KMU-206 MIDTERM II Solutions

Instructor: Selis Önel

May 2, 2008

 

Question 1. (30 pts)

a)      In order to be able to fit a unique polynomial to the three data points given, the equation should have three unknown coefficient that we can determine uniquely. Therefore:

Cp(T) = A + B*T + C*T2

 

b)      Writing the equation at three data points gives:

26.95 = A + B*(300) + C*(300)2

27.40 = A + B*(400) + C*(400)2

28.00 = A + B*(600) + C*(600)2

This linear system of equations can be solved for A, B and C uniquely by using either of the Gauss elimination or Gauss-Jordan methods.

Writing the above equations in matrix form gives:

 

Using the Gauss elimination method:

 

 

Using the Gauss-Jordan method     

 

 

c)      Cp(500)   = 25 + 0.008(500) + (-0.000005)(500)2 =  27.75 J/(gmol.K)

Cp(800)   = 28.20 J/(gmol.K)

Cp(1000) = 28.00 J/(gmol.K)

 

 

 

 

Question 2. (40 pts)

 

Find the positive real zero of f(x) using the bisection method:

f(x) = 2x3+(13/5)x2-4x-(12/5)

a) f(0)= -2.4000,   f(0.5)= -3.5000,   f(1)= -1.8000,    f(1.5)= 4.2000

 

 

b)The graph and the function values calculated above show that there should be a positive root for x values between 1 and 1.5. Therefore, it is possible to select [a,b]=[1,1.5]

c)    The iterations using bisection method in the interval [1,1.5] are:

Step          a              b           m            f(a)         f(b)         f(m)        |b-a|

    1.0000    1.0000    1.5000    1.2500   -1.8000    4.2000    0.5688    0.5000 > 0.01

    2.0000    1.0000    1.2500    1.1250   -1.8000    0.5688   -0.7617    0.2500 > 0.01

    3.0000    1.1250    1.2500    1.1875   -0.7617    0.5688   -0.1345    0.1250 > 0.01

    4.0000    1.1875    1.2500    1.2188   -0.1345    0.5688    0.2075    0.0625

    5.0000    1.1875    1.2188    1.2031   -0.1345    0.2075    0.0341    0.0313

    6.0000    1.1875    1.2031    1.1953   -0.1345    0.0341   -0.0508    0.0156

    7.0000    1.1953    1.2031    1.1992   -0.0508    0.0341   -0.0085    0.0078

bisection has converged as 0.0078<0.01, and x ~ 1.1992

 

Question 3. (15 pts)

 

a)      A is a square diagonal matrix, therefore  |A|=(1000)(1)(0.01)=10

b)      A is a square diagonal matrix, therefore its eigen values can be found immediately using the diagonal elements: λ1=1000,  λ2=1,  λ3 =0.01

c)      For a square symmetric matrix, the condition number can be determined as the ratio of the largest eigen value to the smallest. cond(A)= λmax/ λmin=1000/0.01=100000

As the condition number for this system is too large, the system is ill-conditioned and not stable.

 

Question 4. (15 pts)

 

a)      The characteristic equation (polynomial) for the coefficient matrix A can be obtained by setting det[A-λI] = 0

 

b)      The characteristic equation found above can be simplified as (1-λ)(λ+1)(λ-3) = 0 , where the λ values can be determined as: λ1=1,  λ2=-1,  λ3 = 3

c)      The eigen vectors can be found as: