Question

9.3. Calculate the mass flux of methanol in water at 77°F (2=4.95 x 10-5 ft2/hr). The concentrations of meth- anol at two points 0.1 in. apart are 0.05 and 0.10 lb molelft?.

Answer 1

Solution:

• Diffusivity of methanol in water, D = 4.95 x 10^{​​​​​​-5} ft²/hr
• Concentration of methanol at one point, C₁ = 0.05 lbmol/ft³
• Concentration of methanol at other point, C₂ = 0.10 lbmol/ft³
• Distance between these two points, $\Delta$ y = 0.1 in = 0.01in x(1ft/12in) = 0.008333 ft

From Fick's laws of diffusion, the mass flux of methanol is given by

N=-DC

Na-D%

N=-D_C1-C2 0.008333 ft

(0.05 -0.1) lbmol/ft? N = -(4.95 * 10-5ft/hr) 0.008333ft

N = 0.000297 lbmol/ft²-hr