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Welcome to lecture two of module one, the last lecture we have discussed on various
mechanisms of heat transport. Now, we will discuss little bit on conduction, we will
go in deeper in to conduction. First what we will try to do is that, we will find out
general heat conduction equation.
General heat conduction equation and this heat conduction equation will be developed
based on first principles, basic energy balance equations. In fact, we will see that general
heat conduction equation both for all Cartesian coordinate, symmetrical coordinate and and
spherical coordinate system. What do we understand by the general heat conduction equation is
that, it takes care of a situation, the equation which will be developed that takes care of
the situation, when you can have unsteady state heat conduction, when you can have temperature
variation with time, temperature variation with space, and there is a possibility of
heat generation within the body, through which the conduction phenomena we are discussing.
Now, to start with, we will take the volume elements for different cases. To start with,
we will take the volume element for Cartesian coordinate system. Let us draw a volume element
for the Cartesian coordinate system.
So, let us draw the axis first, so this is your say x, this is y, and this is say z,
and say this is the volume element, and so this is dy, this is dz, and this is dx. And
what is happening is like, so in this phase it is q dot y, and this phase it is q dot
y plus dy. Similarly, say this phase, if this is q dot z, this phase it is q dot z plus
dz. So, what is this q dot y means? q dot y means, that heat is rate of heat transport
at plane y and q dot y plus dy means, rate of heat transfer at plane y plus dy in the
right hand side, q dot z means, rate of heat transfer at the bottom plane andz plus dz
means, rate of heat transfer the top plane.
Similarly, you can have q dot x, and there is the x direction, and we can have q dot
x plus dx, so these are the rate of heat transfers. At the same time we can have, say heat generation
within this volume element and, we will say that q dot gen is the rate of heat generation
per unit volume. per unit volume
So, in to the volume, what is the volume this element? the volume of this differential element
is dx, dy and dz. The volume of the differential element is dx, dy, dz and this is q dot gen
is the volume of the differential element, rate q dot gen into dx, dy, dz, is the rate
of heat generation in the differential element, and we will assume that rate of heat generation
is uniform throughout this volume element.
Now, the based on the first principles with the analysis, the according to the first principles
or the energy balance, we have to do according to energy balance, we will say that rate of
energy conducted in, rate of energy conducted in plus rate of energy heat energy generated
rate of heat energy generated within the volume element, this should be equal to rate of energy
conducted out. Mind it, that we are just taking care of only conduction mode of heat transfer,
there is no other thing is happening in this volume element. The rate of energy conducted
out plus
rate of change in internal energy, or rate of energy accumulation within the volume element.
Energy means, we are considering again that in all the cases we should say the rate of
heat energy, rate of heat energy and rate of heat energy accumulation within the volume
element. This also can be called as rate of change of internal energy of the volume element.
Now, let us tell that term 1 is the rate heat energy conducted in as term 1, then rate of
heat energy generated within the volume element as term 2. Rate of heat energy is conducted
out as term 3, and rate of heat energy accumulation within the volume element as term 4. So, then
we have 1 plus 2 is equal to 3 plus 4. Now, we will find out what is 1, what is 2, what
is 3, and what is 4. Now, term 1 implies what? Term 1 says, that it is q dot x plus q dot
y plus q dot z. All the rate of heat transfer from heat enters, not transferred; it is entering
in to the volume element q dot x, q dot y and q dot z. Then what is this term 2? Term
2 says, rate of heat generated within the volume element, so we have told q dot, it
is already we have written that q dot generated into dx, dy, dz.
Now, what is term 3? Term 3 implies, it is the energy, heat energy conducted out. That
means, q dot x plus dx plus q dot y plus dy plus q dot z plus dz. And then what is term
4? Term 4 is that rate of energy change, rate of internal energy change or rate of energy
accumulation within the volume element. So, how is it? So what is the energy within the
volume element, if rho is the density or the volume element, and dx, dy, dz is the volume
of the volume element? So, rho into dx, dy, dz is the mass, into c p is the specific heat
of the volume element, into m c p, into d t. So, it is, you can say del t by del t.
I am writing del t by del t, because we will see that temperature will be, will try to
see that is the function of time as well as coordinates. In this case, it is a Cartesian
coordinate system, so in this case the coordinates are x, y, z. So, it is a, we are finding out
the general heat conduction equation in Cartesian coordinate system.
Now, so these are the 4 terms. Now, we will see each and every term, even for finer details.
We will go for term 1, we know that q dot x is nothing, but minus k into area, area
here in the case, is dy, dz is the area, dy, dz is the area minus k a, this is as per the
Fourier's law of heat conduction into del t by del x. Similarly, q dot y is minus k
into, the area would be dx, dz into delt by del y, and q dot z is minus k into dx, dy,
del t by del z. Now, if we see that between term 1 and 3, so if we write 1 minus 3, that
means is equal to q dot, it is it will be q dot, x minus q dot, x plus dx, plus q dot
y, minus q dot y plus dy, plus q dot z minus q dot, z plus dz. Now, if we apply rules expansion
we will find between this it is, that this is del del x of q dot x into dx, plus del
del y of q dot y into dy, plus del del z of q dot z into dz.
Now, we know what is qx, what is qy, what is qz. So, if you put over here then what
we get is, we will get as del del x of, minus k dy, dz del t by del x into dx, plus del
del y of minus k into dx, dz, del t by del y into dy, plus del del z of minus k dx, dy,
del t del z into dz. Now, we will assume that, i'm sorry 1 thing i should say in this term,
this will be minus of this, this will be minus of this, and this will be minus of this. So,
this is then becoming minus of this, plus minus of this, plus minus of this. Now, we
will assume that we will assume that k the commeterial is isotropic, that means k is
independent of position, mutual is isotropic, k is independent of position. In that case,
we can take k out of the differential. So, in that case what we will write is, that k,
dx, dy, dz into del square t by del x square, plus k into dx, dy, dz into del square t by
del y square, plus k into dx, dy, dz into del square t by del z square.
That means we can write it as k dx, dy, dz into del square t by del x square, plus del
square t by del y square, plus del square t by del z square, and this is equal to 1,
term 1 minus term 3. So, if we see that previously it was 1 plus 2 is equal to 3 plus 4, so 1
minus 3 is equal to plus 2
Now, if i write 1 minus 3, plus 2 is equal to 4, from the previous case, so 1 minus 3
we have got as k into dx, dy, dz into del square t by del x square, plus del square
t by del y square, plus del square t by del z square. So, this is between the 1 minus
3, then plus 2, we have already seen what is 2. If, we see that 2 is q dot gen dx, dy,
dz. It is q dot gen into dx, dy, dz, and then that is equal to 4, what is 4? Already, you
have seen 4 is rho cp, dx, dy, dz into del t by del t, rho cp, dx, dy, dz into del t
by del t. Now, we can do, if we divide all size by dx, dy, dz, that the volume element,
then what we get is k into del square t by del x square, plus del square t by del y square,
plus del square t by del z square, plus q dot gen by, q dot gen, is equal to rho cp
into del t by del t, or this term is called grade square t, or del square t, which is
called differential operator del. So, grade square t, grade square is equal to, i'll tell
grade square t is equal to q dot, grade square t plus q dot gen by k, equals to 1 by alpha
del t by, this is t, del t. Now, you see this equation here, this is the
vectorial form of general heat conduction equation. It is also called as Fourier field
equation Fourier field equation in rectangular coordinate, this is also sometimes called
as Fourier 2nd law of heat conduction, so like that different names are given. But,
what we find here this, this grade square is that, or sometimes called del square, it
is basically del square by del x square, plus del square, it is a grade operator, del square
operator, y square plus del square by del z square. Now, it is a generalized form of the heat conducted
equation for rectangular coordinate system, and where we can see that, grade square t
term says that temperature distribution within, which is varing with the x, y and z, temperature
distribution within the volume element, q dot gen, q dot gen by k, that is the generation
internal logic. And alpha, alpha is k by rho cp, this is also called thermal diffusivity.
This has got an unit of mitre square per second in si unit.
So, this thermal diffusivity is similar to moment of diffusivity in fluid mechanics,
they have got similar kind of physical significances, like mu by rho nu, This is momentum diffusivity
that is used in fluid mechanics, and small d, capital d, this is called diffusivity,
molecular diffusivity you can say, that is used in mass transfer. So, for heat transfer
operation this is the thermal diffusivity, which is k by rho cpt, k is the thermal conductivity,
rho is the density, and cp is the heat capacity, and in momentum diffusivity that, nu is mu
by rho. It is used for mu is called viscosity, rho is the density of the fluid, and this
momentum diffusivity is applicable for fluid mechanics, or momentum transport is involved
and d is the molecular diffusivity, which is applicable particularly for mass transport,
when the mass transport is involved. Now, so grade square t, plus q dot gen by k into
equal to 1 by alpha del t by del t, this is general conducted equation, which is also
called as Fourier equation in rectangular coordinate system. we will
Similar way, we will try to see that heat conduction equation for cylindrical coordinate
system. So, let us try to draw a figure again for the cylindrical coordinate system, if
we say x, this is y, and this is z, and say yes this is phi, this is say d phi, and say
this is r, and this part is d r, and this part is dz. So, here exactly in the similar
way if we go for the, you see here there are three things, this is dz, this is dr, and
another part is d phi. So, if we go in the similar way, the volume element is, so we
have again the 4 term analogies term 1 plus 2, as we have previously, is equal to 3 plus
4, and again here 1 equals to q dot r, plus q dot phi, plus q dot z, so r phi z is the
coordinate, that is the cylindrical coordinate. And exactly in the similar way, 3 is q dot
r plus d r, plus q dot phi plus d phi, plus q dot z plus dz, and 2 is, what is that? 2
is the generation term. So, we say that this is q dot generation into
the volume element, and the volume element would be dr, dz and rd phi, rd phi is the
small section, that this one is rd phi into dr into dz, so that becomes the volume of
the volume element. Now, your 4th term is again, that is you that is the accumulation
term, or the change of internal energy, rate of change of internal energy. In the 4th term
it should be, that rho cp into dr, dz into rd phi, into del t by del t, again here what
we are doing is, we assuming that t, it is not an assumption, we are calculating that
only, that t as a function of coordinates, as well as time. So now, again exactly in
the similar way 1 minus 3 as we have done, that implies minus del q dot r by del r into dr, plus minus del q dot,
phi by del phi into d phi, plus minus del q dot z by del z into dz.
Now, we have to find out what is this q dot r, q dot phi and q dot z. So, q dot r equals
to again using Fourier first law, or the Fourier law from the phenomenological law. From there,
what we see is that q dot r is equal to, minus k into area, in case of r, it will be rd phi,
into dz, into del t by del r, and del t by del r, then q dot phi will be minus k dr,
dz. This is the area, thermal conductivity into area, into del t by r del phi. And q
dot z equals to, minus k into rd phi, into dr into del t by del z. Now, if we put all
the three values in the previous case, what we are going to get is, we will be getting
k into, r into del del r, of k into r d phi, dz del t by del r into d r, plus. So, this
is 1 minus 3 equals to, plus del, del phi, into k into dr, dz into del t by r del phi,
into d phi, plus del, del z into kr d phi d r into del t by del z, dz.
Now, we again the similar thing that, we will assume that k is independent of the volume
element coordinates. So, under that situation we can take the k out, but 1 thing we will
say you, because r is there r cannot be taken out. So, what we will write is k, this is
equal to k, d phi, dz into dr, this dr is taken into del, del r of r del t by del r,
then we get plus, k into d phi, dz, dr into, we will get del, del phi of del t by, r del
phi, plus k into d phi dr, dz, into r del, del z of del t by del z. So, we can write
it like this k into, if i take 1 r also here, r d phi, dz, dr, then we can write it like
this 1 by r, del, del r of, because we have taken 1 r here, into r del t by del r, plus
so you have taken 1 r here, so it should be 1 by r square, then into del square t by del
phi square, plus del square t by del z square. So, this happens to be the 1 minus 3 term.
Now, if we write 1 minus 3, plus 2 equals to 4, as we have written in the previous case,
then we will get again, if we write that k r d phi, dz, dr k r d phi dz d r into 1 by
r del, del r of 1 by r del, del r of r del t by del r, del t by del r plus 1 by r square,
del square t by del phi square, plus del square t by del z square, if we see 1 by r square,
del square t by del phi square, this, this is this plus the 2nd term is, we have seen
q dot gen this, it is q dot generation into r d phi, dz, dr that will be equal to 4th
term. 4th term will be equal to again rd phi, dz, dr into rho cp into del t by del t.
Thus, if you divide it by the volume element r d phi, dz, dr, as we have done previous
case, then what are you going to get is that 1 by r, del del r of, r del t by del r, plus
1 by r square, del square t by del phi square, plus del square t by del z square. And that
is becoming equal to, q dot generation by k, plus 1 by alpha, this is plus, and this
is equal to 1 by alpha, we have got del t by del t. So, here also we can write, or we
can write that grade square t plus, as we have done previous cases, gen by k is equal
to 1 by alpha, del t by del t. Here also this is the same thing, similar form we could find
out again in case of cylindrical coordinate system, and alpha is again the thermal conductivity,
which is k by rho cp and grade square t, this is the grade square is the operator, this
is a differential operator. In case of cylindrical coordinate system, and we can understand that
grade square is 1 by r, del del r of r, del del r, plus 1 by r square del del phi, del
square by del phi square, plus del square by del z square.
