Unit 11

1. THE STEAM POWER CYCLE J2006/11/1
THE STEAM POWER CYCLE
OBJECTIVES
General Objective : To understand and apply the concept of steam power cycle in
thermodynamics
Specific Objectives : At the end of the unit you will be able to:
 define, derive, calculate and differentiate the following heat
engine cycle:
• Carnot cycle
• Rankine cycle
UNIT 11

2. THE STEAM POWER CYCLE J2006/11/2
11.0 INTRODUCTION
team is the most common working fluid used in heat engine cycles because
of its many desirable characteristic, such as low cost, availability, and high
enthalpy of vaporization. Other working fluids used include sodium, potassium, and
mercury for high-temperature applications and some organic fluids such as benzene
and the freons for low-temperature applications. The majority of this chapter is
devoted to the discussion of steam power plants, which produce most of the electric
power in the world today.
Steam power plants are commonly referred to as coal plants, nuclear plants or
natural gas plants, depending on the type of fuel used to supply heat to the steam. But
the steam goes through the same basic cycle in all of them. Therefore, all can be
analysed in the same manner.
In this chapter it can be shown that there is an ideal theoretical cycle which is the
most efficient conceivable; this cycle is called the Carnot cycle. The highest thermal
efficiency possible for a heat engine in practice is only about half that of the ideal
theoretical Carnot cycle, between the same temperature limits. This is due to the
irreversibilities in the actual cycle, and to the deviations from the ideal cycle, which
are made for various practical reasons. The choice of a power plant in practice is a
What is the most
common working
fluid used in heat
engine cycle?
INPUTINPUT

3. THE STEAM POWER CYCLE J2006/11/3
compromise between thermal efficiency and various factors such as the size of the
plant for a given power requirement, mechanical complexity, operating cost and
capital cost.
Fig 11.0 Model of a steam plant

4. THE STEAM POWER CYCLE J2006/11/4
11.1 The Carnot cycle
From the Second Law of Thermodynamics it can be derived that no heat engine can
be more efficient than a reversible heat engine working between the same temperature
limits. Carnot, a French engineer, has shown in a paper written in 18241
that the most
efficient possible cycle is one in which all the heat supplied is supplied at one fixed
temperature, and all the heat rejected is rejected at a lower fixed temperature. The
cycle therefore consists of two isothermal processes joined by two adiabatic
processes. Since all processes are reversible, then the adiabatic processes in the cycle
are also isentropic. The cycle is most conveniently represented on a T-s diagram as
shown in Fig. 11.1.
Fig 11.1 The Carnot cycle
1
This paper, called ‘Reflection on the Motive Power of Heat’ was written by Carnot before the enunciation of
the First and Second Laws of Thermodynamics. It is a remarkable piece of original thinking, and it laid the
foundations for the work of Kelvin, Clausius and others on the second law and its corollaries.
T
s
1
23
4T1
T2
A B
Boiler
Turbine
Condenser
Compressor
W12
Q23
W34
Q41
1
23
4

5. THE STEAM POWER CYCLE J2006/11/5
A brief summary of the essential features is as follows:
4 to 1: The heat energy is supplied to the boiler resulting in evaporation of the
water, therefore the temperature remains constant.
1 to 2: Isentropic expansion takes place in the turbine or engine.
2 to 3: In the condenser, condensation takes place, therefore the temperature
remains constant.
3 to 4: Isentropic compression of the wet steam in a compressor returns the
steam to its initial state.
The plant required and the numbers referring to the state points for the Carnot cycle is
shown in Fig. 11.1. The steam at the inlet to the turbine is dry saturated. The steam
flows round the cycle and each process may be analysed using the steady flow energy
equation where changes in kinetic energy and potential energy may be neglected.
i.e. h1 + Q = h2 + W
In this statement of the equation the subscripts 1 and 2 refer to the initial and final
state points of the process; each process in the cycle can be considered in turn as
follows:
Boiler:
h4 + Q41 = h1 + W41
Therefore, since W = 0,
Q451 = h1 – h4 (11.1)
Turbine:
The expansion is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2), and h2 can
be calculated using this latter fact. Then
h1 + Q12 = h2 + W12
∴ W12 = ( h1 – h2) (11.2)

