/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The reaction \(A \rightarrow B\)... [FREE SOLUTION] | 91Ó°ÊÓ

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The reaction \(A \rightarrow B\) takes place in two reactors in series. The reactors are well mixed but are not at steady state. The unsteady-state mass balance for each stirred tank reactor is shown below: $$\begin{aligned} \frac{d C A_{1}}{d t} &=\frac{1}{\tau}\left(C A_{0}-C A_{1}\right)-k C A_{1} \\\ \frac{d C B_{1}}{d t} &=-\frac{1}{\tau} C B_{1}+k C A_{1} \\ \frac{d C A_{2}}{d t} &=\frac{1}{\tau}\left(C A_{1}-C A_{2}\right)-k C A_{2} \\\ \frac{d C B_{2}}{d t} &=\frac{1}{\tau}\left(C B_{1}-C B_{2}\right)+k C A_{2} \end{aligned}$$ where \(C A_{0}=\) concentration of \(A\) at the inlet of the first reactor, \(C A_{1}=\) concentration of \(A\) at the outlet of the first reactor (and inlet of the second), \(C A_{2}=\) concentration of \(A\) at the outlet of the second reactor, \(C B_{1}=\) concentration of \(B\) at the outlet of the first reactor (and inlet of the second), \(C B_{2}=\) concentration of \(B\) in the second reactor, \(\tau=\) residence time for each reactor, and \(k=\) the rate constant for reaction of \(A\) to produce \(B\). If \(C A_{0}\) is equal to \(20,\) find the concentrations of \(A\) and \(B\) in both reactors during their first 10 minutes of operation. Use \(k=0.12 /\) min and \(\tau=5\) min and assume that the initial conditions of all the dependent variables are zero.

Short Answer

Expert verified
To find the concentrations of $A$ and $B$ in both reactors during the first 10 minutes of operation, we need to numerically solve the given coupled first-order ordinary differential equations with the initial conditions $C A_{1}(0) = C A_{2}(0) = C B_{1}(0) = C B_{2}(0) = 0$, the inlet concentration of A, $C A_{0} = 20$, the rate constant $k = 0.12$/min, and the residence time $\tau = 5$ min. Using a numerical ODE solver, obtain the solutions for $C A_{1}(t)$, $C B_{1}(t)$, $C A_{2}(t)$, and $C B_{2}(t)$ for $0 \le t \le 10$ min. Analyze and interpret the results to understand the dynamic behavior of the reaction system in the two reactors.

Step by step solution

01

In this problem, we are given four coupled first-order ordinary differential equations (ODEs) that represent the unsteady-state mass balances for the concentrations of A and B in the two reactors: i) \( \frac{d C A_{1}}{d t} = \frac{1}{\tau}\left(C A_{0}-C A_{1}\right)-k C A_{1} \) ii) \( \frac{d C B_{1}}{d t} = -\frac{1}{\tau} C B_{1}+k C A_{1} \) iii) \( \frac{d C A_{2}}{d t} = \frac{1}{\tau}\left(C A_{1}-C A_{2}\right)-k C A_{2} \) iv) \( \frac{d C B_{2}}{d t} = \frac{1}{\tau}\left(C B_{1}-C B_{2}\right)+k C A_{2} \) #Step 2: Define the given parameters#

We are given the following parameter values: - \(C A_{0} = 20\) - \(k = 0.12\) / min - \(\tau = 5\) min We are also given the initial conditions of all dependent variables as zero, which means \(C A_{1}(0) = C A_{2}(0) = C B_{1}(0) = C B_{2}(0) = 0\). #Step 3: Solve the ODE system numerically#
02

In order to find the concentrations of A and B in both reactors during the first 10 minutes of operation, we need to solve the given system of ODEs numerically. Any numerical ODE solver can be used for this purpose, such as Euler's method, Runge-Kutta method, or in-built numerical solvers in computational software like Python or Matlab. Solve the ODE system for the given parameter values and initial conditions, and obtain the numerical solutions for \(C A_{1}(t)\), \(C B_{1}(t)\), \(C A_{2}(t)\), and \(C B_{2}(t)\) for \(0 \le t \le 10\) min. #Step 4: Interpret the results#

After obtaining the numerical solutions for the concentration of A and B in both reactors during the first 10 minutes of operation, analyze and interpret the results. The results should give you insight into the dynamic behavior of the reaction system and how the concentrations of A and B change over time in the two reactors.

