Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 48
Entropy Transport for a Flow Process: Examples In the previous lecture, we were discussing
about entropy transport across a control volume. We will now start solving some problems which
will illustrate those concepts that we developed in the previous lecture. So, we start with this problem: problem 7.1.
A small turbine delivers 150 kilowatt and is supplied with a steam at 700 degree centigrade,
2 megapascal. The exhaust passes through a heat exchanger where the pressure is 100 kilopascal
and it exits as saturated liquid. The turbine is reversible and adiabatic. Find the specific
turbine work and heat transfer in the heat exchanger. So, let me draw a schematic of this problem
in the board. So, this is state 1, after expansion it comes through state 2, and then it passes
through a heat exchanger and comes out at state 3. State 1 is 700 degree centigrade,
2 MPa. It completely defines state 1. So, this is waters, I mean steam basically. State
2, it is 10 kilopascal, that is known. State 3 is also 10 kilopascal, and quality is 0,
saturated liquid. Okay? The turbine is reversible and adiabatic, okay?
Even if it was not told that the turbine is adiabatic, we could have guessed, why? See,
there is a keyword in the problem statement ‘small turbine’; because it is small it
has maybe negligible area surface area for heat transfer to take place, and that means
it can be approximated as adiabatic. But here, explicitly in the problem it is given that
it is reversible and adiabatic. Because it is reversible and adiabatic, if we apply the
second law for the turbine as the control volume…
You can write s2 is equal to s1, reversible and adiabatic. So, if you; so, how do you
know what is s1? So, from table of steam at state 1. Okay? So, once you know this, you
will see that this is, I mean the value s2 this is between sf and sg at 10 kPa. So, this
will be between sf and sg at 10 kPa, if you look into the table, you will get this, this
is very common for practical, from practical situation that turbine exhaust is a two-phase
system. So, from common engineering practice we know this, you can verify this from the
table. So, then, you can write this s2 as 1 minus x2 into sf2 plus x2 into sg2.
So from here, you can find out what is x2, because sf and sg you get from the table at
10 kPa, what is sf and sg. So, you get what is x2. Once you get what is x2, you can calculate
all properties at state two, for example, specific volume or whatever you require, okay?
So, in this case, what do you require? You require is the heat transfer in the heat exchanger,
so there is a Q dot. So, you apply the first law for the heat exchanger as control volume.
If you apply that, you have Q dot plus m dot 2 into h2 plus, you know the kinetic energy,
potential energy terms, for almost all problems that we will be solving those terms are not
important, so to save time, I will just write dot dot, is equal to m dot 3 h3 plus, you
know w dot is 0, heat exchanger has only heat transfer, it exchanges heat with the ambient
in this case. So, Q dot is equal to, and m dot 2 and m dot
3 they are all the same, they are m dot. So, m dot into h3 minus h2. h3 is nothing but
hf at 10 kPa. Right? And, what is h2? h2 is 1 minus x2 into hf2 plus x2 into hg2, where
hf and hg are 10 kPa…at 10 kPa. So, if you substitute these values, you will get what
is Q dot. So, first of all, to get Q dot, see there is another parameter that needs
to be obtained and that is m dot, right? Nobody has given what is m dot here, but what is
given is what is the turbine work, you can indirectly calculate m dot from here. So, if you apply first law for turbine as
control volume, so you have Q dot plus m dot 1 into h1, plus w dot turbine, right? Q dot
turbine is 0. M dot 1 and m dot 2 are m dot. So, w dot turbine is equal to m dot into h1
minus h2. So, this will let you know what is h1 from table, steam table. So, h1 you
know, h2 I have just shown how to calculate, and this turbine power is given as 150 kilowatt.
So, this will give you what is m dot. So, m dot is 1397, sorry, m dot is 0.1074 kg per
second. Okay? So, process has to be separately drawn on a T s diagram.
Now here, there is no information on entropy transport, so, see we have to decouple the
entropy considerations of turbine with the entropy considerations of the heat exchanger.
Nothing is told about the reversibility irreversibility everything about the heat exchanger. So, with
the information given, it is not possible to map the process between 2 to 3 in a T s
diagram, we only know that it’s a constant pressure process, that much we know. But what
happens to the entropy, it depends on the kind of irreversibility and whether it is
isothermal and all these things there that have to be worked out, it is definitely not
isothermal, right? Because, you see, it is…it may be isothermal. It is, because it changes
the… So, in this case, you can assume so at 10,
this is at 10 kPa; at 10 kPa, at constant pressure, it changes phase from two-phase
to saturated liquid. So it is, therefore, a constant temperature process. So far as
the process diagram in a T s diagram goes, that doesn’t take into account what is happening
externally. So, externally it may be reversible or irreversible, you can still draw this process
in a T s diagram considering only the internal part of the process. So, how you can do that?
So, you have this liquid vapour dome, you started with state 1, from state 1, you have
a constant entropy process, till you come to the saturated thing, and then from here
to here…this one. So, this much you can draw in the T s diagram,
what I wanted to mean is that from this you cannot conclude anything about the entropy
transport. The reason is, you have no information on the external irreversibility. So, having
information on the T s diagram will not give you or give you an idea of the entropy transport
simply because this irreversibility associated with this, for that completely information
is not given, but change in entropy between the states you can figure out from the T s
diagram, okay? So, the final answer if you put this m dot here, Q dot will be minus 250
kilowatt, okay? So, we will move ahead to work out the next
problem, let me first erase the board. Problem 7.2: two flows of air are both at 200 kilopascal,
one has 1 kg per second at 400 Kelvin, and the other has 2 kg per second at 290 Kelvin,
okay? The two lines exchange energy through a number of ideal heat engines taking energy
from the hot line and rejecting it to the colder line. The two flows then leave at the
same temperature, assume the whole set up to be reversible, find the exit temperature
and total power out of the heat engines. So, let me draw the schematic of the process,
otherwise it may be difficult for you to follow it. So, there are two streams of air. This is
stream one, and this is another stream another pipeline, this is stream 2. So, here the air
enters at state 1, here the enter—air enters at state 2. Both are 200 kilopascal, but the
temperatures are different: this is 400 Kelvin, and this is 290 Kelvin. When the air comes
out, it comes out at the common exit state, which is p e, T e. P e is 200 kilopascal.
So, here it is m dot is equal to 1 kg per second, here m dot is 2 kg per second. Okay?
I am just trying to draw the maximum information that you get from the given data; then there
is a heat engine, reversible, which takes heat from this, and rejects heat to this,
and in the process, does some net work in a cycle. Okay? The question is, what is the
power output of the heat engine? So, the strategy to the problem will be that
we will find out the power output by considering…okay, first you have to consider what is—first
you have to find out what is the exit temperature. So, power output, to know what is the power
output, if you apply the first law to the entire system, you have to know what is the
exit state, and for that, you have to appeal to the second law for entropy transport.
So, second law, so now what is your control volume, this is very important. So, control
volume includes the two pipelines and the heat engine. Okay? So, for this control volume,
if you apply the second law, steady state steady flow process, summation of m dot e
s e minus summation of m dot i s i is equal to Q dot c v by T summation Q dot by T, c
v we may not write, plus rate of entropy generation. Entropy generation is 0, because everything
is reversible, that is given. What is heat transfer, you may have an illusion by looking
into this and imagining that there is heat transfer, but these are all internal between
the…parts of the system. So, externally for this control volume, there is no heat
transfer, so it is 0. So, m dot e s e summation is m dot 3 s 3 plus
m dot 4 s 4, and m dot i s i summation is m dot 1 s 1 minus m dot 2 s 2, this is 0.
And s 3 is equal to s 4, right? Because state 4 is…state 3 and state 4, they are at the
same uniform state. So, s 3 is equal to s e, s 4 is equal to s e. So, you can write,
and m dot 3 is same as m dot 1, that is the same stream. So, m dot 3 and m dot 1 are the
same. Similarly, m dot 2 and m dot 4 are the same. So, what you can write is m dot 3 into
s e minus s 1 plus m dot 1, sorry, yes, m dot 4 and m dot 2, so plus m dot 2, so m dot
3 is same as m dot 1, we can write this as m dot 1, plus m dot 2 into s e minus s 2,
is equal to 0. You can put this m dot 1 as 1 kg per second,
and this is 2 kg per second; in fact, you could solve this problem if the absolute values
are not given but the ratios are given, because what you require simply is m dot 1 by m dot
2 or m dot 2 by m dot 1. So, this will give you what is s e. You can use ideal gas law
for finding out s 1 and s 2, state these states are completely given, or, because of such
a large variation of temperature, it is better that you look into air table, property table
of air which takes into account variable cp cv, and get s 1 and s 2 from air table.
So, if you get that, you will get what is s e, and combination of s e and p e will identify
state e from the air table, and that will tell you what is T e, given that you already
know p e. So, once you have this information on T e, now you can apply the first law. Because
you know T e, you know the enthalpy at state e; for first law application, you need to
know the enthalpy. So, Q dot plus m dot 1 h 1, I am not writing
kinetic energy potential energy terms anymore, plus m dot 2 h 2, this is a summation of m
dot i h i, is equal to m dot 3 h 3 plus m dot 4 h 4, plus w dot. Q dot is 0, and h 4
and h 3 are h e. From T e, you can calculate what is h e; from the table again. If you
use constant cp, then you can write cp into T e, but it is better to use the variable
cp cv for the large temperature variation that this problem is having.
M dot values you know, so from here you can, h 1 and h 2 from the state…so, this will
give you what is h 1 from table—air table. Will it depend on 200 kPa? No, because enthalpy
of an ideal gas is a function of temperature only. So, typically this is good enough, typically
this is good enough to calculate h 2 from air table, but entropy of an ideal gas is
both function of pressure and temperature. Enthalpy and internal energy are functions
of temperature only, but entropy is a function of both pressure and temperature. So, from here you can calculate what is w
dot, and that is equal to 11.36 kilowatt. We will work out one more problem before we
call it a day in this lecture. So, consider a steam turbine power plant;
so, this is a very basic introduction to power plant engineering for you, and see how we
have slowly graduated or developed from the basics of thermodynamics to come to a stage
when we are now able to analyze thermal power plants. So, look into this. So, you have a
system where there is a boiler, steam generator is like a boiler. So, you supply heat to water,
which is converted into steam in the steam generator. The exit state in the steam generator
is given as state 1, then that steam enters the turbine.
So, from boiler, it enters the turbine, there is some work that is extracted from the stream
and—steam, and it rejects heat to the condenser, by second law of thermodynamics, this heat
rejection must take place in the condenser to make the turbine work in a cycle. Then,
the condensate in the condenser is pumped to the boiler pressure by a pump. You have
to find out the specific turbine work output and the turbine exit state, specific work
means work done rate of work done per unit mass, and the pump work input and enthalpy
at the pump exit state and the thermal efficiency of the cycle.
So, I have discussed this kind of cycle in the context of the Carnot cycle applied for
a flow process, and you can see a kind of application of that, an extension of that
to more realistic cycles, this is not a Carnot cycle, a more realistic cycle where the compressor
is replaced by a pump, and we will try to analyze this cycle. So, I am just drawing the schematic: you have
the steam generator, then it enters the turbine, so this is steam generator, then it enters
the turbine, then it enters the condenser, and then there is a pump. So, this is 1, this
is 2, this is 3, this is 4. So, at one, you have p 1 is equal to 20 mPa and T 1 is equal
to 700 degree centigrade, this is superheated steam. At 2, you have 20 kPa, you do not know
any other property as yet; at 3, you have 20 kPa, and you do not…and 40 degree centigrade,
right? And 40 degree centigrade, yes it is given somewhere, and then, at 4, you have
20 MPa, and no more information is required. So, you have first you start with control
volume as turbine. So, we will start with—so, you have four different entities, each will
be four different control volumes to get unknowns in the properties. So, control volume turbine,
we assume that it is reversible and adiabatic. So, you have s 2 is equal to s 1, s 1 from
table. So, once you get s 1 from table, s 2 now it is in the two-phase region, we have
solved a similar problem, so this is 1 minus x 2 into s f 2 plus x 2 into s g 2. S f and
s g are calculated at 20 kPa, from the table. So, that will give you what is x 2. Once you
calculate what is x 2, you can calculate what is h 2 which is 1 minus x 2 into h f 2 plus
x 2 into hg 2, h f and hg are again from the table. So, h 2 you can calculate. Now, the
specific work output. You can apply the first law of thermodynamics for the turbine. Q dot
plus m dot 1 h 1, here all m dots are same, in the cycle same fluid is circulating, so
we will just write m dot into h 1, neglect changes in kinetic energy and potential energy.
