Conduction Through Plane Wall Numerical Approach HMT Tutorial- 11, Heat and Mass Transfer, GATE Video Lecture | Study Topper Handwritten Notes & Videos for GATE ME - Mechanical Engineering | Best Video for Mechanical Engineering

hello my dear this is welcome arson and
today I am here to share with you one of
the another video tutorial and in this
tutorial I will share with you about
conduction just a second
yeah conduction of conduction and here
say you know it's a heat transfer right
so conduction through a slab through a
slab well I have been relocated my you
know my living place that's why it is it
has taken little bit time to share
videos related to this heat transfer
well so don't worry I'm going to share
with you a lot of more videos every day
that's why we enjoy okay so so what will
happen how to how to solve questions
related to this that is conduction
through a slab and this is most
important for you also
so suppose that we have a slab like this
and the thickness of the slab is let's
say B and it's a t1 temperature is here
and it's t2 right let's say like this
only and heat is transferring through
this wall and we can say also slab now
how would you calculate how do you
calculate heat transfer for this so we
know according to four years law four
years
low that is Q is equal to minus K ay DT
of 1 DX already I have shared with you
what is for your slow and why that what
is the reason behind this minus sign
okay but here we are taking few
assumptions we are taking few
assumptions those assumptions are most
important for solving problem related to
this maybe you will get question related
to days I mean what type of a junction
is to be assumed for solving conduction
through a slab so first is steady state
condition
steady of state is steady second is
one-dimensional big transfer
one-dimensional heat transfer HD and
third exemption is uniform uniform key
value value of K is uniform throughout
the material or throughout the slab
that's what we can calculate where we
can implement this equation to find out
heat transfer rate through this slab
okay so basically where we can exactly
implement this equation and I am sure
with you which we know suppose that we
are having a home and in a desert right
so what what would we do to find out the
heat transfer rate in death in the
desert
from room to from outside to the room or
maybe from room to room to the outside
so most of our time what will happen is
desert outside temperature is more as
compared to inside temperature of the
room so so we can do that we can we can
find out heat transfer rate in that case
also but in in the atrium condition
suppose that right now I am Delhi
so in Delhi also if we see go outside
now say temperature is more as compared
to inside temperature at least in the in
the in the in the evening time I mean I
will you will see one o'clock you can
see or we can we can just you know we
can just see or just fill that situation
well winter season is coming so we the
situation can be reversed right so let
us go for it first to come to the topic
and so here what get doing it Jaitley
why why working on this so we know that
Q is equal to minus ka DT upon X right
so Q and Q is heat transfer right so
suppose the Hatter and DT change in
temperature and what is DX basically
basic DX basically make of it okay so
here DX is equal to
TX is equal to B so minus K ay DT means
a t means T 2 minus T 1 T 2 minus T 1
and BX means B all right so minus k a k
if i will turn if we add pink multi
multiple this like- signature will case
equation what will happen
k a k 1 minus a 2 by B right so we can
we can write like this D by K now so
this is the ultimate equation for heat
transfer rate through this lab hope you
can see this right now okay so you might
from maybe next lecture I will share I
will use another pen that I can easily
understand what I am writing I hope you
can understand now also so this is the
thing now what is it this B by K a is
known as resistance that is also denoted
by rth but here our TS resistance means
thermal resistance okay so this is like
this only right so basically but this
equation is applicable only case of only
in the case of conduction so this is
also important that if you are studying
conduction then you need to understand
this right so now next word next word
here we face another problem in the
complete examination what we see in the
complete domination they don't give the
value of a that means area of this slab
they're not going to go ready
so we face problem in that case you know
why because we are not family ER with
solving numerical problem so I when I
talk to many of my many of my friends
regarding gate examination they told me
that yes if you want to solve if you
want to get good marks in gate or any
other come
examination where numerical problems has
been asked too much in that case you
need to solve problems and previous air
questions is much better for getting
good marks in the complete examination
if in the complete submission they are
asking numerical problems right so here
they are not going to provide to the
value of a but they will just do a trick
what is that they will provide you that
heat tasks for weight per unit area that
means what per unit area that means
already already they are transferring
this area to this side if this is here
we can write the equation like this only
T 1 minus T 2 into H B by K Q so if you
task for this a to this side what will
happen Q by a is equal to P 1 minus T 2
B by K so this is this is a this is the
trick man this is the trick this trick
you must utilized when you are solving a
coercion delete to this or all right so
I hope you understand something from
this video right so this is all about
conduction through a slab now one of the
question has been asked in cage already
they have asked this question but in
Kate they have they have asked this
quotient and if I am NOT wrong this
question has been asked in gate 2006
okay so what about the position let me
just share with you hope so it's it's 96
not 2006 so for a given heat flow and
for the same I am just reading the
question I have a question in front of
my desk so this is the question for a
given heat flow and for the same
thickness the temperature drop across
the material will be maximum for
temperature drop across the material
will be maximum for right so upper first
option second steal third class rule
and forth refractories break alright so
this is awesome and they are asking of
if I will say in one line then
temperature drop maximum temperature
drop now we can only guess the answer if
you know or if we know the value of K
for these materials but a lien on
material for these terms I must say so K
means thermal conductivity I hope you
know this right
so we have to find out first its value K
for this material light so even if we
know the value of K for this material we
can easily solve the question that is
most important okay so you know cover
steel these are metals and metals a very
good conductor of it that means
temperature drop will not be maximum for
these materials at least so a B option
we can sell as it is we don't need to
see these option anymore now you are
having to add or some glass wool and
refractory B right
so if if any how we can come to know how
does these two which one is having less
thermal conductivity then we can easily
mark the right answer so the less
thermal conductivity among them in
between these two metals is colossal
glass will is having minimum number of
thermal conductivity minimum thermal
conductivity that's why temperature drop
will be maximum for the case of class
zone so I hope you understand the
concept right so thanks for watching for
this tour chiral conduction through a
slab I'll share with you another another
topic and that would be much more
interesting and much more familiar
coming much more you know much more
important for the computation well
thanks for watching see you soon in my
next video bye bye