Concepts of Thermodynamics

Prof. Suman Chakraborty

Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur

Lecture – 56
Exergy Analysis: Examples (Contd.)

We continue with problem-solving on reversibility and Exergy. So, the next problem is
problem 8.4.

(Refer Slide Time: 00:27)

A 1 kg block of copper at 350 ºC is quenched in a 10 kg oil bath initially at an ambient

temperature of 20 ºC. Calculate the final uniform temperature, assuming no external heat
transfer and change in availability of the system, copper and oil. So, availability is
exergy, it is the same.
(Refer Slide Time: 01:01)

So, the situation is you have a copper block in oil and copper plus oil is your system. For
solving this problem, you also require the specific heat of copper (CCu) which is 0.42
kJ/kg and specific heat of oil (Coil) which is 1.8 kJ/kg. So, the strategy of solving this
problem is again like you need to calculate the change in availability; for that, you
require the entropies, and for entropy transport, you require the heat transfer.

So, let us calculate the 1st law, or let us calculate the final temperature concerning the 1st
law:

1st law: Q12 = {mCuCCu(T2 – T1,Cu) + moilCoil(T2 – T1,oil)} + W12

= {U2 - U1} + W12

W12 = 0. Heat transfer is also 0.

Q12 = mCuCCu(T2 – T1,Cu) + moilCoil(T2 – T1,oil) + W12 = U2 - U1 + W12

Your system is this one (dotted outline), and this system doesn’t interact with the
surroundings through any heat transfer or any work. So, T1,oil and T1,Cu, these are known;
T1,oil = 20 ºC, T1,Cu = 350 ºC. So, from this equation, you can find out what is T2. So,
mass of copper (mCu) = 1 kg and mass of oil (moil) = 10 kg.
So, you can calculate what is T2. So, T2 = 300.65 K. So, what is required is the total
availability which is A2 – A1. This is nothing but the difference in the availability
between availability of the initial state and final state of copper and oil.

A2 – A1 = mCu(Φ2 – Φ1)Cu + moil(Φ2 – Φ1)oil

where Φ2 = (u2 – T0s2) – (u0 – T0s0)

Similarly, for Φ1, just replace 2 with 1, and again for oil, it is very similar. So, you
essentially need to calculate s2 – s1 for copper and s2 – s1 for oil to calculate this.

T2
s2 – s1 = C.ln( )
T1

P2 P2
There is no question of R.ln( ) because it is incompressible substance; that factor
P1 P1
is not there. So, because you know the temperatures of the final state and initial state,
you can calculate the change in availability using above formula (A2 – A1), and then A2 –
A1 = -44 kJ, that is the answer to this problem.

We will move on to the next problem. So, before moving onto the next problem, a very
important insight about the answer to this problem is Φ2 – Φ1 is negative. So, there is a
decrease in availability or exergy and this is because that the entire work potential
because of this heat transfer is lost due to irreversibility, due to entropy generation.
(Refer Slide Time: 06:41)

Next problem: Steam enters a turbine at 25 MPa, 550 ºC and exits at 5 MPa, 325 ºC at a
flow rate of 70 kg/s. Determine the total power output of the turbine, its isentropic
efficiency and second law efficiency. This is a very straight forward problem, but again,
we will try to develop an insight through this problem.

(Refer Slide Time: 07:21)

So, let me make a schematic; we have solved a very similar problem in the previous
lecture. But I want to get little bit deeper into the aspect of second law efficiency and
isentropic efficiency. So, you have the pressure; pi is the inlet pressure and pe is the exit
pressure of the turbine. The state ‘i’ is given, so, let us draw the state points in this
diagram.

State ‘i’: 25 MPa, 550 ºC

State ‘e’: 5 MPa, 325 ºC

ṁ = 70 kg/s.

Because we intend to also relate it with the isentropic efficiency and a virtual isentropic
process, we will also be concerned about the state ‘es’, which is a hypothetical state.

State ‘es’: se = si, pe = pe

s s

So, in this case, let us see what first you have to calculate. The first thing that you have
to calculate is what is the isentropic efficiency and what is the second law efficiency. So,
to calculate the isentropic efficiency, you also have to know what is the reversible
adiabatic or isentropic work. So, you apply the first law, steady state steady flow,
between ‘i’ and ‘es’. So, you have:

1st law (SSSF) (‘i’ to ‘es’): q + hi = he + ws s

All the terms, we are writing per unit mass.

So, from the state ‘es’, we know how to calculate hes ; in one of our previous problems,
we have addressed it, so, I am not getting into the details. The turbine is adiabatic in this
case.

q + hi = he + ws
s

So, ws = hi - hes

Similarly, if you apply the first law for the process between ‘i’ to ‘e’:

1st law (SSSF) (‘i’ to ‘e’): q + hi = he + w

State ‘e’ is already given. So, he you can find out from the table. So,
w = hi - he

w
So, the isentropic efficiency ηs = , very straight forward, and this is coming out to be
ws
79 %.

