Chapter 24: Fundamentals of Mass Transfer. Diffusivity of Water in Air at 20 Degrees Celsius and 1 Atmosphere

Diffusivity of Water in Air at 1 atmosphere and 20\textdegreeC. A = H_2O and B = Air.

Diffusivity equation. Assuming nonpolar and nonreacting molecules

D_{AB} = \frac{0.001858 T^{3/2} (\frac{1}{M_A} + \frac{1}{M_B})^{1/2}}{P \delta_{AB}^2 \Omega_D}

T = 20 + 273.15 = 293 K ; P = 1 atm ; M_A = 2 M_H+ M_O= 2(1) + 16 = 18;

M_B = \%O_2(M_{O_2}) + \%N_2(M_{N_2}) = 0.21(32) + 0.79(28) = 29

Need \delta_{AB} and \Omega_D, From Table K.2 of the appendix, values are obtained for water and Air

Water: \delta_A, \AA: 2.649 \AA and \frac{\epsilon_A}{\kappa}, K: 356 K

Air: \delta_B, \AA: 3.617 \AA and \frac{\epsilon_A}{\kappa}, K: 97 K

\delta_{AB}

\delta_{AB} = \frac{\delta_A + \delta_B}{2} = \frac{2.649 + 3.617}{2} = 3.13 \AA

Now I need \Omega_D, which can be found on Table K.1. To find \Omega_D, need to calculate \frac{\kappa T}{\epsilon_{AB}}:

\frac{\epsilon_{AB}}{\kappa} = \sqrt{(\frac{\epsilon_{A}}{\kappa})(\frac{\epsilon_{B}}{\kappa})} = \sqrt{(356)(97)} = 186

T = 20 + 273 = 293 K

P = 1 atm

\frac{\epsilon_{AB}}{\kappa T} = \frac{186}{293} = 0.634

For Table K. 1, need the inverse

\frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.634} = 1.58

Must interpolate:

\frac{y - y_0}{x-x_0} = \frac{y_1 - y_0}{x_1-x_0}

y = \frac{y_1 - y_0}{x_1 - x_0}(x-x_0) + y_0

\Omega_D = y; x = 1.58

y = \frac{1.167 - 1.182}{1.60 - 1.55}(1.58 - 1.55) + 1.182 = 1.173 = \Omega_D

Now the diffusivity equation and diffusion of water vapor in air at 20\textdegreeC and 1 atm

D_{AB} = \frac{0.001858 T^{3/2}(\frac{1}{M_A} + \frac{1}{M_B})^{1/2}}{P \delta_{AB}^{2} \Omega_D}

D_{AB} = \frac{0.001858(293)^{3/2}(\frac{1}{18} + \frac{1}{29})^{1/2}}{1(3.13)^{2}(1.173)} = 0.243 \frac{cm^{2}}{s}

Compare to the experimental value from Table J.1 of book

D_{AB}P \frac{cm^{2}atm}{s} = 0.260 \frac{cm^{2}atm}{sec} at T = 298 K

Need to find the experimental diffusivity value at pressure of 1 atm and 298 K

\frac{D_{AB}P}{P} = \frac{0.260\frac{cm^{2}P}{s}}{1 atm} = 0.260 \frac{cm^{2}}{s}

Now, we will use an equation to predict the experimental diffusivity at 20\textdegreeC. The following equation can be used for pressures below 25 atmospheres

\frac{D_{AB, T_2, P_2}}{D_{AB, T_1, P_1}} = (\frac{P_1}{P_2}) (\frac{T_2}{T_1})^{3/2} \frac{\Omega_{D|T_1}}{\Omega_{D|T_2}}; T_2 = 293K; T_1 = 298 K

Need \Omega_{D|T_1}

\frac{\epsilon_{AB}}{\kappa} = 186; \frac{\epsilon_{AB}}{\kappa T} = \frac{186}{298} = 0.624; \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.624} = 1.60

No need to interpolate Table K.1 since 1.60

\Omega_{D|T_1} = 1.167

Diffusion coefficient prediction

D_{AB, T_2, P_2} = (\frac{P_1}{P_2})(\frac{T_2}{T_1})^{3/2}\frac{\Omega_{D|T_1}}{\Omega_{D|T_2}} = (\frac{1 atm}{1 atm}) (\frac{293}{298})^{3/2}(\frac{1.167}{1.173})(0.260) = 0.252 \frac{cm^2}{s}

Percent difference

Percent Difference = \frac{Calculated - Corrected Experimental}{Corrected Experimental} x 100 = \frac{0.243 - 0.252}{0.252} x 100 = -3.57\%

There is nice agreement between calculated and predicted experimental water in air diffusivity values. The final diffusivity values for water in air at 20\textdegreeC and 1 atm are

D_{Calc} = 0.243 \frac{cm^2}{s} and D_{Experimental} = 0.252 \frac{cm^2}{s}

Summary

Nonpolar Hirschfelder, Bird, and Spotz Equation

D_{AB} = \frac{0.001858 T^{3/2}[(\frac{1}{M_A}) + (\frac{1}{M_B})]^{1/2}}{P \sigma_{AB}^{2} \Omega_D} = 0.243 \frac{cm^{2}}{s}

Experimental Value from Table J.1

D_{AB} = 0.260 \frac{cm^{2}}{s}

Predicted Value

D_{AB_{T_2,P_2}} = D_{AB_{T_1, P_1}} (\frac{P_1}{P_2})(\frac{T_2}{T_1})^{3/2}\frac{\Omega_{D|T_1}}{\Omega_{D|T_2}} = 0.252 \frac{cm^2}{s}

References:

Welty, James R.; Wicks, Charles E.; Wilson, Robert E. (1984) Fundamentals of Momentum, Heat, and Mass Transfer, third edition. New York: John Wiley & Sons.

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3 thoughts on “Chapter 24: Fundamentals of Mass Transfer. Diffusivity of Water in Air at 20 Degrees Celsius and 1 Atmosphere

  1. Pingback: Chapman and Enskog Versus Hirschfelder Equation and Compared to Experimental Value at 25C and 1 Atm « Random Ramblings

    • Hello,

      First, thanks for the comment.

      Yeah, I rely upon books for most of my examples.

      I read a journal article about the evaporation rate of sarin equaling water, but the divergent boiling points of each caused me to disagree. As such, I wanted to do a molar flux calculation that compared sarin to water. Although not a perfect calculation, the molar flux calculation verified the vapor pressure ratio, and each suggested that there was a difference in evaporation rates.

      As you know, chemical engineering is quite complicated, and I graduated in 1999! 🙂 As such, my ability to do complex calculations is dependent upon much review. In this case, I had to do a basic review of diffusivity to eventually perform a molar flux calculation.

      Have a nice day.

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