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The first-order reaction $$A \longrightarrow B$$ with \(k=0.8 \mathrm{min}^{-1}\) is carried out in a real reactor with the following RTD function For \(2 \tau \geq t \geq 0\) then \(E(t)=\sqrt{\tau^{2}-(t-\tau)^{2}} \min ^{-1}(\text {hemi circle })\) For \(t>2 \tau\) then \(E(t)=0\) (a) What is the mean residence time? (b) What is the variance? (c) What is the conversion predicted by the segregation model? (d) What is the conversion predicted by the maximum mixedness model?

Short Answer

Expert verified
The mean residence time \(\bar{t}\) is given by \(\bar{t} = \int_0^{2\tau} t\sqrt{\tau^2 - (t-\tau)^2} dt\). The variance \(\sigma^2\) is obtained from \(\sigma^2 = \int_0^{2\tau} (t - \bar{t})^2 \sqrt{\tau^2 - (t-\tau)^2} dt\). The conversion predicted by the segregation model is \(X = 1 - \int_0^{2\tau} e^{-kt} \sqrt{\tau^2 - (t-\tau)^2} dt\). The conversion predicted by the maximum mixedness model is \(X = \int_0^{2\tau} [1- e^{-kt}] \sqrt{\tau^2 - (t-\tau)^2} dt\).

Step by step solution

01

(a) Calculating the Mean Residence Time

The mean residence time, \(\bar{t}\), is calculated using the following expression: \[ \bar{t} = \int_0^\infty tE(t) dt \] Given the RTD function, we split the integral into two parts for the two time intervals: \[ \bar{t} = \int_0^{2\tau} t\sqrt{\tau^2 - (t-\tau)^2} dt + \int_{2\tau}^\infty t\cdot0 dt \] Since the integrand of the second integral is zero it contributes nothing and we get \(\bar{t}\) from the first integral.
02

(b) Calculating the Variance

The variance, \(\sigma^2\), is calculated using the following expression: \[ \sigma^2 = \int_0^\infty (t - \bar{t})^2 E(t) dt \] Like before, split the integral into two parts corresponding to the time intervals: \[ \sigma^2 = \int_0^{2\tau} (t - \bar{t})^2 \sqrt{\tau^2 - (t-\tau)^2} dt + \int_{2\tau}^\infty (t - \bar{t})^2 \cdot0 dt \] Again, the second integral contributes nothing and the variance is obtained from the first integral.
03

(c) Conversion predicted by the Segregation Model

The conversion, \(X\), for a first-order reaction predicted by the segregation model is given by: \[ X = 1 - \int_0^\infty e^{-kt} E(t) dt \] Split the integral as before: \[ X = 1 - \left(\int_0^{2\tau} e^{-kt} \sqrt{\tau^2 - (t-\tau)^2} dt + \int_{2\tau}^\infty e^{-kt} \cdot0 dt\right) \] The contribution from the second integral is zero. Hence, \(X\) is obtained from the subtraction of the first integral from 1.
04

(d) Conversion predicted by the Maximum Mixedness Model

For the maximum mixedness model, the conversion is calculated by considering an infinitesimal volume of fluid that spends time \(t\) in the reactor and behaves as a CSTR operated at the same conversion. Hence, the overall conversion can be calculated as follows: \[ X = \int_0^\infty [1- e^{-kt}] E(t) dt \] The integral can be again split into two parts to add up the results from the two time intervals: \[ X = \int_0^{2\tau} [1- e^{-kt}] \sqrt{\tau^2 - (t-\tau)^2} dt + \int_{2\tau}^\infty [1- e^{-kt}] \cdot0 dt \] The second integral contributes nothing and thus, calculation for the \(X\) is done numerically from the first integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Reaction
A first-order reaction is a chemical process where the rate at which the reaction proceeds is directly proportional to the concentration of a single reactant. This can be represented by the rate law: \[ \text{Rate} = k[A] \] where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant. Such reactions are commonly found in nature and industry, featuring simple rate expressions which make them particularly easy to analyze in Chemical Reaction Engineering.

When modeling first-order reactions within reactors, we look at the effect of time, temperature, and concentration on the conversion rates, leading to design equations where variables like mean residence time and reaction time distribution can significantly influence reactor performance.
Mean Residence Time
Mean residence time (MRT), symbolized as \( \bar{t} \), is a concept in Chemical Reaction Engineering which represents the average amount of time that a molecule spends in a reactor. Calculating the MRT is pivotal for understanding the reactor's behavior and can guide engineers in optimizing reaction conditions.

The MRT is obtained by integrating the product of time and the Reaction Time Distribution (RTD) function, E(t), over all time. It can be mathematically represented as: \[ \bar{t} = \int_0^\infty tE(t) dt \] In practical terms, a longer mean residence time suggests that reactions have more time to proceed to completion. However, too long a residence time could indicate an inefficient process. In essence, it's a delicate balance that must be optimized for each specific reaction and reactor setup.
Segregation Model
The segregation model is a theoretical approach used to predict the behavior of chemical reactors, specifically dealing with non-ideal flow conditions. It assumes that the fluid elements are perfectly segregated, meaning there is no mixing within the elements as they pass through the reactor.

Under this model, each fluid element reacts independently based on the time it resides in the reactor, which is governed by the RTD. The overall conversion is then an average of conversions of individual elements. In our exercise, the conversion for a first-order reaction using the segregation model is calculated by the integral: \[ X = 1 - \int_0^\infty e^{-kt} E(t) dt \] This model is particularly useful for reactions and systems where mixing is limited or undesirable.
Maximum Mixedness Model
In contrast to the segregation model, the maximum mixedness model assumes that mixing within a reactor is instantaneous and occurs to the greatest extent possible. This model reflects the behavior of a Continuous Stirred-Tank Reactor (CSTR), where all molecules are assumed to have the same residence time.

The conversion for a first-order reaction under this model can be calculated using the integral: \[ X = \int_0^\infty [1- e^{-kt}] E(t) dt \] The maximum mixedness model is idealized and often used as a benchmark to compare against more realistic models that account for variance in residence times. It helps to determine the extent of mixing needed to achieve specific conversion goals.
Variance in Chemical Reactors
Variance in chemical reactors refers to the distribution of residence times of the reactants within the reactor. This is critically important because it affects the extent of reaction and the distribution of products. Variance is denoted by \( \sigma^2 \) and it gives a measure of the spread of the RTD around the mean residence time.

The equation used to calculate variance is: \[ \sigma^2 = \int_0^\infty (t - \bar{t})^2 E(t) dt \] A high variance indicates a wide range of residence times, which can be detrimental to reaction selectivity and yield, particularly in reactions with different orders. Understanding and controlling variance is key to designing efficient reactors.
Reaction Time Distribution (RTD)
Reaction Time Distribution (RTD) is a critical function in Chemical Reaction Engineering that describes how long molecules or fluid elements stay within a reactor. It is represented by the function E(t), which, when integrated over time, offers insights on mean residence time and variance.

The RTD is vital for predicting conversion rates and understanding the flow pattern within reactors. For example, it can help engineers discern whether a reactor behaves more like an ideal plug flow reactor (PFR) or a CSTR, which in turn influences reactor design and operation. The E(t) function events presented in our textbook case provide the basis for calculating specific RTD-based metrics such as mean residence time and variance.

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Most popular questions from this chapter

The second-order liquid-phase reaction $$\text { 2A } \stackrel{k_{14}}{\longrightarrow} \mathrm{B}$$ is carried out in a nonideal CSTR. At \(300 \mathrm{K}\) the specific reaction rate is 0.5 \(\mathrm{dm}^{3} / \mathrm{mol} \cdot \mathrm{min.}\) In a tracer test. the tracer concentration rose linearly up to I mg/dm \(^{3}\) at 1.0 minutes and then decreased linearly to zero at exactly 2.0 minutes. Pure A enters the reactor at a temperature of \(300 \mathrm{K}\). (a) Calculate the conversion predicted by segregation and maximum mixedness models. (b) Now consider that a second reaction also takes place $$A+B \stackrel{k_{2} c}{\longrightarrow} C, \quad k_{2 C}=0.12 \mathrm{dm}^{3} / \mathrm{mol} \cdot \mathrm{min}$$ Compare the selectivities \(\tilde{S}_{B / C}\) predicted by the segregation and maximum mixedness models. (C) Repeat (a) for adiabatic operation. Additional information: $$\begin{aligned} C_{P_{A}} &=50 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} \\ \Delta H_{R x 1 \mathrm{A}} &=-7500 \mathrm{J} / \mathrm{mol} \\ C_{\mathrm{A} 0} &=2 \mathrm{mol} / \mathrm{dm}^{3} \end{aligned}$$ $$C_{P_{B}}=100 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}, \quad \mathrm{E}=10 \mathrm{kcal} / \mathrm{mol}$$