6. THE STEAM POWER CYCLE J2006/11/6
Condenser:
h2 + Q23 = h3 + W23
Therefore, since W = 0
Q23 = h3 – h2
i.e. Q23 = - ( h2 – h3 )
∴ Heat rejected in condenser = h2 – h3 (11.3)
Compressor:
H3 + Q34 = h4 + W34
The compression is isentropic ( i.e. s3 = s4 ), and adiabatic ( i.e. Q = 0 ).
∴ W34 = ( h3 – h4 ) = -( h4 – h3 )
i.e. Work input to pump = ( h4 – h3 ) (11.4)
11.1.1 Thermal efficiency of Carnot cycle
The thermal efficiency of a heat engine, defined in chapter 9, was shown to be given
by the equation,
1
2
1
Q
Q
−=η
In the Carnot cycle, with reference to Fig. 11.1, it can be seen that the heat supplied is
given by the area 41BA4,
i.e. Q1 = area 41BA4 = T1(sB- sA)
Similarly the heat rejected, Q2, is given by the area 23AB2,
i.e. Q2 = area 23AB2 = T(sB – sA)
Hence we have
Thermal efficiency of Carnot cycle,
)(
)(
1
1
2
AB
AB
carnot
ssT
ssT
−
−
−=η

7. THE STEAM POWER CYCLE J2006/11/7
i.e.
1
2
1
T
T
carnot −=η
(11.5)
or boilerin thesuppliedHeat
outputNet work
=carnotη
i.e.
41
3421 )()(
hh
hhhh
Carnot
−
−−−
=η (11.6)
11.1.2 The work ratio for Carnot cycle
The ratio of the net work output to the gross work output of the system is called the
work ratio. The Carnot cycle, despite its high thermal efficiency, has a low work
ratio.
Work ratio workgross
net work
= (11.7)
i.e. Work ratio =
)(
)()(
21
3421
hh
hhhh
−
−−−
(11.8)
The work output of the Carnot cycle also can be found very simply from the T-s
diagram. From the first law,
WQ Σ=Σ
therefore, the work output of the cycle is given by
W = Q1 – Q2
Hence for the Carnot cycle, referring to Fig. 11.1,
Wcarnot = area 12341 = (T1 - T2)(sB – sa) (11.9)

8. THE STEAM POWER CYCLE J2006/11/8
This cycle is never used in practice owing to:
1. The difficulty in stopping the condensation at 3, so that subsequent compression
would bring the state point to 4.
2. A very large compressor would be required.
3. Compression of wet steam in a rotary compressor is difficult as the water tends to
separate out.
4. Friction associated with the expansion and compression processes would cause
the net work done to be very small as compared to the work done in the
turbine itself.
The Carnot cycle is modified to overcome the above difficulties and this modified
cycle, known as the Rankine cycle, is widely used in practice.
Thermal efficiency of Carnot cycle,

9. What is the highest possible theoretical efficiency of a Carnot cycle with a hot
reservoir of steam at 200o
C when the cooling water available from condenser is
at 10o
C?
A steam power plant operates between a boiler pressure of 42 bar and a
condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency
and the work ratio for a Carnot cycle using wet steam.
THE STEAM POWER CYCLE J2006/11/9
Example 11.1
Solution to Example 11.1
From equation 11.5,
1
2
1
T
T
carnot −=η
=
273200
27310
1
+
+
−
=
473
283
1−
i.e. Highest possible efficiency = 1 – 0.598
= 0.402 or 40.2 %
Example 11.2
Solution to Example 11.2
A Carnot cycle is shown in the figure given in the next page.
T1 saturation temperature at 42 bar
= 253.2 + 273 = 526.2 K
T2 saturation temperature at 0.035 bar
= 26.7 + 273 = 299.7 K
T
s
1
23
4
526.2
299.7

