Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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416 Chapter 9 Gas Power Systems

EXAMPLE 9.12 Analyzing a Turbojet Engine

Air enters a turbojet engine at 0.8 bar, 240K, and an inlet velocity of 1000 km/h (278 m/s). The pressure ratio across the

compressor is 8. The turbine inlet temperature is 1200K and the pressure at the nozzle exit is 0.8 bar. The work developed

by the turbine equals the compressor work input. The diffuser, compressor, turbine, and nozzle processes are isentropic, and

there is no pressure drop for flow through the combustor. For operation at steady state, determine the velocity at the nozzle

exit and the pressure at each principal state. Neglect kinetic energy at the exit of all components except the nozzle and neg-

lect potential energy throughout.

SOLUTION

Known: An ideal turbojet engine operates at steady state. Key operating conditions are specified.

Find: Determine the velocity at the nozzle exit, in ft/s, and the pressure, in lbf/in.

, at each principal state.

Schematic and Given Data:

Assumptions:

1. Each component is analyzed as a control volume at steady state. The control volumes are shown on the accompanying

sketch by dashed lines.

2. The diffuser, compressor, turbine, and nozzle processes are isentropic.

3. There is no pressure drop for flow through the combustor.

4. The turbine work output equals the work required to drive the compressor.

5. Except at the inlet and exit of the engine, kinetic energy effects can be ignored. Potential energy effects are negligible

throughout.

6. The working fluid is air modeled as an ideal gas.

Analysis: To determine the velocity at the exit to the nozzle, the mass and energy rate balances for a control volume en-

closing this component reduce at steady state to give

0  Q

 W

 m

c1h

 h

2 a

 V

b g1z



= 240°K

= 0.8 bar

= 1000 km/h

= 0.8 bar

= 1200°K

= 240°K

= 0

TurbineCompressor

Diffuser

Nozzle

Combustor

= 8

 Figure E9.12

9.9 Gas Turbines for Aircraft Propulsion 417

where is the mass flow rate. The inlet kinetic energy is dropped by assumption 5. Solving for V

This expression requires values for the specific enthalpies h

and h

at the nozzle inlet and exit, respectively. With the oper-

ating parameters specified, the determination of these enthalpy values is accomplished by analyzing each component in turn,

beginning with the diffuser. The pressure at each principal state can be evaluated as a part of the analyses required to find the

enthalpies h

and h

Mass and energy rate balances for a control volume enclosing the diffuser reduce to give

With h

from Table A-22 and the given value of V

Interpolating in Table A-22 gives p

 1.070. The flow through the diffuser is isentropic, so pressure p

With p

data from Table A-22 and the known value of p

Using the given compressor pressure ratio, the pressure at state 2 is p

 8(1.347 bars)  10.78 bars

The flow through the compressor is also isentropic. Thus

Interpolating in Table A-22, we get h

 505.5 kJ/kg.

At state 3 the temperature is given as T

 1200 K. From Table A-22, h

 1277.79 kJ/ kg. By assumption 3, p

 p

. The

work developed by the turbine is just sufficient to drive the compressor (assumption 4). That is

Solving for h

Interpolating in Table A-22 with h

, gives p

 116.8

The expansion through the turbine is isentropic, so

With p

 p

and p

data from Table A-22

The expansion through the nozzle is isentropic to p

 0.8 bars. Thus

 p

 1116.02

0.8

5.25

 17.68

 110.78 bars2

116.0

238.0

 5.25 bars

 p

 1051 kJ/kg

 h

 h

 h

 1277.79  278.7  505.5

 h

 h

 h



 p

 1.07182 8.56



1.070

0.6355

10.8 bar2 1.347 bars



 278.7 kJ/kg

 240.02 kJ /kg  c

12782

b `

1 N

1 kg

m/s

` `

1 kJ

 h



 121h

 h

❶

418 Chapter 9 Gas Power Systems

From Table A-22, h

 265.8 Btu/lb, which is the remaining specific enthalpy value required to determine the velocity at the

nozzle exit.

Using the values for h

and h

determined above, the velocity at the nozzle exit is

Note the unit conversions required here and in the calculation of V

The increase in the velocity of the air as it passes through the engine gives rise to the thrust produced by the engine. A

detailed analysis of the forces acting on the engine requires Newton’s second law of motion in a form suitable for control

volumes (See Sec. 9.12.1).

 927 m/s



211051  621.32

1 kg

m/s

1 N

b a

1 kJ

 121h

 h

❶

❷

OTHER APPLICATIONS. Other related applications of the gas turbine include turboprop and

turbofan engines. The turboprop engine shown in Fig. 9.22a consists of a gas turbine in which

the gases are allowed to expand through the turbine to atmospheric pressure. The net power

developed is directed to a propeller, which provides thrust to the aircraft. Turboprops are ef-

ficient propulsion devices for speeds of up to about 600 km/ h (400 miles/h). In the turbo-

fan shown in Fig. 9.22b, the core of the engine is much like a turbojet, and some thrust is

obtained from expansion through the nozzle. However, a set of large-diameter blades attached

to the front of the engine accelerates air around the core. This bypass flow provides addi-

tional thrust for takeoff, whereas the core of the engine provides the primary thrust for

cruising. Turbofan engines are commonly used for commercial aircraft with flight speeds of

up to about 1000 km/h (600 miles/h). A particularly simple type of engine known as a ram-

jet is shown in Fig. 9.22c. This engine requires neither a compressor nor a turbine. A suffi-

cient pressure rise is obtained by decelerating the high-speed incoming air in the diffuser

(a)(b)

(c)

Propeller

Fan

By-pass flow

Basic engine-core

Diffuser

Flame holder

Nozzle

 Figure 9.22 Other examples of aircraft engines. (a) Turboprop. (b) Turbofan. (c) Ramjet.

9.10 Combined Gas Turbine–Vapor Power Cycle 419

(ram effect). For the ramjet to operate, therefore, the aircraft must already be in flight at high

speed. The combustion products exiting the combustor are expanded through the nozzle to

produce the thrust.

In each of the engines mentioned thus far, combustion of the fuel is supported by air

brought into the engines from the atmosphere. For very high-altitude flight and space travel,

where this is no longer possible, rockets may be employed. In these applications, both fuel

and an oxidizer (such as liquid oxygen) are carried on board the craft. Thrust is developed

when the high-pressure gases obtained on combustion are expanded through a nozzle and

discharged from the rocket.



9.10 Combined Gas Turbine–Vapor

Power Cycle

A combined power cycle couples two power cycles such that the energy discharged by heat

transfer from one cycle is used partly or wholly as the input for the other cycle. The binary

vapor cycle introduced in Sec. 8.5 is an example of a combined power cycle. In the present

section, a combined gas turbine–vapor power cycle is considered.

The stream exiting the turbine of a gas turbine is at a high temperature. One way the po-

tential (exergy) of this high-temperature gas stream can be used, thereby improving overall

fuel utilization, is by a regenerator that allows the turbine exhaust gas to preheat the air be-

tween the compressor and combustor (Sec. 9.7). Another method is provided by the com-

bined cycle shown in Fig. 9.23, involving a gas turbine cycle and a vapor power cycle. The

two power cycles are coupled so that the heat transfer to the vapor cycle is provided by the

gas turbine cycle, which may be called the topping cycle.

Combustor

TurbineCompressor

gas

vap

Cooling

water

Vapor

cycle

Heat exchanger

Condenser

Gas turbine

Air inlet

Exhaust

Pump

Turbine

 Figure 9.23 Combined

gas turbine–vapor power

plant.

combined cycle

topping cycle

420 Chapter 9 Gas Power Systems

The combined cycle has the gas turbine’s high average temperature of heat addition and

the vapor cycle’s low average temperature of heat rejection, and thus a thermal efficiency

greater than either cycle would have individually. For many applications combined cycles

are economical, and they are increasingly being used worldwide for electric power gener-

ation.

With reference to Fig. 9.23, the thermal efficiency of the combined cycle is

(9.28)

where is the net power developed by the gas turbine and is the net power devel-

oped by the vapor cycle. denotes the total rate of heat transfer to the combined cycle, in-

cluding additional heat transfer, if any, to superheat the vapor entering the vapor turbine. The

evaluation of the quantities appearing in Eq. 9.28 follows the procedures described in the

sections on vapor cycles and gas turbines.

The relation for the energy transferred from the gas cycle to the vapor cycle for the system

of Fig. 9.23 is obtained by applying the mass and energy rate balances to a control volume

enclosing the heat exchanger. For steady-state operation, negligible heat transfer with the sur-

roundings, and no significant changes in kinetic and potential energy, the result is

(9.29)

where and are the mass flow rates of the gas and vapor, respectively.