Then, you have to calculate the second law efficiency. So, interestingly,

w hi  he
ηs = =
ws hi  he, s

We will see how it relates to the second law efficiency. So, second law efficiency, how it
is defined?

w hi  he
ηII = =
w rev (hi  T 0 si )  (he  T 0 se)

Now, by looking into the expression for second law efficiency and isentropic efficiency,
can you tell what is the conceptual difference? The conceptual difference is that in this
case, the comparison is between the same end states; hi - he is the actual work; (hi – T0si)
– (he – T0se) is the reversible work that could be obtained with the same end states, that is
definitely less than the reversible adiabatic work between the states ‘i’ and ‘es’; ‘es’ is a
hypothetical state, it is not the actual end state. In reality, the end states are fixed.

So, ‘i’ and ‘e’ are fixed, ‘es’ is just a hypothetical state, ‘i’ and ‘e’ are the fixed states.
The numerator in second law efficiency is the actual work that you get with these fixed
states and the denominator is, if the entire process was reversible between these two
states, what could have been work that you have obtained, that is different from a
reversible adiabatic process between ‘i’ and ‘es’. So, here you are perhaps comparing one
apple with an orange. hi - he is work between states ‘i’ and ‘e’; hi – he,s is work between
states ‘i’ and ‘es’.

w hi  he
So, there are disparate bases of comparison. For ηII = = , the
w rev (hi  T 0 si )  (he  T 0 se)
basis of comparison is more consistent because you are comparing with the same end
states ‘i’ and ‘e’. (hi – T0si) – (he – T0se) is just the reversible work between the end states
‘i’ and ‘e’, and hi - he is the actual work between the states ‘i’ and ‘e’; this reversible
work need not be adiabatic.

So, then what is that reversible work that you could obtain? To get an insight, imagine
the process like this. So, you could break this up into two parts: one is from ‘i’ to ‘es’,
another from ‘es’ to ‘e’. So, to make a reversible process from ‘es’ to ‘e’, you need a heat
transfer, and that heat transfer is like in this case ∫Tds from ‘es’ to ‘e’. So, look at the
temperature of ‘es’ to ‘e’: this temperature is 325 ºC. So, temperature of ‘es’ to ‘e’ is
around 300, 325.

So, here, if ‘e’ is in the two-phase region, both are 325 ºC, because saturation
temperature is not changing from ‘es’ to ‘e’. So, it is 325 ºC.

So, this state ‘e’, 5 MPa and 325 ºC, I have drawn it conceptually in the 2-phase region,
but it may actually be little bit in the superheated region also. So, what is the saturation
temperature at 5 MPa?

264 ºC, so, ‘e’ is superheated actually. So, normally, the exit state of a turbine will not
very commonly be in a superheated state. So, normally, the quality at state ‘e’ will be
given. So, in this particular problem, it comes to the superheated state. So, it is better that
I show it in superheated state. So, normally the exit state is governed by this pressure pe
and a quality, and the quality is very close to 100 %. In this case, pressure and
temperature both are independently given means it’s not a two-phase region, and the
saturation temperature at 5 MPa is less than 325 ºC.

So, this is little bit superheated; may not be very highly superheated, but little bit
superheated. So, 264 to 325, that is the change in temperature as you go from ‘es’ to ‘e’.
So, from ‘es’ to ‘e’, you require a heat transfer in a reversible process. What will arrange
the heat transfer? You do not have any source or anything. So, you have to arrange that
heat transfer from the ambient. Ambient, the key issue is that ambient is at lower
temperature. So, ambient is at a T0, which is less than this 264 or 325, all this.

So, because T0 is less, so, from a lower temperature to a higher temperature system, you
want to have a heat transfer. But still you have to have this heat transferred to have the
change in state from ‘es’ to ‘e’ take place and that in a reversible manner. So, you have to
arrange for a reversible heat pump across the ambient that will effectively transfer heat
from T0 to this. That reversible heat pump will require a work input.

So, you have to arrange for a reversible heat pump which requires a work input; although
reversible, but still it requires a work input, and that part of the work, therefore, you do
not get as a work potential. That is why, you see, it is not hi - he, but hi – he - T0(si – se).
That ‘-T0(si – se)’ (or ‘-T0(se – si)’, whatever), because si is equal to ses , it is essentially
the heat transfer associated with this process.

So, you are having a basis where the potential is not the full hi – he. There is a less
utilisation possibility because you also are allowing a heat transfer and there has to be a
reversible arrangement to achieve that heat transfer in exchange with the ambient.

