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8
Principles of Solidification
8–10 Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required, and (b) the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm. Solution: From Table 8–1, ∆Tmax = 480oC r* = (2)(255 × 10−7 J/cm2)(1453 + 273) = 6.65 × 10−8 cm (2756J/cm3)(480) V = 45.118 × 10−24 cm3 Vnucleus = (4π/3)(6.65 × 10−8 cm)3 = 1232 × 10−24 cm3 number of unit cells = 1232/45.118 = 27.3 atoms per nucleus = (4 atoms/cell)(27.3 cells) = 109 atoms 8–11 Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate (a) the critical radius of the nucleus required, and (b) the number of iron atoms in the nucleus. Assume that thelattice parameter of the solid BCC iron is 2.92 Å. Solution: r* = (2)(204 × 10−7 J/cm2)(1538 + 273) = 10.128 × 10−8 cm (1737 J/cm3)(420)

ao = 3.56 Å

V = (4π/3)(10.128)3 = 4352 Å3 = 4352 × 10−24 cm3 Vuc= (2.92 Å)3 = 24.897 Å3 = 24.897 × 10−24 cm3 number of unit cells = 4352/24.897 = 175 atoms per nucleus = (175 cells)(2 atoms/cell) = 350 atoms 8–12 Suppose that solid nickel was able to nucleatehomogeneously with an undercooling of only 22oC. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid FCC nickel is 0.356 nm. 87

88

The Science and Engineering of Materials

Instructor’s Solution Manual

Solution:

r* =

(2)(255 × 10−7 J/cm2)(1453 + 273) = 145.18 × 10−8 cm (2756 J/cm3)(22) (see Problem 8–10) cm)3= 1.282 × 10−17 cm3 10−8

Vuc = 45.118 × 10−24 cm3 Vnuc = (4π/3)(145.18 ×

number of unit cells = 1.282 × 10−17 / 45.118 × 10−24 = 2.84 × 105 atoms per nucleus = (4 atoms/cells)(2.84 × 105 cell) = 1.136 × 106 8–13 Suppose that solid iron was able to nucleate homogeneously with an undercooling of only 15oC. How many atoms would have to group together spontaneously for this to occur? Assumethat the lattice parameter of the solid BCC iron is 2.92 Å. Solution: r* = (2)(204 × 10−7 J/cm2)(1538 + 273) = 283.6 × 10−8 cm (1737 J/cm3)(15) (see Problem 8-10)

Vuc = 24.897 × 10−24 cm3

Vnuc = (4π/3)(283.6 × 10−8 cm)3 = 95,544,850 × 10−24 cm3 number of unit cells = 95,544,850/24.897 = 3.838 × 106 atoms per nucleus = (2 atoms/cells)(3.838 × 106 cell) = 7.676 × 106 8–14 Calculate the fractionof solidification that occurs dendritically when iron nucleates (a) at 10oC undercooling, (b) at 100oC undercooling, and (c) homogeneously. The specific heat of iron is 5.78 J/cm3.oC. Solution: f= (5.78 J/cm3.oC)(10oC) c∆T = ∆Hf 1737 J/cm3 = 0.0333

(5.78 J/cm3.oC)(100oC) c∆T = = 0.333 ∆Hf 1737 J/cm3 (5.78 J/cm3.oC)(420oC) c∆T = , therefore, all dendritically ∆Hf 1737 J/cm3 8–28 Calculate thefraction of solidification that occurs dendritically when silver nucleates (a) at 10oC undercooling, (b) at 100oC undercooling, and (c) homogeneously. The specific heat of silver is 3.25 J/cm3.oC. Solution: f= (3.25 J/cm3.oC)(10oC) c∆T = ∆Hf 965 J/cm3 = 0.0337

(3.25 J/cm3.oC)(100oC) c∆T = = 0.337 ∆Hf 965 J/cm3 (3.25 J/cm3.oC)(250oC) c∆T = = 0.842 ∆Hf 965 J/cm3 8–29 Analysis of a nickel castingsuggests that 28% of the solidification process occurred in a dendritic manner. Calculate the temperature at which nucleation occurred. The specific heat of nickel is 4.1 J/cm3.oC. Solution: f= (4.1 J/cm3.oC)(∆T) c∆T = = 0.28 ∆Hf 2756 J/cm3 ∆T = 188oC or Tn = 1453 − 188 = 1265oC

CHAPTER 8 8–31

Principles of Solidification

89

A 2-in. cube solidifies in 4.6 min. Calculate (a) the moldconstant in Chvorinov’s rule and (b) the solidification time for a 0.5 in. × 0.5 in. × 6 in. bar cast under the same conditions. Assume that n = 2. Solution: (a) We can find the volume and surface area of the cube: V = (2)3 = 8 in.3 A = 6(2)2 = 24 in.2 t = 4.6 = B(8/24)2 B = 4.6/(0.333)2 = 41.48 min/in.2 (b) For the bar, assuming that B = 41.48 min/in.2: V = (0.5)(0.5)(6) = 1.5 in.2 A = 2(0.5)(0.5)...
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