# Chapter 24: Fundamentals of Mass Transfer. Example 3

Example 3

Reevaluate the diffusion coefficient of carbon dioxide in air at 20$\textdegree$C 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$\textdegree$C 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$\textdegree$C 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 = $\frac{0.151 - 0.155}{0.155} x 100 = -2.58\%$

# 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.

# Chapter 24: Fundamentals of Mass Transfer. Example 2

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

Example 2

Evaluate the diffusion coefficient of carbon dioxide in air at 20$\textdegree$C and atmospheric pressure. Compare this value with the experimental value reported in Appendix Table J.1.

Will be using the following diffusivity equation:

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

We have temperature and pressure. We can calculate the molecular weights via a periodic chart. $\delta$ and $\Omega$ can be obtained from Tables K.1 and K.2.

From K.2 of the appendix values $\delta$ and $\frac{\epsilon}{\kappa}$ are obtained:

Carbon dioxide: $\delta$ in $\AA$, 3.996 and $\frac{\epsilon_{CO_2}}{\kappa}$ in K, 190

Air: $\delta$ in $\AA$, 3.617 and $\frac{\epsilon_{N_2}}{\kappa}$ in K, 97

$\delta_{AB} = \frac{(\delta_A + \delta_B)}{2} = \frac{(3.996 \AA + 3.617 \AA)}{2} = 3.806 \AA$

$\frac{\epsilon_{AB}}{\kappa}= \sqrt{(\frac{\epsilon_{A}}{\kappa})(\frac{\epsilon_B}{\kappa}}) = \sqrt{(190)(97)} = 136$

T = 20 + 273 = 293 K

P = 1 atm

$\frac{\epsilon_{AB}}{\kappa T} = \frac{136}{293} = 0.463$

$\frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.463} = 2.16$

$\Omega_D (Table K.1) = 1.047$

This value was obtained by interpolation

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

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

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

From Table K.1

$y_i = \Omega$ and $x_i = \frac{\kappa T}{\epsilon_{AB}}$

and

$x = 2.16$

Interpolate

$y = \frac{(1.041-1.057)}{(2.20-2.10)}(2.16 - 2.10) + 1.057 = 1.047 = \Omega_D$

We have all variable except molecular weights. Considering the most prevalent gasses:

$M_{CO_2} = M_C + 2M_O = 12 + 2(16) = 12 + 32 = 44$

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

Now, we have all the information needed to calculate the diffusivity of $CO_2$ in air when using:

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

$D_{AB} = \frac{0.001858(293^{3/2})(\frac{1}{44} + \frac{1}{29})^{1/2}}{1 atm(3.806 \AA)^2 (1.047)} = 0.147 \frac{cm^2}{s}$

Now, we want to compare to the experimental value that is reported in Table J.1

$T, K = 273, D_{AB}P \frac{cm^2 atm}{s} = \frac{0.136 \frac{cm^2 atm}{s}}{1 atm} = 0.136 \frac{cm^2}{s}$

Since the value is reported at 273 K, must use a conversion equation to compare at 293 K

$\frac{D_{AB,T_1}}{D_{AB,T_2}} = (\frac{T_1}{T_2})^{3/2}(\frac{\Omega_{D,T_2}}{\Omega_{D,T_1}})$

at $T_1 = 293 K$ and $\Omega_{D, T_1} = 1.047$

at $T_2 = 273 K$ and $\Omega_{D, T_2} = ?$ from Table Table K.1

$\frac{\epsilon_AB}{\kappa}\frac{1}{T_2} = 136\frac{1}{273} = 0.498$

$\frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.498} = 2.01$

Once again, interpolation of Table K.1 is needed

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

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

$x = 2.01$

$y = \frac{1.057 -1.075}{2.10-2.00}(2.01-2.00) + 1.075 = 1.074 = \Omega_{D, T_2}$

Since we have all the values for the conversion equation

$D_{AB, 293} = (\frac{293}{273})^{3/2}(\frac{1.074}{1.047})(0.136) = 0.155 \frac{cm^2}{s}$

The diffusivity of carbon dioxide in air

Calculated: $0.147 \frac{cm^2}{s}$ and Corrected Experimental: $0.155 \frac{cm^2}{s}$

Percent Difference

$\frac{Calculated - Corrected Experimental}{Corrected Experimental} x 100= \frac{0.147 - 0.155}{0.155} x 100=5.16\%$

# Chapter 24: Fundamentals of Mass Transfer. Example 1

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

Chapter 24: Fundamentals of Mass Transfer, Page 471

Note: LaTex causes certain parts of the text to be raised, italicized, and larger. It is not to the power. I believe the “power” will be recognizable once I point the latter out.

Example 1

The composition of air is often given in terms of only the two principle species in the gas mixture

oxygen, $O_2, y_{O_2} = 0.21$

nitrogen, $N_2, y_{N_2} = 0.79$

Determine the mass fraction of both oxygen and nitrogen and the mean molecular weight of the air when it is maintained at 25$\textdegree$C (298 K) and 1 ATM $(1.013 X 10^5 Pa)$. The molecular weight of oxygen is 0.032 kg/mol and of nitrogen is 0.028 kg/mol.

