Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 41
Entropy Balance for Reversible and Irreversible Processes In the previous lecture, we were discussing
about the change in entropy in various processes adiabatic reversible adiabatic. So, we will generalize that a little bit and
we will discuss about you know a reversible polytropic process. A reversible polytropic
process is defined by a process, governed by the following relationship pV to the power
n equal to constant when n; I give you some examples when n is equal to 1, it is reversible
isothermal process of an ideal gas that is the so called Boyle’s law.
When n is equal to C p by C v that is reversible adiabatic process of an ideal gas, and when
n equal to 0 it is a constant pressure or isomeric reversible isomeric process. So,
if you now try to draw a p-V diagram describing all these processes. So, when n is equal to,
first of all let us take the limit two limits of constant pressure and constant volume there
they are easy to draw. So, constant volume is this one, let us change the axis. So, normally
and this is very important normally we draw volume in the x axis and pressure in the y
axis. The reason is that volume is that independent variable that we can vary in experiments,
independently varying pressure is not something which is you know easy to do.
So, if now we draw constant volume process or constant pressure process let us first
draw. So, this is p equal to constant. So, this is n equal to 0. Similarly, V equal to
constant you can figure out; n is equal to plus minus infinity, V equal to constant then
we will consider reversible isothermal and reversible adiabatic process. So, just let
us try to get the slope of this diagram. So, just take log of both sides, you have l n
p plus n l n V is equal to l n C. So, now if you differentiate both sides
so you have d p by; so you have dp dV is equal to minus n p by V right. Pressure and volume
they are both positive, n is positive right. I mean when we are talking n could be actually
anything, but when we are talking about comparison between isothermal for an ideal gas and adiabatic
for an ideal gas in one case n value is 1 and another case n value is C p by C v both
of those are positive those examples will have n equal to positive.
So; that means, in those examples first of all dp dV is negative; that means, it is quite
obvious from physical intuition that if you increase volume the pressure will reduce for
a given system. So, the slope of this is negative not positive ok; now if n is higher then the
magnitude of the slope is more right. So, let us say, let us consider example of air.
So, C p by C v for air if you consider calorically perfect gas it is about 1.4 and for reversible
isothermal n is equal to 1. So, for reversible adiabatic definitely the value of n should
be greater than 1, because C p is always greater than C v. So, if n is greater than one then
dp dV magnitude for reversible adiabatic process of an ideal gas is greater than the magnitude
of that for reversible isothermal process of an ideal gas.
In other words reversible adiabatic process in this p-V diagram for an ideal gas is steeper
than the reversible isothermal process in the same p-V diagram. So, we can draw may
be, and in both cases. So, this is n is equal to C p by C v and this is n is equal to 1
ok. So, these are these are just some examples you could construct more examples; I would
like to now instead of p-V diagram draw T-s diagram for some of these processes. So, that
you know it will give you a nice visualization of the same processes in the T-s diagram. So, first of all isothermal process, reversible
isothermal process is temperature equal to constant. So, n equal to 1, is this one.
What will be the vertical line in this? s equal to constant that is reversible adiabatic
process n equal to C p by C v, then the two processes that are left are constant pressure
and constant volume. So, let us see which one is steeper in the T-s plane. So, remember
we are considering reversible polytropic process with specific example of ideal gas. So, Tds
is equal to dU plus p d, this is such a useful relationship that you can use for any process
diagram to you know explain the diagram. So, for constant volume this is 0 for constant
volume and for ideal gas this is C v dT; m C v d T. So, d S is equal to m C v. So, dS
dT is equal to m C v by T for constant volume right.
Similarly, for constant pressure we will use the other Tds relationship.
Sir just one thing I , while writing equal to m C v d T you told that it is an ideal
gas. It is an ideal gas, yes.
dU is sensibility even if it is not an ideal gas because it is a constant .
