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Chap5_2pgs

# Chap5_2pgs - a high hydrogen gas from a low hydro-gen gas...

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fltom and lon ]'|overnents in ]'|alerinls 5{ Atoms are found to move from one lattice position to another at the rate of 5 x ld jumps/s at 400oC when the activation energy for their movement is 30,000 caUmol. Calculate the jump rate at 750oC. Solution: Rare = 5 x ld _ coexp[-30,000(1.9E7x673)] x co e*p1-30,000/@fr1iffi = exp(-22'434 + 14'759) 5x ld =exp(-7.675)=4.64x lo4 x *=ffi=l.oSxloejumpvs 5-15 The diffusion coefficient for Cr+3 in CrrO, is 6 x 10-15 crn2/s at 727oC and is I x 104 cm2/s at.140SC. Calculate (a) the activation energy and (b) the constant Do. solution: Orffi=ffi 6 x 1016 = exp[-O(0.000503 - 0.00030)] = exp[4.000203 O] -12.024 =-0.000203 C or Q=59,230 cal/mol (b) I x l0re = Do exp[-59,230(1.987x1673)] = Do exp(-17.818) I x lfre = 1.E28 x l0{ D0

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CHAPTER 5 Atom and lon Movements in Materials 43 0.01 67 x 1024 - 0.0209 x 1024 = -f .68Z;n atoms/cm3 ' cm LclLx = 0.0025 cm S-25 A 0.001-in. BCC iron foil is used to separate
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Unformatted text preview: a high hydrogen gas from a low hydro-gen gas at OSOpC. 5 x ld H atoms/cm3 are in equilibrium with the hot side of the foil, while 2 x 103 H atomVcm3 are in equilibrium with the cold side Determine (a) the concentration gradient of hydrogen and (b) the flux of hydrogen through the foil. 2 x 1 0 3 - 5 x l O E = -1 969 x 108 H atoms/cm3 ' cm Sofution: (a) LclLx = (0.001 in.)(2.54 cm/in.) (b) J = -D(Lcl L,x) = -0.0012 exp[-3600/( I .987X923)](- 1969 x 108) J -0.33 x 108 H abms/cm2 . s 5-43 What temperature is required to obtain O.SO% C at a distance of 0.5 mm beneath the surface af a 0.207o c steel in 2 h, when I .10% c is present at the surface? Assume that the iron is FCC. Solution: l'1 - o'5 ,.rff= o'667 = erf[o'05/ zJ a I O.OS1Z',lfr =0.685 or Jil =0.0365 or Dr=0.00133 r = (2 hX3600 s/h) = 72gg . p = 0.00B3nzQQ = 1.85 x l0-7 =a.23 exp[-32,g0an.gg7n exp(-16,558/7) = 8.043 x 10-7 f=ll80K=907oC...
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Chap5_2pgs - a high hydrogen gas from a low hydro-gen gas...

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