10. THE STEAM POWER CYCLE J2006/11/10
Then from equation 11.5
1
21
T
TT
carnot
−
=η
=
2.526
7.2992.526 −
= 0.432 or 43.2 %
Also,
Heat supplied = h1-h4 = hfg at 42 bar = 1698 kJ/kg
Then,
432.0==
Q
W
carnotη
∴ W = 0.432 x 1698
i.e. W = 734 kJ/kg
To find the gross work of the expansion process it is necessary to calculate h2, using
the fact that s1 = s2.
From the Steam Tables,
h1 = 2800 kJ/kg and s1 = s2 = 6.049 kJ/kg K
From the equation
s2 = 6.049 = sf2 + x2sfg2 = 0.391 + x2 8.13
∴ x2 = 696.0
13.8
391.0049.6
=
−
Then,

11. THE STEAM POWER CYCLE J2006/11/11
h2 = hf2 + x2hfg2 = 112 + 0.696 x 2438
= 1808 kJ/kg
Hence, from equation 11.2,
W12 = (h1 – h2)
= (2800 – 1808)
= 992 kJ/kg
Therefore, using equation 11.7,
Work ratio workgross
net work
=
=
992
734
= 0.739

12. THE STEAM POWER CYCLE J2006/11/12
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
11.1 Describe the suitable components of a simple close cycle steam plant as
illustrated in the figure below.
11.2 A steam power plant operates between a boiler pressure of 40 bar and a
condenser pressure of 0.045 bar. Calculate for these limits the cycle efficiency
and the work ratio for a Carnot cycle using wet steam.
Activity 11A
(a)
(b)
(c)
(d)
W12
Q23
W34
Q41
1
23
4

13. THE STEAM POWER CYCLE J2006/11/13
Feedback To Activity 11A
11.1 The components of a simple close cycle steam plant as illustrated in the figure
are:
a) The boiler, where water is converted into steam at a constant pressure
and temperature by the heat energy is received from the combustion of
the fuel.
b) The engine or turbine, in which the steam expands to a low pressure
causing work energy to be available.
c) The condenser, in which heat energy flows from the low pressure
steam into the condenser cooling water, resulting in the steam being
condensed.
d) The feed pump or the compressor, which returns the water into the
boiler.
11.2 A Carnot cycle is shown in the figure below.
T1 saturation temperature at 40 bar
= 250.3 + 273 = 523.3 K
T2 saturation temperature at 0.045 bar
= 31.0 + 273 = 304.0 K
T
s
1
23
4
523.3
304.0

14. THE STEAM POWER CYCLE J2006/11/14
Then from equation 11.5
1
21
T
TT
carnot
−
=η
=
3.523
3043.523 −
= 0.419 or 41.9 %
Also,
Heat supplied = h1-h4 = hfg at 42 bar = 1714 kJ/kg
Then,
419.0==
Q
W
carnotη
∴ W = 0.419 x 1714
i.e. W = 718.2 kJ/kg
To find the gross work of the expansion process it is necessary to calculate h2, using
the fact that s1 = s2.
From the tables,
h1 = 2801 kJ/kg and s1 = s2 = 6.070 kJ/kg K
From the equation
s2 = 6.070 = sf2 + x2sfg2 = 0.451 + x2 7.98
∴ x2 = 704.0
98.7
451.0070.6
=
−
Then,
h2 = hf2 + x2hfg2 = 130 + 0.704 x 2428
= 1839.3 kJ/kg
Hence, from equation 11.2,
W12 = (h1 – h2)
= (2801 – 1839.3)
= 961.7 kJ/kg
Therefore, using equation 11.7,

