Background Chemistry and Fluid Mechanics As mentioned, this chapter discusses the background knowledge needed in order to understand the subsequent chapters of this book. The student must have already gained this knowledge, but it is presented here as a refresher. Again, the background knowledge to be reviewed includes chemistry and fluid mechanics. Before they are discussed, however, units used in calculations need to be addressed, first. This is important because confusion may arise if there is no technique used to decipher the units used in a calculation. UNITS USED IN CALCULATIONS In calculations, several factors may be involved in a term and it is important to keep tract of the units of each of the factors in order to ascertain the final unit of the term. For example, consider converting 88 kilograms to micrograms. To make this con- version, several factors are present in the term for the calculation. Suppose we make the conversion as follows: (1) where 88(1000)(1000)(1000) is called the term of the calculation and 88 and the 1000s are the factors of the term. As can be seen, it is quite confusing what each of the 1000s refers to. To make the calculation more tractable, it may be made by putting the units in each of the factors. Thus, (2) This second method makes the calculations more tractable, but it makes the writing long, cumbersome, and impractical when several pages of calculations are done. For example, in designs, the length of the calculations can add up to the thickness of a book. Thus, in design calculations, the first method is preferable with its attendant possible confusion of the units. Realizing its simplicity, however, we must create some method to make it tractable. This is how it is done. Focus on the right-hand side of the equation of the first calculation. It is known that the unit of 88 is kg, and we want it converted to µ g . Remember that conversions follow a sequence of units. For example, to convert 88 kg 88 1000()1000()1000()88,000,000,000 µ g== 88 kg 88 kg 1000 g kg   1000 mg g   1000 µ g mg   88,000,000,000 µ g== TX249_Frame_C00 Page 25 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero 26 Physical–Chemical Treatment of Water and Wastewater kg to µ g, the sequence might be any one of the following: (3) (4) In Sequence (3), the conversion follows the detailed steps : first, from kg to g, then from g to mg and, finally, from mg to µ g. Looking back to Equation (1), this is the sequence followed in the conversion. The first 1000 then refers to the g; the second 1000 refers to the mg; and the last 1000 refers to the µ g. Note that in this scheme, the value of a given unit is exactly the equivalent of the previous unit. For example, the value 1000 for the g unit is exactly the equivalent of the previous unit, which is the kg. Also, the value 1000 for the mg unit is exactly the equivalent of its previous unit, which is the g, and so on with the µ g. In Sequence (4), the method of conversion is a short cut . This is done if the number of micrograms in a kilogram is known. Of course, we know that there are 10 6 micrograms in a kilogram. Thus, using this sequence to convert 88 kg to micrograms, we proceed as follows: Example 1 Convert 88,000,000,000 µ g to kg using the detailed-step and the short-cut methods. Solution: Detailed step: Note that the first (10 − 3 ) refers to the mg; the second (10 − 3 ) refers to the g; and the last (10 − 3 ) refers to the kg. There is no need to write the units specifically. Short cut: Note that there are 10 − 6 kg in one µ g. Example 2 Convert 10 cfs to m 3 /d. Solution: cfs is cubic feet per second. 1 s = 1 / 60 min; 1 min = 1 / 60 hr; 1 hr = 1 / 24 d Therefore, kggmg µ g⇒⇒ ⇒ kg µ g⇒ 88 kg 88 10 6 ()88,000,000,000 µ g== 88,000,000,000 µ g 88,000,000,000 10 3– ()10 3– ()10 3– ()88 kg Ans== 88,000,000,000 µ g 88,000,000,000 10 6– ()88 kg Ans== 1ft 1/3.281 m= 10 cfs 10 ft 3 /s 10 1 3.281   3 1 1 60   1 60   1 24   24,462 m 3 /d Ans== = TX249_Frame_C00 Page 26 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero Background Chemistry and Fluid Mechanics 27 Now, let us turn to a more elaborate problem of substituting into an equation. Of course, to use the equation, all the units of its parameters must be known. Accordingly, when the substitution is done, these units must be satisfied. For exam- ple, take the following equation: (5) It is impossible to use the above equation if the units of the factors are not known. Thus, any equation, whether empirically or analytically derived, must always have its units known. In the above equation, the following are the units of the factors: I amperes, A [ C o ] gram equivalents per liter, geq/m 3 Q o cubic meters per second, m 3 /s η dimensionless m dimensionless Now, with all the units known, it is easy to substitute the values into the equation; the proper unit of the answer will simply fall into place. The values can be substituted into the equation in two ways: direct substitution and indirect substitution . Direct substitution means substituting the values directly into the equation and making the conversion into proper units while already substi- tuted. Indirect substitution, on the other hand, means converting the values into their proper units outside the equation before inserting them into the equation. These methods will be elaborated in the next example. Example 3 In the equation I = , the following values for the factors are given: [ C o ] = 4000 mg/L of NaCl; Q o = 378.51 m 3 / d ; η = 0.77, m = 400, and M = 0.90. Calculate the value of I by indirect substitution and by direct substi- tution. Solution: Indirect substitution: In indirect substitution, all terms must be in their proper units before making the substitution, thus: NaCl = 23 + 35.45 = 58.45, molecular mass of sodium chloride Therefore, I 96,494 C o []Q o η m/2() M = 96,494[C o ]Q o η (m/2) M C o [] 4000 mg/L 4000 10 3– () 58.45 0.068 gmols/L=== 0.068 geq/L 68 geq/m 3 == Q o 378.51 m 3 /d 378.51 1 24 60()60() 0.0044 m 3 /s== = TX249_Frame_C00 Page 27 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero 28 Physical–Chemical Treatment of Water and Wastewater And, now, substituting, therefore, Direct substitution: Note that the conversions into proper units are done inside the equation, and that the conversions for [C o ] and Q o are inside pairs of braces {}. Once accustomed to viewing these conversions, you may not need these braces anymore. One last method of ascertaining units in a calculation is the use of consistent units. If a system of units is used consistently, then it is not necessary to keep track of the units in a given calculation. The proper unit of the answer will automatically fall into place. The system of units is based upon the general dimensions of space, mass, and time. Space may be in terms of displacement or volume and mass may be in terms of absolute mass or relative mass. An example of absolute mass is the gram, and an example of relative mass is the mole. The mole is a relative mass, because it expresses the ratio of the absolute mass to the molecular mass of the substance. When the word mass is used without qualification, absolute mass is intended. The following are examples of systems of units: meter-kilogram-second (mks), meter-gram-second (mgs), liter-gram-second (lgs), centimeter-gram-second (cgs), liter-grammoles-second (lgmols), meter-kilogrammoles-second (mkmols), centi- meter-grammoles-second (cgmols), etc. Any equation that is derived analytically does not need to have its units specified, because the units will automatically conform to the general dimension of space, mass, and time. In other words, the units are automatically specified by the system of units chosen. For example, if the mks system of units is chosen, then the measurement of distance is in units of meters, the measurement of mass is in units of kilograms, and the measurement of time is in seconds. Also, if the lgs system of units is used, then the volume is in liters, the mass is in grams, and the time is in seconds. To repeat, if consistent units are used, it is not necessary to keep track of the units of the various factors, because these units will automatically fall into place by virtue of the choice of the system of units. The use of a consistent system of units is illustrated in the next example. I 96,494 C o []Q o η m/2() M 96,494 68()0.0044()0.77()400/2() 0.90 == 122.84 A Ans= I 96,494 C o []Q o η m/2() M = 96,494 4000 10 3– () 58.45 1000()    378.51 1 24 60()60()    0.77()/400/2() 0.90 = 122.84 A Ans= TX249_Frame_C00 Page 28 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero Background Chemistry and Fluid Mechanics 29 Example 4 The formula used to calculate the amount of acid needed to lower the pH of water is Calculate the amount of acid needed using the lgmols system of units. Solution: Of course, to intelligently use the above equation, all the factors should be explained. We do not need to do it here, however, because we only need to make substitutions. Because the lgmols units is to be used, volume is in liters, mass is in gram moles, and time is in seconds. Therefore, the corresponding con- centration is in gram moles per liter (gmols/L). Another unit of measurement of concentration is also used in this equation, and this is equivalents per liter. For the lgmols system, this will be gram equivalents per liter (geq/L). Now, values for the factors need to be given. These are shown below and note that no units are given. Because the lgmols system is used, they are understood to be either geq/L or gmols/L. Again, it is not necessary to keep track of the units; they are understood from the system of units used. Therefore, Note that the values are just freely substituted without worrying about the units. By the system of units used, the unit for [A cadd ] geq is automatically geq/L. GENERAL CHEMISTRY Chemistry is a very wide field; however, only a very small portion, indeed, of this seemingly complex subject is used in this book. These include equivalents and equivalent mass, methods of expressing concentrations, activity and active concen- tration, equilibrium and solubility product constants, and acids and bases. This knowledge of chemistry will be used under the unit processes part of this book. EQUIVALENTS AND EQUIVALENT MASSES The literature shows confused definitions of equivalents and equivalent masses and no universal definition exists. They are defined based on specific situations and are never unified. For example, in water chemistry, three methods of defining equivalent mass are used: equivalent mass based on ionic charge, equivalent mass based on A cadd [] geq A cur [] geq 10 pH– to 10 pHcur– – φ += A cur [] geq 2.74 10 3– () pH to 8.7 pH cur 10.0=== A cadd [] geq 2.74 10 3– () 10 8.7– 10 10– – 0.00422 + 2.74 10 3– () geq/L Ans== TX249_Frame_C00 Page 29 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero 30 Physical–Chemical Treatment of Water and Wastewater acid–base reactions, and equivalent mass based on oxidation–reduction reactions (Snoeyink and Jenkins, 1980). This section will unify the definition of these terms by utilizing the concept of reference species; but, before the definition is unified, the aforementioned three methods will be discussed first. The result of the discussion, then, will form the basis of the unification. Equivalent mass based on ionic charge. In this method, the equivalent mass is defined as (Snoeyink and Jenkins, 1980): (6) For example, consider the reaction, Calculate the equivalent mass of Fe(HCO 3 ) 2 . When this species ionizes, the Fe will form a charge of plus 2 and the bicarbonate ion will form a charge of minus 1 but, because it has a subscript of 2, the total ionic charge is minus 2. Thus, from the previous formula, the equivalent mass is Fe(HCO 3 ) 2 /2, where Fe(HCO 3 ) 2 must be evaluated from the respective atomic masses. The positive ionic charge for calcium hydroxide is 2. The negative ionic charge of OH − is 1; but, because the subscript is 2, the total negative ionic charge for calcium hydroxide is also 2. Thus, if the equivalent mass of Ca(OH) 2 were to be found, it would be Ca(OH) 2 /2. It will be mentioned later that Ca(OH) 2 /2 is not compatible with Fe(HCO 3 ) 2 /2 and, therefore, Equation (6) is not of universal appli- cation, because it ought to apply to all species participating in a chemical reaction. Instead, it only applies to Fe(HCO 3 ) 2 but not to Ca(OH) 2 , as will be shown later. Equivalent mass based on acid–base reactions. In this method, the equivalent mass is defined as (Snoeyink and Jenkins, 1980): (7) where n is the number of hydrogen or hydroxyl ions that react in a molecule. For example, consider the reaction, Now, calculate the equivalent mass of H 3 PO 4 . It can be observed that H 3 PO 4 converts to . Thus, two hydrogen ions are in the molecule of H 3 PO 4 that react and the equivalent mass of H 3 PO 4 is H 3 PO 4 /2, using the previous equation. The number of hydroxyl ions that react in NaOH is one; thus, using the previous equation, the equivalent mass of NaOH is NaOH/1. Equivalent mass molecular weight ionic charge = Fe HCO 3 () 2 2Ca OH() 2 + Fe OH() 2 2CaCO 3 2H 2 O++→ Equivalent mass molecular weight n = H 3 PO 4 2NaOH 2Na + HPO 4 2− 2H 2 O++→+ HPO 4 2− TX249_Frame_C00 Page 30 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero Background Chemistry and Fluid Mechanics 31 Equivalent mass based on oxidation–reduction reactions. For oxidation– reduction reactions, the equivalent mass is defined as the mass of substance per mole of electrons involved (Snoeyink and Jenkins, 1980). For example, consider the reaction, The ferrous is oxidized to the ferric form from an oxidation state of +2 to +3. The difference between 2 and 3 is 1, and because there are 4 atoms of Fe, the amount of electrons involved is 1 × 4 = 4. The equivalent mass of Fe(OH) 2 is then 4Fe(OH) 2 /4. Note that the coefficient 4 has been included in the calculation. This is so, because in order to get the total number of electrons involved, the coefficient must be included. The electrons involved are not only for the electrons in a molecule but for the electrons in all the molecules of the balanced chemical reaction, and, therefore must account for the coefficient of the term. For oxygen, the number of moles electrons involved will also be found to be 4; thus, the equivalent mass of oxygen is O 2 /4. Now, we are going to unify this equivalence using the concept of the reference species. The positive or negative charges, the hydrogen or hydroxyl ions, and the moles of electrons used in the above methods of calculation are actually references species. They are used as references in calculating the equivalent mass. Note that the hydrogen ion is actually a positive charge and the hydroxyl ion is actually a negative charge. From the results of the previous three methods of calculating equivalent mass, we can make the following generalizations: 1. The mass of any substance participating in a reaction per unit of the number of reference species is called the equivalent mass of the substance, and, it follows that 2. The mass of the substance divided by this equivalent mass is the number of equivalents of the substance. The expression molecular weight/ionic charge is actually mass of the substance per unit of the reference species, where ionic charge is the reference species. The expression molecular weight n is also actually mass of the substance per unit of the reference species. In the case of the method based on the oxidation–reduction reaction, no equation was developed; however, the ratios used in the example are ratios of the masses of the respective substances to the reference species, where 4, the number of electrons, is the number of reference species. From the discussion above, the reference species can only be one of two possi- bilities: the electrons involved in an oxidation–reduction reaction and the positive (or, alternatively, the negative) charges in all the other reactions. These species (electrons and the positive or negative charges) express the combining capacity or valence of the substance. The various examples that follow will embody the concept of the reference species. Again, take the following reactions, which were used for the illustrations above: 4Fe(OH) 2 + O 2 + 2H 2 O → 4Fe(OH) 3 Fe(HCO 2 ) 3 + 2Ca(OH) 2 → Fe(OH) 2 + 2CaCO 3 + 2H 2 O 4Fe OH() 2 O 2 2H 2 O 4Fe OH() 3 →++ TX249_Frame_C00 Page 31 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero In the first reaction, the ferrous form is oxidized to the ferric form from an oxidation state of +2 to +3. Thus, in this reaction, electrons are involved, making them the reference species. The difference between 2 and 3 is 1, and since there are 4 atoms of Fe, the amount of electrons involved is 1 × 4 = 4. For the oxygen molecule, its atom has been reduced from 0 to −2 per atom. Since there are 2 oxygen atoms in the molecule, the total number of electrons involved is also equal to 4 (i.e., 2 × 2 = 4). In both these cases, the number of electrons involved is 4. This number is called the number of reference species, combining capacity, or valence. (Number of reference species will be used in this book.) Thus, to obtain the equivalent masses of all the participating substances in the reaction, each term must be divided by 4: 4Fe(OH) 2 /4, O 2 /4, 2H 2 O/4, and 4Fe(OH) 3 /4. We had the same results for Fe(OH) 2 and O 2 obtained before. If the total number of electrons involved in the case of the oxygen atom were different, a problem would have arisen. Thus, if this situation occurs, take the convention of using the smaller of the number of electrons involved as the number of reference species. For example if the number of electrons involved in the case of oxygen were 2, then all the participating substances in the chemical reaction would have been divided by 2 rather than 4. For any given chemical reaction or series of related chemical reactions, however, whatever value of the reference species is chosen, the answer will still be the same, provided this number is used consistently. This situation of two competing values to choose from is illustrated in the second reaction to be addressed below. Also, take note that the reference species is to be taken from the reactants only, not from the products. This is so, because the reactants are the ones responsible for the initiation of the interaction and, thus, the initiation of the equivalence of the species in the chemical reaction. In the second reaction, no electrons are involved. In this case, take the convention that if no electrons are involved, either consider the positive or, alternatively, the negative oxidation states as the reference species. For this reaction, initially consider the positive oxidation state. Since the ferrous iron has a charge of +2, ferrous bicarbonate has 2 for its number of reference species. Alternatively, consider the negative charge of bicarbonate. The charge of the bicarbonate ion is −1 and because two bicarbonates are in ferrous bicarbonate, the number of reference species is, again, 1 × 2 = 2. From these analyses, we adopt 2 as the number of reference species for the reaction, subject to a possible modification as shown in the paragraph below. (Notice that this is the number of reference species for the whole reaction, not only for the individual term in the reaction. In other words, all terms and each term in a chemical reaction must use the same number for the reference species.) In the case of the calcium hydroxide, since calcium has a charge of +2 and the coefficient of the term is 2, the number of reference species is 4. Thus, we have now two possible values for the same reaction. In this situation, there are two alternatives: the 2 or the 4 as the number of reference species. As mentioned previously, either can be used provided, when one is chosen, all subsequent calculations are based on the one particular choice; however, adopt the convention wherein the number of reference species to be chosen should be the smallest value. Thus, the number of reference species in the second reaction is 2, not 4—and all the equivalent masses of the participating substances are obtained by dividing each balanced term of the reaction by 2: Fe(HCO 3 ) 2 /2, 2Ca(OH) 2 /2, Fe(OH) 2 /2, 2CaCO 3 /2 and 2H 2 O/2. TX249_Frame_C00 Page 32 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero Note that Ca(OH) 2 has now an equivalent mass of 2Ca(OH) 2 /2 which is different from Ca(OH) 2 /2 obtained before. This means that the definition of equivalent mass in Equation (6) is not accurate, because it does not apply to Ca(OH) 2 . The equivalent mass of Ca(OH) 2 /2 is not compatible with Fe(HCO 3 ) 2 /2, Fe(OH) 2 /2, 2CaCO 3 /2, or 2H 2 O/2. Compatibility means that the species in the chemical reaction can all be related to each other in a calculation; but, because the equivalent mass of Ca(OH) 2 is now made incompatible, it could no longer be related to the other species in the reaction in any chemical calculation. In contrast, the method of reference species that is developed here applies in a unified fashion to all species in the chemical reaction: Fe(HCO 3 ) 2 , Ca(OH) 2 , Fe(OH) 2 , CaCO 3 , and H 2 O and the resulting equiv- alent masses are therefore compatible to each other. This is so, because all the species are using the same number of reference species. Take the two reactions of phosphoric acid with sodium hydroxide that follow. These reactions