This document discusses various thermodynamic power cycles including: - The Carnot cycle, which is the most efficient but impractical cycle. - Rankine cycles, which are more practical vapor power cycles that use steam as the working fluid. - Simple Rankine cycles involve heating water to steam then expanding it in a turbine before condensing it back to water. - Rankine cycles with superheated steam, which increase efficiency by heating steam above its saturation temperature. - The efficiencies of different cycles are calculated and compared in examples. Superheated steam cycles have higher efficiencies than simple Rankine cycles due to higher average temperatures.

1. THERMODYNAMIC POWER CYCLES -
I
N S Senanayake
ME4001Applied
Thermodynami
cs

2. Content
 Introduction
 Power cycles
 Efficiency definition
 Introduction to Carnot Cycle and Carnot Efficiency
 Vapour power cycles
 Mollier charts (h – s)
 Carnot vapour power cycle
 Simple Rankine Power Cycle
 Rankine Cycle with super heat steam
 Comparison of efficiency for Carnot, Simple Rankine
cycle and Rankine cycle with Super heat steam

3. Introduction
In any thermal power generation plant, heat energy converts
into mechanical work. Then it is converted to electrical
energy by rotating a generator which produces electrical
energy.
Heat is derived from various energy sources like fossil fuels,
nuclear fusion of radioactive elements and geothermal
energy.
Potential energy stored in water also can produce
mechanical work, and then mechanical work to electrical
energy, this conversion of energy is commonly known as
hydropower generation. Kinetic energy of wind and solar
radiation are the other sources of energy that can be used to

4. Power cycles
• Carnot vapour power cycle
• Rankine Cycle
vapour power cycle
• Carnot gas power cycle
• Otto cycle
• Diesel cycle
• Dual cycle
• Brayton cycle
Gas power cycle
Depending on the types of processes involved power cycles
can also be classified as follows:

5. Efficiency
The overall efficiency of a thermal power plant is
the combination of two efficiencies.
 Combustion efficiency or the source efficiency
The proportion of the available energy transferred as
heat to the working fluid.
 Thermodynamic Cycle efficiency
This indicates the proportion of the heat supplied
to system which is finally converted to useful
mechanical work.
As the efficiency can never be 100% according
the Second Law of thermodynamics, our concern
is to raise the efficiency as much as possible. The
efficiency depends on the nature of
thermodynamic processes that constitute a cycle.

6. Carnot Cycle
Carnot Cycle is a reversible
thermodynamic cycle comprising of 4
processes.
1 -2 Reversible Isothermal heat
addition
2 – 3 Reversible adiabatic
expansion (isentropic)
3 – 4 Reversible Isothermal heat
rejection
4 -1 Reversible Adiabatic
compression (isentropic)
The temperature –entropy diagram (T-S
diagram) is the best form of cycle that
gives the maximum efficiency

7. Carnot efficiency
 Corollary 1 of Carnot's Theorem:
All reversible heat engines operating between the
same two heat reservoirs must have the same
efficiency.
 Corollary 2 of Carnot's Theorem:
The efficiency of a reversible heat engine is a function only of
the respective temperatures of the hot and cold reservoirs. It
can be evaluated by replacing the ratio of heat transfers
QL and QH by the ratio of temperatures TL and TH of the
respective heat reservoirs.

8. Carnot efficiency cont..
H
L
H
LH
H
R
th
Q
Q
Q
QQ
Q
W


 1
Thermal efficiency for a reversible heat engine can be
written as follows.
H
L
H
R
th
T
T
Q
W
 1
H
L
H
L
T
T
Q
Q
According to Carnot theorem

9. Vapor power cycles

10. Introduction to vapor power
cycles
 In converting heat to work it is necessary to have a substance
called working fluid to “store” energy and release it in the
form of mechanical work. Such a fluid should not upset the
system and its components that are needed to bring about
this heat and work transfer. You are aware of that it is the
water (or steam) we mostly use as the working substance in
thermal power plant. The thermodynamic processes
undergone by the working substance with the efficiency will
be discussed in detail in the next session.
 Although the Carnot Cycle is applicable to vapor power
cycles, because of the constant temperature heat absorption
and rejection (latent heat), there are practical difficulties to
exactly match the Carnot Cycle.

