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Metallurgy Engineering Interview Questions and Answers

Metallurgy Engineering Interview Questions and Answers

Question - 1 : - What Is Carbide?

Answer - 1 : -

The epsilon carbide is transition carbide of between Fe2C and Fe3C composition, with hexagonal close packing microstructure, which forms over a range of 250400 ºC temperature during lowerbainite transformation. This transformation happens during tempering heat treatment of quenched steels, or during slow cooling in that temperature range.

It is fine dispersed carbide in a ferrite needlelike matrix. Lower bainite has lower tensile strength (about 510%), but usually same hardness as martensite structures, but with a higher toughness.

Question - 2 : - What Are The Properties And Application Of Manganese Steel?

Answer - 2 : -

Steel is an alloy of iron and carbon, consisting of iron phase and iron carbides. Crude steel produced from iron contains an undesirable amount of oxygen and some sulphur. Manganese plays a key role because of two important properties: its ability to combine with sulphur and its powerful deoxidation capacity. When there is insufficient manganese, the sulphur combines with iron to form a low melting point sulphide, which melts at hot rolling temperatures, causing a surface cracking phenomenon known as “hot shortness”. Desulphurization processes reduce the need for manganese in this respect. Some 30 of the manganese used today is still used for its properties as a sulphide former and deoxidant.

The other 70 of the manganese is used purely as an alloying element. These alloying uses depend on the desired properties of the steel being made. Steel, as has been noted, contains iron and carbon. At room temperature, iron crystallizes into a bodycentered cubic structure named alpha iron (ferrite). At a high temperature (above 910 degrees C), the structure is transformed into a facecentered cubic form, which is called a gamma iron (austenite). When the steel is cooled down slowly, the carbon, soluble in austenite, precipitates as iron carbide called cementite, the austenite transforms to ferrite and they precipitate together in a characteristic lamellar structure known as pearlite.

Question - 3 : - What Is The Metallurgical Explanation Of Effects Of Chromium, Nickel, Molybdenum, And Carbon In Stainless Steels?

Answer - 3 : -

Alloying elements in stainless steels can be divided into 2 main categories namely austenite and ferrite stabilizers. Austenite stabilizers must be present in austenitic as well as martensitic STSs (austenite at annealing temperature is the precursor phase for these two categories although for the latter group, it transforms to martensite before cooling down to room temperature). In order to stabilize austenite at annealing temperature, the ratio of austenite to ferrite stabilizers must be high.

The strongest austenite stabilizers are N, C, Ni, Mn, and Cu whereas elements like Cr, Si, Nb, Ti, and Mo are the most important ferrite stabilizers. Niequivalent to Crequivalent ratio is an effective way to quantify the austenite formation tendency of STSs. There are different expressions for Creq and Nieq, one of which looks like this:

Nieq=Ni+0.3Mn+22C+14.2N+Cu, and Creq=Cr+1.37Mo+1.5Si+2Nb+3Ti).

Question - 4 : - What Is The Difference Between Diagrams Of It And Ct In Heat Treatment Of Steels?

Answer - 4 : -

Isothermal Transformation (IT) and Continuous Transformation (CT) diagrams are diagrams used to investigate kinetic aspect of phase transformations and are of extensive use in steels heat treatment. In these diagrams generally called TimeTemperatureTransformation (TTT), the abscissa is time in logarithmic scale and ordinate is temperature. The C shaped curves indicate the onset and the end of diffusion (civilian) transformations e.g. pearlite or bainite formation or precipitation of carbides.

IT diagram shows what happens when steel is held at a constant temperature for a prolonged period. The development of the microstructure with time can be followed by holding small specimens in a lead or salt bath and quenching them one at a time after increasing holding times and measuring the amount of phases formed in the microstructure with the aid of a microscope. An alternative method involves using a single specimen and a dilatometer, which records the elongation of the specimen as a function of time. The basis for the dilatometer method is that the micro constituents undergo different volumetric changes and thus, the onset of transformations could be detected.

Question - 5 : - How Can You Separate Gold From Mercury?

Answer - 5 : -

  • Gold (Au) and mercury (Hg) are two separate metals. Therefore, from pure mercury you cannot extract Gold.
  • Amalgamation is better suited to ores in which the gold occurs in the free and cores state. Fine particles of Gold are better treated by Cyanidation.
  • Gold dissolves in aqua regia. Aqua Regia is a mixture of hydrochloric acid and nitric acid. It can dissolve gold, which single acids alone cannot do. Here we see that each of the acids separately has no effect on the gold but a mixture of the two dramatically reacts with the gold.
To three samples of gold are added

  • concentrated hydrochloric acid
  • concentrated hydrochloric and concentrated nitric acids and
  • concentrated nitric acid. The gold reacts only with the concentrated acid mixture which is referred to as aqua regia (royal water)".

Question - 6 : - What Does Allotropic Mean What Does Iron And Steel Have To Do With It?

Answer - 6 : -

Iron is a metal with polymorphism structure. Each structure stable in the range of temperature, for example deltairon with bcc structure in the range of 15381394C changes to gamma Iron with fcc in the range of 9121394C and gamma iron to AlfaIron etc. For hardening of the metal, this quality of iron is exploited.

Question - 7 : - Why We Are Doing Post Heating In Alloy Steel Welding?

Answer - 7 : -

The preheating is application of heat to a base metal immediately before welding. Preheating helps reduce hardness in the metal.

In addition, the application of heat to the weld immediately after welding is postheating .The Post heating helps reduce stress in the weld metal.

Question - 8 : - When We Do Cardonitriding Or Casehardening Operation For A Plain Carbon Steel The Case Depth, Hardness & Microstructure Are Not Alike Why And What Are The Basic Thing, Which Makes Such A Difference?

Answer - 8 : -

Usually that will depend on the time and temperature, as well as chemical composition of the furnace atmosphere you are using.

If these variables are not kept well controlled, they may cause very different results. Please note that time and temperature are correlated, the more temperature, the less time and lesser diffusion control.

Question - 9 : - What Is Or What Characterize The Thermodynamic State Of Metal (structure Of Metal)?

Answer - 9 : -

In the solid state, metals have a crystalline structure made of metal atoms, which are drawn together by low force vanderwaals interactions. The electrons form a cloud around the atom structure and migrate from one point to the other constantly.

The structured state of the atoms allows for low entropy in this state. Depending on the metal, several different structures may form, and one metal may have more than one structure at different temperatures, since its entropy depends on atom vibration as well, which is connected to the internal energy, reflected as temperature.

Crystalline structures have, usually, a straight correlation of stress in the elastic region. When traction stress is applied, the atoms are forced away from each other, up to a point where it, theoretically, should loose coherence by breaking all interactions at once and forming new surfaces.

This energy level is so high that other mechanisms of energy dissipation happen first, usually connected to defects and dislocations in the crystalline structure. These mechanisms allow for the inducing of surface cracking, or plastic deformation.

Question - 10 : - What Are The Compositions Of Brass, How Can This Metal Be Heat-treated, What Is The Melting Point Of This Metal?

Answer - 10 : -

Brass is an alloy of copper and zinc, with varying degrees of mixing. It is a substitution alloy, which means the copper and zinc elements substitute each other in microstructure matrix positions. This behavior translates into a metal which has a lower melting point in between that of its elements (Cu=1084, 83ºC, Zn=419, 58ºC) in pure state. Usually, the possibility of heat treatment will depend on what are you trying to achieve. For the substitution range of compositions, for example, you cannot obtain hardening from heattreating.

NCERT Solutions


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