FinalCHEE3334Fall1999Solutions

FinalCHEE3334Fall1999Solutions - CHEE 3334 Fall'99 FINAL...

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CHEE 3334 Fall ’99 Michael Nikolaou FINAL EXAM Open-book, open-notes. Total points 100. Please, budget your time ! Do NOT try to read the textbook during the exam! Write neatly! Do NOT use your programmable calculator to find results you are asked to compute in a certain way--after all, the calculator may find the wrong answer or not all answers! Show all work. Return this sheet with your exam . GOOD LUCK! (70 pts.) 1. A continuous-flow industrial fermentation reactor was initially at steady state. Then, an engineer conducted two experiments to determine how the reactor yield deviates from its steady state value as a result of the flowrate to the reactor deviating from its steady state. In each experiment she changed, over time, the flowrate from its steady state value, and measured how the reactor yield responded to that change. The results of her experiments are shown in the following two Figures and corresponding Tables, where k is time in hours, u ( k ) is the difference between the changed flowrate value and its steady-state value, and y ( k ) is the difference between the resulting yield value and its steady-state value, in dimensionless units. 0 .0 0 0 .5 0 1 .0 0 1 .5 0 2 .0 0 0 2 4 6 8 1 0 -2 .0 0 -1 .0 0 0 .0 0 1 .0 0 2 .0 0 0 2 4 6 8 1 0 Table 1 k u ( k ) y ( k ) 0 1 0.00 1 1 1.39 2 1 1.53 3 1 0.95 4 1 1.42 5 1 1.00 6 1 0.69 7 1 1.50 8 1 1.36 9 1 1.39 10 1 0.89 Table 2 k u ( k ) y ( k ) 0 -1 0.00 1 1 -1.54 2 -1 0.75 3 -1 -0.99 4 1 -1.14 5 1 0.85 6 -1 1.45 7 -1 -0.73 8 -1 -1.54 9 1 -1.21 10 1 0.65 The engineer postulated that the dynamic model describing the relationship between u and y is ) 2 ( ) 1 ( ) ( - + - = k bu k au k y (1) and suggested that the least-squares method applied to errors from 2 = k to 10 = k should be used to estimate the unknown parameters. a. On the basis of the data in Table 1 only , estimate the parameters a and b by using the linear least-squares method. Are your estimates unique? { { 1.53 1 1 (2) (1) (0) 0.95 1 1 (3) (2) (1) 0.89 1 1 (10) (9) (8) y y X X y u u a a y u u e b b y u u θ θ � � � � � � � � �� �� � � � � = - = - �� �� � � � � �� �� � � � � � � � � � � � � M M M M M M 142 43 123 142 43 1 442 4 43 The optimal values of the parameters a, b are solutions of the equations ˆ 9 9 10.73 ˆ ˆ 9 9 10.73 T T a X X X y b θ � � � � = = � �
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