oxygen balance of rivers. balance organic matter (c, n) decay sediment demand respiration...
TRANSCRIPT
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OXOXYYGGEEN N BALANCEBALANCE
OF RIVERSOF RIVERS
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BALANCE
ORGANIC MATTER (C, N) DECAYSEDIMENT DEMAND
RESPIRATION
ATMOSPHERIC DIFFUSION
PHOTOSYNTHESISTRIBUTARIES
V dC/dt = IN – OUT + Diffusion – Organic C Decay – Nitrification – Sediment demand +
Photosynthesis – Respiration ± Tributaries
TRIBUTARIESSOURCES
SINKS
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IMPACTS OF WASTE WATER INLETS
• BOD5 emission is increasing, BOD5 concentration is increasing, dissolved
oxygen (DO) concentration is decreasing
• DO: important indicator element of organic pollution
TYPICAL DO CONCENTRATION VALUES
• Raw waste water: O mg/l
• Saturation concentration in unpolluted water (based on Henry’s Law): ~ 10 mg/l
(at 20 °C )
• Protecting fish reproduction: 6 mg/l
• Different sensitivity of the species and age groups: (e.g. trout: 6-7 mg/l, carp: 4
mg/l)
• Water quality standards: criterias according to different water uses
• Classification: in an integrated way (BOD, COD, DO conc., etc.)
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SIMPLE O2 BALANCE
WASTE WATER
ORGANIC CARBON (BOD)
HETEROTROPHIC BACTERIA
(MINERALIZATION)
O2 DIFFUSION
DISSOLVED O2
CO2
TWO VARIABLES IN THE TRANSPORT EQUATION (BOD AND O2)
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MINERALIZATION OF ORGANIC CARBONMINERALIZATION OF ORGANIC CARBON
Time (days)
O2 consumption (BOD, mg/l)
BOD
5
BOD5
Organic C content (L, mg/l)
Organic carbon content: in term of DO consumption (BOD – Biochemical Oxygen Demand) ~ 2.7 organic C
L – Organic C content = remaining oxygen demand
L, BOD
L0 = BOI
Lkdt
dL1
First order equation
L = L0 exp(- k1 t)
BOD5 = BOD - BOD exp(- k1 5) = BOD (1- exp(- k1 5))
BOD = L0 - L0 exp(- k1 t) = L0 (1 - exp(- k1 t))
)5exp(1
1
15 kBOD
BODf
L0
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DECAY COEFFICIENT (kDECAY COEFFICIENT (k11))Characterization of intensity of mineralization processes, constant
Dimension: 1/day
= 1.04
T
k1
20 °C
1
Validity!
DEPENDENCE ON WASTE WATER TREATMENT TECHNOLOGY
3.20.08Biological treatment2.00.15CEPT technology1.60.2Mechanical treatment1.20.35No treatment
fk1(T=20 °C)Technology
DEPENDENCE ON TEMPERATURE
20)(T20CT11 θk(T)k
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OXYGEN REAERATION (ATMOSPHERIC DIFFUSION)OXYGEN REAERATION (ATMOSPHERIC DIFFUSION)
C < Cs, diffusion from the atmosphere, C approximates Cs
C
Cs – saturation concentration (at a given temperature)
Henry’s Law: p = He Cs
p – partial pressure of the gas
He – Henry number, function of (T, P, ionic content, etc.)
T
Cs
Salt content
7.6309201015
14.60Cs (mg/l)T (°C)
Summer hot periods, heat pollution!
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C
V
h
h
CCAD
dt
dCV s
mol
Dmol: Molecular diffusion coefficient (m2/s)
)( CCAKdt
dCV sL
KL: Oxygen transmission coefficient (m/day)
V
AKkCCk
dt
dC Ls 22 )(
k2: Specific oxygen reaeration coefficient (1/day)
OXYGEN REAERATION (ATMOSPHERIC DIFFUSION) contd.OXYGEN REAERATION (ATMOSPHERIC DIFFUSION) contd.
C – DO concentration
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AFFECTING FACTORSWater depth
Flow properties (velocity, turbulence)
EMPIRICAL FORMULA
)'(93.3)(
5.1
5.0
5.1
5.0
2 DobbinsConnorOH
v
H
vDk x
)(026.567.12 Churchill
H
vk
Validity, dimension (m/s and m)!!!
EPA procedure: k2 0.1 .. 100 1/day (nomogram series)
MEASUREMENT
Local experiments with injection of volatile gas (ethilene, propane, propilene, krypton)
OXYGEN REAERATION COEFFICIENT
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FOR A RIVER SECTIONFOR A RIVER SECTION
Q, v
Lb, Cb q, Lw, Cw
Conditions: permanent flow and emission (Q(t), E(t)=const.), far from the source (1D)
)(2
2** CR
x
CD
x
Cv
dt
Cxx
ORGANIC C
)exp( 101x
x v
xkLLLk
dx
dLv
Or:v
xt * Travel time (travelling with the water), assumption: v(t) = const.
