Mechanical Engineering > Topper Handwritten Notes & Videos for GATE ME > Conduction and Convection Through Composite Cylinder HMT Tutorial- 18, Heat and Mass Transfer, Mechanical Engineering, GATE
Conduction and Convection Through Composite Cylinder HMT Tutorial- 18, Heat and Mass Transfer, Mechanical Engineering, GATE Video Lecture - Topper Handwritten Notes & Videos for GATE ME
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FAQs on Conduction and Convection Through Composite Cylinder HMT Tutorial- 18, Heat and Mass Transfer, Mechanical Engineering, GATE Video Lecture - Topper Handwritten Notes & Videos for GATE ME
| 1. What is conduction and convection? |  |
Ans. Conduction is the transfer of heat through a solid material or between two solids in direct contact, due to the collision of particles. Convection, on the other hand, is the transfer of heat through a fluid medium (liquid or gas) by the actual movement of the fluid particles.
| 2. How does heat transfer occur in a composite cylinder? | |
Ans. Heat transfer through a composite cylinder occurs through both conduction and convection. Conduction takes place between the layers of the cylinder, while convection occurs between the outer surface of the cylinder and the surrounding medium (fluid).
| 3. What factors affect the rate of heat transfer in a composite cylinder? | |
Ans. The rate of heat transfer in a composite cylinder is influenced by several factors, including the thermal conductivity of the cylinder material, the temperature difference between the inner and outer surfaces, the thickness of each layer, and the convective heat transfer coefficient between the outer surface and the surrounding fluid.
| 4. How can the rate of heat transfer be calculated in a composite cylinder? | |
Ans. The rate of heat transfer in a composite cylinder can be calculated by considering the conductive heat transfer through each layer of the cylinder using the thermal conductivity and thickness of each layer. Additionally, the convective heat transfer coefficient and the temperature difference between the outer surface and the surrounding fluid need to be taken into account.
| 5. What are the practical applications of analyzing conduction and convection in composite cylinders? | |
Ans. Understanding conduction and convection in composite cylinders is crucial in various engineering applications. It is commonly applied in the design and analysis of heat exchangers, such as shell-and-tube heat exchangers or double-pipe heat exchangers. This knowledge also finds application in the design of thermal insulation systems, where controlling heat transfer through composite cylinders is essential for energy efficiency.
Text Transcript from Video
okay my dear so in this video I am going
to share with you conduction convection
both the processes going on heat
transfer process through a composites in
them alright so as we have drawn here on
in front of you a composite cylinder
right so internal diameter time I mean I
have put it radius into radius of the is
one you can see it's inter it is is r1
and extend radius is R two and here the
most out total radius is R 3 okay now
what we are going to do is saying that
gas is flowing inside the pipe this step
I suppose densify follow pipe and gas is
following if gas is flowing and what
type of gas we are talking about it hot
gases suppose the temperature of hot
gases is T Z okay and here temperature
is let me use different n t1 here
temperature is t2 in the surface and
here temperature is heating and the
outside temperature let us say this is
surrounding at D infinity let's say this
this is T infinity and here from this
whole scenario you can assume it as T Z
as you or it is obvious as T Z greater
than T infinity okay because hot gases
is flowing inside it and this one is T
infinity so you can say that t ze is
greater than T infinity of you
understanding right now most important
thing most important thing here we have
to consider convection as I have already
shared with you in case of convection we
follow one law that is H a TS minus TF
and in case of conduction will follow
this one T D by DX this law is known as
for years law and this one is known as
Newton's law of cooling okay these two
laws are most important and even we we
usually use these two laws only in case
of conduction as well as convection okay
now now what is most important we have
to throw the thermal circuit diagram and
we also need to know what our
assumptions for this case we are
considering for finding out the total
heat transfer okay rate also because
today heat transferred per in a second
we are going to find out and our unit
would be in what so we must say heat
transfer rate so what are the
assumptions we are taking assumption 1
and assumption two first reservation
that must be in steady state second one
dimensional one dimensional a radial
heat transfer radial heat transfer
let me share with you in general we
cannot get these two if we are taking in
a real situation it is really like
impossible to get these two situation
okay one time I saw a radial heat
transfer is almost impossible next being
possible you can say but yes we're
considering without these two
consideration we cannot find out total
heat transfer rate right away so let's
go for it
first of all we have to draw the thermal
thermal circuit tighter
okay now as you know you're having a
convection process to here so we have to
turn convection conduction conduction
again convection okay suppose that this
is T Z and this is infinity therefore
this one with T 1 this one is T 2 this
one is tickling let me share with you
this diagram with this one because I
have drawn this diagram from here okay
so we have is check it with TZ is
easements gas hot gases right and then
this surface T 1 then this one T 2 and
then this one T 3 and then after
technique what what is directly
connected with temperature 3 that is
environment or even surrounding so that
is known as T infinity hope you are
getting right now okay okay now
understand it okay so this here and here
we are having what type of process is
going on 1 to 2 and 2 to 3 1 to 2 and 2
to 3 conduction process is going on from
here to here TZ 2 T 1 and T 3 to
infinity what kind of process is going
on convection process is going on so
