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**Sample text**

K C H A P T E R 2. 05 Therefore, the use of Fourier's law of heat conduction, Eq. 3 Fick's First Law of Diffusion Consider two large parallel plates of area A. The lower one is coated with a material ,4 which has a very low solubility in the stagnant fluid B filling the space between the plates. Suppose that the saturation concentration of Jt is PAo and ,4 undergoes a rapid chemical reaction at the surface of the upper plate and its concentration is zero at that surface. 4. Since the solubility of r is low, an almost a linear distribution is reached under steady conditions.

6 that a straight line represents the data fairly well. The equation of this line can be determined by the method of least squares in the form y-mx+b (2) where y = log CA (3) To determine the values of m and b from Eqs. (A. 6-10) and (A. 6-11) in Appendix A, the required values are calculated as follows: Yi xi x 2. 76) - ( 1 5 0 ) ( - 6 . 0087 ( - 6 . 94 ( 6 ) ( 5 5 0 0 ) - (150) 2 Therefore, Eq. (2) takes the form log CA = -- 0 . 115 e - ~176 (5) or, Differentiation of Eq. 23 m o l / m 4 (6) x--0 Substitution of the numerical values into Eq.

1. flux and the temperature gradient is called the~nal conductivity. In SI units, Q is in W ( J / s ) , A in m 2, d T / d x in K / m , and k in W/( m. K). The thermal conductivity of a material is, in general, a function of temperature. However, in many engineering applications the variation is sufficiently small to be neglected. 2 in Appendix D. The microscopic form of Eq. 1-3) is known as Fourier's law of heat conduction and is given by (2 1-4) dT qu = - k --~y in which the subscript y indicates the direction of the energy flux.