- 2.6 – Water Quality Modelling

(optional)

2-6 WATER QUALITY MODELLING

(optional) Peter Kelderman UNESCO-IHE Institute for Water Education Online Module Water Quality Assessment

Overview

• Some Mathematics (background) • Mass balances • BOD-DO modelling in a river • River Danube calamities model

3

Introduction: general

• What is a model? – a simplified representation of a real system • Answer “why?” questions: – study cause-effect relationships – find out lack of knowledge – gain insight into the system • Answer “what if?” questions: – see effects of "measures" – see effect of changes in conditions – Also used as part of monitoring (complicated processes, calamities,..) 4

Dimension of a model

• Looks at the length co-ordinates x, y z: • If x, y, z are not included in the model  0-D model – Example: pollutant in a completely mixed lake, or: • One length co-ordinate: 1-D model – Example: BOD/DO model along river length (x) • 2-D and 3-D models: two or three length co-ordinates – E.g. “Delft 3-D model” for pollutant concentration in the North Sea at different locations + water depths  very complicated !

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1D model

6

3D model- salinity in an estuary

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Basic terms

• Input data: – geometry input data (“model”) – boundary conditions: values of unknowns at the boundaries of the spatial domain covered by the model – initial conditions: values of the unknowns at the start of the simulation (t=0) – external variables or “forcing functions”: values of quantities affecting the model results, which are not predicted by the model. Their values are taken from outside (

e.g.

meteorology data) – model parameters: acceleration of gravity, bottom friction, decay rate, … 8

Some remarks

• A model can only be as good as the ideas/assumptions behind it: “Garbage in  garbage out” • All essential processes should be included; don’t be over impressed by many sophisticated models  often lot of “window dressing ”. (Example: P models for shallow lakes that ignore P sediment-water exchange  maybe completely wrong results!) • But, keep the model as simple as possible  simple is beautiful • In understanding/evaluating models in practice, look for the theory behind the models; • And look at the real water system behavior!

• The proof of a good model is the agreement between the model and the real situation (calibration/validation).

9

Balances and differential equations

• In (Water quality) modelling, balances (mass, energy, momentum) are commonly used: • Mass balance: “change/sec”

dMass



Mass in



Mass out



processes dt E.g.

:In a lake, if 10 kg/s of a pollutant come in, 5 kg/s go out and 3 kg/s go the bottom  change = +2 kg/s • So modelling works with solving differential equations(exactly or as an approximation (numerical solutions))!

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Simple decay calcultation

In words : “decrease/sec is proportional to amount present” (If radioactivity decreases with 20 g/s for 100 kg, then it will decrease with 10*20=200 g/s for 10*100 kg = 1000 kg) 

dM dt



kM (Note that both sides are positive!)

k = decay rate (1/d) M = mass (g) t = time (d) • Used for many processes in water quality modelling,

e.g.

: • Decay of BOD • Radioactive decay • Dieing of bacteria (

e.g.

E-coli) • • Mortality of algae

etc.

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Decay formulation:

Simple decay calculation (2) 

dM dt



kM Separating variables:

 1

M dM



kdt Take the definite integral on both sides: Will lead to:



M M

 0 1

M dM



t k dt

0  ln  

M M

0    

M M

0   

e



kt Will finally lead to:



M e

 0

kt

12

1 0.8

0.6

0.4

0.2

0 0

Decay Rate

Slow rate (low k value) High rate (high k value) Exercise: Check that it wil take 6.9 days for the Mass to reach 50% of its original value M 0 , for k = 0.1 day -1 

M e

 0

kt

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Micropolllutants in Rivers

W = 500 kg/day Q = 10 (m 3 /s) x Input chlorobenzene concentration = 580 μg/L (check yourself) (rapid) Degradation by volatilization (loss to atmosphere) ; k v = 8.5 day -1 700 Similar to BOD degradation: C poll.

= C 0 exp (-k v t) (t = travel time in river) (13 hours to reduce to 1% of C 0 ) 600 500 400 300 200 100 0 0.2

0.4

Tim e (days)

0.6

0.8

14 1

The essence of water quality modelling!

volume V discharge Q in concentration C in discharge Q out concentration C out reaction rate k

How are the concentrations changing, in space and in time?

