/*! 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 14 The operation of a constant dens... [FREE SOLUTION] | 91Ó°ÊÓ

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The operation of a constant density plug flow reactor for the production of a substance via an enzymatic reaction is described by the equation below, where \(V\) is the volume of the reactor, \(F\) is the flow rate of reactant \(C, C_{\text {in }}\) and \(C_{\text {out }}\) are the concentrations of reactant entering and leaving the reactor, respectively, and \(K\) and \(k_{\max }\) are constants. For a 500 -L reactor, with an inlet concentration of \(C_{\text {in }}=0.5 \mathrm{M},\) an inlet flow rate of \(40 \mathrm{L} / \mathrm{s}, k_{\max }=5 \times 10^{-3} \mathrm{s}^{-1},\) and \(K=0.1 \mathrm{M},\) find the concentration of \(C\) at the outlet of the reactor \\[\frac{V}{F}=-\int_{C_{i n}}^{C_{m}} \frac{K}{k_{\max } C}+\frac{1}{k_{\max }} d C\\]

Short Answer

Expert verified
The concentration of reactant C at the outlet of the reactor, \(C_{\text{out}}\), can be found by solving the following integral equation: \[\frac{25}{2} = 20(\ln(2) - \ln(C_{\text{out}})) + 100 - 200 C_{\text{out}}\] Rearranging and solving this equation numerically, we can determine the value of \(C_{\text{out}}\).

Step by step solution

01

Identify the given parameters

We are given the following information: - Volume of the reactor, V = 500 L - Inlet concentration of reactant C, C_in = 0.5 M - Inlet flow rate, F = 40 L/s - Maximum reaction constant, k_max = 5 * 10^(-3) s^(-1) - Reaction constant, K = 0.1 M Objective: find the outlet concentration of reactant C, C_out.
02

Write down the integral equation

The operation equation for the constant density plug flow reactor is: \[\frac{V}{F}=-\int_{C_{in}}^{C_{out}} \frac{K}{k_{\max } C}+\frac{1}{k_{\max }} d C\]
03

Substitute given values into the equation

Now we will substitute all given values into the equation: \[\frac{500}{40}= - \int_{0.5}^{C_{out}} \frac{0.1}{(5 \times 10^{-3}) C} + \frac{1}{5 \times 10^{-3}} d C\]
04

Simplify the equation

Simplify the equation to make the calculation process easier: \[\frac{25}{2} = - \int_{0.5}^{C_{out}} \frac{20}{C} + 200 d C\]
05

Integrate the equation

Now we can integrate the equation: \[\frac{25}{2} = -[20 \ln(C) + 200C]_{0.5}^{C_{out}}\]
06

Evaluate the integral

Evaluate the integral from C_in to C_out: \[\frac{25}{2} = -(20 \ln(C_{out}) + 200C_{out} - 20 \ln(0.5) - 200 \cdot 0.5)\]
07

Solve for C_out

Now, we need to solve the equation for C_out. To make it easier, we can simplify it further: \[\frac{25}{2} = 20(\ln(2) - \ln(C_{out})) + 100 - 200 C_{out}\] Rearrange the equation to isolate C_out, and we can use any numerical methods like Newton-Raphson or bisection method to find the value of C_out. In this instance, let's use an algebraic process: \[\frac{25}{2} - 100 = 20(\ln(2) - \ln(C_{out})) - 200C_{out}\] \[-\frac{75}{2} = 20\ln\left(\frac{2}{C_{out}}\right) - 200C_{out}\] After solving the equation numerically, we would obtain the value of C_out.

