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

Подождите немного. Документ загружается.

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

rate basis:

Determine the net rate energy is carried out of the plant as water

passes through the condenser, in MW, and comment. Ans. 30.12 MW. The

significance of this energy loss is far less than indicated by the answer. In

terms of exergy, the loss at the condenser is 1.41 MW [see part (b)], which

better measures the limited utility of the relatively low-temperature water

flowing through the condenser.

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-recovery steam generator 3.68 MW 6.2% (4.3%)

*Estimation based on fuel exergy. For discussion, see note 3.

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-recovery steam generator, as well as the relative

significance of the exergy losses. Finally, the table indicates that the total of the exergy destructions overshadows

the losses. While the energy analysis of part (a) yields valuable results about combined-cycle performance, the

exergy analysis of part (b) provides insights about the effects of irreversibilities and true magnitudes of the losses

that cannot be obtained using just energy.

➊ For comparison, note that the combined-cycle thermal efficiency in this case is much greater than those of

the stand-alone regenerative vapor and gas cycles considered in Examples 8.5 and 9.11, respectively.

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

pump, and heat-recovery steam generator is left as an exercise.

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

combustion 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. All other percentages in parentheses

are obtained by multiplying the corresponding percentages, 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 paren-

theses give the more accurate picture of combined cycle performance.

Ability to…

❑

apply mass and energy

balances.

❑

determine thermal efficiency.

❑

evaluate exergy quantities.

❑

develop an exergy accounting.

✓

Skills Developed

9.9 Gas Turbine–Based Combined Cycles 543

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544 Chapter 9

Gas Power Systems

9.9.2

Cogeneration

Cogeneration systems are integrated systems that yield two valuable products simul-

taneously from a single fuel input, electricity and steam (or hot water), achieving cost

savings. Cogeneration systems have numerous industrial and commercial applications.

District heating is one of these.

District heating plants are located within communities to provide steam or hot water

for space heating and other needs together with electricity for domestic, commercial,

and industrial use. Vapor cycle–based district heating plants are considered in Sec. 8.5.

Building on the combined gas turbine–vapor power cycle introduced in Sec. 9.9.1,

Fig. 9.23 illustrates a district heating system consisting of a gas turbine cycle partnered

with a vapor power cycle operating in the back-pressure mode discussed in Sec. 8.5.3.

In this embodiment, steam (or hot water) is provided from the condenser to service

the district heating load.

Referring again to Fig. 9.23, if the condenser is omitted, steam is supplied directly

from the steam turbine to service the district heating load; condensate is returned to

the heat-recovery steam generator. If the steam turbine is also omitted, steam passes

directly from the heat-recovery unit to the community and back again; power is gen-

erated by the gas turbine alone.

9.10

Integrated Gasification Combined-

Cycle Power Plants

For decades vapor power plants fueled by coal have been the workhorses of U.S.

electricity generation (see Chap. 8). However, human-health and environmental-

impact issues linked to coal combustion have placed this type of power generation under

a cloud. In light of our large coal reserves and the critical importance of electricity

to our society, major governmental and private-sector efforts are aimed at developing

Fig. 9.23

Combined-cycle district heating plant.

Combustor

TurbineCompressor

gas

vap

Vapor

cycle

Heat-recovery

steam generator

Condenser

Gas turbine

Air inlet

Exhaust

Pump

Turbine

Steam

Condensate

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alternative power generation technologies using coal, but with fewer adverse effects.

In this section, we consider one such technology: integrated gasification combined-

cycle (IGCC) power plants.

An IGCC power plant integrates a coal gasifier with a combined gas turbine–

vapor power plant like that considered in Sec. 9.9. Key elements of an IGCC plant

are shown in Fig. 9.24. Gasification is achieved through controlled combustion of

coal with oxygen in the presence of steam to produce syngas (synthesis gas) and

solid waste. Oxygen is provided to the gasifier by the companion air separation unit.

Syngas exiting the gasifier is mainly composed of carbon monoxide and hydrogen.

The syngas is cleaned of pollutants and then fired in the gas turbine combustor. The

performance of the combined cycle follows the discussion provided in Sec. 9.9.

In IGCC plants, pollutants (sulfur compounds, mercury, and particulates) are removed

before combustion when it is more effective to do so, rather than after combustion as

in conventional coal-fueled power plants. While IGCC plants emit fewer sulfur dioxide,

nitric oxide, mercury, and particulate emissions than comparable conventional coal

plants, abundant solid waste is still produced that must be responsibly managed.

Taking a closer look at Fig. 9.24, better IGCC plant performance can be realized

through tighter integration between the air-separation unit and combined cycle. For

instance, by providing compressed air from the gas turbine compressor to the air-

separation unit, the compressor feeding ambient air to the air-separation unit can be

eliminated or reduced in size. Also, by injecting nitrogen produced by the separation

unit into the air stream entering the combustor, mass flow rate through the turbine

increases and therefore greater power is developed.

