenmolesNOD

Lmg

molg

molg

Lmg

Lmg

Example Solution

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