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So, to conclude our recapitulation on thermodynamics, finally we will discuss few more important
relations, that are very useful in high speed aerodynamics, and then we will conclude our
revisit of thermodynamics, but as we mentioned many times repeatedly, that this is just a
recapitulation, and for more details, a book on thermodynamics must be consulted, or you
may consult your earlier notes on this course. Now as we have stated that, the specific heats
are very useful quantity in thermodynamics.
Now, we will express these specific heats, using entropy. The specific heats we will
discuss using entropy, in terms of entropy. We have the specific heat as, delta Q by delta
T, and this is delta Q can be expressed as T delta S, reversible. If we consider this
heat is added reversibly, then delta Q is T delta S by delta T.
Or we can write that T, d S, d T. Hence the two principle specific heats C p can be written
as T, d S, d T at constant pressure, and the other one C v is T, d S, d T at constant specific
volume. Now, let us take entropy as a function of temperature and d S pacific volume. We
can do this since we have, mentioned repeatedly that for a pure substance, any state function
can be expressed as function of two other parameters. So in this case you have the expressed
entropy as function of temperature and specific volume. And this then gives us that, the change
in entropy d S is, d S d T at constant volume into d T plus d S d V at constant temperature
into d V. We then can write, d S d T at constant pressure, is d S d T at constant volume, plus
d S d C V at constant temperature, into d v d T at constant pressure. Now, from the
definition of C p, this is what is C p by T, and this is C V by T. So we have this of
course, multiplying by T is C p equal to C v plus T into. Now, from Maxwell’s relation,
using one of the Maxwell’s relation d S, d v at constant temperature is d p d T at
constant volume. We can write T d p d T at constant volume into d v d T at constant pressure.
For you have used, using Maxwell relation d S, d v at constant temperature equal to
d p d T at constant volume. So we get the difference between the two specific heat,
C p minus C v equal to T into d p d T v. The right hand side can be really obtained, if
the thermal equation of state is known.
So, the right hand can be evaluated, from equation of state. We can very easily see,
that if the equation of state is perfect gas. For perfect gas, and we then have C p minus
C v equal to R, a very well known relation, where R is the specific gas constant, of the
perfect gas, which we know that if we consider air, R is usually taken as 287 joule per k
g Kelvin. Let us look to this equation from one more aspect. In this equation we have,
on the right hand side d v d T at constant pressure. Now d v d T at constant pressure,
that is the change in specific volume with temperature. Keeping the pressure constant,
is related to coefficient of thermal expansion. So d v d T of p, present on the right hand side,
is measure of thermal expansion, change in volume due to change in temperature, with
the pressure held constant. So this per unit volume is known as the coefficient of thermal
expansion. So, d v d T at constant pressure per unit volume is the coefficient of thermal
expansion.
So we have coefficient of thermal expansion, we denote it by beta, it is 1 by v, d v d
T at constant pressure, which using Maxwell’s relation becomes minus 1 by v, d S d p at constant temperature. And, using
the preferred fluid dynamic parameter, this is minus rho d S d p at constant temperature,
and in fluid dynamics or the action of gravity, is very important, this term plays a major
role. So the difference in specific heat, now also can be expressed in terms of this coefficient
of thermal expansion, which we can see again that. So C v is T into, that first term d p d T at
constant v. Now let us write once again, T d p d T at constant volume, into d v d T at
constant pressure. We write this as T, using a thermodynamic identity d p d T at constant
v is d p d v at constant temperature, into d v d T at constant pressure. This is a thermodynamic
identity, which says the product of three derivative is minus one, and then. So we replace
this term by this, and then keep the final term d v d T at constant pressure.
So this then becomes, C p minus C v equal to minus T into, d p d v at constant temperature,
into d v d T at constant pressure square. Or in terms of, that is thermal expansion,
this becomes minus beta into v T, d p d v T into, or one of the d v d T I have expressed
using beta v, and which I have said equal to R, if. we see that C p minus C v equal
to R. the relation very well known to you is not applicable for other gases. It is applicable
only for perfect gas. Now, the changing entropy is very important, because as we have seen
from our discussion in second law of thermodynamics, that the change in entropy is decides, whether
a particular process is possible or not. We have seen that in a reversible adiabatic change,
there will be no change in entropy. So in a reversible process, entropy cannot
decrease. So now let us try to find the change in entropy, in terms of other measurable quantities.
And let us say we want to see the change in entropy, in terms of temperature and pressure.
