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We started the discussion on the regenerative feed heating and we also discuss the ideal
regenerative feed heating system, where the feed water heater from the feed pump was heated
reversibly to the temperature equal to saturation temperature to the boiler pressure. This was
the Rankine cycle which was made equivalent to a Carnot cycle. At the same time, we appreciated
the way we understood the philosophy of the regenerative feed heating by which a Rankine
cycle is converted to a Carnot cycle, where heat is added to the maximum temperature of
heat addition. Since the sensible heat part that is the heating
of water from a temperature lower than the saturation temperature at the boiler pressure,
up to the saturation temperature was made possible by internal heat; the heat of the
expanding steam. At the same time, we appreciated the practical difficulty of obtaining it.
Since we cannot place a heat exchanger in the turbine, a reversible heat transfer to
achieve a temperature that is equal to the saturation temperature of the water at the
boiler pressure which is equal to the inlet temperature of the turbine is not possible.
One of you mentioned that the dryness fraction at the end of the expansion in the turbine
was also reduced. Therefore, for all these reasons, this particular ideal regenerative
feed heating system cannot be obtainable in practice; this is purely a concept.
In practice, the feed water is heated, because it was understood that if we can heat the
feed water to some temperature by taking the heat from the steam, then we can increase
the efficiency of the cycle, since the mean temperature of heat addition is increased.
Heaters, the typical heat exchangers are made outside the turbine and the steam from the
turbine at different pressures, while steam is expanding through the turbine blade and
heat from that steam is being transferred to the feed water passing through the heater.
There are two types of such heaters: in one heater, the steam and water are being mixed
together so that they are under the same pressure, otherwise the mixing is not possible. In other
type of heater known as closed heater, the steam and water do not mix with each other
and water is at some pressure and steam at another. There is a separating wall and the
water flows through it. Then steam is condensed on the surface and this is the way they exchange
the heat transfer. The closed heater is beyond the scope of our
study. Now, we will discuss the open heaters which are relatively simpler than the closed
heater. So, we will discuss the closed heater system which is implemented in practice for
regenerative feed heating. Feed means the feed water and regenerative means use of heat
itself to heat the system from one part and to heat the other part of the system. That
is the definition of regeneration. We now come to this; closed heater.
Let us consider this boiler here, and the steam enters at some pressure and temperature.
The boiler pressure and temperature at this point 1 and this steam expands up to the condenser
pressure 2. Let us consider point 2, which means steam after expanding to a pressure
corresponding to 2, some amount of steam is blade for the regenerative feed heating. Let
us consider the analysis for 1kg of flow at the inlet to the turbine. Let the flow rate
be 1kg at 1. 2 correspond to certain intermediate pressure, after which pressure, the steam
is being expanded. Some amount is being blade and let this point 2, 3, 4, 5. Let mass be
m2, and let this mass be m1; which is being blade from point 2.
Let this steam expand to another point 3, where another amount of mass which is m2 is
being blade. Then, the steam expands to the condenser pressure 4. After condensation this
point is 5; it does not matter the nomenclature. Sometimes, people go to this point with 2.
Therefore, at 4 we are having 1 minus m1 minus m2 kg of steam. From 2 to 3 expansion of steam
is taking place whose mass is 1 minus m1 per 1kg of mass at the inlet to the turbine; then
3 m mass is blade. 3 to 4 final expansion takes place 1 minus m1, 2 kg mass. So, it
is 1 minus m1 m2 kg of mass condensed at the condenser pressure. That is 4 to the point
5; that means, it is condensed to the saturated liquid. There are three feed pumps: first
feed pump pressurizes the saturated water to the point 6 which is equal to the pressure
3, because water is being pumped. This is heater range 2. Therefore, 6 is the pressure
and correspond to 3; that means, 4 and 5 are of the same pressure. 5 to 6 is the pressure
rise. If you consider this heater, it receives steam m2 at pressure p3 and temperature t3.
It gets water at same pressure p6 and p3 are same, but at the different temperatures.