So, this way we can find out that general conducted equation in cylindrical coordinate
system also. Now, exactly similar thing we can repeat for spherical coordinate, if we
do it for spherical coordinate
for spherical coordinate system .
Let us draw this first. So, this is y, this is x, and this is z, and you have. So, this
is d theta, and say this is theta, and here we can say this is d phi, this is phi. So,
what are the, you have this part is you can understand, this is your r, this part is d
r, and this part is r d theta, and this part is say r sin theta, and d phi, this is what.
So, under this situation, now again the same thing, same concept we will apply, what we
will get is that 1 minus 3, plus 2 equal to 4, the same termina logic we will apply, and
we we have the equation again that minus del, del r of q dot r into d r, plus minus del,
del theta, q theta into d theta, plus minus del, del phi q dot phi into d phi, this is nothing but, 1 minus
3 equal to this. Now, again we have to put that, what is q
r dot, what is q theta dot, and what is q phi dot. q dot r equals to minus k into r
d theta, r sin theta, d phi, r d theta, r sin theta is the area into del t by del r,
and then q dot theta equals to minus k, into area is d r into d r into r sin theta, d phi
into del t by r del theta, and q dot phi is equal to again, minus k into d r, into here
it we getting r d theta, into de t by r sin theta del phi. Now, once you put these things
here, then claball these together, if we put all these here then we will get all these
expressions in this previous equation, and then what is 2 that is generation term is,
it is equal to q dot generation into dr, rd theta, into r sin theta d phi. So, this is
the volume element this is the volume element into this, so it is the total generation,
and 4 total energy generation, and here 4 term is that change in rate of change in internal
energy or rate of change of that rate of accumulation, that is is again the total volume is dr, rd
theta r sin theta d phi, into rho into cp, into del t by del t.
Now, exactly in the similar way if we do the, if we substitute all the things there and
do the deviations, we will land up with this equation that 1 by r square, del, del r of
r square, del t by del r, plus 1 by r square sin theta, del, del theta of sin theta, into
del t by del theta, plus 1 by r square sin square theta, into del square t by del phi
square, plus q dot gen by k equals to 1 by alpha del t by del t will come down to same
this kind of expression. And this can be written as again the grade square t, plus q dot gen
by k is equal to 1 by alpha, del t by del t.
So, we can understand that for all the cases, even for Cartesian coordinate, or cylindrical
coordinate, or or spherical coordinate, in all the cases we can develop this in to this
kind of equation and heat conducted equation, we can develop. And from there, we can see
that this general heat conducted equation, or what is that? Fourier field equation, this
equation can have different assumptions, or apply different constants in to this equation,
we can simplify this in to some different forms. For example, if we say that it is a
steady state operation, if we say it is a steady state process, steady state heat transfer.
When we say steady state heat transfer that means, we are telling about that temperature
should not change, or the process parameter should not change with time. So, under that
situation what we can say, that grade square t, plus q dot gen by k is equal to zero. So,
this is an equation this is an equation which is called poisson's equation this equation
is called poisson's equation and when there is no heat generation, when there is no heat
generation that means q dot generation equal to zero, when then we have zero and at steady
state and at steady state grade square t equals to zero, this is called laplace equation.
In fact, these of the 2 equations which are the simplified form of the heat transfer situations,
in actually, i have told that grade square t, this equation is the general conducted
equation, but grade square t is equal to, when we write the equation like this grade
square t equal to 1 by alpha del t by del t, then or when we write the del t by del
t is equal to alpha grade square t. Actually, this equation is called Fourier field equation,
not the earlier one. This equation is the generalized equation,
generalized conduction equation which is applicable to all the coordinate system, but when we
say that, no heat generation term then del t by del t is equal to alpha grade square
t, this is called Fourier field equation. When there is heat generation term there,
but it is at steady state situation and is called poisson's equation. And when no heat
generation and steady state, then grade square t is equal to zero, this is called Laplace
equation.
Actually, if we see, initially i have told this equation we should say, that this whole
equation, this equation without this term, when this is not there, when this is equal
to zero, then this whole equation is called Fourier equation in rectangular coordinate,
or Fourier second law of heat conduction.
But, in general this whole equation is called the general heat conduction equation, which
we have seen for all the cases, like in case of cylindrical case coordinates also, we find
this is the general heat conduction equation, and in spherical coordinate system also we
find the general heat conduction equation. But, the Fourier field equation is this equation,
when the generation rate is zero, there is no heat generation within the body.
