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218 hw6 - ME 218 ENGINEERING COMPUTATIONAL METHODS...

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ME 218: ENGINEERING COMPUTATIONAL METHODS Solutions to Home-Work #6 Problem 1: Forward Difference: dT/dt| i = (T i+1 T i )/h where h = time step, dT/dt| i = rate of cooling at the i th step and T i = temperature of the ball at the i th step. Similarly, backward difference: dT/dt| i = (T i T i-1 )/h Solving for the given data: Time (min) T (deg C) Forward diff. Backward diff. 0 90 -8.02 5 49.9 -3.22 -8.02 10 33.8 -1.08 -3.22 15 28.4 -0.44 -1.08 20 26.2 -0.16 -0.44 25 25.4 -0.16
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-9 -8 -7 -6 -5 -4 -3 -2 -1 0 0 10 20 30 40 50 60 70 80 90 100 Temp (deg C) dT/dt (rate of cooling) Forward diff. Backward diff. Plot of dT/dt versus temperature (T) Given, dT/dt = -k(T – T a ) Or, dT/(T – T a ) = -kdt This equation yields an exponential relationship of the temperature with time: (T – T a ) = Ae -kt Thus, performing an exponential curve fit on ( T – T a ) against time, using the technique we are familiar with: (T – T a ) = 67.457*exp(-0.2029t) To solve for the rate of cooling after 30 minutes: Differentiating the above equation: dT/dt = -67.457*0.2029*exp(-0.2029t) = -13.687*exp(-0.2029t) Hence, the rate of cooling at t = 30 minutes is 0.0311 deg C / min. This value is the exact solution based on the exponential fit, and does not depend on whether the forward or the backward difference scheme is used.
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