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

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Gas Power

Systems

When you complete your study of this chapter, you will be able to...

Perform air-standard analyses of internal combustion engines based on the Otto, Diesel,

and dual cycles, including:

sketching p–y and T–s diagrams and evaluating property data at principal states.

applying energy, entropy, and exergy balances.

determining net power output, thermal efficiency, and mean effective pressure.

Perform air-standard analyses of gas turbine power plants based on the Brayton cycle and

its modifications, including:

sketching T–s diagrams and evaluating property data at principal states.

applying mass, energy, entropy, and exergy balances.

determining net power output, thermal efficiency, back work ratio, and the effects of

compressor pressure ratio.

For subsonic and supersonic flows through nozzles and diffusers:

demonstrate understanding of the effects of area change, the effects of back pressure

on mass flow rate, and the occurrence of choking and normal shocks.

analyze the flow of ideal gases with constant specific heats.

LEARNING OUTCOMES

493

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494 Chapter 9

Gas Power Systems

Considering Internal Combustion

Engines

This part of the chapter deals with internal combustion engines. Although most gas

turbines are also internal combustion engines, the name is usually applied to recip-

rocating internal combustion engines of the type commonly used in automobiles,

trucks, and buses. These engines differ from the power plants considered in Chap. 8

because the processes occur within reciprocating piston–cylinder arrangements and

not in interconnected series of different components.

Two principal types of reciprocating internal combustion engines are the spark-

ignition

engine and the compression-ignition engine. In a spark-ignition engine, a mix-

ture of fuel and air is ignited by a spark plug. In a compression-ignition engine, air

is compressed to a high enough pressure and temperature that combustion occurs

spontaneously when fuel is injected. Spark-ignition engines have advantages for appli-

cations requiring power up to about 225 kW (300 horsepower). Because they are

relatively light and lower in cost, spark-ignition engines are particularly suited for use

in automobiles. Compression-ignition engines are normally preferred for applications

when fuel economy and relatively large amounts of power are required (heavy trucks

and buses, locomotives and ships, auxiliary power units). In the middle range, spark-

ignition and compression-ignition engines are used.

9.1 Introducing Engine Terminology

Figure 9.1 is a sketch of a reciprocating internal combustion engine consisting of a

piston that moves within a cylinder fitted with two valves. The sketch is labeled with

some special terms. The bore of the cylinder is its diameter. The stroke is the distance

the piston moves in one direction. The piston is said to be at top dead center when

it has moved to a position where the cylinder volume is a minimum. This minimum

volume is known as the clearance volume. When the piston has moved to the position

of maximum cylinder volume, the piston is at bottom dead center. The volume swept

out by the piston as it moves from the top dead center to the bottom dead center

position is called the displacement volume. The compression ratio r is

defined as the volume at bottom dead center divided by the volume

at top dead center. The reciprocating motion of the piston is converted

to rotary motion by a crank mechanism.

In a four-stroke internal combustion engine, the piston executes four

distinct strokes within the cylinder for every two revolutions of the

crankshaft. Fig. 9.2 gives a pressure–volume diagram such as might be

displayed electronically.

1. With the intake valve open, the piston makes an intake stroke to

draw a fresh charge into the cylinder. For spark-ignition engines, the

charge is a combustible mixture of fuel and air. Air alone is the

charge in compression-ignition engines.

2. With both valves closed, the piston undergoes a compression stroke,

raising the temperature and pressure of the charge. This requires

work input from the piston to the cylinder contents. A combustion

process is then initiated, resulting in a high-pressure, high-temperature

gas mixture. Combustion is induced near the end of the compression

stroke in spark-ignition engines by the spark plug. In compression-

ignition engines, combustion is initiated by injecting fuel into the

hot compressed air, beginning near the end of the compression

stroke and continuing through the first part of the expansion.

spark-ignition

compression-ignition

compression ratio

Spark plug or

fuel injector

Piston

Stroke

Bore

Crank mechanism

Reciprocating

motion

Top

dead center

Bottom

dead center

Rotary

motion

Valve

Clearance

volume

Cylinder

wall

Fig. 9.1 Nomenclature for reciprocating

piston–cylinder engines.

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3. A power stroke follows the compression stroke, during which the gas

mixture expands and work is done on the piston as it returns to bot-

tom dead center.

4. The piston then executes an exhaust stroke in which the burned gases

are purged from the cylinder through the open exhaust valve.

Smaller engines operate on two-stroke cycles. In two-stroke engines, the

intake, compression, expansion, and exhaust operations are accom-

plished in one revolution of the crankshaft. Although internal combus-

tion engines undergo mechanical cycles, the cylinder contents do not

execute a thermodynamic cycle, for matter is introduced with one com-

position and is later discharged at a different composition.

A parameter used to describe the performance of reciprocating pis-

ton engines is the mean effective pressure, or mep. The mean effective

pressure

is the theoretical constant pressure that, if it acted on the piston

during the power stroke, would produce the same net work as actually

developed in one cycle. That is

mep 5

net work for one cycl

displacement volume

(9.1)

For two engines of equal displacement volume, the one with a higher

mean effective pressure would produce the greater net work and, if the

engines run at the same speed, greater power.

AIR-STANDARD ANALYSIS. A detailed study of the performance of a recip-

rocating internal combustion engine would take into account many features. These

would include the combustion process occurring within the cylinder and the effects

of irreversibilities associated with friction and with pressure and temperature gradi-

ents. Heat transfer between the gases in the cylinder and the cylinder walls and the

work required to charge the cylinder and exhaust the products of combustion also

would be considered. Owing to these complexities, accurate modeling of reciprocating

internal combustion engines normally involves computer simulation. To conduct ele-

mentary thermodynamic analyses of internal combustion engines, considerable sim-

plification is required. One procedure is to employ an air-standard analysis having the

following elements:

c A fixed amount of air modeled as an ideal gas is the working fluid. See Table 9.1

for a review of ideal gas relations.

c The combustion process is replaced by a heat transfer from an external source.

c There are no exhaust and intake processes as in an actual engine. The cycle is

completed by a constant-volume heat transfer process taking place while the piston

is at the bottom dead center position.

c All processes are internally reversible.

In addition, in a cold air-standard analysis, the specific heats are assumed constant at

their ambient temperature values. With an air-standard analysis, we avoid dealing with

the complexities of the combustion process and the change of composition during

combustion. A comprehensive analysis requires that such complexities be considered,

however. For a discussion of combustion, see Chap. 13.

Although an air-standard analysis simplifies the study of internal combustion

engines considerably, values for the mean effective pressure and operating tempera-

tures and pressures calculated on this basis may depart significantly from those of

actual engines. Accordingly, air-standard analysis allows internal combustion engines

to be examined only qualitatively. Still, insights concerning actual performance can

result with such an approach.

Top dead

center

Intake

valve

opens,

exhaust

valve

closes

Exhaust valve

opens

Exhaust

Compression

Power

Intake valve

closes

Intake

Volume

Bottom

dead center

Combustion initiated

Fig. 9.2 Pressure–volume diagram for a

reciprocating internal combustion engine.

air-standard analysis:

internal combustion engines

cold air-standard analysis

9.1 Introducing Engine Terminology 495

mean effective pressure

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496 Chapter 9

Gas Power Systems

Ideal Gas Model Review

Equations of state:

py 5 RT (3.32)

pV 5 mRT (3.33)

Changes in u and h:

u(T

) 2 u(T

) 5

(T) dT

(3.40)

h(T

) 2 h(T

) 5

(T ) dT

(3.43)

Constant Specific Heats Variable Specific Heats

u(T

) 2 u(T

) 5 c

2 T

) (3.50)

h(T

) 2 h(T

) 5 c

2 T

) (3.51)

See Tables A-20, 21 for data.

u(T ) and h(T ) are evaluated from appropriate

tables: Tables A-22 for air (mass basis) and

A-23 for other gases (molar basis).

