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# Thermodynamics for Beginners - Chapter 12 THERMODYNAMIC ANALYSES OF POWER PLANTS

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Power plants are where power is produced such as in the electricity generating stations and turbojet engines. The workings of these power plants are complicated. In this chapter, however, we will learn about the basic working principles governing a few simple power plants and about carrying out thermodynamic analyses of simplified power plants to determine vital parameters such as the overall thermal efficiency. This chapter is written to make the students of this book appreciate some real life applications of what they have so far learned in thermodynamics.
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Chapter 12
THERMODYNAMIC ANALYSES
OF POWER PLANTS
It sounded an excellent plan, no doubt, and very neatly and simply
arranged; the only diﬃculty was that she had not the smallest idea
Lewis Carrol (Alice in Wonderland)
Power plants are where power is produced such as in the electricity gen-
erating stations and turbojet engines. The workings of these power plants
are complicated. In this chapter, however, we will learn about the basic
working principles governing a few simple power plants and about carrying
out thermodynamic analyses of simpliﬁed power plants to determine vital
parameters such as the overall thermal eﬃciency. This chapter is written
to make the students of this book appreciate some real life applications of
what they have so far learned in thermodynamics.
274 Chapter 12
12.1 Gas Turbine for
Electric Power Generation
Water, as we have seen in Chapter 1, is used to rotate the shaft of a
turbine in a hydroelectric power station. In a gas turbine used for electric
power generation, the shaft of the turbine is rotated by the gases produced
in the burning of a fuel with air. The basic working principle of a simple
gas turbine used for electric power generation is explained below with the
help of the ﬂow diagram shown in Figure 12.1.
combustion
chamber
compressor turbine
exhaust gases released
into the atmosphere
shaft
fuel
shaft power
for generation
of electricity
atmospheric
air
Figure 12.1 Flow diagram of a simple gas turbine power plant.
gases
compressed
air
Air is drawn from the atmosphere and is compressed to a high pressure
in a compressor. The high pressure air enters a combustion chamber, in
which fuel is sprayed onto the compressed air, and the fuel-air mixture is
burned at constant pressure. The gases leaving the combustion chamber
at high pressure and high temperature are directed towards the turbine
blades so as to rotate the turbine shaft. In general, nearly half of the work
output of the rotating turbine shaft is used to rotate the compressor shaft,
and the rest is used to produce electricity by spinning a coil between the
poles of a magnet or an electromagnet. Thus, part of the heat generated
in burning the fuel in the combustion chamber is converted into useful
electrical energy.
Thermodynamic Analyses of Power Plants 275
The gases leaving the turbine are released into the atmosphere. These
gases contain carbon dioxide, nitrous oxides, sulfur oxides and particulate
matter. The temperature of the exhaust gases can also be very high. Be-
fore releasing these gases into the atmosphere, therefore, it is essential to
make sure that the polluting potential of these exhaust gases is reduced to
the level set by the environmental authority of the country concerned. Even
though there is absolutely nothing one could do about the carbon dioxide,
a product of complete combustion of the fuel used, into the atmosphere.
Example 12.1
Consider the gas turbine power plant shown
in Figure 12.1. Atmospheric air at 1 bar and 300 K is drawn into a com-
pressor at a mass ﬂow rate of 350 kg/s, and is compressed to 5 bar. The
compressed air is heated at constant pressure in the combustion chamber
by burning the fuel injected onto the air ﬂowing through the combustion
chamber. The gases leave the combustion chamber at 1200 K, and en-
ter the turbine of the power plant, rotate the turbine shaft, and leave the
turbine at 1 bar.
(a) Determine the power input to the compressor, power production of
the turbine, and the power available for electricity generation. Also,
determine the back work ratio, deﬁned as the ratio of the compressor
work to the turbine work.
(b) Calculate the heat input to the combustion chamber. Also, determine
the thermal eﬃciency of the power plant, deﬁned as the ratio of the
net work output to the heat input.
(c) Determine the air-fuel ratio, assuming the heating value of the fuel
as 42 MJ/kg.
(d) Discuss the validity of the solutions obtained above.
Solution to Example 12.1
The block diagram of the given problem is shown in Figure 12.2.
276 Chapter 12

 
 

X
X
X
X
X
X
X
X
1234
1bar
300 K
5bar
T2=?
5bar
1200 K
1bar
T4=?
[(Ws)in]comp
[(Ws)out]turb
[Qin]comb
compressor
chamber
combustion
turbine
T
T
T
T
T
T
Figure 12.2 Block diagram for Example 12.1.
Let us work out the problem under ideal conditions making the following
assumptions:
Assumption 1: Assume that the ﬂows through the compressor and the turbine
are adiabatic and reversible. Then, (7.31) could be used to relate the pres-
sure and the temperature at the inlet to the pressure and the temperature
at the exit.
Assumption 2: Neglect the increase in the mass ﬂow rate of gases leaving the
combustion chamber as a result of fuel injection. That is, the mass ﬂow
rates through the compressor, combustion chamber and the turbine are
all 350 kg/s.
Assumption 3: Assume Cpas 1.005 kJ/kg ·Kandγas 1.4 for both the air and
the combustion gases.
Assumption 4: Assume that all the gases involved in the problem behave like
ideal gases, with constant speciﬁc heats.
Assumption 5: Neglect the potential and kinetic energy changes.
(a) First, let us determine the power input to the compressor. Using (10.11),
we have
[( ˙
Ws)in]comp mC
p(T2T1) = 350 ×1.005 ×(T2300) kJ/s
of which T2is unknown. Since the ﬂow through the compressor is assumed to
be reversible and adiabatic, (7.31) gives
T2=T1P2
P1(γ1)
= 300 K×5
10.4/1.4
= 475.2K
Thermodynamic Analyses of Power Plants 277
Therefore, [( ˙
Ws)in]comp = 61.63 MJ/s. That is, the power requirement of
the compressor is 61.63 MW.
Let us now determine the work output of the turbine. Using (10.9), we have
[( ˙
Ws)out]turb mC
p(T3T4) = 350 ×1.005 ×(1200 T4)kJ/s
of which T4is unknown. Since the ﬂow through the turbine is assumed to be
T4=T3P4
P3(γ1)
= 1200 K×1
50.4/1.4
= 757.7K
Therefore, [( ˙
Ws)out]turb = 155.58 MJ/s. That is, the power output of the
turbine is 155.58 MW.
The power available for electricity generation is the diﬀerence between the
power output of the turbine (= 155.58 MW) and the power requirement of the
compressor (= 61.63 MW), which is 93.95 MW.
