Facoltàdi Ingegneria, Università degli Studi di Brescia

NOTES ON THE ENERGY BALANCE IN PHYSICAL LIMNOLOGY AND WATER QUALITY RELATED ISSUES

Prof. Marco Pilotti

Facoltà di Ingegneria, Università

degli Studi di Brescia

PHYSICAL LIMNOLOGY: Heat budget

M. Pilotti - NOTES ON ENGINEERING LIMNOLOGY

Some References

• B. Henderson-Sellers., Engineering limnology, Pitman Advanced Pub. Program,

1984, Boston

• S. Chapra, Surface Water Quality Modeling, Mc Graw Hill, 1997

On this topic and limnology in general master THESIS are available !

PHYSICAL LIMNOLOGY: Heat distribution and water properties


Let us introduce in a pot the same amount of heat from two opposite direction

Will the final T distribution be different ?

If the water is originally quiet,

the hydrostatic equation holds

According to which, surface

where p=C1 are also

characterized by ρ= C2 and

T = C3. The difference is tied

to the stability of the equilibrium.

The final flow field and the heat

distribution are totally different.

The reason is in the equation of state

∫ ⋅S

dSnqrr

0=∇+∇− pzgρ



Water density ρ(p,T,S,s)

P: pressure; T:temperature [°C]; S:dissolved salts; s:suspendend particles

an easy and rough approximation is

That shows that it is primarily a function of T . Accordingly, it is important to introduce the

Thermal Expansivity [K-1]

A first order approximation is

SPSP TT

V

V ,, )1

()1

(∂∂−=

∂∂= ρ

ρα

20 )4(007.01000 CT °−−=ρ

)4(014.01000

1CT °−≅α



For fresh water, Chen_and_Millero (1977) provided an experimental equations with a precision

better than 0.002 Kg/mc in the field of interest of limnology

(0-30 °C, 0-600 ppm, g/kg Total dissolved Solids, 0-180 bars)

ρ0 is the density at the surface

to optain the density at pressure p one applies the correction

to take into account the increase of density when suspended solids are present (C is the

solid concentration in kg/m3)

when sediments are quartz made

( )2654320 lTiThSgTfTeTdTcTbTa +−+−+−+−+=ρ

[ ] ( ) ( )MpLTISHTGTFpETDTCTBTAbar

pp

+−++−+−+−+=

−=

2432

0 1

1)(

εε

ρρ

[ ] CVV

VC

ssss

ss

−=−+−=∆

=

ρρρρρρ

ρ

1)1(

/

C63.0=∆ρ



• The coefficients are:

a := 999.8395; b := 6.7914*1e-02; c := 9.0894*1e-03;

d := 1.0171*1e-04; e := 1.2846*1e-06; f := 1.1592*1e-08;

g := 5.0125*1e-11;

h := 8.181*1e-1; i := 3.85*1e-3; l := 4.96*1e-5;

A := 19652.17; B := 148.113; C := 2.293;

D := 1.256*1e-02; E := 4.18*1e-05; F:= 3.2726;

G := 2.147*1e-4; H := 1.128*1e-04; I := 53.238;

L := 0.313; M := 5.728*1e-3;





• T(z) is possibly the most important physical aspects for lake modelling

• This is a consequence of the EOS of water

• It determines vertical stability, so conditioning vertical dynamics and vertical exchange

of chemical quantities (O2, P,…) and biological activity, which is also conditioned by

light availability

• The evolution into Epilimnion, Metalimnion and Hypolimnion is a consequence of energy

fluxes and thermocline deepening by turbulent mixing and direct absorption of solar

radiation

• Positive energy balance in summer and negative in late summer when convective

cooling start

• Dimictic (2 overturns) and monomictic (1 overturn) lakes

• Polymictic (several times in a year), Oligomictic (rarely), Meromictic (never)



• The other fundamental driver of the lake is wind, that interacts with the T structure

to determine the flow field

Summ

er Winter

Summ

er Winter

Summ

er Winter

Summ

er Winter

Summ

er Winter

Summ

er Winter

Summ

er Winter

Summ

er Winter



• Is Iseo lake getting meromictic ?

