EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 9780100254145
Author: Chapra
Publisher: YUZU
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 28, Problem 7P

The following equations define the concentrations of threereactants:

d c a d t = 10 c a c c + c b d c b d t = 10 c a c c c b d c c d t = 10 c a c c + c b 2 c c

If the initial conditions are c a = 50 , c b = 0 ,  and  c c = 40 , find the concentrations for the times from 0 to 3 s.

Expert Solution & Answer
Check Mark
To determine

To calculate: The concentration for the times from 0 to 3 s. Where the equations for concentrations of three reactants are,

dcadt=10cacc+cbdcbdt=10cacccbdccdt=10cacc+cb2cc

Initials conditions are ca=50,cb=0 and cc=40

Answer to Problem 7P

Solution:

The concentration of the reactants ca,cb and cc at t=0 s are 50,0,40 respectively.

The concentration of the reactants ca,cb and cc at t=1 s are 50,0,5.4135 respectively.

The concentration of the reactants ca,cb and cc at t=2 s are 50,0,5.4135 respectively.

The concentration of the reactants ca,cb and cc at t=3 s are 50,0,5.4135 respectively.

Explanation of Solution

Given information:

The system of equations,

dcadt=10cacc+cbdcbdt=10cacccbdccdt=10cacc+cb2cc

Initial conditions, ca=50,cb=0 and cc=40

Formula used:

To calculate the values of ca,cb and cc: Jacobi matrix is,

J(ca,cb,cc)=[fca(ca,cb,cc)fcb(ca,cb,cc)fcc(ca,cb,cc)gca(ca,cb,cc)gcb(ca,cb,cc)gcc(ca,cb,cc)hca(ca,cb,cc)hcb(ca,cb,cc)hcc(ca,cb,cc)]

Eigen value λ of the matrix A can be calculated as,

|AλI|=0

Calculation:

Consider the system of first order nonlinear differential equation of reactants

dcadt=10cacc+cbdcbdt=10cacccbdccdt=10cacc+cb2cc

To calculate equilibrium points, consider the equations given below:

dcadt=0dcbdt=0dccdt=0

Compare the system of first order nonlinear differential equations with the above equations,

10cacc+cb=010cacccb=010cacc+cb2cc=0

Therefore, the equilibrium point is,

ca=0,cb=0,cc=0

Suppose, the system of non-linear differential equations are equal to some functions, that is,

dcadt=f(ca,cb,cc)dcbdt=g(ca,cb,cc)dccdt=h(ca,cb,cc)

Now, compare these equations with system of non-linear differential equations,

f(ca,cb,cc)=10cacc+cbg(ca,cb,cc)=10cacccbh(ca,cb,cc)=10cacc+cb2cc

Now, find the Jacobian matrix,

J(ca,cb,cc)=[fca(ca,cb,cc)fcb(ca,cb,cc)fcc(ca,cb,cc)gca(ca,cb,cc)gcb(ca,cb,cc)gcc(ca,cb,cc)hca(ca,cb,cc)hcb(ca,cb,cc)hcc(ca,cb,cc)]

Then, the Jacobian matrix at the equilibrium points ca=0,cb=0,cc=0 is,

J(0,0,0)=[fca(0,0,0)fcb(0,0,0)fcc(0,0,0)gca(0,0,0)gcb(0,0,0)gcc(0,0,0)hca(0,0,0)hcb(0,0,0)hcc(0,0,0)]=[010010012]

Now, the linearized system corresponding to nonlinear system of differential equation is,

(cacbcc)=J(0,0,0)(ca0cb0cc0)=[010010012](cacbcc)

Let, A be the coefficient matrix of the above system,

A=[010010012]

Suppose, λ be the eigenvalue of the matrix A.

Thus,

|AλI|=0|0λ1001λ0012λ|=0

Now calculate the determinant as,

(0λ)[(1λ)(2λ)0]1(0(2λ)0)=0λ[(1+λ)(2+λ)]=0λ=2,1,0

Therefore, the eigenvalues of the matrix are λ=2,1,0

Now, find the eigenvector corresponding to each eigenvalue of the matrix.

The eigenvector is,

[AλI]X=0

Where, X=(cacbcc)

Substitute the value of X in [AλI]X=0,

[AλI](cacbcc)=0

Put λ=2 in [AλI](cacbcc)=0,

[A+2I](cacbcc)=0[[010010012]+2[100010001]](cacbcc)=(000)[210010010](cacbcc)=(000)(cacbcc)=(001)

Put λ=1 in [AλI](cacbcc)=0,

[A+I](cacbcc)=0[[010010012]+[100010001]](cacbcc)=(000)[110000011](cacbcc)=(000)(cacbcc)=(111)

Put λ=0 in [AλI](cacbcc)=0,

[A+I](cacbcc)=0[[010010012]+0[100010001]](cacbcc)=(000)[010010012](cacbcc)=(000)(cacbcc)=(100)

Therefore, the eigenvector corresponding to each eigenvalue of the matrix are respectively

X=(001),(111),(100)

Hence, the solution of the system of nonlinear differential equation is,

X=C1X1eλ1t+C2X2eλ2t+C3X3eλ3t=C1(001)e2t+C2(111)et+C3(100)e0×t=C1(001)e2t+C2(111)et+C3(100)

After solve the above equation,

(cacbcc)=C1(001)e2t+C2(111)et+C3(100)=(00C1e2t)+(C2etC2etC2et)+(C300)=(C2et+C3C2etC1e2tC2et)

Then the values of ca,cb and cc are,

ca(t)=C2et+C3cb(t)=C2etcc(t)=C1e2tC2et

The initial conditionsgiven as,

ca(0)=50,cb(0)=0,cc(0)=40

Now, apply the initial condition in the above equations,

ca(0)=C2e0+C3cb(0)=C2e0cc(0)=C1e2×0C2e0

This imply that,

50=C2+C30=C240=C1C2

Then, C1=40,C2=0,C3=50

Substitute, the value of C1,C2 and C3 in ca,cb and cc

ca(t)=50cb(t)=0cc(t)=40e2t

The concentration at t=0 s are,

ca(t)=50cb(t)=0cc(t)=40

Therefore, theconcentration of the reactants ca,cb and cc at t=0 s are 50,0,40 respectively.

Now, the concentration at t=1 s are,

ca(t)=50cb(t)=0cc(t)=40e2=5.4135

Therefore, the concentration of the reactants ca,cb and cc at t=1 s are 50,0,5.4135 respectively.

Now, the concentration at t=2 s are,

ca(t)=50cb(t)=0cc(t)=40e2×2=0.7327

Therefore, the concentration of the reactants ca,cb and cc at t=2 s are 50,0,5.4135 respectively.

Now, the concentration at t=3 s are,

ca(t)=50cb(t)=0cc(t)=40e2×3=0.09916

Use the following MATLAB code to plot the concentrationvalues,

clear;clc;

% enter the value or expression of the concentrations

c_a = 50;

c_b = 0;

% define the range of t

t = linspace(0,3,20);

c_c = 40.*exp(-2.*t);

% plot the results

plot([0,3],[c_a, c_a],'r',[0,3],[c_b, c_b],'g', t, c_c,'k')

legend('c_a','c_b','c_c')

xlabel('t')

Execute the above to obtain the plot as,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 28, Problem 7P

Therefore, the concentration of the reactants ca,cb and cc at t=3 s are 50,0,5.4135 respectively.

Hence, the concentration of the reactants ca and cb is constant through the time t but concentration of the reactant cc is decreasing rapidly with increase in time t.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
In Figure , the radial distribution function P(r) for the 2s state of hydrogen has two maxima. Find the values of r (in terms of a0) where these maxima occur.
1. An uninsulated cylinder with a locked piston containing 0.35 moles of gas is placed in a boiling water bath at TH = 100°C. The volume of the piston is 0.52 liters. This point is labeled as 1 in the diagram below. TH> Tc To Vi V2 (a) The piston is unlocked and the gas is allowed to expand to a volume of 1.14 liters, moving the system from point 1 to point 2. What is the heat flow? What is the work done by the gas? (b) The piston is locked and moved to an ice bath at Tc = 0°C, moving the system from point 2 to point 3. What is the heat flow? What is the work done by the gas? (c) The piston is unlocked to move the system from point 3 to point 4. What is the heat flow? What is the work done by the gas? (d) To move the system from point 4 to point 1, the cylinder is locked and the is gas returned to the boiling water bath. What is the heat flow? What is the work done by the gas? (e) This heat engine is called a Stirling Cycle. What is its efficiency? (f) If a Carnot engine were operated…
3. Platinum and gold are completely soluble in both the liquid and solid states. The melting point of platinum is 3225°F and that of gold is 1945°F. An alloy containing 40% gold starts to solidify at 2910°F by separating crystals of 15 percent gold in solid. An alloy containing 70% gold starts to solidify at 2550°F by separating crystals of 37% gold in solid. Draw the equilibrium diagram to scale on a piece of graph paper and label all points, lines and areas.

Chapter 28 Solutions

EBK NUMERICAL METHODS FOR ENGINEERS

Ch. 28 - An on is other malbatchre actor can be described...Ch. 28 - The following system is a classic example of stiff...Ch. 28 - 28.13 A biofilm with a thickness grows on the...Ch. 28 - 28.14 The following differential equation...Ch. 28 - Prob. 15PCh. 28 - 28.16 Bacteria growing in a batch reactor utilize...Ch. 28 - 28.17 Perform the same computation for the...Ch. 28 - Perform the same computation for the Lorenz...Ch. 28 - The following equation can be used to model the...Ch. 28 - Perform the same computation as in Prob. 28.19,...Ch. 28 - 28.21 An environmental engineer is interested in...Ch. 28 - 28.22 Population-growth dynamics are important in...Ch. 28 - 28.23 Although the model in Prob. 28.22 works...Ch. 28 - 28.25 A cable is hanging from two supports at A...Ch. 28 - 28.26 The basic differential equation of the...Ch. 28 - 28.27 The basic differential equation of the...Ch. 28 - A pond drains through a pipe, as shown in Fig....Ch. 28 - 28.29 Engineers and scientists use mass-spring...Ch. 28 - Under a number of simplifying assumptions, the...Ch. 28 - 28.31 In Prob. 28.30, a linearized groundwater...Ch. 28 - The Lotka-Volterra equations described in Sec....Ch. 28 - The growth of floating, unicellular algae below a...Ch. 28 - 28.34 The following ODEs have been proposed as a...Ch. 28 - 28.35 Perform the same computation as in the first...Ch. 28 - Solve the ODE in the first part of Sec. 8.3 from...Ch. 28 - 28.37 For a simple RL circuit, Kirchhoff’s voltage...Ch. 28 - In contrast to Prob. 28.37, real resistors may not...Ch. 28 - 28.39 Develop an eigenvalue problem for an LC...Ch. 28 - 28.40 Just as Fourier’s law and the heat balance...Ch. 28 - 28.41 Perform the same computation as in Sec....Ch. 28 - 28.42 The rate of cooling of a body can be...Ch. 28 - The rate of heat flow (conduction) between two...Ch. 28 - Repeat the falling parachutist problem (Example...Ch. 28 - 28.45 Suppose that, after falling for 13 s, the...Ch. 28 - 28.46 The following ordinary differential equation...Ch. 28 - 28.47 A forced damped spring-mass system (Fig....Ch. 28 - 28.48 The temperature distribution in a tapered...Ch. 28 - 28.49 The dynamics of a forced spring-mass-damper...Ch. 28 - The differential equation for the velocity of a...Ch. 28 - 28.51 Two masses are attached to a wall by linear...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
What is a Linear Equation in One Variable?; Author: Don't Memorise;https://www.youtube.com/watch?v=lDOYdBgtnjY;License: Standard YouTube License, CC-BY
Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY