Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 55
Exergy Analysis: Examples In our previous lectures, we were discussing
about reversible work, irreversibility and exergy, these 3 important concepts, and now,
we will solve some problems which will illustrate the use of these concepts. So, we’ll start with problem 8.1: a refrigerator
has a freezer at T F and cold space at T C from which energy is removed and rejected
to the ambient at T A as shown in the figure. Assuming that the rate of heat transfer from
the cold space Q dot C is same as that from the freezer Q dot F, find an expression for
minimum power into the heat pump; I mean, which is in this case is a refrigerator, not
a heat pump, but conceptually they work in the same way, you have to find out what is
the minimum power input. So, minimum power refers to a condition when
the—minimum power input refers to a condition when the irreversibilities are minimum, and
that means it is a reversible refrigerator. So, that is the concept that you have to understand.
Maximum power output will correspond to a energy producing device for a reversible heat
engine. Energy absorbing device will require minimum power input to run the device for
a reversible heat pump or a refrigerator. So, this when it is maximum and minimum you
have to understand carefully; when it is energy producing that energy is maximum because of
minimum losses, and when it is energy absorbing, that energy required is minimum to run the
show. So, let me make a schematic of this problem.
So, you have Q dot C with the temperature T C, W dot, Q dot A, Q dot F. So, this is
a cyclic device which is a refrigerator in this case and it is—most important is it
is reversible. Okay? So, what are the basic equations that you need to apply for a reversible
refrigerator in this case? One is cyclic integral of heat equal to cyclic integral of work,
which is the 1st law. So, let us apply the fundamental laws: cyclic
integral of heat equal to cyclic integral of work. So, cyclic integral of heat for this
device, so, we will just right in terms of the rate, so, cyclic integral of the rate
of heat transfer is same as cyclic integral of the rate of work. So, rate of heat transfer
is Q dot C plus Q dot F minus Q dot A, right? This is cyclic integral of heat. Cyclic integral
of work is minus W dot, as per this figure. 2nd law—2nd law doesn’t; see, 1st law
concerns about work, 2nd law concerns about only heat transfer because work is a high
grade energy, it doesn’t participate in entropy transport. So, 2nd law: you have cyclic
integral of del Q over T is equal to 0, for a reversible cycle, right? So, this is Q dot
C divided by T C plus Q dot F divided by T F minus Q dot A by T A is equal to 0. Look
carefully that we have taken the temperatures not of the system boundaries here, but the
reservoirs here. In this way, we have accounted for both internal and external part of the
process, whatever is internal plus whatever is external and ensured that it is both internally
and externally reversible. Okay? So, now, it’s a matter of—so, what is
given in the problem, let us see. You—it is given what is Q dot F and Q dot C, but
Q dot A is not given, so, let us eliminate Q dot A. So, Q dot A from here you can write
Q dot C into T A by T C plus Q dot F into T A by T F. So, then, from this equation,
minus W dot is equal to Q dot C into 1 minus T A by T C plus Q dot F into 1 minus T A by
T F. This is the final expression. So, in this expression, you can substitute Q dot
C is 3 kilowatt; T A, T F, T C all are given, you have to convert those into Kelvin and
just substitute, and this will be 0.504 kilowatt for this.
See, solving a problem is not about getting the final answer, there is also an insight
that needs to be developed. So, let us try to develop an insight of this expression.
If you recall that here there is a system to which you have two heat transfers effectively
done and you are trying to find out an expression for the reversible work. So, you can clearly
see that this is exergy associated with the heat transfer Q dot C, this is exergy associated
with the heat transfer Q dot F. So, this essentially is an expression that gives the reversible
work through exergy balance. Because it’s a reversible system, the entire exergy that
is supplied is equivalent to the reversible work, there is no irreversibility, okay?
So, you can see that the same problem can be perceived from exergy balance and also
directly from a consequence of the 1st law and the 2nd law. Let us go to the next problem,
let me erase this. A steam turbine receives steam at 6 MPa, 800
degree centigrade and it has a heat loss of 49.7 kilo Joule per kg and isentropic efficiency
of 90 percent—I have not yet introduced what is isentropic efficiency. So, I will
do that in context of this problem and then we will discuss further. For an exit pressure
of 15 kilo Pascal and surroundings at 20 degree centigrade, find the actual work and irreversible
work between the inlet and exit. So, this is a very classical situation related to analysis
of steam turbine. I will first draw a schematic of the situation in the board. So, there is a steam turbine. It has a inlet
state i and the exit state e. The inlet state is 6 MPa, 800 degree centigrade. So, if you
draw this in a T-s diagram…sometimes there is a temptation of drawing it in a enthalpy
entropy diagram, h s diagram, the reason is that for 1st law based calculations, you require
enthalpy, it will not really require temperature, so, if there is a diagram that directly gives
enthalpy versus entropy, then that is very convenient. That, in industrial application
perspective, is also known as Mollier chart or Mollier diagram.
Here, we are more concerned about the fundamentals of thermodynamics, so, we will not get into
the use of Mollier chart, but we will—I am just telling that as a passing note because
you may be requiring the h s chart for solving problems when you are participating in more
applied thermodynamics courses; this is a fundamental thermodynamics course.
So, you have the liquid vapour dome. The turbine is grossly specified by the following. So,
in one, you have a pressure that—at which the steam enters
the turbine; this is, say, given by p i and the pressure at which the steam exits from
the turbine. Now, by isentropic efficiency, we want to compare the actual work done by
the turbine vis-a-vis the reversible adiabatic work, that is, if the state e would have been
such that from i to e, it would have been a reversible adiabatic process or an isentropic
process, then what would be the work? So, if it would have been a reversible adiabatic
process, let us call this as e s; this is not the actual state e, but this is the reversible
adiabatic hypothetical state e, so, e s. The actual process will take the point e somewhere
here. Now, the question is, where is the guarantee that this point e will be in the right side
of e s or it could also be in the left side of e s? Okay? So, let us look into that more
careful. So, between i to e s, there is no—so, you
can decompose the change of state from i to e as from i to e s and then from e s to e,
just break it into 2 parts. From i to e s, there is no change in entropy, because it
is a reversible adiabatic process; from e s to e, you can write s e minus s e s is equal
to some q by T plus entropy generation, right? This is what you can write. So, the question
is that this is always positive, right? This could be plus, minus or 0.
So, the net effect could be that this one is plus; when it is trivially plus? It is
trivially plus that this is the case when the turbine is adiabatic, because if—if
the turbine is adiabatic, this is 0, then entropy generation is always positive, so,
e goes to the right side. This is what most commonly happens in industrial practice because
for all practical purposes, the turbine can be treated adiabatic in most of the circumstances.
However, there is no guarantee to that. You could have this Q plus, minus or 0, so, the
net s e minus s e s could be plus, minus or 0. So, technically, this e could be in the
right of e s, it could even coincide with e s and it could be in the left of e s. That
must be conceptually understood. So, the very fact that we have in most of the practical
cases e to the right side of e s should not develop a prejudice in our mind that e will
always be in the right of e s, it depends on the nature of heat transfer across the
turbine, that is what it—it is governed by, because other than heat transfer, the
entropy generation part, that is trivially positive part. Okay.
So, now, coming back to this problem, let us see what is defined—what is given. State
e is 15 kPa which is same as state e s. There is no difference in pressure between e and
e s, so, this is 15 kPa. So, now, if you apply the 1st law for per unit mass flow rate across
the turbine, so, you have Q plus h i—say, 1st law we apply for i to e s. In that case,
the reversible adiabatic Q s is 0, because it is adiabatic. So, I am completing the expression.
So, I am writing this as W s; W s corresponds to reversible adiabatic per unit work or isentropic
work. So, isentropic work will necessarily be h
i minus h e s; this is h e s. So, how do you know what is the state e s? Very simple. The
state e s can be defined by 2 properties: one is s e s is equal to s i and s i is known
from steam table; at this data, you find out what is s i. And p e s
is equal to what? P e s equal to 15 kPa, p e. Okay?
So, s e s is equal to s i and p e s equal to 15 kPa, this will give you what is state
e s; it will be in the 2-phase region typically. So, you can find out the quality by setting
s e s is equal to 1 minus x e s into s f s plus x e s into s g s, where s f and s g are
s f and s g at 15 kPa, okay? So, once you calculate this what is s e s, then you can
calculate what is h e s also by using this interpolation formula for enthalpy. So, h
e s is equal to 1 minus x e s into h f s plus x e s into s g s. Okay?
So, to summarise, by this calculation, you know what is reversible adiabatic work. What
is the actual work? For actual work, you have to use the 1st law between the states i to
e. Now, this turbine is a special turbine where heat transfer is not neglected. So,
Q is there plus h i is equal to h e plus W, right? This Q is given. This Q is—it has
a heat loss of 49.7 kilo Joule per kg. Okay? So, this will tell you—and what is the exit
state? The exit state is governed by what is the actual work. Now, you know that the
actual work by the isentropic or reversible adiabatic work, this is defined as the isentropic
efficiency. This is the definition, please make a note of this. I will talk about the
definition of isentropic efficiency of a work absorbing device also after I complete this
problem. This is the definition of isentropic efficiency of a work producing device, where
the actual work is less than the reversible adiabatic work.