here our th1 are th2 suppose that this
is our ds3 M R T as for so here to here
what is going on conduction here and
here what is going on
convection is also connection understood
now we need to find out thermal
resistance for convection and what is
the basic formula we need to know and
what is the basic formula for finding
out thermal resistance for conduction
and here the same for convection once
you will come to know the basic formula
for finding out thermal resistance then
it becomes quite easier for us to find
out the heat transfer rate okay all
right so let's go wait so what is it
heat has but read for convection is we
are having given we have given a formula
right all right we have given a formula
so here thermal resistance is equal to
word one sorry TS - ting 81 by HP as is
what heat transfer coefficient and a
means area now most important thing is
that we have to know how to find out
area of any cylinders because this is a
case of cylinder here you need to have
basic concept without knowing basic
concept you cannot find out the right
answer so we need to know the area to
find out area of the cylinder the
formula for area to find out area of a
cylinder so if the formula for area for
cylinder is as you know 2 pi RL okay now
here 1 by H 1 suppose that this one is H
1 and here s 2 I will tell you which one
2 pi RL 2 PI R 1 F let's say this here
conduction what we have given Q equal to
minus ka DT by D X DX means thickness
and temperature difference is here so
what will happen B by K a DT DT its T 1
minus T 2 suppose the initial minus
final okay because we have V occur as we
have just multiplying this minus sign
here because in general what is the
difference I mean et needs means
changing temperature so final minus
initial if we will multiple that minus
sign with that then what will happen in
ECL minus final that's why I have said
their initial minus final that means t1
minus t2 are you getting me I hope you
understand okay
know most important things what would be
the formula here for this resistance so
that would be natural log r2 by r1 2 pi
K 1 L here natural log r3 by r2 2 pi K 2
L what is k1 k1 is this thermal
conductivity for this wording what is k2
thermal conductivity for this one it
doesn't matter what kind of body you are
going to get but the the what the the
firm loud is saying okay this you need
to put the formula and bellows you are
going to get answer what required to
find out okay now for here the firm loud
we like this that's this hunter values
will be changed because here is
something wrong and here you know
something because here we are getting
something where we have some other
process and process the same we are
getting different value and we are here
also we are going to get different Bell
so 1 by H 2 into 2 PI R 2 sorry not r2
it's its it will be R 3 because of T we
have our 3 2 PI R 2 R 3 L now now most
important most important what is that we
have to find out rate of a transfer so
we know that we know that Q is equals
rate or heat transfer this is known as
rate of HT heat transfer okay that unit
would be in Joule per second that means
in what that means in what you must not
forget this so Q is equal to
TV - terrified summation of RTS total
resistance so TZ - stay in line 1 by H 1
we'll have to add all these values 1 2 3
4 2 PI R 1 l plus natural log R 2 by R 1
2 pi K 1 L + natural log R 3 by R to 2
pi K 2 L plus 1 divided by H 2 2 PI R 3
l okay so this is the formula for
finding out heat transfer rate of in
composite cylinder conduction as well as
convention if there is conduction and
convection process both the process has
been considered now another problem most
important pay attention what I am going
to share with you because the same
pretty problem already asked in kid
without considered without conservation
of convection that is they ask heat as
per gate per unit length and they
haven't geeked the value of length in
the formula I'm in the cushion so what
you are going to you have to get come on
all these L and you have to transfer
this L this L from here to this side
then what will happen qy l that means
heat transfer rate per unit length that
is the key for solving tricky
comparative examination problems in
general they are gonna not going to SQ
problem from here most of the time
because here it is littered with lengthy
and time taking process
okay i hope you understand the situation
and this problem these must see this
concept thanks for watching see you soon
in my next video bye bye
to share with you conduction convection
both the processes going on heat
transfer process through a composites in
them alright so as we have drawn here on
in front of you a composite cylinder
right so internal diameter time I mean I
have put it radius into radius of the is
one you can see it's inter it is is r1
and extend radius is R two and here the
most out total radius is R 3 okay now
what we are going to do is saying that
gas is flowing inside the pipe this step
I suppose densify follow pipe and gas is
following if gas is flowing and what
type of gas we are talking about it hot
gases suppose the temperature of hot
gases is T Z okay and here temperature
is let me use different n t1 here
temperature is t2 in the surface and
here temperature is heating and the
outside temperature let us say this is
surrounding at D infinity let's say this
this is T infinity and here from this
whole scenario you can assume it as T Z
as you or it is obvious as T Z greater
than T infinity okay because hot gases
is flowing inside it and this one is T
infinity so you can say that t ze is
greater than T infinity of you
understanding right now most important
thing most important thing here we have
to consider convection as I have already
shared with you in case of convection we
follow one law that is H a TS minus TF
and in case of conduction will follow
this one T D by DX this law is known as
for years law and this one is known as
Newton's law of cooling okay these two
laws are most important and even we we
usually use these two laws only in case
of conduction as well as convection okay
now now what is most important we have
to throw the thermal circuit diagram and
we also need to know what our
assumptions for this case we are
considering for finding out the total
