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

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

added

h 5

cycle

5 1 2

2 u

2 h

(9.11)

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

compression ratio.

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

, u

, h

, and

or 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 y

data

To find T

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

5 p

to give

5 r

where r

5 V

, called the cutoff ratio, has been introduced.

Since V

5 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 y

at T

, the temperature T

can be determined by

interpolation once y

is found from the isentropic relationship

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

provided by

5 a

k21

5 r

k21





1constant k2

The temperature T

is found similarly from

5 a

k21

5 a

k21





1constant k2

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

EFFECT OF COMPRESSION RATIO ON PERFORMANCE. As for the

Otto cycle, the thermal 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 thermal efficiency of the Diesel cycle

can be expressed as

h 5 1 2

k21

2 1

k1r

2 12





1cold-air standard basis2

(9.13)

where r is the compression ratio and r

is the cutoff ratio. The derivation is left

as an exercise. This relationship is shown in Fig. 9.6 for k 5 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

cutoff ratio

9.3 Air-Standard Diesel Cycle 503

Fig. 9.6

Thermal efficiency of the

cold air-standard Diesel cycle, k 5 1.4.

(

)

2015105

Compression ratio, r

Thermal efficiency, η (%)

= 2

Diesel

cycle

= 3

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504 Chapter 9 Gas Power Systems

ratio is the same, the thermal efficiency of the cold air-standard Diesel cycle is less

than that of the cold air-standard Otto cycle.

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

Analyzing the Diesel Cycle

c c c c EXAMPLE 9.2 c

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

of 18, the temperature 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 effec-

tive 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 effec-

tive pressure.

Schematic and Given Data:

Analysis:

(a)

The analysis begins by determining properties at each principal state of the cycle. With T

5 300 K, Table

A-22 gives u

5 214.07 kJ/kg and y

5 621.2. For the isentropic compression process 1–2

621.

5 34.51

Interpolating in Table A-22, we get T

5 898.3 K and h

5 930.98 kJ/kg. With the ideal gas equation of state

5 p

5 10.12a

898.3

300

b11825 5.39 MPa

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

5 p

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

Introducing the cutoff ratio, r

5 V

5 r

5 2

898.3

5 1796.6 K

Engineering Model:

The air in the piston–cylinder

assembly is the closed system.

2. The compression and expan-

sion 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.

Fig. E9.2

p = c

= c

s = c

= 0.1 MPa

= 300 K

= 18

––

= 2r

––

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If the mass of air is 0.0123 kg, what is the displacement vol-

ume, in L? Ans. 10 L.

From Table A-22, h

5 1999.1 kJ/kg and y

5 3.97.

For the isentropic expansion process 3–4

Introducing V

5 V

, the compression ratio r, and the cutoff ratio r

, we have

13.9725 35.73

Interpolating in Table A-22 with y

, we get u

5 664.3 kJ/kg and T

5 887.7 K. 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 V

5 V

, the ideal gas equation of state gives

5 p

5 10.1 MPa2a

887.7 K

300 K

b5 0.3 MPa

(b) The thermal efficiency is found using

h 5 1 2

5 1 2

2 u

2 h

5 1 2

664.3 2 214.07

1999.1 2 930.98

5 0.578 157.8%2

mep 5

cycle

2 y

cycle

11 2 1

The net work of the cycle equals the net heat added

cycle

5 1h

2 h

22 1u

2 u

5 11999.1 2 930.9822 1664.3 2 214.072

5 617.9 kJ

The specific volume at state 1 is

M2T

831

28.97

N ? m

kg ? K

b1300 K2

5 0.861 m

Inserting values

mep 5

617.9 kJ

0.86111 2 1

182 m

N ? m

1 kJ

1 MPa

5 0.76 MPa

➊ 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.

➊

Ability to…

❑

sketch the Diesel cycle p–y

and T–s diagrams.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

calculate the thermal

efficiency and mean effective

pressure.

✓

Skills Developed

9.3 Air-Standard Diesel Cycle 505

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506 Chapter 9 Gas Power Systems

dual cycle

9.4 Air-Standard Dual Cycle

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 approx-

imate 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 addition; 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 remainder 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–y diagrams can be interpreted as heat and work, respectively, as in

the cases of the Otto and Diesel cycles.

