1. Dr. Revanuru Subramanyam
CIVL241: Environmental Engineering
Summer BY
Dr. Revanuru Subramanyam
Department of Civil and Environmental Engineering
University of Nizwa

2. Chapter: I Basic Principles of Environmental Engineering (9 hours)
Units
Chemical kinetics
Mass balance.
Energy and materials balance in ecosystem
Course Outcome
Explain the units used in environmental engineering.
Use mass balance and energy balance in environmental process

3. 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.

4. 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).

5. 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, bioaccumulation 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.

6. Fundamental Dimension : quantitative description for basic characteristics, such as force (F), mass (M), length (L), and time (T).
Derived Dimensions : calculated by an arithmetic manipulation of one or more fundamental dimensions. Velocity has dimensions of length per time (L/T) and volume is L3.
Dimensions are descriptive but not numerical. They cannot describe how much, they simply describe what. Length described in units as a meter or yard.

7. Engineering Dimensions and Units
Density : is mass divided by unit volume
Where :
ρ : density
M: mass
V: volume
In SI unit density is kg/m3 while in American system is Ib/ft3
Water density is 1000 kg/m3 or 62.4 Ib/ft3

13. CO Percent = 10/106X100= 0.001%
= 10 X 28 X 103/24.45
= µg/m3

20. Quiz
A water treatment plant has 6 settling tanks that operate in parallel. And each tank has a volume of 40 m3. If the flow to the plant is 35 m3/s. what is the retention time in each of the settling tank? If the entire flow goes first through one tank then the second and so on, what would be the retention time in each tank?

21. Chemical Kinetics

22. 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.

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

24. 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

25. 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

26. 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

27. 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.

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

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.
If the rate-law is known, what are the key parameters?
Solution next

36. 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

37. 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

38. [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

40. 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

42. MASS OR MATERIAL BALANCE
Material balance around a "black box" unit operation is introduced first.
A black box is a device, system or object which can be viewed in terms of its inputs and outputs, without any knowledge of its internal workings.
In all cases the flow is assumed to be at steady state, that is, not changing with time.

43. 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

44. MATERIAL BALANCES WITH A SINGLE MATERIAL
Fig.1 shows a black box into which some material is flowing. All flows into the box are called influents and represented by the letter X.
If no processes are going on inside the box that will either make more of the material or destroy some of it and if the flow is assumed not to vary with time (that is, to be at steady, state).
[ mass per unit time] = [ mass per unit time]
IN OUT
[X0]= [ X1]+ [ X2 ]
Generally convenient to use the volume balance for liquids and the mass balance for solids.
Fig.1

45. Splitting Single-Material Flow Streams
Previous figure illustrates that under steady state conditions and no material being produced or destroyed then the material balance :
[ X0]= [ X1]+ [ X2 ]
The material X can, of course, be separated into more than two fractions, so the material balance can be:
where there are n exit streams, or effluents.

48. Combining Single-Material Flow Streams
A black box can also receive numerous influents and discharge one effluent. as shown in Figure 3.4. If the influents are labeled X1, X2, , X„„ the material balance would yield
steady state condition is remaining for combining single material flow streams .

58. Solution: Q0= Qa + Qm = = 800 cfs Q0 XC0= Qa X Ca+ Qm X Cm C0=(340X X1500)/800= mg/l

59. Ex 6) A stream flowing at 10 m3/s has a tributary feeding into it with a flow of 5 m3/s. the stream's concentration of chloride upstream of the junction is 20 mg/L, and the tributary chloride concentration is 40 mg/L. find the downstream chloride concentration. **Tributary : a river or stream that flows into a larger river or a lake Solution Outline black box for problem under study. So the black box is the stream and concentration plus flow rates feeding into the black box from stream and tributary. while the effluent is chloride concentration and stream discharges

60. Ex 7) 18,925 m3/d of wastewater, with a concentration of 10 mg/L of a conservative pollutant is released into a stream having an upstream flow of 37,850 m3/d and pollutant concentration of 3 mg/L. what is the concentration in ppm just downstream? ** conservative : pollutant which doesn’t growth up or consumed inside the stream meaning no reaction coming out due to such pollutant. Solution Upstream and downstream are black box boundary parameters for current problem. So the influent is coming from upstream and effluent out off the black box at downstream.

61. Ex 8) A river with a 400 ppm of salts ( conservative substance) and an upstream flow of 25 m3/s receives an agriculture discharge of 5 m3/s carrying 200 mg/l of salts. The salts quickly become uniformly distributed in the river. A municipality just down -stream withdraws water and mixes it with enough pure water (no salt) from another source to deliever water having no more than 500 ppm salts to its customers. What would be the mixture ratio of pure water to river water?

62. Ex 9) A lagoon with volume of 1200 m3 has been receiving a steady flow of conservative waste at a rate of 100 m3/day for a long time to assume the steady state conditions apply. The waste entering the lagoon has a concentration of 10 mg/l. assuming completely mixed condition.
What would be the concentration of pollutant in the effluent leaving the lagoon?
If the input waste concentration suddenly increased to 100 mg/l, what would be the concentration in the effluent?

68. Change in stored energy= mc T

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

70. A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Bomb calorimeters have to withstand the large pressure within the calorimeter as the reaction is being measured.

74. ECOLOGY
Ecology 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.

75. 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).

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

86. The large-scale use of pesticides for the control of undesired organisms began during World War II with the invention and wide use of the first effective organic pesticide, DDT.
DDT, however, was cheap, lasting, effective, and it did not seem to harm human beings.
Many years later it was discovered that DDT decomposes very slowly, is stored in the fatty tissues of animals, and is readily transferred from one organism to another through the food chain.
As it moves through the food chain, it is biomagnified, or concentrated as the trophic levels increase.

87. Note that the concentration factor from the DDT level in water to the larger birds is about 500,000.
As the result of these very high concentrations of DDT (and subsequently from other chlorinated hydrocarbon pesticides as well), a number of birds were on the verge of extinction
because the DDT affects their calcium metabolism, resulting in the laying of eggs with very thin (and easily broken) shells.
Then, the subtle effects of chemicals such as DDT on human reproduction systems came into focus, and public concern finally forced DDT to be banned when it was discovered that human milk fed to infants often contained four to five times the DDT content allowable for the interstate shipment of cows’ milk!

88. Nutrient Issues in Surface Water
Nitrogen and phosphorus are essential for healthy plant and animal populations, with each water body requiring the right balance of nutrients to maintain aquatic life.
Elevated concentrations of nutrients can lead to excessive, often unsightly, growth of aquatic plants.
Overgrowth of aquatic plants can clog water-intake pipes and filters and can interfere with recreational activities, such as fishing, swimming, and boating. Subsequent decay of aquatic plants can result in foul odors and taste.
Excessive aquatic plant growth reduces dissolved oxygen in water and alters stream habitat, both of which are critical for fish and other aquatic life.
These problems occur locally and in receiving coastal waters where they can threaten fish and shellfish that are economically and ecologically important.

89. 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.

91. 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: