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Welcome to the first lecture of module 3 on heat transfer, in this lecture we will discuss
on One Dimensional Unsteady State Conduction Heat Transfer, till last lecture we have discussed
on steady state conduction heat transfer, but here we are concentrating on unsteady
state situation.
And as we understand from unsteady state point of view, unsteady state conduction heat transfer
means that heat conduction takes place under transient condition where the temperature
at any location is function of time at the same time it can be function of a location
also at any at any time it can be function of location or position also. Or depending
on the situation it can be one-dimensional, two-dimensional or three-dimensional heat
conduction, we will be mostly concentrating in this lecture on the one dimensional heat
conduction that means temperature is a function of one direction of or rather one coordinate
in the direction of heat transfer, and also temperature will be function of the time of
heat transfer. So, and this unsteady state heat conduction
phenomenon is a very common process common situation in industrial processes is very
frequently encountered typical industrial processes, if we see that it includes like
Heat treatment of metal parts annealing operation then hot rolling of metal or an alloy to a
sheet then Heating a furnace to a steady wall temperature. Then heat gain or loss by loss
by the buildings certain changes of environmental temperature when the suppose, there is a pipeline
there is a transport of transport is taking place of say oil transport or it could be
water transport, then the which is the pipeline are buried under the earth after certain date,
and there can be atmospheric temperature is certainly reduce particularly in the winter
seasons. So, that is also a situation when the temperature is suddenly changed that kind
of situation is a practical example of unsteady state situation which may not be related to
directly to the industry, but which may be related to the common life of human beings.
Similarly, Cooling of I C engines then Start up and start down processes in many of the
industrial processes which involves heat generation or which involves heat transfer in it. Then
all the Start up and shut down processes they do encountered unsteady state situation as
well as there are certain energy recovery from hot gases by thermal regenerators are
used for energy recovery particularly example is that hot energy recovery from fuel gas.
So, hot gas is energy recovery from the hot gases in the thermal regenerators. So, these
are typical examples of several industrial processes, but unsteady state heat transfer
takes place. Now, when we say about that then there are
various models which are used for unsteady state conduction analysis and these unsteady
state conduction analysis is very much needed because we have to know that how is the temperature
profile within a material? How the temperature is varying from one place to the other place
of the material? And once we know the temperature then only we can calculate if, we want how
much is the heat requirement for further heat requirement or how much will be the heat transfer
from the material with environment. So this information time to time information on all
these things are very much necessary and therefore, the analysis of unsteady state heat conduction
is necessary and further there are 2 types of model as I told there one is called lumped
parameter model another is called distributed parameter model.
Now, the question is what is this Lumped parameter model? In the lumped parameter model as you
understand from the name itself the parameter is lumped over the entire body and in this
case of parameter is nothing but the temperature we are interested about the temperature. So
this temperature is lumped over the entire body of concern and therefore, what does it
mean, that mean the temperature is uniform assumed to be uniform it may not be exactly
uniform, but it is assumed to be uniform throughout these body therefore, there is no variation
of temperature within the body of concern or body of study, but temperature is the function
of time at any location of that body. So, this kind of situation that lumped parameter
model kind of situation appears when there will be almost no resistance in the body that
means what, that means the thermal conductivity of the body is very high that means within
the body the heat is being conducted very fast. So, there is no resistance almost there
is no resistance and therefore, it is assume that the temperature variations from one place
to the other place of the body is almost negligible. So this lumped parameter model is applicable
for cases where a material is having very high thermal conductivity and the material
is of smaller of size in general we will discuss this issues little later.
Then, another model is distributed parameter model from the name itself we can understand
that in this case the parameter is distributed over the material that means the temperature
is the parameter over here in present case and the value of temperature is distributed
over the entire body that means from one place to the another place of the body temperature
is varying. At any time from one place to the another place of the body the temperature
is varying and therefore, it is called distributed. At the same time at one particular location
of the body the temperature is function of time. So, that does in case of distributed
parameter model temperature is a function of both time as well as location in the entire
body. There are 2 types of problems we will be discussing because distributed parameter
model is a huge area we do not have we have we do not have enough time to discuss in all
these because, there are many other aspects of heat transfer has to be covered.
So distributed parameter model or heat transfer unsteady state heat transfer, when the distribution
of temperature is there and temperature is a function of time and coordinate there can
be lot of analysis can be done in this area, but what we will do is for some understanding
for certain understanding of undergraduate chemical engineers. So what we will try to
do, we will concentrate on one dimensional heat transfer first of all and second thing
the few cases we will consider and we will see that how the temperature profile is existing
in a body and as well as how to find out the heat flow rate and how to find out a different
location at different time the temperature as well as heat provide for this a subject
to certain conditions subject to certain coordinates. We know that there are 3 coordinates existing
cartesian coordinate cylindrical coordinate and spherical coordinate. So we will see that
in case of cartesian coordinate that means in case of plane wall for example, that building
walls which is plane wall an example, of a plane wall so what happens for that case unsteady
state heat transfer how is the temperature profile and how is the distributed parameters
system model is applied in this case. Similarly, the case of a cylindrical body
for example, a when you know that a drawers then displayed in columns these are a cylindrical
body and if, there is a high possibility that distributed parameter model is applicable
for those kind of situations similarly, for spherical vessels large spherical vessels
also we can have such kind of situations. So we will discuss, in all the 3 different
geometries that how distributor parameter model applies. And also there are two situations
we will discuss as I have shown one is the case for transit heat flow in a semi infinite
solid and in and in infinite body subject to sudden convective environment for heating
or cooling, we will discuss this issue subsequently in different lectures.
Now thus there are 2 problems as I told one is the Transient heat flow in semi infinite
solid and infinite body subjected to certain subjected to sudden convective environment
for heating or cooling. This is a 2 different cases bothers for one dimensional problems
or unsteady state heat transfer and this is the distributed parameter model cases and
now solution for this kind of situation is not very easy these are little complicated.
So therefore, we will be concentrating on 1 dimensional because 3 dimensional cases
are more complicated. So we will be trying to simplify the cases as much as possible
and see just to have an understanding that how the temperature profile is there and what
we will do is that there are a few ways of doing the solution of these particular cases
and what we will do there are two different ways of solving this one is the analytical
solution methods another is the graphical solution practically analytical solution method
means, here the things will be solve by separation of variables method and combination of variable
method. So we will discuss, we will take a one cases
for separation of variables and combination of variables cases how the heat transfer problem
can be solved and then graphical solution actually, in this case also that temperature
profile is being developed by somebody else, he has already worked out and has he has been
given that that temperature where has been given in the form of figures in the form of
graphs and now for our training purpose for our learning purpose the students in fact
in fact for practical application also, we can find out how will be the temperature for
a particular situation and how to find out this temperature from the figures or the chart
available at the same time not only the temperatures at different locations of the body, but the
heat flow rate also can be calculated can be found out with the help of the various
charts or figures available. So, we will see that with the help of certain problems how
to make up these charts. So now, to start to start with what we will
do is we will start the case of unsteady state heat transfer phenomena and we will start
with lumped parameter model now one parameter model as I discussed with you that it is dependent
on it is dependent on the situation when the temperature is lumped or parameter is lumped
in the entire body, so what we will do is let us consider typically.
A body let us consider typically a body this body has a volume is equal to V into volume
and let us write like this way otherwise we will confuse with a velocity so let us, write
this is the volume and this is the area of heat transfer. This is a some object which
is say the object is hot and it is in some outer environment, these environment what
heat is being transferred from these so say if I say, that heat is being to dot the heat is being transferred to the
environment and let us say that environment and temperature is T infinity is the temperature
of environment and T i is the temperature of the body at t equals to zero that means
the initial temperature of the body. Now what will happen that T infinity is less than T i. That means the
heat will be transfer from the body to the environment. Now what we will assume that
as this is the lumped parameter model case
as because, it is a lumped parameter model. So temperature is uniform through this now
what is happening with time what is going to happen that energy that internal energy
of the body is going to come down. So, the body is going to lose energy heat energy or
it is internal energy and it is getting transferred to the environment. So the energy that has
been lost by the body will be reflected with the help of or in terms of decreasing temperature
with time. So what we will write is that, if rho is the density of the body and V is
the volume as I discuss so this is the mass and C p is the specific heat that means which
is the heat requirement per unit k g per unit mass. So, rho V C p m c p into d t d T by
d t that will be the change in internal energy and because, it is going to a lower temperature.
So, d T by d t is a negative quantity those changing internal energy we will write it
as minus of this. So, this becoming the total thing is becoming positive change in internal
energy would be equal to the amount of heat lost which is to the environment and that
is the happening with the help of convection. So it is an example of conduction and convection
phenomenon that is taking place in tandem it is taking place together.
So heat is being conducted from the body to the surface and from here it is the convection
phenomenon that is being happening. Then we will write it that h is the heat transfer
convictive heat transfer coefficient and A is the area of the body that is the area surface
area of the body as we understand that in case of sphere it is 4 pie r square in case
of cylinder it is 2 pie r l into sorry 2 pie r l plus 2 into pie r square that is 2 end,
end of the 2 areas. Similarly, in case of cube it will be 6 into side square like this.
So this will be the surface area of heat transfer, surface area of heat transfer means that the
area which is exposed to the environment for the transfer of the heat. Now so, h A into
at any time T is the temperature of the body minus this is the as per as Newton’s law
of cooling that heat transfer is h A delta T. So, we are writing that T minus infinity
is the same equation. Now this is the very basic equation, for the
equation of energy balance has been happening that rate of change of internal energy is
nothing but equal to the rate of heat transfer by convection from the solid body to the environment.
Now from this if we just simply do little calculations then what we will get that t
d T by say T minus T infinity and if we integrate form T is equal to T i 2 T is equal to T will
be equal to minus h A by rho C p into V into d t and this is equal to T equal to zero or
T is the initial temperature and t equal to T. And what we will assume, that with other
areas naturally volume and area are not functions of time also will assume that h rho and C
p are independent of time. They are constant quantities and thus what we will get is if
we do the integration, we will get that T minus T infinity by T i minus T infinity equals
to e to the power minus h A by rho C p into V into t.
So, this is the very common form of equation. Now we can say that at t equals to say tau
and tau is equal to rho C p V by h A. At t is equal to tau and tau is equal to so, if
we say that then what we will get the T minus T infinity by T i minus T infinity we will
say that it is minus it will be minus t by tau, tau equals to rho C p V by h A and when
t is equal to tau then for this case we will get T minus T infinity by T i minus T infinity
and that is equal to e to the power minus 1 and this value is equals to 0 point 368.
So what does it mean, it means that T minus T infinity equals to 0.368 of T naught minus
T infinity. Initially, this was the temperature difference between the body and environment
and after time constant t equals to tau after the value of tau that value is reduced to
36.8 percent. So at sorry t equal to tau T minus T infinity by T naught minus T infinity
is equal to 0.368 that means at time t equal to tau that temperature difference in the
body is becoming 36.8 percent of the original temperature difference. That is the definition
of a time constant. So tau is called when tau is equal to rho
C p V by h A is called time constant. So time constant is the time. So, the unit of rho
C p V by h A is time it is unit is so time constant is a time, that is needed for the
temperature in the body temperature difference in the body, to reach to 36.8 percent of the
initial temperature difference. So, initial temperature difference was T naught minus
T infinity and after time constant the temperature is T minus T infinity and this T minus T infinity
is becoming 36.8 percent of T naught minus T infinity. So this is called the time constant
and rho C p V by h A is called the time constant. So this one very important information, that
we had develop from this and in future, when we will be starting any first of the processes
we will be learning about this time constant. So this is also view to be a first order process.
Now, drop in temperature in the body is viewed as a first order process now if we go for
further analysis of this say for example, we have got the expression that h A by if
we do a little bit to these rho C p into V that is the inverse of the time constant.
If we write these in a little organized way what we will do is we will get h into V by
A by k into K by rho C p into V by a whole square. This is the expression that we are
going to get and we say that V by A is equal to say l this is a characteristic length and
that we can very easily understand, that V by A l is a characteristic length and because,
V is the volume unit is volume and A is the unit of the area of it is meter square and
V is the meter cube. So, V by a is the unit of length and we will say that V by a is the
characteristic length. So we will write this is as h l by k into k by rho C p, we know
that k is the thermal conductivity of the body and rho c p we know that alpha k by rho
c p is the thermal diffusivity which is meter square per second and V by A we will write
as l square. So then, in that case we can write that t
minus t infinity by t i minus t infinity also can be written as e to the power minus Biot
sorry minus h l by k into alpha by alpha t by l square. So, this is also written as e to the
power minus b i into f o this b i is called Biot number. So, this is your h l by K this
can be written as 1 by K A divided by sorry l by K A divided by one by h a we know that
l by K A is the conductive resistance. We have seen already in case of conduction and
one by h A is called the convective resistance. So, this is Biot number is becoming a ratio
of conductive resistance to convective resistance. This is conductive resistance to convective resistance and both are for
heat transfer conductive resistance for heat transfer and convective resistance for heat
transfer. So, it is a ratio of 2 resistances one is
conductive and Biot number is ratio of conductive resistance to convective resistance for heat
transfer and the important point over here is that, if Biot number is very less then
we can understand that conductive resistance is very small. One another point we should
say here that the conductive resistance for we are discussing here this is called solid
conduction. This is the solid heat conduction heat transfer or heat conduction by the solid
material. In future, we will such cases in the lecture will be find will be coming across
another terminology that is called Nusselt number which is denoted as n u this is also
having a similar physical significance, that it is also can be expressed as conductive
resistance to convective resistance, but the difference between Nusselt number and the
Biot number is that in case of Nusselt number we’re telling out the conductive resistance
of the fluid so this Nusselt number terminology exist, when there is a heat transfer between
a solid and fluid so Nusselt says the conductive resistance of the fluid whereas, Biot number
say that conductive resistance of the solid. So this is the major difference lies between
Nusselt number and the Biot number the other terminology.
That is called Fourier number F o and this is we have seen, that it is expressed as alpha
t by l square. Now, this can be this is usually expressed as this is or this is written as
ah if we say that it is an indication of that degree of penetration of heating or cooling
effect degree of penetration of degree of penetration of heating or cooling effect through
the solid it is also can be expressed as it is also can be expressed as that temperature
penetration depth by body dimension. So, I find that the temperature penetration
depth and it is related to thermal conductivity. So how much thermal profile temperature profile
has been penetrated into the body with time t and l square is the body dimension. So,
this way a Fourier number is also is a dimension less quantity and that is defined. Thus we
can say that T minus T infinity by T i minus T infinity is equal to e to the power minus
Biot into Fourier. And also we have seen that Biot number is as the ratio of thermal conductive
resistance thermal conductive resistance to heat transfer to convective resistance to
heat transfer and also I told, that when the Biot is very small that means conductive resistance
is very small. So what we told at the beginning that lumped parameter model is applicable
when the temperature variation in the body is negligible, that means the thermal conductivity
is very high that means thermal conductive resistance within the body is very low, that
means the Biot number is very low. So it is expected, that lumped parameter model will
be very much applicable when Biot number is very low.
For Biot number is very low that lumped parameter model is very is it can be applied and it
is being reported that if Biot number is less that 0.1 then lumped parameter model can give
reasonable estimate that is with in 5 percent of the error. So if Biot number is less than
point one then lumped parameter model can give an estimate within 5 percent of error. So what is being suggested is that or recommended
is that because, there are always acute amount of uncertainty involved to calculate in the
calculation of value of h convective heat transfer coefficient that uncertainty is sometimes
rise within 25 percent of these 25 percent. So there is a huge uncertainty is there in
calculating convective heat transfer coefficient whereas, the uncertainty while getting using
lumped parameter model, if the Biot number value is much less it is within five percent
therefore, it is advisable that one when trying to find out that heat transfer rate or flow
of heat or the temperature profile in a body first try out whether the lumped parameter
model is applicable or not that will actually make the cases simpler.
Now, what we will do that there is some thermal network for single capacity system here we
will say that is it for single body a single capacity system. So that, we can draw a thermal
network for that just enable us to enable us to electrical system.
if we see that T i is the sum temperature and so this is say T infinity environmental
temperature and there is say switch is there here and this is sorry this is one by h A
this is equal to R resistance and this is the capacitance this equal to rho C p into
V. So this is a thermal network for single capacity
system, if we try to understand here the situation is like this is the switch and what is happening
that there is 2 potentials T infinity and T i. T i is as I told the T i is the initial
temperature of the that means of the hot body and T i is the environment temperature. So
when you put the switch on when you put the switch on the body is charged the capacitor
is charged with this potential. So this is the potential difference for the
capacitor the capacity the capacitor will be charged with the thermal potential of T
i and then once we switch off, then what will happen this energy will be released through
this resistance through in this circuit this energy will be released. Thus the capacitor
is holding that energy it is getting charged holding the energy which is equivalent to
the potential difference of T i minus T infinity at the initial condition when the switch is
on, and once you switch off when the switch is made open then what will happen that this
circuit is disconnected. So this energy will be release through this circuit that means
thus energy will be release to the environment and while resistance is 1 by h A.
So this is way, that in this case also in our present case also that that 1 by h A is
the environmental resistance to from the surface we see here that from the surface of this
it is getting to the environment with the resistance of 1 by h A.
So this is some understanding about that representation of the lumped parameter model system in the
form of thermal network for single capacity system here it is single capacity, because
we have already one body and this may not be stored in that capacitor. Now what we will
do, we will take off problems on this particular issue, we will discuss a problems.
So let us discuss the problems say question one. An aluminum cube of side 5 centimeter
and at 100 degree centigrade temperature is suddenly exposed to air at 25 degree centigrade
the convective heat transfer coefficient is 25 Watt per meter per Kelvin sorry Watt per
meter square per Kelvin rho aluminum is equal to 2700 sorry A L 2700 k g per meter cube
and K aluminum is equal to thermal conductivity 215 Watt per meter per Kelvin and C p specific
heat of aluminum is equal to 0 point 9 into 10 to the power 3 Joule per k g per Kelvin
the problem is calculate the time required for the center temperature to reach 40 degree
centigrade? So this is the problem, the problem is that
there is aluminum cube of side 5 centimeter and it is at 100 degree centigrade temperature
and it is suddenly exposed to air which is at 25 degree centigrade and it is it will
be loosing energy from the cubes to the environment. So it looses the energy, then we have to find
out that it will be at sometime it will be reaching a temperature of 40 degree centigrade.
So what is that time that is the question? Now before we ah proceed for lumped parameter
analysis we have to first check whether the lumped parameter analysis is applicable or
not. So the check point is that we will check the
Biot number, if the Biot number is less than point one we will say that lumped parameter
model is applicable. So we will calculate Biot number, Biot already we know that is
which is h l by K or it is h into V by area by K.
Now here, we have to see that it is a cube. So for cube volume is equal to we know the
length square sorry length cube and what is this area surface area all the surfaces of
the cube is exposed to the environment that mean there are 6 surfaces which are exposed
to the environment. So what we will get is that V is equal to 0 Point 05 whole cube and
the 6 meter cube. 5 centimeter 0.05 meter, so point 0.05 meter cube that is volume and
area is is equal to 6 side square so 0.05 whole square this will be meter square.
So we will get, V by A is equal to if we do the calculations we will get that is equal
to 5 by 6 into 10 to the power minus 2 meter. And then, what is the value of Biot number
if we do the calculation we will see that, it is equal to h equal to 25 into 5 by 6 into
10 to the power minus 2 divided by case for aluminum 215 and the unit of Biot number is
it is a number, so it is unit less and these value is equal to, if we do the calculation
0.00097 which is much less than 0.1. So that, if this so this is the check with this check,
we can say that Biot number is less much less than 0.1.
So lumped parameter model is applicable, that means what we are writing conclusion from
this check is that lumped parameter model is applicable.
So once lumped parameter model is applicable now, we will enter into the equation temporary
for finding out the temperature profile actually, we know the we have to find out the time requirement
for certain change in the temperature. So we need to have the temperature profile, the
temperature profile we have already seen the T minus T infinity. T minus T infinity by
T i minus t infinity and that is equal to e to the power Biot into Fourier and Biot number values we have already
seen. Now for Fourier number we have Fourier number we have it is alpha t by V by A whole
square right alpha t by l square l means V by A.
Now actually, what we have to calculate in this problem is to calculate this temperature
for this calculate sorry we have to calculate this time. So what we will do is we know that
l lawn T minus T infinity by T i minus T infinity is equal to minus lawn this is equal to Biot
into Fourier. So, this value we is known to us for my case it time for time do for the
time which is wanted at that time T is equal to 40 degree centigrade. So, we will get minus
l n this is 40 minus the T infinity is 25. So here, whether it is we are writing in Kelvin
or in centigrade it does not matter because it is finally, we are getting the difference
of the temperatures. So 40 minus 25 by we will have 100 minus 25 and that is equal to
Biot into Fourier. So or minus l n 15 by 75 is equal to Biot into fourier. So we can say
that Fourier number is equal to minus l n this is equal to 1.61 by Biot number and this
is equal to 1.61 by 0.00097.
And then, what we say is that alpha t by V by A whole square and that is equal to 1.61
by 0.00097 from there we can say that all t equals to that is the time requirement,
that is what we have to calculate it is 1.61 by 0.00097 into v by a whole square is 25
by 36 into 10 to the power minus 4 into alpha which will be one by alpha 1 by alpha is equal
to that we have to calculate alpha is equal to we have seen that it is 8.85 into 10 to
the power minus 5. How to calculate alpha? So, we will be calculating
alpha like this alpha is equal to k by rho C p and the values are given to us K is equal
to 215 by rho is equal to 2700 into C p is equal to 900. So this gives that 8.85 into
10 to the power minus 5 meter square per second. So if we put these values here and then we
will get that this value becomes 1302.4 second and that is equal to 21.71 minutes. So that
means, that the cube the aluminum cube will calm down to that temperature of 40 degree
centigrade temperature at 21.71 minute so this is the solved.
Now let us see another problem, it is a question 2 it says that a copper ball of 5 centimeter
diameter of initial temperature 100 degree centigrade is suddenly exposed to ambient
at 25 degree centigrade. The initial cooling rate is 4 degree centigrade per minute. Calculate
the convective sorry heat transfer coefficient? Assume that lumped parameter model is applicable
and rho copper is equal to 8900 k g per meter cube, then C p of copper is equal to say 380
Joule per kg per Kelvin and thermal conductive K for cupper is equal to 370 Watt per meter
per Kelvin So now, the problem is there is a copper ball
and the copper ball is losing energy the as because the temperature of the ball is getting
changed with time we have seen in the previous in the basic equation the temperature is changing
with time. So d T by d t rate of change of temperature will gradually decrease or will
gradually change. Because it is loosing energy, so at very high temperature the rate of change
in temperature will be much lower much higher than that at the very low temperature, because
the driving force will gradually decrease as we know that in our previously
what we have got we have discussed if we see that here, that in this reason there is a
driving force driving force is the T minus T infinity, and these driving force will gradually
decrease time goes on because the hot body will be gradually decreasing so that therefore,
the d T by d t will go on decreasing ok. Therefore, d T by d t is the will be changing
and so initial value is being given to us for this problem. So what we will do is and
it will be a decreasing value it is cooling initial cooling rate is 4 degree centigrade
per minute. So we will convert it as so d T by d t is equal to 4 degree centigrade.
So minus d T by d t it is the cooling is 4 degree centigrade per minute and that is equal
to 4 degree centigrade by 60 per second and now if see our basic equation. So rate of
heat transfer is we know from basic equation.
if we see that minus rho C p V into d T by d t is equal to h into Area into T minus T
infinity and d T by d t is given rho is known to us C p is known to us volume of the copper
ball is known, and then surface area of the copper ball is 4 pie r square that is also
known h is given I have find out sorry h is t is known so h has to be find out.
So that way, we can find out that h is equal to rho into C p into V minus d T by d t by
we have Area into T minus T infinity and here the values are rho is equal to 8900 k g per
meter cube into C p value is 380 Joule. So we have to be very careful about the units
and V is the volume, volume is four third pie into r cube ah it will be, so actually
that let the problem will little be reduced that let us let it to be point capable of
radius 0.5 centimeter is write radius 0.5 centimeter if we write then it will be that
it calculations is accordingly, so 8900 into 380 into four third pie into 0.05 put that
pie r cube as the volume into the cooling rate is 4 by 60 will come. So we will write
first area is again 4 pie into r square r square 0.05 whole square into seventy-five
is the T minus T infinity is 75 into that the 60 will come because of these conversion
of degree centigrade per minute to degree centigrade per second. So, this value it comes
to be around 50 Watt per meter square per Kelvin as this h a delta so it is becoming
fifty Watt per meter square per Kelvin. So this h the convective heat transfer coefficient
between the copper ball and the environment is 50 Watt per meter square per Kelvin.
So, we can calculate even by doing some experiment proper measuring of the temperatures we can
calculate how is the convective heat transfer coefficient using the lumped parameter model.
So, what we will do is we will stop here in this for this lecture in the next lecture
we will discuss on unsteady state heat transfer for distributed parameter model system when
the temperature does not remain uniform in a in a body where the temperature is varying
in with position in a body. So we will discuss in the next lecture, thank you very much.