Question

The constant-pressure specific heat of air at different temperatures are tabulated in the following table: T(K) 1.0 1.1 1.2 Cp (kJ/kg.K) 1.1410 1.1573 1.1722 1.1858 1.1982 1.2095 1.3 1.4 1.5 a) Use a third order interpolation polynomial to find Cp at T = 1.45 K. Show complete solution. b) If you want to develop a model that passes through all these data points using the Newton's difference method, explain briefly without calculations how you will do it. Also, how can you test the sensitivity of your developed model?

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Abdul-Rahim Taysir

Answer 1

Answer a)

Arower a) to the third order interpolation polynomial. y= a tbut cn²+ dx3 uning least square fit mehed, sa že? E lyi-a-62; -Co*;+d4;}) ich least I minima, i=1 n=6 OS is ds db =0 as co . and as dd za de we get: Exy 2 2 Ey= natb{n tokx² + d Ex3 a Ex + b {n²+ cen²+ d Ent Eng a{x² + b{n} + cent td Ens - Eskiy asas tbex" + cen5 td {xo o 3 we have 21 26 - Ex² En² nt M U 9.55 12.375 6.2979 ny n'y ny En = 29.5045 11 1.2 1.3 1.4 1.1410 1.513 1.1722 1.1858 Eas= = 21.78375 1.1902 1.5 1.095 Esy=8.85394 Enery11.30315 Exly = 14.6820 {n= 6.2 Ey 7,064 En2=9.55 En3= 12.375 Ent= 16.2979

Nan in een substitching these valies' owing uring calculator we and (4 4 egin 4 variables on solut get , as -0.00 4 3 38 o 0.016042 c 2,184 178 d -10107305 thes, the third order polynomial becomes! 2 2 y= -0.004338 + (-0 0.016042)n + 2.1841782 + (-1.107305 ) x3 at T= 1.45k n = 1045 y = cps 3 -0.004338 +(-0.016042) (1045) + 20184178 (lo 145)? Answer, 1.1673016 (1.450 cpe 1.198878

Answer b):

To develop a model that passes through all these data points using Newton's difference method, we need to apply the formula of Newton's difference method and we'll get a polynomial, just as we got using least square interpolation method. And we can test the sensitivity by finding the error between actual and the observed answer from the Newton's difference method.

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