1. Conventional Pollutants in Rivers and Estuaries
ORGANIC
MATTER
OXYGEN
PRODUCTION
(plants)
DECOMPOSITION
(bacteria/animals)
Solar
energy
Chemical
energy
CARBON
DIOXIDE
INORGANIC
NUTRIENTS

2. THE DISSOLVED OXYGEN SAG
WWTP
River DO
(mgL-1)
Critical concentration
(decomposition=reaeration)
Distance
Decomposition
dominates
Rearetion
dominates

3. BIOCHEMICAL OXYGEN DEMAND
Experiment
decomposition of carbonaceous matter
C6H12O6+6O2 6CO2+6H2O
Mass balance 
General solution g=g0e -k1t
Oxygen mass balance 
General solution for oxygen

4. BIOCHEMICAL OXYGEN DEMAND (ctd.)
L=rog g
L=L0 exp(-k1t)
BOD=L0-L

5. Streeter-Phelps equation
Steady state for ultimate BOD 
Steady state for DO 
Deficit D=Cs-C 
Solutions:
BOD : D:

6. Streeter-Phelps Equation (Example)
L0=10 mg L-1
ka=2.0 d-1
kd=0.6 d-1
D0 = 0 mgL-1
U=16.4 mi d-1

7. Streeter-Phelps Equation (Code)
xspan=0:100;
%parameter definition
global ka kd U
ka=2.0;
kd=0.6;
U=16.4
y0=[10 0]';
%initial concentrations are given in mg/L
[x,y] = ODE45('dydx_sp',xspan,y0) ;
plot(x, y(:,1),'linewidth',1.25)
hold on
plot(x, y(:,2),'r','linewidth',1.25);
do_an=y0(2)*exp(-ka*x/U)+y0(1)*...
kr/(ka-kd)*(exp(-kd*x/U)-exp(-ka*x/U));
ylabel('mg L^{-1}')
xlabel('Distance (mi)')
legend('BOD', 'DO')
plot(x,do_an,'r+')
function dy=dydx_sp(x, y)
global ka kd U
L=y(1);
D=y(2);
dy=y;
dy(1)=-kd/U*L;
dy(2)=+kd/U*L-ka/U*D;

8. Critical Deficit and Distance
Dc , xc  dDc/dxc=0

9. DC’S DEPENDENCE ON
VARIOUS FACTORS
Increases with W=LQ
Increases with T
Increases with D0
Decreases with Q

10. Stream Re-aeration Formulas
O’Connor-Dobbins
Owens-Edwards-Gibbs
H=1-2.5; U= ; Q=4-36
USGS
ka = re-aeration constant, d-1
U = mean stream velocity, ft-1
H = mean stream depth, ft
Q = flow-rate, ft3s-1
t = travel time, d

11. Sedimentation of BOD L D Steady state for ultimate BOD kd ka L D ks
Steady state for ultimate BOD 
Deficit D=Cs-C 
Solution

12. Estimation of kr and kd in a stream

13. Principle of Superposition
Mass balance for DO deficit
In terms of L and N

14. Diurnal Variations

15. Sensitivity Analysis
First order analysis
y=f(x)
y0=f(x0)

16. Sensitivity Analysis Monte Carlo Analysis 1. Generate dx0 = N(0,x)
2. Determine y=f(x0+dx0)
3. Save Y={Y | y}
4. i=i+1
5. If i < imax go to 1
6. Analyze statistically Y

17. xspan=0:100;
%parameter definition
Lr=zeros(100,101);
Dr=zeros(100,101);
global ka kd U
U=16.4;
y0=[10 0]';
%initial concentrations are given in mg/L
for i=1:100,
ka= *randn;
kd= *randn;
while ka < 0 | kr < 0,
end
[x,y] = ODE45('dydx_sp',xspan,y0) ;
Lr(i,:)=y(:,1)';
Dr(i,:)=y(:,2)';
subplot 211
plot(x, mean(Lr,1),'linewidth',1.25)
hold on
plot(x, mean(Lr,1)+std(Lr,0,1),'--', …
'linewidth',1.25)
plot(x, mean(Lr,1)-std(Lr,0,1),'--', …
ylabel('mg L^{-1}')
title('BOD vs. distance')
subplot 212
plot(x, mean(Dr,1),'r','linewidth',1.25);
plot(x, mean(Dr,1)+std(Dr,0,1),'r--', …
'linewidth',1.25);
plot(x, mean(Dr,1)-std(Dr,0,1),'r--', …
xlabel('Distance (mi)')
title('DO Deficit vs. distance')
print -djpeg bod_mc.jpeg

18. DYNAMIC APPROACH Routing water (St. Venant equations)
Continuity equation
Momentum equation (Local acceleration + Convective acceleration+pressure + gravity + friction = 0
Kinematic wave
Diffusion wave
Dynamic wave

19. KINEMATIC ROUTING Geometric slope = Friction slope Manning’s equation
Express cross section area as a function of flow

20. KINEMATIC ROUTING (ctd)
Express the continuity equation exclusively as a function of Q
Discretize continuity equation and solve it numerically
k+1 t k 1 2 3 4 5 n
n-1 n x

21. KINEMATIC ROUTING (ctd)
Discretize continuity equation and solve it numerically
Example
Q=2.5m3s-1; S0=0.004
B=15m; n0=0.07
Qe= sin(wt); w=2pi(0.5d)-1

22. S0=0.004;
B=15;
n0=0.07;
n=80;
Q=zeros(2,n)+2.5;
dx=1000.; %meters
dt=700.; %seconds
alpha=(n0*B^(2./3.)/sqrt(S0))^(3./5.)
beta=3./5.;
for it=1:150
if it*dt/24/3600 < 0.25
Q(2,1)= * …
sin(2.*pi*it*dt/(0.5*24*3600));
else
Q(2,1)=2.5;
end
for i=2:n
Q(2,i)=(dt/dx*Q(2,i-1)* …
((Q(1,i)+Q(2,i-1))/2.)^(1-beta)...
+alpha*beta*Q(1,i))/…
(dt/dx*((Q(1,i)+Q(2,i-1))/2.)^(1-beta)...
+alpha*beta);
Q(1,:)=Q(2,:);
if floor(it/40)*40==it
x=1:n;
plot(x,Q(1,:));
hold on

23. ROUTING POLLUTANTS Mass conservation Discretized mass balance equation
k+1 t k 1 2 3 4 5 n
n-1 n x

24. ROUTING POLLUTANTS (ctd)
Alternate formulation
Example
u = 1 ms-1
x = 1000 m
t = 500 m

25. ROUTING POLLUTANTS Numerical Example
u=1.; %m/s
dx=1000; %m
dt=500; %s
n=100;
x=1:100;
y=x-20;
c0=exp(-0.015*y.*y);
c1=c0;
plot(x,c0);
hold on
for it=1:120
for i=2:n-1
c1(i)=c0(i)+u*dt/dx* …
(c0(i-1)-c0(i));
end
c0=c1;
if fix(it/40)*40==it
xlabel('x (km)');
ylabel('C mgL^{-1}');

26. ROUTING POLLUTANTS (ctd)
More accurate (second order both time and space) formulation