Changes in s:

s(T

, y

) 2 s(T

, y

) 5

(T )

1 R ln

(6.17)

s(T

, p

) 2 s(T

, p

) 5

(T )

2 R ln

(6.18)

Constant Specific Heats Variable Specific Heats

s(T

, y

) 2 s(T

, y

) 5

1 R ln

(6.21)

s(T

, p

) 2 s(T

, p

) 5

s8(T

) 2 s8(T

) 2 R ln

(6.20a)

s(T

, p

) 2 s(T

, p

) 5

2 R ln

(6.22)

See Tables A-20, 21 for data.

where s°(T) is evaluated from appropriate

tables: Tables A-22 for air (mass basis) and

A-23 for other gases (molar basis).

Relating states of equal specific entropy: Ds 5 0:

Constant Specific Heats Variable Specific Heats — Air Only

5 a

(k 2 1)

(6.43)

5 a

k2 1

(6.44)

5 a

(6.45)

where k 5 c

is given in Tables A-20

for several gases.

(air only) (6.41)

(air only) (6.42)

where p

and y

are provided for air

in Tables A-22.

TABLE 9.1

ENERGY & ENVIRONMENT Nature needed 500 million years to create the

world’s stock of readily accessible oil, but some observers predict we will consume

much of what’s left within the next 50 years. Still, the important issue, they say, is

not when the world runs out of oil, but when production will peak. After that, unless demand is

reduced, oil prices will rise. This will end the era of cheap oil we have enjoyed for decades and

pose challenges for society.

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9.2 Air-Standard Otto Cycle

In the remainder of this part of the chapter, we consider three cycles that adhere to

air-standard cycle idealizations: the Otto, Diesel, and dual cycles. These cycles differ

from each other only in the way the heat addition process that replaces combustion

in the actual cycle is modeled.

The air-standard Otto cycle is an ideal cycle that assumes heat addition occurs instan-

taneously while the piston is at top dead center. The Otto cycle is shown on the p–y

and T–s diagrams of Fig. 9.3. The cycle consists of four internally reversible processes

in series:

c Process 1–2 is an isentropic compression of the air as the piston moves from bot-

tom dead center to top dead center.

c Process 2–3 is a constant-volume heat transfer to the air from an external source

while the piston is at top dead center. This process is intended to represent the

ignition of the fuel–air mixture and the subsequent rapid burning.

c Process 3–4 is an isentropic expansion (power stroke).

c Process 4–1 completes the cycle by a constant-volume process in which heat is

rejected from the air while the piston is at bottom dead center.

Since the air-standard Otto cycle is composed of internally reversible processes,

areas on the T–s and p–y diagrams of Fig. 9.3 can be interpreted as heat and work,

The rate at which any well can produce oil typically rises to a maximum and then, when about

half the oil has been pumped out, begins to fall as the remaining oil becomes increasingly difficult

to extract. Using this model for the world oil supply as a whole, economists predict a peak in oil

production by 2020, or even sooner. If we don’t plan for the future, this could lead to shortages,

higher fuel prices at the pump, and political repercussions.

The reality is that oil has become our Achilles heel with regard to energy. Transportation

accounts for about 70% of current U.S. oil consumption, while nearly 60% of the oil we use is

imported. Domestic oil production peaked in the 1970s and has declined since then. There have

been calls for greater production from our offshore and wilderness areas, but analysts say this

will not be a solution, for oil from many such locations will meet only a few months of current

demand. Moreover, ecologists warn about environmental damage owing to huge oil spills; col-

lateral soil, water, and air pollution; and other adverse effects related to oil extraction, delivery,

and use. Many think a better way forward is to wean ourselves from relying on oil for transporta-

tion by using strategies for economizing fuel (see www.fueleconomy.gov), using biofuels such as

cellulosic ethanol, and driving hybrid-electric or all-electric vehicles.

Otto cycle

2′

3′

v = c

b a

s = c

a b

Fig. 9.3 p–

and T–s diagrams of the

air-standard Otto cycle.

TAKE NOTE...