The water enters, so both of them are mixed together. Therefore, if m2 is mixed with 1
minus m1, m2 we get 1 minus m1 kg of water, which is at the same pressure because mixing
is taking place at the same pressure which is being pumped. 3, 6, this point is 7. After
pumping, this point is 8, This is being pumped to this pressure. This becomes 9 and this
is mixed with m1. So, ultimately this becomes 1kg which is the pump 10. It will be clearer
if we see the diagram here; ts. After expanding to the point 2, we believe, m1, then 2 to
3, is the expansion with 1kg. Then 2 to 3 expansions are 1 minus m1 kg. So, 3 to 4 is:
1 minus m1 minus m2 kg. At 4, this is 1 minus m1 minus m2 kg. So, this comes to the point
5. Now, 5 to 6; this enters 6 which means, this liquid is heated; whereas, m2 steam which
is blade at this pressure is being condensed. The steam is condensed; m2 amount of steam
at this condition enters to this heater. The mass, 1 minus m1 m2 kg of water at condition
6 is the inlet to this heater. So, the outcome of this heater is the point which corresponds
to 7. This point corresponds to point 8, this m1 kg comes 2. Therefore, this point corresponds
9, which is the outlet of this heater. Then this is 10. This line corresponds to the pressure
p2 and this line corresponds to the pressure p; that is, the boiler pressure, and this
line corresponds to the pressure p3, the second blading pressure.
If I write the expression for work of the turbine, work is reduced. For 1kg, if the
work d1 is h1 minus h2, it means the mass is changing. I am writing, work per 1kg of
steam entering to the turbine plus 1 minus m1 into h2 minus h3, because this path is
associated with 1 minus m1 minus m2 into h3 minus h4. Heat is added from outside source,
because 1 kg means h1 minus h2. So, you can find out that it has WT by Q added. You can
find out many things from this analysis. What is Wp? The amount is 1 minus m1 minus m2.
So, this pump handles 1 minus m1 minus m2. Therefore, it is 1 minus m1 minus m2 into
h6 minus h5. This pump handles 1 minus m1 minus m2, it mixes 2. At 7, m2 steam is cooled
and 1 minus m1 minus m2, steam is heated. It is heated to a saturated liquid, and cooled
to saturated liquid; that means it is condensed. It is sensibly heated to the saturated temperature
such that it becomes 1 minus m1. Therefore, this pump handles 1 minus m1 into h8 minus
h7, h8 minus h7 plus 1. 1 minus m1; this is this pump, so 1 minus
m1; this is the point where it is heated. This comes here, now it is 8. These two points
are repeating; that means, the two are the same points and there is no process here.
Therefore, it is heated to 9, 8 to 9 sensible heating to saturated liquid, whereas m1 is
condensed and when they mix together as a result, we get at 9, 1 because m1 is added.
9 is 1 kg here. Therefore, feed pump 3 handles 1kg. This is plus 1 into h10 minus h9 and
this is Wp. This whole problem or the main problem of calculation is m1 m2. The main
parameters are: how to find out m1 m2? If I can find out m1 m2, I can find all these
things. Analysis is based on kg at the entry to the turbine and I can find eta. How to
find out m1 m2 from the energy balance of the heaters?
Let us consider the energy balance of this heater; control volume. What is the energy
balance of the heater, energy coming in m1 h2? When you write the energy balance, you
see the block diagram; not the cycle diagram. You can write the amount, which means this
is 1 minus m1 kg. This line is 1 minus m1 kg that means 1 minus m1 into h8. What is
the outlet outflow? This is 1kg, 1 into h9. From this equation, you will explicitly give
me m1, provided, I know h2, h8, h9. Let us find out h2 is this point. This will be clearly
told in the problem. If I know the blade pressure and it is an isentropic expansion, then I
know h2. If I am told, there is an isentropic efficiency to expansion of this blade pressure
then also I can find out h2. The actual h from the isentropic efficiency usually by
isentropic expansion, I can do it straight; familiar diagram or from steam table, that
means I know h2. This point is h8 and this point is h9. h9
is similar at this pressure the enthalpy of the saturated liquid is h9. So, I will have
to know h8; not temperature h7, because h7 I know h8 do not know it is sub cool state. The work d is p8 minus
p7 into v7. I know p7, because I know the next blading pressure. I know p8 that means
I know this pressure and the specific volume at this blade pressure. I can find out h8
from here. Another way of knowing h8 is there. If you know the temperature for water, you
can calculate that as kilo Joule per kg and you can multiply with cp and find out the
enthalpy of the water. Actually enthalpy of the water is taken 0 at the saturated state,
at 0 degree Celsius. If you consider saturated state at 0 degree Celsius, the pressure is
very low. There is almost 2.0006 bar. So, at this pressure, if you heat the water from
0 degree to any temperature t, at constant pressure, then ht minus h0 is per unit mass
cp into t. If you put the reference h0 as 0 then the
temperature at t, the specific enthalpy is cp into t. Now, you can tell that it is at
this pressure, because our reference is at this pressure. But if I heat the water at
different pressure from 0 degree to any temperature t, then I cannot put the reference enthalpy
at 0 temperature, but the answer is the influence of pressure on change of enthalpy is negligible.
