/*! 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 13 A biofilm with a thickness, \(L_... [FREE SOLUTION] | 91Ó°ÊÓ

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A biofilm with a thickness, \(L_{f}[\mathrm{cm}]\), grows on the surface of a solid (Fig. \(P 28.13\) ). After traversing a diffusion layer of thickness, \(L[\mathrm{cm}],\) a chemical compound, \(A,\) diffuses into the biofilm where it is subject to an irreversible first-order reaction that converts it to a product, \(B\) Steady-state mass balances can be used to derive the following ordinary differential equations for compound \(A\) : $$\begin{aligned} &D \frac{d^{2} c_{a}}{d x^{2}}=0 \quad 0 \leq x

Short Answer

Expert verified
In summary, to compute the steady-state distribution of compound A, we need to solve the given ordinary differential equations using the finite-difference method with centered finite differences. Analyzing the problem in two regions, the diffusion layer and the biofilm, we define the finite-difference equations and boundary conditions for both regions. Then, we create an array of grid points and apply the boundary conditions. By rearranging the finite-difference equations, we obtain a system of linear equations with unknown concentrations, which can be solved using suitable methods. Finally, we compute the steady-state distribution of A at each grid point from x=0 to x=L+Lf. Linear interpolation can be used to find the concentration of A at intermediate values of x.

Step by step solution

01

Analyze the regions and the finite-difference method

We have two regions for our analysis: the diffusion layer (0 ≤ x < L) and the biofilm (L ≤ x < L+Lf). For each region, we need to apply the finite-difference method using centered finite differences with Δx = 0.001 cm.
02

Define the finite-difference equations for both regions

In the diffusion layer (0 ≤ x < L): $$D \frac{c_{a}(x-\Delta x)-2 c_{a}(x)+c_{a}(x+\Delta x)}{(\Delta x)^{2}}=0$$ In the biofilm (L ≤ x < L+Lf): $$D_{f} \frac{c_{a}(x-\Delta x)-2 c_{a}(x)+c_{a}(x+ \Delta x)}{(\Delta x)^{2}}-k c_{a}(x)=0$$
03

Define the grid points for solving the equations

First, we need to find the total number of grid points we need to solve the equation, which can be computed as: Total grid points = (L + Lf) / Δx + 1 = (0.008 + 0.004) / 0.001 + 1 = 13 Now, we will create an array of 13 points representing the concentration of A at each point. The grid points will be numbered from 0 to 12 to make it more convenient, with x = 0 at grid point 0 and x = L+Lf at grid point 12.
04

Apply the boundary conditions and solve the finite-difference equations

Apply the first boundary condition at x=0 (grid point 0): $$c_{a}(0)=c_{a 0}=100 \, \mathrm{mol/L}$$ Apply the second boundary condition at x=L+Lf (grid point 12): $$\frac{c_{a}(12)-c_{a}(11)}{\Delta x}=0$$ Now, by rearranging the finite-difference equations from Step 2, we can obtain an equation for each unknown concentration, making use of the known values of D, Df, k, and Δx. For the diffusion layer (grid points 1 to 7): $$c_{a}(x-1)-2 c_{a}(x)+c_{a}(x+1)=0$$ For the biofilm (grid points 8 to 11): $$c_{a}(x-1)-2 \left(1+\frac{k(\Delta x)^{2}}{D_{f}}\right) c_{a}(x)+c_{a}(x+1)=0$$ Now we have a system of linear equations with 12 unknowns, which can be solved using any suitable method, such as matrix inversion or the Gaussian elimination method.
05

Compute the steady-state distribution of A

After solving the system of linear equations, we will have the steady-state distribution of A at each grid point from x=0 to x=L+Lf. To find the concentration of A at intermediate values of x, linear interpolation can be used between the grid points.

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Key Concepts

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

Understanding Ordinary Differential Equations (ODEs)
Ordinary Differential Equations (ODEs) are a fundamental part of mathematical modeling of systems because they describe how a quantity changes over time or space. ODEs are characterized by the presence of derivatives of a dependent variable, such as concentration, with respect to only one independent variable, such as time or spatial coordinates. In the given exercise, the changes in concentration of a chemical compound, A, within a biofilm, are represented by a pair of ODEs. Each equation applies to a different region; one for the diffusion layer and another for within the biofilm itself. To solve these equations, numerical methods like the finite-difference method are often employed when analytical solutions are either overly complex or not available.

