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hs11201f10

hs11201f10 - EMSE 201 Introduction to Materials Science...

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EMSE 201 – Introduction to Materials Science & Engineering Fall 2010 Case Western Reserve University 1 of 4 Case School of Engineering Solution to Homework #9 Explain how answers were obtained. Give appropriate units for all numerical answers. 1) (5 points) The constant A in Equation 17.2 is 12 π 4 R/5 θ D 3 , where R is the gas constant and θ D is the Debye temperature (K). Estimate θ D for aluminum, given that the specific heat is 4.60 J/kg-K at 15 K. (C&R problem 17.3) Solution (from the publisher’s solution manual; used with permission) Determine the magnitude of A , as A = C v T 3 = c v A Al T 3 = (4.60 J / mol - K )(1 kg /1000 g )(26.98 g / mol ) (15 K ) 3 = 3.68 × 10 -5 J/mol-K 4 (2 pts) As stipulated in the problem statement A = 12 π 4 R 5 θ D 3 Or, solving for θ D θ D = 12 π 4 R 5 A 1/3 = (12)( π ) 4 (8.31 J / mol - K ) (5)(3.68 × 10 5 J / mol - K 4 ) 1/3 = 375 K 2) (6 points) To what temperature must a cylindrical rod of tungsten 15.025 mm in diameter and a plate of 1025 steel having a circular hole 15.000 mm in diameter have to be heated for the rod to just fit into the hole? Assume that the initial temperature is 25°C. ( C&R problem 17.7) Solution (modified from the publisher’s solution manual; used with permission) Solution of this problems requires the use of Equation 17.3a, which is applied to the diameters of both the rod and hole. That is d f d 0 d 0 = α l ( T f T 0 ) Solving this expression for d f yields d f = d 0 1 + α l ( T f T 0 ) Now all we need do is to establish expressions for d f (steel) and d f (W), set them equal to one another, and solve for T f . According to Table 17.1, α l (steel) = 12.0 × 10 -6 ( ° C) -1 and α l (W) = 4.5 × 10 -6 ( ° C) -1 . Thus d f (steel) = d f (W)

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