157 MAss Transfer PArt 2

# 157 MAss Transfer PArt 2 - Part I Calculations Calculations...

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Part I Calculations Calculations of Diameter of the test tube. Method 1) Known volume and height to find radius V= 2∏ r2 h , V= 40 µl and h= 10 mm Therefore, 40* - 10 3 cm3 = ∏ r2 (1cm) r = 0.113 cm D = 0.226 cm Method 2) Known mass and density to find volume V= , m= 0.035 g and ρ = 0.7918 g/ cm3 V= . . 0 0350 7918 cm3 = 44 * - 10 3 cm3 Thus, V= 44 * - 10 3 = cm3 r2 (1cm) r= 0.118 cm D= 0.236 cm Calculating the average of the two we get r= 0.116 cm and D= 0.232 cm. Calculating the Diffusivity, , NA z = ( - ) cDAB z2 z1 ln ( xB2 xB1 ) , NA z = decrease in height of methanol time taken * density * 1Mr xB1 = 1 - xA1 = 1- Pvap,A/ P = 1- 20.05 kPa/ 101 kPa = 0.801 X B2 = 1 (reservoir of air at air column) C A = density / Mr = 0.7918/32.04 mol/cm 3 = 0.0247 mol/ cm 3 C= 1/ xA1 * C A A = 1/(1- 0.801) * 0.0247 mol/cm 3 =0.0308 mol/cm 3 Part II

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Qn 2) Background The question relates concentration and temperature gradient with the following eqn Where X A0 can be calculated as follows, We found the evaporation flux using the following eqn, N A0 = dmdt * * 1MW Area of tray Calculating the concentration and temperature gradient Assuming:
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