Therefore, you can approximately put the value of enthalpy 0 at 0 degree Celsius at any pressure,
which means, at any pressure temperature t is its specific cp into t, is a specific enthalpy.
This is the other way of doing it. In the main calculation, there may be very little
differences. We can find out h8 explicitly, where m1 is known. Similarly, I can know m2
by the energy balance of this heater. If we consider a control volume around this heater,
let us write m2 into h3, so this is the inflow. Another inflow is 1 minus m1 into m1 minus
m2, because water amount is 1 minus m2 minus m2, which is coming at enthalpy 6 and the
outcome is 1 minus m1 into h7. h3 is known, blade pressure is known. By isentropic expansion,
we know h3. Similarly, we can find out h6 from h6 minus
h5, which is p6 minus p5 into v5 clear. We know h6; h7 is again simple one. The saturated
water enthalpy at this pressure that means we will have to very carefully calculate h1.
h8, h9 can be known from the use of steam tables or Mollier diagram and then can find
out m1 m2. The determination of m1, m2 by making use of the energy balance over the
heaters energy balance, enveloping the heater, considering heater as the control volume over
the heater, we can find out m1, m2. When, m1 m2 is known, we can make the analysis of
the cycle parameters on the basis of 1kg. This is also the basis of 1kg and you get
the result now. This is the network output for 1kg of steam flowing through the boiler
or for 1kg of steam at the entry to the turbine. The analysis is as simple as this thing and
this is only in our course. Let us be more friendly or easy with the calculations. If
we follow a straightforward practical problem, an example problem, this theory will be again
repeated through the numerical example.
The net power output of the turbine in an ideal reheat - regenerative cycle, this ideal
does not mean inside the turbine. Reheat ideal means the processes are ideal isentropic expansions
and there is no heat loss from the heater. Heat transfer takes place between the steam
and water that is why it is ideal. Cycle is 100 megawatt the net power out. Steam enters,
there is reheat regenerative. The steam enters the high pressure turbine, at 90 bar 550 degree
Celsius. After expansion to 7 bar, 90 bar is the inlet pressure. Some of the steam goes
to an open heater and the balance is reheated to 400 degree Celsius. This is the inlet temperature,
this is the reheat temperature after which it expands to 0.07, the final pressure is
the condenser pressure. Both reheat and regeneration is there, because
the purpose is different. Reheat purpose is to increase the dryness fraction and purpose
of regeneration is to increase the efficiency. Inlet pressure is 90 bar, whenever any problem
will give this power output; it means, they will tell the value of this steam flow rate.
That means you always analyze for unit mass of this steam at the entry and then divided
it. This is not at the beginning very important. The important parameter is the 90 bar 550
degree Celsius; that means, inlet condition of this steam and the intermediate pressure
7 bar and the steam reheated, some of this steam goes to an open heater and the balance
is reheated to 400 degree and then it expands to this. We have to find out the steam flow
rate. I have told a steam flow rate to the HP turbine. What is the total pump work? All
these things are now known to you. Calculate the cycle efficiency, if there is the 10 degree
raise in the temperature of the cooling water, what is the rate of flow? What is the steam
flow rate to the HP turbine, the total pump work and the cycle efficiency? Let us see
this.
This is the boiler, this is the turbine. Now, boiler 1, which is 90 bar and 550 degree Celsius.
After expansion, this is HP turbine, rather turbine T, this is LP turbine.
The steam is blade of and the rest is going to some stream which is blade of and some
steam is being reheated. Therefore, some steam is blade for this and rest is reheated. The
some is reheated, and some is being blade, so this is 1kg. So, this is being reheated
and goes to LP turbine. So, this is the heater. It is then expanded to condenser. This is
the 0.07 bar. This is the cooling water supply. This is feed pump 1, which feeds the water
to the heater, then this heaters gets some m1 kg. Let m1 kg is being blade and then 1
minus m1 kg goes to the reheater. It then mixes ultimately 1kg is feed pump 2 and then
this goes to the boiler. Let this point be 1 and this point is 2. That means 1 to 2.