So, then, because you know what is W s by substituting in this formula, and isentropic
efficiency is given as 90 percent, so, from this, you know what is W. You substitute that
W here. Once you substitute that W here, you get what is h e. So, what is asked is what
is the reversible work between the inlet and exit? So, to understand what is the reversible
work between the inlet and the exit, you need to also know the entropy of the exit, not
just enthalpy, so, you have to specify state e. How is state e specified? It is specified
by p e and h e. Okay? So, from here, you can get what is s e from table.
So, what is the reversible work? H i minus T 0 s i minus h e minus T 0 s e. Okay? This
is the expression that we have derived; this is a reversible work per unit mass because
of change in state from i to e. So, if you know these values, T 0 is given, T 0 is 20
degree centigrade, that is 293.15 Kelvin. So, this will be 1636.8 kilo Joule per kg.
I just want to make a final note before concluding the solution of this problem.
The final note is that this definition of isentropic efficiency is true for a work producing
device only. For a work absorbing device like a compressor, the reversible adiabatic work
is the minimum work, not the maximum. So, for isentropic efficiency of compressor, it
will be the reversible adiabatic work input in the top, in the numerator, divided by the
actual work input which—I mean, both are in terms of magnitude at the bottom. So, for compressor, let me spell it out…mod
of work input by mod of reversible adiabatic work. The spirit of reversing the expression
is that efficiency we always have an imagination it cannot be more than 100 percent, but it’s
always a comparison between reversible adiabatic with the actual case; which one will be there
at the numerator and denominator depends on which one is more and which one is less. Okay?
So, let us solve another problem, problem 8.3. So, problem 8.3: a 2 kg piece of iron is heated
from room temperature 25 degree centigrade to 400 degree centigrade by a heat source
at 600 degree centigrade. What is the irreversibility in the process? So, this is a very straight
forward typical problem, but let us work it out. So, you have a mass of 2 kg of iron. Your
T 1 is 273.15 plus 25 Kelvin, T 2 is 273.15 plus 25 Kelvin, and the temperature of the
ambient is 273.15 plus 600 Kelvin. So, for knowing the irreversibility, you need to know
what is the entropy generation. To know the entropy generation, you have to know what
is the heat transfer. To know the heat transfer, you have to apply the 1st law. So, we’ll
start with the 1st law for the copper block—or iron block. Q 12 is equal to U 2 minus U 1
plus W 12; copper block doesn’t do any work. So, the heat transfer is simply—and U 2
minus U 1 for a solid object like copper is m C into T 2 minus T 1. The specific heat
of copper, C, is 0.42 kilo Joule per kg Kelvin; this data is required for solving your problem.
So, you get what is Q 12. Then you apply the 2nd law: S 2 minus S 1 is equal to Q 12 by
T 0 plus entropy generation. Okay? So, in this case, we have to clearly see whether—so,
this—here we have to make a decision. What is the decision? The decision is that is this
the ambient temperature or is this the heat source temperature different from the ambient
temperature, right? So, here, it appears that it is not ambient temperature; it is the heat
source temperature at 600 degree centigrade. So, the irreversibility that you calculate,
the reference temperature is always important. You could calculate irreversibility by taking
the heat source as the reference temperature, but that is called as immediate surrounding.
Normal tradition is to calculate irreversibility with respect to the actual ambient temperature
as the reference temperature. So, if we do not choose this to be the ambient
temperature; it is not clear in this problem statement that this is not the ambient temperature.
So, in that case, the heat transfer is taking place with this thermal reservoir not T 0,
right? So, the entropy generation, this will appear in the heat transfer term and S 2 minus
S 1 is mC ln T 2 by T 1. So, this—in this expression, everything is known except entropy
generation which is found out. The irreversibility is simply T 0 into entropy
generation. Here, you have to make a decision that what is this T 0, right? Normally, this
is called as exergy reference temperature which is the ambient temperature. So, in this
problem, the intuition is that the ambient temperature is 25 degree centigrade and not
the 600 degree centigrade, right? So, then, if you put that as T 0, so, this will give
you the irreversibility and the irreversibility in this case is 96.4 kilo Joule, okay?
So, let us stop here for the time being. We will continue with more problem-solving in
the next lecture.
