Water Quality Management
in Rivers
Dissolved Oxygen Depletion
Dissolved Oxygen Depletion
Biochemical Oxygen Demand
Measurement
• Take sample of waste; dilute with oxygen saturated
water; add nutrients and microorganisms (seed)
• Measure dissolved oxygen (DO) levels over 5 days
• Temperature 20° C
• In dark (prevents algae from growing)
• Final DO concentration must be > 2 mg/L
• Need at least 2 mg/L change in DO over 5 days
Example
• A BOD test was conducted in the
laboratory using wastewater being
dumped into a Lake. The samples are
prepared by adding 3.00 mL of
wastewater to the 300.0 mL BOD
bottles. The bottles are filled to capacity
with seeded dilution water.
Example : Raw Data
Time
(days)
0
1
2
3
4
5
Diluted
Blank Seeded
sample
Sample DO
DO (mg/L)
(mg/L)
7.95
8.15
3.75
8.10
3.45
8.05
2.75
8.00
2.15
7.95
1.80
7.90
Example : Calculations
• What is the BOD5 of the sample?
[(DO i  DO f ) - (B i  B f )(1  P)]
BODm 
P
P  dilution factor  volume sample/ total volume
B i , B f  initial and final DO concentrations of the
seeded diluted water (blank)
DOi , DO f  initial and final DO concentrations of the
diluted sample
• Plot the BOD with respect to time.
Example : Time – Concentration
Plot
700
600
BOD (mg/L)
500
400
300
200
100
0
0
1
2
3
time (days)
4
5
6
Conc. (mg/L)
Modeling BOD as a First-order
Reaction
6
5
4
3
2
1
0
Organic matter oxidized
Organic matter remaining
0
10
20
time (days)
30
Modeling BOD Reactions
• Assume rate of decomposition of organic
waste is proportional to the waste that is left in
the flask.
dLt
 - kLt
dt
where Lt  amountof oxygendemandleft aft er t imet
k  t heBOD rat econst ant(t ime-1 )
Solving t hisequat ion yields :
Lt  Lo e  kt
where Lo is t heult imat ecarbonaceous oxygendemand
Conc. (mg/L)
Ultimate BOD
6L
5o
4
3
2
1
0
Lo- Lt
BOD exerted
BODt
L
L remaining
t
0
10
20
time (days)
30
Ultimate Biochemical Oxygen
Demand
Lt = amount of O2 demand left in sample at
time, t
Lo = amount of O2 demand left initially (at time
0, no DO demand has been exerted, so BOD = 0)
At any time, Lo = BODt + Lt (that is the
amount of DO demand used up and the
amount of DO that could be used up
eventually)
Assuming that DO depletion is first order
BODt = Lo(1 - e-kt)
Example
• If the BOD5 of a waste is 102 mg/L and the BOD20
(corresponds to the ultimate BOD) is 158 mg/L,
what is k?

BODt  L0 1  e
 kt
BOD t
1
 e  kt
L0

BOD t
ln
1  L
0



   kt


Example (cont)
 BOD t 

 ln1 
L0 

k
t
 102mg/L 

 ln1 
158mg/L 

k
5 day
k  0.21 day
-1
Biological Oxygen Demand:
Temperature Dependence
• Temperature dependence of biochemical
oxygen demand
As temperature increases, metabolism
increases, utilization of DO also
increases
kt = k20T-20
 = 1.135 if T is between 4 - 20 oC
 = 1.056 if T is between 20 - 30 oC
Example
The BOD rate constant, k, was determined
empirically to be 0.20 days-1 at 20 oC.
What is k if the temperature of the water
increases to 25 oC?
What is k if the temperature of the water
decreases to 10 oC?
Example


k25  0.20 day (1.056)
-1
k25  0.26 day

-1

1020
k10  0.20 day (1.135)
-1
2520
k10  0.056day
-1
Nitrogenous Oxygen Demand
• So far we have dealt only with
carbonaceous demand (demand to oxidize
carbon compounds)
• Many other compounds, such as proteins,
consume oxygen
• Mechanism of reactions are different
Nitrogenous Oxygen Demand
• Nitrification (2 step process)
2 NH3 + 3O2  2 NO2- + 2H+ + 2H2O
2 NO2- + O2  2 NO3– Overall reaction:
NH3 + 2O2  NO3- + H+ + H2O
• Theoretical NBOD =
grams of oxygen used
4 x 16