Of course we can express, as you have already done, entropy is a function of temperature
and volume. Entropy is a function of temperature and pressure. Entropy is a function of pressure
and volume and so on. Now, this gives d S d T at constant pressure into d T plus d S d p at constant temperature into d p.
Now this is, by definition is C p by T, so we have C p by T d T, and this d S d p T,
we can express using Maxwell’s law d S d p to be minus d v d T p into p.
So, we write
d S is C p d T by T minus d v d T at constant pressure. The d v d T at constant pressure
can be again expressed in terms of that coefficient of thermal expansion beta, which is beta into
v. So, this can be expressed as minus beta v d p. Also or in terms of preferred fluid
dynamic parameter, minus beta d p by rho. Multiply this by T, we have T d S equal to
C p d T minus beta T d p by rho, or C p d T minus. And from combine first and second
law, this becomes d e plus p d v. Now often, it is necessary to know the dominance of these two term, which
of these term is dominant, that is which contributes more to the change in entropy, and that can
be expressed. Compare contributions to, you see that the entropy changed, because of this
term as well as this term, and you can compare their relative contribution, and that is the
contribution is C p d T, and the other term contribution is beta v t d p, of course this
term is negative. The beta v T, using this relationship we can
write beta v T in terms of C p minus C v. So this can be written as C p by C p minus C v, into d p d v at temperature,
into d v d T at constant pressure, into d T d p. Dividing this by C v, this becomes gamma by gamma minus one, where gamma is C
p by C v, the ratio of specific heats, and you can also write this expression d v d T
at constant pressure into d T or. So we see that the contribution depends on, the ratio
of the volume change due to temperature, and volume change due to pressure. So the ratio
of the volume change due to temperature, and the volume change due to pressure. So it is
again, so the dominance of the contribution depends on.
So, the change in specific volume due to temperature, and change in specific volume due to pressure,
that is the thermal effect, and for is known as the compressibility effect. So in a particular
process we can check, which of these two terms are, more important than the other as far
as the change in entropy is concerned. We have until now expressed the equation of state,
in terms of say the pressure as a function of specific volume and temperature or internal
energy, as a function of volume and temperature. Now, since any two property can also be expressed
in terms of any other two functions, so the equation of state can also be expressed in
terms of other variables, and particularly if the equation of states are expressed in
terms of entropy, then they are called canonical equation of state.
So these are called the canonical equation of states. E expressed as function of entropy
and specific volume or density, and H expressed as a function of entropy and pressure. Now
this change in entropy will, now further extend the relations, that is some of these derivatives
are can be obtained, if the equation of state is known, and we can express the entropy change
in or more measurable form, if you know the equation of state. See as an example for perfect
gas. So we will consider now change in entropy in perfect gas. Once again let us write entropy is a function of volume
and temperature, so that this is by definition C v by T, so this is C v d T by T. And d S
d v by T using Maxwell’s relation this becomes d p d T at constant volume d v. Now for a
perfect gas, d p d T at constant volume can very easily be evaluated, which is simply
R by v. So this becomes C v d T by T, and d p d T by v is simply R by v d v.
Now this can be integrated between, integrating between initial state one. We have s 2 minus
s 1, from 1 to 2 C v d T by T. To complete this integration we have to know, the relationship
between the specific volume and specific gas constant as a function of temperature, and
volume. However, for a clerical perfect gas, we know that this C p C v R are all constant,
and then we can compute this integration. Similarly, we consider entropy function of
temperature and pressure, and this is by definition is C p by T, so this is C p into d T by T.
And again using a Maxwell’s relation, this becomes minus d v d T at constant pressure
into d p, where this term is replaced by this using Maxwell’s relation, and then for perfect
gas this becomes, and once again integrating similarly,.
Now, let us consider the process is isentropic. We have
C p log, if we consider this relation C p log T 2 by T 1 equal to R log p 2 by p 1.
And this gives p 2 by p 1 equal to T 2 by T 1 to the power C p by R, and if we consider
this relationship, if we consider the earlier relationships, then we have R log v 2 by v
1 equal to minus C v log T 2 by T 1, or in terms of.
So for isentropic processes we can, hence for isentropic processes we have. We have
p 2 p 1 equal to T 2 by T 1 to the power gamma by gamma minus one, equal to rho 2 by rho
1 to the power gamma. And this can be used as isentropic equation of state. Any further
relation if we need, we will discuss at appropriate moment, but with this we conclude our brief
review of thermodynamics. So, let us see what we have done in our introductory lectures
first of all, we have considered what are the essential features of high speed aerodynamics,
and there, where we discussed that, when speeds are large the change in pressure will also
be large, which will cause change in density. And as a result there will be many new phenomena,
and even there will be some phenomena which will need relook into, which will now have
new interpretation, and there will be some results which are, completely contrary to
the results, that you have come across in incompressible flow, all because that the
pressure changes here are very large, and density can no longer be considered independent
of pressure. Also we discussed that at very high temperature, at very high speed.
