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Conventional Pollutants in Rivers and Estuaries. ORGANIC MATTER. OXYGEN. DECOMPOSITION (bacteria/animals ). PRODUCTION (plants). Chemical energy. Solar energy. CARBON DIOXIDE. INORGANIC NUTRIENTS. Principle of Superposition. Mass balance for DO deficit In terms of L and N.

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conventional pollutants in rivers and estuaries
Conventional Pollutants in Rivers and Estuaries

ORGANIC

MATTER

OXYGEN

DECOMPOSITION

(bacteria/animals)

PRODUCTION

(plants)

Chemical

energy

Solar

energy

CARBON

DIOXIDE

INORGANIC

NUTRIENTS

principle of superposition
Principle of Superposition
  • Mass balance for DO deficit
  • In terms of L and N
photosynthesis characteristics
PHOTOSYNTHESIS CHARACTERISTICS
  • Dissolved oxygen deficit mass balance
slide5
function ee=errordef(x)

ka=x(1);

Pm=x(2);

Dm=x(3);

dox=[…]

tt=[…]

n=10;

f=13./24;

bn=1:2;

for i=1:10 bn(i)=cos(i*pi*f)*4*pi/f/((pi/f)^2-(2*pi)^2*i*i);

end

d=1:24;

Tp=1.;

for t=1:24

d(t)=Dm;

for i=1:10

d(t)=d(t)-Pm*bn(i)/(ka^2+(2*pi*i/Tp)^2)^0.5* ...

cos(2*pi*i/Tp*(t/24.-f*Tp/2.)-atan(2*pi*i/ka/Tp));

end

end

ta=tt+273.15;

os=ta;

for i=1:24

os(i)=-139.34411+1.575701e5/ta(i)- …

6.642308e7/ta(i)^2+1.2438e10/ta(i)^3- ...

8.621949e11/ta(i)^4;

end

os=exp(os);

d=os-d;

ee=norm(dox-d);

dynamic approach
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

kinematic routing
KINEMATIC ROUTING
  • Geometric slope = Friction slope
  • Manning’s equation
  • Express cross section area as a function of flow
kinematic routing ctd
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

kinematic routing ctd1
KINEMATIC ROUTING (ctd)
  • Discretize continuity equation and solve it numerically
  • Example
    • Q=2.5m3s-1; S0=0.004
    • B=15m; n0=0.07
    • Qe=2.5+2.5sin(wt); w=2pi(0.5d)-1
slide10

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)=2.5+2.5* …

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);

end

Q(1,:)=Q(2,:);

if floor(it/40)*40==it

x=1:n;

plot(x,Q(1,:));

hold on

end

end

routing pollutants
ROUTING POLLUTANTS
  • Mass conservation
  • Discretized mass balance equation

k+1

t

k

1

2

3

4

5

n

n-1

n

x

routing pollutants ctd
ROUTING POLLUTANTS (ctd)
  • Alternate formulation
  • Example
    • u = 1 ms-1
    • x = 1000 m
    • t = 500 m
slide13

ROUTING POLLUTANTSNumerical 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

plot(x,c0);

end

end

xlabel(\'x (km)\');

ylabel(\'C mgL^{-1}\');

routing pollutants ctd1
ROUTING POLLUTANTS (ctd)
  • Second order (both time and space) formulation
  • Stability condition
routing pollutants ctd2
ROUTING POLLUTANTS (ctd)
  • Numerical oscillations
oxygen balance general numerical approach
OXYGEN BALANCE GENERAL NUMERICAL APPROACH

1

  • Do spatial discretization
  • Route the water for each reach
  • Apply water continuity at junctions
    • Q8+Q15=Q16
  • Route the pollutants for each reach
  • Apply pollutant continuity junctions
  • Solve the production/decomposition

for each grid point

Reach

1

2

3

9

4

10

11

5

12

6

13

14

7

Reach

2

15

8

16

17

18

19

Reach

3

20

21

22

sensitivity analysis
Sensitivity Analysis
  • First order analysis
    • y=f(x)
    • y0=f(x0)
sensitivity analysis1
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
slide19

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=2.0+0.3*randn;

kd=0.6+0.1*randn;

while ka < 0 | kr < 0,

ka=2.0+0.3*randn;

kd=0.6+0.1*randn;

end

[x,y] = ODE45(\'dydx_sp\',xspan,y0) ;

Lr(i,:)=y(:,1)\';

Dr(i,:)=y(:,2)\';

end

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),\'--\', …

\'linewidth\',1.25)

ylabel(\'mg L^{-1}\')

title(\'BOD vs. distance\')

subplot 212

plot(x, mean(Dr,1),\'r\',\'linewidth\',1.25);

hold on

plot(x, mean(Dr,1)+std(Dr,0,1),\'r--\', …

\'linewidth\',1.25);

plot(x, mean(Dr,1)-std(Dr,0,1),\'r--\', …

\'linewidth\',1.25);

xlabel(\'Distance (mi)\')

title(\'DO Deficit vs. distance\')

print -djpeg bod_mc.jpeg

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