will illustrate that the equivalent mass of a given substance depends upon the chemical reaction in which the substance is involved. Consider the positive electric charge and the first reaction. Because Na + of NaOH (remember that only the reactants are to be considered in choosing the reference species) has a charge of +1 and the coefficient of the term is 2, two positive charges (2 × 1 = 2) are involved. In the case of H 3 PO 4 , the equation shows that the acid breaks up into and other substances with one H still “clinging” to the PO 4 on the right-hand side of the equation. This indicates that two H + ’s are involved in the breakup. Because the charge of H + is +1, two positive charges are accordingly involved. In both the cases of Na + and H + , the reference species are the two positive charges and the number of reference species is 2. Therefore, in the first reaction, the equivalent mass of a participating substance is obtained by dividing the term (including the coefficient) by 2. Thus, for the acid, the equivalent mass is H 3 PO 4 /2; for the base, the equivalent mass is 2NaOH/2, etc. In the second reaction, again, basing on the positive electric charges and per- forming similar analysis, the number of reference species would be found to be +3. Thus, in this reaction, for the acid, the equivalent mass is H 3 PO 4 /3; for the base, the equivalent mass is 3NaOH/3, etc., indicating differences in equivalent masses with the first reaction for the same substances of H 3 PO 4 and NaOH. Thus, the equivalent mass of any substance depends upon the chemical reaction in which it participates. In the previous discussions, the unit of the number of reference species was not established. A convenient unit would be the mole (i.e., mole of electrons or mole of positive or negative charges). The mole can be a milligram-mole, gram-mole, etc. The mass unit of measurement of the equivalent mass would then correspond to the type of mole used for the reference species. For example, if the mole used is the gram-mole, the mass of the equivalent mass would be expressed in grams of the substance per gram-mole of the reference species; and, if the mole used is the milligram-mole, H 3 PO 4 2NaOH 2Na + HPO 4 2− 2H 2 O++→+ H 3 PO 4 3NaOH 3Na + PO 4 3− 3H 2 O++→+ HPO 4 2− TX249_Frame_C00 Page 33 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero the equivalent mass would be expressed in milligrams of the substance per milligram- mole of the reference species and so on. Because the reference species is used as the standard of reference, its unit, the mole, can be said to have a unit of one equivalent. From this, the equivalent mass of a participating substance may be expressed as the mass of the substance per equivalent of the reference species; but, because the substance is equivalent to the reference species, the expression “per equivalent of the reference species” is the same as the expression “per equivalent of the substance.” Thus, the equivalent mass of a substance may also be expressed as the mass of the substance per equivalent of the substance. Each term of a balanced chemical reaction, represents the mass of a participating substance. Thus, the general formula for finding the equivalent mass of a substance is Example 5 Water containing 2.5 moles of calcium bicarbonate and 1.5 moles of calcium sulfate is softened using lime and soda ash. How many grams of calcium carbonate solids are produced (a) using the method of equivalent masses and (b) using the balanced chemical reaction? Pertinent reactions are as follows: Solution: number of reference species = 2 Equiv. mass mass of substance number of equivalents of substance = term in balanced reaction number of moles of reference species = Ca HCO 3 () 2 Ca OH() 2 2CaCO 3 ↓ 2H 2 O+→+ CaSO 4 Na 2 CO 3 CaCO 3 Na 2 SO 4 +→+ a() Ca HCO 3 () 2 Ca OH() 2 2CaCO 3 ↓ 2H 2 O+→+ Therefore, eq. mass of CaCO 3 2CaCO 3 2 100== eq. mass of Ca HCO 3 () 2 Ca HCO 3 () 2 2 40 2 1 12 3 16()++[]+ 2 162 2 81== = qeq of Ca HCO 3 () 2 2.5 162() 81 5 geq of CaCO 3 === g of CaCO 3 solids 5 100()500 g Ans== CaSO 4 Na 2 CO 3 CaCO 3 Na 2 SO 4 +→+ TX249_Frame_C00 Page 34 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero [...]