Then this is m1. Let 2 be this point, it may not be dry saturated case, it may be here
that depends upon the condition that will come as solution of the problem. First of
all, draw the diagram in ts, draw the cycle ts diagram for your understanding.
This comes at 1 in the HP turbine, after expanding to 2, some will be going to reheater, and
some is coming to the blade. m1 kg is coming straight for this and 1 minus m1 kg is going.
Start with 1kg here, so 1 minus m1 kg is going, and m1 kg is going to heater. So, 1 minus
1kg is reheated to 400 degree Celsius. That means, the temperature is 400 degree Celsius.
This goes to the LP turbine and expanded to 0.07 bar. 0.07 steam; that is 1 minus m1 kg
is pumped to this pressure p. I am not showing the point. 1 minus m1 kg is being fed to this
heater. This will become m1 kg and 1 minus m1 kg. They come as 1kg. Now, this 1kg of
water is again pumped to the boiler pressure and it goes to the boiler. This 1 to 2 is
this point from 2 m1 kg of mass is being blade, but from point 2 there is reheat which 1 will
be 3? This will be 3, but this 2 to 3 is associated with 1 minus m1 kg. From 3, you should be
within this swat region that is 3 to 4. This point is 4. After 4, it is condensed. So,
this point is 5. Therefore, this is 1 minus m1 kg. This is feed pump 1 process, which
gives 5 to 6, which is the pressure line. It is now heated with this blade, steam and
the outcome is 7. The 7 point corresponds to 1kg, because 1
point m1 meets with m1. Then ¬7 to 8 is this. 8 is the boiler pressure. So, this is the
heating in the boiler. This is the heating. In this case, the average temperature of heat addition is
not only t1; t1 to t10, which is much higher than t1 to t5. The temperature of heat addition
is like that. The problem is clear, in its block diagram. How to solve this problem?
If you open this steam table, it may be a little difficult, therefore always prepare
hs diagram. Let me draw the hs diagram. This is the constant pressure line and these are
the constant temperature lines, the things that are there in the cost and x lines.
First, you select this point 1. Let us consider this temperature as 550 degree Celsius and
this pressure is 90 bar. I must find out the intersection point. I am generating here,
that is the only thing in hs diagram. I just go, expanding it up to what intermediate pressure
7 bar. Let this be 7 bar, this is not in scale. So, 90 bar to 7 bar, because other pressures
come 1 to 2, that means I can generate point 2. Then 2 to 3 is the constant pressure, reheating
to 400 degree Celsius. Let us consider this as 3. This is for 100 degree Celsius. It is
the representation of a process, mass is changing one part of this steam and another part of
this steam is going. Another part of the steam is also cooled down. This is the m kg and
is being cooled like this and this part is not shown in the Mollier diagram. This part
is for m1 and this part is for 1 minus m1, or if this part is m1 then this part is 1
minus m1, it hardly matters. From 3 to 4, let this be the 0.07. The pressure line is
not in scale. 97 and 0.07, which means, I can find out also the dryness fraction, which
will be 4. 1, 2, 3, 4, I will draw. 1, 2 is 1 kg. 2 to
3 is 1 minus m1 kg, where, m1 kg is being cooled in this direction. Since, my intension
is to find out h1, h3, g2, h4, which will be required in the calculation and is directly
read from this ordinate state by drawing these lines. You calculate m1 by the energy balance
for this heater you calculate and then you can find out what is required. What is the
steam flow rate? Means, we have to find out WT minus Wp. WT is h1 minus h2 plus 1 minus
m1 into 3 minus h4 and h1, h2, h3, h4 are directly known. What is Wp? There are two
pumps: one here, another here. This pump handles 1 minus m1 into h5 minus h6 and h6 minus h5
and another one is 1 into h8 minus h7. Now, you see from this diagram, Wt, h1 minus h2
plus 1 minus m1 h3 minus h4. What is h6 and h5? From steam table, you have to calculate
both. Steam table will give the value of h5, h6. You can find out from the same principle;
h6 minus h5 is p6 minus p5 into v5; h6 minus h5 is p6 minus p5 into v5. You know h6, h5,
you know h8, h7, because of the pressure h7. These are from steam table, and again, you
know h8 by the same thing. h8 minus h7 will be p8 minus 7 into v7.
