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

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

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

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

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

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

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;

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

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

Stream Re-aeration Formulas O’Connor-Dobbins Owens-Edwards-Gibbs H=1-2.5; U=0.1-0.5; 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

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

Estimation of kr and kd in a stream

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

Diurnal Variations

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

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

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, 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

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

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 (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

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); Q(1,:)=Q(2,:); if floor(it/40)*40==it x=1:n; plot(x,Q(1,:)); hold on

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

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}');

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