15

BOD – DO in a river

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Oxygen Concentration pattern BOD load oxygen River axis • Major WQ problem, already investigated in 19 th century • Major mathematical description: Streeter-Phelps model (1925) 17

Major Processes • Sources of DO:

• Re-aereation of oxygen from the atmosphere • Oxygen production/consumption by plants, algae

• Sinks of DO:

• Oxidation of BOD

• Sediment oxygen demand • Oxygen consumption by respiration of algae and plants 18

Oxidation of BOD (1)

• Assume CBOD and NBOD (NH 4 + degradation rate.

 NO 3 ) have same • According to first order reaction:

dL

 

k L

1

dt

L = BOD (g/m 3 ) t = time (day) k 1 = BOD degradation rate constant (day -1 ) Solution:

L



L e

 0

Time (days)

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Oxidation of BOD (2)

• • Degradation rate constant k 1 • k 1 generally between 0.1 for BOD : – 0.4 (day -1 ) Dependent on

e.g

. type of organic material, river characteristics, wastewater purification, time after discharge • Increase of about 4% per o C k = 0.2

k = 0.4

Tim e (days)

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Re-aeration of Oxygen

• Driving force: Difference between O 2 saturation O 2 concentration and concentration, in the water:

dc dt



k

2 

c sat



c

 k 2 = re-aeration constant (day -1 ) c = Oxygen concentration(g/m 3 ) c sat = Saturation oxygen concentration(g/m 3 ) air water 21

Re-aeration of Oxygen (2)

• Solution: c = c – {(c sat – c 0 ) exp(-k 2 t)} (If you like: try yourself) k = 0.6

k = 1 Sat

tim e (days)

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Re-aeration in rivers

• Value k 2 dependent on: • Velocity river : k 2 proportional to about V 0.5

• Water depth : k 2 proportional to about 1/H • Temperature: about 2% increase per 0 C 1.5

Empirical graph from different UK USA reseaches Example: • Velocity = 0.6 ft/sec = 0.18 m/sec • Depth = 4.5 ft = 1.4 m • 

k 2 = 1.0 day -1

( see Thomann and M üller (1987)) 23

1 foot= 0.3 m 24

Sediment Oxygen demand

Bottom type and location

Sphaerotilus

(10 g dry wt/m 2 ) Municipal sewage sludge – outfall vicinity Municipal sewage sludge –“aged”, downstream outfall Estuarine mud Sandy bottom Mineral soils

SOD (g/m 2 /day) Range Average

7 2-10 1-2 1-2 0.2-1.0

0.05-0.1

4 1.5

1.5

0.5

0.07

(see Thomann and M üller (1987)) • Especially important for organic-rich sediments • Increase about 7% per o C • SOD can be neglected for deep rivers 25

Streeter-Phelps DO model for a river

BOD U (m/s) V dx x – Ignore SOD – Ignore effect algae and water plants – Ignore differences between CBOD and NBOD degradation • Mass Balance for oxygen over segment dx:

V dc



rear



oxidation



dt

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Model equation

dc



k

2 (

c s dt



c

) 

k

1

L

0 exp( 

k

1

x

)

U



U dc dx Re aeration BOD decay inflow outflow

• (L 0 is the BOD concentration in the river after mixing)

•

• BOD decay term expresses: • Decrease in BOD that has already taken place between the discharge point and the segment V (L = L 0 exp(-k 1 x/U)); and the BOD decay in segment V itself: dL/dt = -k 1 L Last term: “plug-flow” equation 27

Solution

• Assume steady state conditions: dc/dt =0  solution:

c



c s



k

2

k

1

L

0 

k

1  exp( 

k

1

t

)  exp( 

k

2

t

)   (

c s



c

0 ) exp( 

k

2

t

) t=travel time river, equivalent with “distance” D c = “Critical” DO deficit” (see Thomann and M üller (1987)) 28

“Critical distance”

• “Critical distance" x* c or travel time t* c time where the minimum O 2 = x* c /U, i.e. the distance/travel concentration in the river is reached. By differentiation; dc/dx = 0:

t

*

c



k

2 1 

k

1 ln   

k

2

k

1   1  (

c s



c

0 )(

k

2

k

1

L

0 

k

1 )      (k 2  k 1 ; for equal values assume a small difference, say 0.01, in the calculations).

For c s = c o :

t

*

c



k

2 1 

k

1 ln  

k

2

k

1   Example: for k 1 = 0.3 day -1 and k 2 =0.8 day -1 , t c * = 1/0.5 (ln 8/3) = 2.0 days 29

Minimal Oxygen concentration

C s



C

min .



k k

2 1

L

0

e



k

1

t c

•  “oxygen deficit” roughly proportional to BOD loading L 0 !

