Chapter 17, Problem 17.1P

(a)

Interpretation Introduction

Interpretation:

Molal flux of ${\text{CO}}_2$ for the given process of diffusion is to be determined.

Concept Introduction:

The formula to calculate molal flux for the equimolar diffusion of gas A through gas B is:

$N_A=\frac{D_v{\rho }_M}{B_T}\left(y_{Ai}-y_A\right)$ ...... (1)

Here, $B_T$ is the thickness of the film over which the flux is to be determined, $y_A$ is the mole fraction of A at the outer edge of the film, $y_{Ai}$ is the mole fraction of A at the inner edge or the interface of the film, $D_v$ is the volumetric diffusivity of A in the mixture with species B, and ${\rho }_M$ is the molar density of the gas mixture at the given temperature and pressure conditions.

Answer to Problem 17.1P

Molal flux of ${\text{CO}}_2$ for the given process of diffusion is, $N_{{\text{CO}}_2}=1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}$ .

Explanation of Solution

Given information:

Diffusion of ${\text{CO}}_2$ through ${\text{N}}_2$ takes place in one direction.

Pressure, $P=1\text{ atm}$

Temperature, $T=0^{\circ }\text{C}$

Mole fraction of ${\text{CO}}_2$ at point A, $y_A=0.2$

Mole fraction of ${\text{CO}}_2$ at point B, $y_B=0.02$

Distance between point A and B, $B_T=3\text{ m}$

Volumetric diffusivity, $D_v=0.144\frac{{\text{cm}}^2}{\text{s}}$

The whole gas phase is stationary.

Nitrogen diffuses at the same rate as carbon dioxide in opposite direction.

For an ideal gas at standard temperature and pressure $\left(1{\text{ atm, 0}}^{\circ }\text{C}\right)$ , the molar volume is taken as $22.4{\text{m}}^3/\text{kg-mol}$ . The given pressure and temperature of the system is same as STP. Calculate the molar density of the gas as:

$\begin{array}{c}{\rho }_M=\frac{1}{V_M}\end{array}$

  $\begin{array}{c}=\frac{1}{22.4}\frac{\text{kmol}}{{\text{m}}^3}\end{array}$

  $\begin{array}{c}=0.0446\frac{\text{kmol}}{{\text{m}}^3}\end{array}$

Use equation (1) to calculate the molal flux of ${\text{CO}}_2$ as:

$\begin{array}{c}{N}_{C{O}_2}=\frac{D_v{\rho }_M}{B_T}\left(y_A-y_B\right)\end{array}$

  $\begin{array}{c}=\frac{\left(0.144\frac{{\text{cm}}^2}{\text{s}}\times \frac{{\text{10}}^{-4}{\text{ m}}^2}{{\text{cm}}^2}\times \frac{3600\text{ s}}{\text{h}}\right)\left(0.0446\frac{\text{kg-mol}}{{\text{m}}^3}\right)}{3\text{ m}}\left(0.2-0.02\right)\end{array}$

  $\begin{array}{c}=1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The net mass flux for the given process of diffusion is to be determined.

Concept Introduction:

The formula to calculate mass flux for the equimolar diffusion of gas A through gas B is:

$m_A=M_A{N}_A$ ...... (2)

The formula to calculate mass flux for the equimolar diffusion of gas B through gas A is:

$m_B=M_B{N}_B$ ...... (3)

Total mass flux will be:

$m_{Net}=m_A+m_B$ ...... (4)

Here, $M_A\text{ and }{M}_B$ are the molar mass of gases A and B respectively.

Answer to Problem 17.1P

The net mass flux for the given process of diffusion is, $m_{Net}=2.2192\times {10}^{-3}\frac{\text{kg}}{{\text{m}}^2\cdot \text{h}}$ .