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Key Concepts

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

Unsteady-state
Chemical processes like reactions in a reactor can change with time. These changes are known as unsteady-state processes. Unlike steady-state conditions, where variables remain constant over time, in unsteady-state conditions, variables such as concentration or temperature fluctuate.

This concept of unsteady-state is crucial in chemical engineering, as it describes more realistic and dynamic real-world systems. In an unsteady-state mass balance, the input and output flows and the internal changes in the concentration of reactants or products can all change simultaneously. This balance is represented using differential equations.

For example, in our problem, the concentrations of chemicals A and B change over time in two reactors. Thus, the system does not reach a steady-state but continuously evolves, and understanding these time dynamics is essential for process optimization.
First-order ODEs
First-order ordinary differential equations (ODEs) are equations that relate a function with its first derivative.

In our exercise, we encounter four coupled first-order ODEs, representing the rates of change in concentrations of substances A and B in two reactors. The form of a first-order ODE can generally be written as \(\frac{dy(t)}{dt} = f(t, y(t))\). Here, the change in the concentration, \(\frac{dC}{dt}\), is affected by terms that might involve residence time \(\tau\) and rate constants \(k\).

First-order ODEs are fundamental in modeling dynamic systems because they describe how the state of the system evolves based on its current state and input. In chemical engineering, these equations are vital in designing and operating reactors, as they explain how reactants turn into products over time.
Numerical methods
Solving differential equations analytically is not always possible, especially for complex systems.

This is where numerical methods come in handy. Numerical methods provide approximate solutions to ODEs using algorithms rather than finding exact solutions. In our situation, since the ODEs are first-order and coupled, numerical approaches like Euler's method or the Runge-Kutta method are appropriate.
  • Euler's method is simple, taking small steps in time to predict future values based on current rate of change.
  • The Runge-Kutta methods, especially the fourth-order variant, are more accurate, using multiple estimates per step to improve predictions.
  • Computational tools like Python's SciPy library or MATLAB are beneficial here, with built-in functions like `ode45` that manage these calculations efficiently.
Utilizing these methods helps in understanding the dynamics of the system over the specified time frame, giving insights into concentration variations from time zero to ten minutes.
Chemical reactions
Chemical reactions are processes where substances, called reactants, transform into different substances, known as products.

The conversion from A to B in the problem is a straightforward example of a chemical reaction occurring in a reactor. The reaction rate for these transformations is dictated by a rate constant, \(k\).

In our scenario, the reactant A, enters the first reactor and through chemical reactions, forms product B. The rate at which this happens is captured by the rate constant \(k\), and it's applied to the concentrations of A and B to compute their rates of change over time.

Understanding these transformations is essential. It allows engineers to design reactors optimally, ensuring efficient conversion rates and predicting how quickly products can be expected at a given moment. Moreover, analyzing such reactions helps in scaling up from laboratory to industrial scales, maintaining the desired product quality and yield.

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Most popular questions from this chapter

The following ordinary differential equation describes the motion of a damped spring-mass system (Fig. \(\mathrm{P} 28.46\) ): \\[ m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+b x^{3}=0 \\] where \(x=\) displacement from the equilibrium position, \(t=\) time, \(m=1 \mathrm{kg}\) mass, and \(a=5 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2}\). The damping term is nonlinear and represents air damping. The spring is a cubic spring and is also nonlinear with \(b=5 \mathrm{N} / \mathrm{m}^{3}\) The initial conditions are Initial velocity \(\frac{d x}{d t}=0.5 \mathrm{m} / \mathrm{s}\) Initial displacement \(\quad x=1 \mathrm{m}\) Solve this equation using a numerical method over the time period \(0 \leq t \leq 8\) s. Plot the displacement and velocity versus time and plot the phase- plane portrait (velocity versus displacement) for all the following cases: (a) A similar linear equation \\[ m \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+5 x=0 \\] (b) The nonlinear equation with only a nonlinear spring term $$\frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+b x^{3}=0$$ (c) The nonlinear equation with only a nonlinear damping term $$m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+5 x=0$$ (d) The full nonlinear equation where both the damping and spring terms are nonlinear $$m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+b x^{3}=0$$

A spherical ice cube (an "ice sphere") that is \(6 \mathrm{cm}\) in diameter is removed from a \(0^{\circ} \mathrm{C}\) freezer and placed on a mesh screen at room temperature \(T_{a}=20^{\circ} \mathrm{C} .\) What will be the diameter of the ice cube as a function of time out of the freezer (assuming that all the water that has melted immediately drips through the screen)? The heat transfer coefficient \(h\) for a sphere in a still room is about \(3 \mathrm{W} /\left(\mathrm{m}^{2} \cdot \mathrm{K}\right) .\) The heat flux from the ice sphere to the air is given by $$\mathrm{Flux}=\frac{q}{A}=h\left(T_{\mathrm{a}}-T\right)$$ where \(q=\) heat and \(A=\) surface area of the sphere. Use a numerical method to make your calculation. Note that the latent heat of fusion is \(333 \mathrm{kJ} / \mathrm{kg}\) and the density of ice is approximately \(0.917 \mathrm{kg} / \mathrm{m}^{3}\).

