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Direct Method of Interpolation Chemical Engineering Majors Authors: Autar Kaw, Jai Paul http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates http://numericalmethods.eng.usf.edu 1 Direct Method of Interpolation http://numericalmethods.eng.usf.edu What is Interpolation ? Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given. Figure 1 Interpolation of discrete. 3 http://numericalmethods.eng.usf.edu Interpolants Polynomials are the most common choice of interpolants because they are easy to: Evaluate Differentiate, and Integrate 4 http://numericalmethods.eng.usf.edu Direct Method Given ‘n+1’ data points (x0,y0), (x1,y1),………….. (xn,yn), pass a polynomial of order ‘n’ through the data as given below: y a0 a1 x .................... an x . n where a0, a1,………………. an are real constants. Set up ‘n+1’ equations to find ‘n+1’ constants. To find the value ‘y’ at a given value of ‘x’, simply substitute the value of ‘x’ in the above polynomial. 5 http://numericalmethods.eng.usf.edu Example To find how much heat is required to bring a kettle of water to its boiling point, you are asked to calculate the specific heat of water at 61°C. The specific heat of water is given as a function of time in Table 1. Use linear, quadratic and cubic interpolation to determine the value of the specific heat at T = 61°C. Table 1 Specific heat of water as a function of temperature. Temperature, Specific heat, 22 42 52 82 100 4181 4179 4186 4199 4217 T C J C p kg C Figure 2 Specific heat of water vs. temperature. 6 http://numericalmethods.eng.usf.edu Linear Interpolation C p T a 0 a1T C p 52 a 0 a1 52 4186 C p 82 a 0 a1 82 4199 4.19910 3 4200 4195 ys f ( range) f x desired 4190 Solving the above two equations gives, a0 4163.5 a1 0.43333 4.18610 3 4185 50 x s 10 0 60 70 80 x s range x desired Hence C p T 4163.5 0.43333T , 52 T 82. J C p 61 4163.5 0.4333361 4189.9 kg C 7 http://numericalmethods.eng.usf.edu 90 x s 10 1 Quadratic Interpolation C p T a0 a1T a2T 2 Cp42 a 0 a1 42 a 2 42 4179 2 Cp52 a 0 a1 52 a 2 52 4186 2 Cp82 a 0 a1 82 a 2 82 4199 2 Solving the above three equations gives a0 4135.0 8 a1 1.3267 a2 6.6667 10 3 http://numericalmethods.eng.usf.edu Quadratic Interpolation (contd) 4.19910 3 C p T 4135.0 1.3267T 6.6667 10 3 T 2 , 4200 4195 42 T 82 ys 4190 f ( range) C p 61 4135.0 1.326761 6.6667 10 J 4191.2 kg C 3 61 2 f x desired 4185 4180 4.17910 3 4175 40 42 45 50 55 60 65 70 75 80 x s range x desired The absolute relative approximate error obtained between the results from the first and second order polynomial is a 9 4191.2 4189.9 100 0.030063% 4191.2 http://numericalmethods.eng.usf.edu 85 82 Cubic Interpolation C p T a0 a1T a2T 2 a3T 3 Cp42 a 0 a1 42 a 2 42 a3 42 4179 2 3 Cp52 a 0 a1 52 a 2 52 a3 52 4186 2 3 Cp82 a 0 a1 82 a 2 82 a3 82 4199 2 3 Cp100 a 0 a1 100 a 2 100 a3 100 4217 2 a0 4078.0 10 a1 4.4771 3 a2 0.062720 a3 3.1849 10 4 http://numericalmethods.eng.usf.edu Cubic Interpolation (contd) 4.21710 CpT 4078 4.4771T 0.06272T 2 3.1849 10 4 T 3 , 42 T 100 3 4220 4210 ys 4200 f ( range) f x desired 4190 T 61 4078 4.447161 0.0627261 3.1849 10 4 61 J 4191.0 kg C 2 3 4180 4.17910 3 4170 40 50 42 60 70 80 x s range x desired 90 100 100 The absolute relative approximate error obtained between the results from the first and second order polynomial is a 11 4190.0 4191.2 100 0.027295% 4190.0 http://numericalmethods.eng.usf.edu Comparison Table Order of Polynomial J C p T kg C Absolute Relative Approximate Error 12 1 2 3 4189.9 4191.2 4190.0 ---------- 0.030063% 0.027295% http://numericalmethods.eng.usf.edu Additional Resources For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit http://numericalmethods.eng.usf.edu/topics/direct_met hod.html THE END http://numericalmethods.eng.usf.edu