The back work ratio is calculated as
bwr =compressor work input
turbine work output =61.63 MW
155.58 MW =39.6%
Comment: About 39.6% of the work produced by the turbine is consumed by
the compressor to compress the ambient air to 5 bar pressure. Only the remain-
ing 60.4% is available for electricity generation. One of the major drawbacks in
using a gas turbine power plant for electricity generation is the relatively large
part of the turbine work output being consumed by the compressor.
(b) Heat input to the combustion chamber can be calculated by applying (10.3)
to the air ﬂowing through the combustion chamber as
[˙
Qin]comb mC
p(T3T2)
= 350 ×1.005 ×(1200 475.2) kJ/s = 254.95 MW
where the fuel ﬂow rate through the combustion chamber is neglected.
The thermal eﬃciency of the gas turbine plant is calculated as
ηth =net work output of the gas turbine plant
heat input to the gas turbine plant
=155.58 MW 61.63 MW
254.95 MW
=93.95
254.95 =36.9%
278 Chapter 12
Comment: Only 36.9% of the 254.95 MW heat input to the combustion cham-
ber is available for electricity generation. The remaining 63.1%, which is about
161 MW, ends up in the atmosphere with the exhaust gases released into the
atmosphere at 757.7 K by the turbine. Note that for each MW of electric power
generation there is at least 1.7 (=161/93.95) MW of power wasted. All this
waste energy reaching the environment is a source of thermal pollution.
(c) Since the heating value of the fuel is assumed to be 42 MJ/kg, the mass
ﬂow rate of fuel supplied to the combustion chamber can be calculated as
mass ﬂow rate of the fuel =[˙
Qin]comb
heating value of the fuel
=254.95 MJ/s
42 MJ/kg
=6.07 kg/s
Comment: The fuel-air mass ratio is therefore about 0.017. In other words,
the fuel mass ﬂow rate is only about 1.7% of the air mass ﬂow rate.
(d) The solutions obtained above are valid only under the ﬁve assumptions made.
In reality, Assumption 1 hardly holds. That is, the ow through the compressor
and the ﬂow through the turbine are far from reversible and adiabatic in reality.
Therefore, the actual exit temperatures of the air leaving the compressor and
the gases leaving the turbine would be very diﬀerent from those calculated in
part (a). It will cause the actual amount of work input to the compressor to
increase and the actual amount of work produced by the turbine to decrease.
These diﬀerences will cause the back work ratio to increase and the thermal
eﬃciency to decrease.
Assumption 2 is also not valid since the mass ﬂow rate through the turbine
is greater than the mass ow rate through the compressor, and the diﬀerence is
the mass ﬂow of the fuel added to the combustion chamber. However, since the
fuel mass ﬂow rate is only about 1.7% of the air mass ﬂow rate, as calculated
in part (c), Assumption 2 would not have introduced any appreciable error. For
the same reason, treating the combustion gases as air, as per Assumption 3,
would not have introduced any appreciable error, either. Assumptions 4 and 5,
however, would introduce some bias in the solutions obtained.
Nevertheless, the solutions obtained above under ideal conditions certainly
give a very good idea about the performance of the gas turbine. For example,
the eﬃciency of the gas turbine in reality will always be less than, and never
more than, 36.9%. And, more than 63.1% of the heat input to the combustion
chamber will be lost to the environment, causing thermal pollution.
Thermodynamic Analyses of Power Plants 279
Example 12.2
Rework Example 12.1 with a regenerator
unit included in the gas turbine power plant as shown in Figure 12.3.
combustion
chamber
compressor turbine
gases
shaft
fuel
atmospheric
air
Figure 12.3 Flow diagram of a gas turbine power plant with regeneration.
regenerator
T1
T2T5T3
T4
T6
exhaust
gases
A regenerator is used to recover part of the energy lost to the envi-
ronment with the exhaust gases leaving the turbine at high temperatures.
It is simply a heat exchanger in which the compressed air is preheated by
the hot gases leaving the turbine as shown in the ﬁgure. Let us assume
that adiabatic conditions prevails at the regenerator and that there is no
pressure drop across it.
(a) Determine the power input to the compressor, power production of
the turbine, and the power available for electricity generation. De-
termine also the back work ratio.
(b) Calculate the heat input to the combustion chamber, and the thermal
eﬃciency of the gas turbine plant.
Solution to Example 12.2
(a) The pressure and the temperature of the air entering the compressor, the
pressure of the air leaving the compressor, the temperature of the gases entering
the turbine, and the pressure of the gases leaving the turbine are all the same as in
Example 12.1. Hence, the exit temperature of the air leaving the compressor,
280 Chapter 12
the work input to compressor, the exit temperature of the gases leaving the
turbineandtheworkproducedbytheturbineareallthesameasthosecalculated
in the Solution to Example 12.1.
That is, T2= 475.2 K, [( ˙
Ws)in]comp = 61.63 MW, T4= 757.7 K, and
[( ˙
Ws)out]turb = 155.58 MW. Therefore, the power available for electricity gener-
ation and the back work ratio are the same as those calculated in the Solution
to Example 12.1, which are 93.95 MW and 36.9%, respectively.
(b) The use of regenerator would cause the heat input to the combustion cham-
ber to be lower than that was calculated in the Solution to Example 12.1,
since the compressed air is preheated by the hot gases leaving the turbine in the
regenerator before it enters the combustion chamber.
Since the regenerator is a heat exchanger, assuming adiabatic condition,
(10.19) can be used to get
˙mgases
˙mair
=(Cp)air(T5T2)
(Cp)gases(T4T6)
Since the mass ﬂow rate of the gases is assumed to be the same as that of
the air, Cpisassumedtobethesameforbothairandgases,T2= 475.2 K and
T4= 757.7 K, we get
T5475.2K= 757.7KT6
where T5and T6are unknown.
For the heat to ﬂow from the gases to the air within the heat exchanger,
T6should be higher than T2(= 475.2 K). If, for example, we take T6to be
500 K then T5would be 732.9 K. The heat input to the air ﬂowing through the
combustion chamber can then be determined as
[˙
Qin]comb mC
p(T3T5)
= 350 ×1.005 ×(1200 732.9) kJ/s = 164.30 MJ/s
which is about 90.65 MW less than the heat input to the combustion chamber
without a regenerator as in the Example 12.1.