Show that for a first-order reaction \(A \longrightarrow B\) the exit concentration maximum mixedness equation $$\frac{d C_{\mathrm{A}}}{d \lambda}=k C_{\mathrm{A}}+\frac{E(\lambda)}{1-F(\lambda)}\left(C_{\mathrm{A}}-C_{\mathrm{A} 0}\right)$$ is the same as the exit concentration given by the segregation model $$C_{\mathrm{A}}=C_{\mathrm{A} 0} \int_{0}^{\infty} E(t) e^{-k t} d t$$ [Hint: Verify $$C_{A}(\lambda)=\frac{C_{A 0} e^{k \lambda}}{1-F(\lambda)} \int_{\lambda}^{\infty} E(t) e^{-k t} d t$$ is a solution to Equation (P13-3.1).]

The third-order liquid-phase reaction \(A \stackrel{k_{1}}{\longrightarrow} B\) was carried out in a reactor that has the following RTD $$E(t)=0 \quad \text { for } \quad t<1 \text { min }$$ $$\begin{aligned}&E(t)=1.0 \quad \text { for } \quad 1 \leq t \leq 2 \min\\\ &E(t)=0 \quad \text { for } \quad t>2 \min\end{aligned}$$ The entering concentration of \(A\) is 2 mol/dm \(^{3}\). (a) For isothermal operation, what is the conversion predicted by 1) a CSTR, a PFR, an LFR, and the segregation model, \(X_{\mathrm{seg}}\). 2) the maximum mixedness model, \(X_{\mathrm{MM}}\). Plot \(X\) vs. \(z(\text { or } \lambda\) ) and explain why the curve looks the way it does. (b) For isothermal operation, at what temperature is the discrepancy between \(X_{\mathrm{sc}_{\mathrm{S}}}\) and \(X_{\mathrm{MM}}\) the greatest in the range \(300

A step tracer input was used on a real reactor with the following results: For \(t \leq 10\) min, then \(C_{T}=0\) For \(10 \leq t \leq 30\) min, then \(C_{T}=10 \mathrm{g} / \mathrm{dm}^{3}\) For \(t \geq 30\) min, then \(C_{T}=40 \mathrm{g} / \mathrm{dm}^{3}\) The second-order reaction \(A \rightarrow B\) with \(k=0.1 \mathrm{dm}^{3} / \mathrm{mol} \cdot\) min is to be carried out in the real reactor with an entering concentration of \(\mathrm{A}\) of \(1.25 \mathrm{mol} / \mathrm{dm}^{3}\) at a volumetric flow rate of \(10 \mathrm{dm}^{3} / \mathrm{min}\). Here \(k\) is given at \(325 \mathrm{K}\). (a) What is the mean residence time \(t_{m} ?\) (b) What is the variance \(\sigma^{2} ?\) (c) What conversions do you expect from an ideal PFR and an ideal CSTR in a real reactor with \(t_{m} ?\) (d) What is the conversion predicted by (1) the segregation model? (2) the maximum mixedness model? (e) What conversion is predicted by an ideal laminar flow reactor? (f) Calculate the conversion using the segregation model assuming \(T(K)=325-500 X\).

Consider a PFR. CSTR, and LFR. (a) Evaluate the first movement about the mean \(m_{1}=\int_{0}^{\infty}(t-\tau) E(t) d t\) for a PFR, a CSTR, and a laminar flow reactor. (b) Calculate the conversion in each of these ideal reactors for a second- order liquid-phase reaction with \(\mathrm{Da}=1.0\left(\tau=2 \min \text { and } k C_{\mathrm{A} 0}=0.5 \mathrm{min}^{-1}\right)\).

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