Fundamentals of Materials Science and Engineering Solutions Ch 11-12

CH A PTE R 11 PHA SE TRA NSFORMATIONS 11.4 This problem gives us the value of y (0.40) at some time t (200 min), and also the value of n (2.5) for the recrystallization of an alloy at some temperature, and then asks that we determine the rate of recrystallization at this same temperature. It is first necessary to calculate the value of k in E quation (11.1) as k=− =− ln(1 − y) tn ln(1 − 0.4) = 9.0 × 10−7 (200 min) 2.5 A t this point we want to compute t0.5 , the value of t for y = 0.5, also using E quation (11.1). Thus t 0.5 = = − ln(1 − 0.5) k ln(1 − 0.5) − .0 × −7 9 10 1/n 1/2.5 = 226.3 min A nd, therefore, from E quation (11.2), the rate is just rate = 11.7 1 1 = = 4.42 × 10−3 (min) −1 t 0.5 226.3 min This problem asks us to consider the percent recrystallized versus logarithm of time curves for copper shown in Figure 11.2. (a) The rates at the different temperatures are determined using E quation (11.2), which rates are tabulated below: Temperature ( ◦ C) R ate (min) −1 135 119 113 102 88 43 0.105 4.4 × 10−2 2.9 × 10−2 1.25 × 10−2 4.2 × 10−3 3.8 × 10−5 43 (b) These data are plotted below as ln rate versus the reciprocal of absolute temperature. -2 ln Rate (1/min) -4 -6 -8 -10 -12 0.0024 0.0026 0.0028 0.0030 0.0032 1/T (1/K) The activation energy, Q, is related to the slope of the line drawn through the data points as Q = −Slope(R ) where R is the gas constant. The slope of this line is −1.126 × 104 K, and thus Q = −(−1.126 × 104 K)(8.31 J/mol-K) = 93,600 J/mol (c) A t room temperature (20◦ C), 1/T = 3.41 × 10−3 K −1 . E xtrapolation of the data in the plot to this 1/T value gives ∼ −12.8 ln(rate) = or ∼ e −12.8 = 2.76 × 10−6 (min) −1 rate = But since rate = 1 t 0.5 then t 0.5 = 1 ∼ 1 = rate 2.76 × 10−6 (min) −1 ∼ = 3.62 × 105 min ∼= 250 days 11.15 Below is shown an isothermal transformation diagram for a eutectoid iron-carbon alloy, with a time-temperature path that will produce (a) 100% coarse pearlite. 44 11.18 Below is shown an isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy, with a time-temperature path that will produce (b) 50% fine pearlite and 50% bainite. 45 11.20* Below is shown a continuous cooling transformation diagram for a 1.13 wt% C iron-carbon alloy, with a continuous cooling path that will produce (a) fine pearlite and proeutectoid cementite. 11.34 This problem asks for estimates of R ockwell hardness values for specimens of an iron-carbon alloy of eutectoid composition that have been subjected to some of the heat treatments described in Problem 11.14. (b) The microstructural product of this heat treatment is 100% spheroidite. A ccording to Figure 11.22(a) the hardness of a 0.76 wt% C alloy with spheroidite is about 87 H R B. (g) The microstructural product of this heat treatment is 100% fine pearlite. A ccording to Figure 11.22(a), the hardness of a 0.76 wt% C alloy consisting of fine pearlite is about 27 H R C. 11.37 For this problem we are asked to describe isothermal heat treatments required to yield specimens having several Brinell hardnesses. (a) From Figure 11.22(a), in order for a 0.76 wt% C alloy to have a R ockwell hardness of 93 H R B, the microstructure must be coarse pearlite. Thus, utilizing the isothermal transformation diagram for this alloy, Figure 11.14, we must rapidly cool to a temperature at which coarse pearlite forms (i.e., to about 675◦ C), allowing the specimen to isothermally and completely transform to coarse pearlite. A t this temperature an isothermal heat treatment for at least 200 s is required. 11.D 1 This problem inquires as to the possibility of producing an iron-carbon alloy of eutectoid composition that has a minimum hardness of 90 H R B and a minimum ductility of 35% R A . If the alloy is possible, then the continuous cooling heat treatment is to be stipulated. A ccording to Figures 11.22(a) and (b), the following is a tabulation of R ockwell B hardnesses and percents reduction of area for fine and coarse pearlites and spheroidite for a 0.76 wt% C alloy. Microstructure HRB % RA Fine pearlite Coarse pearlite Spheroidite >100 93 88 22 29 68 Therefore, none of the microstructures meets both of these criteria. Both fine and coarse pearlites are hard enough, but lack the required ductility. Spheroidite is sufficiently ductile, but does not meet the hardness criterion. 46 CH A PTE R 12 ELECTRICAL PROPERTIES 12.5 (a) In order to compute the resistance of this copper wire it is necessary to employ E quations (12.2) and (12.4). Solving for the resistance in terms of the conductivity, R= ␳l l = A ␴A From Table 12.1, the conductivity of copper is 6.0 × 107 (-m) −1 , and R = 2m l = ␴A [6.0 × 107 (-m) −1 ]( ␲) = 4.7 × 10  −3 3 × 10−3 m 2 2 (b) If V = 0.05 V then, from E quation (12.1) I= V 0.05 V = 10.6 A = 4.7 × 10−3  R (c) The current density is just J= I I = A ␲ (d) The electric field is just 10.6 A 2 = 2d E= ␲ 6 2 2 = 1.5 × 10 A /m −3 3 × 10 2 m V 0.05 V = 2.5 × 10−2 V/m = 2m l 12.13 (a) The number of free electrons per cubic meter for copper at room temperature may be computed using E quation (12.8) as n= = ␴ |e|␮e 6.0 × 107 (-m) −1 (1.602 × 10−19 C)(0.0030 m 2 /V-s) = 1.25 × 1029 m −3 (b) In order to calculate the number of free electrons per copper atom, we must first determine the number of copper atoms per cubic meter, N Cu. From E quation (5.2) N Cu = NA ␳ ′ A Cu 47 Note: in the above expression, density is represented by ␳ ′ in order to avoid confusion with resistivity which is designated by ␳ . Thus N Cu = (6.023 × 1023 atoms/mol)(8.94 g/cm 3 )(106 cm 3 /m 3 ) 63.55 g/mol = 8.47 × 1028 m −3 The number of free electrons per copper atom is just n 1.25 × 1029 m −3 = 1.48 = 8.47 × 1028 m −3 N 12.18 This problem asks for us to compute the room-temperature conductivity of a two-phase Cu-Sn alloy. It is first necessary for us to determine the volume fractions of the ␣ and phases, after which the resistivity (and subsequently, the conductivity) may be calculated using E quation (12.12). Weight fractions of the two phases are first calculated using the phase diagram information provided in the problem. We might represent the phase diagram near room temperature as shown below. A pplying the lever rule to this situation W␣ = C − Co 37 − 8 = = 0.784 C − C␣ 37 − 0 W = Co − C␣ C − C␣ = 8−0 = 0.216 37 − 0 We must now convert these mass fractions into volume fractions using the phase densities given in the problem. (Note: in the following expressions, density is represented by ␳ ′ in order to avoid confusion with resistivity which is designated by ␳ .) U tilization of E quations (10.6a) and (10.6b) 48 leads to V␣ = = W␣ ␳␣ ′ W␣ W + ␳␣ ′ ␳′ 0.784 8.94 g/cm 3 0.216 0.784 + 8.94 g/cm 3 8.25 g/cm 3 = 0.770 V = W ␳′ W␣ W + ␳␣ ′ ␳′ 0.216 8.25 g/cm 3 = 0.216 0.784 + 8.94 g/cm 3 8.25 g/cm 3 = 0.230 Now, using E quation (12.12) ␳ = ␳␣V␣ + ␳ V = (1.88 × 10−8 -m)(0.770) + (5.32 × 10−7 -m)(0.230) = 1.368 × 10−7 -m Finally, for the conductivity ␴= 1 ␳ = 1 = 7.31 × 106 (-m) −1 1.368 × 10−7 -m 12.30 (a) In this problem, for a Si specimen, we are given p and ␴, while ␮h and ␮e are included in Table 12.2. In order to solve for n we must use E quation (12.13), which, after rearrangement, leads to n= = ␴ − p|e|␮h |e|␮e 103 (-m) −1 − (1.0 × 1023 m −3 )(1.602 × 10−19 C)(0.05 m 2 /V-s) (1.602 × 10−19 C)(0.14 m 2 /V-s) = 8.9 × 1021 m −3 (b) This material is p-type extrinsic since p (1.0 × 1023 m −3 ) is greater than n (8.9 × 1021 m −3 ). 