Chapter 17, Problem 17.2P

Interpretation Introduction

Interpretation:

The actual conversion that is obtained if laminar is the real state of flow having negligible diffusion is to be calculated.

Concept introduction:

The conversion, X can be defined as the moles of any species A that are reacted per mole of A fed in the reactor.

The full form of PFR is Plug-Flow Reactor which consists of a cylindrical pipe and it is used for the gas-phase reactions.

Answer to Problem 17.2P

The actual conversion that is obtained if laminar is the real state of flow having negligible diffusion is $78.2\%$.

Explanation of Solution

It is given that an irreversible first order reaction is taking place in long cylindrical reactor.

The given conversion if the reactor is PFR is $86.5\%$ or $0.865$.

The given rate law for the above first order reaction is given below.

    $-r_{\text{A}}=k{C}_{\text{A}}$                                                                                          (1)

Where,

• $r_{\text{A}}$ is the rate of disappearance of reactant A.
• $k$ is the rate constant.
• $C_{\text{A}}$ is the final concentration of A.

If $C_{\text{A}}=C_{\text{A0}}\left(1-X\right)$ then substitute the value of $C_{\text{A}}$ in equation (1).

    $-r_{\text{A}}=k{C}_{\text{A0}}\left(1-X\right)$                                                                              (2)

Where,

• $C_{\text{A0}}$ is the initial concentration of A.
• $X$ is the conversion of the reaction.

The mole balance equation for PFR is given as follows.

    $\frac{dX}{dV}=\frac{-r_{\text{A}}}{F_{\text{A0}}}$                                                                                            (3)

Where,

• $F_{\text{A0}}$ is the molar flow rate.
• $-r_{\text{A}}$ is the rate of disappearance of A.
• $V$ is the volume of the reactor.
• $X$ is the conversion.

If the value $F_{\text{A0}}=C_{A0}{v}_0$ then substitute the values of $-r_{\text{A}}$ and $F_{\text{A0}}$ in equation (3).

    $\begin{array}{c}\frac{dX}{dV}=\frac{k{C}_{\text{A0}}\left(1-X\right)}{C_{A0}{v}_0}\\ \frac{dX}{dV}=\frac{k\left(1-X\right)}{v_0}\\ \frac{dX}{\left(1-X\right)}=\frac{k}{v_0}dV\end{array}$

The integration of the above equation is done as follows.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{dX}{\left(1-X\right)}}=\frac{k}{v_0}{\displaystyle \underset{0}{\overset{V}{\int }}dV}\\ -\ln \left(1-X\right)=\left(\frac{k}{v_0}\right)V\\ -\ln \left(1-X\right)=\left(\frac{V}{v_0}\right)k\end{array}$                                                                         (4)

The ratio of volume of the reactor to the volumetric flow rate is equal to space time as shown below.

    $\frac{V}{v_0}=\tau$

Substitute $\frac{V}{v_0}=\tau$ in equation (4).

    $\begin{array}{c}-\ln \left(1-X\right)=\tau k\\ \left(1-X\right)=e^{-\tau k}\\ X=1-e^{-\tau k}\end{array}$                                                                           (5)

Substitute $X$ as $0.865$ in the above equation.

    $\begin{array}{c}0.865=1-e^{-\tau k}\\ 0.865-1=-e^{-\tau k}\\ 0.135=e^{-\tau k}\\ \tau k=2\end{array}$

The value $\tau k=Da$, where, $Da$ is Damkohler number. Thus, the value of $Da$ is $2$.

The equation for the mean conversion of the laminar is expressed as follows.

    $\overline{X}={\displaystyle \underset{0}{\overset{\infty }{\int }}X\left(t\right)}E\left(t\right)$

Substitute the value of $X\left(t\right)$ as $1-e^{-\tau k}$ from equation (5) to the above equation.

    $\overline{X}={\displaystyle \underset{0}{\overset{\infty }{\int }}1-e^{-\tau k}}E\left(t\right)$

For laminar flow rate,

If

$t<\tau /2$ then the value of $E\left(t\right)$ is 0.

$t\ge \tau /2$ then the value of $E\left(t\right)$ is $\frac{{\tau }^2}{2t^3}$.

Substitute the value of $E\left(t\right)$ as $\frac{{\tau }^2}{2t^3}dt$ in the above equation.

    $\begin{array}{c}\overline{X}={\displaystyle \underset{\tau /2}{\overset{\infty }{\int }}1-e^{-\tau k}}\frac{{\tau }^2}{2t^3}dt\\ =1-{\displaystyle \underset{\tau /2}{\overset{\infty }{\int }}{e}^{-\tau k}}\frac{{\tau }^2}{2t^3}dt\\ =1-{\displaystyle \underset{\tau /2}{\overset{\infty }{\int }}{e}^{-Da}}\frac{{\tau }^2}{2t^3}dt\end{array}$

According to the Hilder approximation the above equation solved as follows.

    $\overline{X}=\frac{\left(4+Da\right)e^{0.5Da}+Da-4}{\left(4+Da\right)e^{0.5Da}+Da}$

Substitute the value of $Da$ as $2$ in the above equation.

    $\begin{array}{c}\overline{X}=\frac{\left(4+2\right)e^{0.5\times 2}+2-4}{\left(4+2\right)e^{0.5\times 2}+2}\\ =\frac{14.3098}{18.3098}\\ =0.7815\\ \approx 0.782or78.2\%\end{array}$

Therefore, the obtained conversion for the laminar is $78.2\%$.

Conclusion

The actual conversion that is obtained if laminar is the real state of flow having negligible diffusion is $78.2\%$.