# soln02 - 10.34 Fall 2006 HW Set#2 Solutions Problem#1 This...

This preview shows pages 1–4. Sign up to view the full content.

10.34 – Fall 2006 HW Set #2 Solutions Problem #1: This was a straightforward linear regression problem that was well-outlined in the problem statement in the textbook. The regression fits the data almost exactly. All of the rate data should have been fit to only two parameters, as you would not expect the values of k 2 and K m to change if the initial enzyme concentration were to change. k2 = 104.5041 g_S/min/g_E Km = 8.5924 g_S/L 0 5 10 15 20 25 30 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.34 Fall 2006 - HW Set#2, Problem 1 S concentration [g/L] Coversion Rate of S [g s /L/min] Experimental Fitted - E 0 = 0.005 g E /L Fitted - E 0 = 0.01 g E /L Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem #2: Part A: This was a simple program writing exercise with no results needed for part A. A simple algorithm for this task would look like this if N data = N param : function a_vec = calc_poly_coeff(x,f) for i=1:length(x) for j=1:length(x) X_mat(j,i) = x(j)^(i-1); end end a_vec = X_mat\f; Part B: The purpose of this part was to write a program that utilizes the result in part A that fit a set of Cv data. The best fit parameters for the 3 rd order polynomial and the condition number of the polynomial system matrix are: The condition number is: 1.568e+010 Polynomial coefficient values: a0 = 2.96 cal mole -1 K -1 a1 = 0.016278 cal mole -1 K -2 a2 = -1.1278e-005 cal mole -1 K -3 a3 = 3.0247e-009 cal mole -1 K -4 Part C: This part was about comparing the results of the polynomial form and the Pade form, and determining which had more accurate interpolation/extrapolation abilities. Linearized form: CT a T + T 2 2 ( ) () = + a C T a TCTa ( ) ⎤⋅ V 0 1 V V 2 V 3 Pade coefficient values: a0 = 2.6799 cal mole -1 K -1 a1 = 0.021745 cal mole -1 K -2 a2 = 1.3466e-006 K -2 a3 = 0.0014574 K -1 The first figure shows the interpolation ability of the two. Both will produce essentially the same quality of interpolation (although this will not always be the case… especially since polynomials Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
can often “snake” through the points instead of having a nice, monotonic increase as in the above case.). However, when one looks to extrapolate the data to high temperatures, the results are wildly different. The polynomial case diverges quickly after the last data point and predicts an unphysical Cv value above ~1700 K. The Pade form proves to be well-behaved over range plotted, mainly due to the formulation of the equation such that the correct asymptote is reached at high temperatures (it can be readily seen that C v (T Æ inf) = C v,inf .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

soln02 - 10.34 Fall 2006 HW Set#2 Solutions Problem#1 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online