218 hw6 - ME 218: ENGINEERING COMPUTATIONAL METHODS...

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Unformatted text preview: ME 218: ENGINEERING COMPUTATIONAL METHODS Solutions to Home-Work #6 Problem 1: Forward Difference: dT/dt|i = (Ti+1 – Ti)/h where h = time step, dT/dt|i = rate of cooling at the ith step and Ti = temperature of the ball at the ith step. Similarly, backward difference: dT/dt|i = (Ti – Ti-1)/h Solving for the given data: Time (min) 0 5 10 15 20 25 T (deg C) 90 49.9 33.8 28.4 26.2 25.4 Forward diff. -8.02 -3.22 -1.08 -0.44 -0.16 -8.02 -3.22 -1.08 -0.44 -0.16 Backward diff. Forward diff. 0 0 -1 10 20 30 40 50 Backward diff. 60 70 80 90 100 -2 -3 dT/dt (rate of cooling) -4 -5 -6 -7 -8 -9 Temp (deg C) Plot of dT/dt versus temperature (T) Given, dT/dt = -k(T – Ta) Or, dT/(T – Ta) = -kdt This equation yields an exponential relationship of the temperature with time: (T – Ta) = Ae-kt Thus, performing an exponential curve fit on (T – Ta) against time, using the technique we are familiar with: (T – Ta) = 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. Numerical Solutions: Forward Difference: dT/dt varies linearly with T. From the best fit line, the linear dependence is as follows: dT/dt = -0.1234T + 3.0515 Hence, at T = 25.4 deg C corresponding to t = 25 minutes, dT/dt = -0.1234*25.4 + 3.0515 = -0.08286 But, dT/dt| T = 25.4 = (Tt = 30 - Tt = 25)/5, hence, Tt = 30 = 5 dT/dt| T = 25 + Tt = 25 = 5*(-0.08286) + 25.4 = 25 deg C at T = 25 deg C, t = 30 minutes, dT/dt = -0.1234*25 + 3.0515 = -0.0335 deg C/min Backward Difference: From the best fit line, dT/dt = -0.3218T + 7.9529 dT/dt| t = 30 = (Tt = 30 - Tt = 25)/5 Thus, at t = 30 minutes, there are two simultaneous equations given by: dT/dt = -0.3218T + 7.9529, and dT/dt = 0.2T – 0.2*25.4 = 0.2T – 5.08. Solving this, we get T = 25 deg C and dT/dt = 0.08 deg C/min at t = 30 minutes. This result is consistent since the value of the derivative at the ith point in a backward difference scheme is the same as the value of the derivative at the (i-1)th point in a forward difference scheme. Problem 2 Flux, F = (1/A)*dm/dt, where A = contact area of transdermal patch = 10 cm2 Thus, over a given time interval h, the mass of insulin delivered by the trapezoidal rule = A*h*(Fi+1 + Fi)/2 Solving the above equation for the data given: Time (hr) 0 1 2 3 4 5 10 15 20 24 Flux (mg/cm2/hr) h (hr) 15 14 12 11 9 8 5 2.5 2 1 (Fi+1 + Fi)/2 14.5 13 11.5 10 8.5 6.5 3.75 2.25 1.5 Mass of insulin delivered (mg) 145 130 115 100 85 325 187.5 112.5 60 Sum = 1260 1 1 1 1 1 5 5 5 4 For the total mass delivered in 24 hours, the mass delivered in each time interval is summed up to yield 1260 mg. ...
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This note was uploaded on 02/11/2009 for the course ME 218 taught by Professor Unknown during the Fall '08 term at University of Texas.

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