Chapter 16, Problem 16.1P

(a)

Interpretation Introduction

Interpretation:

The fraction of a fluid that spends $9\text{minutes}$ or more than $9\text{minutes}$ in a reactor is to be stated.  The fraction of a fluid that spends $2\text{minutes}$ or less than $2\text{minutes}$ in a reactor is to be stated.

Concept introduction:

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

The time of residence of fluid inside the reactor, determines the rate of conversion of the reactant or fluid, depending upon the concentration and mixing of the fluid inside the reactor.

Answer to Problem 16.1P

The fraction of a fluid that spends $9\text{minutes}$ or more than $9\text{minutes}$ in a reactor is $0.0968$. The fraction of a fluid that spends $2\text{minutes}$ or less than $2\text{minutes}$ in a reactor is $0.0677$.

Explanation of Solution

The given $E\left(t\right)$ versus time graph is shown as follows.

Figure 1

The expression to calculate the fluid present in the reactor is as follows.

    $\varphi ={\displaystyle \underset{t_1}{\overset{t_2}{\int }}E\left(t\right)dt}$                                                                                     (1)

Where,

• ${\displaystyle \underset{0}{\overset{\infty }{\int }}E\left(t\right)dt}$ corresponds to the area covered within $E\left(t\right)$ curve.
• $t_1$ and $t_2$ are the time intervals.

The area that is covered between $9\text{minutes}$ to $15\text{minutes}$ is shown as follows.

Figure 2

Substitute the limits $t_1$ as $9\min$ and $t_2$ as $15\min$ in equation (1).

    $\begin{array}{c}\varphi ={\displaystyle \underset{9}{\overset{15}{\int }}E\left(t\right)dt}\\ ={\displaystyle \underset{9}{\overset{13.5}{\int }}E\left(t\right)dt}+{\displaystyle \underset{13.5}{\overset{15}{\int }}0dt}\\ ={\displaystyle \underset{9}{\overset{13.5}{\int }}E\left(t\right)dt}\end{array}$

The Simpson’s $3/8$ rule is used to calculate the area under the curve as follows.

    ${\displaystyle \underset{X_0}{\overset{X_3}{\int }}E\left(t\right)dt}=\frac{3h}{8}\left[f\left(X_0\right)+3f\left(X_1\right)+3f\left(X_2\right)+f\left(X_3\right)\right]$                      (2)

Where,

• $h$ is equal to $\frac{X_3-X_0}{3}$
• $X_0$ is the lower limit.
• $X_3$ is the upper limit.

For the limits, $X_0=9$ and $X_3=13.5$, the value of $h$ is calculated as follows.

    $\begin{array}{c}h=\frac{13.5-9}{3}\\ =1.5\end{array}$

According to the graph, the values of $f\left(X_0\right)$ is $0.044$, $f\left(X_1\right)$ is $0.03$, $f\left(X_2\right)$ is $0.012$ and $f\left(X_3\right)$ is $0.002$.

Substitute the values of $h$ as $1.5$, $f\left(X_0\right)$ as $0.044$, $f\left(X_1\right)$ as $0.03$, $f\left(X_2\right)$ as $0.012$ and $f\left(X_3\right)$ as $0.002$ in equation (2).

    $\begin{array}{c}{\displaystyle \underset{9}{\overset{13.5}{\int }}E\left(t\right)dt}=\frac{3\times 1.5}{8}\left[0.044+3\times 0.03+3\times 0.012+0.002\right]\\ =0.5625\left[0.172\right]\\ =0.0968\end{array}$

Thus, the fraction of a fluid that spends between $9\text{minutes}$ to $15\text{minutes}$ in a reactor is $0.0968$.

The area that is covered between less than $2\text{minutes}$ is shown as follows.

Figure 3

Substitute the limits $t_1$ as $0\min$ and $t_2$ as $2\min$ in equation (1).

    $\varphi ={\displaystyle \underset{0}{\overset{2}{\int }}E\left(t\right)dt}$

The Simpson’s $1/3$ rule is used to calculate the area under the curve as follows.