As a basis for our calculations, consider 1 mole of the gas mixture;

moles of oxygen present = (1 mol) $(y_{O_2})$ = (1 mol)(0.21) = 0.21 mol oxygen

mass of oxygen present = (0.21 mol) $(M_{O_2})$ = (0.21 mol) $(0.032 \frac{ kg}{mol})$ = 0.00672 kg oxygen

moles of nitrogen present = (1 mol) $(y_{N_2})$ = (1 mol)(0.79) = 0.79 mol nitrogen

mass of nitrogen present = (0.79 mol) $(M_{N_2})$ = (0.79) $(0.028 \frac{kg}{mol})$ = 0.0221 kg nitrogen

total mass present = kg oxygen + kg nitrogen = 0.00672 kg oxygen + 0.0221 kg nitrogen = 0.0288 kg

$w_{O_2} = \frac{0.00672 kg}{0.0288 kg} = 0.23$

$w_{N_2} = \frac{0.0221 kg}{0.0288 kg} = 0.77$

Since 1 mole of the gas mixture has a mass of 0.0288 kg, the mean molecular weight of the air must be 0.0288: $M_{air}$ = 0.0288 kg/mol. When one takes into account the other constituents that are present in air, the mean molecular weight of air is often rounded off to $0.029 \frac{kg}{mol}$.

This problem could also be solved using the ideal gas law, PV = nRT. At ideal conditions, 0$\textdegree$C or 273 K and 1 atm or $1.013 X 10^5 Pa$ pressure, the gas constant is evaluated to be

Remember, we want to calculate the mass fraction of oxygen and nitrogen, and the mean molecular weight of air. Which to calculate first is the key

PV = nRT

R = $\frac{PV}{nT}$

n = 1 mole; T = 273 K; P = $1.015 X 10^5 Pa$; V = $22.4 m^3$

R = $\frac{1.013 X 10^5 Pa x 22.4 m^3}{1 kg mol x 273 K}$ = $8.314 X 10^3 \frac{Pa m^3}{(kg mol) K}$

R = $8.314 X 10^3 \frac{(Pa) (m^3)}{(kg mol) (K)} X \frac{1 kg mole}{1000 moles}$ = $8.314 \frac{(Pa)(m^3)}{(mol) (K)}$

The of volume the gas mixture, at 298 K, is

V= $\frac{nRT}{P} = \frac{(1 mol) (8.314 \frac{Pa m^3}{mol K})(298 K)}{1.013 X 10^5 Pa} = 0.0245 (m^3)$

The concentrations are (1 mole basis)

$c_{O_2} = \frac{moles O_2}{volume of gas mixture} = \frac{0.21 mol}{0.0245 m^3} = 8.57 \frac{mol O_2}{m^3}$

$c_{N_2} = \frac{moles N_2}{volume of gas mixture} = \frac{0.79 mol}{0.0245 m^3} = 32.3 \frac{mol N_2}{m^3}$

$c = \sum_{i=1}^{n} c_i = 8.57 \frac{mol O_2}{m^3} + 32.3 \frac{mol N_2}{m^3} = 40.9 \frac{mol}{m^3}$

The total density, $\rho$ is

mass of $O_2$ = $(mol O_2) (Molecular Weight O_2) = (0.21 mol O_2)(0.032 \frac{kg}{mol}) = 0.00672 kg O_2$

mass of $N_2$ = $(mol N_2)(Molecular Weight N_2) = (0.79 mol N_2)(0.028 \frac{kg}{mol}) = 0.0221 kg N_2$

total mass = 0.00672 kg + 0.0221 kg = 0.0288 kg

total density = $\rho = \frac{total mass}{total volume} = \frac{0.0288 kg}{0.0245 m^3} = 1.180 \frac{kg}{m^3}$

and the mean molecular weight of the mixture is

$M = \frac{\rho}{c} = \frac{1.180\frac{kg}{m^3}}{40.9 \frac{mol}{m^3}} = 0.288 \frac{kg}{mol}$

As a side note, density, $\rho_i$, can be used to calculate mass percent of $O_2$ and $N_2$

$\rho_{O_2} = (c_{O_2})(M_{O_2}) = (8.57 \frac{mol O_2}{m^3})(0.032 \frac{kg}{mol})= 0.27 \frac{kg O_2}{m^3}$

$w_{O_2} = \frac{\rho_{O_2}}{\rho_total} = \frac{0.27 \frac{kg O_2}{m^3}}{1.180 \frac{kg}{m^3}} = 0.23$

$\rho_{N_2} = (c_{N_2})(M_{N_2}) = (32.3 \frac{mol N_2}{m^3})(0.028 \frac{kg N_2}{mol N_2} = 0.904 \frac{kg N_2}{m^3}$

$w_{N_2} = \frac{\rho_{N_2}}{\rho_total} = \frac{0.904 \frac{kg N_2}{m^3}}{1.180 \frac{kg}{m^3}}= 0.77$