11. Enthalpy changes
In all thermodynamic power cycles we are
concerned with changes in enthalpy rather than
the enthalpy as a property. Therefore it is
convenient to use enthalpy-entropy diagram,
commonly known as Mollier diagram of steam
(h-s chart) in our calculations to determine heat
and work transfer, rather than using tables of
thermodynamic properties.

12. The enthalpy-entropy diagram

13. Carnot vapour power cycle
Carnot cycle is an ideal cycle having the highest thermodynamics efficiency
TH= T1= T2
TL = T3= T4

14. Carnot vapour power cycle…
Processes : all processes are reversible
1 – 2 : isothermal heat addition in the boiler
2 – 3 : adiabatic expansion in the turbine
3 – 4 : isothermal heat rejection in the condenser
4 – 1 : adiabatic compression in the feed pump
q12 = h2 – h1
- w23 = h3 – h2
q34 = h4 – h3
- w41 = h1 – h4

15. Carnot vapour power cycle…
addedHeat
Net Work
th
   
12
1432
hh
hhhh
th



4123 wwNet work  3412 qqheatNet 

















H
L
th
T
T
1
q
q
1
q
qq
q
ww
12
34
12
3412
12
4123

Thermal efficiency can also be expressed as:
Thermal efficiency

16. Net work done
T
q
s  sTq 
12BAArea,)( 1212  ssTq H
43BAArea,)( 2134  ssTq L
3412 qqheatNet 
43BAArea,)( 1234  ssTq L
   1234AreaTTheatNet 12LH  ss
For cycle net heat = net work
Net work

17. Example 1
Steam power plant operates between a boiler pressure
of 4MPa and condenser pressure of 50kPa. The steam
at 300oC enters the turbine. Assuming the cycle to be
Carnot cycle, Determine the thermal efficiency.
Also find the turbine work, feed pump work and heat
added.

18. solution

19. Thermal efficiency







H
L
th
T
T
1
From steam tables, the saturation temperature at 50kPa is 81.32
38.0
273300
27332.81
1 







th
Turbine work
- w23 = h3 – h2
h2 = 2961.7kJ/kg
State 2 is in the superheated
region as T2 is greater than
sat. temperature at 4MPa,
which is 250.35oC

20. h3 = hf3 + x3hfg3
hf3 = 340.54kJ/kg
hfg3 = 2304.7kJ/kg
x3 = ?
s3 = sf3 + x3sfg3
sf3 =1.0912kJ/kg
sfg3 = 6.5019kJ/kg
s3 = s2 = 6.3639kJ/kg
From steam tables at 50kPa
To find out dryness at state 3, consider the entropy
From steam tables at 50kPa
6.3639 = 1.0912 + x3 (6.5019)
x3 = 0.81

21. h3 = hf3 + x3hfg3
h3 = 340.54 + 0.81(2304.7) = 2207.35kJ/kg
Turbine work, w23 = (h2 – h3) = 2961.7 – 2207.35 =754.35kJ/kg
Heat added
q12 = h2 – h1
h1 =1344.8kJ/kg
h2 = 2961.7kJ/kg
For saturated liquid at 300oC
q12 = 2961.7 – 1344.8 = 1616.9kJ/kg

22. Feed pump work - w41 = h1 – h4
h1 =1344.8kJ/kg
h4 = hf4 + x4hfg4
hf4 = 340.54kJ/kg
hfg4 = 2304.7kJ/kg x4 = ?
From steam tables at 50kPa
To find out dryness at state 4, consider the entropy
sf4 =1.0912kJ/kg
sfg4 = 6.5019kJ/kg
s4 = s1 = 3.2548kJ/kg
From steam tables at 50kPa
3.2548 = 1.0912 + x4 (6.5019)
x4 = 0.33

