Chapter_5_Solutions-_Study_Guide_for_Exam_2
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322
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Chemistry
Date
Dec 6, 2023
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A sheet of steel 1.6 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion
coef±cient for nitrogen in steel at this temperature is 7.0 × 10
m
/s, and the diffusion ²ux is found to be 1.0 × 10
kg/m
-s. Also, it is known that the
concentration of nitrogen in the steel at the high-pressure surface is 4.6 kg/m
. How far into the sheet from this high-pressure side will the concentration be 3.5
kg/m
? Assume a linear concentration pro±le.
m
-11
2
-7
2
3
3
A sheet of steel 1.6 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coef±cient
for nitrogen in steel at this temperature is 7.0 × 10
m
/s, and the diffusion ²ux is found to be 1.0 × 10
kg/m
-s. Also, it is known that the concentration of nitrogen in
the steel at the high-pressure surface is 4.6 kg/m
. How far into the sheet from this high-pressure side will the concentration be 3.5 kg/m
? Assume a linear concentration
pro±le.
Solution
Video Solution
(https://education.wiley.com/content/Callister_Mat_Sci_and_Eng_10e/media/simulations/video_solutions/Callister_10e_Video_Solution_5.9.html)
This problem is solved using Equation 5.2 in the following form:
If we take
C
to be the point at which the concentration of nitrogen is 4.6 kg/m
, then it becomes necessary to solve for
x
, as
Assume
x
is zero at the surface, in which case
= 7.7 × 10
m = 0.77 mm
Hint
Solution
Assistance Used
-11
2
-7
2
3
3
A
3
B
A
-4
Attempts: 2 of 3 used
7.7E-4
5.3 Practice Question 1
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The puri±cation of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of
hydrogen that pass per hour (in kg/h) through a 3.4-mm thick sheet of palladium having an area of 0.25 m
at 500°C. Assume a
diffusion coef±cient of 6.0 x 10
m
/s, that the concentrations at the high- and low-pressure sides of the plate are 3.3 and 0.36
kg/m
(kilogram of hydrogen per cubic meter of palladium), and that steady-state conditions have been attained.
kg/h
2
-8
2
3
The puri±cation of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of hydrogen
that pass per hour (in kg/h) through a 3.4-mm thick sheet of palladium having an area of 0.25 m
at 500°C. Assume a diffusion
coef±cient of 6.0 x 10
m
/s, that the concentrations at the high- and low-pressure sides of the plate are 3.3 and 0.36 kg/m
(kilogram of
hydrogen per cubic meter of palladium), and that steady-state conditions have been attained.
S
olution
The diffusion ²ux, through and perpendicular to a unit cross-sectional area,
J
, may be calculated using Equation 5.1, that is
The diffusion ²ux is proportional to the concentration gradient according to Equation 5.2 as
Equating these two expressions for
J
to one another as
and then solving for the mass (
M
) gives
= 0.0467 kg/h
Hint
Solution
Assistance Used
2
-8
2
3
Attempts: 2 of 3 used
4.67E-2
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Determine the carburizing time (in s) necessary to achieve a carbon concentration of 0.44 wt% at a position 1.0 mm into an iron-carbon alloy that initially contains
0.031 wt% C. The surface concentration is to be maintained at 1.2 wt% C, and the treatment is to be conducted at 1240°C. Assume that
D
= 5.1 × 10
m
/s and
Q
= 154 kJ/mol. You will ±nd the table below useful.
z
erf
(
z
)
z
erf
(
z
)
z
erf
(
z
)
0
0
0.55
0.5633
1.3
0.9340
0.025
0.0282
0.60
0.6039
1.4
0.9523
0.05
0.0564
0.65
0.6420
1.5
0.9661
0.10
0.1125
0.70
0.6778
1.6
0.9763
0.15
0.1680
0.75
0.7112
1.7
0.9838
0.20
0.2227
0.80
0.7421
1.8
0.9891
0.25
0.2763
0.85
0.7707
1.9
0.9928
0.30
0.3286
0.90
0.7970
2.0
0.9953
0.35
0.3794
0.95
0.8209
2.2
0.9981
0.40
0.4284
1.0
0.8427
2.4
0.9993
0.45
0.4755
1.1
0.8802
2.6
0.9998
0.50
0.5205
1.2
0.9103
2.8
0.9999
s
0
-5
2
d
Determine the carburizing time (in s) necessary to achieve a carbon concentration of 0.44 wt% at a position 1.0 mm into an iron-carbon alloy that initially
contains 0.031 wt% C. The surface concentration is to be maintained at 1.2 wt% C, and the treatment is to be conducted at 1240°C. Assume that
D
= 5.1 ×
10
m
/s and
Q
= 154 kJ/mol. You will ±nd the table below useful.
z
erf
(
z
)
z
erf
(
z
)
z
erf
(
z
)
0
0
0.55
0.5633
1.3
0.9340
Hint
Solution
Assistance Used
0
-5
2
d
3.35
0.025
0.0282
0.60
0.6039
1.4
0.9523
0.05
0.0564
0.65
0.6420
1.5
0.9661
0.10
0.1125
0.70
0.6778
1.6
0.9763
0.15
0.1680
0.75
0.7112
1.7
0.9838
0.20
0.2227
0.80
0.7421
1.8
0.9891
0.25
0.2763
0.85
0.7707
1.9
0.9928
0.30
0.3286
0.90
0.7970
2.0
0.9953
0.35
0.3794
0.95
0.8209
2.2
0.9981
0.40
0.4284
1.0
0.8427
2.4
0.9993
0.45
0.4755
1.1
0.8802
2.6
0.9998
0.50
0.5205
1.2
0.9103
2.8
0.9999
Solution
We are asked to compute the carburizing (i.e., diffusion) time required for a speci±c nonsteady-state diffusion situation. It is ±rst necessary to use Equation
5.5:
wherein,
C
= 0.44,
C
= 0.031,
C
= 1.2, and
x
= 1.0 mm = 1.0 × 10
m. Thus,
or
By linear interpolation using data provided in the problem statement
z
erf(
z)
0.65
0.6420
z
0.6500
0.70
0.6778
which leads to the following:
From which
From the problem statement, data for the diffusion of carbon into iron are as follows:
D
= 5.1 ×10
m /s
Q
= 154,000 J/mol
Therefore, we calculate the value of the diffusion coef±cient at 1240°C (1513 K) using Equation 5.8 as follows:
x
0
s
-3
0
-5
2
d
Thus,
Solving for
t
yields
t
= 2.34 × 10
s = 0.650 h
3
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5.4 Practice Question 2
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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to
0.38 wt% at a point 1.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 7.0 mm
position for an identical steel and at the same carburizing temperature.
h
For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to 0.38
wt% at a point 1.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 7.0 mm position for an
identical steel and at the same carburizing temperature.
Solution
This problem calls for an estimate of the time necessary to achieve a carbon concentration of 0.38 wt% at a point 7.0 mm from
the surface. From Equation 5.6b,
But since the temperature is constant, so also is
D
constant, which means that
or
Thus, if we assign
x
= 1.6 mm,
x
= 7.0 mm, and
t
= 11 h, then
from which
t
= 211 h
Hint
Solution
Assistance Used
1
2
1
2
Save for Later
Attempts: 1 of 3 used
Submit Answer
3.40
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