PHYSICAL LIMNOLOGY: Heat budget and T(z)


][JTmCQU p∆=∆=∆ Q is the variation of heat content of a mass m and is an

extensive property. Its value depends on the specific heat

Cp [J/(kgK)]. Note the value of Cp below. (1) is the amount

of heat needed to cause a 1 °C rise of 1 m3 of substance

∫∫∫∫∫∫ +⋅+⋅+⋅=+WSS

n

WWW

hdWdSnqdSvdWvgdWvDt

DudW

Dt

D rrrrrr σρρρ 2

2

1

To understand better and to generalize we must start from the I law of thermodynamics for

a control system contained within a volume W limited by a surface S (e.g., a pool of water)

(symbols as in Hydraulics Class: u internal energy for unit mass, v local velocity,

σn local stress associated to direction n, q vector of energy flux along the boundary and h

internal source of thermal energy

U = internal energy ρ ρ ρ ρ Cp λ λ λ λ (1)

kg/m3 J/(kg K)))) W/(m K)))) Jair 1.164 1012 0.0255 1177.968water 998.2 4182 0.604 4174472brick 1800 840 0.6-1.4 1512000cast iron 7272 420 62 3054240

Thermal conductivity



Considering that u when p

is constant is tied through

the specific heat, one can write

that is a simple thermal balance

or the control system in W

∫∫∫∫ +⋅+⋅=∂∂

WSSW

hdWdSnqdSnvuudWt

rrrrρρ

∫ ⋅S

dSnqrr

When the velocity and the power of shear stress are negligible this equation can be simplified

and rewritten (Reynolds’ Theorem) as

∫∫∫∫ +⋅+⋅=∂∂

WSS

p

W

p hdWdSnqdSnvTCTdWCt

rrrrρρ

Rate of change of thermal energy

within the control volume

Flux of thermal energy through the boundary connected to a flux of mass

Flux of heat through the boundary not connected to mass flux; [q]=W/m2

(e.g., through convective transfer, evaporation, solar radiation)

Internal heat source (absorption of shortwave radiation)

∫∂∂

W

pTdWCt

ρ

∫ ⋅S

p dSnvTCrrρ

∫W

hdW

(1)

(2)

(3)

(4)

(5)

(6)



∫ ⋅=S

N dSnqrrφ

• In reality term (5) is made up of different contributions that must be considered

separately. We shall neglect exchange at the bottom and we shall integrate these

contributions on a daily basis.

• This terms can be grouped as Radiation terms and Nonradiation terms

• Radiation terms: Solar Shortwave, Atmospheric Longwave, Lake Longwave

• Nonradiation terms: Conduction and convection, Evaporation and Condensation

• Conduction and convection: Sensible heat

• Evaporation and Condensation: Latent heat

• They can be grouped also depending on their dependence on water T



∫ ⋅=S

N dSnqrrφ

pcerrN φφφφφφφ +−−−+= 120

2rφ1rφeφ

cφ

pφ

: net Short Wave radiation, incident - reflected

: net incoming Long Wave radiation from atmosphere molecules and clouds

: Long Wave radiation loss from the lake

: evaporation (latent heat)

: conduction and convection (sensible heat)

: precipitation

The net flux is made up of

)1(0 SS A−= φφ

PHYSICAL LIMNOLOGY: Heat budget - radiation


]/[),0,(

18

),,(

24

1

521

2

1

mWTTE

dehc

TE kT

hc

εσ

λλπλλ λ

λ

λ

=∞

−=

−

∫

h : Planck’s constant

c : light speed

σ : Stefan Boltzmann constant

(5.67x10-8 W/(m2K4))

ε : body emissivity

PHYSICAL LIMNOLOGY: Heat budget -net Short Wave radiation


The lake is driven by incoming shortwave radiation and by wind

The average daily SW at the top of atmosphere can be

computed in a very precise way as a function of the latitude

and of the period of the year

The average daily SW at the ground depends

on transmission through the atmosphere, and depends

on the ditribution of aerosol and cloud type and cover.

Accordingly, it is very difficult to compute it without

accurate information on the atmosphere

The net SW entering the lake depends on reflectivity As.

