phase dtransformation diagram and heat
treatment.
Can you answer parts a b and c please. need it done in the next 30 mins
Homework Answers
a) Here, ti get 
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phase dtransformation diagram and heat
treatment.
Can you answer parts a b and c please. need...
Round to the nearest 5%--no need for something like 43.2...that would become 45. If there's none of that phase form...
Round to the nearest 5%--no need for something like 43.2...that
would become 45. If there's none of that phase formed, enter a 0
(zero). If it looks close enough then it is
800 1400 Eutectoid temperature 700 1200 1000 500 800 400 600 300 M(start) 200 400 M + A M( 5096) M19096) 100 10 102 03 104105 Time (s) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for...
Make a copy of the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition provided...
Make a copy of the isothermal transformation diagram for an
iron–carbon alloy of eutectoid composition provided overleaf and
then sketch and label time–temperature paths on this diagram to
produce the following microstructures:
a) 25% martensite and 75% austenite
b) 100% fine pearlite
c) 25% pearlite, 25% bainite, 45% martensite, and 5%
austenite.
d) 100% tempered martensite.
800 A 1400 Eutectoid temperature 700 A 1200 A P 600 1000 500 800 AXTIS 400 A 600 300 M(start 50% 200 400 M+A...
Question 3 (Total 17 marks)- °C 1100 Atomic percent silver 40 50 10 20 30 60...
Question 3 (Total 17 marks)- °C 1100 Atomic percent silver 40 50 10 20 30 60 70 80 90 F 1084.5 1900 1000 L 900 1. + 961.93) 1700 1600 L+B 1400 91.2 800 a B 780° 7.9 71.9 700 1200 600 -- 1000 500 850 400 Cu 10 20 30 70 SO 90 Ag 40 SO 60 Weight percent silver Above is a phase diagram of Cu - Ag.- (a) Which phase(s) exist at 80%wt Ag, 20%wt Cu at...
Describe the heat treatment scheume that would provide the following property changes to 1080 steel. Refer...
Describe the heat treatment scheume that would provide the
following property changes to 1080 steel. Refer to TTT diagram
below.
a) 100% pearlite to a miture of 50% pearlite and 50%
martensite
b) Mixture of 75% pearlite and 25% martensite to 100%
martensite
800 1400 Eutectoid temperature 700 1200 600 1000 500 800 400 600 300 M(start) M(50%) M(90%) 50% 400 200 M + A 100 200 0 10-1 102 Time (s) 10 103 10 10
800 A 1400 Eutectoid temperature 700 A 1200 P 600 1000 500 A 800 400 A 600 300 M(start) 50% 400 200 M+ A M(50%) M(...
800 A 1400 Eutectoid temperature 700 A 1200 P 600 1000 500 A 800 400 A 600 300 M(start) 50% 400 200 M+ A M(50%) M(90%) 100 200 105 102 103 104 10-1 10 1 Time (s) Temperature (°C) Temperature (°F) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for the following heat treatment in the boxes below. Rapidly cool to 625°C, hold for 10s, rapidly cool to 450°C,...
Q2. Below is shown the isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy. List...
Q2. Below is shown the isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy. List the microconstituent(s) present for the heat treatment labeled on this diagram. It is not necessary to state the proportion(s) of the microconstituents. 6 options are given for each case. Pick the correct components and briefly explain your answer. (15 points) 900 1600 HIV 500 ZA+B (c) Temperature (°C) (6) A 6 Temperature (°F) Mstart) 300M(50%) ©) b) (90%) TTTT (b) (c) '(a) 103 (b)...
a) Based on time-temperature-transformation (TTT) diagram in Figure 2, briefly explain and label the time-temperature paths...
a) Based on time-temperature-transformation (TTT) diagram in Figure 2, briefly explain and label the time-temperature paths to produce the following microstructures (a), (b), (c). (d), 100% coarse pearlite 50% bainite 50% martensite 50% fine 100% martensite perlite and 50% martensite (10 marks) 800 727 700 Coarse pearlite 600 a + Fe,C Fine pearlite 500 400 300 200 100 1 see I min 1 hour I day 10 102 10) 10s Time, seconds Figure 2: Time-Temperature Transformation (TTT) diagram for eutectoid...
800 1400 Eutectoid temperature 700 A 1200 600 1000 500 B 800 400 A 600 300 M(start 50% 200 400 M+A M(50%) M(90%) 1...
800 1400 Eutectoid temperature 700 A 1200 600 1000 500 B 800 400 A 600 300 M(start 50% 200 400 M+A M(50%) M(90%) 100 200 C 10-1 102 103 1 10 104 105 Time (s) Temperature (C) Temperature (°F) Rapidly cool to 625°C, hold for 10 s, rapidly cool to 450°C, hold for 10s, then quench to room temperature. % Pearlite 100 % Bainite % Martensite 100 % Austenite 0 % Tempered Martensite 10 Enter answers as a whole number...
Instructions: For the following problems utilize the listed TTT Diagram from your Equation Bookle...
Instructions: For the following problems
utilize the listed TTT Diagram from your Equation Booklet. The
treatment begins with austenitizing temperature. RT = Room
Tempearture. Use 1 for 0% Transformed as an
answer!
--Given Values--
Time Temperature Transformation Diagram: = 1335 Steel
Heat Treatment: = Quench to 904 F hold for 12 seconds then quench
to 700F hold for 30 seconds then quench to RT then reheat to 1200 F
for 18 hrs
Problem 1 Part A:
For the diagram and...
PLEASE SHOW ALL STEPS AND LOGICAL REASONING. Figure 14: Carbon Steels: 1050 800 700 1 8...
