chapter 9 gas power cycles. 9-1 basic considerations in the analysis of power cycles air standard...

CHAPTER

9

Gas Power Cycles

9-1 Basic Considerations in the Analysis of Power Cycles

Air Standard Cycles

The Spark Ignition Engine

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-1Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy.

• Reciprocating internal combustion engines.

• Reciprocating compressors

• Reciprocating steam engines

General classes of engines

Double action reciprocating engine.Double action reciprocating engine.

Single vs. double action engines

Single action reciprocating engineSingle action reciprocating engine.

Indicator diagrams

An analog instrument that measures pressure vs. An analog instrument that measures pressure vs. Displacement in a reciprocating engine. The graph Displacement in a reciprocating engine. The graph is in terms of P and V.is in terms of P and V.

pp

VV

Area is proportional to the work done per cycle

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

FIGURE 9-2The analysis of many complex processes can be reduced to a manageable level by utilizing some idealizations.


FIGURE 9-6P-v and T-s diagrams of a Carnot cycle.

9-2 The Caront Cycle and Its Value in Engineering


FIGURE 9-7A steady-flow Carnot engine.


FIGURE 9-8T-s diagram for Example 9–1.

9-3 Air Standard Assumptions

• The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas.

• All the processes that make up the cycle are internally reversible.

• The combustion process is replaced by a heat-addition process from an external source.

• The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state.

9-4 An Overview of Reciprocating Engines

Air Standard Cycles

The Spark Ignition Engine


FIGURE 9-10Nomenclature for reciprocating engines.


FIGURE 9-11Displacement and clearance volumes of a reciprocating engine.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-12The net work output of a cycle is equivalent to the product of the mean effective pressure and the displacement volume.

Indicator Diagrams

Work per cycle is represented in terms of a Work per cycle is represented in terms of a mean effective pressuremean effective pressure and the displacement. and the displacement.

pp

VV

MEP = Mean effective pressureMEP = Mean effective pressure

X = DisplacementX = Displacement

Indicator DiagramsTo compute the work per cycle, use the MEP and To compute the work per cycle, use the MEP and the displacement.the displacement.

Work = MEP x A x Displacement.Work = MEP x A x Displacement.

MEP = kMEP = kiiY, where kY, where kii is a constant is a constant

Y = the average ordinate on the indicator diagram.Y = the average ordinate on the indicator diagram.X = displacementX = displacementA = area of cylinder heat.A = area of cylinder heat.

YAkWork i

Ideal indicator diagramSpark ignition engine - The Otto cycle

Processes:a-b Intakeb-c Compressionc-d Combustion (spark ignited)d-e Power Strokee-f Gas Exhaustb-a Exhaust Stroke

a b

c

d

e

pp

VV

Clearance volumeClearance volume

Displacement


FIGURE 9-13Actual and ideal cycles in spark-ignition engines and their P-v diagrams.


FIGURE 9-14Schematic of a two-stroke reciprocating engine.

The spark ignition engine - The Otto cycle

The Otto cycle

Displacement

The air standard Otto The air standard Otto cycle approximates the cycle approximates the automotive internal automotive internal combustion engine and combustion engine and aircraft engines.aircraft engines.Assume a quasistatic Assume a quasistatic process, isentropic process, isentropic compression, and compression, and isentropic exhaust.isentropic exhaust.

The air standard Otto The air standard Otto cycle approximates the cycle approximates the automotive internal automotive internal combustion engine and combustion engine and aircraft engines.aircraft engines.Assume a quasistatic Assume a quasistatic process, isentropic process, isentropic compression, and compression, and isentropic exhaust.isentropic exhaust.

pp

VV

1 6

3

4

5

s = Constant

2

The Otto cycle

This is the T-s diagram forthe fixed mass system.

ssV = Constant

TT

6

3

4

5

2

pp

VV

1 6

3

4

5

P-V diagram for the actual variable masssystem.

s = Constant

2

Real Process

• 1-2 Intake of fuel/air mixture at P1

• 2-3 Compression to P3

• 3-4 Combustion, V = cons. and increasing P

• 4-5 Expansion

• 5-6 Exhaust at constant V and falling pressure

• 6-1Exhaust at P1

Real Process

• 1-2 Intake of fuel/air mixture at P1

• 2-3 Compression to P3

• 3-4 Combustion, V = cons. and increasing P

• 4-5 Expansion

• 5-6 Exhaust at constant V and falling pressure

• 6-1Exhaust at P1

Air Cycle Approximation

(Constant mass, closed internally reversible, cycle)

• 2-3 Isentropic compression

• 3-4 Heat addition at constant volume

• 4-5 Isentropic expansion

• 5-6 Heat rejection at constant volume

Air Cycle Approximation

(Constant mass, closed internally reversible, cycle)

• 2-3 Isentropic compression

• 3-4 Heat addition at constant volume

• 4-5 Isentropic expansion

• 5-6 Heat rejection at constant volume

The air-standard Otto cycle

The reversible Otto cycle

• Internally reversible processes• Constant specific heats

– k = Cp/Cv = Constant

• Polytropic compression and expansion, PVk = Const.• Treat air as an ideal gas.

Thermodynamic analysis for the reversible Otto cycle

TT

6

3

4

5

2

V = Constantss

QH

QC

34

65

65

34

1

)(

)(

TT

TT

Q

QQ

TTCQ

TTCQ

H

CH

vL

vH

(1)

Thermodynamic analysisfor the air-standard Otto Cycle

6

5

3

4

5

665

3

443

43

43

6543

ln,ln

T

T

T

T

T

TCs

T

TCs

T

dtC

T

Qs

ss

vv

v

path

rev

(2)

Thermodynamic Analysisfor the air-standard Otto Cycle

1

4

5

5

4

1

3

2

2

332

6

5

3

4

k

k

v

v

T

T

v

v

T

Tss

T

T

T

T

kv

k

vr

v

vr

v

v

T

T

T

TT

T

TT

T

T

T

T

1

3

6

1

6

3

3

6

6

65

3

34

6

5

3

4

1

11

Thermal efficiency

Thermal efficiency

kv v

rv

vr 1

3

6 1

Key assumptions: (1) Internally reversible processesKey assumptions: (1) Internally reversible processes (2) Constant specific heats (2) Constant specific heats

Important consequence: (1) Efficiency is independent ofImportant consequence: (1) Efficiency is independent of working fluidworking fluid (2) Efficiency independent of (2) Efficiency independent of temperaturestemperatures

rv

Efficiency of Air Standard Otto CycleEfficiency of Air Standard Otto Cyclek = 1.3 = Ck = 1.3 = Cpp/C/Cvv

TT

WNET3

44

55

2,6

ss

QQHH

QQCC

k = 1.3

k = 1.4

Air Standard Otto Cycle

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Pressure Ratio, Pv

Effici

ency

Series1Series2

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-18The thermal efficiency of the Otto cycle increases with the specific heat ratio k of the working fluid.

The Otto cycle with real gases

• Variable specific heats.• Efficiency based on internal energies obtained from the gas tables

which take into account variable specific heats.

),,(

1

2

34

65

TOT

Hv m

QTr

uu

uu

TT

3

4

5

2,6

V = Constantss

QH

QC

9-6 Diesel Cycle:

Air Standard Cycles

The Compression Ignition Engine

The compression ignition engine - theDiesel cycle

Air Standard CompressionIgnition Engine - The Diesel cycle

• An IC power cycle useful in many forms of automotive transportation, railroad engines, and ship power plants.

• Key assumptions are,– Constant specific heats, the ideal gas, and internally reversible processes.

Ideal Indicator DiagramCompression Ignition Engine

Processes:Processes:a-b Intakea-b Intakeb-c Compressionb-c Compressionc-d Combustionc-d Combustiond-e Power Stroked-e Power Strokee-f Gas Exhauste-f Gas Exhaustb-a Exhaust Strokeb-a Exhaust Stroke

Processes:Processes:a-b Intakea-b Intakeb-c Compressionb-c Compressionc-d Combustionc-d Combustiond-e Power Stroked-e Power Strokee-f Gas Exhauste-f Gas Exhaustb-a Exhaust Strokeb-a Exhaust Stroke

Displacement


a b

c d

e

pp

VV

Ideal Indicator DiagramCompression Ignition Engine

Processes:Processes:a-b a-b Combustion Combustion (P = Const.)(P = Const.)b-c b-c ExpansionExpansion

(s = Const)(s = Const)c-d c-d Exhaust Exhaust

(V = Const)(V = Const)d-e d-e Exhaust Exhaust

(P = Const)(P = Const)e-d e-d Intake Intake

(P = Const)(P = Const)d-a d-a Compression Compression

(s = Const)(s = Const)

Displacement


VV

a b

c

de

pp


FIGURE 9-21T-s and P-v diagrams for the ideal Diesel cycle.

Advantages of the Diesel Cycle

• Eliminates pre-ignition of the fuel-air mixture when compression ratio is high.