    ${\displaystyle \underset{X_0}{\overset{X_2}{\int }}E\left(t\right)dt}=\frac{h}{3}\left[f\left(X_0\right)+3f\left(X_1\right)+f\left(X_2\right)\right]$                                        (3)

Where,

• $h$ is equal to $\frac{X_2-X_0}{2}$
• $X_0$ is the lower limit.
• $X_2$ is the upper limit.

For the limits, $X_0=0$ and $X_2=2$, the value of $h$ is calculated as follows.

    $\begin{array}{c}h=\frac{2-0}{2}\\ =1\end{array}$

According to the graph, the values of $f\left(X_0\right)$ is $0$, $f\left(X_1\right)$ is $0.04$ and $f\left(X_2\right)$ is $0.083$.

Substitute the values of $h$ as $1$, $f\left(X_0\right)$ as $0$, $f\left(X_1\right)$ as $0.04$, and $f\left(X_2\right)$ as $0.083$ in equation (3).

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{2}{\int }}E\left(t\right)dt}=\frac{1}{3}\left[0+3\times 0.04+0.083\right]\\ =\frac{1}{3}\left[0.203\right]\\ =0.0677\end{array}$

Thus, the fraction of a fluid that spends less than $2\text{minutes}$ in a reactor is $0.0677$.

(b)

Interpretation Introduction

Interpretation:

The change in the value of $E(t)$ if ${\tau }_{\text{p}}$ is reduced by $50\%$ and ${\tau }_{\text{s}}$ increased by $50\%$ is to be stated.  The fraction of a fluid that spends $2\text{minutes}$ or less than $2\text{minutes}$ in a reactor is to be stated.

Concept introduction:

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

The full form of CSTR is Continuous-Stirred Tank Reactor. This reactor has its application in the industrial processes and used for liquid-phase reactions.

The processing time for one reactor volume of fluid, depending on a set of inlet conditions of reactant is represented by the space time, $\tau$. It is the ratio of volume of the reactor to the flow rate entering the reactor.

Answer to Problem 16.1P

The change in the value of $E(t)$ if ${\tau }_{\text{p}}$ is reduced by $50\%$ and ${\tau }_{\text{s}}$ increased by $50\%$ is $\frac{e^{\left(\frac{-\left(t-0.5\right)}{1.5}\right)}}{1.5}$. The fraction of a fluid that spends $2\text{minutes}$ or less than $2\text{minutes}$ in a reactor is $0.0677$.

Explanation of Solution

The second-order liquid-phase reaction is given.

For the second order reaction which takes place in a real CSTR and in a real PFR possesses the value of $\begin{array}{c}{\tau }_{\text{p}}={\tau }_{\text{s}}\\ =1\end{array}$.

The given condition is space time of PFR, ${\tau }_{\text{p}}$, is decreased by $50\%$ and the space time of CSTR, ${\tau }_{\text{s}}$, is increased by $50\%$.

The exit function $E(t)$ for the combined reactors, PFR and CSTR, is given below.

    $\begin{array}{c}E(t)=0t<{\tau }_{\text{p}}\\ =\frac{e^{\left(\frac{-\left(t-{\tau }_{\text{p}}\right)}{{\tau }_{\text{s}}}\right)}}{{\tau }_{\text{s}}}t\ge {\tau }_{\text{p}}\end{array}$                                                                (4)

If ${\tau }_{\text{p}}=1$ for the given combination of PFR and CSTR and it is reduced to $50\%$ then the value of ${\tau }_{\text{p}}$ becomes $0.5$.

If ${\tau }_{\text{s}}=1$ for the given combination of PFR and CSTR and it is increased to $50\%$ then the value of ${\tau }_{\text{s}}$ becomes $1+0.5=1.5$.

Substitute the value of ${\tau }_{\text{s}}$ as $1.5$ and ${\tau }_{\text{p}}$ as $0.5$ in equation (4).

    $E(t)=\frac{e^{\left(\frac{-\left(t-0.5\right)}{1.5}\right)}}{1.5}$

Thus, the value of $E(t)$ becomes $\frac{e^{\left(\frac{-\left(t-0.5\right)}{1.5}\right)}}{1.5}$ for the given conditions.

As calculated above, the fraction of a fluid that spends less than $2\text{minutes}$ in a reactor is $0.0677$.