Ohio Northern University
CE 3231 - Introduction to Environmental Engineering and Science
Introduction
Chemistry, Microbiology & Material Balance
Water & Air Pollution
Dissolved Oxygen Sag Curve
When organic carbon is placed into a stream environment, the dissolved
oxygen drops following a characteristic pattern. The resulting “sag curve”
is able to be modeled to predict the minimum oxygen concentration
expected and the point along the river stretch where the deficit will occur.
Env Risk Management
Readings for This Class:
5.5-5.6
Lecture 25
DO Sag Curve – Streeter Phelps Model
(Water Quality V)
BOD and DO
 Definition of BOD
 Calculating BODx from equation
DO Sag Curves
Simple Plug Flow River Model
Objective: derive a mathematical model to predict DO downstream: DO(x,t)
Wastewater discharge
(High BOD, Low DO)
Assumptions:
1. completely and uniformly mixed in the cross flow and
vertical direction
2. neglect pollutant dispersion
Formulation: Mass balance
DO replenish from air
A control volume in the river
A
A: cross section area (m2)
u: flow velocity
u
DOin
Δx
DOout
DO consumed due to oxidation
Formulation: Mass balance
Formulation: Mass balance
Formulation: Mass balance
 k d L0

 kd t
 krt
 krt
D 
e
e
 D0e

 kr  kd



Critical Point
• Time/Distance @ DO=Domin
– Minimal occurs when dDO/dt=0
– Taking the sag equation with respect to t and setting
it equal to zero, then solve for t
 kr 
kr  kd  
tc 
ln 
1  D0
 

kr  kd
k d L0  
 kd 
1
Distance downstream where DOmin occurs = tc x u
u: flow speed of the river
Streeter Phelps
solved for Dmax
Streeter Phelps
solved for critical
time
ö
æ k L
-kd t
-kr t
-kr t
d 0
+ D0e ÷÷
e -e
D max = çç
ø
è kr - kd
(
)
 kr 
kr  kd  
tc 
ln 
1  D0
 

kr  kd
k d L0  
 kd 
1
In-class Problem
The initial BOD of a river just below a sewage outfall is 25 mg/L. The oxygen deficit just
upstream from the outfall is 2 mg/L. The deoxygenation rate coefficient kd is 0.4/day, and
the reacation rate coefficient kr is 0.7/day. The river is flowing at a speed of 20 miles/day.
You have been asked to determine the impact of this single source on the river:
a) Find the critical distance downstream at which DO is a minimum
a) Find the minimum DO
 kr 
kr  kd  
tc 
ln 
1  D0
 

kr  kd
k d L0  
 kd 
1
Solution (a)
æ 0.7 æ
1
0.7 - 0.4 öö
tc =
ln ç ç1- 2×
÷÷ = 1.65days
0.7 - 0.4 è 0.4 è
0.4 × 25 øø
dc = tc × u =1.65days× 20miles / day = 33miles
In-class Problem
The initial BOD of a river just below a sewage outfall is 25 mg/L. The oxygen deficit just
upstream from the outfall is 2 mg/L. The deoxygenation rate coefficient kd is 0.4/day, and
the reacation rate coefficient kr is 0.7/day. The river is flowing at a speed of 20 miles/day.
You have been asked to determine the impact of this single source on the river:
a) Find the critical distance downstream at which DO is a minimum
a) Find the minimum DO
ö
æ k L
-kd t
-kr t
-kr t
d 0
+ D0e ÷÷
e -e
D max = çç
ø
è kr - kd
)
(
Solution (b)
æ 0.4 × 25 -0.4×1.65 -0.7×1.65
ö
-0.7×1.65
D max = ç
e
-e
+ 2× e
÷ = 7.36
è 0.7 - 0.4
ø
(
)
DO min = DOsat - Dmax =10 - 7.36 = 2.65mg / L
In-class Problem
The initial BOD of a river just below a sewage outfall is 25 mg/L. The oxygen deficit just
upstream from the outfall is 2 mg/L. The deoxygenation rate coefficient kd is 0.4/day, and
the reacation rate coefficient kr is 0.7/day. The river is flowing at a speed of 20 miles/day.
You have been asked to determine the impact of this single source on the river:
a) Find the critical distance downstream at which DO is a minimum
a) Find the minimum DO
Streeter Phelps
solved for Dmax
Streeter Phelps
solved for critical
time
ö
æ k L
-kd t
-kr t
-kr t
d 0
+ D0e ÷÷
e -e
D max = çç
ø
è kr - kd
(
)
 kr 
kr  kd  
tc 
ln 
1  D0
 

kr  kd
k d L0  
 kd 
1
What happens when there is no initial deficit?
What happens where the organic loading increases?
What happens if the river increases velocity?