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

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376 Chapter 9 Gas Power Systems

2′

3′

= c

v = c

= c

s =

 Figure 9.3 p–v and T–s diagrams of

the air-standard Otto cycle.

diagrams of Fig. 9.3. The cycle consists of four internally reversible processes in series:

 Process 1–2 is an isentropic compression of the air as the piston moves from bottom

dead center to top dead center.

 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.

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

 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–v diagrams of Fig. 9.3 can be interpreted as heat and work, respectively. On

the T–s diagram, area 2–3–a–b–2 represents the heat added per unit of mass and area

1–4–a–b–1 the heat rejected per unit of mass. On the p–v diagram, area 1–2–a–b–1 repre-

sents the work input per unit of mass during the compression process and area 3–4–b–a–3

is the work done per unit of mass in the expansion process. The enclosed area of each fig-

ure can be interpreted as the net work output or, equivalently, 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

(9.2)

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

heat and work. Thus, W

m is a positive number representing the work input during com-

pression and Q

m is a positive number representing the heat rejected in Process 4–1. The

net work of the cycle is expressed as

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

cycle





 1u

 u

2 1u

 u

cycle





 1u

 u

2 1u

 u

 u

 u

 u

 u

 u

 u

 u

 u

METHODOLOGY

UPDATE

When analyzing air-

standard cycles, it is

frequently convenient to

regard all work and heat

transfers as positive quan-

tities and write the energy

balance accordingly.

9.2 Air-Standard Otto Cycle 377

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.

(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. 6.44 apply for the isentropic

processes 1–2 and 3–4

(9.4)

(9.5)

where r denotes the compression ratio. Note that since V

 V

and V

 V

, r  V

V



V

. The parameter v

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

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

based on Eq. 6.46 would be used for the isentropic processes in place of Eqs. 9.4 and 9.5,

respectively

(9.6)

(9.7)

where k is the specific heat ratio, k  c

c

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 1–2–3– 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 1–2–3–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

On rearrangement

From Eqs. 9.6 and 9.7 above, T

T

 T

T

,so

Finally, introducing Eq. 9.6

(9.8)

h  1 

k1

1constant k2

h  1 



 1



 1

h  1 

 T

 a

k1



k1

1constant k2

 a

k1

 r

k1

1constant k2

 v

b rv

 v

b

h 

 u

2 1u

 u

 1 

 u

378 Chapter 9 Gas Power Systems

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  1.4, representing

ambient air.

The foregoing discussion suggests that it is advantageous for internal combustion en-

gines to have high compression ratios, and this is the case. The possibility of autoignition,

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

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

companying combustion compresses the remaining charge. Autoignition can occur if the

temperature of the unburned mixture becomes too high before the mixture is consumed by

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

pression 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. Fuels formulated with tetraethyl lead are resistant to

autoignition and thus allow relatively high compression ratios. The unleaded gasoline in

common use today because of environmental concerns over air pollution limits the com-

pression ratios of spark-ignition engines to approximately 9. Higher compression ratios

can be achieved in compression-ignition engines because air alone is compressed. Com-

pression ratios in the range of 12 to 20 are typical. Compression-ignition engines also can

use less refined fuels having 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 (%)

 Figure 9.4 Thermal

efficiency of the cold air-

standard Otto cycle, k 

1.4.

Critical shortages, but not

higher prices, may be avoided by

expensive methods of squeezing

more oil out of existing wells.

Also, unconventional petroleum

resources locked in oil shale and

tar sands could be tapped. Al-

though vast in extent, the costs

of extracting these resources

would be great. Extensive envi-

ronmental damage may result,

including the effects of air emis-

sions, liquid effluents, and vast

amounts of spent shale and sand that would remain after oil

has been removed.

End of Cheap Oil in View, Experts Say

Thermodynamics in the News...

Nature needed 500 million years to create the world’s stock of

readily accessible oil, but some observers predict we will con-

sume most of what’s left within the next 50 years. Still, the im-

portant issue, they say, is not when the world runs out of oil, but

when production will peak. After that, production must decline,

and unless demand is reduced, oil prices will rise. This will end

the era of cheap oil we enjoy and pose challenges for society.

The rate any well can produce oil typically rises to a max-

imum and then, when about half the oil has been pumped out,

begins to fall as the remaining oil becomes increasingly diffi-

cult to extract. Using this model for the world oil supply as a

whole, economists predict a peak in oil production by 2020,

or even as soon as 2010. That will lead to shortages and higher

prices at the pump, which will have global economic and po-

litical repercussions.

EXAMPLE 9.1 Analyzing the Otto Cycle

The temperature at the beginning of the compression process of an air-standard Otto cycle with a compression ratio of 8 is

300K, the pressure is 1 bar, and the cylinder volume is 560 cm

. The maximum temperature during the cycle is 2000K.

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 379

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:

Assumptions:

1. 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

 300 K, Table A-22 gives u

 214.07 kJ/kg and v

 621.2.

For the isentropic compression Process 1–2

Interpolating with v

in Table A-22, we get T

 673 K and u

 491.2 kJ/kg. With the ideal gas equation of state

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

 p

( p

p

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

At T

 2000 K, Table A-22 gives u

 1678.7 kJ/kg and v

 2.776.

For the isentropic expansion process 3–4

 v

 2.776182 22.21

 p

 117.95 bars2 a

2000 K

673 K

b 53.3 bars

 p

 11 atm2 a

673°K

300°K

b 8  17.95 bars



621.2

 77.65

= c

s = c

= 2000°Κ

= 8

––

= 300°Κ

 Figure E9.1

380 Chapter 9 Gas Power Systems

Interpolating in Table A-22 with v

gives T

 1043 K, u

 795.8 kJ/kg. The pressure at state 4 can be found using the isen-

tropic relationship p

 p

p

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

 V

, the ideal gas

equation of state gives

(b) The thermal efficiency is

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

Inserting values into the expression for W

cycle

The displacement volume is V

 V

, so the mean effective pressure is given by

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

perature. 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  1.4 in the following

table:

Cold Air-Standard Analysis,

Parameter Air-Standard Analysis k  1.4

673 K 689 K

2000 K 2000 K

1043 K 871 K

 0.51 (51%) 0.565 (56.5%)

mep 8.04 bars 7.05 bars

 8.04 bars



0.394 kJ

1560 cm

211  1



1 m

1 kJ

b a

1 bar

N/m

mep 

cycle

 V



cycle

11  V



 0.394 kJ

cycle

 16.5  10

4

kg2311678.7  795.82 1491.2  214.0724 kJ/kg

 6.5  10

4



11 bar21560 cm

8.314

28.97

b 1300 K2

1 m

b a

N/m

1 bar

b a

1 kJ

m 



M2T

cycle

 m 31u

 u

2 1u

 u

 1 

795.8  214.07

1678.7  491.2

 0.51 151%2

h  1 



 1 

 u

 p

 11 bar2 a

1043 K

300 K

b 3.48 bars

❶

9.3 Air-Standard Diesel Cycle 381

p = c

= c

s =

s = c

a bb

 Figure 9.5 p–v and T–s

diagrams of the air-standard

Diesel cycle.



9.3 Air-Standard Diesel Cycle

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

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

shown on p–v 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 isen-

tropic 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 piston 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, ar-

eas on the T–s and p–v diagrams of Fig. 9.5 can be interpreted as heat and work, respec-

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

area 1–4–a–b–1 is the heat rejected per unit of mass. On the p–v diagram, area 1–2–a–b–1

is the 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

(9.9)

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

Introducing Eq. 9.9 and solving for the heat transfer

(9.10)  h

 h

 1u

 u

2 p1v

 v

2 1u

 pv

2 1u

 pv

m1u

 u

2 Q

 W





p dv  p

 v

Diesel cycle

382 Chapter 9 Gas Power Systems

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

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

(9.11)

As for the Otto cycle, the thermal efficiency of the Diesel cycle increases with the com-

pression ratio.

To evaluate the thermal efficiency from Eq. 9.11 requires values for u

, u

, h

, and h

equivalently the temperatures at the principal states of the cycle. Let us consider next how

these temperatures are evaluated. For a given initial temperature T

and compression ratio r,

the temperature at state 2 can be found using the following isentropic relationship and v

data

To find T

, note that the ideal gas equation of state reduces with p

 p

to give

where r

 V

V

, called the cutoff ratio, has been introduced.

Since V

 V

, the volume ratio for the isentropic process 3–4 can be expressed as

(9.12)

where the compression ratio r and cutoff ratio r

have been introduced for conciseness.

Using Eq. 9.12 together with v

at T

, the temperature T

can be determined by interpo-

lation once v

is found from the isentropic relationship

In a cold air-standard analysis, the appropriate expression for evaluating T

is provided

The temperature T

is found similarly from

where Eq. 9.12 has been used to replace the volume ratio.

EFFECT OF COMPRESSION RATIO ON PERFORMANCE. As for the Otto cycle, the ther-

mal efficiency of the Diesel cycle increases with increasing compression ratio. This can be

brought out simply using a cold air-standard analysis. On a cold air-standard basis, the ther-

mal efficiency of the Diesel cycle can be expressed as

(9.13)

h  1 

k1

 1

k 1r

 12

1constant k2

 a

k1

 a

k1

1constant k2

 a

k1

 r

k1

1constant k2



 r



h 

cycle



 1 



 1 

 u

 h

 u

 u

cutoff ratio

9.3 Air-Standard Diesel Cycle 383

where r is the compression ratio and r

the cutoff ratio. The derivation is left as an exer-

cise. This relationship is shown in Fig. 9.6 for k  1.4. Equation 9.13 for the Diesel cycle

differs from Eq. 9.8 for the Otto cycle only by the term in brackets, which for r

 1 is

greater than unity. Thus, when the compression ratio is the same, the thermal efficiency of

the cold air-standard Diesel cycle would be less than that of the cold air-standard Otto

cycle.

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

(

)

2015105

Compression ratio, r

Thermal efficiency, η (%)

= 2

Diesel

cycle

= 3

 Figure 9.6 Thermal efficiency of the cold air-

standard Diesel cycle, k  1.4.

EXAMPLE 9.2 Analyzing the Diesel Cycle

At the beginning of the compression process of an air-standard Diesel cycle operating with a compression ratio of 18, the tem-

perature is 300 K and the pressure is 0.1 MPa. The cutoff ratio for the cycle is 2. Determine (a) the temperature and pressure

at the end of each process of the cycle, (b) the thermal efficiency, (c) the mean effective pressure, in MPa.

SOLUTION

Known: An air-standard Diesel cycle is executed with specified conditions at the beginning of the compression stroke. The

compression and cutoff ratios are given.

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

pressure.

Schematic and Given Data:

p = c

s = c

= 0.1 MPa

= 300 K

= 18

––

= 2r

––

 Figure E9.2

384 Chapter 9 Gas Power Systems

Assumptions:

1. 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 properties at each principal state of the cycle. With T

 300 K, Table A-22 gives

 214.07 kJ/kg and v

 621.2. For the isentropic compression process 1–2

Interpolating in Table A-22, we get T

 898.3 K and h

 930.98 kJ/kg. With the ideal gas equation of state

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

 p

( p

p

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

Introducing the cutoff ratio, r

 V

V

From Table A-22, h

 1999.1 kJ/kg and v

 3.97.

For the isentropic expansion process 3–4

Introducing V

 V

, the compression ratio r, and the cutoff ratio r

, we have

Interpolating in Table A-22 with v

, we get u

 664.3 kJ/kg and T

 887.7 K. The pressure at state 4 can be found using

the isentropic relationship p

 p

p

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

 V

, the ideal

gas equation of state gives

(b) The thermal efficiency is found using

mep 

cycle



 v



cycle



11  1



 1 

664.3  214.07

1999.1  930.98

 0.578 157.8%2

h  1 



 1 

 u

 h

 p

 10.1 MPa2 a

887.7 K

300 K

b 0.3 MPa



13.972 35.73



 r

 21898.32 1796.6 K



 p

 10.12 a

898.3

300

b 1182 5.39 MPa



621.2

 34.51

❶

9.4 Air-Standard Dual Cycle 385

The pressure–volume diagrams of actual internal combustion engines are not described

well by the Otto and Diesel cycles. An air-standard cycle that can be made to approximate

the pressure variations more closely is the air-standard dual cycle. The dual cycle is shown

in Fig. 9.7. As in the Otto and Diesel cycles, Process 1–2 is an isentropic compression.

The heat addition occurs in two steps, however: Process 2–3 is a constant-volume heat ad-

dition; Process 3–4 is a constant-pressure heat addition. Process 3–4 also makes up the

first part of the power stroke. The isentropic expansion from state 4 to state 5 is the re-

mainder of the power stroke. As in the Otto and Diesel cycles, the cycle is completed by

a constant-volume heat rejection process, Process 5–1. Areas on the T–s and p–v diagrams

can be interpreted as heat and work, respectively, as in the cases of the Otto and Diesel

cycles.

The net work of the cycle equals the net heat added

The specific volume at state 1 is

Inserting values

This solution uses the air tables, which account explicitly for the variation of the specific heats with temperature. Note

that Eq. 9.13 based on the assumption of constant specific heats has not been used to determine the thermal efficiency.

The cold air-standard solution of this example is left as an exercise.

 0.76 MPa

mep 

617.9 kJ/kg

0.86111  1



182 m

/kg

1 kJ

1 MPa

N/m





M2T



8314

28.97

b 1300 K2

N/m

 0.861 m

/kg

 617.9 kJ/kg

 11999.1  930.982 1664.3  214.072

cycle





 1h

 h

2 1u

 u



9.4 Air-Standard Dual Cycle

= c

p = c

= c

s = c

s =

 Figure 9.7

p–v and T–s diagrams of the air-standard dual cycle.

dual cycle

❶