THERMODYNAMICS

1. Power Cycles
10.1 INTRODUCTION
Various thermodynamic cycles are developed by scientists and researchers to achieve a predetermined
objective. From this point of view, they are classified into two different categories: power cycles and
refrigeration cycles.
On the basis of the phase of a working fluid, cycles are further classified into two: gas cycles and
vapour cycles.
Cycles can also be classified into thermodynamic (closed) and mechanical (open) cycles. In ther-
modynamic cycles, the working fluid is returned to the initial state at the end of the cycle and is re-
circulated, whereas in mechanical cycles, the working fluid is renewed at the end of each cycle.
Since the working substance in a vapour power cycle is steam/water, it is often referred as the
steam power cycle. Steam has many desirable characteristics such as high enthalpy of evaporation,
low cost, easy availability.
The four basic components of a steam power plant are boiler, turbine, condenser and pump—all
are put in a logical sequence as shown below in Fig. 10.1.
Water at an ambient temperature is supplied to the boiler by means of a pump. Heat is added (Q1) to
the water in the boiler to convert the working substance to steam. Steam so produced, having high
pressure and temperature, is allowed to expand in the turbine. Expansion of the steam in the turbine
causes the turbine blades to be rotated thus giving external work as output (WT). This mechanical work
is converted to the electrical energy by coupling a generator with the rotor of the turbine. The pressure
and temperature of the steam after expansion by turbine comes down and the condition of steam
becomes wet. However, to facilitate complete conversion of steam into water, further cooling is re-
quired at constant pressure and temperature (by rejecting latent heat). This is accomplished in the
condenser. Condenser is nothing but a heat exchanger that uses two fluids having different temperature
for mutual interaction of heat. The steam rejects latent heat (Q2) for complete conversion to water. This
water at low pressure and temperature is fed back to the boiler for further use by means of a feed pump,
requiring some work (WP) to be expended. This is how the cycle is continued.
CHAPTER
10

2. 10.2 Engineering Thermodynamics and Fluid Mechanics
Turbine
Boiler
Condenser
Pump
WT
WP
Q1
Q2
1
2
3
4
Figure 10.1 Schematic arrangement of simple steam power plant
10.2 CARNOT VAPOUR POWER CYCLE
A Carnot vapour power cycle comprises of two reversible isothermal and two reversible adiabatic
processes as shown in Fig. 10.2. Dry saturated steam enters the turbine and expands reversibly and
adiabatically to condenser pressure. The steam is then condensed at constant pressure and temperature.
The wet steam leaving the condenser is then compressed reversibly and adiabatically into the boiler
pressure. The saturated liquid is then evaporated in the boiler to complete the cycle.
P
1
2
3
4
V S
T
1
2
3
4
Figure 10.2 Carnot Vapour Cycle on P–v and T–s diagrams

3. Power Cycles 10.3
A Carnot cycle is not practicable for a steam power plant because of the following drawbacks:
(i) The isentropic compression process involves the compression of wet steam to a saturated
liquid. It is not practical to design a pump that can handle wet steam.
(ii) The rate of delivery of work is less, because of very large pump work.
(iii) The turbine that takes saturated vapour at the inlet produces wet steam with low quality. Thus the
turbine has to handle steam with high moisture content. The impingement of liquid droplets on the
turbine blades causes the erosion of the blades.
10.3 RANKINE CYCLE
The Rankine cycle is an ideal cycle for vapour cycles. The cycle is shown in Fig. 10.3 on P–v, T–s,
and h–s diagrams. The Rankine cycle comprises of the following processes:
4–1: Reversible constant pressure heat addition in a boiler
1–2: Reversible adiabatic expansion in a turbine
2–3: Reversible constant pressure heat rejection in a condenser
3–4: Reversible adiabatic compression in a pump
Dry saturated steam enters the turbine and expands reversibly and adiabatically to condenser pres-
sure. The steam is then condensed at constant pressure and temperature to a saturated liquid. The
saturated liquid leaving the condenser is then pumped reversibly and adiabatically into the boiler
pressure. The compressed liquid is first heated to the saturation temperature at boiler pressure and
then evaporated to the state 1 to complete the cycle.
p1
1
2
3
4
p2
h
S
P
V
T
S
1
2
3 4
1
2
3
4
Figure 10.3 Rankine Cycle on P–v, T–s and h–s diagram
10.4 RANKINE CYCLE EFFICIENCY
In the thermodynamic analysis of power cycles, our main objective is to estimate the thermal effi-
ciency. The thermal efficiency measures how successfully the energy input to the working fluid
passing through the boiler is converted to net work output.

