Radial Conduction Through Composite Cylinders HMT Tutorial- 17, Heat and Mass Transfer, Mechanical Engineering, GATE Video Lecture | Study Topper Handwritten Notes & Videos for GATE ME - Mechanical Engineering | Best Video for Mechanical Engineering

all right my dear this is Joelle and
today in this video I am going to share
with you
radial conduction heat transfer HT means
heat transfer through composite slab
alright so in my previous couple of
videos I have she will share with you
conduction of heat in radial hollow
cylinder as well as conduction of heat
in slab composite slab and convection
also but today in this video I am going
to share with you this one so if we are
talking about
composites cylinder then this would be
like this right this one I am will have
this is a hollow cylinder inside even is
hollow right and we're having this with
composite this means this is let me use
another pane that's why you can
understand by the way right so this is
our one suppose that this is r1 this is
r2
this one is r3 okay so this is the
composition der and suppose that length
of the cylinder is L for this one t one
and for this one is K 2 I hope you know
what is K 1 K 1 K 2 thermal conductivity
right so what exactly we are assuming
here at the mission because without
assumptions we cannot assumptions
assumptions only one assumption we are
taking a two young son as a mission I
must say one is steady state condition
and read one dimensional a radial heat
transfer that's it
so we are having we are assuming two
things steady state and one-dimensional
heat transfer okay now first of all as I
hope I have already shared with you he
transferred radially conduction through
a hollow cylinder and here we are having
composite cylinder missed most important
thing is that we have to know how to
tour the thermal circuit diagram once
you we will come to know how to draw
thermal circuit diagram then solving
such type of problem is just like that
you don't need to put or do in order to
had it and one good name is my dear yes
after this will solve a gate problem
related to this topic only that's why
you will have a numerical approach to
with unical approach okay so let's start
this thermal circuit this is the thermal
circuit here right and suppose that we
are having temperature T 1 here T 2 here
and we are having two resistances over
here so as you know we have also seen in
the hollow cylinder I mean corrosion
resistance is word I am sorry L LM are
towards r1 2 by K 1 okay we have already
known come to know about it this is
registered side and for this case T 2 R
3 R 5 r 2 this is natural log 2 pi K 12
K 2 for this one and K 1 for this one
right so we are having two resistances
because of two materials we have Peter
to put it here it could be three four
before by two at least two because this
is a case of composite slab that's why
there should be at least two right
I hope you understand this okay no if we
wanted to find out he does perfect for
this for this for this diagram then what
would be the next step that is important
right
as you know Q is equal to Q is equal to
K Q is equal to delta t by r th this is
the simplest one right you know that
simplest one now if Q is equal to delta
T V by RT H then what we'll have to do
to find out it okay so here is it delta
t delta p RTS means natural log r1 by r2
okay 2 pi k 1 l plus natural log r3 by
r2 2 pi well here what is the meaning of
Delta T's that is most important once we
will come to know the meaning of Delta
T's then things become much easier Delta
P means t1 minus t2 so this is all about
this is all about it that what I mean
how is it possible for us to find out it
does pervade Q for the case of composite
slab okay now now now most important
things so we are having a formula to
finding out it does for rate for
composite slab we can use this formula
we can use this concept to solve
numerical problems doesn't matter where
it would be air score it has been asked
already I hope you are getting me right
so let's go for the numerical problem
all right Mike so today we caught this
question and this question has been
asked in gate 2000 through if I'm not
wrong now we're having options here 1 2
3
right so this is a question of you know
selling that it's this question of
cylinder but the tempo it's level right
so let's do it
first of all diagram is most important
at least if you are beginner if you are
not a beginner then if you can easily
draw any diagram you don't need to have
fun you can easily solve the question
but beginner then if you draw a diagram
then it becomes quite easier for you to
understand the situation and you can
easily solve the quotient without making
any mistake okay we have already drawn
the diagram now what we have is an
exactly we have given two diameters
internal diameter as or external
diameter and of course we are you having
will have also given asbestos thickness
so we have giving this one is three
centimeter insulated three centimeter
asbestos thickness and the internal
diameter is how much even two centimeter
two centimeter and the external diameter
we are given five centimeter this one is
5 centimeter okay this one is 2
centimeter now remember my dear before
solving any numerical problem at least
forget you have to put your
concentration unit on unit because unit
is most important for get remember that
okay so you should not make mistake even
in that case right if you make mistake
in that case you are going to solve the
question but yes you will not mark the
if you're not wrong mark the right
answer that is the main problem ok so
here we got all those things what we
required now what we have given were to
find out that is heat transfer rate per
unit length that means we have to find
out qil as you know we are having this
so we can draw the network title quite
easily because we have just understand
this so t1 t2 and t3 okay now here are
th 1 and R th 2 so our tears 102 B
natural rock R 2 by R 1
so let's say D 2 by D 1 so 5 by 2
don't need to see the what see them unit
because we are having division so
doesn't matter it will not matter and 2
pi k k is how much steel 19 and L we
don't have that forever now natural log
now this is important here probably most
probably you were make going to make
mistake okay
most probably you are going to make
mistake so what if what about that
mistake that is you are going to make
mistake here you're going to put three
but it's not three it's total we have to
add right to add it so three plus five
plus three so six and five 11 okay so
natural log 11 y r2 r3 by r3 by r2 our
domains how much files to PI K 0.2 right
so he calls two resistances here now
what we required nothing
we have even already given him the
defense outer inner most and outer most
surfaces difference between innermost
and ultimate surface temperature is must
be 600 degree centigrade so now have
already given T 1 minus T 3 is equal to
how much 600 T is scintillators so you
don't need to see around that so Q is
equal to 600 degree centigrade 5 by 2 2
pi 19 into L plus natural log 11 by 5 2
pi into 0.2 into L will have to we can
send this L to this side
after taking common right okay right so
Q by L is equal to 600 natural log 5 by
2 to PI into 1911 by 5 2 pi into 0.2 as
simple as that so this is what we need
to find out we called all values here
now what we have to do we have to use
calculator well I have already
calculated this problem so I know the
answer
so after calculation we are going to get
values about 9 44.7 to what because this
is it this is here Q what L meter per
meter right so unit is matching as we
have given in off sense right so what
would the awesome 944 0.72 so yeah C is
the right answer so this is how we can
solve problem if you get in any
summation may be it may be complete exam
in either complete examination may be
university examination then you can
easily solve the problem I hope you
understand the concept behind this topic
hence watching see you soon in my next
video bye bye