23. h4 = hf4 + x4hfg4
h4 = 340.54 + 0.33 (2304.7) = 1101.09kJ/kg
- w41 = h1 – h4 = 1344.8 – 1101.09 = 243.71kJ/kg
Net work output = w23 + w41 = 754.35kJ/kg - 243.71kJ/kg =
510.65kJ/kg
31.0
1616.9
65.510
addedheat
net work
th
Net work
RatioWork 

24. Rankine cycle
 Rankine cycle is ideal cycle for vapour power
cycles
 The Rankine Cycle is similar to the Carnot
Cycle, but the Rankine Cycle is much more
practical because the working fluid typically
exists as a single phase (liquid or vapor) for
the two pressure change processes.

25. Processes in simple Rankine
cycle
 5 to 1: water is heated until it reaches
saturation (phase change / boiling
point) in a constant-pressure
process.
 1 to 2: Once saturation is reached,
further heat transfer takes place at
constant pressure, until the working
fluid is completely vaporized.
 2 to 3: The vapour is expanded
isentropically through a turbine to
produce work. The vapour (steam)
pressure falls as it passes through
the turbine and exits at low pressure.
 3 to 4: The working fluid is routed
through a condenser, where it
condenses (phase change) into liquid
(water).

26. Rankine cycle on h- s

27. Rankine cycle with superheat steam
To ensure that steam
leaving the turbine is
sufficiently dry, in
practical steam power
plant superheat steam is
used instead of
saturated steam to the
turbine input. However,
we need to ensure that
wet steam enters the
condenser to get the
maximum turbine output.

28. Advantages of super heating
• Average temperature at which heat is supplied
increased by superheating the steam and
therefore ideal cycle efficiency is increased.
• The specific steam consumption is markedly
reduced as the net work per unit mass of steam
is high.
• Since the requirement of steam per unit of
energy (kg/kWh) is less, the complexity added by
introducing the superheating system is
compensated by a smaller boiler.

29. There is an additional reason for using superheat
steam in practice. If you closely look at the T-S
diagram below, higher boiler pressures tends to give
wetter steam at turbine exhaust, which is not
desirable. Therefore superheating becomes necessary
to get dried steam at the turbine exhaust.

30. Effect of super heating

31. Example 2
Steam turbine plant operates on Rankine Cycle
with steam entering the turbine at 40bar, 350oC
and leaving at 0.05bar. Steam leaving the turbine
condenses to saturated liquid inside condenser.
Feed pump pumps saturated liquid into the boiler.
Determine the net work per kg of steam and the
cycle efficiency assuming all processes to be ideal.

32. Solution

33. From steam tables
h4 = hf4 = 137.75kJ/kg
At 0.05bar (5kPa) and saturated liquids4 = sf4 = 0.4762kJ/kg
v4 = vf4 = 0.001005m3/kg
h2 = 3093.3kJ/kg
At 40bar (4MPa) and 350oC
s2 = 6.5843kJ/kg
To find h3
s3 = sf3 + x3sfg3
sf3 = 0.4762kJ/kg
sfg3 = 7.9176kJ/kg
at 0.05 bar
s3 = s2
6.5843 = 0.4762 + x3 (7.9176)
x3 = 0.77

34. h3 = hf3 + x3hfg3
hf3 = 137.75kJ/kg
hfg3 = 2423.0kJ/kg
at 0.05 bar
h3 = 137.75 + 0.77(2423.0) =2003.46kJ/kg
To find h1
Consider the pump work w14 = h1 – h4 = vf4 (p2 – p1)
h1 = vf4 (p2 – p1) + h4 = 0.001005(4000 – 0.005)+137.75=141.78kJ/kg
Turbine work (w23) = h2 – h3 =(3093.3 - 2003.46) = 1089.84kJ/kg
Feed pump work (w41) = h1 – h4 =(141.78 – 137.75) = 4.03kJ/kg