(see Thomann and M üller (1987) 30

Characteristics river/BOD load: • k 1 = 0.3 day -1

Calculation Example

C s



C

min .



k k

2 1

L

0

e



k

1

t c

• k 2 = 0.9 day -1 • L 0 = 30 mg/L (BOD conc. in river after mixing) • c s = 8.0 mg/L (DO saturation concentration) • t c = 2.5 days Minimum DO will be: 3.3 mg O 2 /L Till what value should L 0 be reduced to reach DO standard = 5.0 mg/L?

(Answer: L 0 < 19 mg/L) 31

BOD-DO models in practice

• Often much more complicated: • A wide variety of point and non-point sources...

• Different tributaries and river branches, all with different k 1 and k 2 values ..

• Take into account SOD, effect algae, differences between CBOD and NnBOD.......

•  BOD-DO models such as “SOBEK”, or “Mike11 Ecolab” 32

Numerical modelling

• Streeter-Phelps model could mathematically be solved exactly • For more complicated equations, it will be impossible to find exact solutions of the differential equations  “numerical solutions ” • Approximation dC/dt step Δt’’  ΔC/Δt :”change of C during (small) time • With this kind of numerical modelling, the trend in C can then be followed (time)step by step. See

e.g.

Euler method 33

Example: Black River USA

• Main river with two tributaries • River system was divided into 59 segments, all with their specific values for „input parameters" K-CBOD, K NBOD, K 2 , K-SOD • Euler method was used to model DO over the whole river 34

Calibration and Verification

– Calibration of a model: to fit the model with the real situation/data, by “trying out” (realistic !) values of input parameters k

etc.

1 , k 2 , k SOD , Calibration Verification

Not ok Ok

– Verification/validation: to check that the model then also works for another, independent data set 35

Calibration and Verification (2) • Black River BOD-DO model: • Calibration with the help of data set August 14, 1973 (:“very good!”) • Verification with data set November 1, 1973 (:“unusually good result!”) • Finally the model can then be applied; in this case for finding out effect of dam construction 36

Modelling accidental spills

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Accidental spills

• Short event of intensive pollution, due to an accident; examples: – Sandoz, Rhine, 1986: some 20 tonnes of pesticides were spilt into the river Rhine, due to extinguishing water for fire at Sandoz factory  massive fish kills – Baia Mare, Danube, 2000 (see later) • accidental spills are important, because they attract a lot of publicity and concern in the society • often the starting point of action plans to improve pollution situation • Modelling: the peaks of discharge will gradually get wider (non ideal river plug-flow due to back-currents, shallow/deep, etc.) 38

Sandoz, measured concentration downstream

ref: Dietrich (2008), lecture notes Sandoz Accident, University of Konstanz, Germany http://www.umwelttoxikologie.uni-konstanz.de/Lehre/Lecture__Environ__Tox__I_/02_29042008OkotoxIBS4Sem.pdf

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Accidental spill model (basic version)

• 1-dimensional • advection diffusion equation 

Qc



DA



c



x

• Analytical (exact) solution, for a mass M of pollutants released at t=0 at x=0 (Taylor formula): 

A M

4 

Dt

exp    

x



Ut

 2 4

Dt

• Q, U = river discharge/flow; D = “mixing; diffusion” parameter 40

Case Study: Danube River

The Danube River is 2857 km long and the basin covers 817,000 square km in 18 countries in the heart of central Europe. The basin is characterised by large socio-economical differences. It stretches out from rich Western-European states to some relatively poor former Soviet Union Republics. The river has a number of very large tributaries. The Danube water is used extensively by the 85 million inhabitants of the basin. The basin includes many important natural areas, including the Danube delta - the second largest wetland area in Europe. 41

Accidental Spill

42

Cyanide Spill in Danube river

• Source: mining waste in Rumania • 100,000 m 3 wastewater release due to breaking dike in mine • Much damage to Danubecosystem • Danube Basin Alarm Model (DBAM) was used to predict (maximum) levels of cyanide in the river, travel time, restoration..

• No cyanide decay assumed over time scale used  “worst case scenario” (is often done in modelling).

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Results from “hindcasting” application

some extremes missed and time differences ” : “ quite ok, but 12 10 8 6 4 2 0 31-jan 2-feb 4-feb 6-feb Balsa computed Kiskore computed Tiszasziget computed 8-feb 10-feb Balsa observed Kiskore observed Tiszasziget observed 12-feb 14-feb 44