CE 326 Principles of Environmental Engineering
BOD and DO Sag Equations
Typical Wastewater Characteristics
Constituent
Weak
Medium
Strong
BOD5 (Biological Oxygen Demand)
100
200
300
COD (Chemical Oxygen Demand)
250
500
1000
SS – Suspended solids
100
200
350
Total dissolved solids
200
500
1000
Total Nitrogen
20
40
85
Ammonia
12
25
50
Total phosphorus
4
8
15
Regulated
standards
BOD
 wastewater especially those with organic components when discharged into streams/rivers will have an effect on
the water quality of the stream/river by depleting the ___________________________ in water.
 Microorganisms degrade organic matter using DO to form CO2, H2O, new cells and other waste products.
 aquatic life needs oxygen, eg.,
trout needs at least ___________ mg/L of DO in water
carp needs at least ___________mg/L of DO in water
Water saturated with DO at 15 o C is approximately ___________ mg/L
 wastes that deplete DO in water are called __________________________ wastes.
 the portion of the waste that is oxygen-demanding can be quantified as the biochemical oxygen demand
If oxygen is consumed for - the oxidation of carbon only – termed as carbonaceous BOD (__________)
- the oxidation of nitrogen – termed as nitrogeneous BOD (_____________)
 For a given waste there is a maximum oxygen demand possible for the waste.
This maximum demand is termed as the __________________ BOD (BODL, Lo).
Example, if a waste has an ultimate BOD of 50 mg/L ==> it means that the degradable fraction needs 50
mg of DO for every liter of the waste for it to be degrade biologically.
 For a given waste, the change in BODL with time and the oxygen used or demand may be plotted as shown:
 Assuming that the ultimate BOD is degraded exponentially, the amount of BODL remaining after time t is given by
where k is the reaction rate constant or degradation rate constant (d-1) (determined through experiments)
The amount of BODL degraded is equal to Lo – Lt as shown in the dashed line. This is equal to amount of oxygen
consumed. Note that the amount of oxygen consumed is the biochemical oxygen demand for the waste at time t.
Therefore
The biochemical oxygen demand at time t (BODt) is given by the product of the ultimate BOD and
the factor (1 – e –kt). The industry standard is 5 days, i.e., BOD5 or 5-day BOD:
BOD5 = Lo (1 – e –5k)
It takes about 20 – 30 days for a highly biodegradable material such as municipal waste to completely degrade, i.e.,
to reach the ultimate BOD.
Note that some books expressed BODt as
BODt = Lo(1 - 10-Kt)
Need to be careful with the K values used
Nitrogeneous oxygen demand (NBOD)
For municipal waste, given by the conversion of NH4+ to nitrate NO3- :
Microorganisms responsible are nitrobacter and nitrosomonas
Example:
A waste with a BODL of 280 mg/L. Find the BOD5 if the k value at 20o C is 0.0536 day-1
BOD5 = 280(1-e-0.0536(5))
= ______________________ mg/L
Temperature effects
 higher temperature, degradation rate is faster
The degradation rate k related to the change in temperature by the following equation:
kT = k20 (1.1.35) T – 20
kT = k20 (1.056) T –20
20o C > T > 4o C
30o C > T > 20o C
Example: What is the BOD5 at 10o C for the above sample?
BOD Test
- Test conducted using a special 300 mL BOD bottle
- Maximum DO in water is between 8 to 10 mg/L
- If the waste has a BOD greater than 5 mg/L, the sample must be diluted accordingly so that the change in DO
concentration in the bottle is more than 2 mg/L but the final DO must be at least greater than 1 mg/L.
- A given volume of the sample is added into the bottle.
- The bottle is inoculated with microorganism – usually settled wastewater from a sewage treatment plant.
- The sample is diluted to 300 mL using dilution water containing inorganic nutrients (N, P and trace metals).
- The initial DO is measured (DOsi)
- The bottle is capped and incubated in the dark at 20o C for 5 days or t days.
- At the end of 5 or t days, the DO is measured (DOsf)
- The BOD is computed using
BODt = (DOsi - DOsf) x (Volume of diluted sample (300 mL)/Volume of sample
- If blank samples were used, change in DO in the blank samples after time t should be accounted for.
