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MATSE 259
Solutions to classwork #4
1. A plate of iron is exposed to a carburizing (carbon-rich)
atmosphere on oneside and a decarburizing (carbon-deficient)
atmosphere on the other side at
700C. If a condition of steady state is reached, calculate the
diffusion flux ofcarbon through the plate if the concentrations of
carbon at positions of 5 and 10mm beneath the carburizing surface
are 1.2 and 0.8 kg/m3 respectively. Assumea diffusion coefficient
of 3 x 10-11 m2/s at this temperature.
Ficks first law is utilized to determine the diffusion flux:
J = -DBA
BA
xx
CC
= -(3x10-11m2/s)
)m10(5x10
0.8)kg/m(1.223
3
= 2.4x10-9 kg/m2-s
2. The diffusion coefficients for copper in aluminum at 500 and
600C are 4.8 x10-14 and 5.3 x 10-13 m2/s, respectively. Determine
the approximate time at 500Cthat will produce the same diffusion
result (in terms of concentration of Cu atsome specific point in
Al) as a 10-h heat treatment at 600C.
This is a diffusion problem in which the following equation may
be employed
Dt
x2= constant
The composition in both diffusion situations will be equal at
the same position
(i.e. x is also a constant), thus
Dt = constant
at both temperatures. That is,
(Dt)500 = (Dt)600or
t500 =500
600
D
(Dt)=
/sm4.8x10
/s)(10h)m(5.3x10
214
213
= 110.4 h
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3. The purification of hydrogen gas by diffusion through a
palladium sheet wasdiscussed in Section 5.3. Compute the number of
kilograms of hydrogen thatpass per hour through a 5-mm thick sheet
of palladium having an area of 0.20 m2
at 500 C. Assume a diffusion coefficient of 1.0 x 10o -8 m2 /s,
that the
concentrations at the high- and low-pressure sides of the plate
are 2.4 and 0.6 kgof hydrogen per cubic meter of palladium, and
that steady-state conditions havebeen attained.
This problem calls for the mass of hydrogen, per hour, that
diffuses through a Pd
sheet. It first becomes necessary to employ both Equations
(5.1a) and (5.3).
Combining these expressions and solving for the mass yields
M = JAt = - DAtCx
= - (1.0 x 10-8 m2/s)(0.2 m2)(3600 s/h)
0.6 - 2.4 kg/m
3
5 x 10-3 m
= 2.6 x 10-3 kg/h
4. A sheet of BCC iron 1mm thick was exposed to a carburizing
gas atmosphereon one side and a decarburizing gas atmosphere on the
other side at 725C.After having reached steady state, the iron was
quickly cooled to room
temperature. The carbon concentrations at the two surfaces of
the sheet weredetermined to be 0.012 and 0.0075 wt%. Compute the
diffusion coefficient if thediffusion flux is 1.4x10-8 kg/m2-s.
Hint: Convert the concentrations from weightpercent to kilograms of
carbon per cubic meter of iron.
This problem calls for computation of the diffusion coefficient
for a steady-state
diffusion situation. Let us first convert the carbon
concentrations from wt% to kg
C/m3. Assuming that the addition of carbon does not increase the
volume of the
alloy, for 0.012 wt% C
CC" =Fe
C
100/
C
=3cm100/7.87g/
0.012= 9.44 x 10-4 g/cm3 = 0.944 kg/m3
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Similarly, for 0.0075 wt% C
CC" =Fe
C
100/
C
=3cm100/7.87g/
0.0075 = 5.90 x 10-4 g/cm3 = 0.590 kg/m3
Now, using a form of Equation (5.3)
D = - J
xA - xBCA - CB
= - (1.40 x 10-8 kg/m2-s)
-10-3
m0.944 kg/m3 - 0.590 kg/m3
= 3.95 x 10-11 m2/s
5. Cite the values of the diffusion coefficients for the
interdiffusion of carbon in
both -iron (BCC) and -iron (FCC) at 900 oC from Table 5.2 on
page 101 in thetext book. Which is larger? Explain why this is the
case.
We are asked to compute the diffusion coefficients of C in both
and iron at900 C. Using the data in Table 5.2,o
D = (6.2 x 10-7 m2/s) exp
-80000 J/mol
(8.31 J/mol-K)(1173 K)
= 1.69 x 10-10
m2/s
D= (2.3 x 10-5 m2/s) exp
-148000 J/mol
(8.31 J/mol-K)(1173 K)
= 5.86 x 10-12
m2/s
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The D for diffusion of C in BCC iron is larger, the reason being
that theatomic packing factor is smaller than for FCC iron (0.68
versus 0.74); thismeans that there is slightly more interstitial
void space in the BCC Fe, and,
therefore, the motion of the interstitial carbon atoms occurs
more easily.
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