1. CIVL241: Environmental Engineering
Fall-2012/13
CIVL241: Environmental Engineering

2. Environmental engineering has a long history, although the phrase “environmental engineering” is relatively new.
The roots of environmental engineering reach back to the beginning of civilization. Providing clean water and managing wastes became necessary whenever people congregated in organized settlements.
It was not until the mid-1700s that engineers who built facilities for the civilian population began to distinguish themselves from the engineers primarily engaged in matters of warfare, and the term “civil engineering” was born.

3. From that time on, civil engineers had to more than just provide an adequate supply of water; they now had to make sure the water would not be a vector for disease transmission.
Public health became an integral concern of the civil engineers entrusted with providing water supplies to the population centers, and the elimination of waterborne disease became the major objective in the late 19th century.
The civil engineers entrusted with the drainage of cities and the provision of clean water supplies became public health engineers (in Britain) and sanitary engineers (in the United States).

4. Sanitary engineers have achieved remarkable reductions in the transmission of acute disease by contaminated air or water.
However, more complex and chronic problems such as climate change; depleting aquifers; indoor air pollution; global transport of persistent, bioaccumulating and toxic chemicals; synergistic impacts of complex mixtures of human-made chemicals from household products and pharmaceuticals in wastewater effluents, rivers and streams; endocrine-disrupting chemicals; and a lack of information on the effect on human and environmental health and safety of rapidly emerging new materials, such as nanoparticles.

5. Basic Principles of Environmental Engineering
UNIT-I
Basic Principles of Environmental Engineering

7. The metric system is based on powers of 10 so it is much easier to convert units, often just by moving the decimal point.
2) Once you remember the meaning of the prefixes, you can easily convert mass, volume and distance measurements. No further conversion factors need to be memorized except the specific power of 10. But for the English System you have to remember 5280 ft = 1 mile, 4 quarts = 1 gallon, 3 feet = 1 yard, 16 oz. = 1 lb, etc.
3) The metric system is a international standard used by nearly all the countries of the world. There are actually only *three* countries (i.e. United States, Liberia and Burma) that are still using the English system.

20. Chemical Kinetics

21. Kinetics is the study of how fast chemical reactions occur
Chemical Kinetics
Kinetics is the study of how fast chemical reactions occur
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.

22. A B
time
rate = -
D[A] Dt
rate =
D[B] Dt

23. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time
13.1

24. Factors that Affect Reaction Rate
Temperature
Collision Theory: When two chemicals react, their molecules have to collide with each other with sufficient energy for the reaction to take place.
Kinetic Theory: Increasing temperature means the molecules move faster.
Concentrations of reactants
More reactants mean more collisions if enough energy is present
Catalysts
Speed up reactions by lowering activation energy
Surface area of a solid reactant
Bread and Butter theory: more area for reactants to be in contact
Pressure of gaseous reactants or products
Increased number of collisions

25. Differential Rate Laws
Dependence of reaction rate on the concentrations of reactants is called the rate law, which is unique for each reaction.
For a general reaction, a A + b B + c C  products
the rate law has the general form
order wrt A, B, and C, determined experimentally reaction rate = k [A]X [B]Y [C]Z the rate constant
For example, the rate law is
rate = k [Br-] [BrO3-] [H+] for
5 Br- + BrO H+  3Br2 + 3 H2O
The reaction is 1st order wrt all three reactants, total order 3.
Use differentials to express rates

26. Variation of Reaction rates and Order
2nd order, rate = k [A]2
rate
First order, rate = k [A]
k = rate, 0th order
[A]
[A] = ___?
The variation of reaction rates as functions of concentration for various order is interesting.
Mathematical analysis is an important scientific tool, worth noticing.

27. Differential Rate Law determination
Estimate the orders and rate constant k from the results observed for the reaction? What is the rate when [H2O2] = [I-] = [H+] = 1.0 M?
H2O2 + 3 I- + 2 H+  I H2O
Exprmt [H2O2] [I-] [H+] Initial rate M s e e e e-6
Learn the strategy to determine the rate law from this example.
Figure out the answer without writing down anything.
Solution next

28. Differential Rate Law determination - continue
Estimate the orders from the results observed for the reaction
H2O2 + 3 I- + 2 H+  I H2O
Exprmt [H2O2] [I-] [H+] Initial rate M s e e-6 1 for H2O e-6 1 for I e-6 0 for H+
1.15e-6 = k [H2O2]x [I-]y [H+]z
1.15e-6 k (0.010)x(0.010)y(0.0050)z  exprmt = = e k (0.020)x(0.010)y(0.0050)z  exprmt
x = 1
( )x

29. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16
rate = k [S2O82-]x[I-]y
y = 1
x = 1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k =
rate
[S2O82-][I-] =
2.2 x 10-4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s
13.2

30. First-Order Reactions
rate = -
D[A] Dt
[A] = [A]0e-kt
rate = k [A]
[A] is the concentration of A at any time t
ln[A] - ln[A]0 = - kt
[A]0 is the concentration of A at time t=0
13.3

31. The reaction 2A B is first order in A with a rate constant of 2
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
[A] = [A]0e-kt
[A]0 = 0.88 M
ln[A] - ln[A]0 = - kt
[A] = 0.14 M
ln[A]0 - ln[A] = kt ln
[A]0
[A] k = ln
0.88 M
0.14 M
2.8 x 10-2 s-1 =
ln[A]0 – ln[A] k
t =
= 66 s
13.3

32. First-Order Reactions
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2 ln
[A]0
[A]0/2 k = t½
Ln 2 k =
0.693 k =
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½
Ln 2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)
13.3

34. 1st order reaction calculation
N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.
If the rate-law is known, what are the key parameters?
Solution next

35. 1st order reaction calculation
N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.
Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or = 1.0 e – k t apply [A]o = [A] e– k t ln 0.9 = ln 1.0 – k 30 s – = 0 – k * k = s – t½ = / k = 197 s apply k t ½ = ln 2
[A] = 1.0 e – *500 = Percent decomposed: 1.0 – = or 82.7 %
After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed. After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.
Apply integrated rate law to solve problems

36. Second-Order Reactions
rate = -
D[A] Dt
[A] is the concentration of A at any time t
rate = k [A]2
[A]0 is the concentration of A at time t=0
Half life for second order 1
[A] -
[A]0
= kt
t½ = t when [A] = [A]0/2
t½ = 1
k[A]0

37. [A] - [A]0 = kt Zero-Order Reactions D[A] rate = - rate = k [A]0 = kDt
rate = k [A]0 = k
[A] is the concentration of A at any time t
[A] - [A]0 = kt
[A]0 is the concentration of A at time t=0
Half life for zero order
t½ = t when [A] = [A]0/2
t½ =
[A]0 2k
13.3

39. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] - [A]0 = - kt t½
Ln 2 k = 1
rate = k [A]
ln[A] - ln[A]0 = - kt 1
[A] -
[A]0
= kt
t½ = 1
k[A]0 2
rate = k [A]2
13.3

41. MASS OR MATERIAL BALANCE
The accounting of all mass in a chemical/pharmaceutical process is referred to as a mass (or material) balance.

42. Uses
‘day to day’ operation of process for monitoring operating efficiency
Making calculations for design and development of a process i.e. quantities required, sizing equipment, number of items of equipment

57. Calorimeter, the standard means of measuring the heat energy value of materials when they combust.

61. ECOLOGY
Ecology, the topic of this chapter, is the study of plants, animals, and their physical environment; that is, the study of ecosystems and how energy and materials behave in ecosystems.
DEFINITION OF ECOLOGY
The branch of biology that deals with the relations of organisms to one another and to their physical surroundings.
The study of the interaction of people with their environment.

62. ECOSYSTEM
An ecosystem is a community of living organisms (plants, animals and microbes) in conjunction with the nonliving components of their environment (things like air, water and mineral soil), interacting as a system.
These components are regarded as linked together through nutrient cycles and energy flows.
As ecosystems are defined by the network of interactions among organisms, and between organisms and their environment, they can come in any size but usually encompass specific, limited spaces (although it is sometimes said that the entire planet is an ecosystem).

71. Facultative microorganisms use oxygen when it is available but can use anaerobic reactions if it is not available.

73. ASSIGNMENT-II

74. Water can hold only a limited amount of a gas; the amount of oxygen that can be dissolved in water depends on the water temperature, atmospheric pressure, and the concentration of dissolved solids. The saturation level of oxygen in deionized water at one atmosphere and at various temperatures is shown in Table 8.2.

76. OXYGEN SAG CURVE
The shape of the oxygen sag curve, as shown in Figure 8.10, is the result of adding the rate of oxygen use (consumption) and the rate of supply (re-oxygenation).
If the rate of use is great, as in the stretch of stream immediately after the introduction of organic pollution, the dissolved oxygen level drops because the supply rate cannot keep up with the use of oxygen, creating a deficit.
The deficit (D) is defined as the difference between the oxygen concentration in the stream water (C) and the total amount the water could hold, or saturation (S). That is: