AQUATIC WATER QUALITY MODELLING

Slides:



Advertisements
Similar presentations
Equilibrium 1994A Teddy Ku A MgF 2(s) Mg 2+ (aq) + 2F - (aq) In a saturated solution of MgF 2 at 18 degrees Celsius, the concentration of Mg 2+
Advertisements

Introduction to Environmental Engineering Lecture 15 Water Supply and Groundwater.
Benno Rahardyan FTSL-ITB
Chemical kinetics: a branch of chemistry which deal about the rate of reaction. Rate of reaction: The change in concentration of reactant or product.
Sewage and Effluent Treatment 2-4 November 2002 Seán Moran -The first few slides.
Stone media TF design Example 1
Biological waste water treatment
Lecture 13: Introduction to Environmental Engineering
ACTIVATED SLUDGE PROCESS AND KINETICS OF ASP
Chapter 14.  Equilibrium occurs when there is a constant ratio between the concentration of the reactants and the products. Different reactions have.
Watershed & Water Quality Modeling Technical Support Center WASP7 Course Dissolved Oxygen Processes Processes and Equations Implemented in WASP7 Eutrophication.
CEE Fall, 2007 CEE 5134 Deoxygenation – Reaeration and the The Streeter-Phelps Equation Thomas J. Grizzard 25 October, 2007.
Chapter 13 Chemical Kinetics
The Ultimate BOD A much better nickname. Example of a BOD determination 200 mL of waste water was collected, aerated and seeded with bacteria. The dissolved.
CHEMISTRY 121/122 Solubility Equilibrium. What is a solution?  A solution is a mixture in which a solid has been dissolved into a liquid, usually water.
Human Influence on Ecosystems. Effects of Pesticides on Ecosystems Rachel Carson Silent Spring Birth of the Environmental Movement.
CE Introduction to Environmental Engineering and Science
Water Quality Management in Rivers
WASTEWATER. Wastewater -used water (from human activity) -contains pollutants.
Fermentation Kinetics of Yeast Growth and Production
Q 6. How to measure BOD? 1. Measure the volume of a sample of water in a dark bottle. Measure its oxygen level by oxygen meter / chemical method. 2. Incubate.
DO NOW 1.Get out your calendar 2.Put your Water Filtration Lab in the bin 3.List and explain the steps of water filtration.
ISOTHERMAL REACTOR DESIGN
Scheme of the equilibrium Environmental Compartments Model.
Temperature and DO Temperature  A measure of heat Dissolved Oxygen (DO)  The concentration of oxygen (gas) which is dissolved in water. Both are important.
Types of Water Pollution Sewage Disease-causing agents Sediment pollution Inorganic plant and algal nutrients Organic compounds Inorganic chemicals Thermal.
Environmental Modeling Steven I. Gordon Ohio Supercomputer Center June, 2004.
AMBIENT AIR CONCENTRATION MODELING Types of Pollutant Sources Point Sources e.g., stacks or vents Area Sources e.g., landfills, ponds, storage piles Volume.
Environmental Modeling Chapter 7: Dissolved Oxygen Sag Curves in Streams Copyright © 2006 by DBS.
BsysE595 Lecture Basic modeling approaches for engineering systems – Summary and Review Shulin Chen January 10, 2013.
CE 548 Introduction to Process Analysis and Selection
AQUATIC WATER QUALITY MODELLING August 8, 2007 Research Professor Tom Frisk Pirkanmaa Regional Environment Centre P.O.Box 297, FIN Tampere, Finland.
Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp.
Modeling Chemical Systems BIOE Chemical Systems Every physiologic systems depends on multiple chemical reactions Examples include hormone-receptor.
- Equilibrium Constant - 1.  Please READ pp. 439 – 440  Equilibrium constant (K eq ) is the value obtained from the mathematical combination of equilibrium.
© 2014 Carl Lund, all rights reserved A First Course on Kinetics and Reaction Engineering Class 30.
Chapter 13 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there.
OXYGEN BALANCE OF RIVERS. BALANCE ORGANIC MATTER (C, N) DECAY SEDIMENT DEMAND RESPIRATION ATMOSPHERIC DIFFUSION PHOTOSYNTHESIS TRIBUTARIES V dC/dt = IN.
Convection: Internal Flow ( )
Urban Water Global water aspects
CE Introduction to Environmental Engineering and Science Readings for This Class: O hio N orthern U niversity Introduction Chemistry, Microbiology.
Synthetic UH Definition: Synthetic Hydrograph is a plot of flow versus time and generated based on a minimal use of streamflow data. Example: A pending.
An Introduction To Modeling of Surface Waters For TPDES Permits Mark A. Rudolph, P.E. TCEQ Water Quality Division.
9 th Grade Science Chapter 24 Section 1. Chemical Changes  Chemical change: Atoms change bonds and form new materials  Chemical reaction: Process of.
SEWAGE CHARACTERISTICS. Composition >99.0% Water Solids 70% Organic 30% Inorganic Sewerage characteristics can be divided into three broad categories:-
Biochemical Oxygen Demand (BOD) and Dissolved Oxygen (DO) 1. Background Information.
Section 8.2—Equilibrium Constant How can we describe a reaction at equilibrium?
© 2015 Carl Lund, all rights reserved A First Course on Kinetics and Reaction Engineering Class 37.
Temperature. What is Temperature? Temperature is the measure of the average kinetic energy in a system. May be measured in Celsius (°C) or Fahrenheit.
1 Chapter 6 Time Delays Time delays occur due to: 1.Fluid flow in a pipe 2.Transport of solid material (e.g., conveyor belt) 3.Chemical analysis -Sampling.
Happy Days video (2:30).  Consider the balanced redox reaction of potassium manganate(VII) with ammonium iron(II) sulfate. 5Fe 2+ + MnO H.
Chapter 6, Section 2 Describing Chemical Reactions.
- 2.2 – ORGANIC MATTER (Diederik Rousseau UNESCO-IHE Institute for Water Education Online Module Water Quality Assessment 2.
Pages , Sections 18.1, 18.2, and 18.4 (excluding , Section 18.3)
Dissolved oxygen (DO) in the streams
CHAPTER 2 MASS BALANCE and APPLICATION
Chapter 13 Chemical Kinetics CHEMISTRY. Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of.
Sanitary Engineering Lecture 4
In  Out  Generation or Removal  Accumulation
Oxygen Sag Curve By- Prajyoti P. Upganlawar
Conventional Pollutants in Rivers and Estuaries
Time Delays Chapter 6 Time delays occur due to: Fluid flow in a pipe
Section 1 Review What constitutes a reversible reaction?
Redox Titrations.
Introduction If nontoxic organic pollutants get discharged into a river, lake or stream, they should be pretty harmless, right?
Chapter 6: Chemistry in Biology
Find: Dc mg L at 20 C [mg/L] Water Body Q [m3/s] T [C] BOD5 DO
Example A city of 200,000 people discharges 37.0 cfs of treated sewage having an ultimate BOD of 28.0 mg/L and 1.8 mg/L DO into a river with a flow of.
Dissolved Oxygen Processes
Equilibrium Law & the Equilibrium Constant
Presentation transcript:

AQUATIC WATER QUALITY MODELLING August 8, 2007 Research Professor Tom Frisk Pirkanmaa Regional Environment Centre P.O.Box 297, FIN-33101 Tampere, Finland E-mail tom.frisk@ymparisto.fi Phone +358 500 739 991

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.

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)

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 L0 - 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.

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.

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

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

-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.

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

In the following table the dependence of f on decomposition rate coefficient (in lab) is presented: K1* (d-1) f 0.05 3.39 0.1 1.9 0.15 1.4 0.2 1.33 0.3 1.14 0.4 1.06

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.

The saturation concentration of oxygen is dependent on temperature. E.g. the following third degree polynom can be used Cs = 14.6221 - 0.407699 T + 0.00812934 T2 - 0.0000794448 T3 (3.10) where: T = temperature (°C)

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 = 1.047 ja for the reaeration coefficient K2 the value of Θ = 1,024 is often used.

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.

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* 86400 s d-1)  = 8.68 mg l-1 According to Eq. (3.10) at temperature 14°C   Cs = 10.29 mg l-1

Reaction rate coefficients at temperature 14°C : 14-20 K1(14°C ) = 0.15 d-1 *1.047 = 0.1139 d-1 K2(14°C ) = 0.08 d-1 *1.024 = 0.0694 d-1 Travel time to the distance of 10 km is τ = 10· 103 m/(30 m3 s-1· 86400 s d-1/2000 m2) = 7.716 d

Introduction to Eq. (3.4) gives C= 10.29 mg l-1 - (0.1139 d-1 * 8.68 mg l-1/( 0.0694 d-1 – 0.1139d-1 ) * -7.716 d*0.1139 d-1 -7.716 d * 0.0694 d-1 (e - e ) - -7.716 d * 0.0694 d-1 (10.29 mg l-1 – 9.2 mg l-1)e = 10.29 + (-22.217)*(0.4152 – 0.5854) – 1.09 * 0.5854 = 10.29 – 3.781 – 0.6381 = 5.871 mg l-1 = 5.9 mg l-1

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. 1.24 and 2.17) Consideration of benthic oxygen demand Consideration of sedimentation of BOD

- Consideration of nitrification - Consideration of photosynthesis and respiration - New state variables - Consideration of dispersion (Eqs. 1.27 and 1.28) - Inclusion of non-steady hydraulics. Description of chemical and biological processes will be presented in Chapters 4 and 5.