*)exp(* 101 tkLLLk
dt
dL
Calculation of L0 (1D): Instant mixing !!! qQ
qLQLL szvh
0
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DISSOLVED OXYGEN
LkCCkdt
dCorLkCCk
dx
dCv ssx 1212 )(
*)(
D = Cs - C oxygen deficit, assumption: Cs = const.
LkDkdt
dD12*
*)exp(*)exp(*)exp(*)( 2021012
1 tkDtktkLkk
ktD
Q, v
Lb, Cb q, Lw, Cw
FOR A RIVER SECTION contd.FOR A RIVER SECTION contd.
00
0
CCD
qCQCC
S
szvh
*)(*)( tDCtC S
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BOD (L)
x, t*
Lb
L0
DO (C)
x, t*
Cb
C0
Cs
Cmin
xcrit, t*crit
D0
Dmax
Q, v
Lb, Cb q, Lw, Cw
FOR A RIVER SECTION contd.FOR A RIVER SECTION contd.
BOD AND DO PROFILESL
C
Exponential decrease
Oxygen sag
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LOCALIZATION OF THE CRITICAL DISTANCE
LkDkdt
dD12*
012 LkDkMinimum
*)exp(max 102
1krtkL
k
kD
10
120
1
2
12
)(1ln
1*
kL
kkD
k
k
kktkr
0 2
1.5 – 2 days
Role of dilution: L0, D0 Dmax, Cmin !!!
More than one pollution source: superposition (because of the linear basic equations)
Regulation: iterative calculations (efficiency of removal, decay coefficient)
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MORE SOURCESMORE SOURCES
Q, v
Lb, Cb q1, Lw,1, Cw,1
x, t*
x, t*
L
Lb,1
L0,1
CCb,1
C0,1
Cs
Cmin
xcrit, t*crit
D0,1
Dmax
Lb,2
q2, Lw,2, Cw,2
Cb,2
L0,2
D0,2
Superposition
C0,2
L
C
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REAERATION
0
1
2
34
56
7
8
9
0 100 200 300 400 500
Distance (km)
Diss
olve
d ox
ygen
(mg/
l)
STREETER & PHELPS STREETER & PHELPS MODEL MODEL (1925(1925, OHIO RIVER, OHIO RIVER))
BOD DO
EMISSION
Condition for planning: permanent low flow period
Impact assessment of flow dynamics (tcrit*, Dmax)
LkCCkdt
dCDO s 12 )(
*:
Lkdt
dLBOD 1*
:
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ECOLOGICAL IMPACTSECOLOGICAL IMPACTS
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Example: Impact of wastewater discharge on the river Example: Impact of wastewater discharge on the river oxygen concentration oxygen concentration (assumption: 1 D, steady state)(assumption: 1 D, steady state)
Wastewater data: PE 120 000BOD5 : 600 mg/l k1 = 0.42 1/dayKjeldahl N: 120 * 4.57 = 548 mg/lq = 120 000 * 0.1 = 12000 m3/day = 0.14 m3/s
Stream: Background concentration: Lb = 5 mg/l, Cb = 8 mg/lT = 25 C, v = 0.5 m/s, Q = 15 m3/s, Cs = 8.4 mg/lk2 = 0.7 1/nap
Initial concentration:L0 = 16.6 mg/l, D0 = 0.47 mg/l
Critical distance:tcrit = 1.9 nap, xcrit = 82 km
Cmin = 3.6 mg/l
Cmin (mg/l)
0
12
34
56
7
0 100 200 300 400 500 600 700 800Q/q
C
C + N
Dilution effect
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Oldott oxigén szint a kritikus helyen (mg/l)
0
1
2
3
4
5
6
7
Nincs tisztítás Nagyterhelésű
biológiai Kisterhelésű
nitrifikációval Totál oxidáció
Q/q=1000 Q/q=100 Q/q=10
Minimum oxygen contrentation (mg/l)
No treatment High loaded Low loaded Total oxidationactivated sludge activated sludge
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STREETER-PHELPS (1925)
LkCCkdt
dCII s 12 )(
*.
Lkdt
dLI 1*.