4. 10.4 Engineering Thermodynamics and Fluid Mechanics
Considering unit mass of working fluid and neglecting the changes in kinetic and potential energy,
the first law of thermodynamics applied to each of the steady-flow devices (Fig. 10.1) gives
Boiler:
+ =
1 4 1
Q h h
or = -
1 1 4
Q h h (10.1)
Turbine:
= +
1 2 T
h h W
or = -
1 2
T
W h h (10.2)
Condenser:
= +
2 2 1
h Q h
or = -
2 2 3
Q h h (10.3)
Pump:
+ =
3 4
P
h W h
or = -
4 3
P
W h h (10.4)
The efficiency of the Rankine cycle is given by
- - -
-
h = = =
-
1 2 4 3
1 1 1 4
( ) ( )
net T P
W h h h h
W W
Q Q h h
(10.5)
The way to determine the enthalpy change across the pump is to use the thermodynamic property
relation Tds = dh – vdP for an isentropic process. It becomes
=
dh vdP
or D = Ú
h vdP
The pump handles liquid which is incompressible (specific volume is independent of pressure).
Therefore, one can write
D = D
h v P
or - = -
4 3 3 1 2
( )
h h v P P (10.6)
Note that the work required in the pumping process (Pump work) is quite small compared to the
turbine work and is sometimes neglected. Then, we have @
4 3
h h .
The efficiency of the Rankine cycle then becomes
-
h =
-
1 2
1 4
h h
h h
(10.7)

5. Power Cycles 10.5
Example 10.1 A steam power plant is designed to operate on
Rankine cycle. Steam enters the turbine as
saturated vapour at 30 bar and leaves as saturated
liquid in the condenser at 10 kPa. The mass flow
rate of steam is 1 kg/s. Determine the net power
output of the cycle and the thermal efficiency of
the Rankine cycle.
Solution
The T-s diagram of the Rankine cycle is shown in the Fig 10.4.
From the saturated steam table based on pressure (Appendix 1.2),
it is found that at 30 bar
= =
1 1
2804.1 kJ/kg, 6.1878 kJ/kgK
h s ,
Similarly from Appendix 1.2, we get at 10 kPa
= = = =
3
0.6491 kJ/kgK, 8.1510 kJ/kgK, 191.8 kJ/kg
f g f
s s h h ,
= = = 3
3
2392.8 kJ/kg, 0.00101 m /kg
fg f
h v v
To locate state 2, we recognize that = =
1 2 6.1878 kJ/kgK.
s s Hence,
( )
=
=
= + -
10
10
1 2
P kPa
P kPa
f g f
s s x s s
( )
= + -
2
6.1878 0.6491 8.1510 0.6491
x
=
2 0.7383
x
This allows us to find the specific enthalpy at the exit from the turbine, h2 to be
= + = + ¥ =
2 2 191.8 0.7383 2392.8 1958.4 kJ/kg
f fg
h h x h
The specific work output from the turbine is
= - = - =
1 2 2804.1 1958.4 845.7 kJ/kg
T
W h h
The pump work requirement for this ideal cycle is (refer to Fig. 10.1)
( ) ( )
3 1 2 0.00101 30 100 10 3.02 kJ/kg
P
W v P P
= - = ¥ - =
Net work output of the cycle is = - = - =
845.7 3.02 842.68 kJ/kg
net T P
W W W
Net power output of the cycle is = ¥ mass flow rate of steam
net
W
= ¥ =
842.68 1 842.68 kW
The specific enthalpy at the pump outlet, state 4, is the inlet specific enthalpy h3 plus WP. Thus,
4 3 191.8 3.02 194.82 kJ/kg
P
h h W
= + = + =
P2 = 10 kPa
P1 = 30 bar
T
S
1
2
3
4
Figure 10.4

