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Please do both Eulers method and midpoint method. Please do not solve with runge kutta thank you. Problem 2 - (Hand-written answers) A ball bearing at 1200K is allowed to cool down in air at an ambient temperature of 300K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by dT dt 2.2067 x 10-(T-81 x 108) Determine the approximate temperature of the ball at time t480 seconds using: (a) Eulers method, with h-240 seconds, h120 seconds, and h60 seconds. (b) Midpoint method, with h240 seconds, h 120 seconds, and h60 seconds.

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Answer #1

Analytical solution de=-2.2067x 10-12(gt-81x10) f(r,θ)--2.2067x 10-12 (θ-81x 10°) =1200 +/(0.1200)x10 1200+(-2 2067x10- (1200-81x10))x10 1154.42 K For--1,5-10,9 =115442 K = 1 154.42+ 10.1 154.42)× 10 = 115442+2.2067 x10-12 (1154.41-81x10): 10 1115.40 K For = 2,t,-20,a, 1 1 1 5.40 K Similarly, we can find the temperature up tot For, i 470,4-10, 646.51K 480s s(o.eo)a 646.51+f (470.646.51)x10 2646 51 +(-2.2067×10-12(646 51°-81x10))x10 642.84 K Numerical or Exact solution: The numerical solution of the ordinary differential equation is.giucn by the solution of a n linear equation as: 0 925931n θ-300-1.85 19tm (0003338)-0.22067 x10-3-29282 8+300 The solution to this non-linear equation at 480 seconds is: θ-300 θ 300 -1.85 19tan-1 (0.003336)-0. 22067 x 10 x 480-2 9282 0.925931n- θ= 647.57 K The plot for both the solutions is; 1400 1200 1000 800 600 400 200 Exact Solution h-10 100 200 300 400 500 Time, tsec)

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