Quiz 1-Problem 1

# Quiz 1-Problem 1 - 4.6 80 32.32 30.81 1.51 4.7 90 30.39...

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t T T obtained numerically k= 0.02 0 68 68 We put the initial value T0= 68 10 60.5 59.84 Ta= 20 20 54.16 53.07 ∆t 10 30 48.82 47.45 40 44.32 42.78 50 40.52 38.91 60 37.31 35.69 70 34.6 33.03 80 32.32 30.81 90 30.39 28.97 100 28.77 27.45 110 27.4 26.18 I PICKED THIS NUMBER 120 26.24 25.13 130 25.27 24.26 Since k= is in units of 1/min 140 24.44 23.53 then t is in minutes!!!!! 150 23.75 22.93 160 23.16 22.43 Looks like we have picked a value of ∆t sufficie have passed the point of maximum error t T T Num Difference Relative difference (% 0 68 68 0 0 10 60.5 59.84 0.66 1.1 20 54.16 53.07 1.1 2 30 48.82 47.45 1.38 2.8 40 44.32 42.78 1.54 3.5 50 40.52 38.91 1.61 4 60 37.31 35.69 1.62 4.3 70 34.6 33.03 1.58

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Unformatted text preview: 4.6 80 32.32 30.81 1.51 4.7 90 30.39 28.97 1.42 4.7 100 28.77 27.45 1.32 4.6 110 27.4 26.18 1.22 4.4 120 26.24 25.13 1.11 4.2 130 25.27 24.26 1.01 4 140 24.44 23.53 0.91 3.7 150 23.75 22.93 0.81 3.4 160 23.16 22.43 0.73 3.1-10 10 3 20 25 30 35 40 45 50 55 60 65 70 t T T k T T a i i i ∆--2245 + ) ( 1 kt a a e T T T T--+ = ) ( e here iently large to see that numerically we We calculate such error below %) The maximum difference is around t=100 min!!! 30 50 70 90 110 130 150...
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## This note was uploaded on 08/31/2011 for the course ENGR 3723 taught by Professor Staff during the Summer '11 term at Oklahoma State.

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Quiz 1-Problem 1 - 4.6 80 32.32 30.81 1.51 4.7 90 30.39...

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