290 likes | 487 Views
Conventional Pollutants in Rivers and Estuaries. ORGANIC MATTER. OXYGEN. DECOMPOSITION (bacteria/animals ). PRODUCTION (plants). Chemical energy. Solar energy. CARBON DIOXIDE. INORGANIC NUTRIENTS. THE DISSOLVED OXYGEN SAG. WWTP. River. DO (mgL -1 ). Critical concentration
E N D
Conventional Pollutants in Rivers and Estuaries ORGANIC MATTER OXYGEN DECOMPOSITION (bacteria/animals) PRODUCTION (plants) Chemical energy Solar 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) • 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; • 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+')
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 kd ka L D 0 ks • Steady state for ultimate BOD • Deficit D=Cs-C • Solution
Principle of Superposition • Mass balance for DO deficit • In terms of L and N
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, 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
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); end Q(1,:)=Q(2,:); if floor(it/40)*40==it x=1:n; plot(x,Q(1,:)); hold on end end
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 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 (ctd) • More accurate (second order both time and space) formulation