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Chapter 28: Problem 40

Just as Fourier's law and the heat balance can be employed to characterize temperature distribution, analogous relationships are available to model field problems in other areas of engineering. For example, electrical engineers use a similar approach when modeling electrostatic fields. Under a number of simplifying assumptions, an analog of Fourier's law can be represented in one- dimensional form as \\[ D=-\varepsilon \frac{d V}{d x} \\] where \(D\) is called the electric flux density vector, \(\varepsilon=\) permittivity of the material, and \(V=\) electrostatic potential. Similarly, a Poisson equation for electrostatic fields can be represented in one dimension as $$\frac{d^{2} V}{d x^{2}}=-\frac{\rho_{v}}{\varepsilon}$$ where \(\rho_{v}=\) charge density. Use the finite-difference technique with \(\Delta x=2\) to determine \(V\) for a wire where \(V(0)=1000, V(20)\) \(=0, \varepsilon=2, L=20,\) and \(\rho_{v}=30\).

Short Answer

Expert verified
Using the finite-difference technique with the given conditions, we discretize the one-dimensional Poisson equation and apply the boundary conditions \(V(0)=1000\) and \(V(20)=0\). Formulating a system of linear equations based on the discretized equation, we solve for the values of \(V(2), V(4), \cdots, V(18)\). Finally, we obtain the electrostatic potential at all points in the wire, given by \(V = [V(0), V(2), V(4), \cdots, V(18), V(20)]\).

Step by step solution

01

Discretize the equation

We will discretize the given Poisson equation using the second-order central difference formula. Given that \(\frac{d^{2} V}{d x^{2}} \approx \frac{V_{i-1}-2V_{i}+V_{i+1}}{(\Delta x)^2}\), our discretized equation becomes: $$\frac{V_{i-1}-2V_{i}+V_{i+1}}{(2)^2}=-\frac{30}{2}$$ #Step 2: Include boundary conditions
02

Apply boundary conditions

We know \(V(0) = 1000\) and \(V(20) = 0\). We will use these boundary conditions to modify the discretized equation. Following the one-dimensional finite-difference technique, we can set up a system of linear equations for resolving the unknowns \(V(2), V(4), \cdots V(18)\). #Step 3: Formulate the system of linear equations
03

Write the linear equations

Using the discretized equation, we can create the following system of linear equations to be solved: $$V_{-1}-2V_{2}+V_{4}=-30$$ $$V_{0}-2V_{4}+V_{6}=-30$$ $$V_{2}-2V_{6}+V_{8}=-30$$ $$\cdots$$ $$V_{14}-2V_{16}+V_{18}=-30$$ $$V_{16}-2V_{18}+V_{20}=-30$$ The boundary conditions mentioned earlier can be substituted in these equations as \(V(0) = 1000\) and \(V(20) = 0\). This provides us with the linear equations with the constants on the right-hand side: $$V_{-1}-2V_{2}+V_{4}=-30-1000$$ $$V_{0}-2V_{4}+V_{6}=-30+2000$$ $$V_{2}-2V_{6}+V_{8}=-30$$ $$\cdots$$ $$V_{14}-2V_{16}+V_{18}=-30$$ $$V_{16}-2V_{18}+V_{20}=-30$$ #Step 4: Solve the system of linear equations
04

Solve the linear system

Next, solve the system of linear equations using any standard linear equation solving method, such as Gaussian elimination, matrix inversion, or an iterative method like Jacobi or Gauss-Seidel. This will give us the values of \(V(2), V(4), \cdots, V(18)\). #Step 5: Find the electrostatic potential values
05

Compute the electrostatic potential

With the values of \(V(2), V(4), \cdots, V(18)\) obtained from solving the linear system, we now have the electrostatic potential at all points in the wire: $$V = [V(0), V(2), V(4), \cdots, V(18), V(20)]$$

