Prince Sattam Bin Abdulaziz University

COLLEGE OF ENGINEERING

CIVIL ENGINEERING DEPARTMENT CE 4471

Water and Wastewater Treatment

Lecture # 5

Sunday

Dr. Feraih Alenazey

f.alenazey@psau.edu.sa

Course Structure

The course is structured in a lecture format. PowerPoint presentations and other audiovisual aids will be used in delivering the course content.

External experts in environmental engineering will be invited to give a lecture on their area of expertise.

Each student will be required to deliver a PowerPoint presentation during the semester relating to water and wastewater treatment.

Filed visits will be organized to visit places that will help the students to explore current environmental issues as well as environmental challenges in our country.

Text book

Ronald L. Droste,

"Theory and Practice of Water and

Wastewater Treatment",

J. Wiley, Last Edition

Course Assessment

Task % Description

Participation in excursion visits & report

5 1 or 2 excursion visits will be organised during

the semester to visit places or organisations in

the field of environmental engineering.

Oral presentations & attendance 5 Each student should present 1 power point

presentation per semester (3 to 6 power point

slides).

2 Mid-Term exams 40 This task will be discussed in the class.

Quizzes & Assignments 10 This task will be discussed in the class.

Final exam 40 This task will be discussed in the class.

What is Wastewater?

The term 'Wastewater' as it applies to the area of water can be defined as 'water that has been used in homes, industries, and businesses that is not for reuse unless it is treated'.

The term 'Wastewater' as it applies to the area of the environment can be defined as 'Water that has been used and contains dissolved or suspended waste materials

Why do we treat water and wastewater?

The main objective of the wastewater treatment is to remove the impurities such as suspended solids, orginice and and pathogenic organisms.

To convert wastewater - which is water no longer suitable for reuse - into an discharge that can be either returned to the water cycle with minimal environmental issues or reused.

Typical municipal wastewater as represented by its solids content.

Total Solids ( 400 –1200 mg/L)

Inorganic (ash)

50% Organic (volatile)

50%

Dissolved inorg. 35%

Suspended (>1 ) 45%

(15% inorg) (30% inorg) Dissol.organic

20%

Non-settleableInorganic

40%

Settleable (>10)

25%(10% inorg)(15% organic)

Non-setteableOrganic

35%

Major Topics in Wastewater Treatment

a) Chemistry

b) Characteristics of wastewater.

C) Water quality and standards.

b) Water and wastewater treatment

processes .

Why Chemistry is important for wastewater

Chemistry is the MOST practical of the sciences.

Chemistry is rooted in materials and their properties.

Types of Chemistry

Chemistry has several different branches of study, each of which supports environmental science in a slightly different way:

Organic Chemistry Inorganic Chemistry Physical Chemistry Analytical Chemistry Biochemistry

Organic Chemistry

• Organic Chemistry is all about compounds containing carbon and a handful of other types of non-metal atoms (O, H, N, S, P, Cl, Br, F).

• Because of the way carbon bonds to form molecules, there are actually more organic molecules than any other kind.

Inorganic Chemistry

Inorganic chemistry studies all of the molecules that aren’t organic. Most metals and metal compounds fall into this group.

Physical Chemistry

• Physical chemistry studies the physical properties of all molecules. Physical properties include things like boiling point, vapor pressure, solubility, interaction with other molecules.

• Knowledge of physical properties helps us to understand the interaction with water and water systems, the interaction with other materials in the water shed. It also gives us clues as to ways to remove the contaminants and to test for the contaminants.

Biochemistry

Biochemistry incorporates much of the other

four sub-disciplines of chemistry but focuses

specifically on the molecules associated with

living organisms. In doing so, it uses

information from organic, inorganic, physical,

and analytical chemistry.

Environmental Chemistry

• Environmental chemistry is a broad subject that, like biochemistry, straddles all aspects of chemistry.

• it focuses on all molecules and interactions within the environment.

Basic Chemistry Concentrations and other units of measure

Environmental engineering is concerned with the quality of environmental fluids as determined by the nature and amount of contaminants within the fluids.

This section defines the most important measures for quantifying the amount species in environmental fluids

Atomic numbers and atomic weights

Molecule = combination of atoms Atom = nucleus + orbiting electrons Nucleus = protons + neutrons Protons have a positive charge; neutrons have no charge Atomic number = number of protons in nucleus Atomic weight = sum of masses of protons and neutrons

in nucleus

Concentration of substance The concept of concentration exists to answer the question:

How much of the “stuff” is there?