15. THE STEAM POWER CYCLE J2006/11/15
Work ratio workgross
net work
=
=
7.961
2.718
= 0.747

16. THE STEAM POWER CYCLE J2006/11/16
11.2 Rankine Cycle
Many of the impracticalities associated with the Carnot cycle can be eliminated by
condensing it completely in the condenser, as shown schematically on a T-s diagram
in Fig. 11.3. The cycle that results is the Rankine cycle, is the ideal cycle for vapour
power plants. The ideal Rankine cycle does not involve any internal irreversibility
and consists of the following processes:
4,5 to 1: Constant pressure heat addition in a boiler
1 to 2: Isentropic expansion taking place in the turbine or engine
2 to 3: Constant pressure heat rejection in the condenser
3 to 4: Isentropic compression of water in the feed pump
INPUTINPUT
In the Rankine cycle, the exhaust
steam is completely condensed into
water in the condenser.

17. THE STEAM POWER CYCLE J2006/11/17
From a comparison made between Fig. 11.1 and Fig. 11.2, the similarities between
the Carnot and the Rankine cycles can be clearly seen. In the Rankine cycle, the
exhaust steam is completely condensed into water in the condenser. It actually
follows the isentropic expansion in the turbine. This water is then pumped into the
boiler by a boiler feed pump. After the feed pump, since the water is not at the
saturation temperature corresponding to the pressure, some of the heat energy
supplied in the boiler is taken up by the water as sensible heat before evaporation can
begin. This results in the boiler process being no longer completely isothermal; the
process is, therefore, irreversible, causing the Rankine cycle to be an irreversible
cycle and to have a lower efficiency than the Carnot cycle.
Fig 11.2 The Rankine cycle
Boiler
Turbine
Condenser
Pump W12
Q23
W34
1
23
4
5
Q451
T
s
1
23
4
T1
T2
A B
5
p1
p2

18. THE STEAM POWER CYCLE J2006/11/18
The plant required and the numbers referring to the state points for the Rankine cycle
is shown in Fig. 11.2. The steam at the inlet to the turbine may be wet, dry saturated,
or superheated, but only the dry saturated condition is shown in Fig. 11.2. The steam
flows round the cycle and each process may be analysed using the steady flow energy
equation where changes in kinetic energy and potential energy may be neglected.
i.e. h1 + Q = h2 + W
In this statement of the equation the subscripts 1 and 2 refer to the initial and final
state points of the process; each process in the cycle can be considered in turn as
follows:
Boiler:
h4 + Q451 = h1 + W
Therefore, since W = 0,
Q451 = h1 – h4 (11.10)
Turbine:
The expansion is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2 ), and h2 can
be calculated using this latter fact. Then
h1 + Q12 = h2 + W12
∴ W12 = ( h1 – h2) (11.11)
Condenser:
h2 + Q23 = h3 + W23
Therefore, since W = 0
Q23 = h3 – h2
i.e. Q23 = - ( h2 – h3 )
∴ Heat rejected in condenser = h2 – h3 (11.12)

19. THE STEAM POWER CYCLE J2006/11/19
Pump:
h3 + Q34 = h4 + W34
The compression is isentropic ( i.e. s3 = s4 ), and adiabatic ( i.e. Q = 0 ).
∴ W34 = ( h3 – h4 ) = -( h4 – h3 )
i.e. Work input to pump = ( h4 – h3 ) (11.13)
This is the feed pump term, and as the quantity is small as compared to the
turbine work, W12. The feed pump is usually neglected, especially when the
boiler pressures are low.
The net work done in the cycle, W = W12 + W34
i.e. W = ( h1 – h2) – ( h4 – h3 ) (11.14)
Or, if the feed pump work is neglected,
W = ( h1 – h2 ) (11.15)
11.2.1 Thermal efficiency of Rankine cycle
Rankine efficiency,
boilerin thesuppliedHeat
outputNet work
=Rη (11.16)
i.e.
41
3421 )()(
hh
hhhh
R
−
−−−
=η
or
)()(
)()(
3431
3421
hhhh
hhhh
R
−−−
−−−
=η (11.17)
If the feed pump term, (h4 – h3) is neglected, equation (11.17) becomes
)(
)(
31
21
hh
hh
R
−
−
=η
(11.18)

