Page 258 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Coagulation 213

Since amps ohms ¼ volts, and converting from SI to trad- 3. Apply the Equation 9.14 from Hunter
itional units
m
EM (9:14)
e o D
z ¼
6
2
m 2 10 mm 10 cm 4p(0:895 10 3 Ns=m )
2
7
¼ 0:061 10 EM
s V m m ¼ (8:854 10 12 F=m) (78:36)
Ns
mm=s ¼ 1:29 10 6 EM
¼ 0:61 10 2 Fm
V=cm
Recall that EM was given in traditional units, that is,
mm=s=V=cm; now convert to SI
Discussion
The main idea of this example was to illustrate how elec- 6 Ns mm cm m m
trophoretic data are converted to EM. In modern instru- z ¼ 1:29 10 Fm EM V s 10 mm 10 cm
6
2

ments, all of this is done by software.
Nm
EM
FV
¼ 0:0129
Example 9.5 Derivation of Factor to Convert EM Now recall: Nm ¼ Ws and F ¼ As=V
(mm=s=V=cm) to z in mV
Ws
The problem of conversion of EM to z is confounding in z ¼ 0:0129 As EM
that the older equation, that is, Equation 9.13 is not dimen- V V
sionally homogeneous and does not yield the correct
result numerically. The modification by Hunter, however, After canceling units
is dimensionally homogeneous and does give the correct
conversion factor numerically. W
A
¼ 0:0129 EM
Given
Equations 9.13 and 9.14 give equation to convert EM to Now recall, W ¼ V A, to give
zeta potential. A more expedient conversion is given as
Equation 9.15, which is valid for 258C (the viscosity of EM
water, a term in both equations, is a function of tempera- z ¼ 0:0129 V D
ture). The factor in Equation 9.15 is actually 12.9, and is
rounded off to 13 and converts EM in (mm=s=V=cm) to z in and to convert to mV,
mV. Such an equation is useful since electrokinetic data
are given often in EM. Therefore, merely multiplying EM 1000 mV
by 13 gives z in mV. z ¼ 0:0129 V V EM
¼ 12:9 EM (mV)
Required
Derive the factor, 12.9 in Equation 9.15.
Discussion
Solution The outcome starting with Equation 9.14 shows that the
1. The three referenced data are equation is homogeneous with respect to units and yields
a. D ¼ the dielectric constant for water at 258C the accepted empirical factor for the EM to z conversion,
(dimensionless, i.e., no units) that is, 12.9.
¼ 78.36 from Lide (1996, pp. 6–18)
2
b. m (258C) ¼ viscosity of water at 258C(N s=m )
¼ 0.89 10 3 Ns=m 2 Example 9.6 Calculation of Zeta Potential from EM
c. e o ¼ permittivity of vacuum (F=m)
¼ 9.854187817 10 12 F=m (Lide, 1996, back Given
cover) Let an EM measurement be 1.20 mm=s=V=cm at 258C.
2. Start with Equation 9.13, which is attributed to Required
Helmholtz–Smoluchowski and Debye–Hückel, Calculate zeta potential
that is,
Solution
As noted, Black and Willems (1961, p. 592), gave a rule
4pm
EM (9:13) of thumb to convert EM in (mm=s=V=cm) to z in mV as,
z ¼
D z ¼ 13 EM (also confirmed in Example 9.5). Applying this
rule of thumb to the problem at hand
Calculation of the collection of terms: 4pm=
2
D ¼ 4p (0.89 10 3 Ns=m )=78.36 ¼ 1.43 10 4 z ¼ 13 EM (9:15)
2
Ns=m . The number sought is 13. The discrepancy
is not reconcilable in either numerical value or 13 ( 1:20 mm=s=V=cm)
units. ¼ 15:6mV