When you know WT, Wp, you can find out this for 1kg of steam going through the total power,
which is 100 megawatt. Why did you get this steam flow rate? What is the total pump work?
For 1kg of hp turbine flow this is the power. So far, it has reversed 100 megawatt. You
have to find out the steam flow rate to the hp turbine.
For 1kg of steam flow to the hp turbine, you get the power for 100 megawatt power steam
flow rate, which means 100 divided by specific work output. Specific means, for 1kg flow
through hp turbine. What is steam flow rate to the hp turbine? What is the total pump
work? Total pump work is this. You have to find out m1 by making this energy balance
across the heater for 1kg, but when you know this kg of flow rate, you can find out by
multiplying total pump work. If you do on the basis of 1kg, then you will land up to
the answer that this is the total pump work for 1kg of steam flow through hp turbine.
The actual answer will be the total mass flow m1; this m1 is for 1kg. Therefore, the pump
work is for 1kg of flow to the hp turbine. I have already found out the steam flow rate
through the hp turbine. Multiply the total pump work like the total network is given.
For cycle efficiency, you do not require the mass flow rate WT minus Wp divided by h1 minus
h8. What is that heat transfer? If there is a 10 degree raise in the temperature of the
cooling water, what is the rate of flow of cooling water in the condenser?
The rate of flow of cooling water in the condenser is the condenser heat balance; m dot cp into
this delta t is equal to the heat lost in the condenser; that means, h4 minus h5 is
equal to mass flow rate. This part is very simple m dot into cp into delta t which is
10 degree equals to h4 minus h5 into 1 minus m1 into h4 minus h5 into the mass flow rate
of
hp turbine hp m dot hp turbine. Since I will have to find out the mass flow rate of the
cooling water, if you suppress it, the result is the mass flow rate of the cooling water
per unit steam flowing through the hp turbine. Next part, if the velocity of steam flowing
from the LP turbine to the condenser and is limited to a maximum of 130 meter per second.
Find the diameter of the connecting pipe. Mass flow rate is rho Av; rho in fluid mechanics
is written like this, but here, we write like the small v this V is the velocity. You know,
V4 velocity is the given velocity of flow. So, you can find out the area. So, that is
the diameter of
the connecting pipe. The value of m dot is 1 minus m1 into the mass flow rate of this
thing; that means, this is per unit mass flow rate to hp turbine. m is the fraction per
unit mass; so, 1 minus 1 into mass. flow rate of the hp turbine everything is known, that
is the actual mass flow. I end the discussion on this cycle. I think the ideal cycles have to start, but I do not know whether
I will get time today. How to calculate the mean temperature of heat addition without
regeneration or with regeneration for any heat transfer process? Again, I repeat, for
any heat addition process.
Let us consider a heat transfer process; in general, a heat addition process, where the
system changes from state 1 to state 2, where the temperature increases, heat is added.
Let us describe a reversible heat addition process. In a TS diagram heat addition is
Q1, where t2 is the final temperature, and t2 is greater than t1. Heat added Q1 is equal
to h2 minus h1, which is not read in this axis. So, in hs axis, we can read this as
a varying temperature. Heat addition process is not an isothermal heat addition process,
heat is added such that system increases its temperature from 1, 2, and obviously, there
is an increase in entropy, a mean temperature Tm. Tm is defined as h2 minus h1. This is
the question which can come in the examination also. s2 minus s1; which means that this graphically
can be explained like that. Let this area be ab 1, 2 ba is the area which
represents the heat transfer during the process 1 to 2, because we know delta Q is Tds. Therefore,
Q is the integral of Tds 1 to2, which is the area under this curve. If, I draw a rectangle
with this ordinate steam, the value of Tm will be a projection like this. If I can construct
an ordinate such that this area is now ab cd and if I can draw a parallel line to abcd
so that the area abcd becomes equal to area 1 2 ab, then this ordinate we will read as
mean temperature. That means this is the mean temperature of heat addition.