heat transfer okay rate also because
today heat transferred per in a second
we are going to find out and our unit
would be in what so we must say heat
transfer rate so what are the
assumptions we are taking assumption 1
and assumption two first reservation
that must be in steady state second one
dimensional one dimensional a radial
heat transfer radial heat transfer
let me share with you in general we
cannot get these two if we are taking in
a real situation it is really like
impossible to get these two situation
okay one time I saw a radial heat
transfer is almost impossible next being
possible you can say but yes we're
considering without these two
consideration we cannot find out total
heat transfer rate right away so let's
go for it
first of all we have to draw the thermal
thermal circuit tighter
okay now as you know you're having a
convection process to here so we have to
turn convection conduction conduction
again convection okay suppose that this
is T Z and this is infinity therefore
this one with T 1 this one is T 2 this
one is tickling let me share with you
this diagram with this one because I
have drawn this diagram from here okay
so we have is check it with TZ is
easements gas hot gases right and then
this surface T 1 then this one T 2 and
then this one T 3 and then after
technique what what is directly
connected with temperature 3 that is
environment or even surrounding so that
is known as T infinity hope you are
getting right now okay okay now
understand it okay so this here and here
we are having what type of process is
going on 1 to 2 and 2 to 3 1 to 2 and 2
to 3 conduction process is going on from
here to here TZ 2 T 1 and T 3 to
infinity what kind of process is going
on convection process is going on so
here our th1 are th2 suppose that this
is our ds3 M R T as for so here to here
what is going on conduction here and
here what is going on
convection is also connection understood
now we need to find out thermal
resistance for convection and what is
the basic formula we need to know and
what is the basic formula for finding
out thermal resistance for conduction
and here the same for convection once
you will come to know the basic formula
for finding out thermal resistance then
it becomes quite easier for us to find
out the heat transfer rate okay all
right so let's go wait so what is it
heat has but read for convection is we
are having given we have given a formula
right all right we have given a formula
so here thermal resistance is equal to
word one sorry TS - ting 81 by HP as is
what heat transfer coefficient and a
means area now most important thing is
that we have to know how to find out
area of any cylinders because this is a
case of cylinder here you need to have
basic concept without knowing basic
concept you cannot find out the right
answer so we need to know the area to
find out area of the cylinder the
formula for area to find out area of a
cylinder so if the formula for area for
cylinder is as you know 2 pi RL okay now
here 1 by H 1 suppose that this one is H
1 and here s 2 I will tell you which one
2 pi RL 2 PI R 1 F let's say this here
conduction what we have given Q equal to
minus ka DT by D X DX means thickness
and temperature difference is here so
what will happen B by K a DT DT its T 1
minus T 2 suppose the initial minus
final okay because we have V occur as we
have just multiplying this minus sign
here because in general what is the
difference I mean et needs means
changing temperature so final minus
initial if we will multiple that minus
sign with that then what will happen in
ECL minus final that's why I have said
their initial minus final that means t1
minus t2 are you getting me I hope you
understand okay
know most important things what would be
the formula here for this resistance so
that would be natural log r2 by r1 2 pi
K 1 L here natural log r3 by r2 2 pi K 2
L what is k1 k1 is this thermal
conductivity for this wording what is k2
thermal conductivity for this one it
doesn't matter what kind of body you are
going to get but the the what the the
firm loud is saying okay this you need
to put the formula and bellows you are
going to get answer what required to
find out okay now for here the firm loud
we like this that's this hunter values
will be changed because here is
something wrong and here you know
something because here we are getting
something where we have some other
process and process the same we are
getting different value and we are here
also we are going to get different Bell
so 1 by H 2 into 2 PI R 2 sorry not r2
it's its it will be R 3 because of T we
have our 3 2 PI R 2 R 3 L now now most
important most important what is that we
have to find out rate of a transfer so
we know that we know that Q is equals
rate or heat transfer this is known as
rate of HT heat transfer okay that unit
would be in Joule per second that means
in what that means in what you must not
forget this so Q is equal to
TV - terrified summation of RTS total
resistance so TZ - stay in line 1 by H 1
we'll have to add all these values 1 2 3
4 2 PI R 1 l plus natural log R 2 by R 1
2 pi K 1 L + natural log R 3 by R to 2
pi K 2 L plus 1 divided by H 2 2 PI R 3
l okay so this is the formula for
finding out heat transfer rate of in
composite cylinder conduction as well as
convention if there is conduction and
convection process both the process has
been considered now another problem most
important pay attention what I am going
to share with you because the same
pretty problem already asked in kid
without considered without conservation
of convection that is they ask heat as
per gate per unit length and they
haven't geeked the value of length in
the formula I'm in the cushion so what
you are going to you have to get come on
all these L and you have to transfer
this L this L from here to this side
then what will happen qy l that means
heat transfer rate per unit length that
is the key for solving tricky
comparative examination problems in
general they are gonna not going to SQ
problem from here most of the time
because here it is littered with lengthy
and time taking process
okay i hope you understand the situation
and this problem these must see this
concept thanks for watching see you soon
in my next video bye bye
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