Embed Size (px)
Citation preview
Assoc. Prof. Kozet YAPSAKLI
What is BOD?
Biochemical Oxygen Demand
It is just what it sounds like, it is the oxygen required by biochemical processes.
What is BOD? In the presence of free oxygen, aerobic bacteria use the
organic matter in wastewater as “food”.
The BOD test is an estimate of the “food” available in the sample.
The more “food” present in the waste, the more Dissolved Oxygen (DO) will be required.
The BOD test measures the strength of the wastewater by measuring the amount of oxygen used by the bacteria as they stabilize the organic matter under controlled conditions of time and temperature.
BOD testused to determine pollutional strength of domestic and industrial wastes in terms of oxygen that they will require if discharged into natural watercources
5
wastewater
Org. matter
O2
DO
How is BOD different than dissolved oxygen?
BOD isn’t really there.
BOD is the amount of oxygen that would be consumed given sufficient time, bacteria and oxygen to completely decompose the organic matter.
How is BOD different than dissolved oxygen?
“Dissolved oxygen” is there. It is a measure of how much oxygen is dissolved in a water sample. It is a measure of oxygen content.
BOD is the amount of oxygen that would be consumed to completely decompose the organic matter in a water sample. It is not an indication of oxygen content. It is an indication of the amount of organic material present.
The Reaction Complete oxidation (combustion) of organic materials
yield identical products no matter what the organic starting material.
CnHaObNc + O2 → CO2 + H2O + NH3
(unbalanced)
The Balanced Equation
CnHaObNc + (n+a/4 - b/2 -3c/4) O2 →
n CO2 + (a/2 – 3c/2)H2O + c NH3
Note: Don’t get too hung up on the numbers, we don’t use the stoichiometry very often.
How would you determine BOD? Add bacteria.
Add oxygen quantitatively.
Wait until all of the organic material is gone.
Report on the amount of oxygen used.
What is the problem with that little scheme?
What bacteria? There are millions of different kinds.
How long? What if it takes 10 years for all the organic material to disappear? What if some of it NEVER decomposes?
Other factors: Temperature? Amount of light present? Concentration of oxygen?
Source: www.dnr.wi.gov
BOD - loss of biodegradable
organic matter (oxygen demand) Lo
Lt
L o
r B
OD
rem
ain
ing
Time
Lo-Lt = BODt
BOD Bottle
BOD Bottle
BOD Bottle
BOD Bottle
BOD Bottle
Ex: Nitrogenous Oxygen Demand
21
Direct Method
Adjust the sample to 20 °C , aerate with diffused air to increase or decrease dissolved gas content to near saturation.
Fill BOD bottles. Measure D.O. immediately in the first bottle.
Incubate 5 day Measure D.O. BOD =DO1-DO2
Dilution Method
Control all environmental conditions in the bioassay test.
Free of toxic materials
Favourable pH and osmotic cond.
Presence of available nutrients
Standard Temp.
Presence of mixed organisms of soil origin
24
Industrial wastes may be free of m.o. and nutrients
Domestic ww Contain org. Nutrients N, P,
Dilution water used in BOD must compensate these deficiences.
Dilution water
Natural surface water
Tap water=>possibility of toxicity from chlorine residuals.
Synthetic dilution water prepared from distilled, demineralised water
Dilution water must be free of toxic subs. Chlorine, Chloroamines, copper.
25
Seeding: (maintain necessary microorganism)
Domestic wastewaters
27
Dilution water
Seed Nutrient pH buffer Nitrification inhibitor
Blank (dilution water)
Diluted sample
Allyl-thiourea (ATU)
Dilutions of wastes
Set three dilutions
If strength is known two dilutions
Some casesup to 4 dilutions
Samples should deplete at least 2 mg/L D.O.
At least 0.5 mg/L of D.O. should remain at the end of incubation.
DO1-DO2=2-7 mg/L
28
Sample calculation
Determine the 5-day BOD for a 15 ml sample that is diluted with dilution water to a total volume of 300 ml when the initial DO concentration is 8 mg/l and after 5 days, has been reduced to 2 mg/l.