Alternatively and relaibly, SW can

be measured with a net radiometer

As a first guess, use As=0.06. However it varies significantly

depending on the height of the sun on the horizon

Sφ

)1(0 SS A−= φφ



The fraction β of SW radiation captured in the first layer

is about 0.4 - 0.5. More realistically, it depends on

water turbidity (sediments, fito- zoo plankton)

Where η is the extinction coefficient, that is a measure

of the rate of absorption of light under the surface

(Beer’s law)

η is seldom measured. Usually it is estimated on the

basis of Sechi disk depth d, on the basis of some

empirical correlations

614.0)ln(265.0 += ηβ

( ) zez ηφβφ −−= 01)(

73.0/1.1

/7.1

d

d

=

=

ηη



PHYSICAL LIMNOLOGY: Heat budget -Long Wave radiation


Due to its temperature, almost all the electromagnetic energy emitted by the lake has a

wave length λ > 4µm (Longwave)

The overall radiative loss of the lake is

σ : Stefan Boltzmann constant (5.67x10-8W/(m2K4))

εw: water emissivity (0.97)

In a similar way, the incoming longwave radiation is the reradiated energy emitted by

clouds and atmosphere

εa: air emissivity

Ta: effective air T

Provided that clouds and aerosols at different height (and so at different T) contribute, the

T of the air is an effective one.

Several empirical relationship have been proposed as a function of relative humidity for

cloudless skies or of Cloud Cover.

41 wwr Tσεφ =

4aar Tσεφ =

PHYSICAL LIMNOLOGY: Heat budget -Evaporative flux


( )[ ] ]/[1000273365.29.2500 kgJTL

EL

wV

Ve

⋅−−== ρφ

Evaporation E [m/s] causes a significative energy loss, as recognised by Dalton (1802), who

observed that it is a function of wind speed and that more water can be evaporated if the

air mass is relatively dry. When the air gets in contact with water, its umidity grows and

its efficiency in absorbing vapor decreases

LV is the Latent Heat of Vaporizati

Several formulas are available for evaluating E, most with structure

esw = saturated vapour pressure at T of water

ea actual vapour pressure at T of air

W = wind speed

]/[))(( smeewfCE asw −⋅=





ea , the actual vapour pressure at T of air is a function of the saturated vapour pressure

at T of air, esa, through the relative humidity Ψ

]/[10171.2)( 2)91.33/(415711 mNeTe wTwsw

−−⋅=

pressurevapoursaturated

pressurevapour=Ψ

PHYSICAL LIMNOLOGY: Heat budget -Sensible heat


Sensibl heat is mostly due to convective removal. It is related to the Latent Heat energy by

Use of the Bowen’s ratio

p is the atmospheric pressure [N/m2].

The order of magnitude of this term is usually less than both φe and the radiative terms.

It can be negative, that is, it can be an energy gain for the lake

asw

awec ee

TTp

−−⋅⋅= −31061.0φφ

PHYSICAL LIMNOLOGY: Heat budget -measure of radiation


PHYSICAL LIMNOLOGY: Heat budget -variability of the contributions during the year


pcerrN φφφφφφφ +−−−+= 120



Let us now suppose that our control system is a completely stirred tank (CST), so that

variables vary only in time. If Cp and ρ are kept constant, Equation (2) can be written as

( ) ( ) ( ) hWqSQTCTQCWTdt

dC outpinpp ++−= ρρρ

( )inp TQCρ

That must be solved along with the mass balance equation, where Q is the volumetric flow

discharge entering or flowing out of the lake ( [Q]=m3/s )

outpTQCρ Outflow of thermal energy connected to the volumetric flow

discharge flowing out of the lake with the temperature of the lake

ERQQdt

dWoutin −+−=

Sources of thermal energy connected to the volumetric flow

discharge entering the lake; [Q]=m3/s

And where R is the rainfall and E the evaporation from the surface. In the following these two

terms will be supposed equal, for simplicity’s sake, and so can be dropped.

qS Energy exchange across the lake’s boundary (surface, sediments)

(7)

(8)

(9)

(10)

(11)



PHYSICAL LIMNOLOGY: Heat budget - example - Lake Iseo


Let us now suppose Iseo lake is at steady state. It is straightforward to obtain from (7) the

equilibrium temperature of the lake

( )( )

( )( ) ( ) ][KT

QC

hWqS

Q

TQ

QC

hWqSTQC

outpout

in

outp

inp =++=++

ρρρ

( ) ( ) ( )( )

( )( )

CQ

TQ

sCmTQ

dtTQTQTQ

out

in

in

CanaleOglio

in

°==

°=

++= ∫

87.95.60

95.596

95.596

3653

365

1

smQ

sm

dtQQ

sm

dtQQ

Out

Canale

Canale

Oglio

Oglio

3

3

365

1

3

365

1

5.60

9.30365

6.29365

=

==

==

∫

∫

Note that here it is unrelevant whether °C or K

are used on both sides. We shall suppose h =0;

