Question

For a steel alloy it has been determined that a carburizing heat treatment of 11 h duration at 822°C will raise the carbon concentration to 0.49 wt% at a point 2.7 mm from the surface. Estimate the time necessary to achieve the same concentration at a 5.7 mm position for an identical steel and at a carburizing temperature of 1130°C. Assume that D0 is 2.4 × 10-5 m2/s and Qd is 112 kJ/mol.

Answer #2

answered by: anonymous

Determine the carburizing time (in s) necessary to achieve a
carbon concentration of 0.44 wt% at a position 1.9 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 1020°C. Assume that D0 = 5.1 x 10-5 m2/s and
Qd =154 kJ/mol. You will find the table below useful.

Problem: Utilize the Tabulation Error
Function Value Table from your Examination
Booklet.
--Given Values--
Concentration of Nitrogen (wt%): = 0.38
Position in mm = 8.5
Diffusion Coefficient for Nitrogen in Iron at 700℃ = 9.70E-10
Nitrogen in Iron Heat Treatment Time (hours): = 13
Carburizing Time (hrs): = 10
Steel Alloy Case Depth (mm): = 3
Steel Alloy New Case Depth (mm) = 8.8
Activation Energy kJ/mol: = 107
Temperature for Problem 3 (Celsius): = 821
- = Data for...

Determine the carburizing time (in s) necessary to achieve a
carbon concentration of 0.44 wt% at a position 1.6 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 1180°C. Assume that D0 =
5.1 x 10-5 m2/s and Qd
=154 kJ/mol. You will find the table below useful.
z
erf(z)
z
erf(z)
z
erf(z)
0
0
0.55
0.5633
1.3...

Problem: Determine the carburizing time
necessary to achieve a carbon concentration listed below at a
position listed below into an iron-carbon alloy that initially
contains 0.11 wt% C. The surface concentration is to be maintained
at 1.2 wt% C, and the treatment is to be conducted at the
temperature listed below. Assume that Do = 6.2x10-5
m2/s and Qd = 166 kJ/mol. Utilize the
Tabulation Error Function Value Table from your
Examination Booklet.
--Given Values--
Position (mm) = 3.7
Temperature...

After a 34-hour heat treatment at 1080°C (and subsequent cooling
to room temperature), the concentration of A is 3.4 wt% at the
13.6-mm position within metal B. If another heat treatment is
conducted on an identical diffusion couple, this time at 890°C for
34 hours, at what position will the composition be 3.4 wt% A? The
preexponential and activation energy values for the diffusion of A
in B are 3.4 × 10-5 m2/s and 156 kJ/mol, respectively.

Directions: Additional information to complete these problems
can be found in your Examination Booklet. Write all Engineering
Notation to the tenth place only.
--Given Values--
Temperature 1 in Celsius = 803
Thickness of Palladium Sheet in mm = 2.4
Area of Palladium sheet (m2) = 0.33
Thickness of Sheet of Steel (mm) = 4.69
Diffusion Flux (kg/(m2-s)) = 6.26E-07
Temperature 2 in Celsius = 1049
Problem 1:
Calculate the value of the diffusion coefficient D (in
m2/s) at Temperature 1...

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