Transcript Biological Foundations - Civil and Environmental Engineering | SIU

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
Problem
Strategy
Solution
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
• Review and understand the terms of the governing equation
BOD t 
DO i  DO f
 Vs

 Vb




DO i  DO f
P
Standard Bottle: 300 ml
P = 10/300 =0.033
BOD 5 
DO i  DO f
P

9 .0  1 .8
0 . 033
 218
mg
L
BOD (mg/L)
Typical Curve
BOD5
Time (days)
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
BOD (mg/L)
nitrogenous
BOD5
Time (days)
Lo
carbonaceous
Problem
Strategy
Solution
If the BOD3 of a waste is 75 mg/L and
k=0.345 day-1, what is the ultimate BOD?
For some of you there may be a confusion as to
which equation to use:

BODt  Lo 1  e
 kt

Lt  Loe
 kt
Recall the equation for BODt
BOD t 
DO i  DO f
 Vs

 Vb




DO i  DO f
P
The amount of DO measured will decrease over
time. Does BOD increase or decrease over time?
Lo
O xygen con su m ed , B O D t
B O D re m ain in g, L t
400
300
O XY G E N C O N SU M E D
200
100
O XYG EN DEM AND R EM AINING
0
0
5
10
15
20
T im e , da ys
OXYGEN CONSUMED

BODt  Lo 1  e
 kt

OXYGEN DEMAND REMAINING
Lt  Loe kt
Want to use the equation that shows an increase with time!
25

75  L o 1  e
  0 . 345  3 
 0 . 645 L o
L o  116 mg / L

Problem
Strategy
Solution
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?
Review and understand the equations needed for the solution




 kt
BODt  Lo 1  e
P

400
Lo
B O D rem ain in g, L t
 Vs

 Vb

DO i  DO f
O xygen con su m ed , B O D t
BOD t 
DO i  DO f
300
yt
200
BO D t
100
Lt
0
0
5
10
T im e , da ys
15
20
25
a) What is the 5 day BOD?
BOD
5

DO i  DO
P
f

93
0 . 03
 200
mg
L
b) What is the ultimate BOD?
Lo 
BOD
1 e
5
 kt

200
1 e
  0 . 22  5 
 300
mg
L
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
• Balance the equation
• Determine the MW of compound and O2
• Calculate ThOD
ThOD
mg L  
mg
L
chemical
MW chemical

# moles oxygen
# moles chemical
 32
g
mol
oxygen
1. Balance the following equation
C 6 H 12 O 6  __ O 2  __ CO 2  __ H 2 O
C 6 H 12 O 6  6 O 2  6 CO 2  6 H 2 O
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
ThOD
mg L  
400
mg
180
g

L
1 moles glu cos e
mol
 426 . 7 mg
6 moles oxygen
L
 32
g
mol
oxygen
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]
• Balance two equations
• Determine the MW of both compounds
• Calculate ThOD for both, then add
ThOD

mg
L

mg
L
chemical
MW chemical

# moles oxygen
# moles chemical
 32
g
mol
oxygen
1. Write the balanced equation for the oxidation of
ethanol (often written EtOH) to the end products
of CO2 and H2O.
CH 3 CH 2 OH  3O 2  2 CO 2  3 H 2 O
MW EtOH = 46 g/mol
2. ThOD of EtOH is calculated as follows:
ThOD

mg
L

30
mg
46
g
L
mol
 62 . 6 mg
L

3 moles oxygen
1 moles EtOH
O2
 32
g
mol
oxygen
3. Calculate the ThOD for wastewater containing
40 mg/L sucrose [C6H12O6]
C 6 H 12 O 6  6 O 2  6 CO 2  6 H 2 O
MW Sucrose = 180 mg/L
ThOD

mg
L

40
180
mg
L
g
mol
 42 . 7 mg
L

6 moles oxygen
1 moles sucrose
O2
 32
g
mol
oxygen
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
Example
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.
1. As in the previous example, write the
balance equation, but include NH3 as an end
product.
C 2 H 5 O 2 N  ? O 2  ? CO 2  ? H 2 O  ? NH 3
2. Balanced equation:
C 2 H 5 O 2 N  1 12 O 2  2 CO 2  H 2 O  NH
3
3. The molecular weight of the acid is 75 g/mol. The
amount of oxygen required to oxidize the
carbonaceous portion is:
ThOD

mg
L

25
mg
75
g
 16
L

mol
mg
L
O2
1 . 5 moles oxygen
1 moles acid
 32
g
mol
oxygen
4. One mole of ammonia is produced for each mole
of acid oxidized. The equation for oxidation of the
ammonia is:

N H 3  2 O2  N O3  H 2 O  H
ammonia
nitrate

Example
Solution
5. To determine the nitrogenous oxygen demand:
NOD

mg
L

25
mg
75
g
L

1 moles ammonia
mol
 21 . 3 mg
2 moles oxygen
L
O2
 32
g
mol
oxygen
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