The flow is associated with large change in kinetic energy, which is comparable to the
internal energy, and internal energy and kinetic energy are coupled, and transform from one
to the other will change the other considerably. And, this implied that the thermodynamics
is an inherent part of high speed aerodynamics. High speed aerodynamically problem cannot
be handled, without thermo dynamical consideration. And due to this, we thought brief review of
the thermodynamically concepts, and certain thermodynamically relations are necessary
and useful. And we have discussed, the basic thermo dynamical relationship, we have discussed
about first law, we have discussed about second law. We have discussed the Maxwell’s equation
of state, we have discussed the concept of internal energy entropy. The equation of state,
even what change the equation of state needs in case of a real gas. Also we have defined
the coefficient of compressibility and compressibility factor, using the thermo dynamical relationship.
And we have seen that the perfect gases are those gases, for which the compressibility
factor is one, and we mentioned that air under ordinary conditions can be considered as a
perfect gas, and many of these equations can be simplified. The equation of state can expressed,
in simple mathematical relation, given by p rho equal to R T, sorry p equal to rho R
T or p v equal to R T, for v the specific volume. And also we have obtained certain
relationship for perfect gas. We have also discussed that this perfect gas assumption
for air is, quite valid as long as the temperature does not increase considerably, where air
must be considered the mixture of gases with their with chemical and thermal reactions,
need to be considered. Also again this perfect gas laws are not valid, if the pressure is
very high and temperature is very low. So that the gases deviates from the usual low
density approximation, and in that case those real gas effects need to be considered.
Particularly that all gases liquefies that effect need to be considered. Well in high
temperature this effect is not necessary, but the effect that are to be considered are,
the chemical and thermal reactions, excitation of the vibration mode and many chemical reactions,
but as you have mentioned at that time, and again repeating that, in this course of high
speed aerodynamics. We will not consider this high temperature very high speed flows, and
consequently the thermodynamics is also not discussed to consider those effects; that
is we have not considered the equations for mixtures. What we have considered, is thermodynamics
of a pure substance, where any state variable can be or state function can be expressed
as, any other two state, as a function of any other two state variables.
Using this we have found that, for isentropic processes that equation of state can be also
expressed in an alternative form, which is given here, that T as rho to the power gamma
p is proportional to rho to the power gamma, or p is proportional to rho to the power gamma
or p proportional to T to the power gamma by gamma minus one, or T proportional to rho
to the power one by gamma minus one, rho proportional to T to the power one by. In particular we
have, while discussing the entropy, we discussed certain points that, that is the main role
of the second law of thermodynamics, which clearly states, whether a particular process
will be possible or not. So the second law of thermodynamics is excludes possibility,
in addition to defining an absolute scale of temperature, and the concept of entropy.
In so discussing we have also mentioned about that, heat is a degraded form of energy, because
the heat cannot entirely become converted into work, while useful work can be converted
completely into heat. And in our subsequent discussion on high speed
aerodynamics, we will see that the replication of second law will come quite often, to check
whether a particular process or particular result that we will be obtaining, are really
possible in nature or not. As an example at this stage, we will later on see that, the
conservation laws predicts, that it is possible to have a shock, which will change the flow
from subsonic to supersonic, or from supersonic to subsonic. However only, the none of the
conservation laws, that is the conservation of energy, momentum or mass, can definitely
tell us whether the processes are feasible or not, only with the application of second
law we will be seeing that, the shock really can occur, only if the flow is supersonic.
So even though, second law does not give answers and governing equation as the first law.
The first law of course, gives us the most important governing equation being the conservation
of energy. It supplements the conservation of mass, and conservation of momentum, which
are regularly used in incompressible flow, but the second law does not give us any such,
new conservation equation, but it decides whether the results of the governing equations
or the conservation laws, are really possible in nature. So with this, we will conclude
our review and next we will start discussion on the subject of high speed aerodynamics
or gas dynamics. Of course, we will start with the simplest possible flow problem, which
is of course, the one dimensional flow. So from next lecture onwards
we will discuss about, one dimensional compressible flows, or
one dimensional gas dynamics. Thank you.