The thermal eﬃciency of the gas turbine plant is calculated as
ηth =net work output of the gas turbine plant
heat input to the gas turbine plant
=155.58 MW 61.63 MW
164.30 MW =93.95
164.30 =57.2%
Thermodynamic Analyses of Power Plants 281
Comment: When using a regenerator to preheat the compressed air by the hot
gases leaving the turbine, the heat requirement of the combustion chamber is
reduced. So that the thermal eﬃciency of the gas turbine power plant increases
to 57.2% from 36.9%, obtained without the regenerator in the Solution to
Example 12.1. That is, about 57.2% of the 164.30 MW heat input to the
combustion chamber is available for electricity generation. The remaining 42.8%
of the heat input, which is about 70 MW, reaches the atmosphere with the ex-
haust gases released into the atmosphere at 500 K by the regenerator. Note that
for each MW of electric power generation there is about 0.75 (=70/93.95) MW
of power wasted into the environment. Thermal pollution caused by the gas
turbine power plant with a regenerator is therefore considerably lower than the
thermal pollution caused by the gas turbine power plant without a regenerator.
12.2 Gas Turbine for Jet Propulsion
Gas turbine is an essential part of a turbojet engine commonly used for
aircraft propulsion. A turbojet engine consists of a diﬀuser, compressor,
combustion chamber, turbine and a nozzle, as shown in the schematic of a
turbojet engine in Figure 12.4.
diﬀuser compressor
combustion
chamber
tur-
bine nozzle
Figure 12.4 Schematic of a turbojet engine.
1
234
5
6
- - - - -
282 Chapter 12
The inlet of a turbojet engine is shaped into a diﬀuser in which the
ambient air entering the turbojet engine is slowed down to near zero speed.
Reduction in the speed of the air ﬂowing through the diﬀuser is accom-
panied by increase in the air pressure. The slightly compressed air enters
the compressor in which it is compressed to a high pressure. The exit
pressure of the compressor can be 7 to 25 times higher than the inlet pres-
sure. The high pressure air leaving the compressor enters the combustion
chamber, where fuel is sprayed onto the compressed air and the resultant
mixture is burned. The hot products of combustion at a high pressure and
a temperature enter the turbine, and sets the turbine shaft on rotation.
The compressor-combustion chamber-turbine combination of the tur-
bojet engine functions in a manner very similar to that of the gas turbine
power plant discussed in Section 12.1. In the turbojet engine, however, tur-
bine produces just enough power to drive the compressor shaft and some
other devices such as a small generator. Therefore, almost all the work
output of the turbine can be considered as being consumed by the com-
pressor in the turbojet engine, unlike in the case of the gas turbine power
plant in which a good portion of the turbine power output is used for the
generation of electricity.
The gases leaving the turbine of the turbojet engine enter a nozzle, and
attain high speed as they ﬂow through the nozzle which is accompanied
by a pressure reduction. The high speed gases leaving the turbojet engine
imparts a thrust on the aircraft so as to propel the aircraft forward. The
thrust equation for a turbojet engine can be derived from the Newton’s
second law, for the case where the inlet pressure is the same as the exit
pressure, as
Net Thrust mair (Vexit −V
inlet)(12.1)
where Vexit is the exit velocity of the gases leaving the nozzle relative to
the aircraft and Vinlet is the inlet velocity of the air entering the diﬀuser
relative to the aircraft. In Figure 12.4, the inlet is marked by 1 and the exit
is marked by 6. In deriving (12.1), we have omitted the fact that the mass
ﬂow rate of the gases leaving the turbojet engine diﬀers from the mass ﬂow
rate of air entering the engine by the mass ﬂow rate of the fuel supplied to
the combustion chamber. Since the fuel-air mass ratio used in the turbojet
engine is usually very small, the above omission does not introduce any
appreciable error in the calculation of the net thrust.
Thermodynamic Analyses of Power Plants 283
Example 12.3
Air enters a diﬀuser of the turbojet engine
of an aircraft ﬂying at a speed of 300 m/s at an altitude where the pressure
is 0.95 bar and the temperature is 10oC. Air leaving the diﬀuser with a
negligibly small speed enters the compressor which has a pressure ratio of
9:1. Each kg of compressed air entering the combustion chamber receives
about 650 kJ of heat as it passes through the combustion chamber at
constant pressure. The gases leaving the combustion chamber enter the
turbine at a high temperature and a pressure, set the turbine shaft on
rotation, and leave the turbine. The gases leaving the turbine pass through
the nozzle where they achieve high speeds. The turbojet engine is so
designed that the pressure of the gases at the exit of the nozzle falls back
to 0.95 bar. Assume that the ﬂow within the turbojet engine except in
the combustion chamber is reversible and adiabatic and that all the work
output of the turbine is used to drive the compressor. Neglect the speed
of the ﬂow through the turbojet engine except at the inlet and the exit of
the engine. Also, neglect the eﬀect of fuel ﬂow rate.
(a) Determine the pressures and the temperatures at the inlets and the
outlets of the diﬀuser, compressor, combustion chamber, turbine and
the nozzle of the turbojet engine.
(b) Determine the speed of the gases at the nozzle exit.
(c) Compute the forward thrust imparted on the turbojet engine per kg
of the air ﬂowing through the engine.
Solution to Example 12.3
Let us use the schematic of the turbojet engine shown in Figure 12.4 and the
labels 1, 2, 3, 4, 5 and 6 on it to mark the states of the ﬂuid entering/leaving
the diﬀuser, compressor, combustion chamber, turbine and the nozzle of the
turbojet engine.
Flow through the Diﬀuser:
At the diﬀuser inlet, P1=0.95bar,T1= 283 K and c1= 300 m/s, where
c1denotes the speed of the air at the diﬀuser inlet. At the diﬀuser exit, c2=0,
where c2denotes the speed of the air at the diﬀuser exit.
The steady ﬂow energy equation applicable for the adiabatic ﬂow through a
284 Chapter 12
diﬀuser is used to determine T2as
T2=T1+c2
1c2
2
2×Cp
= 283 K+3002m2/s2
2×1005 J/kg ·K= 327.8K
Since the ﬂow through the diﬀuser is assumed to be reversible and adiabatic,
(7.31) gives
P2=P1T2
T1γ/(γ1)
=0.95 bar ×327.8
283 1.4/0.4
=1.59 bar
P2=1.59barandT2= 327.8 K at the diﬀuser exit, which is also the
compressor inlet. The speed of the ﬂow is assumed to be negligible at this cross-
section.
Flow through the Compressor:
The compressor has a pressure ratio of 9:1, so that
P3=9×P2=9 ×1.59 bar =14.31 bar
Since the ﬂow through the compressor is assumed to be reversible and adi-
abatic, (7.31) gives
T3=T2P3
P2(γ1)
= 327.8K×9×P2
P20.4/1.4
= 614.1K
P3= 14.31 bar and T3= 614.1 K at the compressor exit, which is also the
combustion chamber inlet. The speed of the ﬂow is assumed to be negligible at
this cross-section.