S = 61*106 m2



2

365

1 12.130365 m

WdtSWSW == ∫

2

365

1 5.44365 m

WdtLWLW −== ∫

2

365

1

2

365

1

05.46365

54.3365

mWdtLH

LatentHeat

mWdtSH

atSensibleHe

−==

−==

∫

∫

Note that net LW is properly given by

If one uses ε= 0.97 and the time series of water T at the

surface, one can compute the effective Ta of atmosphere;

Ta=278.24 K=5.08 °C

( )365

365

1

44 dtTTLW

aw∫ −−=

εσ



Regarding heat geothermal flux, Livingstone, (1997) computed for Lake Geneve and Zurich

0.07 - 0.13 W/m2. So this contribution can be disregarded.

At steady state the Lake Iseo temperature would be

( )( )

( )[ ]( ) ( )448-448-

644

24.278101.3284-44.4024.278101.3284-45.9187.95.6041822.998

106124.27853.80

87.9 −⋅=−⋅+=⋅⋅⋅−−

+=+

= eqeq

eq

out

pin

eq TT

T

Q

CqS

TQ

T

εσρ

That has solution T = 16.03 °C. Under this condition the whole energy entering from the

lake surface would be taken away by the outflow and lost as LW radiation and Latent Heat.

This result is in contradiction with the volume weighed yearly average Iseo lake that is

T is about 7.4 °C.

Note that equation (7) at equilibrium is not a function of W (if h= 0). Being perfectly stirred,

all the lake is supposed to be at the same T

Iseo lake is not a CSTR (completely stirred tank reactor) and is never at equilibrium

PHYSICAL LIMNOLOGY: Heat budget and T vertical profile


Let us consider a cilindrical lake with area A and adiabatic boundary. The upper

surface is limited by the atmosphere. No other thermal input is present.

Is it possible to determine the evolution in time of the thermal profile T(z) ?

Let us start from equation (2), that we simplify with the same ipothesis that led to

eq. (1)

Now we apply a consequence of mass conservation and of Reynolds Theorem (see

hydraulics class)

But these terms can be rewritten as

∫∫∫ +⋅=WSW

p hdWdSnqTdWCDt

D rrρ

∫∫∫ +⋅=WSW

p hdWdSnqdWDt

DTC

rrρ

∫∫∫ +⋅∇−=

∇+∂∂

WSW

p hdWdSnTdWTvt

TC

rλρ(I1)



Where we have considered the definition of material derivative

and the Fick’s law for molecular diffusion (see tipical values before)

Now let us apply gradient’s theorem (again, see hydraulics class)

And accordingly

Tvt

T

Dt

DT ∇+∂∂=

Tq ∇−= λr

∫∫ ∇=⋅∇−WS

dWTdivdSnT )(λλ r

∫ =

−∇−

∇+∂∂

W

p dWhTdivTvt

TC 0)(λρ

∫∫∫ +∇=

∇+∂∂

WWW

p hdWdWTdivdWTvt

TC )(λρ



Which is clearly equivalent to the PDE

Let us now suppose that λ is constant and let us remember that our system is 1D with

variation only as a function of z. Accordingly v has only the vertical component w (should be 0)

This is a parabolic convection-diffusion PDE that, to be solved, requires the flow field (w)

and appropriate initial condition T0(z) and boundary conditions. The natural boundary

conditions are related to the heat flux for z= 0 and z=Zmax; alternatively, the latter can be

substituted by T(t, Zmax) (T at the bottom) that usually can be regarded as constant.

( )hTdivC

Tvt

T

hTdivTvt

TC

p

p

+∇=∇+∂∂

=−∇−

∇+∂∂

)(1

0)(

λρ

λρ

pp C

h

z

T

Cz

Tw

t

T

ρρλ +

∂∂=

∂∂+

∂∂

2

2

(I2)



Equation (I2) can be integrated numerically to produce T(z) as required.

In order to do it, we must approximate numerically (I2): although in this case this is not a

difficult task we shall pursue a different, more productive road

Equations (I1) and (I2) seems physically equivalent. However, (I1) is a property of a finite

volume and (I2) is a punctual properties (!). From the mathematical point of view they are

not equivalent. E.g., (I2) requires that T(z) belongs to C2 in z and (I1) doesn’ t.