6. 10.6 Engineering Thermodynamics and Fluid Mechanics
To calculate the thermal efficiency, we must know the boiler heat input. It is
= - = - =
1 1 4 2804.1 194.82 2609.28 kJ/kg
Q h h
The thermal efficiency of the Rankine cycle is then calculated to be
h = = =
1
842.68
0.3230 or 32.3%
2609.28
net
W
Q
Example 10.2 A thermal power plant is to be operated on an ideal Rankine cycle. Steam enters the
turbine at 30 bar and 400°C and leaves as saturated liquid in the condenser at 10 kPa.
The mass flow rate of steam is 1.5 kg/s. Determine the net power output of the cycle
and the thermal efficiency of the Rankine cycle.
Solution
The T-s diagram of the Rankine cycle is shown in the Fig 10.5.
P2 = 10 kPa
P1 = 30 bar
T
S
1
2
3
4
400 °C
Figure 10.5
From the superheated steam table (Appendix 1.3), it is found that at 30 bar and 400°C
= =
1 1
3230.9 kJ/kg, 6.9212 kJ/kgK
h s ,
From the saturated steam table based on pressure (Appendix 1.2), it is found that at 10 kPa
= = = =
3
0.6491 kJ/kgK, 8.1510 kJ/kgK, 191.8 kJ/kg
f g f
s s h h
= = = 3
3
2392.8 kJ/kg, 0.00101 m /kg
fg f
h v v
To locate state 2, we recognize that = =
1 2 6.9212 kJ/kgK.
s s Hence,
( )
=
=
= + -
10
10
1 2
P kPa
P kPa
f g f
s s x s s
( )
= + -
2
6.9212 0.6491 8.1510 0.6491
x
=
2 0.8361
x

7. Power Cycles 10.7
This allows us to find the specific enthalpy at the exit from the turbine, h2 to be
= + = + ¥ =
2 2 191.8 0.8361 2392.8 2192.42 kJ/kg
f fg
h h x h
The specific work output from the turbine is
= - = - =
1 2 3230.9 2192.42 1038.48 kJ/kg
T
W h h
The pump work requirement for this ideal cycle is (refer to Fig. 10.1)
( ) ( )
3 1 2 0.00101 30 100 10 3.02 kJ/kg
P
W v P P
= - = ¥ - =
Net power output of the cycle is = ¥ mass flow rate of steam
net
W
( )
= - ¥ mass flow rate of steam
T P
W W
( )
= - ¥ =
1038.48 3.02 1.5 1553.19 kW
The specific enthalpy at the pump outlet, state 4, is the inlet specific enthalpy h3 plus Wp. Thus,
4 3 191.8 3.02 194.82 kJ/kg
P
h h W
= + = + =
To calculate the thermal efficiency, we must know the boiler heat input. It is
= - = - =
1 1 4 3230.9 194.82 3036.08 kJ/kg
Q h h
The thermal efficiency of the Rankine cycle is then calculated to be
-
h = = =
1
1038.48 3.02
0.3411 or 34.11%
3036.08
net
W
Q
Example 10.3 An ideal Rankine cycle operating between temperature of 500°C and 50°C. Calculate
the cycle efficiency and the quality of steam at the turbine outlet if the pump outlet
pressure is 2 MPa.
Solution The T-s diagram of the Rankine cycle is shown in the Fig 10.6.
50 °C
P1 = 2 MPa
T
S
1
2
3
4
500 °C
Figure 10.6