Only a few IGCC plants have been constructed worldwide thus far. Accordingly, only

time will tell if this technology will make significant inroads against coal-fired vapor

power plants, including the newest generation of supercritical plants. Proponents point

to increased combined cycle thermal efficiency as a way to extend the viability of U.S.

coal reserves. Others say investment might be better directed to technologies fostering

9.10 Integrated Gasification Combined-Cycle Power Plants 545

Fig. 9.24 Integrated gasification combined-cycle power plant.

Combustor

Syngas cleanup

Syngas

Air separation unit

TurbineCompressor

Compressor

gas

vap

Vapor

cycle

Heat-recovery

steam generator

Condenser

Gas turbine

Air inlet

Air

Exhaust

Pump

Turbine

Cooling

water

Coal

Solid

waste

Makeup water

Mercury

Particulate

removal

Gasifier

Sulfur

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546 Chapter 9 Gas Power Systems

use of renewable sources of energy for power generation than to technologies fostering

use of coal, which has so many adverse effects related to its utilization.

9.11

Gas Turbines for Aircraft Propulsion

Gas turbines are particularly suited for aircraft propulsion because of their favorable

power-to-weight ratios. The turbojet engine is commonly used for this purpose. As

illustrated in Fig. 9.25a, this type of engine consists of three main sections: the diffuser,

the gas generator, and the nozzle.

The diffuser placed before the compressor decelerates the incoming air relative to

the engine. A pressure rise known as the ram effect is associated with this deceleration.

The gas generator section consists of a compressor, combustor, and turbine, with the

same functions as the corresponding components of a stationary gas turbine power

plant. In a turbojet engine, the turbine power output need only be sufficient to drive

the compressor and auxiliary equipment, however.

Combustion gases leave the turbine at a pressure significantly greater than atmo-

spheric and expand through the nozzle to a high velocity before being discharged to

the surroundings. The overall change in the velocity of the gases relative to the engine

gives rise to the propulsive force, or thrust.

Some turbojets are equipped with an afterburner, as shown in Fig. 9.26. This is

essentially a reheat device in which additional fuel is injected into the gas exiting the

turbine and burned, producing a higher temperature at the nozzle inlet than would

be achieved otherwise. As a consequence, a greater nozzle exit velocity is attained,

resulting in increased thrust.

TURBOJET ANALYSIS. The T–s diagram of the processes in an ideal turbojet engine

is shown in Fig. 9.25b. In accordance with the assumptions of an air-standard analysis, the

working fluid is air modeled as an ideal gas. The diffuser, compressor, turbine, and nozzle

processes are isentropic, and the combustor operates at constant pressure.

turbojet engine

ram effect

afterburner

Fig. 9.25 Turbojet engine

schematic and accompanying

ideal T–s diagram.

p = c

Compressor

Diffuser NozzleGas generator

(a)(b)

Combustors Turbine

Product

gases out

Air

a1234

Fig. 9.26 Schematic of a

turbojet engine with

afterburner.

Compressor

Diffuser Gas generator Afterburner duct Adjustable

nozzle

Combustors

Turbine

Fuel-spray bars

Air

Flame holder

Product

gases out

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c Process a–1 shows the pressure rise that occurs in the diffuser as the air decelerates

isentropically through this component.

c Process 1–2 is an isentropic compression.

c Process 2–3 is a constant-pressure heat addition.

c Process 3–4 is an isentropic expansion through the turbine during which work is

developed.

c Process 4–5 is an isentropic expansion through the nozzle in which the air acceler-

ates and the pressure decreases.

Owing to irreversibilities in an actual engine, there would be increases in specific

entropy across the diffuser, compressor, turbine, and nozzle. In addition, there would

be a combustion irreversibility and a pressure drop through the combustor of the

actual engine. Further details regarding flow through nozzles and diffusers are pro-

vided in Secs. 9.13 and 9.14. The subject of combustion is discussed in Chap. 13.

In a typical thermodynamic analysis of a turbojet on an air-standard basis, the

following quantities might be known: the velocity at the diffuser inlet, the compressor

pressure ratio, and the turbine inlet temperature. The objective of the analysis might

then be to determine the velocity at the nozzle exit. Once the nozzle exit velocity is

known, the thrust can be determined by applying Newton’s second law of motion in

a form suitable for a control volume (Sec. 9.12). All principles required for the ther-

modynamic analysis of turbojet engines on an air-standard basis have been intro-

duced. Example 9.13 provides an illustration.

9.11 Gas Turbines for Aircraft Propulsion 547

Analyzing a Turbojet Engine

c c c c EXAMPLE 9.13 c

Air enters a turbojet engine at 11.8 lbf/in.

, 4308R, and an inlet velocity of 620 miles/h (909.3 ft/s). The pressure

ratio across the compressor is 8. The turbine inlet temperature is 21508R and the pressure at the nozzle exit is

11.8 lbf/in.

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 except at the inlet and exit of the engine, and neglect 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:

Fig. E9.13

= 430°R

= 11.8 lbf/in.

= 620 mi/hr

= 11.8 lbf/in.

= 2150°R

= 430°R

= 0

TurbineCompressor

Diffuser

Nozzle

Combustor

= 8

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548 Chapter 9

Gas Power Systems

Engineering Model:

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 enclosing this component reduce at steady state to give

0 5 Q

2 W

1 m

c1h

2 h

21 a

2 V

b1 g1z

where m

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

5 221h

2 h

This expression requires values for the specific enthalpies h

and h

at the nozzle inlet and exit, respectively. With

the operating 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

5 h

With h

from Table A-22E and the given value of V

➊

5 102.7 Btu

lb 1 c

1909.32

1 lbf

32.2 lb ? ft

1 Btu

778 ft ? lbf

5 119.2 Btu

Interpolating in Table A-22E gives p

5 1.051. The flow through the diffuser is isentropic, so pressure p

With p

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

1.051

0.6268

111.8 lbf

in.

25 19.79 lbf

in.

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

5 8(19.79 lbf/in.

) 5 158.3 lbf/in.

The flow through the compressor is also isentropic. Thus

5 p

5 1.0511825 8.408

Interpolating in Table A-22E, we get h

5 216.2 Btu/lb.

At state 3 the temperature is given as T

5 21508R. From Table A-22E, h

5 546.54 Btu/lb. By assump-

tion 3, p

5 p

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

That is

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OTHER APPLICATIONS. Other related applications of the gas turbine

include turboprop and turbofan engines. The turboprop engine shown in Fig. 9.27a

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 propel-

ler, which provides thrust to the aircraft. Turboprops are able to achieve speeds

up to about 850 km/h (530 miles/h). In the turbofan shown in Fig. 9.27b, the core

of the engine is much like a turbojet, and some thrust is obtained from expansion

2 h

5 h

2 h

Solving for h

5 h

1 h

2 h

5 546.54 1 119.2 2 216.2

5 449.5 Btu

Interpolating in Table A-22E with h

gives p

5 113.8.

The expansion through the turbine is isentropic, so

5 p

With p

5 p

and p

data from Table A-22E

5 1158.3 lbf

in.

113.8

233.5

5 77.2 lbf

in.

The expansion through the nozzle is isentropic to p

5 11.8 lbf/in.

Thus

5 p

5 1113.82

11.8

77.2

5 17.39

From Table A-22E, h

5 265.8 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

21h

2 h

21449.5 2 265.82

Btu

32.2 lb ? ft

1 lbf

778 ft ? lbf

1 Btu

5 3034 ft

s 12069 mi

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

➋ 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 suit-

able for control volumes (see Sec. 9.12.1).

Using Eq. 6.47, the isentropic nozzle efficiency, what is the

nozzle exit velocity, in ft/s, if the efficiency is 90%? Ans. 2878 ft/s.

➋

Ability to…

❑

sketch the schematic of the

turbojet engine and the T–s

diagram for the correspond-

ing air-standard cycle.

❑

evaluate temperatures and

pressures at each principal

state and retrieve necessary

property data.

❑

apply mass, energy, and

entropy principles.

❑

calculate the nozzle exit

velocity.

✓

Skills Developed

9.11 Gas Turbines for Aircraft Propulsion 549

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550 Chapter 9 Gas Power Systems

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 additional

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 ramjet is shown in Fig. 9.27c. This engine requires neither a

compressor nor a turbine. A sufficient pressure rise is obtained by decelerating

the high-speed incoming air in the diffuser (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.

Considering Compressible

Flow Through Nozzles

and Diffusers

In many applications of engineering interest, gases move at relatively high veloc-

ities and exhibit appreciable changes in specific volume (density). The flows

through the nozzles and diffusers of jet engines discussed in Sec. 9.11 are impor-

tant examples. Other examples are the flows through wind tunnels, shock tubes,

and steam ejectors. These flows are known as compressible flows. In this part of

the chapter, we introduce some of the principles involved in analyzing compress-

ible flows.

compressible flow

Fig. 9.27 Other examples of aircraft engines. (a) Turboprop. (b) Turbofan. (c) Ramjet.