Flow through the Combustion Chamber:
The ﬂow through the combustion chamber is assumed to be at constant
pressure. So that
P4=P3=14.31 bar
650 kJ of heat as it passes through the combustion chamber. Neglecting the
eﬀect of fuel ﬂow rate, applying the steady ﬂow energy equation to the ﬂow
through the combustion chamber gives
650 kJ/kg =Cp(T4T3)
from which T4can be found as
T4=T3+650 kJ/kg
Cp
=614.1+ 650
1.005 K= 1260.9K
Thermodynamic Analyses of Power Plants 285
P4= 14.31 bar and T4= 1260.9 K at the combustion chamber exit, which
is also the turbine inlet. The speed of the ﬂow is assumed to be negligible at
this cross-section.
Flow through the Turbine:
Since all the work output of the turbine is assumed to be used to drive the
the compressor, we have
[( ˙
Ws)out]turbine =[(˙
Ws)in]compressor
Using the steady ﬂow energy equation to both the turbine and the compres-
sor,wecanexpandtheaboveequationto
˙mgases (Cp)gases (T4T5)= ˙mair (Cp)air (T3T2)
Since the eﬀect of fuel ﬂow rate is ignored, the mass ﬂow rates of the gases
and of air can be taken as equal. The values of Cpfor air and for the gases can
be taken as equal as well. Therefore, we can calculate T5using
T5=T4T3+T2= (1260.9614.1 + 327.8) K= 974.6K
Since the ﬂow through the turbine is assumed to be reversible and adiabatic,
(7.31) gives
P5=P4T5
T4γ/(γ1)
=14.31 bar ×974.6
1260.91.4/0.4
=5.81 bar
P5=5.81barandT5= 974.6 K at the turbine exit, which is also the nozzle
inlet. The speed of the ﬂow is assumed to be negligible at this cross-section.
Flow through the Nozzle:
The pressure of the gases at the exit of the nozzle falls back to 0.95 bar.
Therefore
P6=0.95 bar
(7.31) gives
T6=T5P6
P5(γ1)
= 974.6K×0.95
5.81 0.4/1.4
= 580.9K
Steady ﬂow energy equation is applied to the ﬂow through the nozzle to
determine c6as
c6=2×Cp(T5T6)
=2×1005 ×(974.6580.9) J/kg = 890 m/s
286 Chapter 12
P6=0.95barandT6= 580.9 K at the nozzle exit. The speed of the ﬂow
at the nozzle exit is 890 m/s.
Comment: Such high speed at the nozzle exit is achieved under the assumed
ideal conditions, such as the ﬂow within the turbojet engine except in the com-
bustion chamber being reversible and adiabatic.
Forward thrust imparted on the turbojet engine:
Since the exit pressure is the same as the inlet pressure, the forward thrust
imparted on the turbojet engine per kg of the air ﬂowing through the engine is
calculated using (12.1) as
Net Thrust
˙mair
=(c6c1) = (890 300) m/s = 590 N per kg/s
12.3 Steam Turbine for
Electric Power Generation
In a gas turbine used for electric power generation, the shaft of the
turbine is rotated by the gases produced in the burning of a fuel with air.
In a steam turbine used for electric power generation, superheated steam
is used to rotate the shaft of the turbine. The basic working principle of a
simple steam turbine used for electric power generation is explained below
with the help of the ﬂow diagram shown in Figure 12.5.
Saturated water entering the pump is compressed to a high pressure and
the compressed water is fed to the steam generator, which is sometimes
referred to as the boiler. The compressed water is heated to superheated
steam state in the steam generator, which is a large heat exchanger where
the heat is transferred from the hot combustion gases to the water. The
superheated steam leaving the steam generator at a high pressure and a
temperature enters the turbine, where it expands rotating the turbine shaft.
The work output of the rotating turbine shaft is used to produce elec-
tricity by spinning a coil between the poles of a magnet or an electromagnet.
Thus, part of the heat transferred from the hot combustion gases to the
Thermodynamic Analyses of Power Plants 287
water in the steam generator is converted into useful electrical energy. The
wet steam leaving the turbine is condensed to saturated water state in a
condenser, which is also a large heat exchanger where the heat is trans-
ferred from the steam to cooling water. The saturated water leaving the
condenser enters the pump, thus making a cyclic ﬂow through the pump,
steam generator, turbine and condenser.
The combustion gases leaving the steam generator, rich in pollutants
such as the greenhouse gas carbon dioxide, enter the atmosphere. The
cooling water leaving the condenser at an elevated temperature is cooled
in the cooling towers, by transferring considerable amount of heat to the
atmosphere, and thereby causing thermal pollution.
turbine
pump
2
4
Figure 12.5 Flow diagram of a simple steam turbine power plant.
steam generator
condenser
cooling water
3
hot gases
1


Example 12.4
In the steam turbine power plant, shown in
Figure 12.5, assume that saturated water at 0.08 bar ﬂowing at a rate of
50 kg/s is compressed by the pump to 70 bar. The compressed water is
fed to the steam generator where it is converted to superheated steam at
450C at the same pressure. The superheated steam expanding through
the turbine, leaves the turbine as wet steam at 0.08 bar. The wet steam is
288 Chapter 12
condensed to saturated water state in the condenser, thus completing the
cycle.
(a) Determine the enthalpies of the ﬂow at the inlets and the outlets
of the turbine and the pump, assuming reversible adiabatic ﬂows
through the turbine and the pump.
(b) Determine the power production at the turbine, power requirement
of the pump, and the back work ratio, deﬁned as the ratio of the
pump work to the turbine work.
(c) Determine the heat input at the steam generator, and the thermal
eﬃciency of the steam turbine plant, deﬁned as the ratio of net work
outtoheatin.
(d) Determine the heat rejected at the condenser.
(e) Calculate the cooling water mass ﬂow rate if the cooling water tem-
C less than the temperature
of the wet steam condensing in the condenser. Take cooling water
Cpas 4.2 kJ/kg ·K.
Solution to Example 12.4
(a) Let us use the ﬂow diagram of the simple steam turbine power plant shown in
Figure 12.5 and the labels 1, 2, 3 and 4 on it to mark the states of water/steam
entering or leaving the components of the steam turbine power plant.
At state 3, which is the turbine inlet, we have superheated steam at 70
bar and 450C. The speciﬁc enthalpy found from a Superheated Steam Table is
h3= 3287 kJ/kg.
At state 4, which is the turbine outlet, we have wet steam at 0.08 bar. Since
the ﬂow through the turbine is assumed to be reversible adiabatic, s4=s3=
sat 70 bar and 450C=6.632 kJ/kg ·K.
For state 4 at 0.08 bar and s4= 6.632 kJ/kg ·K, the dryness fraction can
be calculated using
x4=s4sf
sfg
=6.632 0.593
7.634 =0.7911
The speciﬁc enthalpy is then
h4=hf+x4×hfg = 174 + 0.7911 ×2402 = 2074 kJ/kg
At state 1, which is the pump inlet, we have saturated water at 0.08 bar.
The speciﬁc enthalpy is h1= 174 kJ/kg.
Thermodynamic Analyses of Power Plants 289
At state 2, which is the pump outlet, we have compressed water at 70
bar. Since the ﬂow through the pump is assumed to be reversible adiabatic,
s2=s1=sfat 0.08 bar =0.593 kJ/kg ·K.
State 2 at 70 bar and s2= 0.593 kJ/kg ·K is a compressed water state since
sfat 70 bar is 3.122 kJ/kg ·K. We should therefore be able to ﬁnd h2using
a Compressed Water Table. However, since data for compressed water is not
easily found, let us use the following convenient but an approximate method to
ﬁnd h2.
The work input to a pump with adiabatic ﬂow is expressed by (10.11). The
work input to a pump with reversible ﬂow is approximated by (11.28). Since the
given ﬂow is reversible adiabatic, we combine (10.11) and (11.28) to get h2as
follows:
˙m(h2h1)˙mv
1(P2P1)
h2h1+v1(P2P1) (12.2)
where h1is known, v1=vfat 0.08 bar = 0.0010084 m3/kg, P1=0.08barand
P2= 70 bar. Therefore, we have
h2174 kJ/kg +0.0010084 m3/kg ×(7000 8) kPa = 181 kJ/kg
Summarizing the results obtained above, we have
State 1 2 3 4
Condition saturated compressed superheated wet
water water steam steam
P(in bar) 0.08 70 70 0.08
T(in C) 41.5 -450 41.5
h(in kJ/kg) 174 181 3287 2074
x0 - - 79.11%
(b) The power production of the turbine must be determined. Using (10.9) for
the ﬂow through an adiabatic turbine, we have
[( ˙
Ws)out]turb m(h3h4)=50×(3287 2074) kJ/s =60.65 MW
The power requirement of the pump must be determined. Using (10.11) for
the ﬂow through an adiabatic pump, we have
[( ˙
Ws)in]pump m(h2h1)=50×(181 174) = 0.35 MW
The back work ratio is calculated as
bwr =pump work input
turbine work output =0.35 MW
60.65 MW =0.6%
290 Chapter 12
Comment: The power consumption of the pump to compress water from 0.08
bar to 70 bar is only about 0.6% of the power production of the turbine. That is,
almost all the work produced by the turbine is available for electricity generation.
It is an advantage that a steam turbine power plant has over a gas turbine power
plant.
(c) Heat input to the steam generator is calculated by applying (10.3) to the
ﬂuid ﬂowing through the steam generator as
[˙
Qin]steam generator m(h3h2)
=50×(3287 181) kJ/s = 155.30 MW
The thermal eﬃciency of the steam turbine plant is calculated as
ηth =net work out
heat in
=60.65 MW 0.35 MW
155.30 MW =60.30
155.30 =38.8%
Comment: Only about 38.8% of the 155.30 MW heat input to the steam
generator is available for electricity generation.
(d) Heat rejected at the condenser is calculated by applying (10.3) to the ﬂuid
ﬂowing through the condenser as
[˙
Qout]condenser m(h4h1)
=50×(2074 174) kJ/s =95.00 MW
Comment: The percentage of the heat input in the steam generator that is
rejected by the condenser is calculated using (95.00/155.30)×100, which equals
61.2%. In other words, for each MW of electric power generation there is at
least about 1.6 (=95.00/60.30) MW of power is wasted. All this waste energy
reaching the environment is a source of thermal pollution.
(e) Steam condenses in the condenser at the saturated temperature at 0.08 bar,
which is 41.5C. The cooling water temperature is therefore raised from 32Cto
39C. Assume that the cooling water absorbs all that 95.00 MW of heat rejected
at the condenser. The mass ﬂow rate of the cooling water is calculated as
˙mcooling water =95.00 MW
4.2×(39 32) kJ/kg
= 3231 kg/s =11.63 ×106kg/h
Thermodynamic Analyses of Power Plants 291
Example 12.5
Consider the schematic of the steam turbine
power plant shown in Figure 12.6. Saturated water at 0.08 bar ﬂowing at
a rate of 50 kg/s is compressed reversibly and adiabatically by a pump to
70 bar. The compressed water is fed to a steam generator where it is con-
verted to superheated steam at 450C at the same pressure. The steam
ﬂows reversibly and adiabatically through a HP (high pressure) turbine,
and leaves the turbine at 7 bar. The steam leaving the turbine is reheated
to 425C, and fed to the LP (low pressure) turbine. The steam ﬂowing
reversibly and adiabatically through the LP turbine, leaves it as wet steam
at 0.08 bar. The wet steam is condensed to saturated water state in the
condenser, thus the cycle is completed.
pump
2
6
Figure 12.6 Flow diagram of a steam turbine with a reheater.
steam generator
condenser
cooling water
3
reheater
hot gases
hot gases
1


HP
turbine LP
turbine
5
4
(a) Determine the enthalpies of the ﬂow at the inlets and the outlets of
the turbines and the pump.
(b) Determine the combined power production at the HP and the LP
turbines, power requirement of the pump, and the back work ratio.
(c) Determine the total heat input to the steam generator and the re-
heater, and the thermal eﬃciency of the steam turbine plant.
(d) Determine the heat rejected at the condenser.
292 Chapter 12
Solution to Example 12.5
At state 1, which is the pump inlet, we have saturated water at 0.08 bar.
Therefore, h1= 174 kJ/kg.
At state 2, which is the pump outlet, we have compressed water at 70 bar.
Since the ﬂow through the pump is reversible adiabatic, we get h2181 kJ/kg,
as worked out in the Solution to Example 12.4.
At state 3, which is the HP turbine inlet, we have superheated steam at 70
bar and 450C. Therefore, h3= 3287 kJ/kg.
At state 4, which is the HP turbine outlet, we have steam at 7 bar. Since the
ﬂow through the turbine is reversible adiabatic, s4=s3=sat 70 bar and 450C=
6.632 kJ/kg ·K. For a state at 7 bar and s4= 6.632 kJ/kg ·K, the dryness frac-
tion can be calculated using
x4=s4sf
sfg
=6.632 1.992
4.717 =0.9837
The speciﬁc enthalpy is then
h4=hf+x4×hfg = 697 + 0.9837 ×2067 = 2730 kJ/kg
At state 5, which is the LP turbine inlet, we have superheated steam at 7
bar and 425C. Therefore, h5= 3322 kJ/kg.
