Solidification Time in Metal Casting (GATE Previous Year Question with Solution), GATE Video Lecture - Mechanical Engineering

hello my dear Susan camasta and in this
video tutorial I am going to share with
you
solidification time yeah this is a this
is a topic of casting you must in order
to validate soil have landed this page
and you know in this topic I am going to
share with your principal pran CI b le
principle or rule right so this is about
it I mean this is a little principle and
this rule is a rule which is known as HB
u this is a little bit difficult
spelling because it's named cauri no
principal this principal is known as
coin of principle okay I hope you you
got it right now okay
so according to this principle what this
principle basically stayed I had to
share with you this principle states
that notification time is directly
proportional to the shape of the casting
casting volume divided by surface area
surface area is whole Square and we can
also write this TF is equal to K volume
by surface area whole square this claim
is known as solidification
solidification factor
he is known as solidification factor one
of the most important thing in this
video I am going to share with your
daddy is what is the unit of K you need
to know so how you can able to find out
it's unit that I am going to share with
you so where we will forget in exams so
this is solidification time right that
means in second if you are talking about
as I unit second clearly vit volume
meter cube surface area meter square
this oh sorry
okay so this one is on cancel a mobile
image so M Square and K is equal to
second per meter square so what is the
unit of K second per meter square
remember this int okay is second per
meter square
I hope you got AC units right so now
let's go for few shapes okay shape like
what type of shape we want it to cost
based on that we can find out so
deviation time
that means sorry vacation time is also
depend on the castings I mean the shape
of the cost with the ship which for we
say we wanted to cost okay so let's go
for it
first let's go for Q I know you can see
for cube cube volume my surface area
it's q AQ surface area six a square so
it's a by six right now second let's go
for Q you know what is cube let me share
with you first
probably you forward okay I know you
know this is a cube right who is it
whose land breadth and height three all
three of same that is known as cube now
let's go for its sphere say suppose that
this is here and this is diameter and
its hub pod is known as radiance so
volume by surface area so the volume of
the sphere is
4 by 3 PI R cube the surface area is 4
PI R square and after calculation what
we get 5 per year concern for forged
campuses R so R by 3 that means dy 6
this is for sphere I'm just trying to
share with you relation between volume
by surface area of different shapes
that's why one will get quotients in the
examination it will become quite easier
for you to solve the question right away
I mean to throw the question of also in
a very less time that's why I am doing
all these now let's go for cylinder ok
cylinder this one is for cylinder right
so in this cylinder it says diameter is
b and height of the cylinder is H so
volume by surface area volume printer
what is the volume of Trinder PI by 4
into d square into H okay now surface
area 2 pi by 4 into d square plus PI
into D into action this is the formula
now most important thing is we are
considering a cylinder whose height is
equal to diameter then the case what
would be the case for this if height is
equal to diameter then what would be the
value of surface volume by surface area
ratio that is interesting right so let's
put just put the value and get the
answer so PI by 4 d square into D
divided by 2 into PI by 4 into d square
PI d square because at equal to D so
what we got here PI by 4 into DQ it's to
be come on come on to PI d square plus 2
pi d square what we got here
therefore PI by 4 into d to be per cube
2 by 3 PI d square pi pi get canceled -
why - so the remaining would be DQ y 6
now most important thing we can notice
here
oh no it's not DQ basically you know t
DQ over 66 d square so dy 6 for vini
many now let me write first in figure dy
6 we are getting for cylinder if H is
equal to D and if swaram by surface is
automatically a ratio is d by 6 now look
at the volume my surface area ratio for
its sphere that means volume and surface
area ratio for its sphere and cylinder
both both for both it's same for good is
same so don't need to be confused in it
because it's very simple so this concept
actually based on this concern after in
gate examination they have asked you a
question for two marks just after
sharing me to another shape I'm going to
share with you that question how to
forgive this I will share with you how
to solve that quotient that's why you
can understand you have understood
hopefully going up sensible and good you
can also understand how to solve problem
based on these 10 people this is the
main motto of creating this video I hope