As witnessed by relations such as Eqs. 9.28 and 9.29, combined cycle performance can

be analyzed using mass and energy balances. To complete the analysis, however, the second

law is required to assess the impact of irreversibilities and the true magnitudes of losses.

Among the irreversibilities, the most significant is the exergy destroyed by combustion. About

30% of the exergy entering the combustor with the fuel is destroyed by combustion irre-

versibility. An analysis of the gas turbine on an air-standard basis does not allow this exergy

destruction to be evaluated, however, and means introduced in Chap. 13 must be applied for

this purpose.

The next example illustrates the use of mass and energy balances, the second law, and

property data to analyze combined cycle performance.

 h

2 m

 h

vap

gas

h 

gas

 W

vap

than before. Advanced

blade cooling meth-

ods, new alloys, and

unique single crystal

blade construction al-

low the turbines to

withstand higher tem-

peratures. The high

thermal efficiency also

results in the lowest

carbon dioxide pro-

duction per megawatt

of electricity of any gas

turbine in the world.

Nitric oxide (NO

)

emission levels are

half the average of other turbines now in use.

New Power Plant Breaks Imagined Barrier

Thermodynamics in the News...

The new General Electric H System

is the first combined-

cycle power plant to achieve 60% thermal efficiency, once the

“four minute mile” of power plant technology. The installa-

tion is in Wales, where the 480-megawatt system provides

power to adjacent chemical manufacturing plants and to the

energy grid.

Development efforts began in the early 1990s, when the

best available technology combining gas and steam turbines

achieved thermal efficiencies of only 50%. Each percentage in-

crease is estimated to save $15–20 million dollars of natural

gas over the life of a typical system, according to industry

sources. Electric rates with the H System

are said to be 30%

less than were previously available.

The efficiency gains are achieved because natural gas

is burned at a temperature nearly 300F (167C) higher

9.10 Combined Gas Turbine–Vapor Power Cycle 421

EXAMPLE 9.13 Combined Cycle Exergy Accounting

A combined gas turbine–vapor power plant has a net power output of 45 MW. Air enters the compressor of the gas turbine

at 100 kPa, 300 K, and is compressed to 1200 kPa. The isentropic efficiency of the compressor is 84%. The condition at the

inlet to the turbine is 1200 kPa, 1400 K. Air expands through the turbine, which has an isentropic efficiency of 88%, to a

pressure of 100 kPa. The air then passes through the interconnecting heat exchanger and is finally discharged at 400 K. Steam

enters the turbine of the vapor power cycle at 8 MPa, 400C, and expands to the condenser pressure of 8 kPa. Water enters

the pump as saturated liquid at 8 kPa. The turbine and pump of the vapor cycle have isentropic efficiencies of 90 and 80%,

respectively.

(a) Determine the mass flow rates of the air and the steam, each in kg/s, and the net power developed by the gas turbine and

vapor power cycle, each in MW.

(b) Develop a full accounting of the net rate of exergy increase as the air passes through the gas turbine combustor. Discuss.

Let T

 300 K, p

 100 kPa.

SOLUTION

Known: A combined gas turbine–vapor power plant operates at steady state with a known net power output. Operating pres-

sures and temperatures are specified. Turbine, compressor, and pump efficiencies are also given.

Find: Determine the mass flow rate of each working fluid, in kg/s, and the net power developed by each cycle, in MW.

Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor and discuss the results.

Schematic and Given Data:

Combustor

TurbineCompressor

gas

t – W

vap

t – W

Vapor

cycle

Heat exchanger

Condenser

Gas turbine

Air inlet

Exhaust

Pump

Turbine

out

= 1400 K

= p

= 1200 kPa

= 400 K

= p

= 100 kPa

= p

= 8 kPa

= 400°C

= 8 MPa

= 300 K

= 100 kPa

= 88%

= 90%

= 80%

= 84%

Assumptions:

1. Each component on the accompanying

sketch is analyzed as a control volume at steady

state.

2. The turbines, compressor, pump, and inter-

connecting heat exchanger operate adiabati-

cally.

3. Kinetic and potential energy effects are neg-

ligible.

4. There are no pressure drops for flow through

the combustor, interconnecting heat exchanger,

and condenser.

5. An air-standard analysis is used for the gas

turbine.

6. T

 300 K, p

 100 kPa.

