Contact angle

2.6 The WU method

In his observations on interfacial tension WU also started with the polar and disperse fractions of the surface energy of the participating phases. However, in contrast to FOWKES and OWENS, WENDT, RABEL and KAELBLE, who used the geometric mean of the surface tensions in their calculations, WU used the harmonic mean. In this way he achieved more accurate results, in particular for high-energy systems.

At least two test liquids with known polar and disperse fractions are required for this method; at least one of the liquids must have a polar fraction >0.

WU’s initial equation for the interfacial tension between a liquid and a solid phase is as follows:

$\gamma_{12} = \sigma_l + \sigma_s - 4 (\frac {\sigma_l^D \cdot \sigma_s^D}{\sigma_l^D + \sigma_s^D} + \frac {\sigma_l^P \cdot \sigma_s^P}{\sigma_l^P + \sigma_s^P})$ (Equation 41)

If YOUNG’s equation is inserted in Equation 41

$\sigma_s = \gamma_{sl} + \sigma_s \cdot \cos\theta$ (Equation 42)

then the following relationship is obtained:

$\sigma_l (\cos\theta + 1) - 4 (\frac{\sigma_l^D \cdot \sigma_s^D}{\sigma_l^D + \sigma_s^D} + \frac{\sigma_l^P \cdot \sigma_s^P}{\sigma_l^P + \sigma_s^P}) = 0$ (Equation 43)

In order to determine the two required quantities $\sigma_s D$ and $\sigma_s P$ , WU determined the contact angles for each of two liquids on the solid surface and then, based on Equation 43, he drew up an equation for each liquid. After a factor analysis the resulting equations were as follows:

$(b_1 + c_1 - a_1) \sigma_s D \sigma_s P + c_1 (b_1 - a_1) \sigma_s D + b_1 (c_1 - a_1) \sigma_s P - a_1 b_1 c_1 = 0$ (Equation 44)

$(b_2 + c_2 - a_2) \sigma_s D \sigma_s P + c_2 (b_2 - a_2) \sigma_s D + b_2 (c_2 - a_2) \sigma_s P - a_2 b_2 c_2 = 0$ (Equation 45)

The variables $a_1, b_1, c_1$ for the first liquid and $a_2, b_2, c_2$ for the second liquid express the following terms:

$a_1$ = $\frac {1}{4} \sigma_{l,1} (\cos\theta_1 + 1)$
$b_1$ = $\sigma^D_{l,1}$
$c_1$ = $\sigma^P_{l,1}$
$a_2$ = $\frac {1}{4} \sigma_{l,2} (\cos\theta_2 + 1)$
$b_2$ = $\sigma^D_{l,2}$
$c_2$ = $\sigma^P_{l,2}$

The solution of the equations produces the surface energy of the solid $\sigma_s$ and its polar end disperse components $\sigma_s^P$ and $\sigma_s^D$ . However, the following point must be taken into consideration: as quadratic equations are involved this means that two solutions are obtained for both $\sigma_s^P$ and $\sigma_s^D$ only one of these solutions describes the actual surface energy.

2.6.1 Selecting the correct solution

The user now has the problem that, from the two solutions obtained above, the one which supplies the physically correct result for the system under investigation must be selected. This is very easy when one of the solutions has a negative sign. As negative values for the surface energy do not make sense from a physical point of view, in this case the second solution (with a positive sign) provides the result of the measurement.

For example:

	Solution 1	Solution 2
Surface energy of the solid	35.2 mN/m	15.7 mN/m
Disperse fraction	37.2 mN/m	12.2 mN/m
Polar fraction	-2.0 mN/m	3.5 mN/m

In this example Solution 1 can be rejected as it supplies a negative polar fraction of the surface energy. Solution 2 is the correct result. In such a case the DSA1 program automatically ignores the negative solution and presents the positive solution as the result.

However, it is often the case that both solutions make sense from a physical point of view. In such cases the decision can be simplified by including further information:

Which of the two solutions has the order of magnitude which is to be expected from a knowledge of the properties of the substance?
Which of the two solutions agrees best with the results obtained with other pairs of liquids?
Which of the two solutions is closest to results obtained by calculations according to FOWKES, or OWENS, WENDT, RABEL and KAELBLE?

	Liquid 1	Liquid 2
1^st pair	water	diiodomethane
1^nd pair	water	ethylene glycol
3^rd pair	water	benzyl alcohol
4^th pair	diiodomethane	ethylene glycol
5^th pair	diiodomethane	benzyl alcohol
6^th pair	ethylene glycol	benzyl alcohol

This means that the 4 test liquids supply 6 part-results; as described above, each of these results has two solutions. This means that the choice of the right solution must be made for each individual pair of liquids. The pairing of two purely disperse liquids ( $\sigma^P_l = 0$ ) produces no solution for the equation system; they are not included in the calculation.

The final result of the surface energy determination is the arithmetic mean of the selected part-results.

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