20. THE STEAM POWER CYCLE J2006/11/20
When the feed pump term is included, it is necessary to evaluate the quantity, W34.
From equation (11.13)
Pump work = -W34 = νf (p4 – p3) (11.19)
11.2.2 The work ratio for Rankine cycle
It has been stated that the efficiency of the Carnot cycle is the maximum possible, but
that the cycle has a low work ratio. Both efficiency and work ratio are criteria of
performance. The work ratio is defined by
Work ratio workgross
net work
=
i.e. Work ratio =
)(
)()(
21
3421
hh
hhhh
−
−−−
or Work ratio =
)(
)(v)(
21
34f21
hh
pphh
−
−−−
(11.20)
11.3 Specific steam consumption
Another criterion of performance in steam plant is the specific steam consumption. It
relates the power output to the steam flow necessary to produce steam. The steam
flow indicates the size of plant with its component part, and the specific steam
consumption is a means whereby the relative sizes of different plants can be
compared.
The specific steam consumption is the steam flow in kg/h required to develop 1
kW,
i.e. W x (specific steam consumption, s.s.c.) = 1 x 3600 kJ/h
(where W is in kJ/kg),

21. A steam power plant operates between a boiler pressure of 42 bar and a
condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency,
the work ratio, and the specific steam consumption for a Rankine cycle with
dry saturated steam at entry to the turbine.
THE STEAM POWER CYCLE J2006/11/21
i.e. s.s.c. = kg/kwh
3600
W
or. s.s.c. = kg/kwh
)()(
3600
3421 hhhh −−−
(11.21)
Example 11.3
Solution to Example 11.3
The Rankine cycle is shown in the figure below.
As in example 11.2
h1= 2800 kJ/kg and h2 = 1808 kJ/kg
Also, h3 = hf at 0.035 bar = 112 kJ/kg
Using equation 11.18, with v = vf at 0.035 bar
T
s
1
23
4
526.2
299.7
A B
5
42bar
0.035 bar

22. THE STEAM POWER CYCLE J2006/11/22
Pump work = -W34 = νf (p4 – p3)
= 0.001 x ( 42 – 0.035) x 102
= 4.2 kJ/kg
Using equation 11.11
W12 = h1 – h2 = 2800 – 1808 = 992 kJ/kg
Then using equation 11.17
)()(
)()(
3431
3421
hhhh
hhhh
R
−−−
−−−
=η
=
)2.4()1122800(
)2.4()992(
−−
−
= 0.368 or 36.8 %
Using equation 11.7
Work ratio workgross
net work
=
=
992
4.2-992
= 0.996
Using equation 11.21
s.s.c. =
)()(
3600
3421 hhhh −−−
=
)2.4()992(
3600
−
= 3.64 kg/kW h

23. THE STEAM POWER CYCLE J2006/11/23
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
11.3 Based on the diagram below, describe the four-stage processes that represent a
steam plant operating on an ideal Rankine cycle.
11.4 A steam power plant operates between a boiler pressure of 40 bar and a
condenser pressure of 0.045 bar. Calculate for these limits the cycle
efficiency, the work ratio, and the specific steam consumption for a Rankine
cycle with dry saturated steam at entry to the turbine.
Activity 11B
Boiler
Turbine
Condenser
Pump
W12
Q23
W34
1
23
4
5
Q451

