Describe suitable compositions and heat treatments to manufacture the following
microstructures:
(i) 20wt% pearlite dispersed in a matrix of ferrite
(ii) 5wt% graphite dispersed in a matrix of pearlite.
Solution
(i) 20wt% pearlite dispersed in a matrix of ferrite:
For pearlite formation in ferrite we need to use hypoeutectoid steels.
Full Annealing:- It is a heat treatment used to produce soft, coarse pearlite in steels by
austenitizing & then furnace cooling. Cooling rates are slow to allow formation of pearlite in
matrix of ferrite. It is a process of slowly increasing the temperature nearly to 50 ºC above the
Austenitic temperature line A3 or line ACM in the case of Hypoeutectoid steels.
The temperature is kept constant for some time so that all the material to transform into
Austenite. I Then the material is slowly cooled at the rate of about 20 ºC/hr in a furnace to about
50 ºC into the Ferrite range. Cooling can also be done at room temperature air using natural
convection. The grain structure has coarse Pearlite with excess ferrite. The steel becomes soft
and ductile.
Normalizing:- It uses austenitizing and then air cooling which produces a fine pearlite structure.
The process involves increasing the temperature to nearly 60 º C above line A3 or line ACM in
the Austenite range. Its temperature is kept constant for some time to complete Austenite
conversion fully. It is then removed from the furnace and cooled at room temperature under
natural convection (air cooling). As a result we get grain structure of fine Pearlite with excess of
Ferrite. End material is much softer due to presence of pearlite.
(ii) 5wt% graphite dispersed in a matrix of pearlite:
For graphite to be dispersed in pearlite matrix we need to perform annealing (as mentioned
above) for Grey cast iron material. The cooling rate must be kept moderate to form pearlite
matrix..
Done Homework 4 e beaker in the illustration below contains two solut.pdf
1. Done Homework 4 e beaker in the illustration below contains two solutions of salt with different
concentrations (measured by molarity Mo The two solutors are separated by membranethatus
ermeable to both salt and water. A 0.2 M 0 0.9 M B salt salt Which of the following will occur in
this container? Net diffusion of water from B to A and of salt from A to B. Net diffusion of water
from A to B and of salt from B to A. water across the membrane, but not of salt.
Solution
Option III
Net diffusion of water will occur across the membrane.
The salt solution in A is hypotonic with molarity of 0.2 M and that of B is hyper tonic with
molarity 0.9 M.
So the water always travels from high concentration gradient to low concentration gradient
(osmosis). So water will travel from A to B and no movement of salt takes place.