8. 10.8 Engineering Thermodynamics and Fluid Mechanics
From the superheated steam table (Appendix 1.3), it is found that at 2 MPa and 500°C
= =
1 1
3467.6 kJ/kg, 7.4317 kJ/kgK
h s
From the saturated steam table based on temperature (Appendix 1.1), it is found that at 50°C
= = = =
3
0.7036 kJ/kgK, 8.0771 kJ/kgK, 209.3 kJ/kg
f g f
s s h h
= = = =
3
3 2
2382.8 kJ/kg, 0.001012 m /kg, 0.01235 MPa
fg f
h v v P
To locate state 2, we recognize that Hence,
( )
10 kPa
10 kPa
1 2
P
P
f g f
s s x s s
=
=
= + -
( )
= + -
2
7.4317 0.7036 8.0771 0.7036
x
=
2 0.9125
x
This allows us to find the specific enthalpy at the exit from the turbine, h2 to be
= + = + ¥ =
2 2 209.3 0.9125 2382.8 2383.8 kJ/kg
f fg
h h x h
The specific work output from the turbine is
= - = - =
1 2 3467.6 2383.6 1084 kJ/kg
T
W h h
The pump work requirement for this ideal cycle is (refer to Fig. 10.1)
( ) ( ) 3
3 1 2 0.001012 2 0.01235 10 2.01 kJ/kg
P
W v P P
= - = ¥ - ¥ =
The specific enthalpy at the pump outlet, state 4, is the inlet specific enthalpy h3 plus WP. Thus,
4 3 209.3 2.01 211.31 kJ/kg
P
h h W
= + = + =
The thermal efficiency of the Rankine cycle is then calculated from equation (10.5) as
- -
h = = = =
- -
1 1 4
1084 2.01
0.3323 or 33.23%
3467.6 211.31
net T P
W W W
Q h h
Example 10.4 A steam power plant is designed to operate on Rankine cycle. Steam enters the
turbine at 4 MPa and 500°C and leaves as saturated liquid in the condenser at 10 kPa.
Calculate (a) the thermal efficiency with pump work included, (b) the thermal
efficiency neglecting pump work, and (c) the percentage error in efficiency
neglecting pump work.
Solution The T-s diagram of the Rankine cycle is shown in the Fig 10.7.

9. Power Cycles 10.9
P2 = 10 kPa
P1 = 4 MPa
T
S
1
2
3
4
500 °C
Figure 10.7
From the superheated steam table (Appendix 1.3), it is found that at 4 MPa and 500°C
= =
1 1
3445.3 kJ/kg, 7.0901 kJ/kgK
h s
From the saturated steam table based on pressure (Appendix 1.2), it is found that at 10 kPa
= = = =
3
0.6491 kJ/kgK, 8.1510 kJ/kgK, 191.8 kJ/kg
f g f
s s h h
= = = 3
3
2392.8 kJ/kg, 0.00101 m /kg
fg f
h v v
To locate state 2, we recognize that = =
1 2 7.4317 kJ/kgK
s s . Hence,
( )
=
=
= + -
10
10
1 2
P kPa
P kPa
f g f
s s x s s
( )
= + -
2
7.0901 0.6491 8.1510 0.6491
x
=
2 0.8586
x
This allows us to find the specific enthalpy at the exit from the turbine, h2 to be
= + = + ¥ =
2 2 191.8 0.8586 2392.8 2246.26 kJ/kg
f fg
h h x h
The specific work output from the turbine is
= - = - =
1 2 3545.3 2246.26 1199.04 kJ/kg
T
W h h
(a) The pump work requirement for this ideal cycle is (refer to Fig. 10.1)
( ) ( )
3
3 1 2 0.00101 4 10 10 4.03 kJ/kg
P
W v P P
= - = ¥ ¥ - =
The specific enthalpy at the pump outlet, state 4, is the inlet specific enthalpy h3 plus WP. Thus,
4 3 191.8 4.03 195.83 kJ/kg
P
h h W
= + = + =
The thermal efficiency of the Rankine cycle is then calculated from equation (10.5) as
h = = = =
- -
1 1 4
1199.04
0.369 or 36.9%
3445.3 195.83
net net
W W
Q h h

10. 10.10 Engineering Thermodynamics and Fluid Mechanics
(b) Neglecting the pump work, we have ª =
4 3 191.8 kJ/kg
h h
Then the thermal efficiency of the Rankine cycle becomes
h = = = =
- -
1 1 4
1199.04
0.3685 or 36.85%
3445.3 191.8
net net
W W
Q h h
(c) The percentage error in efficiency neglecting pump work is
-
= =
0.369 0.3685
0.001357 or 0.1357%
0.3685
Example 10.5 A thermal power plant is to be operated on an ideal Rankine cycle. Steam enters into
the turbine at 2 MPa, 360°C and leaves as saturated liquid in the condenser at 8 kPa.
The pump feeds the water back into the boiler. Assume ideal processes, find, per kg
of steam, the net work and the cycle efficiency.
Solution The T-s diagram of the Rankine cycle is shown in the Fig 10.8.
P2 = 8 kPa
P1 = 2 MPa
T
S
1
2
3
4
360 °C
Figure 10.8
At 2 MPa and 360 °C steam is in superheated state. However, at that condition, properties are not
given in the steam table (Appendix 1.3). Properties are then obtained by linear interpolation of the
properties at 2 MPa , 350°C and 2 MPa, 400°C.
From the superheated steam table (Appendix 1.3), it is found that
at 2 MPa and 350 °C 3137.0 kJ/kg, 6.9563 kJ/kgK
h s
= =
at 2 MPa and 400 °C 3247.6 kJ/kg, 7.1271 kJ/kgK
h s
= = ,
Using linear interpolation, specific enthalpy and entropy at 2 MPa and 360°C are found to be
1 1
3137.0 6.9563 360 350
3247.6 3137.0 7.1271 6.9563 400 350
h s
- - -
= =
- - -
1 1
3159.12 kJ/kg, 6.9905 kJ/kgK
h s
= =
From the saturated steam table based on pressure (Appendix 1.2), it is found that at 8 kPa
3
0.5924 kJ/kgK, 8.2295 kJ/kgK, 173.9 kJ/kg
f g f
s s h h
= = = =

11. Power Cycles 10.11
3
3
2403.1 kJ/kg, 0.001008 m /kg
fg f
h v v
= = =
To locate state 2, we recognize that 1 2 7.4317 kJ/kgK
s s
= = . Hence,
( )
10
10
1 2
P kPa
P kPa
f g f
s s x s s
=
=
= + -
( )
2
6.9905 0.5924 8.2295 0.5924
x
= + -
2 0.8378
x =
This allows us to find the specific enthalpy at the exit from the turbine, h2 to be
2 2 173.9 0.8378 2403.1 2187.22 kJ/kg
f fg
h h x h
= + = + ¥ =
The specific work output from the turbine is
1 2 3159.12 2187.22 971.9 kJ/kg
T
W h h
= - = - =
The pump work requirement for this ideal cycle is (refer to Fig. 10.1)
( ) ( )
3
3 1 2 0.001008 2 10 8 2.008 kJ/kg
P
W v P P
= - = ¥ ¥ - =
Net work output of the cycle is 971.9 2.008 969.892 kJ/kg
net T P
W W W
= - = - =
The specific enthalpy at the pump outlet, state 4, is the inlet specific enthalpy h3 plus Wp. Thus,
4 3 173.9 2.008 175.908 kJ/kg
P
h h W
= + = + =
To calculate the thermal efficiency, we must know the boiler heat input. It is
1 1 4 3159.12 175.908 2983.212 kJ/kg
Q h h
= - = - =
The thermal efficiency of the Rankine cycle is then calculated to be
1
969.892
0.3251 or 32.51%
2983.212
net
W
Q
h = = =
Example 10.6 A steam power plant is designed to operate on Rankine cycle. Steam enters into the
turbine at 2 MPa, 400°C and leaves as saturated liquid in the condenser at 10 kPa.
The mass flow rate of steam is 1 kg/s. Find out the power developed by the turbine
and the efficiency of the cycle. Assume the efficiencies of the turbine and the pump
as 0.85 and 0.8 respectively.
Solution The T-s diagram of the Rankine cycle is shown in the Fig 10.9.