Is it possibly better to solve eq. (I1) ? To decouple the problem from the flow field, let us

suppose that w= 0, as it should be strictly from continuity.

where is the thermal diffusivity [m2/s].

∫∫∫ +⋅∇−=∂∂

W pS pW

dWC

hdSnT

CTdW

t ρρλ r

pCρλ

(I3)



To start with, (I3) can be applied to any of the arbitrary control volumes

into which our original water column is decomposed

The water column is decomposed into arbitrary i control volumes (CV),

which have the upper and lower surface in common with the preceeding

and following CV.

Where we consider the following rule for n

Now the problem is that of computing terms (A), (B) and (C). This will be

done in the next Classwork where equation (I3) will be solved, in order to

determine the T(z,t) profile .pCρ

λ

∫∫∫∫ +

∇−∇−=

∂∂

Wi pSi pSi pWi

dWC

hTdS

CTdS

CTdW

t ρρλ

ρλ

21

(A) (B) (C)

STREETER-PHELPS: Organic production-decomposition

M. Pilotti - NOTES ON WATER QUALITY MODELING

Molecular weightsH 1C 12N 14O 16P 31S 32

Ca 40

Carbon dioxide +

(1)

(2)

STREETER-PHELPS: Organic production-decomposition


6x32 mg of Oxygen are needed in order to decompose (12x6+1*12+16x6) mg of

glucose:

Accordingly, if one has 100 g of Glucose, than 106.67 g of Oxygen are needed for

its consumption ( Note that (6x32)/(106x12+263+110x16+16x14+31)=0.964 )

Let us suppose that one can more easily measure C than glucose. One could say

that in order to decompose (12x6) mg of C-equivalent sugar, one needs 6x32 mg

of Oxygen

( Note that considering the more accurate reaction: (107x32)/(106x12)=2.69 )

More generally, one does not exactly know the composition of pollutants without a

carefull chemical analysis. However one can easily measure experimentally, the

overall mg of Oxygen needed to decompose 1 mg of organic pollutant: it’s the

BOD (Biochemical Oxygen Demand)

Why Oxygen consumption in water ?

mgG

mgOrOG 0667.1

180

192==

mgC

mgOrOC 667.2

72

192==

Molecular weightsH 1C 12N 14O 16

STREETER-PHELPS: Organic production-decomposition: Glucose in a CSTR


Example: 4 mg of glucose within a volume V of 1 liter of water in a closed reactor

+ bacteria+ initial Oxygen concentration O0 of 8 mg/L. Let us study glucose

consumption.

G is the glucose concentration; at t= 0 G=G0=4 mg/L

Let us suppose a linear kinetic with constant k = 0.1 [day -1]

Being the bottle closed, Oxygen will decrease according to the kinetic

kteGGkVGdt

dGV −=−= 0;

)1(

;;;

00

0

ktOG

ktOGOGOG

eGrOO

ekGrdt

dOkGr

dt

dO

dt

dGVr

dt

dOV

−

−

−−=

−=−==

The overall amount of Oxygen

needed is rOG G0= 4.27 mg O2.

What would happen if we had 10

mg of glucose ?

37.0k

1tif 1 ≅== −− ee tk

STREETER-PHELPS: Organic production-decomposition: CBOD in a closed CSTR


However, sewage is not made up simply of sugar. It is not practical to think at a detailed analysys

of its component and a detailed determination of the decompostion rate. Traditionally, a lumped

approach was favoured.

L = amount of Oxidizable matter expressed in terms of Oxygen equivalent [mgO/L], to be

determined experimentally. Given that it might be easy to measure Corg, the organic carbon

concentration, sometimes

BOD(t) = y(t) = L0 -L(t) is the Oxygen consumed.

Accordingly, L0 can be seen either as the initial amount of oxidizable matter or as the final BOD,

BOD∞, when all the organic matter has been consumed.