At state 6, which is the LP turbine outlet, we have wet steam at 0.08
bar. Since the ﬂow through the turbine is reversible adiabatic, s6=s5=
sat 7 bar and 425C=7.710 kJ/kg ·K. For a state at 0.08 bar and s6= 7.710
kJ/kg ·K, the dryness fraction can be calculated using
x6=s6sf
sfg
=7.710 0.593
7.634 =0.9323
The speciﬁc enthalpy is then
h6=hf+x6×hfg = 174 + 0.9323 ×2402 = 2413 kJ/kg
Summarizing the results obtained above, we have
State 1 2 3 4 5 6
Condition sat. comp. super. wet super. wet
water water steam steam steam steam
P(in bar) 0.08 70 70 7 7 0.08
T(in C) 41.5 -450 165 425 41.5
h(in kJ/kg) 174 181 3287 2730 3322 2413
x0 - - 98.37% -93.23%
Note: ‘sat.’ stands for saturated, ‘comp.’ for compressed, and ‘super.’ for superheated.
Thermodynamic Analyses of Power Plants 293
(b) Using (10.9) for the ﬂow through the adiabatic HP turbine, we have
[( ˙
Ws)out]HP turbine m(h3h4)=50×(3287 2730) kJ/s
=27.85 MW
Using (10.9) for the ﬂow through the adiabatic LP turbine, we have
[( ˙
Ws)out]LP turbine m(h5h6)=50×(3322 2413) kJ/s
=45.45 MW
The combined power production at the HP and the LP turbines is the sum-
mation of the power productions of the HP and the LP turbines, which is 73.30
MW.
Using (10.11) for the ﬂow through an adiabatic pump, we have
[( ˙
Ws)in]pump m(h2h1)=50×(181 174) = 0.35 MW
The back work ratio is calculated as
bwr =pump work input
combined turbine work output =0.35 MW
73.30 MW =0.5%
(c) Heat input to the steam generator is calculated by applying (10.20) to the
ﬂuid ﬂowing through the steam generator as
[˙
Qin]steam generator m(h3h2)
=50×(3287 181) kJ/s = 155.30 MW
Heat input to the reheater is calculated by applying (10.20) to the ﬂuid
ﬂowing through the reheater as
[˙
Qin]reheater m(h5h4)
=50×(3322 2730) kJ/s =29.60 MW
The total heat input to the steam generator and the reheater is the summa-
tion of the heat inputs to the steam generator and the reheater, which is 184.90
MW.
The thermal eﬃciency of the steam turbine plant is calculated as
ηth =net work out
heat in
=73.30 MW 0.35 MW
184.90 MW =72.95
184.90 =39.5%
294 Chapter 12
Comment: Only about 39.5% of the 184.90 MW combined heat input to the
steam generator and the reheater is available for electricity generation.
(d) Heat rejected at the condenser is calculated by applying (10.21) to the ﬂuid
ﬂowing through the condenser as
[˙
Qout]condenser m(h6h1)
=50×(2413 174) kJ/s = 111.95 MW
Comment: The percentage of the combined heat input that is rejected by
the condenser is calculated using (111.95/184.90)×100, which equals 60.5%. In
other words, for each MW of electric power generation there is at least about
1.5 (=111.95/72.95) MW of power is wasted. All this waste energy reaching
the environment is a source of thermal pollution.
Example 12.6
Consider the schematic of the steam turbine
power plant shown in Figure 12.7. saturated water at 7 bar ﬂowing at a
rate of 50 kg/s is compressed adiabatically by a pump to 70 bar, and is fed
to a steam generator where it is converted to superheated steam at 450C
at the same pressure. The superheated steam ﬂows adiabatically through
the 1st stage turbine, and leaves it at 7 bar. A part of the steam leaving
the 1st stage turbine is fed to an open feedwater heater operated at 7 bar.
(An open feedwater heater is simply a mixing chamber in which a hot ﬂuid
stream and a cold ﬂuid stream mix with each other to form a ﬂuid stream
at intermediate temperature.) The remaining steam is fed to the 2nd stage
turbine, through which it ﬂows adiabatically, and leaves it as steam at 0.08
bar. It is condensed to saturated water state at the same pressure in the
condenser, compressed adiabatically by a second pump to 7 bar, and fed
to the open feedwater heater, thus the cycle is completed.
(a) Determine the enthalpies of the ﬂow at the inlets and the outlets of
the turbines and the pumps. Assume reversible ﬂows through the
pumps and the turbines.
(b) Determine the mass ﬂow rate of the steam diverted to the open
feedwater heater from the exit of the 1st stage turbine, assuming
adiabatic conditions at the open feedwater heater.
Thermodynamic Analyses of Power Plants 295
(c) Determine the combined power production at the turbines, combined
power requirement of the pumps, and the back work ratio.
(d) Determine the total heat input to the steam generator, and the ther-
mal eﬃciency of the steam turbine plant.
(e) Determine the heat rejected at the condenser.
1st pump
2
5
Figure 12.7 Flow diagram of a steam turbine with an open
feedwater heater.
steam generator
condenser
cooling water
3
hot gases


1st
turbine 2nd
turbine
4
4
2nd pump
6


open
feedwater
heater
4
7
1
Solution to Example 12.6
At state 1, which is the inlet of the ﬁrst pump, we have saturated water at
7 bar. Therefore, h1= 697 kJ/kg.
At state 2, which is the outlet of the ﬁrst pump, we have compressed water
at 70 bar. Since the ﬂow through the pump is assumed to be reversible adiabatic,
h2is calculated using (12.2) as
h2h1+v1(P2P1) = 697 + 0.0011082 ×(7000 700) = 704 kJ/kg
Atstate3,whichistheinletofthe1
st stage turbine, we have superheated
steam at 70 bar and 450C. At state 4, outlet of the 1st stage turbine, we
have steam at 7 bar. The ﬂow through the turbine is assumed to be reversible
296 Chapter 12
adiabatic. Therefore, as found in the Solution to Example 12.5,h3= 3287
kJ/kg, x4= 0.9837, and h4= 2730 kJ/kg.
The wet steam leaving the 1st stage turbine is divided into two streams, one
entering the 2nd stage turbine and the other entering the open feedwater heater.
All these streams are assumed to be at state 4.
