# 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$\textdegree$C. 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$\textdegree$C 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$\textdegree$C. 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$\textdegree$C 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.

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

1. thanks a lot , i was looking for the water diffusivity to calculatr the flux across membrane distillation

• 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.