Chapter 24: Fundamentals of Mass Transfer. Example 3

Example 3

Reevaluate the diffusion coefficient of carbon dioxide in air at 20\textdegreeC and atmospheric pressure using the Fuller, Schettler, and Giddings equation and compare the new value with the one reported in example 2.

The equation is

D_{AB} = \frac{0.001T^{1.75}(\frac{1}{M_A}\frac{1}{M_B})^{1/2}}{P[(\sum \nu)_A^{1/3} + (\sum \nu)_B^{1/3}]^2}

Molecular weight carbon dioxide

M_A = M_{CO_2} = M_C + 2 M_{O} = 12 + 2(16) = 44

Molecular weight Air

M_B = M_{Air}; 21\% \frac{mole O_2}{moles} O_2; 79\% N_2 \frac{mole N_2}{moles}

1 mole basis

0.21 \frac{mole O_2}{moles}(1 mole) = 0.21 mole O_2

0.79 \frac{mole N_2}{moles}(1 mole) = 0.79 mole N_2

Total mass

mass_{O_2} = 0.21 mole O_2 (M_{O_2}) = 0.21 mole O_2(32 \frac{g O_2}{mole O_2}) = 6.72 g O_2

mass_{N_2} = 0.79 mole N_2 (M{_{N_2}} = 0.79 mole N_2(28 \frac{g N_2}{mole N_2} = 22.1 g N_2

Total Mass = 6.7 + 22.1 = 29 grams

Molecular Weight of Air = \frac{29 grams}{1 mole}= 29 \frac{g}{mole}

\nu terms from book. For carbon dioxide and Air, the terms were already calculated (Page 491) and included as “Simple Molecules”.

 (\sum \nu)_A = (\sum \nu)_{CO_2 } = \nu_{CO_2} = 26.9

(\sum \nu)_B = (\sum \nu)_{Air} = \nu_{Air} = 20.1

We have all the needed variables to use the Fuller, Schettler, and Giddings equation

D_{AB} = \frac{0.001T^{1.75} (\frac{1}{M_{CO_2}}+\frac{1}{M_{Air}})^{1/2}}{P[(\nu_{CO_2})^{1/3} + (\nu_{Air})^{1/3}]^{2}} = \frac{0.00143(293 K)^{1.75}(\frac{1}{44} + \frac{1}{29})^{1/2}}{1 atm[(26.9)^{1/3} + (20.1)^{1/3}]^{2}} =

D_{AB} = \frac{7.1}{32.7} = 0.151 \frac{cm^2}{s}

Compared to Hirschfelder, Bird, and Spoz at 20\textdegreeC and 1 atm

D_{AB} = \frac{0.001858T^{3/2}[\frac{1}{M_A} + \frac{1}{M_B}]^{1/2}}{P \delta_{AB}^2 \Omega_D} = 0.147 \frac{cm^2}{s}

Compared to the Experimental at 20\textdegreeC and 1 atm

D_{AB_{T_2, P_2}} = \frac{P_1}{P_2}(\frac{T_2}{T_1})^{3/2}\frac{\Omega_{D|T1}}{\Omega_{D|T2}} = 0.155 \frac{cm^2}{s}

Percent Difference from Predicted Experimental

Precent Difference = \frac{0.151 - 0.155}{0.155} x 100 = -2.58\%

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