Only NPTEL lectures that is worth seeing.

Is there any other college also,where each and every student is studying material science from Prof.Rajesh Prasad NPTEL Lectures?

@hamza_ME_

3 жыл бұрын

SMME

@basavasangu310

3 жыл бұрын

Mvit

@Iam_abhishekkr

3 жыл бұрын

IITM

@shikhartaori895

3 жыл бұрын

mnnit

@063_ayushkumar5

3 жыл бұрын

Nit srinagar

Thank you. This is greatly appreciated.

What a great video sir! Thanks!

Great lecture as always

Best teacher I’ve ever seen, thank you so much!

Great video, Sir. Thanks.

Awesome video

Excellent explanation Sir 🙏🙏

Very educational video, thank you!

sir you are amazing

Best...

Thank you so much sir.

Dear sir, The lectures are nice and well explained by you. Can you please guide me for single crystal growth of incongrent melting materials?

well explained....... thanks

thanks a lot sir

Thank you for your beautiful explanation sir. Sir, On phase diagram the transformation of γ into Pearlite is shown exactly at 725 C. But, upon looking at T-T-T diagram we found out that the process actually begins and ends at 2 two different temperatures. Considering T-T-T diagram to be more practical, why didn't we represent this process as γ => (γ+Pearlite) => Pearlite, just like remaining processes with (γ+Pearlite) to be existing in some range of temperature (little though).

@rajeshprasadlectures

3 жыл бұрын

This is a question of thermodynamics vs. kinetics. Phase diagram represents thermodynamics. It tells us what will happen in equilibrium. To attain equilibrium we have to wait for a sufficiently long time. If one waits for a sufficiently long time then for all temperatures at or below the eutectoid temperature (725) one always gets pearlite. This is what is indicated in the phase diagram. However, any transformation takes time. This is not shown by the phase diagram. This is where TTT diagram comes in with the time axis. It thus represents kinetics. It then tells you how much time is required to transform austenite into pearlite. Of course, it also tells you that if you do not allow equilibrium to be attained you may get various metastable phases, like martensite, as well.

@lsrinivasan242

2 жыл бұрын

The diagram represented in phase diagram is slow equilibrium process. But we dont cool slowly for getting various steels. He have told this at 3:51

thankyou so much sir......

Sir we've already seen that for an alloy, transformation occur at a range of temp.(except for eutectic composition)...but in TTT diagram phase transformation start and finished at the same temp...HOW IS THIS POSSIBLE??

@kartikvyakaranam1710

3 жыл бұрын

but it doesn't. if you look at when it starts, it is at a higher temperature that when it stops.

mst content

Sir, at 18:40, we have gone past both the C curves and we ended up at slightly beside the nose of C curves, which is at higher temperature than room temperature (maybe around 300 C). We can't keep the material at same temperature throughout it's life. At some point, we will cool it to room temp. In such a case doesn't the grain structure of the material change further? Thank you for your time and patience.

@rajeshprasadlectures

3 жыл бұрын

All transformation lines (the C-curves and the horizontal Ms and Mf) of the TTT diagram represent transformation of austenite. They are thus applicable to untransformed austenite only. After the austenite has transformed to pearlite or bainite then crossing these lines leads to no further transformation.

@suriyaprakash8663

3 жыл бұрын

@@rajeshprasadlectures so once the transformation ends ,( the cooling curve crosses tf line) irrespective of fast or slow cooling rate we get same structure ??

sir when i am drawing graph of D=Do*exp(-Q/RT) it is not coming c curve. sir does my observation is true or not if it is true then why not it is following diffusion equation as this process involve diffusion

@introductiontomaterialsscience

5 жыл бұрын

The diffusion does control the rate of the transformation. But it is not the only factor. The other factor is the driving force. Thus although diffusion is high close to the eutectoid temperature but the driving force is very low. This combines to give an overall low rate. At very low temerature, driving force become high but the diffusion is very slow. Thus one again gets a low rate for overall transformation. A compromise is obtained at the intermediate temperature giving the maximum rate of transformation.

What if a cooling process intersects the start curve at two points and goes to martensite region making a straight line without touching the stop curve? Would it be martensite?

@introductiontomaterialsscience

3 жыл бұрын

Since austenite-to-pearlite transformation has started there will be some pearlite already formed and remain in the final product. But at the same time since the stop curve is not reached, there is still some untransformed austenite when the cooling curve intersects the start curve again. This untransformed austenite will transform to martensite. Thus the final product will be a mixture of pearlite and martensite.