So, specific work output is w dot by m dot is equal to h 1 minus h 2. So, this is 1 5
6 9 kilo joule per kg, this is the turbine specific work. So, control volume turbine.
So, next we will apply control volume pump. 20 kPa, 40 degree centigrade, this, if you
look into the table, this is very close to saturated liquid, it may be little bit compressed
also. Ideally, it is—it is supposed to be saturated liquid or close to that, it can
even be compressed liquid and that is what is desirable, because the pump handles only
liquid, but not two-phase mixture. So, it is liquid. So, enthalpy at state 3,
you can say roughly it is h f at 40 degree centigrade. Not h f at 20 kPa, because the
enthalpy at this state is very close to enthalpy of saturated liquid at this temperature. Right?
Because, if you now change the state little bit from the saturated liquid state, the pressure—the
pressure dependence may be more severe. So, this state if you approximate it as a
saturated liquid state, the property from which you calculate that should be primarily
from temperature and not from pressure, okay? So, h 3 is h f. So, if—even if it deviates
a little bit from the saturated liquid, then this it will still be approximately this.
And v 3 is again will be roughly v f at 40 degree centigrade, this is roughly like 1
by density of water at 40 degree centigrade. Now, the work done for the pump, see this
is single inlet, single exit, reversible steady state steady flow, this formula we have derived,
not pdv, vdp. So, this—this is per unit mass, so w pump.
So, this is—now this v is roughly a constant because the pump is handling liquid; because
it is handling liquid, the liquid is roughly incompressible. So, its density is not changing;
true from here to here its density is changing, but this is what is engineering approximation:
for all practical purposes, the density is not changing substantially so that you can
approximate this by this v 3. The negative work indicates that you are inputting work.
So, this pump work—so, what you require—so, this is minus 20.1 kilo joule per kg. Okay?
And then, if you apply first law for the pump, Q dot plus m dot h 3 is equal to m dot h 4
plus w dot pump. So, there is negligible heat transfer. So,
h 4 is equal to h 3 minus w p, w p is w dot p by m dot; because w p is already negative,
negative negative makes it positive, so h 4 is greater than h 3. So, h 4 will be 187.6
kilo joule per kg. Then, you apply control volume for steam generator.
So, you know enthalpy here, you know enthalpy here. So, Q dot H plus m dot h 1 is equal
to m dot—sorry, m dot h 4, your inlet is 4, exit is one, work is 0 for boiler. So,
Q dot H by m dot, so this is specific heat transfer to the boiler that is h 1 minus h
4. So, p 1 and T 1 means you can get h 1 from
table. So, this is 3621.5 kilo joule per kg. So now, for efficiency of the cycle, what’ll
you—what you require? For efficiency of the cycle, you require the net work output
by q H, right? This is the definition of efficiency of a cycle. So, what is the net work output?
Net work output is w turbine minus w pump. This is output power, this is the input power,
mod of output power minus mod of input power, that is the net power or net work, divided
by q H. So, this will be 42.8 percent for this problem.
So, you get an idea. So, this is a very practical problem, typical efficiency of thermal power
plants will be around 40 percent, thermal efficiency, and this is what we—you get
from the numerical data here. Thank you very much. We will continue with
more problem-solving in the next lecture.
