C let nv1 nv at 1120k and nv2 nv at 1000k engineering

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Unformatted text preview: ition, thus ⎛2 ⎞ rinterstitial = ⎜ − 1⎟ R ⎝3 ⎠ b. c. Let NV1 = Nv at 1120˚K and NV2 = Nv at 1000˚K Engineering Material ⎜−⎟ − ⎜ R ⎜T T ⎟⎟ N V1 Ne = 5= = e⎝ ⎝ 1 2 ⎠ ⎠ ⎛Q⎞ ⎜− V ⎟ NV 2 ⎜ ⎟ Ne⎝ RT2 ⎠ R ln 5 (8.31 J / mol˚ K ) ln 5 ⇒ QV = − 1 1 = − = 116300 J / mol 1 1 − − T1 T2 1120 1000 ⎛Q ⎜− V ⎜ RT ⎝ 1 ⎞ ⎟ ⎟ ⎠ ⎛Q⎛ 1 V ⎜ ⎞ 1 ⎞⎟ Problem 4 a. ⎛2 ⎞ rinterstital = ⎜ − 1⎟ R ⎝3 ⎠ with R = 0.124 nm we get rinterstital=4 .0 1 = 0.077nm ⇒ rC a r b o n 0.077 = rinterstital 0.0192 this explains the low solubility of Carbon in α-Fe b. c. 4R = 2 R + 2rinterstitial 2 2R ⇒ rinterstitial = − R = ( 2 − 1) R = 0.0526nm 2 r 0.77 ⇒ Carbon = = 1.46 rinterstitial 0.0526 a=...
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