Concept explainers
Temperatures are measured at various points on a heatedplate (Table P20.60). Estimate the temperature at (a)
TABLE P20.60 Temperatures
|
|
|
|
|
|
|
100.00 | 90.00 | 80.00 | 70.00 | 60.00 |
|
85.00 | 64.49 | 53.50 | 48.15 | 50.00 |
|
70.00 | 48.90 | 38.43 | 35.03 | 40.00 |
|
55.00 | 38.78 | 30.39 | 27.07 | 30.00 |
|
40.00 | 35.00 | 30.00 | 25.00 | 20.00 |
(a)
To calculate: The value of temperature at
100.00 | 90.00 | 80.00 | 70.00 | 60.00 | |
85.00 | 64.49 | 53.50 | 48.15 | 50.00 | |
70.00 | 48.90 | 38.43 | 35.03 | 40.00 | |
55.00 | 38.78 | 30.39 | 27.07 | 30.00 | |
40.00 | 35.00 | 30.00 | 25.00 | 20.00 |
Answer to Problem 60P
Solution:
The value of temperature at
Explanation of Solution
Given Information:
The data is provided as,
100.00 | 90.00 | 80.00 | 70.00 | 60.00 | |
85.00 | 64.49 | 53.50 | 48.15 | 50.00 | |
70.00 | 48.90 | 38.43 | 35.03 | 40.00 | |
55.00 | 38.78 | 30.39 | 27.07 | 30.00 | |
40.00 | 35.00 | 30.00 | 25.00 | 20.00 |
Formula used:
The zero-order Newton’s interpolation formula:
The first-order/linear Newton’s interpolation formula:
The second- order/quadratic Newton’s interpolating polynomial is given by,
Where,
The first finite divided difference is,
And, the n th finite divided difference is,
Calculation:
To calculate the temperature
First use the linear interpolation formula and arrange the points as close to about
The values are,
And,
First calculate
Put in above equation,
Similarly for quadratic interpolation,
Now calculate
Put in quadratic interpolation equation,
Now, do it for cubic interpolation by the use of the formula,
Now calculate
Put in cubic interpolation equation,
And, the error is calculated as,
Similarly the other dividend can be calculated as shown above,
Therefore, the difference table can be summarized for
Order | Error | |
0 | 38.43 | 6.028 |
1 | 44.458 | |
2 | 43.6144 | |
3 | 43.368 | |
4 | 43.48045 |
Since the minimum error for order third, therefore, it can be concluded that the value of temperature at
(b)
To calculate: The value of temperature at
100.00 | 90.00 | 80.00 | 70.00 | 60.00 | |
85.00 | 64.49 | 53.50 | 48.15 | 50.00 | |
70.00 | 48.90 | 38.43 | 35.03 | 40.00 | |
55.00 | 38.78 | 30.39 | 27.07 | 30.00 | |
40.00 | 35.00 | 30.00 | 25.00 | 20.00 |
Answer to Problem 60P
Solution:
The value of temperature at
Explanation of Solution
Given Information:
The data is provided as,
100.00 | 90.00 | 80.00 | 70.00 | 60.00 | |
85.00 | 64.49 | 53.50 | 48.15 | 50.00 | |
70.00 | 48.90 | 38.43 | 35.03 | 40.00 | |
55.00 | 38.78 | 30.39 | 27.07 | 30.00 | |
40.00 | 35.00 | 30.00 | 25.00 | 20.00 |
Formula used:
The zero-order Newton’s interpolation formula:
The first-order/linear Newton’s interpolation formula:
The second- order/quadratic Newton’s interpolating polynomial is given by,
Where,
The first finite divided difference is,
And, the n th finite divided difference is,
Calculation:
To calculate the temperature
Since, this is a two-dimensional interpolation, therefore one way is to use cubic interpolation along the y direction for specific values of x and then go along the x direction for values of y obtained from the previous analysis.
First use the linear interpolation formula and arrange the points as close to about
The values are,
And,
First calculate
Put in above equation,
Similarly for quadratic interpolation,
Now calculate
Put in quadratic interpolation equation,
Now, do it for cubic interpolation by the use of the formula,
Now calculate
Put in cubic interpolation equation,
And, the error is calculated as,
Similarly the other dividend can be calculated as shown above,
Therefore, the difference table can be summarized for
Order | Error | |
0 | 64.49 | |
1 | 59.0335 | |
2 | 58.411 | |
3 | 58.47032 |
Now, do this for
The values are,
And,
First calculate
Put in above equation,
Similarly for quadratic interpolation,
Now calculate
Put in quadratic interpolation equation,
Now, do it for cubic interpolation by the use of the formula,
Now calculate
Put in cubic interpolation equation,
Similarly, for
Now for the calculation for
The values are,
And,
First calculate
Put in above equation,
Similarly for quadratic interpolation,
Now calculate
Put in quadratic interpolation equation,
Now, do it for cubic interpolation by the use of the formula,
Now calculate
Put in cubic interpolation equation,
And, the error is calculated as,
Similarly the other dividend can be calculated as shown above,
Therefore, the difference table can be summarized for
Order | Error | |
0 | 47.15 | |
1 | 46.4885 | |
2 | 46.0479875 | |
3 | 46.140425 |
Hence, the value of temperature at
This problem can also be solved with MATLAB as it contains the predefined function interp2.
The MATLAB code is as shown below,
The output in the command window is,
For more accuracy, the result can also be obtained from the bicubic interpolation as shown below,
Finally, the interpolation can also be implemented with the use of splines as shown below,
Hence, it can be concluded that the result is similar to that obtained from the calculation.
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Chapter 20 Solutions
Numerical Methods for Engineers
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