 4.57 g O 2 /g N
grams of nitrogen oxidized
14
Nitrogenous Oxygen Demand
Nitrogenous oxygen demand
• Untreated domestic wastewater
ultimate-CBOD = 250 - 350 mg/L
ultimate-NBOD = 70 - 230 mg/L
Total Kjeldahl Nitrogen (TKN) = total
concentration of organic and ammonia
nitrogen in wastewater: 15 - 50 mg/L as N
Ultimate NBOD  4.57 x TKN
Other Measures of Oxygen
Demand
Chemical Oxygen Demand
• Chemical oxygen demand - similar to BOD
but is determined by using a strong
oxidizing agent to break down chemical
(rather than bacteria)
• Still determines the equivalent amount of
oxygen that would be consumed
• Value usually about 1.25 times BOD
Water Quality
Management in Rivers
Dissolved Oxygen Depletion
Dissolved Oxygen Sag Curve
Mass Balance Approach
• Originally developed by H.W. Streeter and
E.B. Phelps in 1925
• River described as “plug-flow reactor”
• Mass balance is simplified by selection of
system boundaries
• Oxygen is depleted by BOD exertion
• Oxygen is gained through re-aeration
Steps in Developing the DO Sag
Curve
1. Determine the initial conditions
2. Determine the re-aeration rate from
stream geometry
3. Determine the de-oxygenation rate from
BOD test and stream geometry
4. Calculate the DO deficit as a function of
time
5. Calculate the time and deficit at the
critical point
Selecting System Boundaries
Initial Mixing
Qw = waste flow (m3/s)
DOw = DO in waste (mg/L)
Lw = BOD in waste (mg/L)
Qr = river flow (m3/s)
DOr = DO in river (mg/L)
Lr = BOD in river (mg/L)
Qmix = combined flow (m3/s)
DO = mixed DO (mg/L)
La = mixed BOD (mg/L)
1. Determine Initial Conditions
a.
Initial dissolved oxygen concentration
Qw DOw  Qr DOr
DO 
Qw  Qr
b. Initial dissolved oxygen deficit
D  DOs  DO
where D = DO deficit (mg/L)
DOs = saturation DO conc. (mg/L)
Qw DOw  Qr DOr
Da  DOs 
Qmix
1. Determine Initial Conditions
c.
DOsat is a function of temperature. Values
can be found in Table.
Initial ultimate BOD concentration
Qw Lw  Qr Lr
La 
Qw  Qr
2. Determine Re-aeration Rate
a.
O’Connor-Dobbins correlation
3.9u1/ 2
kr 
h3 / 2
where kr = reaeration coefficient @ 20ºC (day-1)
u = average stream velocity (m/s)
h = average stream depth (m)
b. Correct rate coefficient for stream temperature
T 20
kr  kr ,20
where Θ = 1.024
Determine the De-oxygenation Rate
a.
rate of de-oxygenation = kdLt
where kd = de-oxygenation rate coefficient
(day-1)
Lt = ultimate BOD remaining at
time (of travel downstream) t
b. If kd (stream) = k (BOD test)
Lt  L0e
and
 kd t
rate of deoxygentation  kd L0ekd t
3. Determine the De-oxygenation Rate
c.
However, k = kd only for deep, slow moving
streams. For others,
u
kd  k  
h
where η = bed activity coefficient (0.1 – 0.6)
d. Correct for temperature
T 20
kr  kr ,20
where Θ = 1.135 (4-20ºC) or 1.056 (20-30ºC)
4. DO as function of time
• Mass balance on moving element
dD
 k d Lt  k r D
dt
• Solution is


 
kd La kd t kr t
Dt 
e e
 Da e kr t
kr  kd
5. Calculate Critical time and DO
 kr 
k r  k d 
1

tc 
ln  1  Da
kr  kd  kd 
k d La 


kd La kd tc kr tc
 k r tc
Dc 
e
e
 Da e
kr  ka