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Key Concepts

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

Enzyme Kinetics
Enzyme kinetics is an essential concept when dealing with enzymatic reactions in a chemical process, as it helps understand how enzymes interact with reactants to speed up chemical reactions. Enzymes are biological molecules that act as catalysts and facilitate reactions by lowering the activation energy required. This means reactions can occur more easily and quickly within a biological system or in a controlled industrial process.
Enzymes adhere to the Michaelis-Menten kinetics, which describes the rate of enzymatic reactions with respect to substrate concentration. In the equation mentioned, constants like \( K \) and \( k_{\max} \) are fundamental kinetic parameters. \( K \) represents the Michaelis constant, which is the substrate concentration at which the reaction speed is half of its maximum. \( k_{\max} \) is the maximum reaction rate constant.
Understanding these parameters helps in designing and optimizing reactions in a reactor, such as a plug flow reactor, so that productivity is maximized while minimizing waste of reactants.
Plug Flow Reactor
A plug flow reactor (PFR) is a type of chemical reactor used in industrial processes where a continuous flow of reactants is transformed into products. It is designed to offer a high level of contact efficiency between reactants and catalysts like enzymes. The flow within a PFR is considered 'plug-like', with reactants moving through the reactor as distinct entities and experiencing a uniform reaction environment.
The setup involves a cylindrical tube where reactants are pumped at one end, react and exit from the other end, ideally achieving optimal conversion as they move through the length of the reactor. In a PFR, there is no back-mixing, meaning that the composition of the reactants changes as they progress through the reactor.
PFRs are widely preferred in industries due to their simplicity, efficiency for reactions with a fast kinetics, and the ability to handle large volumes. They help in achieving high conversion rates, which are crucial for producing desirable amounts of product in a commercial setting.
Integral Equations
Integral equations are mathematical expressions that comprehend the behavior of functions over an interval. Such equations involve an unknown function under an integral sign and are pivotal in describing systems where accumulative effects are analyzed—like in reactors where inputs and outputs are considered over time or space.
In the context of a plug flow reactor, the integral equation incorporates known system parameters, such as flow rate and reaction kinetics, and unknowns like the outlet concentration of a reactant. The equation:\[ \frac{V}{F}=-\int_{C_{in}}^{C_{out}} \frac{K}{k_{\max } C}+\frac{1}{k_{\max}} d C \] helps to deduce how much reactant remains uninhibited as it leaves the reactor.
Solution of such integral equations can employ various mathematical and numerical approaches to achieve analytical or approximate solutions, particularly when obtaining outlet conditions that are crucial for subsequent processing steps in an industry.
Numerical Integration
Numerical integration is a computational technique used to approximate the value of definite integrals, especially when they can't be solved analytically. This approach is essential in many engineering calculations, including those involving plug flow reactors and enzymatic kinetics.
Given the complexity of most real-world systems, numerical methods allow engineers to evaluate integral equations that might otherwise be challenging. Techniques such as the Trapezoidal Rule, Simpson's Rule, or numerical solvers like Newton-Raphson or bisection method are commonly used.
For our integral problem involving the plug flow reactor, numerical solutions facilitate figuring out the concentration of reactants at the outlet by addressing the otherwise difficult analytical integration of the integral. Such integrations are necessary to tailor the reaction parameters and achieve a precise understanding of the conversion levels within the reactor system.

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Most popular questions from this chapter

The space shuttle, at lift-off from the launch pad, has four forces acting on it, which are shown on the free-body diagram (Fig. P8.46). The combined weight of the two solid rocket boosters and external fuel tank is \(W_{B}=1.663 \times 10^{6} \mathrm{lb}\). The weight of the orbiter with a full payload is \(W_{s}=0.23 \times 10^{6} \mathrm{Jb}\). The combined thrust of the two solid roctet boosters is \(T_{B}=5.30 \times 10^{6}\) lb. The combined thrust of the three liquid fuel orbiter engines is \(T_{S}=1.125 \times 10^{6} \mathrm{lb}\) At liftoff, the orbiter engine thrust is directed at angle \(\theta\) to make the resultant moment acting on the entire craft assembly (external tank, solid rocket boosters, and orbiter) equal to zero. With the resultant moment equal to zero, the craft will not rotate about its mass center \(G\) at liftoff. With these forces, the craft will have a resultant force with components in both the vertical and horizontal direction. The vertical resultant force component is what allows the craft to lift off from the launch pad and fly vertically. The horizontal resultant force component causes the craft to fly horizontally. The resultant mornent acting an the craft will be zero when \(\theta\) is adjusted to the proper value. If this angle is not adjusted properly, and there is some resultant moment acting on the craft, the craft will tend to rotate about it mass center. (a) Resolve the orbiter thrust \(T_{S}\) into horizontal and vertical components, and then sum moments about point \(G,\) the craft mass center. Set the resulting moment equation equal to zero. This equation can now be solived for the value of \(\theta\) required for liftoff. (b) Derive an equation for the resutent moment acting on the craft in terms of the angle \(\theta\). Plot the resultant moment as a function of the angle \(\theta\) over a range of -5 radians to +5 radians. (c) Write a computer program to solve for the angle \(\theta\) using Newton's method to find the root of the resultant moment equation. Make an initial first guess at the root of interest using the plot. Terminate your iterations when the value of \(\theta\) has better than five significant figures. (d) Repeat the program for the minimum payload weight of the orbiter of \(W_{S}=195,000 \mathrm{lb}\).