Explanation of Solution

From part (a), the molar flux of ${\text{CO}}_2$ through ${\text{N}}_2$ is calculated as:

$N_{{\text{CO}}_2}=1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}$

Molar flux of $N_2$ is equal as that of ${\text{CO}}_2$ but in opposite direction. Thus,

$N_{{\text{N}}_2}=-1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}$

Molar masses of CO₂ and N₂ are taken as:

$\begin{array}{c}{M}_{{\text{CO}}_2}=44\frac{\text{kg}}{\text{kg-mol}}\end{array}$

  $\begin{array}{c}{M}_{{\text{N}}_2}=28\frac{\text{kg}}{\text{kg-mol}}\end{array}$

Use equations (2) and (3) to calculate the mass flux of CO₂ and N₂ as:

$\begin{array}{c}{m}_{{\text{CO}}_2}=M_{{\text{CO}}_2}{N}_{{\text{CO}}_2}\end{array}$

  $\begin{array}{c}=\left(44\frac{\text{kg}}{\text{kg-mol}}\right)\left(1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\right)\end{array}$

  $\begin{array}{c}=6.1028\times {10}^{-3}\frac{\text{kg}}{{\text{m}}^2\cdot \text{h}}\end{array}$

  $\begin{array}{c}{m}_{{\text{N}}_2}=M_{{\text{N}}_2}{N}_{{\text{N}}_2}\end{array}$

  $\begin{array}{c}=\left(28\frac{\text{kg}}{\text{kg-mol}}\right)\left(-1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\right)\end{array}$

  $\begin{array}{c}=-3.8836\times {10}^{-3}\frac{\text{kg}}{{\text{m}}^2\cdot \text{h}}\end{array}$

Use equation (4) to calculate the net mass flux as:

$\begin{array}{c}{m}_{Net}=m_{{\text{CO}}_2}+m_{{\text{N}}_2}\end{array}$

  $\begin{array}{c}=\left(6.1028\times {10}^{-3}\frac{\text{kg}}{{\text{m}}^2\cdot \text{h}}\right)+\left(-3.8836\times {10}^{-3}\frac{\text{kg}}{{\text{m}}^2\cdot \text{h}}\right)\end{array}$

  $\begin{array}{c}=2.2192\times {10}^{-3}\frac{\text{kg}}{{\text{m}}^2\cdot \text{h}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The speed of the observer at which the net mass flux relative to the observer becomes zero is to be determined.

Concept Introduction:

The formula that relates the molar flux of A with its concentration and velocityis:

$N_A=c_A{u}_A$ ...... (5)

The formula that relates the molar flux of B with its concentration and velocity is:

$N_B=c_B{u}_B$ ...... (6)

Concentration of a species is defined as:

$c_i=y_i{\rho }_M$ ...... (7)

Answer to Problem 17.1P

The speed of the observer from point A at which the net mass flux relative to the observer becomes zero is, $u_{0,A}=4.43\times {10}^{-7}\frac{\text{m}}{\text{s}}$ .

The speed of the observer from point B at which the net mass flux relative to the observer becomes zero is, $u_{0,B}=4.89\times {10}^{-7}\frac{\text{m}}{\text{s}}$ .

Explanation of Solution

From part (a), the molar flux of ${\text{CO}}_2$ through ${\text{N}}_2$ is calculated as:

$N_{{\text{CO}}_2}=1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}$

Molar flux of $N_2$ is equal as that of ${\text{CO}}_2$ but in opposite direction. Thus,

$N_{{\text{N}}_2}=-1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}$

Calculate the concentration of CO₂ and N₂ at point A and at point B as:

$\begin{array}{c}{c}_{{\text{CO}}_2,A}=y_A{\rho }_M\end{array}$

  $\begin{array}{c}=\left(0.2\right)\left(0.0446\frac{\text{kmol}}{{\text{m}}^3}\right)\end{array}$

  $\begin{array}{c}=8.92\times {10}^{-3}\frac{\text{kmol}}{{\text{m}}^3}\end{array}$

  $\begin{array}{c}{c}_{{\text{N}}_2,A}=\left(1-y_A\right){\rho }_M\end{array}$

  $\begin{array}{c}=\left(1-0.2\right)\left(0.0446\frac{\text{kmol}}{{\text{m}}^3}\right)\end{array}$

  $\begin{array}{c}=3.568\times {10}^{-2}\frac{\text{kmol}}{{\text{m}}^3}\end{array}$

