Conventional Pollutants in Rivers and Estuaries

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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
2
Principle of Superposition

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

3
Diurnal Variations
4
PHOTOSYNTHESIS CHARACTERISTICS

• Dissolved oxygen deficit mass balance

5
for i124 os(i)-139.344111.575701e5/ta(i)-
6.642308e7/ta(i)21.2438e10/ta(i)3- ...
8.621949e11/ta(i)4 end osexp(os) dos-d ee
norm(dox-d)

• function eeerrordef(x)
• kax(1)
• Pmx(2)
• Dmx(3)
• dox
• tt
• n10
• f13./24
• bn12
• for i110 bn(i)cos(ipif)4pi/f/((pi/f)2-(2p
  i)2ii)
• end
• d124
• Tp1.
• for t124
• d(t)Dm
• for i110
• d(t)d(t)-Pmbn(i)/(ka2(2pii/Tp)2)0.5
  ...
• cos(2pii/Tp(t/24.-fTp/2.)-atan(2pii
  /ka/Tp))
• end

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DYNAMIC APPROACH

• Routing water (St. Venant equations)
• Continuity equation
• Momentum equation (Local acceleration
  Convective accelerationpressure gravity
  friction 0

Kinematic wave
Diffusion wave
Dynamic wave
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KINEMATIC ROUTING

• Geometric slope Friction slope
• Mannings equation
• Express cross section area as a function of flow

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KINEMATIC ROUTING (ctd)

• Express the continuity equation exclusively as a
  function of Q
• Discretize continuity equation and solve it
  numerically

k1
?t
k
1
2
3
4
n
n-1
n
5
?x
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KINEMATIC ROUTING (ctd)

• Discretize continuity equation and solve it
  numerically
• Example
• Q2.5m3s-1 S00.004
• B15m n00.07
• Qe2.52.5sin(wt) w2pi(0.5d)-1

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S00.004 B15 n00.07 n80 Qzeros(2,n)2.5 d
x1000. meters dt700. seconds alpha(n0B(2.
/3.)/sqrt(S0))(3./5.) beta3./5. for it1150
if itdt/24/3600 sin(2.piitdt/(0.5243600)) else
Q(2,1)2.5 end for i2n
Q(2,i)(dt/dxQ(2,i-1) ((Q(1,i)Q(2,i-1))/2.)
(1-beta)... alphabetaQ(1,i))/ (dt/dx
((Q(1,i)Q(2,i-1))/2.)(1-beta)...
alphabeta) end Q(1,)Q(2,) if
floor(it/40)40it x1n
plot(x,Q(1,)) hold on end end
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ROUTING POLLUTANTS

• Mass conservation
• Discretized mass balance equation

k1
?t
k
1
2
3
4
n
n-1
n
5
?x
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ROUTING POLLUTANTS (ctd)

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

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ROUTING POLLUTANTSNumerical Example
u1. m/s dx1000 m dt500 s n100 x1100
yx-20 c0exp(-0.015y.y) c1c0 plot(x,c0) ho
ld on for it1120 for i2n-1
c1(i)c0(i)udt/dx (c0(i-1)-c0(i)) end
c0c1 if fix(it/40)40it
plot(x,c0) end end xlabel('x
(km)') ylabel('C mgL-1')
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ROUTING POLLUTANTS (ctd)

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

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ROUTING POLLUTANTS (ctd)

• Numerical oscillations

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OXYGEN BALANCE GENERAL NUMERICAL APPROACH

• Do spatial discretization
• Route the water for each reach
• Apply water continuity at junctions
• Q8Q15Q16
• Route the pollutants for each reach
• Apply pollutant continuity junctions
• Solve the production/decomposition
• for each grid point

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Reach 1
2
3
9
4
10
11
5
12
6
13
14
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Reach 2
15
8
16
17
18
19
Reach 3
20
21
22
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Sensitivity Analysis

• First order analysis
• yf(x)
• y0f(x0)

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Sensitivity Analysis

• Monte Carlo Analysis
• 1. Generate dx0 N(0,?x)
• 2. Determine yf(x0dx0)
• 3. Save YY y
• 4. ii1
• 5. If i
• 6. Analyze statistically Y

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xspan0100 parameter definition Lrzeros(100,10
1) Drzeros(100,101) global ka kd
U U16.4 y010 0' initial concentrations are
given in mg/L for i1100, ka2.00.3randn
kd0.60.1randn while ka
ka2.00.3randn kd0.60.1randn 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