Other oxygen demand measurements
COD – chemical oxygen demand - the amount of oxygen needed to chemically oxidize the organics and other
compounds in water. A strong oxidizing agent, dichromate is usually used. Note BOD relies on
microorganism to oxidize the organics and not all organics are oxidized by the microorganisms.
COD test is an attractive as it takes 2 hours as opposed to 5 days for BOD
For municipal waste water BOD5  ____________________COD
ThOD – Theoretical oxygen demand – the amount of oxygen required to oxidize a substance to carbon dioxide and
water as calculated by stoichiometry if the chemical composition of the substance is known.
Example: acetic acid oxidizes to CO2 and water as shown
CH3COOH + 2O2 = = > 2CO2 + 2H2O
The ThOD for 1 mole/L of acetic acid (___________g/L) is equal to 2 moles/L (_______ g/L) of oxygen.
Changes in DO in streams and rivers
Let us now examine how the DO in a stream will change when oxygen-demanding waste is discharged into the
river. Assume wastewater with biodegradable waste is discharged into a stream. The following will occur in a
stream.
● The waste will be degraded and the DO in the water will be consumed.
Oxygen in the water will be depleted at a rate defined as the _______________________ rate.
The deoxygenation rate can be quantified by using the deoxygenation rate constant, kd (day-1)
● The DO in the stream is replenished by transfer of oxygen from air to water.
Oxygen transfer rate is given by the _________________________ rate constant, kr (day-1)
Before we proceed to develop equations for the depletion of DO and the change of BODL in a stream, mass balances
are used to find the BODL, DO and temperature after mixing.
River
Flow rate = Qr
Ultimate BOD = Lr
Dissolved oxygen = DOr
Temperature of water = Tr
Q, La, DOo, T
Flow rate of waste = Qw
Ultimate BOD of waste = Lw
Dissolved oxygen of waste = DOw
Temperature of waste = Tw
The flow rate after mixing is given by :
Q = Q r + Qw
The BODL after mixing is given by mass balances:
Mass Input = Mass Output
Similarly for DO and temperature after mixing we have:
Using a controlled volume as shown in the figure below, a mass balance can be written for the DO with distance:
Rate of DO
change in water =
at distance X
from point of
mixing
Input
DO
-
Output
DO
+
Oxygen transfer
from air to
water
Oxygen from air to water
La, DOo, T
Oxygen consumed in river
L (?), DO (?)
waste
Distance X
The equation is given by:
(see Equation 4-40 on page 310)
where
D = DOs - DO = oxygen deficit
Da = DOs – DOo = initial oxygen deficit
DOs = saturated dissolved oxygen concentration (mg/L) at temperature T
DOo = initial DO in river, i.e., after mixing at the point of discharge
DO = dissolved oxygen in stream at downstream distance x or time of travel, t
t = distance downstream/velocity of stream = x/u
u = speed of water in stream
The change of DO in the stream may be plotted as shown below:
Oxygen depleted
because of
waste
The minimum DO in the stream is called the critical point or critical deficit. The time, t c, and distance xc, at which it
will occur downstream after being discharged is given by:
tc = xc/u
xc = critical distance from point of discharge
The critical deficit or critical DO can be found by substituting tc into the earlier equation.
Both kr and kd are determined experimentally or estimated using the following equations:
kd = k + u  /H
where
kr = 3.9 u0.5/H1.5
u = speed of the stream (m/s)
H = average depth of stream (m)
 = bed activity coefficient = 0.1 for stagnant and deep water, 0.6 for rapid flowing stream
k = BOD rate constant at 20o C (d-1)
Most of the kd and kr values reported are at 20o C. To adjust the kd and kr for other temperatures, the following is
used:
Example: Problem 4 –17
Watertown’s wastewater is discharged into Green River. Determine whether or not the discharge of wastewater will
reduce the DO below the state standard of 5.0 mg/L at Smithville, 5.79 m downstream. Given:
Parameter
Wastewater
Green River
Flow (m3/s)
0.280
0.877
BODL at 28o C (mg/L)
6.44
7.00
DO (mg/L)
1.00
6.50
Kd at 28o C (d-1)
-0.199
Kr at 28o C (d-1)
-0.370
speed of stream (m/s)
-0.650
Temperature o C
28o C
28o C
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