35. Net work = w23 – w41
Net work = 1089.84 – 4.03 = 1085.81kJ/kg
3678.0
2951.52
1085.81
addedheat
Net work
Efficiency 
Heat added= h2 – h 1 =3093.3 - 141.78 =2951.52kJ/kg

36. Exercise
Calculate the ideal efficiency and specific steam
consumption for the following cases with 30bar
and 0.04bar turbine and condenser pressure
respectively
Carnot cycle
Simple Rankine cycle
Rankine Cycle with super heat steam at450oC

37. Comparison of different
cycles
Performance indicator Cycle Name
Carnot Simple
Rankine
Rankine with
super heat
Ideal efficiency, h 0.404 0.35 0.375
Specific steam consumption
(kg/kWh)
4.97 3.84 2.98

38. Isentropic efficiency - Turbine
The desired output from a turbine is the work
output. Hence, the definition of isentropic
efficiency of a turbine is the ratio of the actual
work output of the turbine to the work output of
the turbine if the turbine undergoes an isentropic
process between the same inlet and exit
pressures.
WorkIsentropic
WorkTurbineActual
T 

39. The isentropic efficiency of
turbine can be written as
12
12
hh
hh
s
a
T



h1 = enthalpy at the inlet
h2a = enthalpy of actual process at the exit
h2s = enthalpy of isentropic process at the
exit
12
12
TT
TT
s
a
T




40. Isentropic efficiency -
compressor
The isentropic efficiency of a compressor or pump is
defined as the ratio of the work input to an isentropic
process, to the work input to the actual process between
the same inlet and exit pressures.
WorkActual
WorkCompressorIsentropic
C 

41. The isentropic efficiency of
compressor can be written as
12
12
hh
hh
a
s
C



h1 = enthalpy at the inlet
h2a = enthalpy of actual process at the exit
h2s = enthalpy of isentropic process at the
exit
12
12 )(
hh
ppv
a
C




42. Example
If the isentropic efficiency for turbine is 0.8 redo
the example 1
3a

43. solution
From steam tables h2 = 3093.3kJ/kg
At 40bar (4MPa) and 350oC
s2 = 6.5843kJ/kg
h4 = hf4 = 137.75kJ/kg
At 0.05bar (5kPa) and saturated liquids4 = sf4 = 0.4762kJ/kg
v4 = vf4 = 0.001005m3/kg
To find h3
s3 = sf3 + x3sfg3
sf3 = 0.4762kJ/kg
sfg3 = 7.9176kJ/kg
at 0.05 bar
s3 = s2
6.5843 = 0.4762 + x3 (7.9176)
x3 = 0.77

44. h3 = hf3 + x3hfg3 hf3 = 137.75kJ/kg
hfg3 = 2423.0kJ/kg
at 0.05 bar
h3 = 137.75 + 0.77(2423.0) =2003.46kJ/kg
32
32
hh
hh a
T



84.089,1
3.3093
46.20033.3093
3.3093
8.0 33 aa hh 



 43.221,23 ah
To find h1
Consider the pump work
h1 = vf4 (p2 – p1) + h4 = 0.001005(4000 – 0.005)+137.75=141.78kJ/kg
Turbine work (w23) = h2 – h3a =(3093.3 – 2221.43) = 871.87kJ/kg
Feed pump work (w41) = h1 – h4 =(141.78 – 137.75) = 4.03kJ/kg

45. Net work = w23 – w41
Net work = 871.87 – 4.03 = 730.09kJ/kg
2473.0
2951.52
730.09
addedheat
Net work
Efficiency 
Heat added= h2 – h 1 =3093.3 - 141.78 =2951.52kJ/kg