you are getting me right now
okay so let's go for that is you know
that last one last shape so I am talking
about in the last shape that is let's
say it's a slab okay what about slabs L
is length is there business here and
hydrosphere
so for slab volume by surface area ratio
lb edge to lb be h plus HL this would be
the ratio volume by surface area so this
is all about the concept of
socialization time and based on Cohen
offs principle okay now let's go for
that problem which has been asking gate
I hope you are interested to learn that
problem right so let's go for that
problem and solve it that would be quite
easier and quite helpful for you
now here is the problem so what we have
given in this problem let me just share
with you in sand casting to earth is a
sphere a sphere and cylinder of equal
volume so both the volumes are required
now what I have share with you right now
only volume if volumes are equal so what
is the formula for volume for the sphere
4 by 3 PI R cube what is the formula for
volume 4 cylinder PI by 4 d square H
okay so we should write here volume of
square 4 by 3 PI R cube volume of
cylinder 5 by 4 d square into H these
two formulas we require to solve the
problem there so I have share with you
all formulas now net separately task
from the sand molten metal under
identified entacle condition the height
and diameter of the cylinder required so
here the condition H is equal to D in
case of cylinder only therefore this
equation become PI by 4 bu okay now what
we have to find out the ratio of the
polarization time and is fair to that of
the cylinder that means we have to find
out TC t s by T C so dilution time of
the sphere to the cylinder then we have
to find out TS by TC okay now let's go
for that how to find out TS by TC Patty
is also important once you will come to
know how to find out when things become
easier for you right so we have to first
know the relation between spear diameter
as well as 3 in the diameter once will
come to no relation between sphere
diameter and critical diameter then
things become will be easier for us okay
then we can find out area and all that
that's right this row 4 is now so what
we got basically here here is that so
what we can write here that I am 4 PI D
by 2 whole to the power cube right so PI
by 3 and 4 by 3 PI D cube by 8 so this
will become 5 d qy
six by two u by six so and in this
quotient one state equal volume that
means PI by 4 DQ is equal to PI by six D
u now this D is for sphere these these
four cylinders that we have to denote so
we can write PI by 6 t s q is equal to
PI by 4 d C cubed 5 PI here cancel 2 to
the 4 2 theta R this one so DF is equal
to DX q is equal to 3 by 2 DC's Q
because I am sending this C to that side
so because of the rule of the noble
cause multi-mission we can write that
now DF is equal to 3 by 2 DP cube and we
can write 3 by 2 DZ cube this is cute
and because they see is cube that means
it will be P so what we got here till
now DF is equal to we have to finish
calculator right we need calculator 3 by
2 this one we have to find out cube root
so 1 point 1 1 point 1 4 5 ok this is or
1 point 1 4 4 7 because we have to take
up to 4 decimal value website we can get
very good term right answer right so D
is equal to 1 point 1 4 4 7 DC we got
the relation between diameter of the
sphere and diameter of the cylinder now
what we have to do we have to find out
we have to do what TS by TC now I have
already share with you the relation what
will happen in case of cylinder if H is
equal to DB and volume wise surface area
ratio is equal to D by 6 so in J's of D
by 6 DP y 6 okay now in case of a sphere
B by 6 so D F by 6 got it
how then now you can understand how that
the what I have shared with you in the
earlier is useful right now probably you
have landed in this place on this space
right now only then progressing what is
going on here but if you watch the video
from beginning then you can understand
what right now I am talking about it's
really important it's really
that concept what I am sharing with you
right now now what you got here d/f by 6
6 by DC 6 if you cancel what we have
right now DF by DC now what is the value
of DF DX ok now it is the value of D yes
you can okay okay now what we have to
just write square because because we
know that P has devotional to volume by
surface area is whole square so okay so
what we got here DF by DZ now now D is
what is the value of the X what is the
value of d SB is equal to 1 point 1 4 4
7 by DC and DC so this means we get
concern from where we chose this so 1
point 1 4 7 square what is answer 1
point 1 4 for 7 whole square is equal to
how much we are getting here 1 point 3 1
0 3 okay its ratio that's why I do it it
is use less confidence one point three
one he is the right answer so this is
how you can solve problem related to
modulation time based on going of
principal I hope you got some inside
information about this topic and you
have enjoy the video thanks for watching
see you soon in my next video bye bye