When dealing with the finite-difference method for ODEs, the idea is to approximate the continuous derivatives by differences at discrete points. Selecting adequate step sizes, such as \( \Delta x \) in the exercise, is crucial for balancing accuracy and computational efficiency. If the steps are too large, the solution may not be sufficiently accurate, while too small steps might require more computation time without significant gain in precision.
Diffusion Processes
Diffusion is a process by which particles spread out as a result of random motion from regions of higher concentration to regions of lower concentration. The given problem involves diffusion of a chemical within a biofilm, which is a common scenario in fields such as biology, environmental science, and chemical engineering. In the exercise, two different diffusion coefficients, \( D \) and \( D_{f} \) for the diffusion layer and the biofilm respectively, account for differing diffusion rates in the two media. Understanding how diffusion occurs and is modeled mathematically is critical to solving related problems effectively.

Different scenarios call for different types of diffusion models. For steady-state problems where the concentration does not change with time, ODEs like those provided manage to encapsulate the diffusion behavior within the system. Numerical methods come into play when the equations get too complicated for a simple analytical approach.
Steady-State Mass Balances
Mass balances are equations that describe the conservation of mass in a system. A steady-state mass balance implies that the amount of mass entering a system equals the amount leaving, such that there is no accumulation of mass within the system over time. The ODEs in the exercise represent a steady-state condition for the diffusing compound A. The first equation evidences no accumulation in the diffusion layer, while the second shows a balance between diffusion and reaction in the biofilm.

In essence, the exercise challenges the student to use steady-state mass balances to predict how compound A's concentration changes throughout the system. This is a typical scenario in process engineering and environmental systems where understanding the distribution and reaction of substances is necessary to design and operate processes like waste treatment or chemical manufacturing.
Numerical Methods
Numerical methods are a branch of mathematics and computer science that create, analyze, and implement algorithms for obtaining numerical solutions to problems for which analytical solutions are difficult or impossible to obtain. The finite-difference method used in the exercise is one such numerical method, which replaces continuous derivative terms with finite difference approximations. To solve the given ODEs, the exercise divides into a set of algebraic equations to be solved iteratively or using matrix methods.

The finite-difference method is particularly useful in handling boundary-value problems, such as those in the exercise, where information about the solution is known at more than one point. By applying this method together with suitable boundary conditions and a well-chosen grid, students can derive the concentration distribution across the biofilm layer and diffusion layer – a handy technique for many scientific and engineering calculations.

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Most popular questions from this chapter

Seawater with a concentration of \(8000 \mathrm{g} / \mathrm{m}^{3}\) is pumped into a well-mixed tank at a rate of \(0.6 \mathrm{m}^{3} / \mathrm{hr}\). Because of faulty design work, water is evaporating from the tank at a rate of \(0.025 \mathrm{m}^{3} / \mathrm{hr}\). The salt solution leaves the tank at a rate of \(0.6 \mathrm{m}^{3} / \mathrm{hr}\) (a) If the tank originally contains \(1 \mathrm{m}^{3}\) of the inlet solution, how long after the outlet pump is turned on will the tank run dry? (b) Use numerical methods to determine the salt concentration in the tank as a function of time.

The rate of cooling of a body can be expressed as \\[ \frac{d T}{d t}=-k\left(T-T_{a}\right) \\] where \(T=\) temperature of the body \(\left(^{\circ} \mathrm{C}\right), T_{a}=\) temperature of the surrounding medium \(\left(^{\circ} \mathrm{C}\right),\) and \(k=\) the proportionality constant \(\left(\min ^{-1}\right) .\) Thus, this equation specifies that the rate of cooling is proportional to the difference in temperature between the body and the surrounding medium. If a metal ball heated to \(90^{\circ} \mathrm{C}\) is dropped into water that is held at a constant value of \(T_{a}=20^{\circ} \mathrm{C}\), use a numerical method to compute how long it takes the ball to cool to \(40^{\circ} \mathrm{C}\) if \(k=0.25 \mathrm{min}^{-1}\).

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.

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 \(k_{h}=0.1 / \mathrm{d}\).

The following equation can be used to model the deflection of a sailboat mast subject to a wind force: \\[ \frac{d^{2} y}{d z^{2}}=\frac{f}{2 E I}(L-z)^{2} \\] where \(f=\) wind force, \(E=\) modulus of elasticity, \(L=\) mast length and \(I=\) moment of inertia. Calculate the deflection if \(y=0\) and \(d y / d z=0\) at \(z=0 .\) Use parameter values of \(f=60, L=30, E=\) \(1.25 \times 10^{8},\) and \(I=0.05\) for your computation.

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