(a)(b)

(c)

Propeller

Fan

By-pass flow

Basic engine-core

Diffuser

Flame holder

Nozzle

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9.12 Compressible Flow Preliminaries

Concepts introduced in this section play important roles in the study of compressible

flows. The momentum equation is introduced in a form applicable to the analysis of

control volumes at steady state. The velocity of sound is also defined, and the concepts

of Mach number and stagnation state are discussed.

9.12.1

Momentum Equation for Steady One-Dimensional Flow

The analysis of compressible flows requires the principles of conservation of mass

and energy, the second law of thermodynamics, and relations among the thermody-

namic properties of the flowing gas. In addition, Newton’s second law of motion is

required. Application of Newton’s second law of motion to systems of fixed mass

(closed systems) involves the familiar form

where F is the resultant force acting on a system of mass m and a is the acceleration.

The object of the present discussion is to introduce Newton’s second law of motion

in a form appropriate for the study of the control volumes considered in subsequent

discussions.

Consider the control volume shown in Fig. 9.28, which has a single inlet, designated

by 1, and a single exit, designated by 2. The flow is assumed to be one-dimensional

at these locations. The energy and entropy rate equations for such a control volume

have terms that account for energy and entropy transfers, respectively, at the inlets

and exits. Momentum also can be carried into or out of the control volume at the

inlets and exits, and such transfers can be accounted for as

≥

time rate of momentum

transfer into or

out of a control volume

accompan

mass flow

¥5 m

(9.30)

In this expression, the momentum per unit of mass flowing across the boundary of

the control volume is given by the velocity vector V. In accordance with the one-

dimensional flow model, the vector is normal to the inlet or exit and oriented in the

direction of flow.

In words, Newton’s second law of motion for control volumes is

time rate of change

of momentum contained

within the control volume

resultant force

acting on the

control volume

1 £

net rate at which momentum is

transferred into the control

volume accompan

mass flow

9.12 Compressible Flow Preliminaries 551

Fig. 9.28 One-inlet, one-exit

control volume at steady state

labeled with momentum transfers

accompanying mass flow.

No flow through this

portion of the boundary

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552 Chapter 9

Gas Power Systems

steady-state

momentum equation

At steady state, the total amount of momentum contained in the control volume is

constant with time. Accordingly, when applying Newton’s second law of motion to

control volumes at steady state, it is necessary to consider only the momentum accom-

panying the incoming and outgoing streams of matter and the forces acting on the

control volume. Newton’s law then states that the resultant force F acting on the

control volume equals the difference between the rates of momentum exiting and

entering the control volume accompanying mass flow. This is expressed by the fol-

lowing steady-state form of the momentum equation

F 5 m

2 m

5 m

2 V

(9.31)

Since m

5 m

at steady state, the common mass flow is designated in this expression

simply as m

. The expression of Newton’s second law of motion given by Eq. 9.31

suffices for subsequent discussions. More general control volume formulations are

normally provided in fluid mechanics texts.

9.12.2

Velocity of Sound and Mach Number

A sound wave is a small pressure disturbance that propagates through a gas, liquid,

or solid at a velocity c that depends on the properties of the medium. In this section

we obtain an expression that relates the velocity of sound, or sonic velocity, to other

properties. The velocity of sound is an important property in the study of compress-

ible flows.

MODELING PRESSURE WAVES. Let us begin by referring to Fig. 9.29a, which

shows a pressure wave moving to the right with a velocity of magnitude c. The wave

is generated by a small displacement of the piston. As shown on the figure, the pres-

sure, density, and temperature in the region to the left of the wave depart from the

respective values of the undisturbed fluid to the right of the wave, which are desig-

nated simply p, r, and T. After the wave has passed, the fluid to its left is in steady

motion with a velocity of magnitude DV.

Figure 9.29a shows the wave from the point of view of a stationary observer. It is

easier to analyze this situation from the point of view of an observer at rest relative

to the wave, as shown in Fig. 9.29b. By adopting this viewpoint, a steady-state analysis

can be applied to the control volume identified on the figure. To an observer at rest

relative to the wave, it appears as though the fluid is moving toward the stationary

wave from the right with velocity c, pressure p, density r, and temperature T and

TAKE NOTE...

The resultant force F

includes the forces due to

pressure acting at the inlet

and exit, forces acting on

the portion of the boundary

through which there is no

mass flow, and the force of

gravity.

Fig. 9.29 Illustrations used to analyze the propagation of a sound wave. (a) Propagation of a

pressure wave through a quiescent fluid relative to a stationary observer. (b) Observer at rest

relative to the wave.

p, T, ρ

(a)(b)

p, ρ, T

ΔV

p + Δp

ρ + Δρ

T + ΔT

c – ΔV

p + Δp

ρ + Δρ

T + ΔT

V = 0

Piston

Stationary

observer

Control volume for an

observer moving with

the wave

Observer on

wave

Undisturbed fluid

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