12.38 For this problem, we are given conductivity values at two different temperatures for an intrinsic semiconductor, and are then asked to determine its band gap energy. It is possible, using 49 E quation (12.18), to set up two independent equations with C and E g as unknowns. A t 20◦ C Eg 2kT ln ␴ = C − Eg (2)(8.62 × 10−5 eV/atom-K)(293 K) ln[1.0 (-m) −1 ] = C − or C = 19.80E g A t 373 K Eg (2)(8.62 × 10−5 eV/atom-K)(373 K) ln[500 (-m) −1 ] = C − 6.21 = C − 15.55E g From these two expressions E g = 1.46 eV 12.45* In this problem we are asked to determine the magnetic field required to produce a H all voltage of −1.0 × 10−7 V, given that ␴ = 1.5 × 107 (-m) −1 , ␮e = 0.0020 m 2 /V-s, Ix = 45 A , and d = 35 mm. Combining E quations (12.21) and (12.23b), and after solving for B z , we get Bz = = |V H |␴d I x ␮e (|−1.0 × 10−7 V|)[1.5 × 107 (-m) −1 ](35 × 10−3 m) (45 A )(0.0020 m 2 /V-s) = 0.58 tesla 12.52* We want to compute the plate spacing of a parallel-plate capacitor as the dielectric constant is increased from 2.5 to 4.0, while maintaining the capacitance constant. Combining E quations (12.29) and (12.30) yields C= r oA l Now, let us use the subscripts 1 and 2 to denote the initial and final states, respectively. Since C1 = C2 , then r1 o A l1 = r2 o A l2 A nd, solving for l2 l2 = r2 l1 r1 = (4.0)(1 mm) = 1.6 mm 2.5 12.58* (a) We want to solve for the voltage when Q = 3.5 × 10−11 C, A = 160 mm 2 , l = 3.5 mm, and r = 5.0. Combining E quations (12.27), (12.29), and (12.30) yields Q = V r o 50 A l A nd, solving for V V= = Ql r oA (3.5 × 10−11 C)(3.5 × 10−3 m) (5.0)(8.85 × 10−12 F/m)(160 mm 2 )(1 m 2 /106 mm 2 ) = 17.3 V (b) For this same capacitor, if a vacuum is used Ql oA V= (3.5 × 10−11 C)(3.5 × 10−3 m) (8.85 × 10−12 F/m)(160 × 10−6 m 2 ) = = 86.5 V (e) The polarization is determined using E quations (12.35) and (12.6) as P= = o( r − 1) V l (8.85 × 10−12 F/m)(5.0 − 1)(17.3 V) 3.5 × 10−3 m = 1.75 × 10−7 C/m 2 12.D 2 This problem asks that we determine the electrical conductivity of an 80 wt% Cu-20 wt% Z n alloy at −150◦ C using information contained in Figures 12.8 and 12.35. In order to solve this problem it is necessary to employ E quation (12.9) which is of the form ␳ total = ␳ t + ␳ i since it is assumed that the alloy is undeformed. Let us first determine the value of ␳ i at room temperature (25◦ C), a value which will be independent of temperature. From Figure (12.8), at 25◦ C and for pure Cu, ␳ t (25) = 1.75 × 10−8 -m. Now, since it is assumed that the curve in Figure 12.35 was generated also at room temperature, we may take ␳ as ␳ total (25) at 80 wt% Cu-20 wt% Z n which has a value of 5.3 × 10−8 -m. Thus ␳ i = ␳ total(25) − ␳ t (25) = 5.3 × 10−8 -m − 1.75 × 10−8 -m = 3.55 × 10−8 -m Finally, we may determine the resistivity at −150◦ C, ␳ total (−150), by taking the resistivity of pure Cu at −150◦ C from Figure 12.8, which gives us ␳ t (−150) = 0.55 × 10−8 -m. Therefore ␳ total(−150) = ␳ i + ␳ t (−150) = 3.55 × 10−8 -m + 0.55 × 10−8 -m = 4.10 × 10−8 -m A nd, using E quation (12.4) the conductivity is calculated as ␴= 1 ␳ = 1 = 2.44 × 107 (-m) −1 4.10 × 10−8 -m 51