At state 5, which is the outlet of the 2nd stage turbine, we have steam at
0.08 bar. Since the ﬂow through the 2nd stage turbine is reversible adiabatic,
s5=s4=s3=sat 70 bar and 450C=6.632 kJ/kg ·K. For state 5 at 0.08
bar and s5= 6.632 kJ/kg ·K, the dryness fraction can be calculated using
x5=s5sf
sgsf
=6.632 0.593
7.634 =0.7911
The speciﬁc enthalpy is then
h5=hf+x5×hfg = 174 + 0.7911 ×2402 = 2074 kJ/kg
At state 6, which is the inlet of the second pump, we have saturated water
at 0.08 bar. Therefore, h6= 174 kJ/kg.
At state 7, which is the outlet of the second pump, we have is compressed
water at 7 bar. Since the ﬂow through the pump is assumed to be reversible
adiabatic, h2is calculated using (12.2) as
h7h6+v6(P7P6) = 174 + 0.0010084 ×(700 8) = 174.7kJ/kg
Summarizing the results obtained above, we have
State 1 2 3 4 5 6 7
Condition sat. comp. super. wet wet sat. comp.
water water steam steam steam water water
P(in bar) 770 70 70.08 0.08 7
T(in C) 165 -450 165 41.5 41.5 -
h(in kJ/kg) 697 704 3287 2730 2074 174 174.7
x0 - - 98.37% 79.11% 0 -
Note: ‘sat.’ stands for saturated, ‘comp.’ for compressed, and ‘super.’ for superheated.
(b) Take the mass ﬂow rate of the steam diverted to the open feedwater heater
from the exit of the 1st stage turbine as mokg/s. The steam entering the 2nd
stage turbine is therefore (50 - mo) kg/s, which is of course the same amount
pumped into the open feedwater heater by the second pump.
Energy balance over the adiabatic open feedwater heater, which in eﬀect is
an adiabatic mixing chamber, is expressed by (10.16). Applying (10.16) to the
adiabatic open feedwater heater of Figure 12.7, we have
(50 mo)×(h1h7)=mo×(h4h1)
Thermodynamic Analyses of Power Plants 297
which gives
mo=h1h7
h4h7
×50 kg/s
Substituting the known enthalpies, we get
mo=697 174.7
2730 174.7×50 kg/s =0.2044 ×50 kg/s =10.22 kg/s
(c) Using (10.9) for the ﬂow through the adiabatic 1st stage turbine, we have
[( ˙
Ws)out]1st stage turbine m(h3h4)
=50×(3287 2730) kJ/s
=27.85 MW
Using (10.9) for the ﬂow through the adiabatic 2nd stage turbine, we have
[( ˙
Ws)out]2nd stage turbine m(h4h5)
=(50mo)×(2730 2074) kJ/s
=26.10 MW
The combined power production of the two turbines is the summation of the
power productions of the 1st stageandthe2
nd stage turbine, which is = (27.85
+ 26.10) MW = 53.95 MW.
Using (10.11) for the ﬂow through an adiabatic ﬁrst pump, we have
[( ˙
Ws)in]1st pump m(h2h1)
=50×(704 697) = 0.35 MW
Using (10.11) for the ﬂow through an adiabatic second pump, we have
[( ˙
Ws)in]2nd pump m(h7h6)
=(50mo)×(174.7174) = 0.03 MW
The back work ratio is calculated as
bwr =combined pump work input
combined turbine work output =(0.35 + 0.03) MW
53.95 MW =0.7%
(d) Heat input to the steam generator is calculated by applying (10.20) to the
ﬂuid ﬂowing through the steam generator as
[˙
Qin]steam generator m(h3h2)
=50×(3287 704) kJ/s = 129.15 MW
298 Chapter 12
The thermal eﬃciency of the steam turbine plant is calculated as
ηth =net work out
heat in
=combined turbine work - combined pump work
heat in
=53.95 MW (0.35 + 0.03) MW
129.15 MW =53.57
129.15 =41.5%
(e) Heat rejected at the condenser is calculated by applying (10.21) to the ﬂuid
ﬂowing through the condenser as
[˙
Qout]condenser m(h5h6)
=(50mo)×(2074 174) kJ/s =75.58 MW
Comment: For each MW of power produced by the turbine, about 1.4 MW
of power is wasted. All this waste energy reaching the environment is a source
of thermal pollution.
12.4 Gas Turbine - Steam Turbine
Combined Power Plant
One of the major drawbacks in using a gas turbine power plant for
electric power generation in comparison to a steam turbine power plant is
the back work ratio of the gas turbine plant (see, Example 12.1)being
considerably higher than the back work ratio of the steam turbine power
plant (see, Example 12.4). Nearly half of the power produced by the
gas turbine is consumed by the compressor, which is used to compress the
atmospheric air to a high pressure. In the steam turbine power plant, the
pump, which is used to compress the water at a vacuum pressure to a
high pressure, consumes less than about 1% of the power produced by the
turbine.
Besides, in a gas turbine power plant, the gases produced in the burning
of a fuel with air, after expanding through the turbine, are emitted into the
Thermodynamic Analyses of Power Plants 299
atmosphere at a high temperature. Also, the mass ﬂow rate of these hot
exhaust gases is large owing the high air-fuel ratio used in the gas turbine.
In a steam turbine power plant, on the other hand, hot gases produced
by the combustion is used to superheat steam (see, Figure 12.5), which
expands through the turbine causing the turbine shaft to rotate.
It is therefore advisable to combine the gas turbine and the steam tur-
bine such that the hot exhaust gases leaving the gas turbine power plant
can be used to generate the superheated steam required by the steam tur-
bine power plant. Such is practised in power plants known as the combined
gas turbine - steam turbine power plants, and the demand for which is
increasing dramatically worldwide over the recent years. The basic working
principle of a combined simple gas turbine - steam turbine power plant used
for electric power generation is shown in the ﬂow diagram of Figure 12.8.
combustion
chamber
compressor gas
turbine
gases
shaft
fuel
shaft power
for generation
of electricity
air
Figure 12.8 Flow diagram of a simple gas turbine - steam turbine
combined power plant.
steam
turbine
shaft power
for generation
of electricity

 HRSG
pump
cooling water
exhaust gases
to stack
1
23
4
5
6
78
9
condenser
300 Chapter 12
The crucial feature of a combined power plant is the heat recovery
steam generator, abbreviated HRSG, shown in Figure 12.8. HRSG is basi-
cally a heat exchanger in which the hot gases leaving the gas turbine heats
the compressed water supplied by the pump to superheated steam required
by the steam turbine. Since the heat input to the HRSG is the heat that
would have otherwise been lost to the environment, the overall thermal
eﬃciency of a combined plant in general is high.
