OXOXYYGGEEN N BALANCEBALANCE

OF RIVERSOF RIVERS

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

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.)

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)

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

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

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!

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

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

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

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

qQ

qCQCC

S

szvh

*)(*)( tDCtC S

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

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)

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

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*

:

ECOLOGICAL IMPACTSECOLOGICAL IMPACTS

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

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

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

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

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:

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

)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

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

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

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

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