Definition: The concentration of a substance is the “amount”of it per “amount” of containing material (air, water, soil).

It can be expressed in various bases : molar, molal, mole fraction or mass

concentration Molar : # of moles / volume of solution Molal : # of moles / 1000g of solution Mole fraction: number of mole / total number of moles of all substances

xi = ni /Σni

(n is the number of mole of i, Σni total number of mole for all substances)

Mass concentration

• Concentration is expressed in terms of part per million(ppm) or mg/L (1 ppm = 1 mg/L)

• Sometimes by: ppb part per billion 1 part in 109 ppt part per trillion 1 part in 1012

where the “part” usually stands for “mole”.

ppm= mass of substance / mass of solution

Examples: If Today’s carbon dioxide concentration in the

atmosphere is reported to be about 390 ppm??

This means that there are 390 moles of CO2 per million

moles of air

Unit conversion

It is often necessary to switch units, for example, to pass from a chemical reaction (in which amounts are most naturally expressed in moles) to a mass budget (in which amounts are most naturally expressed in grams).

Examples 1.1 (page 5)

Unit conversion

Example 2: A water sample contains 20 mg of MgSO4

per liter. Calculate :

(a) Mole concentration in moles/L?

(a)Mole fraction of MgSO4 per liter. (Assume water density 1000 g/ L)?

Unit conversion

Chapter 1, Question 2 , page 31

Chapter 1

Ions, Molecules and Bonding

Molecules are group of atoms bonded together.

Species and products of reactions dissolved in water are incomplete molecules.

Some atoms or groups of atoms(radicals) are able to completely give up or acquire one or more electrons

Charged species are called ions.

Ions, Molecules and Bonding Charged species are called ions. Examples: H+ simple proton OH– hydroxyl or hydroxide ion HCO3 – bicarbonate CO32 – carbonate NH4+ ammonium NO2 – nitrite NO3 – nitrate PO43– phosphate SO42– sulfate

Note that some have positive charges and some negative charges; some have single charges, others double charges.

Oxidation Number The oxidation number of an atom in a compound depends on the

number of electrons that are associated with it.

The oxidation number of an element is the number of electrons that need to be added to the element to make a neutral atom

Thus. It can vary depending on the compound

The increase in oxidation state of an atom through a chemical reaction is known as an oxidation; a decrease in oxidation state is known as a reduction. Such reactions involve the formal transfer of electrons

For pure elements, the oxidation state is zero, So, pure sodium (Na), oxygen (O2) or sulfur (S8) all have zero oxidation numbers.

RULES FOR ASSIGNING OXIDATION NUMBERSPage (6 and 7)

1- The algebraic sum of the oxidation numbers of elements in a compound is zero.

H2SO4

2(+1) + S + 4(-2)= 0 , then S = +6

2- The oxidation number of all elements in the free state zero. identical bonded atoms has no contribution to the oxidation number

3- Alkali metal (first column in the P.T) assumes to have +1 oxidation number in their compound . Alkaline metal (second column in the P.T) assumes to have +2 oxidation number in their compound .

4- The oxidation number of oxygen is -2 except in peroxide, it is -1

5- The oxidation number of hydrogen has an oxidation state of +1, except when bonded to metallic hydrides such as NaH and CaH2 , then it is -1

6- The oxidation state for a pure ion is equivalent to its ionic charge.

Example1

In Na2SO4, what is the oxidation number of sulfur?

The oxidation number of one atom of Na is +1. There are 2 atoms

So, 2(+1) = +2One atom of oxygen is assigned -2. There are 4 atoms

So, 4(-2) = -8Since the positive sum plus the negative sum must equal 0

(+2) + x + (-8) = 0

x = +6 , The sulfur must be using a +6 oxidation state.

Example2

in NCL3, what is the oxidation number of N?