For internally reversible

processes of closed sys-

tems, see Secs. 2.2.5 and

6.6.1 for discussions of

area interpretations of

work and heat transfer on

p–y and T–s diagrams,

respectively.

9.2 Air-Standard Otto Cycle 497

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498 Chapter 9 Gas Power Systems

respectively. On the T–s diagram, area 2–3–a–b–2 represents heat added per unit of

mass and area 1–4–a–b–1 the heat rejected per unit of mass. On the p–y diagram,

area 1–2–a–b–1 represents work input per unit of mass during the compression pro-

cess and area 3–4–b–a–3 is work done per unit of mass in the expansion process. The

enclosed area of each figure can be interpreted as the net work output or, equiva-

lently, the net heat added.

CYCLE ANALYSIS. The air-standard Otto cycle consists of two processes in which

there is work but no heat transfer, Processes 1–2 and 3–4, and two processes in which

there is heat transfer but no work, Processes 2–3 and 4–1. Expressions for these

energy transfers are obtained by reducing the closed system energy balance assuming

that changes in kinetic and potential energy can be ignored. The results are

5 u

2 u



5 u

2 u

5 u

2 u

5 u

2 u

(9.2)

Carefully note that in writing Eqs. 9.2, we have departed from our usual sign conven-

tion for heat and work. Thus, W

/m is a positive number representing the work input

during compression and Q

/m is a positive number representing the heat rejected in

Process 4–1. The net work of the cycle is expressed as

cycle

5 1u

2 u

22 1u

2 u

Alternatively, the net work can be evaluated as the net heat added

cycle

5 1u

2 u

22 1u

2 u

which, on rearrangement, can be placed in the same form as the previous expression

for net work.

The thermal efficiency is the ratio of the net work of the cycle to the heat added.

h 5

2 u

5 1 2

2 u

(9.3)

When air table data are used to conduct an analysis involving an air-standard Otto

cycle, the specific internal energy values required by Eq. 9.3 can be obtained from

Table A-22 or A-22E as appropriate. The following relationships based on Eq. 6.42

apply for the isentropic processes 1–2 and 3–4

5 y

(9.4)

5 y

5 ry

(9.5)

where r denotes the compression ratio. Note that since V

5 V

and V

5 V

, r 5

5 V

The parameter y

is tabulated versus temperature for air in Tables A-22.

When the Otto cycle is analyzed on a cold air-standard basis, the following expres-

sions based on Eq. 6.44 would be used for the isentropic processes in place of Eqs. 9.4

and 9.5, respectively

5 a

k21

5 r

k21





1constant k2

(9.6)

5 a

k21



1constant k2

(9.7)

where k is the specific heat ratio, k 5 c

TAKE NOTE...

When analyzing air-standard

cycles, it is frequently con-

venient to regard all work

and heat transfers as

positive quantities and

write the energy balance

accordingly.

compression ratio

Otto_Cycle

A.27 – Tabs a & b

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EFFECT OF COMPRESSION RATIO ON PERFORMANCE. By referring to

the T–s diagram of Fig. 9.3, we can conclude that the Otto cycle thermal efficiency

increases as the compression ratio increases. An increase in the compression ratio

changes the cycle from 1–2–3–4–1 to 1–29–39–4–1. Since the average temperature of

heat addition is greater in the latter cycle and both cycles have the same heat rejection

process, cycle 1–29–39–4–1 would have the greater thermal efficiency. The increase in

thermal efficiency with compression ratio is also brought out simply by the following

development on a cold air-standard basis. For constant c

, Eq. 9.3 becomes

h 5 1 2

2 T

On rearrangement

h 5 1 2

2 1

From Eqs. 9.6 and 9.7 above, T

5 T

h 5 1 2

Finally, introducing Eq. 9.6

h 5 1 2

k21



1cold-air standard basis2

(9.8)

Equation 9.8 indicates that the cold air-standard Otto cycle thermal efficiency is a

function of compression ratio and k. This relationship is shown in Fig. 9.4 for k 5 1.4,

representing ambient air.

The foregoing discussion suggests that it is advantageous for internal combustion

engines to have high compression ratios, and this is the case. The possibility of autoigni-

tion, or “knock,” places an upper limit on the compression ratio of spark-ignition engines,

however. After the spark has ignited a portion of the fuel–air mixture, the rise in pressure

accompanying combustion compresses the remaining charge. Autoignition can occur if

the temperature of the unburned mixture becomes too high before the mixture is con-

sumed by the flame front. Since the temperature attained by the air–fuel mixture during

the compression stroke increases as the compression ratio increases, the likelihood of

autoignition occurring increases with the compression ratio. Autoignition may result in

high-pressure waves in the cylinder (manifested by a knocking or pinging sound) that

can lead to loss of power as well as engine damage.

Owing to performance limitations, such as autoignition, the compression ratios of spark-

ignition engines using the unleaded fuel required today because of air pollution concerns

are in the range 9.5 to 11.5, approximately. Higher compression ratios can be achieved in

compression-ignition engines because only air is compressed. Compression ratios in the

range 12 to 20 are typical. Compression-ignition engines also can use less refined fuels hav-

ing higher ignition temperatures than the volatile fuels required by spark-ignition engines.

In the next example, we illustrate the analysis of the air-standard Otto cycle. Results

are compared with those obtained on a cold air-standard basis.

1062840

Compression ratio, r

η (%)

Fig. 9.4 Thermal efficiency

of the cold air-standard Otto

cycle, k 5 1.4.

Analyzing the Otto Cycle

c c c c EXAMPLE 9.1 c

The temperature at the beginning of the compression process of an air-standard Otto cycle with a compression

ratio of 8 is 5408R, the pressure is 1 atm, and the cylinder volume is 0.02 ft

. The maximum temperature during

the cycle is 36008R. Determine (a) the temperature and pressure at the end of each process of the cycle, (b) the

thermal efficiency, and (c) the mean effective pressure, in atm.

9.2 Air-Standard Otto Cycle 499

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500 Chapter 9

Gas Power Systems

SOLUTION

Known:

An air-standard Otto cycle with a given value of compression ratio is executed with specified conditions

at the beginning of the compression stroke and a specified maximum temperature during the cycle.

Find: Determine the temperature and pressure at the end of each process, the thermal efficiency, and mean

effective pressure, in atm.

Schematic and Given Data:

Engineering Model:

The air in the piston–cylinder assembly is the closed system.

2. The compression and expansion processes are adiabatic.

3. All processes are internally reversible.

4. The air is modeled as an ideal gas.

5. Kinetic and potential energy effects are negligible.

Analysis:

(a)

The analysis begins by determining the temperature, pressure, and specific internal energy at each principal

state of the cycle. At T

5 5408R, Table A-22E gives u

5 92.04 Btu/lb and y

5 144.32.

For the isentropic compression Process 1–2

4.32

5 18.04

Interpolating with y

in Table A-22E, we get T

5 12128R and u

5 211.3 Btu/lb. With the ideal gas equation of

state

5 p

5 11 atm2

12128R

5408R

8 5 17.96 at

The pressure at state 2 can be evaluated alternatively by using the isentropic relationship, p

5 p

)

Since Process 2–3 occurs at constant volume, the ideal gas equation of state gives

5 p

5 117.96 atm2

36008R

12128R

5 53.3 at

At T

5 36008R, Table A-22E gives u

5 721.44 Btu/lb and y

5 0.6449.

For the isentropic expansion process 3–4

5 y

5 0.64491825 5.16

= c

s = c

= 3600°R

= 8

––

= 540°R

Fig. E9.1

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Interpolating in Table A-22E with y

gives T

5 18788R, u

5 342.2 Btu/lb. The pressure at state 4 can be found

using the isentropic relationship p

5 p

)

or the ideal gas equation of state applied at states 1 and 4. With

5 V

, the ideal gas equation of state gives

5 p

5 11 atm2 a

18788R

5408R

5 3.48 at

(b) The thermal efficiency is

h 5 1 2

5 1 2

2 u

5 1 2

342.2 2 92.04

721.44 2 211.

5 0.51 151%2

cle

5 m

2 u

where m is the mass of the air, evaluated from the ideal gas equation of state as follows:

m 5

114.696 lbf

in.

144 in.

10.02 ft

1545

28.97

ft ? lbf

lb ? 8R

15408R2

5 1.47 3 1

Inserting values into the expression for W

cycle

cle

5 11.47 3 10

lb231721.44 2 342.222 1211.3 2 92.0424 Btu

2 Btu

The displacement volume is V

2 V

so the mean effective pressure is given by

mep 5

cycle

2 V

cycle

1 2 V

0.382 Btu

0.02 ft

1 2 1

778 ft ? lbf

1 Btu

1 ft

144 in.

5 11

8.03

➊ This solution utilizes Table A-22E for air, which accounts explicitly for the variation of the specific heats with

temperature. A solution also can be developed on a cold air-standard basis in

which constant specific heats are assumed. This solution is left as an exercise,

but for comparison the results are presented for the case k 5 1.4 in the fol-

lowing table:

Cold Air-Standard Analysis,

Parameter Air-Standard Analysis k 5 1.4

12128R 12418R

36008R 36008R

18788R 15678R

h 0.51 (51%) 0.565 (56.5%)

mep 8.03 atm 7.05 atm

➊

Ability to…

❑

sketch the Otto cycle p–y

and T–s diagrams.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

calculate thermal efficiency

and mean effective pressure.

✓

Skills Developed

Determine the heat addition and the heat rejection for the cycle,

each in Btu. Ans. Q

5 0.750 Btu, Q

5 0.368 Btu.

9.2 Air-Standard Otto Cycle 501

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502 Chapter 9

Gas Power Systems

Diesel cycle

9.3 Air-Standard Diesel Cycle

The air-standard Diesel cycle is an ideal cycle that assumes heat addition occurs dur-

ing a constant-pressure process that starts with the piston at top dead center. The

Diesel cycle is shown on p–y and T–s diagrams in Fig. 9.5. The cycle consists of four

internally reversible processes in series. The first process from state 1 to state 2 is the

same as in the Otto cycle: an isentropic compression. Heat is not transferred to the

working fluid at constant volume as in the Otto cycle, however. In the Diesel cycle,

heat is transferred to the working fluid at constant pressure. Process 2–3 also makes

up the first part of the power stroke. The isentropic expansion from state 3 to state

4 is the remainder of the power stroke. As in the Otto cycle, the cycle is completed

by constant-volume Process 4–1 in which heat is rejected from the air while the pis-

ton is at bottom dead center. This process replaces the exhaust and intake processes

of the actual engine.

Since the air-standard Diesel cycle is composed of internally reversible processes,

areas on the T–s and p–y diagrams of Fig. 9.5 can be interpreted as heat and work,

respectively. On the T–s diagram, area 2–3–a–b–2 represents heat added per unit of

mass and area 1–4–a–b–1 is heat rejected per unit of mass. On the p–y diagram,

area 1–2–a–b–1 is work input per unit of mass during the compression process. Area

2–3–4–b–a–2 is the work done per unit of mass as the piston moves from top dead

center to bottom dead center. The enclosed area of each figure is the net work output,

which equals the net heat added.

CYCLE ANALYSIS. In the Diesel cycle the heat addition takes place at constant

pressure. Accordingly, Process 2–3 involves both work and heat. The work is given by

p dy 5 p

2 y

(9.9)

The heat added in Process 2–3 can be found by applying the closed system energy

balance

2 u

2 W

Introducing Eq. 9.9 and solving for the heat transfer

5 1u

2 u

21 p1y

2 y

25 1u

1 py

22 1u

1 py

5 h

2 h

(9.10)

where the specific enthalpy is introduced to simplify the expression. As in the Otto

cycle, the heat rejected in Process 4–1 is given by

5 u

2 u

Fig. 9.5 p–y and T–s

diagrams of the air-standard

Diesel cycle.

p = c

= c

s = c

aa bb

Diesel_Cycle

A.28 – Tabs a & b

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