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MatS_3011_S12_HW_4 - x between 0 and 2 cm(i.e take t=1 hour...

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University of Minnesota Department of Chemical Engineering and Materials Science MatS 3011: Introduction to Materials Science and Engineering Spring Semester 2012 Homework #4 (Due February 29, 2012) 1) (15 pts) Callister 4.22 2) (10 pts) Callister 5.6 3) (20 pts) (a) Callister 5.10. You can do this by substituting the solution into Equation 5.4b and showing that the solution satisfies Equation 5.4b. Differential equations have different solutions depending on the boundary conditions. The kind of boundary condition used to solve the equation depends on the physical situation. This particular solution in 5.10 is obtained when “stuff” (call it A) diffuses into a material, B, from a limited amount of A at x=0. For example, x=0 may be the surface of the material and we may have put a layer of the diffusing stuff A on the surface of B. (b) The solution in 5.10 is different than the solution shown in Equation 5.5. Plot the concentration profile solution in Problem 5.10 for t=1 hour, t=2 hours and t=3 hours for
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Unformatted text preview: x between 0 and 2 cm. (i.e. take t=1 hour plot C vs. x. Repeat for t= 2 hours and t=3 hours. Put the plots on the same graph.) Take D=10-5 cm 2 /s and B=0.036 g/cm 2 . (c) The boundary conditions that result in Equation 5.5 are such that a constant concentration of stuff (A) is maintained at its surface. Plot the concentration profile solution given by Equation 5.5 for t=1 hour, t=2 hours and t=3 hours for x between 0 and 2 cm. As in (b) put these 3 plots on the same graph (but not on the graph you produced in (b)). You should end up with two graphs (one for (b) and one for (c)) each with 3 curves. Take D=10-5 cm 2 /s and C s =0.036 g/cm 3 and C o =0 g/cm 3 . (d) Describe in words what you see in plots you produced in (b) and (c). Give your physical interpretation of why these plots look the way they do. 4) (10 pts) Callister 5.11. 5) (15 pts) Callister 5.15. 6) (15 pts) Callister 5.21. 7) (15 pts) Callister 5.26....
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