  $\begin{array}{c}{c}_{{\text{CO}}_2,B}=y_B{\rho }_M\end{array}$

  $\begin{array}{c}=\left(0.02\right)\left(0.0446\frac{\text{kmol}}{{\text{m}}^3}\right)\end{array}$

  $\begin{array}{c}=8.92\times {10}^{-4}\frac{\text{kmol}}{{\text{m}}^3}\end{array}$

  $\begin{array}{c}{c}_{{\text{N}}_2,B}=\left(1-y_B\right){\rho }_M\end{array}$

  $\begin{array}{c}=\left(1-0.02\right)\left(0.0446\frac{\text{kmol}}{{\text{m}}^3}\right)\end{array}$

  $\begin{array}{c}=4.3708\times {10}^{-2}\frac{\text{kmol}}{{\text{m}}^3}\end{array}$

Calculate the velocities of CO₂ and N₂ at point A and at point B. Use the numerical values of the flux.

$\begin{array}{c}{u}_{{\text{CO}}_2,A}=\frac{N_{{\text{CO}}_2}}{c_{{\text{CO}}_2,A}}\end{array}$

  $\begin{array}{c}=\frac{1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\times \frac{\text{h}}{3600\text{ s}}}{8.92\times {10}^{-3}\frac{\text{kmol}}{{\text{m}}^3}}\end{array}$

  $\begin{array}{c}=4.32\times {10}^{-6}\frac{\text{m}}{\text{s}}\end{array}$

  $\begin{array}{c}{u}_{{\text{N}}_2,A}=\frac{N_{{\text{N}}_2}}{c_{{\text{N}}_2,A}}\end{array}$

  $\begin{array}{c}=\frac{1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\times \frac{\text{h}}{3600\text{ s}}}{3.568\times {10}^{-2}\frac{\text{kmol}}{{\text{m}}^3}}\end{array}$

  $\begin{array}{c}=1.0798\times {10}^{-6}\frac{\text{m}}{\text{s}}\end{array}$

  $\begin{array}{c}{u}_{{\text{CO}}_2,B}=\frac{N_{{\text{CO}}_2}}{c_{{\text{CO}}_2,B}}\end{array}$

  $\begin{array}{c}=\frac{1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\times \frac{\text{h}}{3600\text{ s}}}{8.92\times {10}^{-4}\frac{\text{kmol}}{{\text{m}}^3}}\end{array}$

  $\begin{array}{c}=4.32\times {10}^{-5}\frac{\text{m}}{\text{s}}\end{array}$

  $\begin{array}{c}{u}_{{\text{N}}_2,B}=\frac{N_{{\text{N}}_2}}{c_{{\text{N}}_2,B}}\end{array}$

  $\begin{array}{c}=\frac{1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\times \frac{\text{h}}{3600\text{ s}}}{4.3708\times {10}^{-2}\frac{\text{kmol}}{{\text{m}}^3}}\end{array}$

  $\begin{array}{c}=8.815\times {10}^{-7}\frac{\text{m}}{\text{s}}\end{array}$

The mass flux of CO₂ and N₂ through a reference plane in terms of the velocity of the moving observer $\left(u_0\right)$ is written as:

$\begin{array}{c}{m}_{{\text{CO}}_2}=c_{{\text{CO}}_2}{M}_{{\text{CO}}_2}\left(u_{{\text{CO}}_2}-u_0\right)\end{array}$

  $\begin{array}{c}{m}_{{\text{N}}_2}=c_{{\text{N}}_2}{M}_{{\text{N}}_2}\left(u_{{\text{N}}_2}-u_0\right)\end{array}$

Since the velocity of the observer at the point of zero net mass flux is to be calculated, mass flux of CO₂ must be equal to the mass flux of N₂. Thus,

$\begin{array}{c}{m}_{{\text{CO}}_2}=m_{{\text{N}}_2}\end{array}$

  $\begin{array}{c}{c}_{{\text{CO}}_2}{M}_{{\text{CO}}_2}\left(u_{{\text{CO}}_2}-u_0\right)=c_{{\text{N}}_2}{M}_{{\text{N}}_2}\left(u_{{\text{N}}_2}-u_0\right)\end{array}$

