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Water and Wastewater Treatment: Lecture #5

This lecture provides an overview of water and wastewater treatment processes, including the importance of chemistry in wastewater treatment. Topics covered include the characteristics of wastewater, water quality and standards, and different branches of chemistry relevant to environmental science.

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Water and Wastewater Treatment: Lecture #5

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  1. Prince Sattam Bin Abdulaziz UniversityCOLLEGE OF ENGINEERINGCIVIL ENGINEERING DEPARTMENT CE 4471Water and Wastewater Treatment Lecture # 5 Sunday Dr. Feraih Alenazey f.alenazey@psau.edu.sa

  2. 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.

  3. Text book Ronald L. Droste, "Theory and Practice of Water and Wastewater Treatment", J. Wiley, Last Edition

  4. Course Assessment

  5. 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

  6. 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.

  7. Typical municipal wastewater as represented by its solids content.

  8. Total Solids ( 400 –1200 mg/L) Inorganic (ash) 50% Organic (volatile) 50% Suspended (>1 m) 45% (15% inorg) (30% inorg) Dissolved inorg. 35% Dissol. organic 20% Settleable (>10m) 25%(10% inorg) (15% organic) Non-settleable Inorganic 40% Non-setteable Organic 35%

  9. Major Topics in Wastewater Treatment a) Chemistry b) Characteristics of wastewater. C) Water quality and standards. b) Water and wastewater treatment processes .

  10. Why Chemistry is important for wastewater Chemistry is the MOST practical of the sciences. Chemistry is rooted in materials and their properties.

  11. 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

  12. 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.

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

  14. 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.

  15. 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.

  16. 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.

  17. 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

  18. 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

  19. 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, Σnitotal number of mole for all substances)

  20. 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

  21. 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)

  22. Unit conversion • Example 2: • A water sample contains 20 mg of MgSO4 per liter. Calculate : • Mole concentration in moles/L? • Mole fraction of MgSO4 per liter. (Assume water density 1000 g/ L)?

  23. Unit conversion Chapter 1, Question 2 , page 31

  24. 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.

  25. 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.

  26. 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.

  27. 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 -2except 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.

  28. 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) = +2 • One atom of oxygen is assigned -2. There are 4 atoms So, 4(-2) = -8 • Since 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.

  29. 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

  30. 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

  31. 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

  32. Answers • 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.

  33. 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.

  34. 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.

  35. 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

  36. 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.

  37. 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

  38. 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

  39. 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.

  40. Example 2 ( example 1.3 , page 9)

  41. 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.

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

  43. Example 1.4 (page 11-12)

  44. Equilibrium (1.6, page 13)

  45. Conductivity (1.7.1 , page 14)

  46. Ionic Strength (1.7.2, page 15)

  47. Example

  48. Chemical kinetics

  49. Example 1.6 (page 19)

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