Note that 0.05 <= k1 <= 0.5 [d-1], but faster for rivers. Accordingly, measuring L0 or BOD∞ would

take too long, so that usually BOD5 is measured and BOD∞ computed as

−=−=

=

=

−=−

−

tk

tk

eLkLkdt

dOeLL

dt

dL

dt

dO

VLkdt

dLV

1

1

011

01 ( )

−−=−==

−

−∞

−

)1(

1;1

11

00

0tk

tktk

eLOO

eBODBODeLL

orgorgOC CCrL 667.2==

( )155 1/ keBODBOD −

∞ −=

STREETER-PHELPS: BOD and flowing water


Let us consider a flowing body of water studied according to the 1-D approximation that we studied

for Open Channel Flow. Let us consider the meaning of the material derivative

0=∂∂+

∂∂=

x

LU

t

L

Dt

DL Water flows without varying the amount of

Oxidizable material

The amount of Oxidizable material decreases

both due to decomposition and due to settling.

LkkDt

DLsd )( +−=

The equation above can be rewritten for Steady State as

Accordingly, if one knows the initial concentration at x =0, that is the junction point between the water

treatment plant and the river coming from upstream

One can easily obtain the L(x) profile along the river

This equation can be used also for evaluating the removal rate, because one has simply to measure

L(x) as a function of distance

Lkkx

LU sd )( +−=

∂∂

RWTP

RRWTPWTP

QQ

QLQLL

++=0

xkk sdeLxL )(0)( +−=

)(ln

1)( 0

xL

L

xkk sd =+

STREETER-PHELPS: BOD and flowing water with reareation


Let us consider a flowing body of water studied according to the 1-D approximation that we studied

for Open Channel Flow. Let us now consider Oxygen reareation. Here we shall make use of the

symbol d/dt but we make reference to the lagrangian meaning D/Dt

LkLkkdt

dLrsd −=+−= )(

The amount of Oxidizable material decreases due to decomposition and

settling

Let us consider Oxygen dynamics: note that O decreases only due to

decomposition (kd in place of kr !); On the other hand we have a

reareation as (e.g. O‘Connor-Dobbins formula, with U [m/s], Y [m])

the problem analitically simplifies if we introduce the Oxygen deficit D; let

us suppose that at station 0, L=L0 and D =D0

)( OOkLkdt

dOSad −+−=

dOdDOOD S −=−= );(

+=+

=

−+=

−=−

−

tkda

tk

ad

r

r

r

eLkDkdt

dDeLL

DkLkdt

dD

Lkdt

dL

0

0

+−−

=

=−−−

−

tktktk

ra

d

tk

aar

r

eDeekk

LktD

eLtL

00

0

)()(

)(

][93.3 15.1

5.0−= d

Y

Uka

Oxygen sag curve


If we keep in mind that the time derivative from which we started are substantial derivative, we can

immediately pass from time to space

If U, the water velocity, is constant, then t=x/U, so that the space distribution of L and D is

Finally, if u=u(x) and U is the space average velocity then x =tU and

But if u=u(x) possibly also the coefficients vary in space so that a numerical solution is advisable

∫=L

udxL

U0

1

+−−

=

=−−−

−

tktktk

ra

d

tk

aar

r

eDeekk

LktD

eLtL

00

0

)()(

)(

+−−

=

=−−−

−

U

xk

U

xk

U

xk

ra

d

U

xk

aar

r

eDeekk

LkxD

eLxL

00

0

)()(

)(

∫= L

udx

xLt

0

+−−

=

=

∫

−∫

−∫

−

∫

−

LaLaLr

Lr

udx

xLk

udx

xLk

udx

xLk

ra

d

udx

xLk

eDeekk

LktD

eLtL

000

0

00

0

)()(

)(



STREETER-PHELPS: effects on the environment of sewage treatment plant effluent


Accordingly, there is a critical station where D is maximum and Oxigen content minimum. This

happens in correspondence of a critical travel time, tc

that can be easily obtained by setting so obtaining

By substituting within D(t) one gets the critical deficit as

The analysis is simplified if D0=0

tktktk

ra

d aar eDeekk

LktD −−− +−

−= 0

0 )()(

( )

−−−

=0

01ln1

Lk

kkD

k

k

kkt

d

ra

r

a

rac

0)( =

dt

tdD

( ) ra

r

kk

k

d

ra

r

a

a

dc Lk

kkD

k

k

k

LkD

−−

−−=0

00 1

ra

r

kk

k

r

a

a

dc

r

a

rac k

k

k

LkD

k

k

kkt

−−

=

−= 0;ln

1




D has been introduced to simplify the equation: D= Os-O. However, Os is a function of several

variables, most noticeably of T. In fresh water at 1 ATM, the relationship holds (APHA, 1992)

Where T is in Kelvin.