  $\begin{array}{c}{u}_0=\frac{c_{{\text{CO}}_2}{M}_{{\text{CO}}_2}{u}_{{\text{CO}}_2}-c_{{\text{N}}_2}{M}_{{\text{N}}_2}{u}_{{\text{N}}_2}}{c_{{\text{CO}}_2}{M}_{{\text{CO}}_2}+c_{{\text{N}}_2}{M}_{{\text{N}}_2}}\end{array}$

  $\begin{array}{c}{u}_0=\frac{N_{{\text{CO}}_2}{M}_{{\text{CO}}_2}-N_{{\text{N}}_2}{M}_{{\text{N}}_2}}{c_{{\text{CO}}_2}{M}_{{\text{CO}}_2}+c_{{\text{N}}_2}{M}_{{\text{N}}_2}}\end{array}$

  $\begin{array}{c}{u}_0=\frac{N_{{\text{CO}}_2}\left(M_{{\text{CO}}_2}-M_{{\text{N}}_2}\right)}{c_{{\text{CO}}_2}{M}_{{\text{CO}}_2}+c_{{\text{N}}_2}{M}_{{\text{N}}_2}}\end{array}$

Now, use the above equation to calculate the observer velocity for point A as well as point B as:

  $u_{0,A}=\frac{N_{{\text{CO}}_2}\left(M_{{\text{CO}}_2}-M_{{\text{N}}_2}\right)}{c_{{\text{CO}}_2,A}{M}_{{\text{CO}}_2}+c_{{\text{N}}_2,A}{M}_{{\text{N}}_2}}$

  $=\frac{\left(1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\times \frac{\text{h}}{3600\text{ s}}\right)\left(\left(44\frac{\text{kg}}{\text{kmol}}\right)-\left(28\frac{\text{kg}}{\text{kmol}}\right)\right)}{\left(8.92\times {10}^{-3}\frac{\text{kmol}}{{\text{m}}^3}\right)\left(44\frac{\text{kg}}{\text{kmol}}\right)+\left(3.568\times {10}^{-2}\frac{\text{kmol}}{{\text{m}}^3}\right)\left(28\frac{\text{kg}}{\text{kmol}}\right)}$

  $\begin{array}{c}=4.43\times {10}^{-7}\frac{\text{m}}{\text{s}}\\ u_{0,B}=\frac{N_{{\text{CO}}_2}\left(M_{{\text{CO}}_2}-M_{{\text{N}}_2}\right)}{c_{{\text{CO}}_2,B}{M}_{{\text{CO}}_2}+c_{{\text{N}}_2,B}{M}_{{\text{N}}_2}}\end{array}$

  $=\frac{\left(1.387\times {10}^{-4}\frac{\text{kg-mol}}{{\text{m}}^2\cdot \text{h}}\times \frac{\text{h}}{3600\text{ s}}\right)\left(\left(44\frac{\text{kg}}{\text{kmol}}\right)-\left(28\frac{\text{kg}}{\text{kmol}}\right)\right)}{\left(8.92\times {10}^{-4}\frac{\text{kmol}}{{\text{m}}^3}\right)\left(44\frac{\text{kg}}{\text{kmol}}\right)+\left(4.3708\times {10}^{-2}\frac{\text{kmol}}{{\text{m}}^3}\right)\left(28\frac{\text{kg}}{\text{kmol}}\right)}$

  $=4.89\times {10}^{-7}\frac{\text{m}}{\text{s}}$

(d)

Interpretation Introduction

Interpretation:

The speed of the observer at which the nitrogen is stationary relative to the observer is to be determined.

Concept Introduction:

The formula that relates the molar flux of A with its concentration and velocity is:

$N_A=c_A{u}_A$ ...... (5)

The formula that relates the molar flux of B with its concentration and velocity is:

$N_B=c_B{u}_B$ ...... (6)

Concentration of a species is defined as:

$c_i=y_i{\rho }_M$ ...... (7)

Answer to Problem 17.1P

The speed of the observer from point A at which the nitrogen is stationary relative to the observer is, $u_{0,A}=1.0789\times {10}^{-6}\frac{\text{m}}{\text{s}}$ .

