Thermo-Module-problem-set_v3
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CALPHAD Lab
First step is to download Thermocalc. To get most recent Thermocalc demo go here:
https://www.thermocalc.com/academia/students/educational-package-for-
students/
Useful links:
1.
Thermocalc help:
file:///Applications/Thermo-Calc-academic-
2017b.app/Contents/Resources/HTML5/Content/Welcome.htm
2.
Save projects often. Mine tend to freeze and I need to restart the Thermo-calc
software.
3.
Use “Project” pane features to reuse settings and “Clone” feature to perform
similar actions. Right clicking gives options. All of problem 1 and 2 can each be in
their own projects, giving two projects that contain the whole lab.
1.
Martensitic stainless steels offer a combination of high mechanical strength and
corrosion resistance. To increase the hardness of the surface, the process of
“carburization”, or annealing in the presence of a carbon-rich gas, can be
performed, which will increase the carbon concentration in the steel near the
surface. However, the increase in hardness must be balanced against a loss of
corrosion resistance. This exercise will help understand the reasons for this
trade-off.
The most basic martensitic stainless steel is Type 410, which has a
nominal composition of (by mass percent)
Cr 13.5
C
0.15.
Fe 86.35
Usually when specifying steel compositions this would be written Fe-13.5Cr-
0.15C, where composition is written prior to each element, and the Fe
composition is omitted since it is the main constituent. 410 steels may also
include a small amount of Mn which is primarily used to combine with and
remove oxygen impurities; it can be neg
lected
for the purposes of this exercise. Carburization causes an increase in carbon
concentration near the surface, leading to the formation of chromium-rich
carbide phases within the FCC matrix, including M
23
C
6
, M
7
C
3
, and M
3
C (where M
stands for metal). M
23
C
6
and M
7
C
3
increase hardness, but the formation of M
3
C
(also called cementite) reduces strength. In the following, you can label phases in
picture by right-clicking but it takes a lot of time as you need to do each one
separately so don’t worry about labeling them unless I ask explicitly.
1.1. Using Thermo-calc with the database FEDEMO do the following.
1.1.1.
Plot an isopleth of this steel for mass percent carbon varying from 0-5
mass-%. An isopleth keeps all compositions constant except one alloying
element and the host, in this case C and Fe, respectively. You can do this
with the “Property Model” tool and select “Phase Diagram” in the
“Equilibrium Calculator” to do this. Put plot below as Figure 1.1.1. Please
label the phases that occur at 1230K. Note that the FCC Fe phase forms
at this temperature.
1.1.2.
Plot the phase fractions as a function of mass percent carbon at 1230
K in the range 0-5 mass-% carbon. You can do this with “Property
Model” in the “Equilibrium Calculator”. Put plot below as Figure 1.1.2.
1.2. Using the results of part (a), at a temperature of 1230 K, determine the range
of mass percent carbon that allows M
23
C
6
and/or M
7
C
3
to form, but prevents
cementite from forming (carburization will not raise C content dramatically
which is why you can limit your search to <5 mass % C). Put text below as
Response 1.2.
From about 0.2% C to 3.5% C.
1.3. Corrosion resistance in stainless steels is significantly enhanced by the
presence of Cr in the FCC matrix phase.
After carburization, it has been
found that corrosion resistance at the surface decreases. Explain why this
occurs using a plot of Cr content of the FCC phase versus carbon
concentration. Again use a temperature of 1230 K and up to 5 mass % C. You
can get Cr content as the output variable with a “Property Model” in the
“Equilibrium Calculator” but you must set the output in the “Plot Renderer
Y-axis tab. Put plot below as Figure 1.3. Put text below as Response 1.3.
The corrosion resistance decreases because as the carbon
concentration increases, the Cr content in FCC matrix decreases as
seen in the plot.
2.
Kinetics of carbide precipitation in steels. Open example: P_03_Precipitation_Fe-
C-Cr_TTT_Cementite-M7C3-M23C6.tcu. You can find this under File/Example
Files/Precipitation Module – TC-Prisma.
2.1. First, write down the composition of this example. Put text below as
Response 2.1.
Fe-12Cr-0.1C
2.2. Perform a calculation at thermodynamic equilibrium (use the
“Project Pane” to find the preset “Equilibrium Calculation”) to determine
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what phases are present at 1000K. Put the plot below as Figure 2.2. Put text
below as Response 2.2. Note that the point of this is to understand where
our kinetic simulation should end up when we run for long enough time,
which is equilibrium.
Figure 2.2
BCC_A2 and M23C6 are present at 1000K
2.3. Calculate a TTT diagram and determine how long it takes to form M23C6 at
1000K (try to be accurate within a factor of 2x or so but don’t worry about
being precise) (use the “Project Pane” to find the preset “Precipitation
Calculation”). Keep all the precipitate phases even though we are just asking
about M23C6 as they could influence the behavior of M23C6. Set the
temperature step size to 50 to make the calculation faster. Set all the
interfacial energies to be “user defined” (somewhat strangely, this calculates
defaults and please use these defaults. This gives a value of 0.252 J/m2 for
M23C6). Set the max time to 10
5
seconds. Put the TTT plot below as Figure
2.3. Put text below as Response 2.3. Note that this calculation can take 2-
10min to run depending on the computer.
Figure 2.3
About 860 seconds.
2.4. Change the interfacial E by ~10% (reduce by 0.02 J/m2) for M23C6 to see
impact on time. Determine how long it takes to form M23C6 at 1000K with
the new interfacial energy (try to be accurate within a factor of 2x or so but
don’t worry about being precise)? This demonstrates how sensitive the
results are to the exact interfacial energy. Why do you think the results are so
sensitive? Put plot
below as Figure 2.4. Put text below as Response 2.4.
Figure 2.4
About 43.4 seconds with the new interfacial energy.
2.5. Perform an isothermal annealing simulation at 1000 K for10
5
s.
Use the
original interfacial energies, Adjust plot renderer settings to get clear plots
of average radius and number density and volume fraction vs. time for all 3
precipitates.
Set axes to log if needed to see all data.
Put plots below as
Figure 2.5a (radius), 2.5b (number density), and 2.5c (volume fraction).
Please put results for each precipitate type on the same plot, e.g., Figure 2.5a
should have radius of cementite, M7C3, M27C6). What is happening over
long times to these precipitates? Put text below as Response 2.5.
Figure 2.5a
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Figure 2.5b
Figure 2.5c
Over a long period of time, both cementite and M7C3 reduced in size
while M23C6 becomes the primary precipitate.
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- Chrome File Edit View History Bookmarks People Tab Window Help McGraw-Hill Campus - ALEKS Science - CHM1045 GEN CHEM 1 BLENDED 669113 A bconline.broward.edu/d21/le/content/466883/fullscreen/12868783/View McGraw-Hill Campus - ALEKS Science O GASES Interconverting pressure and force A chemistry graduate student is designing a pressure vessel for an experiment. The vessel will contain gases at pressures up to 470.0 MPa. The student's design calls for an observation port on the side of the vessel (see diagram below). The bolts that hold the cover of this port onto the vessel can safely withstand a force of 2.80 MN. pressure vessel bolts side View port Calculate the maximum safe diameter w of the port. Round your answer to the nearest 0.1 cm. O cm Explanation Check O2021 McGraw-Hill Education. All Rights Reserved. Terms of Use FEBarrow_forwardThermodynamics Please READ THE PROBLEM! I’m tired of posting the same problem over and over. please Read. This is what the asking for please: -1. sketch and label all your answers on the nozzle -2. Sketch and label all your answers on the process on a P-V diagram.... Thank youarrow_forwardCourse Home llege.com/course.html?courseld=17313546&OpenVellumHMAC=1c89e19b153e443490bb4df0da3b2ded#10001 to Review | Constants pour unistur very unu sıyın mm nyurve. Fv = 390 N Sur Previous Answers Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Eigure 1). Suppose that an 78.6-kg climber, who is 1.88 m tall and has a center of gravity 1.0 m from his feet, rappels down a vertical cliff with his body raised 40.4° above the horizontal. He holds the rope 1.54 m from his feet, and it makes a 20.7° angle with the cliff face. ✓ Correct Part D Figure 1 of 1 What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff? Express your answer using two significant figures. {—| ΑΣΦ ? fs= Submit Provide Feedback Next > P Pearson Copyright © 2022 Pearson…arrow_forward
- ECO 5. AUTOMOTIVE. The power an engine produces is called horsepower. In mathematical terms, one horsepower is the power needed to move 550 pounds one foot in one second, or the power needed to move 33,000 pounds one foot in one minute. Power, in physics, is defined simply as the rate of doing work. The formula below gives the horsepower at 5,252 radians per second. https://philkotse.com/toyota-corona-ior-sale-in-baguio/1991-for-sale-in-aid7017151 625T 1313 where H is the horsepower and T is the torque a. Find the inverse of the model. b. If a taxi produces a horsepower of 200, what is the torque it generates? Solve here:arrow_forwardStage Design A cable is used to support an actor as he swings onto the stage. Now suppose the tension in the cable is 920 N as the actor reaches the lowest point. What diameter should an 11-m-long steel cable have if we do not want it to stretch more than 0.50 cm under these conditions? R. Actor Sandbag SOLUTION Conceptualize Look back at the Example "A Grand Entrance," where we analyzed a cable used to support an actor as he swung onto the stage. We ignored any stretching of the cable there, but we wish to address this phenomenon in this example. Categorize We perform a simple calculation involving the equation F A Y = AL Li so we categorize this example as --Select--- v problem. Solve the Young's modulus equation for the cross-sectional area of the cable: FL; A = YAL Assuming the cross section is circular, find the diameter of the cable from d = 2r and A = ar (Use the following as necessary: F, Y, Li, AL, and r.): FL A d = 2r = 2 - = 2 Substitute numerical values. (Enter your answer…arrow_forwardProjects A and B are mutually exclusive. The minimum attractive rate of return (MARR) is 12%. Using rate of return analysis, which project should be selected? If the image fails to load here, go to https://www.dropbox.com/s/ld6wctqieu8jgwp/ROR.jpg Year 0 1 2 3 4 ROR A - $750 $200 $200 $200 $600 17.68% B - $1,150 $300 $350 $400 $700 16.44% O Project A O Project B O Both Project A and B O Select none of the project. O Insufficient information to make a decision. B-A - $400 $100 $150 $200 $100 13.69%arrow_forward
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