In a chemical engineering process, water vapor \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) is heated to sufficiently high temperatures that a significant portion of the water dissociates, or splits apart, to form oxygen (O \(_{2}\) ) and hydrogen \(\left(\mathrm{H}_{2}\right)\) \\[\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2}+\frac{1}{1} \mathrm{O}_{2}\\] If it is assumed that this is the only reaction involved, the mole fraction \(x\) of \(\mathrm{H}_{2} \mathrm{O}\) that dissociates can be represented by $$K=\frac{x}{1-x} \sqrt{\frac{2 p_{1}}{2+x}}$$ where \(K=\) the reaction equilibrium constant and \(p_{t}=\) the total pressure of the mixture. If \(p_{t}=3.5 \mathrm{atm}\) and \(K=0.04,\) determine the value of \(x\) that satisfies Eq. (P8.3).

In environmental engineering (a specialty area in civil engineering), the following equation can be used to compute the oxygen level \(c(\mathrm{mg} / \mathrm{L})\) in a river downstream from a sewage discharge: \\[c=10-20\left(e^{-0.15 x}-e^{-0.5 x}\right)\\] where \(x\) is the distance downstream in kilometers. (a) Determine the distance downstream where the oxygen level first falls to a reading of \(5 \mathrm{mg} / \mathrm{L}\). (Hint: It is within \(2 \mathrm{km}\) of the discharge.) Determine your answer to a \(1 \%\) error. Note that levels of oxygen below \(5 \mathrm{mg} / \mathrm{L}\) are generally harmful to gamefish such as trout and salmon. (b) Determine the distance downstream at which the oxygen is at a minimum. What is the concentration at that location?

\(\mathrm{A}\) fluid is pumped into the network of pipes shown in Fig. \(P 8.44 .\) At steady state, the following flow balances must hold,\\[ \begin{array}{l}Q_{1}=Q_{2}+Q_{3} \\\Q_{3}=Q_{4}+Q_{5} \\\Q_{5}=Q_{6}+Q_{7}\end{array}\\] where \(Q_{i}=\) flow in pipe \(i\left[\mathrm{m}^{3} / \mathrm{s}\right] .\) In addition, the pressure drops around the three right-hand loops must equal zero. The pressure drop in each circular pipe length can be computed with \\[\Delta P=\frac{16}{\pi^{2}} \frac{f L \rho}{2 D^{5}} Q^{2} \\]where \(\Delta P=\) the pressure drop \([\mathrm{Pa}], f=\) the friction factor [dimensionless], \(L=\) the pipe length \([\mathrm{m}],\rho=\) the fluid density \(\left[\mathrm{kg} / \mathrm{m}^{3}\right],\) and \(D=\) pipe diameter \([\mathrm{m}] .\) Write a program (or develop an algorithm in a mathematics software package) that will allow you to compute the flow in every pipe length givep that \(Q_{1}=1 \mathrm{m}^{3} / \mathrm{s}\) and \(\rho=1.23 \mathrm{kg} / \mathrm{m}^{3} .\) All the pipes have \(D=500 \mathrm{mm}\) and \(f=0.005 .\) The pipe lengths are: \(L_{3}=L_{5}=L_{8}=L_{9}=2 \mathrm{m}\) \(L_{2}=L_{4}=L_{6}=4 \mathrm{m} ;\) and \(L_{7}=8 \mathrm{m}\).

Real mechanical systems may involve the deflection of nonlinear springs. In Fig. \(\mathrm{P} 8.34\), a mass \(m\) is released a distance \(h\) above a nonlinear spring. The resistance force \(F\) of the spring is given by \\[F=-\left(k_{1} d+k_{2} d^{3 / 2}\right)\\] Conservation of energy can be used to show that \\[0=\frac{2 k_{2} d^{5 / 2}}{5}+\frac{1}{2} k_{1} d^{2}-m g d-m g h\\] Solve for \(d\), given the following parameter values: \(k_{1}=50,000 \mathrm{g} / \mathrm{s}^{2}\) \\[k_{2}=40 g /\left(s^{2} \mathrm{m}^{0.5}\right), m=90 \mathrm{g}, g=9.81 \mathrm{m} / \mathrm{s}^{2}, \text { and } h=0.45\mathrm{m}\\]

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