D0 = 8 D5 = 2 P = 15 ml/300ml = 0.05 BOD (mg/l) = _______ = 120
0.05
8 - 2
If all the DO is used up the test is invalid, as in B above
Need for a blank The dilution water containing the seeding material
will contain organic materialaddition of the diluting water to the sample will increase the amount of oxidizable organic matter.
A correction must be applied BOD analysis of the seeded dilution water
Calculation of BOD
f: fraction of the seed in the incubated sample
Calculation of BOD
Reaction rates-BOD Kinetics
It’s all about the kinetic model BOD reactions have “1st order kinetics”
“1st order kinetics” refers to the dependence on concentration
Rate = - Δ[react] = ∆ [prod] = k [react]1
∆time ∆time
The Rate Law Rate = - Δ[react] = ∆ [prod] = k [react]1
∆time ∆time
What’s that “k”?
k is called the “rate constant”, it’s the only thing that is constant in this kinetic scheme.
A word about k The “rate constant”, k :
Depends on the reaction
Depends on the type of bacteria (in the case of BOD)
Depends on the temperature
Rate of reaction is proportional to the concentration of food:
It is customary to describe biodegradable organic matter in terms of its equivalent oxygen consumption potential:
In many cases, the interest is in how much oxygen has been consumed, rather than how much BOD remains:
In the BOD test, it is y which is measured rather than Lt
Ultimate BOD If you determine the BOD after 5 days, this is called
“the 5 day BOD” (BOD5). If you determine the BOD after 20 days, this is called “the 20 day BOD” (BOD20). These are really BOD exerted values.
The “ultimate BOD” is the amount of oxygen
required to decompose all of the organic material after “infinite time”. This is usually simply calculated from the 5 day data. (Who can wait for infinity?)
k´ of raw domestic ww 0,40 per day
Why do such differences in reaction rates occur?
The nature of the organic matter
The ability of organisms present to utilize the organic matter.
Glucose high k Lignin, Synthetic detergents slowly attacked by bacteria
Rate of hydrolysis and diffusion
Some industrial wastes particularly containing synthetic chemicals with structure not found in natural compıundslag period
How to overcome: • Seed from the river where the ww is
discharged • Use adapted seed developed in the lab
BODL and Theoretical Oxygen Demand
Ultimate BOD or L0 = ThOD considered to be equal for a biodegradable substance
Oxidation of glucose
C6H12O6 + 6O2 6CO2+ 6H2O
180 192
192 g O2 /mole of glucose
OR
1.065 mg O2 /mg glucose
300 mg/L of glucose ThOD=
Experimentally;
BOD measurements (20 day)
BOD(L)= 250-285 mg/L
85% of theoretical amount
Not all the glucose converted to CO2 and water
300*192/180=320 mg/L
Org. Matter Food source
Energy Growth
Part of org. matter Converted to cell tissue Cell tissue will remain unoxidizedtill endogeneous
respiration
When bacteria die they become food material for others.
Further transformation to CO2 , H2O and cell tissue
Living bacteria + Dead ones Food for higher organisms
Protozoans
56
A certain amount of organic matter remains in these transformations.
Resistant to further biological attack.
Humus amount of org. matter corresponding to discrepancy between total BOD and ThOD.
57
Analysis of BOD Data
Calculation of k value is needed to obtain L0 using BOD5.
k and L0 are determined from a series of BOD measurements.
Methods:
Least squares
Thomas Method
Methods of moments
Daily-difference method
Rapid ratio method
Fujimoto Method
59
Least Squares Method
y6=
22=
Thomas Method
Thomas Method
Thomas Method
Thomas Method-Example
From a BOD test we have the following data:
Time, Days BOD, mg/L
2 10
4 16
6 20
Slope = 0.545
Intercept = 0.021
k1 = 6(0.021/0.545) = 0.64 day-1
L = 1/[6(0.021)(0.545)2] = 26.7 mg/L
k’
L0
