1. AQUATIC WATER QUALITY MODELLING
August 8, 2007
Research Professor Tom Frisk
Pirkanmaa Regional Environment Centre
P.O.Box 297, FIN Tampere, Finland
Phone

2. 3. RIVER MODELS
3.1 Streeter-Phelps model
Streeter and Phelsps presented in 1925 a model for
predicting dissolved oxygen concentration in rivers
The model is based on the assumption that dissolved
oxygen concentration is dependent on two independent
processes: decomposition of organic matter and reaeration.
Both processes are described as first order reactions.
The model is based on plug flow hydraulics.

3. The basic equations of the model can be written as follows:dL
---- = -K1L (3.1) dτ dC
---- = -K1L + K2(Cs - C) (3.2) dτ
where:
L = concentration of organic matter measured as ultimate
BOD (M L-3), τ = travel time (T), K1 = decomposition rate
coefficient of organic matter (T-1), C = dissolved oxygen
concentration (M L-3), K2 = reaeration coefficient (T-1),
Cs = saturation concentration of dissolved oxygen (M L-3)

4. The solution of Eq. (3.1) wit the initial condition
L=L0 when τ=0 is the following:
-K1τ
L = L0 e (3.3)
When Eq. (3.3) is taken into account the following solution is gained to Eq. (3.2) with the initial condition
C=C0 when τ=0:
K1 L K1τ - K2τ K2τ
C = Cs (e e ) - (Cs -C0)e (3.4)
K2 - K1
Reaction rate coefficients K1, K2 and saturation concentration
of dissolved oxygen Cs are dependent on temperature.

5. If the values of the reaction rate coefficients K1 and K2
are equal Eq. (3.4) cannot be used because of division by
zero. The following equation must be used instead:
- K1τ K1τ
C = Cs - K1τ L0 e (Cs -C0)e (3.5)
In the model, organic matter is expressed as ultimate BOD. Ultimate BOD can be estimated on the basis of shorter-term BOD measurements (in Nordic countries BOD7 , in other countries most often BOD5) applying first order kinetics.
BOD7 is oxygen consumption during seven days due to
decomposition of organic matter.

6. Fig. 3.1
(Kylä-
Harakka
1979)
C = Oxygen sag curve
A = contribution of decomposition
B = contribution of reaeration

7. Ultimate BOD (L) is a measure of organic matter indicating the amount of oxygen (mg l-1) which is needed in total decomposition. The concentration of organic matter in the beginning of the BOD test is = L0 and after seven days = L(7).
If we assume first order kinetics we can write:
-K1* ∙7d
L(7) = L0 e (3.6)
where:
K1* = decomposition rate coefficient in lab (T-1)
and

8. -K1* ∙7d
BOD7 = L0 - L(7) = L0(1 – e ) (3.7)
Ultimate BOD can now be calculated on the basis of BOD7:
BOD7
L0 = (3.8)
-K1* ∙7d
1 - e
Eq. (3.8) can also be written as follows:
L0 = f · BOD7 (3.9)
where: f = the stoichiometric coefficient between ultimate
BOD and BOD7.

9. Fig. 3.2. BOD curve. L0 = ultimate BOD, L(7) = BOD which is left
after seven days
(Kylä-Harakka 1979)

10. In the following table the dependence of f on decomposition
rate coefficient (in lab) is presented:
K1* (d-1) f

11. Derivation of Eqs. (3.6) - (3.8) is based on first order
kinetics with no lag phase in the beginning of
decomposition.
In reality, in the BOD test only readily decomposing
organic compounds affect. Thus ultimate BOD can be
defined as the total concentration of readily decomposing
organic matter.
Decomposition rate coefficient in the lab is not the same as
in the river. This is because there are different temperatures
and, in general, the conditions in the test bottles are
different than in natural waters.

12. The saturation concentration of oxygen is dependent on
temperature. E.g. the following third degree polynom
can be used
Cs = T T2
T3 (3.10)
where:
T = temperature (°C)

13. The dependence of reaction rates is according to Streeter
and Phelps (1925) the following:
T - Ts
K(T) = K(Ts) Θ (3.11)
where:
K(T) = reaction rate coefficient at temperature T
Ts = standard temperature (can be selected, most often 20°C)
Θ = an empirical constant, characteristic of each reaction
rate coefficient
For decomposition rate coefficient K1 the value of Θ = ja for the reaeration coefficient K2 the value of Θ = 1,024 is often used.

14. The model is based on PFR hydraulics and travel time
can be calculated as the ratio of distance (x) and velocity
of flow (u):
τ = x/u (3.12)
Example 3.1.
The discharge of the river is = 30 m3 s-1 and the loading
of organic matter as BOD7 = 15 t d-1. The average cross-
sectional area is = 2000 m2. Calculate dissolved oxygen
concentration at the distance of 10 km from the waste-
water pipe at temperature 14°C using the following values
for the coefficients at standard temperature (20°C):
K1 = 0.15 d-1 and K2 = 0.08 d-1.

15. Dissolved oxygen concentration above the wastewater
pipe is = 9.2 mg l-1. It is assumed that BOD7 in the river
water above the wastewater pipe is = 0. The ratio
Ultimate BOD/BOD7 is assumed to be = 1.5.
The mixing concentration of ultimate BOD is first calculated:
L0 = 1.5 *15 t d-1*106 g t-1/(30 m3 s-1* s d-1)
 = 8.68 mg l-1
According to Eq. (3.10) at temperature 14°C
Cs = mg l-1

16. Reaction rate coefficients at temperature 14°C :
14-20
K1(14°C ) = 0.15 d-1 * = d-1
K2(14°C ) = 0.08 d-1 * = d-1
Travel time to the distance of 10 km is
τ = 10· 103 m/(30 m3 s-1· s d-1/2000 m2) = d

17. Introduction to Eq. (3.4) gives
C= mg l-1 -
( d-1 * 8.68 mg l-1/( d-1 – d-1 ) *
d* d d * d-1
(e e ) -
d * d-1
(10.29 mg l-1 – 9.2 mg l-1)e =
( )*( – ) – 1.09 *
= – – = mg l-1 = 5.9 mg l-1

18. 3.2 Additions to the River Model
There are many additions made to the Streeter-Phelps
model, and there are several modifications of the river
models one of the most famous of which is called QUAL II,
existing as many different versions.
The most important additions made to the model are the
following:
Consideration of additional discharge and input (Eqs and 2.17)
Consideration of benthic oxygen demand
Consideration of sedimentation of BOD

19. - Consideration of nitrification
- Consideration of photosynthesis and respiration
- New state variables
- Consideration of dispersion (Eqs and 1.28)
- Inclusion of non-steady hydraulics.
Description of chemical and biological processes will
be presented in Chapters 4 and 5.