# Structure.questions XIII answer - exists At the crack tip R...

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MTLE-2100 Structure of Engineering Materials, Questions XIII-Answer 1. Vapor pressure of a sphere with radius, r, is given by ln(p 1 /p 0 ) = (2 γ /r)(M/RT ρ ) Calculate the increase in the vapor pressure, p 1 /p 0 , for Al 2 O 3 particles with radius 1000A and 100A at 1850 ° C. Assume that its surface energy is 1 J/m 2 , density 4 g/cm 3 and molecular weight is 102 g/mol (0.102 kg/mol). r=1000A = 10 -7 m, 100A = 10 -8 m, R=8.314 J/K mol. Density = 4g/cm 3 = 4x10 +3 kg/m 3 ln (p 1 /p 0 ) = (2x 1/r(m) )(0.102/8.314 x 2123x4x10 3 ) = 2.9x10 -9 /r [(J/m 2 m)(kg/mol)/(J/mol)(kg/m 3 )] = no unit when r has a unit of m. For r=1000A = 10 -7 m, ln (p 1 /p 0 ) = 2.9 x10 -2 p 1 /p 0 = 1.029 r=100A = 10 -8 m , ln (p 1 /p 0 ) = 2.9 x10 -1 p 1 /p 0 = 1.34 2. Solubility of a surface with principal radii, R 1 and R 2 is given by ln(S 1 /S 0 ) = γ (1/R 1 +1/R 2 )(M/RT ρ ). Predict what happens at a crack tip of SiO 2 glass when the glass is immersed in a hot water where there is a finite solubility of SiO 2

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Unformatted text preview: exists. At the crack tip R 1 = , R 2 has small negative value. Consequently, ln(S 1 /S ) would have a large negative value. S 1 /S &amp;lt;&amp;lt;1. Solubility at the crack tip is small. 3.Calculate the solubility change S 1 /S for silica spheres with radius 1000A and 100A at room temperature in water. Assume that the surface energy is 0.5J/m 2 , density is 2.3g/cm 3 . The molecular weight of SiO 2 is 60 g/mol. ln(S 1 /S ) = (1/R 1 +1/R 2 )(M/RT ) = (2 /r)(M/RT ) =(1/r)(0.060/8.314x 300x2.3x10 3 ) =1.05x10-8 /r For r=1000A = 10-7 m, ln (S 1 /S ) = 0.105 S 1 /S 0 = 1.11 r=100A = 10-8 m , ln (S 1 /S ) = 1.05 S 1 /S 0 = 2.86 4. Calculate the excess pressure a sphere of SiO 2 with radius 100 A is subjected to. P = 2 /r P = 2x 1 (J/m 2 )/ 10-8 m = 2x10 8 (J/m 3 ) = 2x10 8 (N/m 2 ) = 200MPa....
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Structure.questions XIII answer - exists At the crack tip R...

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