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Key Concepts

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

Poisson Equation
The Poisson Equation is a fundamental partial differential equation in physics and engineering that describes how a potential field responds to the presence of a source. In electrostatics, it connects the electrostatic potential \(V\) with the charge density \(\rho_v\). Typically represented as:\[ \frac{d^{2} V}{d x^{2}} = -\frac{\rho_v}{\varepsilon} \]Here, \(\varepsilon\) is the permittivity of the material, essentially a measure of how an electric field affects and is affected by a dielectric medium. The equation tells us how the second derivative of the potential \(V\) relates to the charge density \(\rho_v\), showing that the curvature of the potential field increases with higher charge density. This relationship is crucial for understanding how electrostatic forces distribute within a given space.
Electrostatic Fields
Electrostatic fields are vector fields surrounding charged particles or objects, representing the force experienced by other charges within the field. These fields are generated by static charges and do not change with time.- **Properties**: - Electrostatic fields originate from positive charges and terminate at negative charges. - They obey the inverse square law, meaning the field strength decreases with the square of the distance from the source charge. - The lines of force never cross each other and are perpendicular to the surface of a charged conductor at equilibrium.- **Role in Poisson Equation**: - The potential \(V\) from the Poisson Equation helps define the behavior of these fields. - By understanding the potential, one can determine the electric field as the negative gradient of the potential.
Boundary Conditions
In mathematics and physics, boundary conditions are constraints necessary for solving differential equations, like the Poisson Equation, especially in finite domains. For electrostatic problems:- **Types of Boundary Conditions**: - **Dirichlet Boundary Condition**: Values of the function (potential \(V\)) are provided at the boundaries (e.g., \(V(0) = 1000, V(20) = 0\)). - **Neumann Boundary Condition**: The derivative (or gradient) of the potential is specified at the boundary. This condition might describe a situation where the electric field intensity is known.- **Importance**: - They ensure a unique solution to the differential equation by constraining the potential \(V\) within given limits. - Proper application allows for the accurate modeling of physical scenarios, such as the potential along a wire in our example.
Discretization
Discretization is the process of transforming continuous models and equations, like differential equations, into discrete counterparts. This is essential when solving Poisson's Equation numerically using methods such as the finite difference method. - **Finite Difference Method**: - Approximates derivatives by differences and divides the domain into small segments of equal length, termed as grid points. - For example, the second derivative in Poisson's Equation is discretized using a central difference approximation. - **Why Discretize**: - It enables the solving of complex equations that aren't manageable analytically by converting them into algebraic equations. - Makes it possible to apply numerical strategies and computational algorithms on discrete points, facilitating efficient computation even for large systems.

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Most popular questions from this chapter

The following ODEs have been proposed as a model of an epidemic: \\[ \begin{array}{l} \frac{d S}{d t}=-a S I \\ \frac{d I}{d t}=a S I-r I \\ \frac{d R}{d t}=r I \end{array} \\] where \(S=\) the susceptible individuals, \(I=\) the infected, \(R=\) the recovered, \(a=\) the infection rate, and \(r=\) the recovery rate. A city has 10,000 people, all of whom are susceptible. (a) If a single infectious individual enters the city at \(t=0\) compute the progression of the epidemic until the number of infected individuals falls below \(10 .\) Use the following parameters: \(a=0.002 /\) (person \(\cdot\) week ) and \(r=0.15 /\) d. Develop time series plots of all the state variables. Also generate a phase plane plot of \(S\) versus \(I\) versus \(R\) (b) Suppose that after recovery, there is a loss of immunity that causes recovered individuals to become susceptible. This reinfection mechanism can be computed as \(\rho R,\) where \(\rho=\) the reinfection rate. Modify the model to include this mechanism and repeat the computations in (a) using \(\rho=0.015 / \mathrm{d}\).

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 forced damped spring-mass system (Fig. \(\mathrm{P} 28.47\) ) has the following ordinary differential equation of motion: \\[ m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+k x=F_{o} \sin (\omega t) \\] where \(x=\) displacement from the equilibrium position, \(t=\) time, \(m=2 \mathrm{kg}\) mass, \(a=5 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2},\) and \(k=6 \mathrm{N} / \mathrm{m} .\) The damping term is nonlinear and represents air damping. The forcing function \(F_{o} \sin (\omega t)\) has values of \(F_{o}=2.5 \mathrm{N}\) and \(\omega=0.5 \mathrm{rad} / \mathrm{sec} .\) The initial conditions are Initial velocity \(\frac{d x}{d t}=0 \mathrm{m} / \mathrm{s}\) Initial displacement \(x=1 \mathrm{m}\) Solve this equation using a numerical method over the time period \(0 \leq t \leq 15\) s. Plot the displacement and velocity versus time, and plot the forcing function on the same curve. Also, develop a separate plot of velocity versus displacement.

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.

For a simple \(\mathrm{RL}\) circuit, Kirchhoff's voltage law requires that (if Ohm's law holds) \\[ L \frac{d i}{d t}+R i=0 \\] where \(i=\) current, \(L=\) inductance, and \(R=\) resistance. Solve for \(i\) if \(L=1, R=1.5,\) and \(i(0)=0.5 .\) Solve this problem analytically and with a numerical method. Present your results graphically.
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