So, this is the way, now what we can say that whenever, we start with any coordinate system
for learning the conduction equation, what we understand is that we have to start with
the very we can start with the very best equation, that is that generalized conduction equation.
And then, depending upon the constants, depending upon the applications we can reduce this to
simpler forms, and apply for our problems. One thing i should say here
that, Laplace or the del term, or the grade square term, here is 1 by r square del, del
r of r square del, del r, plus 1 by r square sin theta, del del theta of sin theta, del
del theta, then plus 1 by r square sin square theta, del square by del phi square. So, this
is called the Laplace operator del square, in case of spherical coordinate system.
Now, we will try to see some, discuss some small problems related to this. Very the first
thing what we will say that, there is small thing that, what are the assumptions behind
this generalconduction equation whatever, we have developed? What are the assumptions,
main assumptions, or the major assumptions behind the derivation of the general heat
conduction equation? Yes, so we can understand that this is basically two things we have
assumed, 1st thing is that, k thermal conductivity this thermal conductivity has been assumed
to the independent of the coordinates, as well as temperature. So, k is particularly
independent of coordinate that we have assumed, and another important assumption is, that
we have assumed that heat generation is uniform within the volume element. Now, we will see
some other problems; let us suppose that we have this equation of general conduction equation
as general form we have found out. Now, if we say that the reduce,
So, what is the general form of equation we have found? We have found the general form
equation the grade square t, plus q dot gen by k is equal to 1 by alpha, del t by del
t. Now, the question is, now what happens to
this equation in Cartesian one dimensional
heat conduction equation in a plane wall at steady state, so what happens to 1 dimensional
heat conduction equation in a plane wall at steady state how this equation should be reduced,
so what we will do that? So, let us consider just any situation say, a plane wall, and
this is side one, and this is the side two, and let us consider this is an x direction.
So, we are telling one dimensional that means, the heat is conducted along x direction. So,
then what is happening is, one dimensional heat conduction situation in plane wall at
steady state without heat generation. So, what happens now, if we see that our basic
equation is this, grade square t plus q dot gen by k, is equal to 1 by alpha, del t by
del t. Now, what we will do is that it is told, that it is steady state, so once it
is steady state, so this term goes zero because it is steady state. And then, there is no
heat generation, so this term is zero because no heat generation, correct. Then what is
left is, so this is we can say that grade square t is equal to zero.
Now, the question is that if we remember, what is this grade square term, then we will
find out that del square t by del x square, plus del square t by del y square, plus del
square t by del z square equal to zero. Now, we are telling that it is a one dimensional
heat conduction that means, there is no heat transport along y and z direction, that means
there is no gradeient along y z direction, if so if there is no gradeient, because we
know that if there is a heat transfer as together in y direction, then q dot y has to be linked
with, is equal to minus some k into del t by del y. So, this part is, if there is no
conduction and then this has to be zero, and if this is zero so del square t by del y square
has to be zero. So, in along y direction there is nothing, at the same time del square t
by del z square equals to zero. So, there is the gradeient of temperature is zero, so
there second order derivatives are also getting zero, so for 1 dimensional heat conduction
this two was zero. So, that means what is happening is, then we are reducing this problem,
in to this kind of situation, where we can say that del t by del x, del square t by del
x square is equal to zero, and this implies if we do the integradeion twice, we will find
that t is equal to c 1 x plus c 2 ,where c 1 and c 2 are the integradeion constants.
Now, this c 1 and c 2, that means what we can say, from the basic equation under this
conditions being applied we are able to we are able to get the temperature profile we
are able to get the temperature profile within this, and we can see that the temperature
profile is a linear profile, say t is equal to c 1 x plus c 2, it is a linear temperature
profile that we are able to get. And we can find out the values of c 1 and c 2, based
on the conditions at, if we put conditions at station one and station two putting conditions
at stations 1 and stations 2 we can get c 1 and c 2 values. So, this is the way, we
can simplify starting from the conduction equation for any problem given to us, we can
simplify this for our understating, for getting a temperature profile exactly, the similar
situation can be applied, similar things can be applied for cylindrical coordinate system,
similar thing can be applied for Cartesian coordinate system this can be applied. And
in due course of time we will see that, how this equation can be applied and to get some
pictures, some understanding, our heat transfer phenomena that is taking place by conduction
mode, in a Cartesian coordinate system, in a rectangular coordinate system, or in a spherical
in a cylindrical coordinate system, or in a spherical coordinate system.
So, just now we will stop here with this.