The speed of the observer from point B at which the nitrogen is stationary relative to the observer is, $u_{0,B}=83.815\times {10}^{-7}\frac{\text{m}}{\text{s}}$ .

Explanation of Solution

For the nitrogen phase to appear stationary to the observer, the speed of the observer must be equal to the speed of nitrogen and in the direction of the diffusion of nitrogen.

At point A,

$\begin{array}{c}{u}_{o,A}=u_{{\text{N}}_2,A}\end{array}$

  $\begin{array}{c}=1.0798\times {10}^{-6}\frac{\text{m}}{\text{s}}\end{array}$

At point B,

$\begin{array}{c}{u}_{o,B}=u_{{\text{N}}_2,B}\end{array}$

  $\begin{array}{c}=8.815\times {10}^{-7}\frac{\text{m}}{\text{s}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Molal flux of CO₂ relative to the observer in part (d) is to be estimated.

Concept Introduction:

The molal flux of A through a reference plane in terms of the velocity of the moving observer $\left(u_0\right)$ is written as:

$N_A=c_A\left(u_A-u_0\right)$ ...... (8)

Answer to Problem 17.1P

Molal flux of CO₂ relative to the observer in part (d) at point A is, $N_{{\text{CO}}_2,A}=4.82\times {10}^{-8}\frac{\text{kmol}}{{\text{m}}^2\cdot \text{s}}$ .

Molal flux of CO₂ relative to the observer in part (d) at point B is, $N_{{\text{CO}}_2,B}=9.93\times {10}^{-8}\frac{\text{kmol}}{{\text{m}}^2\cdot \text{s}}$ .

Explanation of Solution

Velocity of the observer at points A and B from part (d) is:

$\begin{array}{c}{u}_{o,A}=1.0798\times {10}^{-6}\frac{\text{m}}{\text{s}}\end{array}$

  $\begin{array}{c}{u}_{o,B}=8.815\times {10}^{-7}\frac{\text{m}}{\text{s}}\end{array}$

The molar flux of CO₂ through a reference plane in terms of the velocity of the moving observer $\left(u_0\right)$ with respect to stationary nitrogen is written as:

$\begin{array}{c}{N}_{{\text{CO}}_2}=c_{{\text{CO}}_2}\left(u_{{\text{CO}}_2}-\left(-u_0\right)\right)\end{array}$

  $\begin{array}{c}{N}_{{\text{CO}}_2}=c_{{\text{CO}}_2}\left(u_{{\text{CO}}_2}+u_0\right)\end{array}$

At point A, the molar flux of CO₂relative to the observer is calculated as:

$\begin{array}{c}{N}_{{\text{CO}}_2,A}=c_{{\text{CO}}_2,A}\left(u_{{\text{CO}}_2,A}+u_{0,A}\right)\end{array}$

  $\begin{array}{c}=\left(8.92\times {10}^{-3}\frac{\text{kmol}}{{\text{m}}^3}\right)\left(4.32\times {10}^{-6}\frac{\text{m}}{\text{s}}+1.0798\times {10}^{-6}\frac{\text{m}}{\text{s}}\right)\end{array}$

  $\begin{array}{c}=4.82\times {10}^{-8}\frac{\text{kmol}}{{\text{m}}^2\cdot \text{s}}\end{array}$

At point B, the molar flux of CO₂ relative to the observer is calculated as:

$\begin{array}{c}{N}_{{\text{CO}}_2,B}=c_{{\text{CO}}_2,B}\left(u_{{\text{CO}}_2,B}+u_{0,B}\right)\end{array}$

  $\begin{array}{c}=\left(8.92\times {10}^{-4}\frac{\text{kmol}}{{\text{m}}^3}\right)\left(4.32\times {10}^{-5}\frac{\text{m}}{\text{s}}+8.815\times {10}^{-7}\frac{\text{m}}{\text{s}}\right)\end{array}$

  $\begin{array}{c}=9.93\times {10}^{-8}\frac{\text{kmol}}{{\text{m}}^2\cdot \text{s}}\end{array}$