Sir c atoms in austenite have the composition of 0.8%C so how in transformation they divide into 0.02 & 6.67 %C even they have only 0.8%C only

@introductiontomaterialsscience

6 жыл бұрын

Let us apply the lever rule. You have austenite of 0.8%C.This decomposes into ferrite of 0.02%C and cementite of 6.67%C. So if you apply the lever rule on a tie line at a temperature just below the eutectoid temperature you can find that fraction of cementite in pearlite is only (0.8-0.02/6.67-0.02)= 0.78/6.65=0.12.Thus only 12 % of the whole mass becomes cementite 0f 6.67%C. the available carbon for rest of 88% mass is much lower and becomes ferrite of 0.02%.

@introductiontomaterialsscience

5 жыл бұрын

Total amount of carbon is constant before and after the decomposition of austenite. In the undecomposed asutenite has a uniform concentration of 0.8%. But after decomposition the alpha regions reject carbon to have a lower concentration of 0.02%. The carbon rejected by alpha is accepted by cementite raising its concentration to 6.67. the overall average concentration of the decopmosed austenite will still be 0.8 as in the case of undecomposed austenite.

Sir, what happens once the coarse Pearlite is formed after Annealing i.e. above the nose temperature( say 550℃) ? What happens when we cool this pearlite to room temperature ? Why doesn't it transforms to bainite or martensite on further cooling ?? Please, answer.

@introductiontomaterialsscience

4 жыл бұрын

Austenite is an unstable phase. Thus it wants to transform to stable phases alpha and cementite. Pearlite is already a mixture of these stable phases. Therefore once it forms it will not transform to any other phase.

@Vickysahu0092

4 жыл бұрын

@@introductiontomaterialsscience Thank you very much sir.

Sir at 13:34 you said if we quench stable austenite to temp below 725°c and keep it there it will follow a straight isothermal line,but sire its an alloy so how can it go through phase transformation under constant temperature?the temperature must change during its phase change,no?

@introductiontomaterialsscience

3 жыл бұрын

If a phase is unstable at a given temperature, it can in principle, transform isothermally to the stable phase. Under normal uncontrolled conditions, the temperature may change. However, one can try to perform experiments under controlled conditions of constant temperature. This is usually attained by using very thin samples in a constant temperature bath.

@ankushsaha007

3 жыл бұрын

@@introductiontomaterialsscience thank u sir🙏

thnks prof

Sir formula for annealing amorphous metal time calculation

when log t tends to very large then t tends to minimum then how you said the transformation will take more time

@piyush4338

4 жыл бұрын

log(t) tends to large means time taken is also large. You answered your question at the starting itself ;)

Sir, among normalizing and annealing which will have greater transformation temp?? Sir, from your diagram it looks that normalizing will have greater transformation temp??But you mention that annealing will have greater transformation temp??

@introductiontomaterialsscience

4 жыл бұрын

Since normalising involves a faster cooling rate, the normalising curve cuts the start C-curve at a lower temperature than the annealing curve.

Sir. gamma plus pearlite.. what you have written.. where is it present in the phase diagram ??

@rajeshprasad101

4 жыл бұрын

It will not be present in the phase diagram. But in the TTT diagram it will be present for those times where some gamma has transformed to pearlite and some gamma is yet to transform.Phase diagram shows only the transformed equilibrium phases. And nowhere in the phase diagram gamma is in equilibrium with pearlite.

At 3Am, a night before viva

very bad explanation, you just cite what is told in books, with no additional examples and intuition and the interpretation , WE NEED INTERPRETATION AND APPLICATIONS of graphs and NOT EXPLANATION of a graph , because explanation of a graph sound like shit ; zero intuition, you say if we make a curve like this, but wtf, you cannot create a curve!, you have to tell what that curve is telling you at certain situations. Because at a test , all what matters is that if you can APPLY the graph

@gauravmishra1508

3 жыл бұрын

the course is designed for Undergrad students.. so chill

@jitendrajangir39

3 жыл бұрын

The Course is not for material selection engineers/professionals, for that there are standards like ASTM, ASME etc. You need that and for understanding those standards, you need these kind of background.

Pdf drive.google.com/file/d/1ES4srV6hGV_Faljg2CWYJrikBNqT0ClD/view?usp=drivesdk