Cycle Analysis

Since the dual cycle is composed of the same types of processes as the Otto and

Diesel cycles, we can simply write down the appropriate work and heat transfer

expressions by reference to the corresponding earlier developments. Thus, during the

isentropic compression process 1–2 there is no heat transfer, and the work is

5 u

2 u

As for the corresponding process of the Otto cycle, in the constant-volume portion

of the heat addition process, Process 2–3, there is no work, and the heat transfer is

5 u

2 u

In the constant-pressure portion of the heat addition process, Process 3–4, there is

both work and heat transfer, as for the corresponding process of the Diesel cycle

5 p1y

2 y



and





5 h

2 h

During the isentropic expansion process 4–5 there is no heat transfer, and the work is

5 u

2 u

Finally, the constant-volume heat rejection process 5–1 that completes the cycle

involves heat transfer but no work

5 u

2 u

Fig. 9.7 p–y and T–s diagrams of the air-standard dual cycle.

v = c

p = c

v = c

s = c

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

added

h 5

cycle

m 1 Q

5 1 2

m 1 Q

5 1 2

2 u

21 1h

2 h

(9.14)

The example to follow provides an illustration of the analysis of an air-standard

dual cycle. The analysis exhibits many of the features found in the Otto and Diesel

cycle examples considered previously.

9.4 Air-Standard Dual Cycle 507

c c c c EXAMPLE 9.3 c

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

temperature is 300 K and the pressure is 0.1 MPa. The pressure ratio for the constant volume part of the heating

process is 1.5:1. The volume ratio for the constant pressure part of the heating process is 1.2:1. Determine (a) the

thermal efficiency and (b) the mean effective pressure, in MPa.

SOLUTION

Known:

An air-standard dual cycle is executed in a piston–cylinder assembly. Conditions are known at the begin-

ning of the compression process, and necessary volume and pressure ratios are specified.

Find: Determine the thermal efficiency and the mep, in MPa.

Schematic and Given Data:

Analyzing the Dual Cycle

Analysis: The analysis begins by determining properties at each principal state of the cycle. States 1 and 2 are

the same as in Example 9.2, so u

5 214.07 kJ/kg, T

5 898.3 K, u

5 673.2 kJ/kg. Since Process 2–3 occurs at

constant volume, the ideal gas equation of state reduces to give

5 11.521898.325 1347.5 K

Interpolating in Table A-22, we get h

5 1452.6 kJ/kg and u

5 1065.8 kJ/kg.

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.

Fig. E9.3

p = c

= c

s = c

= 0.1 MPa

= 300 K

= 18

––

= 1.2

––

= 1.5,

––

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508 Chapter 9 Gas Power Systems

Evaluate the total heat addition and the net work of the cycle,

each in kJ per kg of air.

Ans. Q

/m 5 718 kJ/kg, W

cycle

/m 5 456 kJ/kg.

Since Process 3–4 occurs at constant pressure, the ideal gas equation of state reduces to give

5 11.2211347.525 1617 K

From Table A-22, h

5 1778.3 kJ/kg and y

5 5.609.

Process 4–5 is an isentropic expansion, so

5 y

The volume ratio V

required by this equation can be expressed as

With V

5 V

, V

5 V

, and given volume ratios

5 18 a

1.2

b5 15

Inserting this in the above expression for y

5 15.609211525 84.135

Interpolating in Table A-22, we get u

5 475.96 kJ/kg.

(a) The thermal efficiency is

h 5 1 2

m 1 Q

5 1 2

2 u

21 1h

2 h

5 1 2

1475.96 2 214.07

11065.8 2 673.221 11778.3 2 1452.62

5 0.635163.5%2

(b) The mean effective pressure is

mep 5

cycle

2 y

cycle

1 2 1

The net work of the cycle equals the net heat added, so

mep 5

2 u

21 1h

2 h

22 1u

2 u

11 2 1

The specific volume at state 1 is evaluated in Example 9.2 as y

5 0.861 m

/kg. Inserting values into the above

expression for mep

mep 5

311065.8 2 673.221 11778.3 2 1452.622 1475.96 2 214.0724a

N ? m

1 kJ

1 MPa

0.86111 2 1

182 m

5 0.56 MPa

Ability to…

❑

sketch the Dual cycle p–y

and T–s diagrams.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

calculate the thermal

efficiency and mean effective

pressure.

✓

Skills Developed

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Considering Gas Turbine

Power Plants

This part of the chapter deals with gas turbine power plants. Gas turbines tend to be

lighter and more compact than the vapor power plants studied in Chap. 8. The favor-

able power-output-to-weight ratio of gas turbines makes them well suited for trans-

portation applications (aircraft propulsion, marine power plants, and so on). In recent

decades, gas turbines also have contributed an increasing share of U.S. electric power

needs, and they now provide about 22% of the total (see Table 8.1).