A nonisothermal batch reactor can be described by the following equations: $$\begin{aligned} &\frac{d C}{d t}=-e^{(-10 /(T+273))} C\\\ &\frac{d T}{d t}=1000 e^{(-10 /(T+273))} C-10(T-20) \end{aligned}$$ where \(C\) is the concentration of the reactant and \(T\) is the temperature of the reactor. Initially the reactor is at \(15^{\circ} \mathrm{C}\) and has a concentration of reactant \(C\) of \(1.0 \mathrm{gmol} / \mathrm{L}\). Find the concentration and temperature of the reactor as a function of time.

In Prob. \(28.30,\) a linearized groundwater model was used to simulate the height of the water table for an unconfined aquifer. A more realistic result can be obtained by using the following nonlinear ODE: \\[ \frac{d}{d x}\left(K h \frac{d h}{d x}\right)+N=0 \\] where \(x=\) distance \((\mathrm{m}), K=\) hydraulic conductivity \([\mathrm{m} / \mathrm{d}], h=\) height of the water table \([\mathrm{m}],\) and \(N=\) infiltration rate \([\mathrm{m} / \mathrm{d}] .\) Solve for the height of the water table for the same case as in Prob. 28.30 That is solve from \(x=0\) to \(1000 \mathrm{m}\) with \(h(0)=10 \mathrm{m}, h(1000)=5 \mathrm{m}\) \(K=1 \mathrm{m} / \mathrm{d},\) and \(N=0.1 \mathrm{m} / \mathrm{d} .\) Obtain your solution with (a) the shooting method and (b) the finite-difference method \((\Delta x=100 \mathrm{m})\).

Bacteria growing in a batch reactor utilize a soluble food source (substrate) as depicted in Fig. P28.16. The uptake of the substrate is represented by a logistic model with Michaelis-Menten limitation. Death of the bacteria produces detritus which is subsequently converted to the substrate by hydrolysis. In addition, the bacteria also excrete some substrate directly. Death, hydrolysis and excretion are all simulated as first-order reactions. Mass balances can be written as \(\frac{d X}{d t}=\mu_{\max }\left(1-\frac{X}{K}\right)\left(\frac{S}{K_{s}+S}\right) X-k_{d} X-k_{e} X\) \(\frac{d C}{d t}=k_{d} X-k_{h} C\) \(\frac{d S}{d t}=k_{e} X+k_{h} C-\mu_{\max }\left(1-\frac{X}{K}\right)\left(\frac{S}{K_{s}+S}\right) X\) where \(X, C,\) and \(S=\) the concentrations \([\mathrm{mg} / \mathrm{L}]\) of bacteria, detritus, and substrate, respectively; \(\mu_{\max }=\) maximum growth rate [/d], \(K=\) the logistic carrying capacity \([\mathrm{mg} / \mathrm{L}] ; K_{s}=\) the Michaelis-Menten half-saturation constant \([\mathrm{mg} / \mathrm{L}], k_{d}=\) death rate \([/ \mathrm{d}] ; k_{e}=\) excretion rate \([/ \mathrm{d}] ;\) and \(k_{h}=\) hydrolysis rate \([/ \mathrm{d}] .\) Simulate the concentrations from \(t=0\) to 100 d, given the initial conditions \(X(0)=1 \mathrm{mg} / \mathrm{L}, S(0)=\) \(100 \mathrm{mg} / \mathrm{L}\) and \(C(0)=0 \mathrm{mg} / \mathrm{L} .\) Employ the following parameters in your calculation: \(\mu_{\max }=10 / \mathrm{d}, K=10 \mathrm{mg} / \mathrm{L}, K_{s}=10 \mathrm{mg} / \mathrm{L}, k_{d}=\) \(0.1 / \mathrm{d}, k_{e}=0.1 / \mathrm{d},\) and \(k_{h}=0.1 / \mathrm{d}\).

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