(hi – T0si) – (he – T0se) is the work for any reversible process between the inlet and the
exit, with no heat transfer. If there is a heat transfer, there is an associated reversible
T0
work which is plus Q(1 - ). But here, we are assuming what? The work potential is
TH
due to the change in state between ‘i’ to ‘e’ without any heat transfer. But, even in this
process, if you break it into two parts, the second part(‘es’ to ‘e’) will require an internal
heat transfer and that can be arranged only through an ambient by employing a reversible
heat pump.

So, si, se, all these things you can calculate. If you calculate all these values, you will see
that the second law efficiency ηII = 88 %. See, the isentropic efficiency (ηs) is 79 %. That
means, the denominator of ηs appears to be more than the denominator of ηII.

So, denominator of ηII already takes into account the less potential because of the need of
this employment of the heat pump to have the heat transfer across the states ‘i’ and ‘e’.
In the case of ηs, it doesn’t take that into account. So, the potential that you are
considering as hi - he,s is not the true potential. The potential that you are considering
between the end states (denominator of ηII), this is the true potential. So, you have to
compare it with the true potential and not the hypothetical potential and that is why the
second law efficiency is much much more scientifically correct to assess the
performance of a device as compared to the isentropic efficiency.
We will solve a couple of more problems. So, the next problem: air flows into a heat
engine at ambient conditions of 100 kPa

(Refer Slide Time: 23:19)

and 300 K as shown in the figure. Energy is supplied as 1200 kJ/kg of air from a 1500 K
heat source and in some part of the process, a heat transfer loss of 300 kJ/kg of air
occurs. The air leaves the heat engine at 100 kPa, 800 K. Find the first law efficiency and
the second law efficiency.

(Refer Slide Time: 24:13)

So, let us calculate this. So, you have a heat engine. It is interacting with various thermal
reservoirs. So, you have a qH of 1200 kJ from TH at 1500 K. You have -qL, with the
magnitude of qL as 300 kJ. So, magnitude is 300 kJ; minus because heat is transferred
from the engine. This is what is given in the schematic of the problem, so, I am trying to
be consistent with the schematic of the problem. And then this TM = 750 K. There is a
work done also.

State ‘i’: p0 = 100 kPa, T0 = 300 K

State ‘e’: p0 = 100 kPa, T0 = 800 K

So, you have to calculate the first law and second law efficiency. So, the first law
efficiency will be like what amount of heat transferred to the system is converted into
work. So, in this case:

1st law:   Q   W
So, always start with the basic equation.

We can use the steady state steady flow equation here. This is a flow process, so, we can
use the steady state steady flow.

(Refer Slide Time: 27:19)

But it’s a heat engine at the end, so, it will work in a cycle. So, but you know the heat
engine itself allows some flow to take place. So,

1st law (SSSF): qH – qL + mihi = mehe + w

So, for everything, we can calculate in terms of 1 kg of air. So, qH is 1200, ‘-qL’ is -300
(qL is 300, so,’-qL’ is -300). mi is 1 kg, me is 1 kg. hi - he will be Cp(Ti - Te).

So, you will get ‘w’ from here. So, the first law efficiency essentially talks about what is
the work which you get as a function of what heat is supplied. So:

w
ηI =
qH

So, this is 31.52 %. The second law efficiency compares the ‘w’ with the reversible
work. So, what is the reversible work? The reversible work is associated with the two
heat transfers (qH & qL) and then the change of state (‘i’ to ‘e’). So,

T0 T0
wrev = qH(1 - ) – qL(1 - ) + mi(hi – T0si) – me(he – T0se)
TH TM

So, mi(hi – T0si) – me(he – T0se) is the work potential due to the change in state and qH(1 -
T0 T0
) – qL(1 - ) is the work potential due to heat transfer, so, everything is known. The
TH TM
CpdT
change in entropy you can calculate by the formula  T
from 300 K to 800 K; -

P2
R.ln( ) = 0 because the pressure doesn’t change. So, the answer to the second law
P1
efficiency:

w
ηII = = 67.2 %
w rev

So, ηII is more practical because it is comparing ‘w’ with a possible reversible process
considering exactly the same inputs, and ηI considers ‘w’ as compared to ‘qH’, but you
also have qL which is obvious and that will make your efficiency apparently low. But
you are comparing your actual work with the maximum possible work within the given
constraints, not with the qH. So, work is compared with work, not work is compared with
heat.

So, we will work out one more problem, possibly the final problem in this chapter before
we move on to another chapter.

(Refer Slide Time: 32:35)

So, problem 8.7: A heat exchanger brings 10 kg/s water from 100 ºC to 500 ºC at 2000
kPa using air coming in at 1400 K and leaving at 460 K. What is the second law
efficiency? This is a very typical problem because, so far, we have calculated second law
efficiency with—by comparing work with work, but heat exchanger doesn’t do any
work. So, how do we rate its second law performance? So, we will learn that through this
example.
(Refer Slide Time: 33:17)

So, you have a heat exchanger in which you have water from state 1 to state 2;

State 1: 100 ºC, 2 MPa

State 2: 500 ºC, 2 MPa

ṁw = 10 kg/s

There is a line of air. Say, you have 3 and 4; this is air, this is water, this is air. ṁair you
do not know, that you have to calculate from energy balance, but this is 1400 Kelvin and
this is 460 Kelvin.

State 3: 1400 K

State 4: 460 K

So, what is happening, you can clearly see, that this water is heated by air. So, what is
the purpose, you have to understand this carefully. The exergy utilised depends on the
purpose. So, here the purpose is that by using this hot air stream, the cold water is heated.
So, first of all, we have to find out what is the mass flow rate of air, so, we can apply the
first law for the heat exchanger as a whole.

1st law (for heat exchanger as a whole):

Q̇ + ṁwh1 + ṁah3 = ṁwh2 + ṁah4 +Ẇ

Student: Sir, when the water is getting heated by the (Refer Time: 35:39) temperature of
the water is more than temperature of air? State 2 and state 4.

It may be a counter flow heat exchanger. So, just to correspond to the proper physical
configuration, entry of air is 3 and exit is 4. So, T3 is 1400 Kelvin, T4 is 460 Kelvin and
you can see that in all cases now, the hotter fluid has the higher temperature, so, it
doesn’t disobey the second law. Alteration of schematic was necessary because if we
kept 3 on left and 4 on right, then at some place, the hotter fluid will have a lower
temperature than the colder fluid; that will violate the second law.

So, this is called as counter flow heat exchanger, instead of a parallel flow, so, exit of
water line is same as entrance of air line. So, it is not given, but from the data, this is
what we can draw consistent with the physical picture. T3 > T2, T4 > T1. So, there will be
a flow of heat always from the air to the water; had it been drawn in the other way, that
would have not happened.

Just conceptually, the figure needs to be altered, but the equation remains the same. So,
you have the work done as 0:

Q̇ + ṁwh1 + ṁah3 = ṁwh2 + ṁah4 +Ẇ

The heat transfer as 0 because there is no external heat transfer:

Q̇ + ṁwh1 + ṁah3 = ṁwh2 + ṁah4 +Ẇ

ṁw is known. So, from above equation, you can find out what is ṁair. So, ṁair(h3 - h4) =
∫CpdT from T3 to T4. So, if you do that, you will get ṁair = 28.939 kg/s.

Now, the conceptual part of the problem; so, how to calculate the second law efficiency.
So, look at the fundamental definition of the second law efficiency. Second law
efficiency is the exergy utilised. It doesn’t define it in terms of heat, work, etc.; it defines
it in terms of exergy. So,

ηII = (exergy utilised) / (net exergy supplied)

So, exergy utilised depends on purpose. So, what is the purpose here? The purpose is to
heat this water. So, exergy utilised = ṁw(ψ2 – ψ1); there is an increase in exergy. That is
the work potential of the work. Water is elevated because of this heat transfer, from the
air to the water. What is the net exergy supplied? So, it is ṁa(ψ3 – ψ4).

ma ( 2  1)
ηII =
mw( 3  4)

So, exergy of the air is lost or sacrificed at the expense of increasing the exergy of the
water. So, net exergy is balanced but with irreversibility, so, it is not exactly balanced
because whatever is the exergy input to the system is equal to the exergy output plus
irreversibility; there is no work done here. So, the exergy that is supplied by the air is not
been able to completely converted into the exergy gain of water and that is because of
the irreversibilites associated with the process. Had that been the case where the exergy
supplied by air is fully converted into exergy gain by water, then ηII would have been
100 %, but it is not 100 %.

So, if you calculate ηII, this will be 61.4 %, and just to remind you about the expressions:

ψ2 = (h2 – T0s2)

ψ1 = (h1 – T0s1)

ψ3 = (h3 – T0s3)

ψ4 = (h4 – T0s4)

So, to summarise, we have solved quite a few problems on the second law analysis and
exergy and we could generalize the understanding not just for work producing and work
absorbing devices, but for cases where no direct work output is involved, but still the
exergy can be used as a parameter to understand the device performance, because the
balance between the exergy supplied and exergy utilised, that imbalance, rather, talks
about the irreversibilities in the system which are nothing but the inevitable losses in a
practical device.

Thank you very much. We will continue with a new chapter in the next class.