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Conventional Pollutants in Rivers and EstuariesPowerPoint Presentation

Conventional Pollutants in Rivers and Estuaries

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Conventional Pollutants in Rivers and Estuaries

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ORGANIC

MATTER

OXYGEN

DECOMPOSITION

(bacteria/animals)

PRODUCTION

(plants)

Chemical

energy

Solar

energy

CARBON

DIOXIDE

INORGANIC

NUTRIENTS

- Mass balance for DO deficit
- In terms of L and N

- Dissolved oxygen deficit mass balance

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=1:10bn(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);

- Routing water (St. Venant equations)
- Continuity equation
- Momentum equation (Local acceleration + Convective acceleration+pressure + gravity + friction = 0

Kinematic wave

Diffusion wave

Dynamic wave

- Geometric slope = Friction slope
- Manning’s equation
- Express cross section area as a function of flow

- 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

- 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

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

- Mass conservation
- Discretized mass balance equation

k+1

t

k

1

2

3

4

5

n

n-1

n

x

- Alternate formulation
- Example
- u = 1 ms-1
- x = 1000 m
- t = 500 m

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}');

- Second order (both time and space) formulation
- Stability condition

- Numerical oscillations

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

- First order analysis
- y=f(x)
- y0=f(x0)

- 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

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