... n3 ] d A (63) Considering the x and y directions, respectively, and following similar steps: ∂S1 ∫V dV ∂x ∂S2 ∫V dV ∂y = ˆ ˆ ∫A S 3 [ n ⋅ n1 ] d A (64) = ˆ ˆ ∫A S 2 [ n ⋅ n2 ] d A (65) ˆ S1 and S2 are the scalar components of S on the x and y axes, respectively, and n 1 ˆ and n 2 are the unit vectors on the x and y axes, respectively Adding Eqs (63), (64), and (67) produces the Gauss–Green... S3 n3 (56) S1, S2, and S3 are the scalar components of S in the x, y, and z directions, respectively, ˆ ˆ ˆ and n 1 , n 2 , and n 3 and are the unit vectors in the x, y, and z directions, respectively © 2003 by A P Sincero and G A Sincero TX249_Frame_C00 Page 58 Friday, June 14, 2002 1:47 PM 58 Physical–Chemical Treatment of Water and Wastewater ˆ Refer to Figure 1 As shown, dA has the unit vector n... 59 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 59 ˆ where n 3 is the unit vector in the positive z direction  z2 ∂ S 3  ˆ ˆ ∫A  ∫z1 dz [ n ⋅ n3 ] d A = ∂z   ˆ ˆ ∫A { S 3 ( x, y, z2 ) } [ n ⋅ n3 ] d A2 2 ˆ ˆ – ∫ { S 3 ( x, y, z 1 ) } [ ( – n ⋅ n 3 ) ] d A 1 A1 A1 and A2 are the bounding surface areas on the negative z direction and positive z ˆ ˆ ˆ direction, respectively... divergence theorem © 2003 by A P Sincero and G A Sincero TX249_Frame_C00 Page 57 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 57 the volume, it will form a closed surface around the volume Now, consider a function 3 such as f(A) = A + 3A that applies on the surface of the volume This function can be integrated over the closed surface and, to do this, the symbol ͛A is used The... H 2 CO 3 , HCO 3 , and CO 3 are formed Cations + − will then interact with these species and, along with the H and OH that always ∗ exist in water solution, complete the equilibrium of the system H 2 CO 3 is a mixture of CO2 in water and H2CO3 (note the absence of the asterisk in H2CO3) The CO2 ∗ in water is written as CO2(aq) and H 2 CO 3 is carbonic acid © 2003 by A P Sincero and G A Sincero TX249_Frame_C00... Friday, June 14, 2002 1:47 PM 56 Physical–Chemical Treatment of Water and Wastewater FLUID MECHANICS Of the fluid mechanics applications used in this book, the Reynolds transport theorem is a bit sophisticated Hence, a review of this topic is warranted This theorem will be used in the derivation of the amount of solids deposited onto a filter and in the derivation of the activated sludge process These topics...TX249_Frame_C00 Page 35 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 35 number of reference species = 2 CaCO 3 Therefore, eq mass of CaCO 3 = = 50 2 CaSO 4 40 + 32 + 4 ( 16 ) 136 eq mass of CaSO 4 = = = ... respectively, the activity coefficients of the aluminum ion and the hydrogen ion and the complexes 2+ 4+ 5+ − 4+ Al(OH) , Al7 (OH) 17 , Al13 (OH) 34 , Al(OH) 4 , and Al 2 (OH) 2 K sp,Al(OH) 3 is the solubility product constant of the solid Al(OH)3(s) and Kw is the ion product of water KAl(OH)c, K Al7 (OH) 17 c , K Al13 (OH) 34 c , K Al(OH) 4 c , and K Al2 (OH) 2 c are, respectively, the equi2+ 4+ 5+ −... HPO 4 γ FeIII [ Fe ] 2 3 (43) γ PO4 , γ Al , γ H , γ HPO 4 , and γ H2 PO4 are, respectively, the activity coefficients of the phosphate, aluminum, hydrogen, hydrogen phosphate, and dihydrogen phosphate ions K HPO 4 , K H2 PO4 , and K H3 PO4 are, respectively, the equilibrium constants of the hydrogen phosphate and dihydrogen phosphate ions and phosphoric acid γFeIII is the activity coefficient of the... H3 O + H2 O (48) + On the left-hand side of the equation, H3O donates its proton to H2O, with this H2O + forming the H3O on the right-hand side of the equation By virtue of the donation + of its proton, the H3O on the left transforms to the H2O on the right-hand side of + the equation H2O on the left is a base and H3O on the right is its conjugate acid + On the other hand, the H2O on the right is the . Background Chemistry and Fluid Mechanics As mentioned, this chapter discusses the background knowledge needed in order to understand the subsequent chapters of. TX249_Frame_C00 Page 26 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero Background Chemistry and Fluid Mechanics 27 Now, let us turn to a more elaborate problem of substituting. Ans= TX249_Frame_C00 Page 28 Friday, June 14, 2002 1:47 PM © 2003 by A. P. Sincero and G. A. Sincero Background Chemistry and Fluid Mechanics 29 Example 4 The formula used to calculate the amount of acid