The oxidation number of one atom of Cl is -1. There are 3 atoms of Cl .Thus 3(-1) + x= 0 , x= +3

in NH3, what is the oxidation number of N?Since the positive sum plus the negative sum must equal 0

x + 3(+1) = 0

x = -3

Example 1.2 (page 8)

what are the oxidation numbers of each atom in the following:

a) Na2HPO4,

b)

The oxidation number of one atom of H is +1. There are 6 atoms of H. from rule 1, the 3 carbon atoms must have oxidation number that sum to -6. From rule 2, C-C bonds make no contribution to the oxidation number . The oxidation number of C atom from left to right are: -2, -1 and -3

Exercise in the class

In each of the following examples, indicate the oxidation state of each element present:

1) H2O

2) ZnSO3

3) ZNSO4

4) Br2

5) BH3

Answers

1) H2O oxygen is the most electronegative element. Because it typically has

a charge of -2 , its oxidation state is -2. For the two hydrogen atoms to cancel it out and get a total 0 charge for the whole molecule, they each have to have an oxidation state of +1.

3) ZnSO3

Oxygen has a -2 oxidation state, and since the sulfite ion has a net -2 charge, this requires that sulfur have a +4 oxidation state. Zinc, as is customary for it, has a +2 oxidation state.

4) ZnSO4

It sure looks like #7, but that extra oxygen makes things a little more challenging. Again, oxygen has a -2 oxidation state, but since the sulfate ion has a -2 charge, the sulfur has to have a +6 oxidation state to make everything work out. Zinc, again, does the +2 oxidation state thing.

Answers

Br2

All pure elements have a zero oxidation state, as is the case for bromine here.

BH3

Hydrogen is actually more electronegative than boron, giving it a -1 oxidation state. In order for the boron to cancel out the three negative hydrogens, it must have a +3 oxidation state.

1.4 Balancing Reaction Types of chemical reactions:

1. Synthesis

Generic: A + B → AB

Examples: CaO + H2O → Ca(OH)2 CO2 + H2O→ H2CO3

2. Decomposition

Generic: AB → A + B

Example: H2CO3 → H+ + HCO3–

3. Single-replacement reaction

Generic: A + BC → AC + B Example :

2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)

4. Double-replacement reaction

Generic: AB + CD → AD + CB Example: HCl + NaOH → H2O + NaCl

5. Combustion / Thermal oxidation

Generic: CmHn + … O2 → n CO2 + (n/2)H2O

Example: CH4 + 2 O2 → CO2 + 2 H2O

Stoichiometry Stoichiometry is the application of mass balance to chemical

transformation.

Atoms are conserved, and when combinations of atoms disintegrate, new combinations form with the same atoms.

In other words molecules exchange among themselves bits and pieces, without any net loss or gain.

Example 1

Example: Oxidation of glucose C6H12O6 + (???) O2 → (???) CO2 + (???) H2O

First, equilibrate the C’s and H’s before and after:

C6H12O6 + (???) O2 → 6 CO2 + 6 H2O

Thus, we need 6x2 + 6x1 = 18 O’s on the right; already 6 on left, need 12 more:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Example 1 Use of stoichiometry to make material budgets

Take oxidation of glucose again:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

From this, we note that it takes 6 molecules of oxygen (O2) to oxidize 1 molecule of glucose (C6H12O6).

Thus, 6 moles of oxygen are needed to oxidize 1 mole of glucose.

MW of oxygen = 2x16 = 32 g/mol

MW of glucose = 6x12 + 12x1 + 6x16 = 72 + 12 + 96 = 180 g/mol

Example 1

So, it takes 6x32 = 192 grams of oxygen to oxidize 180 grams of glucose.

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Check values on the right:

Production is 6 moles of carbon dioxide

(CO2) → 6x(12+32) = 6x44 = 264 grams

6 moles of water

(H2O) → 6x(2+16) = 6x18 = 108 grams

Total on right = 264 + 108 = 372 grams while total on left is = 180 + 192 = 372 grams.

Example 2 ( example 1.3 , page 9)

Oxidation-Reduction Reactions Oxidation and reduction (redox) reactions involve transfer of

electrons Most of the reactions and their equations are based upon a

rearrangement of ions between the components of a mixture.

redox reaction. Redox stands for reduction and oxidation.

oxidation is "a reaction in which a substance combines with oxygen , giving up electrons

reduction is a reaction in which a substance receiving electrons

if a substance combined with hydrogen, it was reduced, while if it gave off hydrogen it was oxidized.

Oxidation-Reduction Reactions(https://youtu.be/OE0MMIyMTNU )

Example 1.4 (page 11-12)

Equilibrium (1.6, page 13)

Conductivity (1.7.1 , page 14)

Ionic Strength (1.7.2, page 15)

Example

Chemical kinetics

Example 1.6 (page 19)