12. 10.12 Engineering Thermodynamics and Fluid Mechanics
P2 = 10 kPa
P1 = 2 MPa
T
S
1
2
3
4
400 °C
2s
4s
Figure 10.9
From the superheated steam table (Appendix 1.3), it is found that at 2 MPa, 400°C
1 1
3247.6 kJ/kg, 7.1271 kJ/kgK
h s
= = ,
From the saturated steam table based on pressure (Appendix 1.2), it is found that at 10 kPa
3
0.6491 kJ/kgK, 8.1510 kJ/kgK, 191.8 kJ/kg
f g f
s s h h
= = = =
3
3
2392.8 kJ/kg, 0.00101 m /kg
fg f
h v v
= = =
To locate state 2, we recognize that 1 2 7.1271 kJ/kgK.
s s
= = Hence,
( )
10
10
1 2
P kPa
P kPa
f g f
s s x s s
=
=
= + -
( )
2
7.1271 0.6491 8.1510 0.6491
s
x
= + -
2 0.8645
s
x =
This allows us to find the specific enthalpy at the exit from the turbine, h2 to be
2 2 191.8 0.8635 2392.8 2257.98 kJ/kg
s f s fg
h h x h
= + = + ¥ =
The specific work output from the turbine is
( ) ( )
1 2 1 2 0.85 3247.6 2257.98 841.18 kJ/kg
T T s
W h h h h
h
= - = - = - =
The pump work is
( ) ( )
10
3
1 2 0.00101 2 10 10
2.51 kJ/kg
0.8
P kPa
f
P
P
v P P
W
h
=
- ¥ ¥ -
= = =
This allows us to find h4 to be
4 3 191.8 2.51 194.31 kJ/kg
P
h h W
= + = + =
The heat input is found using 1 1 4 3247.6 194.31 3053.29 kJ/kg
Q h h
= - = - =
Net work output of the cycle is 841.18 2.51 838.67 kJ/kg
net T P
W W W
= - = - =

13. Power Cycles 10.13
Net power output of the cycle is mass flow rate of steam
net
W
= ¥
838.67 1 838.67 kW
= ¥ =
The thermal efficiency of the Rankine cycle is then calculated to be
1
838.67
0.2747 or 27.47%
3053.29
net
W
Q
h = = =
SUMMARY
A cycle which continuously converts heat into work is called a power cycle.
A cycle which produces refrigeration effect is called a refrigeration cycle.
The Carnot vapour cycle serves as an ideal cycle for vapour power cycle, but not
practicable for a steam power plant.
The Rankine cycle is an ideal cycle for vapour power cycle.
The thermal efficiency of the Rankine cycle is given by
h = net
1 1
T P
W W W
=
Q Q
REVIEW QUESTIONS
10.1 What are the four basic components of a steam power plant working on Rankine cycle? Show by a
block diagram.
10.2 Draw the nature of P–V and T–S plots of a Rankine cycle (with saturated steam at turbine inlet).
10.3 Draw the nature of P–V and T–S plots of a Rankine cycle (with superheated steam at turbine inlet).
10.4 Why is a Carnot cycle not practicable for a steam power plant?
NUMERICAL PROBLEMS
10.1 Steam is the working fluid in a Rankine cycle. Steam enters the turbine as saturated vapour at 40 bar and
leaves as saturated liquid in the condenser at 10 kPa. Determine the thermal efficiency of the Rankine
cycle.
10.2 A steam power plant is operated on an ideal Rankine cycle. Steam enters the turbine as saturated
vapour at 150 bar and leaves as saturated liquid in the condenser at 20 kPa. Determine the thermal
efficiency of the Rankine cycle.
10.3 A steam power plant is operated on an ideal Rankine cycle. Steam enters the turbine at 20 bar, 400 ºC
and leaves as saturated liquid in the condenser at 8 kPa. Determine the thermal efficiency of the Rankine
cycle.
10.4 A steam power plant operates between the pressures of 10 kPa and 2 MPa with a maximum temperature of
400°C. What is the maximum efficiency possible from the power cycle?
10.5 A steam power plant operates on a Rankine cycle with a condenser outlet temperature of 80°C and
boiler outlet temperature of 500°C. The pump outlet pressure is 2 MPa. Determine the maximum
possible thermal efficiency of the cycle.

14. 10.14 Engineering Thermodynamics and Fluid Mechanics
MULTIPLE-CHOICE QUESTIONS
10.1 Thermal power plant works on
(a) Carnot cycle (b) Otto cycle (c) Rankine cycle (d) Joule cycle
10.2 Carnot cycle is
(a) a reversible cycle (b) an irreversible cycle
(c) a semi-reversible cycle (d) an adiabatic irreversible cycle
10.3 The working substance for a Carnot cycle is
(a) atmospheric air (b) air-fuelmixture (c) ideal gas (d) steam