4

11

3

10

2

75

s

10*8.621949-

10*1.2438+

10*6.642308-

10*1.575701+-139.34411)ln(O

aaaa TTTT=

Accordingly, when using Streeter Phelps

model, one has to pay attention at junctions,

where an Oxygen balance must be

accomplishedSource river

Q (m3/s) 0.463 5.787T [°C] 28 20DO [mg/L] 2 7.5DO sat. [mg/L] 7.827 9.092DO deficit [mg/L] 5.827 1.592

LmgDOLmgO

LmgO

CT

deficitS /093.7987.8;/987.8)59.20(

/093.7787.5463.0

5.7787.52463.0

59.20787.5463.0

20787.528463.0

22

2

2

−==

=+

⋅+⋅=

°=+

⋅+⋅=



Let us consider a river where geometrical features change with distance from the source and where

a concentrated effluent enters at the upstream end point. What will be the O(x) pattern along the

river ? See Classwork 10

0

200

400

600

800

1000

1200

1400

1600

0 10000 20000 30000 40000 50000 60000 70000 80000

x [m]

A [

km2 ]

NITROGEN: effects on Water Quality


•Effects on Oxygen content: nitrification decreases available Oxygen

•Effects on Eutrophication: availability of primary components

•Nitrate pollution: NO3-in drinking water

•Ammonia toxicity: ammonia gas (NH3) toxicity for fish

NITROGEN: effects on Oxygen Content


• Autotrophic bacteria assimilate the ammonia through nitrification

mgN

mgOr

mgN

mgOr

Oi

Oa

14.114

325.0

43.314

325.1

=⋅=

=⋅=

• accordingly, a more complete assesment of Oxygen consumption must take into account also

Nitrogen nitrification. If we consider only the nitrogen arising from organic matter decay

•The primary consumption is given by 107x32/(106x12) but in addition 16 atoms of N are produced

every 106 of C. Accordingly a more complete figure is

mgC

mgO5.38.069.257.4

12106

1416

12106

32107 =+=⋅⋅⋅+

⋅⋅ So Oxygen consumption due to nitrification accounts

for 0.8/2.69=0.297 of the overall Oxygen

consumption.

mgN

mgOrrr OiOaOn 57.4=+=

NITROGEN: CBOD and NBOD


• Bod can be due both directly to organic matter decomposition (CBOD) and to the following

nitrification (NBOD). However, often N in nitrogen comes from non organic sources. Accordingly,

CBOD and NBOD could have similar values (e.g., 220 mg/L for untreated sewage in develop

countries)

•However, a Streeter Phelps model for NBOD would be unsatisfactory because it is too far from the

actual process that leads to Oxygen consumption due to Nitrogen. The process has a dynamics in

time

•For instance, in a sewage we may have only Organic matter and in such a case nitrification take

longer because first we have to produce Ammonium first.

In another sewage we may have the same amount of nitrogen but directly as Ammonium. In that

case the process is faster.

•Finally, nitrification is strongly conditioned by other cofactors, such as Oxygen concentration or

pH, that can slow down or hinder the process.

•Accordingly, a more detailed representation than a Streeter Phles model is preferred for

nitrification

NITROGEN: CBOD and NBOD


NITROGEN: Modeling Nitrification


•Assuming first order kinetics, the nitrification process can be represented as

And the Oxygen deficit connected to Nitrogen can be represented as

=

−=

−=

−=

iinn

iinaaii

aaiooaa

ooao

Nkdt

dN

NkNkdt

dN

NkNkdt

dN

Nkdt

dN Organic matter (No) decay

Ammonium (Na) production and decay as nitrite

Nitrite (Ni) production and decay as nitrate

Nitrate (Nn) growth

DkNkrNkrdt

dDaiinoiaaioa −+=

Typical values of rates might be kin = 0.75 d-1 , ka = 0.75 d-1 , koa = 0.25 d-1 , kai = 0.25 d-1.

Although the above equations are amenable of analytical solution, a numerical approach is

straightforward and allows to take into account the effect of limiting cofactor. For instance,

the effect of Oxygen deficiency can be taken into account multiplying each of the nitrification

rates kai and kin with f=1-e-0.6O

(See Classwork 11)

M. Pilotti - NOTES ON STREETER-PHELPS MODEL

STREETER-PHELPS: Interconnections

Documents

Facoltàdi Ingegneria, Università degli Studi di Brescia