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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

how to set about it.

−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(T2−T1) = 350 ×1.005 ×(T2−300) 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(γ−1)/γ

= 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(T3−T4) = 350 ×1.005 ×(1200 −T4)kJ/s

of which T4is unknown. Since the ﬂow through the turbine is assumed to be

reversible and adiabatic, (7.31) gives

T4=T3P4

P3(γ−1)/γ

= 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(T3−T2)

= 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 ﬂow 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 ﬂow 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(T5−T2)

(Cp)gases(T4−T6)

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

T5−475.2K= 757.7K−T6

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(T3−T5)

= 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

1−c2

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/(γ−1)

=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(γ−1)/γ

= 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

Each kg of compressed air entering the combustion chamber receives about

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(T4−T3)

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 (T4−T5)= ˙mair (Cp)air (T3−T2)

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=T4−T3+T2= (1260.9−614.1 + 327.8) K= 974.6K

Since the ﬂow through the turbine is assumed to be reversible and adiabatic,

(7.31) gives

P5=P4T5

T4γ/(γ−1)

=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

Sincetheﬂowthroughthenozzleisassumedtobereversibleandadiabatic,

(7.31) gives

T6=T5P6

P5(γ−1)/γ

= 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(T5−T6)

=2×1005 ×(974.6−580.9) 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

=(c6−c1) = (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

450◦C 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-

perature is raised from 32◦Ctoabout2.5

◦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 450◦C. 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 450◦C=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=s4−sf

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(h2−h1)≈˙mv

1(P2−P1)

h2≈h1+v1(P2−P1) (12.2)

where h1is known, v1=vfat 0.08 bar = 0.0010084 m3/kg, P1=0.08barand

P2= 70 bar. Therefore, we have

h2≈174 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(h3−h4)=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(h2−h1)=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(h3−h2)

=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(h4−h1)

=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 41.5◦C. The cooling water temperature is therefore raised from 32◦Cto

39◦C. 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 450◦C 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 425◦C, 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 h2≈181 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 450◦C. 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 450◦C=

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=s4−sf

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 425◦C. 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 425◦C=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=s6−sf

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(h3−h4)=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(h5−h6)=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(h2−h1)=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(h3−h2)

=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(h5−h4)

=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(h6−h1)

=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 450◦C

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

h2≈h1+v1(P2−P1) = 697 + 0.0011082 ×(7000 −700) = 704 kJ/kg

Atstate3,whichistheinletofthe1

st stage turbine, we have superheated

steam at 70 bar and 450◦C. 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 450◦C=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=s5−sf

sg−sf

=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

h7≈h6+v6(P7−P6) = 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)×(h1−h7)=mo×(h4−h1)

Thermodynamic Analyses of Power Plants 297

which gives

mo=h1−h7

h4−h7

×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(h3−h4)

=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(h4−h5)

=(50−mo)×(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(h2−h1)

=50×(704 −697) = 0.35 MW

Using (10.11) for the ﬂow through an adiabatic second pump, we have

[( ˙

Ws)in]2nd pump =˙m(h7−h6)

=(50−mo)×(174.7−174) = 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(h3−h2)

=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(h5−h6)

=(50−mo)×(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.

Assume adiabatic conditions at the HRSG and reversible adiabatic ﬂows

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.Theﬂows

through the compressor, pump and the turbines are assumed to be reversible and

adiabatic.

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 (T4−T5)= ˙mwater/steam (h8−h7)

Substituting the known numerical values in the above expression, we get

350 ×1.005 ×(484.7−T5)=50 ×(3287 −181)

Thermodynamic Analyses of Power Plants 303

where h8= 3287 kJ/kg for superheated steam at 70 bar at 450◦Candh7=

181 kJ/kg for compressed water at 70 bar at 41.8◦C. Therefore, we get T5=

43◦C.

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

484.7◦C

70 bar

450◦C

70 bar

41.8◦C

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 43◦C.

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.