 Figure E9.13

422 Chapter 9 Gas Power Systems

Analysis: The property data given in the table below are determined using procedures illustrated in previous solved exam-

ples of Chaps. 8 and 9. The details are left as an exercise.

Gas Turbine Vapor Cycle

State h (kJ/kg) s (kJ/kg K) State h (kJ/kg) s (kJ/kg K)

1 300.19 1.7020 6 183.96 0.5975

2 669.79 2.5088 7 3138.30 6.3634

3 1515.42 3.3620 8 2104.74 6.7282

4 858.02 2.7620 9 173.88 0.5926

5 400.98 1.9919

(a) To determine the mass flow rates of the vapor, and the air, begin by applying mass and energy rate balances to

the interconnecting heat exchanger to obtain

Mass and energy rate balances applied to the gas turbine and vapor power cycles give the net power developed by each,

respectively

With

Solving for and inserting  45 MW  45,000 kJ/s and we get

and

Using these mass flow rate values and specific enthalpies from the table above, the net power developed by the gas tur-

bine and vapor power cycles, respectively, is

(b) The net rate of exergy increase of the air passing through the combustor is (Eq. 7.36)

 m

 h

 T

1s°

 s°

 R ln p



 E

 m

 h

 T

 s

vap

 a15.6

b a1023.5

1 MW

kJ/s

` 15.97 MW

gas

 a100.87

b a287.8

1 MW

kJ/s

` 29.03 MW

 10.15472m

 15.6 kg/s

 100.87 kg/s



45,000 kJ/s

5311515.42  858.022 1669.79  300.1924 0.1547 313138.3  2104.742 1183.96  173.88246 kJ/kg



 0.1547,W

net

 m

e31h

 h

2 1h

 h

24

31h

 h

2 1h

 h

24f

net

 W

gas

 W

vap

 m

31h

 h

2 1h

 h

gas

 m

31h

 h

2 1h

 h



 h



858.02  400.98

3138.3  183.96

 0.1547

0  m

 h

2 m

 h

9.10 Combined Gas Turbine–Vapor Power Cycle 423

With assumption 4, we have

The net rate exergy is carried out by the exhaust air stream at 5 is

The net rate exergy is carried out as the water passes through the condenser is

The rates of exergy destruction for the air turbine, compressor, steam turbine, pump, and interconnecting heat exchanger

are evaluated using respectively, as follows:

Air turbine:

Compressor:

Steam turbine:

Pump:

 0.02 MW

 115.621300210.5975  0.59262`

 m

 s

 1.71 MW

 115.621300216.7282  6.36342`

 m

 s

 2.83 MW

 1100.87213002c12.5088  1.70202

8.314

28.97

ln a

1200

100

bd`

 m

1s°

 s°

 R ln p



 m

 s

 3.42 MW

 a100.87

b 1300 K2

c12.7620  3.36202

 a

8.314

28.97

b ln a

100

1200

1 MW

kJ/s

 m

1s°

 s°

 R ln p



 m

 s

 T

 1.41 MW

 a15.6

bc12104.74  173.882

 300 K16.7282  0.59262

d `

1 MW

kJ/s

f 8

 E

 m

 h

 T

 s

 1.39 MW

 a100.87

b31400.98  300.192 30011.9919  1.702024

1 MW

kJ/s

f 5

 E

 m

 h

 T

as°

 s°

 R ln

 59,480

1 MW

kJ/s

` 59.48 MW

 a100.87

c11515.42  669.792

 300 K13.3620  2.50882

 E

 m

 h

 T

as°

 s°

 R ln

❶

424 Chapter 9 Gas Power Systems

Heat exchanger:

The results are summarized by the following exergy rate balance sheet in terms of exergy magnitudes on a rate basis:

Net exergy increase of the gas passing

through the combustor: 59.48 MW 100% (70%)*

Disposition of the exergy:

• Net power developed

gas turbine cycle 29.03 MW 48.8% (34.2%)

vapor cycle 15.97 MW 26.8% (18.8%)

Subtotal 45.00 MW 75.6% (53.0%)

• Net exergy lost

with exhaust gas at state 5 1.39 MW 2.3% (1.6%)

from water passing through condenser 1.41 MW 2.4% (1.7%)

• Exergy destruction

air turbine 3.42 MW 5.7% (4.0%)

compressor 2.83 MW 4.8% (3.4%)

steam turbine 1.71 MW 2.9% (2.0%)

pump 0.02 MW — —

heat exchanger 3.68 MW 6.2% (4.3%)

*Estimation based on fuel exergy. For discussion, see note 2.

