Hello Sir. Hope you are fine. I have a doubt in this lecture. At the time when temperature of the outside surface will be just above Ms, the temperature of inside will be quite higher (somewhere around nose or above nose). If the quenching of specimen is interrupted here and temp is kept constant, the curve of outside will definitely follow the non-transforming region, but the curve of inside will cross the transformation region and hence may transform into pearlite. Sir it's a request to clear this doubt because I am preparing for my exams.
In Nacl, corners as well as face centres are occupied by the same ion, say Cl- ion. So it is FCC. In CsCl corners are occupied by Cl- but body centres by Cs+. So it is not BCC.
is it the potential energy of the system that is decreasing? Because according to the law of conservation of mechanical energy if an isolated system is subject only to conservative forces, then the mechanical energy is constant.
High or low temperature for a material is relative to its melting point. For snow even a temperature of -20 C is rather close to its meting point and so it feels hot!
1. Advantage of TTT diagram over Iron-Iron carbon diagram - Has time axis 2. Disadvantage of TTT diagram over Iron carbon diagram - TTT diagram for only single composition
Why coarse grain in annealing and fine grain in normalising 19:20 -Annealing occurs at higher range of temperature and Tempering occurs at lower range of temperature. - At higher temperature I. E. Annealing - growth rate dominate nucleation, so coarse pearlite are formed - At lower temperature I. E. Normalising - nucleation dominate growth rate, so more nucleation occurs than growth, so fine pearlite formed.
Bro both HT process starts above A1 even in Hyper eutectoid steel it starts cooling above Acm temperature for normalising but for annealing is starts from A1 ...But the difference is cooling rate which determines the final grain size😅
Why C curve? 11:12 to 12:57 -Initially (near TE) Driving force is less but atomic mobilisation is high -Finally (near T=0C) Driving force is high but atomic mobilisation is very less
how can liquid exist below its freezing temperature i mean by that time everything would have been converted solid itself right how will nucleation occur below Tm as what i understood was freezing will occur till complete liquid changes to solid and until then T will remain constant at Tm
What you are stating is the condition of equilibrium solidification where the entire solidification can happen at the melting point. But this requires an extremely slow cooling. If you cool rapidly it is possible to retain liquid, for some time, in metastable state below the meting point before it starts freezing.
@24:72 this fraction is valid only at that particular temperature right .i.e, the lever should be made corresponding to the equilibrium temperature in question right?
Спасибо. Очень наглядно и понятно. Один только вопрос - если в роли матрицы будет выступать композитный полимер, выплавленный из LFT-G -гранул (композитные гранулы с длинными волокнами углерода/базальта/стекла), а в него будет добавляться керамический нано-аддитив по типу нитридных или карбидных нановискеров, то это будет гибридный нанокомпозит с несколькими фазами (нано- и макро-фазы)?? Или же это будет полимерно-волоконно-керамический нанокомпозит, как пример GFRP-LFT-G-nSi3N4 (матрица - сплавление в лист гранул LFT-G TPU/RTPU + LGF, а добавляемый наноаддитив - нановискеры нитрида кремния)?
This is good question. I think we can say that this is an experimental observation. The atomic mechanisms of plastic deformation are consistent with this fact. So in slip one part of the crystal slides over the other part and so the volume remains constant. In twinning, the crystal reorients itself, again keeping the volume constant.
I have made two mistakes in this video : 1. I have considered Trigonal and Rhombohedral as synonyms for the same crystal system. However, as per the International Tables of Crystallography Vol A it.iucr.org/A/ the two terms are NOT synonymous. Crystal System should only be called Trigonal and not Rhombohedral. 2. I have shown only one Bravais lattice in the trigonal crystal system. But in reality there are two Bravais lattices in the trigonal system: hexagonal prinmitive (hP) and hexagonal rhombohedral (hR). And so there is no trigonal P Bravais lattice as shown in the table in this video. I plan to make a video to clarify this. Wil share the link once I do it. You can check en.wikipedia.org/wiki/Crystal_system.
