ROSE-HULMAN INSTITUTE OF TECHNOLOGY

ME 323 Computer Applications II --- Spring 2008–2009

Class

Date

Answers (Numerical values only)

Due Date

Lab 1

3-11

Problem 1

(a) The total fluid volume is 8.48230e+001 cubic meters at a fluid level of 3.000e+000 meters.

(b) The total fluid volume is 2.09440e+002 cubic meters at a fluid level of 5.500e+000 meters.

End of Lab 1

Lab 2

3-18

Problem 1

Number of rectangles

Error

10 -2.9868e+000

100 -3.0179e-002

1000 -3.0182e-004

10000 -3.0181e-006

100000 -2.9253e-008

Problem 2

Number of trapezoids

Error

10 6.0005e+000

100 6.0361e-002

1000 6.0364e-004

10000 6.0365e-006

100000 6.1294e-008

Problem 3

Number of sub-intervals

Error

10 8.9585e-003

100 9.0537e-007

1000 8.5976e-011

10000 8.6658e-011

100000 9.2930e-010

End of Lab 2

Lab 3

3-25

Problem 1

Lower bound of searching interval = 0.0001

Upper bound of searching interval = 0.1

Error tolerance for solution = 0.000001

Solution to the equation is found at x = 1.9988996e-002.

It takes 16 iterations to converge with an accuracy of 7.6218e-007.

Problem 2

What is the fluid volume, in cubic meter = 360

What is the error tolerance, in meter = 0.001

Solution to the equation is found at x = 9.1412354e+000.

It takes 13 iterations to converge with an accuracy of 6.1035e-004

End of Lab 3

Lab 4

4-1

Friction factor in pipe flow analysis

Initial guess of solution = 0.01

Error tolerance for solution = 10^(-6)

Iteration x f error

---------------------------------------------------------

1 1.0000000e-002 3.18e+000 5.96e-003

2 1.5959584e-002 9.23e-001 3.44e-003

3 1.9396281e-002 1.18e-001 5.85e-004

4 1.9980782e-002 1.69e-003 8.77e-006

5 1.9989550e-002 -1.18e-005 -6.13e-008

A solution is found to be at x = 1.9989488e-002.

It takes 5 iterations to converge with an accuracy of 1.0000e-006.

Inverse problem of tank exercise

What is the fluid volume, in cubic meter = 300

What is the error tolerance, in meter = 10^(-6)

Initial guess of solution = 5.0

Iteration x f error

---------------------------------------------------------

1 5.0000000e+000 -1.16e+002 2.30e+000

2 7.3016437e+000 -2.31e+000 5.17e-002

3 7.3533799e+000 4.65e-003 -1.05e-004

4 7.3532748e+000 -3.35e-005 7.58e-007

Fluid level is found to be 7.3532755e+000 m for a volume of 3.00e+002 m^3.

It takes 4 iterations to converge with an accuracy of 7.5826e-007 m.

End of Lab 4

Lab 6

4-22

Case 1:

Initial guess of solution = [2.0 0.0]'

Error tolerance for solution = 10^(-6)

Iteration dx(1) dx(2) f(1) f(2)

------------------------------------------------------------------

1 0.00e+000 -6.39e+000 0.00e+000 6.39e+000

2 -4.15e-001 3.06e+000 4.08e+001 0.00e+000

3 -3.72e-001 1.26e+000 9.56e+000 5.56e-001

4 -1.82e-001 3.12e-001 1.73e+000 3.00e-001

5 -2.64e-002 2.17e-002 1.30e-001 5.24e-002

6 -3.95e-004 1.08e-004 1.17e-003 9.71e-004

7 -7.90e-008 2.63e-009 1.68e-007 2.13e-007

Solution to the equations is at x = 1.004169e+000, y = -1.729637e+000.

It takes 7 iterations to converge with an accuracy of 1.0000e-006.

Case 2:

Initial guess of solution = [-2.0 0.0]'

Error tolerance for solution = 10^(-6)

Iteration dx(1) dx(2) f(1) f(2)

------------------------------------------------------------------

1 0.00e+000 8.65e-001 0.00e+000 -8.65e-001

2 1.77e-001 -2.39e-002 7.48e-001 0.00e+000

3 7.14e-003 -3.39e-003 3.18e-002 2.24e-003

4 1.42e-005 -6.44e-006 6.25e-005 4.13e-006

5 5.55e-011 -2.55e-011 2.44e-010 1.65e-011

Solution to the equations is at x = -1.816264e+000, y = 8.373678e-001.

It takes 5 iterations to converge with an accuracy of 1.0000e-006.

End of Lab 6

Lab 7

4-29

Error tolerance for solution (in K) = 0.001

Number of evenly-spaced sections = 4

Iteration max error

-----------------------------------

1 5.96e+002

2 9.62e+000

3 1.48e-002

4 3.41e-008

T(1) = 6.000000e+002 K.

T(2) = 5.937795e+002 K.

T(3) = 5.893895e+002 K.

T(4) = 5.867763e+002 K.

T(5) = 5.859086e+002 K.

T(6) = 5.867763e+002 K.

It takes 4 iterations to converge with an accuracy of 1.0000e-003.

End of Lab 7

Lab 8

5-6

Activity 1:

Time step       y(x = 5)       Error at x = 5
       1      285.434237005     -5.39208e+000
     0.1      290.825180205     -1.13800e-003
    0.01      290.826318083     -1.22652e-007
   0.001      290.826318205     -1.70530e-011
0.0001      290.826318205     -1.48930e-011 (may differ depending on machine precision)

Activity 2:

Step size for integration = 0.10

Arrival time = 7.500000 sec, Range = 148.320403587 m.

End of Lab 8

Lab 9

5-13

Case 1:

Horizontal location of target (in m) = 100

Vertical location of target (in m) = 30

Initial guess of launch angle (in deg) = 45

Error tolerance for solution (in deg) = 0.001

Time step for integration (in sec) = 0.001

 Iteration       angle (deg)         f (m)           error (deg) 
 -------------------------------------------------------------------
     1         4.5000000e+001    5.2971260e+001    -2.7017235e+001
     2         1.7982765e+001   -1.7066823e+000     9.3348619e-001
     3         1.8916251e+001    4.6178411e-003    -2.5086618e-003
     4         1.8913743e+001    3.4611180e-006    -1.8803026e-006

Launch angle = 18.9137408 degrees to hit a target at x = 100.0000 m, y = 30.0000 m.

Case 2:

Horizontal location of target (in m) = 140

Vertical location of target (in m) = 20

Initial guess of launch angle (in deg) = 45

Error tolerance for solution (in deg) = 0.001

Time step for integration (in sec) = 0.001

 Iteration       angle (deg)         f (m)           error (deg) 
 -------------------------------------------------------------------
     1         4.5000000e+001   -1.4855484e+001    -8.0215790e-001
     2         4.4197842e+001   -3.8436720e+000    -3.0471401e-001
     3         4.3893128e+001   -5.5332264e-001    -4.9831670e-002
     4         4.3843296e+001   -5.1859136e-002    -4.7654872e-003
     5         4.3838531e+001   -4.4078581e-003    -4.0582886e-004

Launch angle = 43.8381251 degrees to hit a target at x = 140.0000 m, y = 20.0000 m.

End of Lab 9