The subtotals given in the table under the net power developed heading indicate that the combined cycle is effective in

generating power from the exergy supplied. The table also indicates the relative significance of the exergy destructions in the

turbines, compressor, pump, and heat exchanger, as well as the relative significance of the exergy losses. Finally, the table

indicates that the total of the exergy destructions overshadows the losses.

The development of the appropriate expressions for the rates of entropy production in the turbines, compressor, pump, and

heat exchanger is left as an exercise.

In this exergy balance sheet, the percentages shown in parentheses are estimates based on the fuel exergy. Although com-

bustion is the most significant source of irreversibility, the exergy destruction due to combustion cannot be evaluated using

an air-standard analysis. Calculations of exergy destruction due to combustion (Chap. 13) reveal that approximately 30% of

the exergy entering the combustor with the fuel would be destroyed, leaving about 70% of the fuel exergy for subsequent

use. Accordingly, the value 59.48 MW for the net exergy increase of the air passing through the combustor is assumed to be

70% of the fuel exergy supplied. The other percentages in parentheses are obtained by multiplying the corresponding per-

centages, based on the exergy increase of the air passing through the combustor, by the factor 0.7. Since they account for

combustion irreversibility, the table values in parentheses give the more accurate picture of combined cycle performance.

 3.68 MW

 1300 K2

ca100.87

b 11.9919  2.76202

 a15.6

b 16.3634  0.59752

d `

1 MW

kJ/s

 T

 s

2 m

 s

❶

❷



9.11 Ericsson and Stirling Cycles

Significant increases in the thermal efficiency of gas turbine power plants can be achieved

through intercooling, reheat, and regeneration. There is an economic limit to the number of

stages that can be employed, and normally there would be no more than two or three. Nonethe-

less, it is instructive to consider the situation where the number of stages of both intercool-

ing and reheat becomes indefinitely large. Figure 9.24a shows an ideal closed regenerative

gas turbine cycle with several stages of compression and expansion and a regenerator whose

effectiveness is 100%. Each intercooler is assumed to return the working fluid to the

9.11 Ericsson and Stirling Cycles 425

temperature T

at the inlet to the first compression stage and each reheater restores the work-

ing fluid to the temperature T

at the inlet to the first turbine stage. The regenerator allows

the heat input for Process 2–3 to be obtained from the heat rejected in Process 4–1. Ac-

cordingly, all the heat added externally would occur in the reheaters, and all the heat rejected

to the surroundings would take place in the intercoolers.

In the limit, as an infinite number of reheat and intercooler stages is employed, all the heat

added would occur when the working fluid is at its highest temperature, T

, and all the heat

rejected would take place when the working fluid is at its lowest temperature, T

. The limiting

cycle, shown in Fig. 9.24b, is called the Ericsson cycle. Since irreversibilities are presumed

absent and all the heat is supplied and rejected isothermally, the thermal efficiency of the

Ericsson cycle equals that of any reversible power cycle operating with heat addition at the

temperature T

and heat rejection at the temperature . This expression

was applied previously to evaluate the thermal efficiency of Carnot power cycles. Although the

details of the Ericsson cycle differ from those of the Carnot cycle, both cycles have the same

value of thermal efficiency when operating between the temperatures T

and T

STIRLING CYCLE. Another cycle that employs a regenerator is the Stirling cycle, shown on

the p–v and T–s diagrams of Fig. 9.25. The cycle consists of four internally reversible processes

in series: isothermal compression from state 1 to state 2 at temperature T

, constant-volume

heating from state 2 to state 3, isothermal expansion from state 3 to state 4 at temperature T

and constant-volume cooling from state 4 to state 1 to complete the cycle. A regenerator whose

effectiveness is 100% allows the heat rejected during Process 4–1 to be used as the heat in-

put in Process 2–3. Accordingly, all the heat added to the working fluid externally would take

: h

max

 1  T



p = c

(a)(b)

p = c

 Figure 9.24 Ericsson cycle as a limit of ideal gas turbine operation using

multistage compression with intercooling, multistage expansion with reheating,

and regeneration.

T = c

v = c

 Figure 9.25 p–v and T–s diagrams of the Stirling cycle.

Ericsson cycle