PLEASE SHOW ALL STEPS AND LOGICAL REASONING.
Figure 14: Carbon Steels: 1050 800 700 1 8 23 26 32 36 42 48 52 I2 60 IO A F C 800 3oo 600 Mgo 20040O T DIAGRAM 0200 Estimated 62 IO io Io TIME SECONDS Time Temperature Transformation Diagram: 1050 Steel Heat Treatment: = Quench to 900 F hold for 3 seconds then quench to RT Problem 1 Part A For the diagram and heat treatment listed above give the percentage...
Round to the nearest 5%--no need for something like 43.2...that
would become 45. If there's none of that phase formed, enter a 0
(zero). If it looks close enough then it is
800 1400 Eutectoid temperature 700 1200 1000 500 800 400 600 300 M(start) 200 400 M + A M( 5096) M19096) 100 10 102 03 104105 Time (s) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for...
Make a copy of the isothermal transformation diagram for an
iron–carbon alloy of eutectoid composition provided overleaf and
then sketch and label time–temperature paths on this diagram to
produce the following microstructures:
a) 25% martensite and 75% austenite
b) 100% fine pearlite
c) 25% pearlite, 25% bainite, 45% martensite, and 5%
austenite.
d) 100% tempered martensite.
800 A 1400 Eutectoid temperature 700 A 1200 A P 600 1000 500 800 AXTIS 400 A 600 300 M(start 50% 200 400 M+A...
Question 3 (Total 17 marks)- °C 1100 Atomic percent silver 40 50 10 20 30 60 70 80 90 F 1084.5 1900 1000 L 900 1. + 961.93) 1700 1600 L+B 1400 91.2 800 a B 780° 7.9 71.9 700 1200 600 -- 1000 500 850 400 Cu 10 20 30 70 SO 90 Ag 40 SO 60 Weight percent silver Above is a phase diagram of Cu - Ag.- (a) Which phase(s) exist at 80%wt Ag, 20%wt Cu at...
Describe the heat treatment scheume that would provide the
following property changes to 1080 steel. Refer to TTT diagram
below.
a) 100% pearlite to a miture of 50% pearlite and 50%
martensite
b) Mixture of 75% pearlite and 25% martensite to 100%
martensite
800 1400 Eutectoid temperature 700 1200 600 1000 500 800 400 600 300 M(start) M(50%) M(90%) 50% 400 200 M + A 100 200 0 10-1 102 Time (s) 10 103 10 10
800 A 1400 Eutectoid temperature 700 A 1200 P 600 1000 500 A 800 400 A 600 300 M(start) 50% 400 200 M+ A M(50%) M(90%) 100 200 105 102 103 104 10-1 10 1 Time (s) Temperature (°C) Temperature (°F) Using the above TTT diagram for the eutectoid composition of steel, enter the relative amount of each microstructure formed for the following heat treatment in the boxes below. Rapidly cool to 625°C, hold for 10s, rapidly cool to 450°C,...
Q2. Below is shown the isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy. List the microconstituent(s) present for the heat treatment labeled on this diagram. It is not necessary to state the proportion(s) of the microconstituents. 6 options are given for each case. Pick the correct components and briefly explain your answer. (15 points) 900 1600 HIV 500 ZA+B (c) Temperature (°C) (6) A 6 Temperature (°F) Mstart) 300M(50%) ©) b) (90%) TTTT (b) (c) '(a) 103 (b)...
a) Based on time-temperature-transformation (TTT) diagram in Figure 2, briefly explain and label the time-temperature paths to produce the following microstructures (a), (b), (c). (d), 100% coarse pearlite 50% bainite 50% martensite 50% fine 100% martensite perlite and 50% martensite (10 marks) 800 727 700 Coarse pearlite 600 a + Fe,C Fine pearlite 500 400 300 200 100 1 see I min 1 hour I day 10 102 10) 10s Time, seconds Figure 2: Time-Temperature Transformation (TTT) diagram for eutectoid...
800 1400 Eutectoid temperature 700 A 1200 600 1000 500 B 800 400 A 600 300 M(start 50% 200 400 M+A M(50%) M(90%) 100 200 C 10-1 102 103 1 10 104 105 Time (s) Temperature (C) Temperature (°F) Rapidly cool to 625°C, hold for 10 s, rapidly cool to 450°C, hold for 10s, then quench to room temperature. % Pearlite 100 % Bainite % Martensite 100 % Austenite 0 % Tempered Martensite 10 Enter answers as a whole number...
Instructions: For the following problems
utilize the listed TTT Diagram from your Equation Booklet. The
treatment begins with austenitizing temperature. RT = Room
Tempearture. Use 1 for 0% Transformed as an
answer!
--Given Values--
Time Temperature Transformation Diagram: = 1335 Steel
Heat Treatment: = Quench to 904 F hold for 12 seconds then quench
to 700F hold for 30 seconds then quench to RT then reheat to 1200 F
for 18 hrs
Problem 1 Part A:
For the diagram and...
PLEASE SHOW ALL STEPS AND LOGICAL REASONING.
Figure 14: Carbon Steels: 1050 800 700 1 8 23 26 32 36 42 48 52 I2 60 IO A F C 800 3oo 600 Mgo 20040O T DIAGRAM 0200 Estimated 62 IO io Io TIME SECONDS Time Temperature Transformation Diagram: 1050 Steel Heat Treatment: = Quench to 900 F hold for 3 seconds then quench to RT Problem 1 Part A For the diagram and heat treatment listed above give the percentage...
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