No, dU is m C v d T it is true only for ideal gas, because the definition of dU is based
on so if you if you write. So, the, what is the definition of C v? So, ok; so, what you
are saying is for if the volume is constant yes, if the volume is constant then it is
true. If the volume is constant that you can then you can write dU equal to m C v d T for
any substance, but if the volume is not constant. So, this x, this equation we are writing for
any substance you know and then you are applying it for a constant volume process.
So, before applying it to a constant volume process if we write this it is better to commit
that we are writing for ideal gas, if we from the beginning commit that it is constant volume
process then it can be generalized ok. So, then
it is 0 for constant pressure process. So, dS dT for constant actually this should be
partial derivates you know, I mean I have written ordinary derivates just for your simplicity
in understanding. But because you have kept volume as constant; that means, you know you
are differentiating S with respect to T with a constraint v equal to constant.
So, let us be technically a bit correct d H is m C p d T; so, m C p by T. In fact, for
constant volume and constant pressure processes you can use this for any substance what you
know this we are comparing for ideal gases. So, ideal gas has a special case will also
work fine. So, now out of these two first of all these
are positive or negative these slopes? These slopes are positive because C p C v are positive
and T is absolute temperature which is positive right. So, absolute temperature is positive.
So, there are many nice interpretations about it, recall this just I am giving as a side
note the efficiency of a Carnot cycle is 1 minus T L by T H right, this we derived in
one of our previous lectures. Now, by second law of thermodynamics the efficiency
of Carnot cycle should be less than 100 percent Kelvin planck statement. So, if that would
be less than 100 percent T L cannot be less than 0. So, that is why the concept of absolute
temperature comes, it cannot have a negative value. So, attaining an absolute 0 is a physical
impossibility may be you can go to very close to absolute 0, but you cannot attain absolute
0 here. Because if you run a hypothetical Carnot cycle with T L is absolute 0 then its
efficiency will be 100 percent and that will violate the Kelvin planck’s statement.
So, this is just different approach of looking into the fact that why second law of thermodynamics
does not allow you to attain absolute 0 ok. So, the moral of the story which we want to
use here is that the temperature is positive. So, both are positive, but you have because
C p is greater than C v del S del T at constant pressure is greater than del S del T at constant
volume. In this diagram what we require is del T del S, not del S del T because its T-s
diagram not s T diagram. So, in terms of del T del S you will have the del T del S at constant
volume greater than del T del S at constant pressure; that means, constant volume line
is steeper than the constant pressure line. So, constant volume is that yellow line and
constant pressure is the pink line. So, this is v equal to constant this is p equal to
constant ok. So, these are you know nice physics visualizations of very common processes that
we come across in thermodynamics ok. Now we will go to the next agenda item which is change
in entropy in an irreversible process. So, when we say change in entropy in an irrreversible
process what we essentially intent to mean is as follows. So, let us consider say at
T-s diagram you have 1 to 2 are reversible path, from 2 to 1 you construct a reversible
path. So, this is reversible A, this is reversible B, but you also construct an alternative irreversible
path which is path C. From the Clausius inequality for reversible
cycle cyclic integral of del Q by T equal to 0 ; that means, 1 to 2 del Q by T via path
A and 2 to 1. So, let us write it reversible by path B this is 0 for irreversible cycle
an any if one of the processes in the cycle becomes irreversible it becomes completely
irreversible cycle. So, cyclic integral of del Q by T is less than 0 so; that means,
del Q by T from 1 A to 2A by reversible path plus irreversible path, this is less than
0 right. So, now if you combine this two so you can
write that from these two combinations del Q by T integral along irreversible path 2
C to 1 C must be less than del Q by T integral from. So, this is B right, I have mistakenly
written as A; 2 B as to 1 B this is reversible because this is reversible, we can write this
thing as equal to d S between 2 to 1. The formula del Q by T equal to d S we can apply
only for reversible process and because it is between 2 to 1 and it does not depend on
path the same thing we can say that this is d S from 2 B to 1B right because it is no
more path dependent. So, to summarize that we can write; so, this
even if we write in place of C that works, in place of B if you write C that also works.
The reason is between point 2 to 1 it does not matter whether its path B or path C, it
is the change in property. So, to summarize we have C is any arbitrary path. So, we can replace,
see in working with mathematical expressions we are more comfortable with equality rather
than inequality. So, we can replace this inequality with an equality by writing that d S for any
arbitrary path is equal to del Q by T plus a positive number, this is called as entropy
generation. So, this is greater than 0 it is 0 when the equality holds; that means,
it is 0 for reversible process. What is this T? This T is the temperature
of the system boundary across which this del Q is interpreted. Now what is interpreted
as the system it depends on what is your domain of interest just like in mechanics what is
your free body diagram it depends on you are representing forces on what, may be one part
or the entire system. So, here also what is your system it depends on whether it is the
machine itself or it is only the thermal reservoir or it is a combination of the machine and
the thermal reservoir based on that this T has to be appropriately substituted because
it is the temperature of that system boundary for which you are writing this equation ok.
Sir, entropy generation is the path function right, it depends on which path you .
Entropy generation is the path function yes. I have written d S.
So, you can write just write because you know this is, I mean it is a not the property.
So, it is just you know just make sure that these are these are differences, but this
is the path function you can better write this as integral of, I mean that makes the
real sense because differentially you do not write it you write this one, right. So, this is S 2 minus S 1. So, writing this
as d S gen is little bit uncomfortable because you know del Q depends on the path. So, entropy
generation depends on the path. In fact, it is very practical because more irreversible
the path is more is the entropy generation. So, entropy generation is a physical measure
of the extent of reversibility, more irreversibility it is more irreversible the process is more
is the entropy generation. So, here we have introduced entropy generation
as or term which is positive just to give the flavor of a mathematical equality instead
of a mathematical inequality, but let us try to give a physical interpretation to this.
So, we will discuss something which is called as lost work. So, lost work is a concept which if expanded
in English will be lost opportunity of doing work ok. So, now, when we say lost opportunity
of doing work, what we mean? The concept is something like this in a practical process
we always have an opportunity of deriving work because of a change in thermodynamics
state. Now, if the process is reversible that work derived is maximum, if it is irreversible
a part of the potential of deriving work out of the change in thermodynamics state is lost
or destroyed, we will prove that how it happens. So, let us assume, let us consider two examples
one is you have a piston cylinder where the gas is moving in a quasi equilibrium process.
So, if the gas if the gas is expanding in a quasi equilibrium process because of the
heat transfer then you have maximum work done which is p d V right. Let
us consider another very interesting arrangement same gas, but now partitioned in two places,
on one side there is gas and another side there is vacuum right. So, before the process
starts both of these hold the same potential of doing work, but if you now puncture this
membrane then this gas fills up this entire volume and no work is done.
So, here delta w is equal to 0 because why no work is done, because it is free expansion.
So, I mean this is this is even a very philosophical way of looking into life, if we do hard work
with resistance then only you know we utilize our potential. If we do not do hard work and
do not face any resistance and have a happy go life then we do not utilize our potential.
So, in this case the potential full p d V is utilized in this case it is not utilized
at all, reality in life is something in between these two. So, reality is that you have p
d V is equal to del W plus delta lost work ok. So, this is any case in between the work
is between p d V and 0. So, p d V is maximum the actual is less than that and the balance
is a lost opportunity of doing work right. So, now you can write T d s is equal to dU
plus del W sorry p d, this you can use for any process right
then for p d V you can write del W plus del lost work and from first law you know that
del Q is equal to dU plus del W. So, you can write T d s is equal to del Q plus delta lost
work. So, d S is equal to del Q by T plus delta lost work by T compare this with d S
is equal to delta Q by T plus delta S generation. So, this is nothing, but entropy generation
because entropy generation is positive the lost work is also positive and this term is
0 if it is reversible ; that means, with every irreversible process we have a lost opportunity
of doing work, this is a physical interpretation of the entropy generation which is associated
with the lost opportunity of doing work. Let us stop here today we will continue with discussions
on entropy and possibly some problem solving in the next lecture.
Thank you very much.