EXTENSIONS
1. Separation of dissolved and particulate organic matter fractions
2. Sediment oxygen demand
3. Nitrification
4. Photosynthesis, respiration
DETAILED DESCRIPTION OF DO BALANCEDETAILED DESCRIPTION OF DO BALANCE
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SEPARATION OF DISSOLVED AND PARTICULATE ORGANIC MATTER FRACTIONS
LfkfH
vVLVkALv
dt
dLV dp
sdps )( 11
Lp = fp L particulate (settling due to gravity)
Ld = fd L dissolved (biological decay)
dpdps kkfkf
H
vk 1
'1
tkkLL dp )(exp0
t*
L0 settling
decay
Lkkdt
dLdp )(
L
Extension of DO equation: dC/dt = - kd L
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SEDIMENT OXYGEN DEMANDSEDIMENT OXYGEN DEMANDCAUSES
- Settling particles of the waste water- Dead aquatic animals and plants and leaves at the bottom- Algae settling
IMPACTS OF SEDIMENT HAVING HIGH ORGANIC C CONTENT- Upper part of the sediment: aerob decomposition oxygen abstraction from
pore water hihgh concentration gradient diffusion- Lower part: continuous oxygen lack, anaerob conditions CO2, CH4, H2S
formation- Gas formation rising bubbles, sediment flotation- Aesthetic problems
DESCRIPTIONconstant, area-specific demand – S (g O2 / m2,day)
H
S
dt
dCSA
dt
dCV sed
0.05-0.1 (0.07)River mouth sediments
0.2-1 (0.5)Sandy sediments
1-2 (1.5)Sediments far from the source
2-100 (4)Sediments near the pollution source
S (g O2 / m2,day)Sediment character
Extension of DO equation:
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NITRIFICATIONNITRIFICATION
5 20 days
BOD
LC
LN KJELDAHL-N (Organic N, NH4-N, NO2-N)
Two steps:
Nitrosomonas 2NH4+ + 3O2 2NO2
- + 2H2O + 4H+
Nitrobacter 2NO2- + O2 2NO3
-
3.43 g O2
1.14 g O2
4.57 g O2
LN = 4.57 Kjeldahl-N
CONDITIONSNitrification (obligate aerob, autotrophic) bacteria,Non-acid environment (pH > 6),Presence of oxygen, DO > 1-2 mg/l,Absence of toxic substancesSimplest description: LC+N = LC + LN – integrated BODExtension of DO equation: dC/dt = - k1 LC+N
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)exp(0 tkLLLkdt
dLN
NNNN
N
1
2
SIMPLE (TN)
DETAILED (N forms)
N1 N2 N3
Settling Denitrification
Assimilation by plants
Hydrolysis, Ammonification
Nitrification
OO22
3332233
2221122
1111
NkNkdt
dN
NkNkdt
dN
Nkdt
dN
N1 – organic N,
N2 – NH4-N
N3 – NO2-N, NO3-N
N1 N2 N3
Extension of DO equation: dC/dt = - k23 4.57 N2
NITRIFICATION contd.NITRIFICATION contd.
Extension of DO equation: dC/dt = - kN LNLN = 4.57 TKN
t t t
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PHOTOSYNTHESIS, RESPIRATIONPHOTOSYNTHESIS, RESPIRATION
6CO2 + 6H20 C6H12O6 + 6O2
Light, chlorophyll
PHOTOSYNTHESIS (P, m O2 / m3, day)
RESPIRATION (R, mg O2 / m3, day)
t (h)
P, R
24
t (h)
O2
24
Cs
supersaturation
C
t1 t2
Pa
Pm Daily average oxygen production:
ma Pf
P2
24
12 ttf
photoperiod
aa RPdt
dCExtension of DO equation:
Measuring: method of „dark-light bottle”Calculation: based on the Chl-a content
Respiration of aquatic plants
Ra
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BASIC DIFFERENTIAL EQUATIONSBASIC DIFFERENTIAL EQUATIONS
ORGANIC CARBON DECAY
NITRIFICATION (simple description)
OXYGEN CONCENTRATION
aaN
NC
ds RPH
SLkLkCCk
dt
dCIII )(
*. 2
Cdp
C
Lkkdt
dLI
*.
NN
N
Lkdt
dLII
*.
*)(exp0 tkkLL dpCC
*)exp(0 tkLL NNN
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OXYGEN DEFICIT AND DISSOLVED OXYGEN CONCENTRATIONOXYGEN DEFICIT AND DISSOLVED OXYGEN CONCENTRATION
*)exp(*)( 20 tkDtD INITIAL DEFICIT
*)exp(*)exp( 22
0 tktkkk
kL d
d
dC ORGANIC CARBON DECAY
*)exp(*)exp( 22
0 tktkkk
kL N
N
NN NITRIFICATION
*)exp(1 22
tkk
SSEDIMENT OXYGEN DEMAND
*)exp(1 22
tkk
Pa PHOTOSYNTHESIS
*)exp(1 22
tkk
Ra RESPIRATION
*)(*)( tDCtC s OXYGEN CONCENTRATION
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CALCULATION OF ANAEROB CONDITIONSCALCULATION OF ANAEROB CONDITIONS
LkCCkdt
dCs 12 )(
* Lk
dt
dL1*
High waste loads Temporary or permanent anaerob conditionsAnaerob decay, gas formation, dissolution of metals
C
t*
L
t*
x1
1. Start of anaerob stage: x1 (C ~ 0)
2. Anaerob stage (dC/dt = 0, C = 0):
x1
L1
ss CkCCkdt
dL22 )(
*
v
xxCkLL s
121
3. End of anaerob stage: x2
ss Ck
kLCkLk
dt
dL
1
22221*
s
s
Ck
CkkL
k
vxx
2
211
112
x2
L2
x2
CskLk 21
Linear function