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Xl2 = 2[(0.152)"' + (0.210)-1]"1 = 0.176 W/(m K)

At 273 K, V (methanol) = 39.6 cm3/mol and V (benzene) = 88.9 cm3/mol. If the weight fraction methanol in the mixture is 0.4, the mole fraction is 0.619. Then, with Eq. (10-12.19),

<t> (benzene) = 1 - 0.414 = 0.586 Using Eq. (10-12.20),

X„ = (0.414)2(0.210) + (0.586)2(0.152) + (2)(0.414)(0.586)(0.176) = 0.174 W/(m K)

0.174 - 0.170 „„ „„, Error =-—-X 100 = 2%

Example 10-15 Estimate the liquid thermal conductivity of a mixture of benzene (1) and methyl formate (2) at 323 K by using the method of Baroncini et al. At this temperature, the values of \L for the pure components are X! = 0.138 and X2 = 0.179 W/(m-K) [7], solution We will estimate the values of Xm at 0.25, 0.50, and 0.75 weight fraction benzene. First, however, we need to determine A, and A2. Although Eq. (10-9.2) and Table 10-6 could be used, it is more convenient to employ the pure component values of XL with Eq. (10-9.1). From Appendix A, TCI = 562.2 K and Tc2 = 487.2 K, so Tn = 323/562.2 = 0.575 and Tr2 = 323/487.2 = 0.663. Then, with Eq. (10-9.1), for benzene,

Similarly Ao = 0.252. [Note that, if Eq. (10-9.2) and Table 10-6 had been used, we would have A, = 0.176 and A2 = 0.236.]

We have selected components 1 and 2 to agree with the criterion A, < A2.

Consider first a mixture containing 0.25 weight fraction benzene, i.e., wi = 0.25 and w2 = 0.75. Then, the mole fractions are x, = 0.204 and x2 = 0.796. Thus,

m 502.5

Calculated results for this and other compositions are shown below with the experimental values and percent errors.

Benzene-Methyl Formate Mixtures; T = 323 K

Weight fraction Mole fraction Xm, calc., Am, exp., Percent benzene benzene TCm, K W/(m-K) W/(m K) error

Example 10-16 Use Rowley's method to estimate the thermal conductivity of a liquid mixture of acetone (1) and chloroform (2) that contains 66.1 weight percent of the former. The temperature is 298 K. As quoted by Jamieson et al. [64], Rodriguez [137] reports A, = 0.161 W/(m K), X2 = 0.119 W/(m K), and for the mixture, Xm = 0.143 W/(m K).

solution First, we need the NRTL parameters for this binary at 298 K. Nagata [114) suggests G12 = 1.360 and G2, = 0.910. From Appendix A, M, = 58.08 and M2 = 119.38 g/mol. Using Eqs. (10-12.15) and (10-12.16),

_ /0.285\2 (0.285)(1.360) + 0.715 ~ \ 0.715 J 0.285 + (0.715X0.910) = 0.187

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