SolutionsHW11

# SolutionsHW11 - HW 11 Problem 18.14 a To Find(a Determine 0...

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HW 11 Problem 18.14 a. To Find: (a) Determine ρ 0 and a in Equation 18.10 for pure copper, using the data in Fig.18.8 (b) Determine ‘A’ in Equation 18.11 for nickel as an impurity in copper, using the data in Figure 18.8 (c) Estimate the electrical resistivity of copper containing 1.75 at% Ni at 100°C using the results obtained in parts (a) and (b) b. Given: Fig.18.8. c. Assumptions: 1. Fig.18.8 is accurate. 2. The alloys are undeformed (hence, d = 0) d. Solution: (a) Two simultaneous equations are set up using two resistivity values (labeled t 1 and t 2 ) at two corresponding temperatures ( T 1 and T 2 ). Thus, from equation 18.10: t 1 = 0 + aT 1 t 2 = 0 + aT 2 a = t 1   t 2 T 1 T 2 2 1 2 1 1 1 0 = T T T t t t or 0 = t 2 T 2 t 1   t 2 T 1 T 2 From equation 18.9, total = t + i + d There are no impurities in Cu => i = 0 Copper is undeformed => d = 0

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Hence, the plot of total versus temperature in Fig.18.8 is equivalent to the variation of t versus temperature (only) for pure copper. From Figure 18.8, T 1 = –150 C, T 2 = –50 C gives t 1 = 0.6 10 -8 ( -m), and t 2 = 1.25 10 -8 ( -m): a = t 1   t 2 T 1 T 2 = ( 0.6 10 -8 ) ( 1.25 10 -8 ) - m 150 C ( 50 C) = 6.5 10 -11 ( -m)/ C 0 = t 1 T 1 t 1   t 2 T 1 T 2 = ( 0.6 10 -8 ) ( 150) ( 0.6 10 -8 ) ( 1.25 10 -8 ) - m 150 C ( 50 C) = 1.58 10 -8 ( -m) (b) The variation of total with temperature is plotted for three values of c i viz. 0.0112, 0.0216, and 0.0332 in Figure 18.8. The curves in the figure allow us to tabulate values of total at some temperature for all values of c i . The following table has values of total at 0 C for all three values of c i . From equation 18.9, total = t + i + d The alloys are undeformed, hence, d = 0. i = total - t …(1) From equation 18.10, t = 0 + aT T = 0 => t = 0 = 1.58 10 -8 ( -m) …(2) (from part a) i = total - 1.58 10 -8 ( -m) …(3) The values of i are noted in the table next to the corresponding values of total.
The value of A may be determined from equation 18.11: i = Ac i (1 c i ) => A = i / [ c i(1- c i) ] c i 1 – c i total ( -m) i ( -m) A ( -m) 0.0112 0.989 3.0 10 -8 1.42 10 -8 1.28 10 -6 0.0216 0.978 4.2 10 -8 2.62 10 -8 1.24 10 -6 0.0332 0.967 5.5 10 -8 3.92 10 -8 1.22 10 -6 The average value of A is 1.25 10 -6 ( -m). c) From equation 18.11,

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SolutionsHW11 - HW 11 Problem 18.14 a To Find(a Determine 0...

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