Today’s electric power-producing gas turbines are almost exclusively fueled by

natural gas. However, depending on the application, other fuels can be used by gas

turbines, including distillate fuel oil; propane; gases produced from landfills, sewage

treatment plants, and animal waste (see BIOCONNECTIONS, Sec. 7.3); and syngas

(synthesis gas) obtained by gasification of coal (see Sec. 9.10).

Diesel-powered vehicles are likely to be more prev-

alent in the United States in coming years. Diesel

engines are more powerful and about one-third more fuel-

efficient than similar-sized gasoline engines currently dominating

the U.S. market. Increased fuel efficiency is achieved by diesels

through advanced engine control and fuel-injection technology.

Diesel-powered vehicles must meet the same emissions standards

as gasoline vehicles. Ultra-low sulfur diesel fuel, commonly avail-

able today at U.S. fuel pumps, and improved exhaust treatment

make this possible.

Biodiesel also can be used as fuel. Biodiesel is domestically pro-

duced from nonpetroleum, renewable sources, including vegetable

oils (from soybeans, rapeseeds, sunflower seeds, and jatropha

seeds), animal fats, and algae. Waste vegetable oil from industrial

deep fryers, snack food factories, and restaurants can be converted to

biodiesel. Biodiesel is biodegradable and safer to handle than petro-

leum diesel. The news about biodiesel is not all good, however. When

compared to petroleum diesel, some say biodiesel is more expensive,

has more nitrogen oxide (NO

) emissions, and may have greater

impact on engine durability.

More Diesel-Powered Vehicles on the Way?

9.5 Modeling Gas Turbine Power Plants

Gas turbine power plants may operate on either an open or closed basis. The open

mode pictured in Fig. 9.8a is more common. This is an engine in which atmospheric

air is continuously drawn into the compressor, where it is compressed to a high

9.5 Modeling Gas Turbine Power Plants 509

Fig. 9.8

Simple gas turbine. (a) Open to the atmosphere. (b) Closed.

Heat exchanger

Net

work

out

TurbineCompressor

out

Combustion

chamber

Fuel

Net

work

out

TurbineCompressor

Products

Air

(b)(a)

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510 Chapter 9

Gas Power Systems

pressure. The air then enters a combustion chamber, or combustor, where it is mixed

with fuel and combustion occurs, resulting in combustion products at an elevated

temperature. The combustion products expand through the turbine and are subse-

quently discharged to the surroundings. Part of the turbine work developed is used

to drive the compressor; the remainder is available to generate electricity, to propel

a vehicle, or for other purposes.

In the closed mode pictured in Fig. 9.8b, the working fluid receives an energy input

by heat transfer from an external source, for example a gas-cooled nuclear reactor.

The gas exiting the turbine is passed through a heat exchanger, where it is cooled

prior to reentering the compressor.

An idealization often used in the study of open gas turbine power plants is that

of an air-standard analysis. In an air-standard analysis two assumptions are always

made:

c The working fluid is air, which behaves as an ideal gas.

c The temperature rise that would be brought about by combustion is accomplished

by a heat transfer from an external source.

With an air-standard analysis, we avoid dealing with the complexities of the com-

bustion process and the change of composition during combustion. Accordingly, an

air-standard analysis simplifies study of gas turbine power plants considerably, but

numerical values calculated on this basis may provide only qualitative indications of

power plant performance. Still, we can learn important aspects of gas turbine opera-

tion using an air-standard analysis; see Sec. 9.6 for further discussion supported by

solved examples.

air-standard analysis:

gas turbines

ENERGY & ENVIRONMENT Natural gas is widely used for power

generation by gas turbines, industrial and home heating, and chemical processing.

The versatility of natural gas is matched by its relative abundance in North

America, including natural gas extracted from deep-water ocean sites and shale deposits.

Pipeline delivery of natural gas across the nation and from Canada has occurred for decades.

Importation by ship from countries such as Trinidad, Algeria, and Norway is a more recent

development.

Turning natural gas into liquid is the only practical way to import supplies from overseas.

The liquefied natural gas (LNG) is stored in tanks onboard ships at about 21638C (22608F). To

reduce heat transfer to the cargo LNG from outside sources, the tanks are insulated and the

ships have double hulls with ample space between them. Still, a fraction of the cargo evaporates

during long voyages. Such boil-off gas is commonly used to fuel the ship’s propulsion system

and meet other onboard energy needs. When tankers arrive at their destinations, LNG is con-

verted to gas by heating it. The gas is then sent via pipeline to storage tanks onshore for dis-

tribution to customers.

Shipboard delivery of LNG has some minuses. Owing to cumulative effects during the LNG

delivery chain, considerable exergy is destroyed and lost in liquefying gas at the beginning of the

chain, transporting LNG by ship, and regasifying it when port is reached. If comparatively warm

seawater is used to regasify LNG, environmentalists worry about the effect on aquatic life nearby.

Many observers are also concerned about safety, especially when huge quantities of gas are

stored at ports in major urban areas. Some say it would be better to use domestic supplies more

efficiently than run such risks.

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9.6 Air-Standard Brayton Cycle

A schematic diagram of an air-standard gas turbine is shown in Fig. 9.9. The direc-

tions of the principal energy transfers are indicated on this figure by arrows. In

accordance with the assumptions of an air-standard analysis, the temperature rise

that would be achieved in the combustion process is brought about by a heat trans-

fer to the working fluid from an external source and the working fluid is considered

to be air as an ideal gas. With the air-standard idealizations, air would be drawn into

the compressor at state 1 from the surroundings and later returned to the surround-

ings at state 4 with a temperature greater than the ambient temperature. After inter-

acting with the surroundings, each unit mass of discharged air would eventually

return to the same state as the air entering the compressor, so we may think of the

air passing through the components of the gas turbine as undergoing a thermody-

namic cycle. A simplified representation of the states visited by the air in such a

cycle can be devised by regarding the turbine exhaust air as restored to the compres-

sor inlet state by passing through a heat exchanger where heat rejection to the sur-

roundings occurs. The cycle that results with this further idealization is called the

air-standard

Brayton cycle.

9.6.1

Evaluating Principal Work and Heat Transfers

The following expressions for the work and heat transfers of energy that occur at

steady state are readily derived by reduction of the control volume mass and energy

rate balances. These energy transfers are positive in the directions of the arrows in

Fig. 9.9. Assuming the turbine operates adiabatically and with negligible effects of

kinetic and potential energy, the work developed per unit of mass flowing is

5 h

2 h

(9.15)

where m

denotes the mass flow rate. With the same assumptions, the compressor work

per unit of mass flowing is

5 h

2 h

(9.16)

Brayton cycle

9.6 Air-Standard Brayton Cycle 511

Fig. 9.9

Air-standard gas turbine

cycle.

Heat exchanger

TurbineCompressor

out

cycle

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512 Chapter 9 Gas Power Systems

The symbol W

denotes work input and takes on a positive value. The heat added to

the cycle per unit of mass is

5 h

2 h

(9.17)

The heat rejected per unit of mass is

out

5 h

2 h

(9.18)

where Q

out

is positive in value.

The thermal efficiency of the cycle in Fig. 9.9 is

h 5

2 W

2 h

22 1h

2 h

(9.19)

The back work ratio for the cycle is

bwr 5

2 h

(9.20)

For the same pressure rise, a gas turbine compressor would require a much greater

work input per unit of mass flow than the pump of a vapor power plant because the

average specific volume of the gas flowing through the compressor would be many

times greater than that of the liquid passing through the pump (see discussion of

Eq. 6.51b in Sec. 6.13). Hence, a relatively large portion of the work developed by

the turbine is required to drive the compressor. Typical back work ratios of gas tur-

bines range from 40 to 80%. In comparison, the back work ratios of vapor power

plants are normally only 1 or 2%.

If the temperatures at the numbered states of the cycle are known, the specific

enthalpies required by the foregoing equations are readily obtained from the ideal

gas table for air, Table A-22 or Table A-22E. Alternatively, with the sacrifice of some

accuracy, the variation of the specific heats with temperature can be ignored and the

specific heats taken as constant. The air-standard analysis is then referred to as a cold

air-standard analysis. As illustrated by the discussion of internal combustion engines

given previously, the chief advantage of the assumption of constant specific heats is

that simple expressions for quantities such as thermal efficiency can be derived, and

these can be used to deduce qualitative indications of cycle performance without

involving tabular data.

Since Eqs. 9.15 through 9.20 have been developed from mass and energy rate bal-

ances, they apply equally when irreversibilities are present and in the absence of

irreversibilities. Although irreversibilities and losses associated with the various power

plant components have a pronounced effect on overall performance, it is instructive

to consider an idealized cycle in which they are assumed absent. Such a cycle estab-

lishes an upper limit on the performance of the air-standard Brayton cycle. This is

considered next.

9.6.2

Ideal Air-Standard Brayton Cycle

Ignoring irreversibilities as the air circulates through the various components of the

Brayton cycle, there are no frictional pressure drops, and the air flows at constant

pressure through the heat exchangers. If stray heat transfers to the surroundings

are also ignored, the processes through the turbine and compressor are isentropic.

The ideal cycle shown on the p–y and T–s diagrams in Fig. 9.10 adheres to these

idealizations.

back work ratio

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