Example 12.7
Combine the gas turbine power plant of Ex-
ample 12.1 and the steam turbine power plant of Example 12.4 using
a heat recovery steam generator as shown in Figure 12.6. The mass ﬂow
rate of air through the gas turbine is 350 kg/s and the mass ﬂow rate of
water through the steam turbine is 50 kg/s. The properties at the states
marked 1 to 9 in the ﬁgure are as tabulated below:
State 1 2 3 4 5 6 7 8 9
Condition air air gases gases gases sat. comp. super. wet
water water steam steam
P(in bar) 1 5 5 1 1 0.08 70 70 0.08
T(in C) 27 202.2 927 484.7 T541.5 41.8 450 41.5
Note: ‘sat.’ stands for saturated, ‘comp.’ for compressed, and ‘super.’ for superheated.
through the compressor, pump and the turbines. Ignore the fuel ﬂow rate.
Take Cpand γfor air and the gases as 1.005 kJ/kg ·K and 1.4, respectively.
(a) Determine the combined power production at the turbines, combined
power requirement of the compressor and the pump, and the back
work ratio.
(b) Determine the heat input to the combined power plant, and the
overall thermal eﬃciency of the plant.
(c) Determine the net energy rejection to the environment by the com-
bined power plant.
Thermodynamic Analyses of Power Plants 301
(d) Determine the heat rejection by the condenser and the loss of energy
to the environment with the exhaust leaving the HRSG.
(e) Determine the temperature of the exhaust gases.
Solution to Example 12.7
(a) Data for the gas turbine power plant are the same as in Example 12.1,and
for the steam turbine power plant are the same as in Example 12.4.Theows
through the compressor, pump and the turbines are assumed to be reversible and
The compressor work input and the gas turbine power output are, therefore,
the same as that are evaluated in the Solution to Example 12.1,whichare
61.63 MW and 155.58 MW, respectively.
The pump work input and the steam turbine power output are, therefore,
the same as that are evaluated in the Solution to Example 12.4,whichare
0.35 MW and 60.65 MW, respectively.
The combined power production at the turbines is the summation of the
power produced at the gas and the steam turbines, which is (155.58 + 60.65)
MW = 216.23 MW
The combined power requirement of the compressor and the pump is the
summation of the power inputs to the compressor and the pump, which is (61.63
+ 0.35 ) MW = 61.98 MW.
The back work ratio is calculated using
bwr =pump work input + compressor work input
combined turbine work output
=61.98 MW
216.23 MW =28.7%
Comment: About 28.7% of the combined power production of the turbines is
consumed by the compressor and the pump. The remaining 71.3% is available
for electricity generation. That is, the back work ratio is reduced when a gas
turbine is used combined with the steam turbine.
(b) Heat is supplied to the combined power plant of Figure 12.8 only at the
combustion chamber. Therefore, the heat input to the combined power plant
is the same as that is evaluated in the Solution to Example 12.1,whichis
254.95 MW.
302 Chapter 12
The overall thermal eﬃciency of the combined power plant is calculated
using
ηth =net work output of the combined power plant
heat input to the combined power plant
=216.23 MW 61.98 MW
254.95 MW
=154.25
254.95 =60.5%
Comment: The thermal eﬃciency of the combined power plant, 60.5%, is
much higher than the thermal eﬃciency of the gas turbine power plant, 36.9%,
or the steam turbine power plant, 38.8%, when operated separately. It is the
principal feature for which the combined power plant is becoming popular as a
mode of electricity generation.
(c) The total heat supply to the combined power plant is 254.95 MW. The net
work output of the plant is 154.25 MW. The net energy rejection to the environ-
ment by the plant is therefore calculated as (254.95 MW - 154.25 MW), which
is 100.70 MW.
Comment: That is, 39.50% of the heat input to the combined power plant is
lost to the environment. In other words, for each MW of power produced, there
is about 0.65 (=100.70/154.24) MW of power wasted. This amount is far less
than the amount of power wasted per MW of power produced in the cases where
the gas turbine and the steam turbine are operated separately. It is yet another
favourable feature that promotes the use of combined power plant for electricity
generation.
(d) Energy rejection at the condenser is the same as that is evaluated in the
Solution to Example 12.4, which is 95.00 MW. The loss of energy to the en-
vironment with the exhaust leaving the HRSG is therefore calculated as (100.70
MW - 95.00 MW), which is 5.7 MW.
(e) To determine the temperature of the exhaust gases leaving the HRSG, which
is T5, let us redraw the HRSG part of the combined power plant as in Figure
12.9, with the pressure and temperature data displayed on it.
Treating the HRSG as an adiabatic heat exchanger, we can write the steady
ﬂow energy equation over it to get
˙mgases (Cp)gases (T4T5)= ˙mwater/steam (h8h7)
Substituting the known numerical values in the above expression, we get
350 ×1.005 ×(484.7T5)=50 ×(3287 181)
Thermodynamic Analyses of Power Plants 303
where h8= 3287 kJ/kg for superheated steam at 70 bar at 450Candh7=
181 kJ/kg for compressed water at 70 bar at 41.8C. Therefore, we get T5=
43C.
Figure 12.9 HRSG unit of the combined power plant.
exhaust gases
to stack
4
5
78
from the
gas turbine
to the
steam turbinefrom the
pump
1bar
T5
1bar
484.7C
70 bar
450C
70 bar
41.8C
Comment: Note that, in the HRSG, for the heat to ﬂow from the hot gases to
the water/steam mixture, T4must be greater than T8and T5must be greater
than T7, and these conditions are satisﬁed in the case considered. However, in
reality, it is impractical to reduce the temperature of the exhaust gases leaving
the HRSG to such low values as 43C.
12.5 Minimizing the Heat Loss
from Power Plants
It is of importance to note that none of the power plant studied in this
chapter succeeds in converting all heat supplied to the power plant into
work. That is, the thermal eﬃciency of no engine is 100%. Ideally, we
would have liked to reach 100% thermal eﬃciency, since it would result in
no waste of the precious heat energy, which we obtain mostly by burning
304 Chapter 12
the fossil fuels that generates carbon dioxide and other greenhouse gases
that are responsible for global warming and the resulting climate change.
Besides, the heat that is lost to the environment from the power plants
causes thermal pollution.
We could have reached 100% thermal eﬃciency, if all heat gained by
the working ﬂuid were converted into net work. According to the ﬁrst law,
it is possible to convert all heat into work, since the ﬁrst law states that
energy cannot be created or destroyed but it can be changed from one form
to another.
But, centuries of experience has shown that it is impossible to convert
all heat into work in engines of the kinds studied in this chapter. This
observation has been generalized into a fundamental law, which, indeed, is
the second law of thermodynamics, an introduction to which is given in the
next chapter.