24. THE STEAM POWER CYCLE J2006/11/24
Feedback To Activity 11B
11.3 The ideal Rankine cycle does not involve any internal irreversibility and
consists of the following four-stage processes:
Stage 1.
4 to 1: Constant pressure heat addition in a boiler.
Stage 2.
1 to 2: Isentropic expansion that takes place in the turbine or engine.
Stage 3.
2 to 3: Constant pressure heat rejection in the condenser.
Stage 4
3 to 4: Isentropic compression of water in the feed pump
11.4 The Rankine cycle is shown in the figure below.
As in activity 11.2
h1= 2801 kJ/kg and h2 = 1839.3 kJ/kg
T
s
1
23
4
523.3
3040.
A B
5
40bar
0.045 bar

25. THE STEAM POWER CYCLE J2006/11/25
Also, h3 = hf at 0.045 bar = 130 kJ/kg
Using equation 11.18, with v = vf at 0.045 bar
Pump work = -W34 = νf (p4 – p3)
= 0.001 x ( 40 – 0.045) x 102
= 4.0 kJ/kg
Using equation 11.11
W12 = h1 – h2 = 2801 – 1839.3 = 961.7 kJ/kg
Then using equation 11.17
)()(
)()(
3431
3421
hhhh
hhhh
R
−−−
−−−
=η
=
)0.4()1302801(
)0.4()7.961(
−−
−
= 0.359 or 35.9 %
Using equation 11.7
Work ratio workgross
net work
=
=
961.7
4.0-961.7
= 0.996
Using equation 11.21
s.s.c. =
)()(
3600
3421 hhhh −−−
=
)0.4()7.961(
3600
−
= 3.76 kg/kW h

26. THE STEAM POWER CYCLE J2006/11/26
You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback to Self-Assessment on the
next page. If you face any problem, discuss it with your lecturer. Good luck.
1. Explain why the Rankine cycle and not the Carnot cycle is taken as the ideal
cycle for steam plant. Sketch the T-s diagram for these cycles when using
steam as the working fluid.
2. What is the highest thermal efficiency possible for a Carnot cycle operating
between 210o
C and 15o
C.
3. A steam power plant operates between a boiler pressure of 30 bar and a
condenser pressure of 0.04 bar. Calculate for these limits the cycle efficiency,
the work ratio and the specific steam consumption for a Carnot cycle using
wet steam.
4. A steam power plant operates between a boiler pressure of 30 bar and a
condenser pressure of 0.04 bar. Calculate for these limits the cycle efficiency,
the work ratio, and the specific steam consumption for a Rankine cycle with
dry saturated steam at entry to the turbine.
5. In a steam power plant, dry saturated steam enters the turbine at 47 bar and is
expanded isentropically to the condenser pressure of 0.13 bar. Determine the
Rankine cycle efficiency when
a) the feed pump work is neglected
b) the feed pump work is taken into account
SELF-ASSESSMENT

27. THE STEAM POWER CYCLE J2006/11/27
Have you tried the questions????? If “YES”, check your answers now.
1. Carnot cycle is never used in practice owing to:
1. The difficulty in stopping the condensation at 3, so that subsequent
compression would bring the state point to 4.
2. A very large compressor would be required.
3. Compression of wet steam in a rotary compressor is difficult as the water
tends to separate out.
4. Friction associated with the expansion and compression processes would
cause the net work done to be very small as compared to the work done in the
turbine itself.
The Carnot cycle is modified to overcome the above difficulties and this modified
cycle, known as the Rankine cycle, is widely used in practice.
Feedback to Self-Assessment
T
s
1
23
4T
1
T
2
T
s
1
23
4
T
1
T
2
A B
5
p1
p2
Carnot cycle Rankine cycle

28. THE STEAM POWER CYCLE J2006/11/28
2. 40.37 %
3. 40.4 %, 0.771, 4.97 kg/kW h
4. 34.6 %, 0.997, 3.88 kg/kW h
5. 33.67 %, 33.55 %
CONGRATULATION
S!!!!…May success be
with you always…