The physical definition is the mean temperature of heat addition in any varying temperature
process is a constant temperature. The temperature at which heat was added is constant; that
means, the temperature at which if this heat was added could have given the same entropy
change. The same heat added could be expressed as Tm into s2 minus s1, because this is the
area dc ba at constant temperature heat addition is t into s2 s1. This is the way, one can
calculate the mean temperature of heat addition. Similar is the case for heat rejection. If
this could have been a heat rejection, the same diagram from 2 to 1, we could have found
out like this. Therefore, this is the way one can find out the mean temperature of heat
addition. I will close the lecture, next class I will
start the air standard cycle. There is some time so I will not start any calculations,
but I shall tell you what is an air standard cycle or gas power cycle not air standard
gas power cycle. I have already discussed earlier that a power
cycle means a thermodynamic cycle, where power is being obtained at the cost of heat energy;
this is a power cycle. When the working fluid changes its phase during operations in different
process of the cycle, this is known as vapor power cycle.
The example is the thermal power plant, where the working system water changes its phase.
A large class of practical power plant uses gas. Therefore, the theoretical cycles of
this power plant will be gas power cycles. For example; gas turbine power plant, where
the air is burnt. Fuel is burnt in the air and products of combustion in gaseous phase
works and gives you the power in a gas turbine plant, aircraft engine, automotive engines
that means your automobiles. We get mechanical power and working system is a gas. The thermodynamic
cycles which gives mechanical power is the power cycles, the cluster of heat and the
system is gas; we call it gas power cycle. Basic definition of a thermodynamic cycle
is that the initial state and final state should be same; that means, all the state
properties and state variables have to be identical. Mass is one important property.
If the mass of the system changes or even the constituents of this system changes during
the cyclic process, even the pressure, temperature comes to the same value it does not follow
ideals in actual in the definition of a cycle, because the cycle means the mass has to be
same. Therefore, in all practical power plants, they do not actually reveal a thermodynamic
cycle, because mass is changing. Some part is air and some part is product of combustion,
where fuel has been burnt so that mass changes and moreover the products of combustion, the
gas composes as they are different from that of the air. Where is the mixture of air and
burnt products of the fuel? These are not actually cycles by definition.
There are other reasons by which the definitions of cycles are highlighted, why the basic definitions
of heat engine cycles are there? There is an external source from which heat is added
to the working fluid, which is being closely followed by a power plant that steam is working
fluid, or water is the working fluid where heat is added from outside. The heat source
is the gas which is known as flue gas. The product of combustion, when the fuel which
is largely coal is burnt or sometimes the liquid fuel is burnt and products of combustion
are not the orking system. But most of the gas power cycles, for example gas turbine
cycles, as you know, these are now today popular knowledge that in your automobile cycles also
the products of combustion itself is the working system.
There is no heat addition from outside; rather, a chemical energy is being into thermal energy
and this thermal sort of heat, cannot be carried. Actually, heat energy which can be transferred
because of a temperature difference, but that carries the thermal energy which acts just
like heat and because of the processes, we get work. This is the actual cyclic operations.
Do not confirm to an actual thermodynamic cycles.
These power plants are being approximated by theoretical cycles which follow all the
definitions of proper thermodynamic cycles and where air is used as the working fluid
throughout the cycle and heat is added from outside and rejected to the atmosphere. Burning
of the fuel will generate the thermal energy; we loosely tell the generation of heat, because
heat can be generated. Heat is energy intrinsic, but generation of internal energy, high temperature
is being simulated by a process where air is the working fluid and it gets heat from
the outside. These are the assumptions which are made and
air where an ideal gas is considered to be the working fluid of thermodynamic cycles,
and they approximate or they represent the theoretical cycles of these power plants known
as air standard cycles. Then, what is the difference between air standard cycles and
actual cycle? All of you know that automobile petrol engine runs with auto cycles, but auto
cycle is the air standard cycles, where the working fluid is air and the process is ideal.
In actual automobile cycles, the cycle operations which are performed in practice, the working
fluid and then fuel is burnt in the cycle. Fluid becomes, the working system becomes
the products of combustion in the air and also the processes are not ideal. These are
always there. Process cannot be intrinsic or constant pressure cannot maintain. There
is a pressure loss. These are being approximated by the cycles which have all ideal processes.
Heat transfer takes place by heating from an external source and cooling to an external
sink. Air is the working fluid behaving as an ideal gas with constant specific heat.
Those cycles are known as air standard cycles and they are the theoretical cycles for those
gas power plants. We will discuss those air standard cycles in next classes.
Thank you.