Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is $2.243\times {10}^{-2}\text{moles}/\text{s}$.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with $25$ weight percent silver.

The inner radius of the tube is $5.11\times {10}^{-4}\text{m}$.

The outer radius is $6.35\times {10}^{-4}\text{m}$.

The length of the tube is $5.81\text{m}$.

The inside high pressure is $1\times {10}^6\text{N}/{\text{m}}^{\text{2}}$.

The outside low pressure is $1\times {10}^4\text{N}/{\text{m}}^{\text{2}}$.

The temperature is $673\text{K}$.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

$J_{\text{H}}=\left[4.86\times {10}^{-7}\left(\frac{p_{\text{h}}^{\frac{\text{1}}{\text{2}}}-p_{\text{l}}^{\frac{\text{1}}{\text{2}}}}{r_2-r_1}\right)\frac{\text{moles}}{{\text{m}}^{\text{2}}\cdot \text{s}}\right]\left[\exp \left(-\left(\frac{6.6\times {10}^3\text{J}/\text{mole}}{RT}\right)\right)\right]$   ....... (I)

Here, $J_{\text{H}}$ is the flux of hydrogen ions, $p_{\text{h}}$ insider high pressure, $p_{\text{h}}$ is the outside high pressure, $r_2$ is the outer radius, $r_1$ is the inner radius, $R$ is the universal gas constant and $T$ is the temperature.

The effective area of the tube is given by,

$A=\frac{2\pi l\left(r_2-r_1\right)}{\ln \left(\frac{r_2}{r_1}\right)}$   ....... (II)

Here, $A$ is the effective area of the tube and $l$ is the length of the tube.

The hydrogen ions produced per second is given by,

${\left(J_{\text{H}}\right)}_{\text{PS}}=J_{\text{H}}A$   ....... (III)

Here, ${\left(J_{\text{H}}\right)}_{\text{PS}}$ are the hydrogen ions produced per second.

Substitute $\left[4.86\times {10}^{-7}\left(\frac{p_{\text{h}}^{\frac{\text{1}}{\text{2}}}-p_{\text{l}}^{\frac{\text{1}}{\text{2}}}}{r_2-r_1}\right)\frac{\text{moles}}{{\text{m}}^{\text{2}}\cdot \text{s}}\right]\left[\exp \left(-\left(\frac{6.6\times {10}^3\text{J}/\text{mole}}{RT}\right)\right)\right]$ for $J_{\text{H}}$ and $\frac{2\pi l\left(r_2-r_1\right)}{\ln \left(\frac{r_2}{r_1}\right)}$ for $A$ in equation (III).

${\left(J_{\text{H}}\right)}_{\text{PS}}=\left[\begin{array}{l}\left[4.86\times {10}^{-7}\left(\frac{p_{\text{h}}^{\frac{\text{1}}{\text{2}}}-p_{\text{l}}^{\frac{\text{1}}{\text{2}}}}{r_2-r_1}\right)\frac{\text{moles}}{{\text{m}}^{\text{2}}\cdot \text{s}}\right]\\ \left[\exp \left(-\left(\frac{6.6\times {10}^3\text{J}/\text{mole}}{RT}\right)\right)\right]\left[\frac{2\pi l\left(r_2-r_1\right)}{\ln \left(\frac{r_2}{r_1}\right)}\right]\end{array}\right]$   ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute $1\times {10}^6\text{N}/{\text{m}}^{\text{2}}$ for $p_{\text{h}}$, $1\times {10}^4\text{N}/{\text{m}}^{\text{2}}$ for $p_{\text{l}}$, $6.35\times {10}^{-4}\text{m}$ for $r_2$, $5.11\times {10}^{-4}\text{m}$ for $r_1$, $673\text{K}$ for $T$, $5.81\text{m}$ for $l$ and $8.31\text{J}/\text{mole}\cdot \text{K}$ for $R$ in equation (IV).

${\left(J_{\text{H}}\right)}_{\text{PS}}=\left[\begin{array}{l}\left[4.86\times {10}^{-7}\left\{\frac{{\left(1\times {10}^6\text{N}/{\text{m}}^{\text{2}}\right)}^{\frac{1}{2}}-{\left(1\times {10}^4\text{N}/{\text{m}}^{\text{2}}\right)}^{\frac{1}{2}}}{\left(6.35\times {10}^{-4}\text{m}\right)-\left(5.11\times {10}^{-4}\text{m}\right)}\right\}\frac{\text{moles}}{{\text{m}}^{\text{2}}\cdot \text{s}}\right]\\ \left[\exp \left\{-\left(\frac{6.6\times {10}^3\text{J}/\text{mole}}{\left(8.31\text{J}/\text{mole}\cdot \text{K}\right)\left(673\text{K}\right)}\right)\right\}\right]\\ \left[\frac{2\pi \left(5.81\text{m}\right)\left\{\left(6.35\times {10}^{-4}\text{m}\right)-\left(5.11\times {10}^{-4}\text{m}\right)\right\}}{\ln \left(\frac{6.35\times {10}^{-4}\text{m}}{5.11\times {10}^{-4}\text{m}}\right)}\right]\end{array}\right]$

$=\left[\begin{array}{l}\left[4.86\times {10}^{-7}\left(\frac{900}{1.24\times {10}^{-4}}\right)\frac{\text{moles}}{{\text{m}}^{\text{2}}\cdot \text{s}}\right]\\ \left[\exp \left(-\left(\frac{6.6\times {10}^3}{5592.63}\right)\right)\right]\left[\frac{45.27\times {10}^{-4}{\text{m}}^{\text{2}}}{0.219}\right]\end{array}\right]$

$=\left[\left[3.527\frac{\text{moles}}{{\text{m}}^{\text{2}}\cdot \text{s}}\right]\left[0.3072\right]\left[0.0207{\text{m}}^{\text{2}}\right]\right]$

$=2.243\times {10}^{-2}\text{moles}/\text{s}$

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is $2.243\times {10}^{-2}\text{moles}/\text{s}$.