• All heat transfer to a constant mass system.

The air-standard compression ignition engine

ab

c

dV = ConstantV = Constant

p = Constantp = Constant

QQHH

QQCC

TT

ss

WWoutout

WWinin

TT

)(

1

dcvC

abpH

H

C

TTCQ

TTCQ

Q

Q

)(

)(1

)(

)(1

ab

dc

abp

dcv

TTk

TT

TTC

TTC

Thermal efficiency of the air-Thermal efficiency of the air-standard Diesel cyclestandard Diesel cycle

ab

c

dV = ConstantV = Constant

p = Constantp = Constant

QQHH

QQCC

ss

VV

Compression RatioCompression Ratio

Expansion RatioExpansion Ratio

Cut Off RatioCut Off RatioDisplacement


a b

c

de

pp a

dv

v

vr

b

ce

v

vr

a

bc

v

vr

Key parameters forKey parameters forthe Diesel cyclethe Diesel cycle

Thermal efficiency

a b

1

111

1c

kc

kv rk

r

r

Displacement


c

de

pp

VV

a bpp

cd

e

Displacement


VV

Thermal EfficiencyThermal Efficiency

1

111

1c

kc

kv rk

r

r

Compression Ratio: rv vd

va

Expansion Ratio: re vc

vb

Cut Off Ratio: rc vb

va

Air-standard Diesel cycleAir-standard Diesel cycle

• At rc = 1, the Diesel and Otto cycles have the same efficiency.– Physical implication for the Diesel cycle: No change in volume when heat

is supplied.

– A high value of k compensates for this.

• For rc > 1, the Diesel cycle is less efficient than the Otto cycle.

Important features of the Diesel cycle

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-22Thermal efficiency of the ideal Diesel cycle as a function of compression and cutoff ratios (k = 1.4).

Efficiency Comparisons(Approximate)

rrvv

rc = 1

rc > 1

• Efficiency is based on internal Efficiency is based on internal energies and enthalpies obtained energies and enthalpies obtained from the gas tables which take into from the gas tables which take into account variable specific heats.account variable specific heats.

ba

ba

hh

uu

1

Effect of variable specific heats

Key terms and concepts

Air-standard Diesel CycleExpansion ratioPressure ratio

Cut of ratioCompression ignition

Comparison of the Otto and Diesel cycles

Comparison No. 1: (a) Same inlet state (P,V)Comparison No. 1: (a) Same inlet state (P,V) (b) Same compression (b) Same compression

ratio, rratio, rvv

(c) Same Q(c) Same QHH

Key factor: Constant volume heat addition of the Otto Key factor: Constant volume heat addition of the Otto cycle cycle vs. constant pressure heat addition of vs. constant pressure heat addition of the Diesel cycle.the Diesel cycle.


VV

pp

b

c

d

a

DisplacementDisplacement

Otto cycle with a specified Otto cycle with a specified inlet condition at “a” with inlet condition at “a” with a given compression ratio a given compression ratio rrvv = v = vaa/v/vbb

Otto cycle with a specified Otto cycle with a specified inlet condition at “a” with inlet condition at “a” with a given compression ratio a given compression ratio rrvv = v = vaa/v/vbb

VV

pp

b

c

c*

d*d

a

Displacement

Otto and Diesel cycles with same compression inlet conditions at “a” and thesame compression ratio, rv.

Otto and Diesel cycles with same compression inlet conditions at “a” and thesame compression ratio, rv.

c

Diesel Cycle

1

111

1c

kc

kv rk

r

r

kv

r 11Otto Cycle

pp

b c*d*d

aVV

First Law analysis of the heat addition process

WQdU Otto: Heat addition with V = 0 in process b c.

W = 0, and P and T increaseOtto: Heat addition with V = 0 in process b c.

W = 0, and P and T increase

Diesel: Heat addition with P = Constant in process b c*. dW > 0, and P and T lower that in the Otto cycle.

Diesel: Heat addition with P = Constant in process b c*. dW > 0, and P and T lower that in the Otto cycle.

V = Const.V = Const.V = Const.V = Const.

TT

ss

d*

da

b

TTcc

TTc*c*P = Const.P = Const.P = Const.P = Const.

c c*

Otto CycleOtto CycleOtto CycleOtto Cycle

Diesel CycleDiesel CycleDiesel CycleDiesel Cycle

TT

ss

d*da

b

Tc

Tc*c*

cThe areas under The areas under the process paths the process paths b b c and b c and b c* c* are equal under the are equal under the assumption ofassumption ofequal heat addition, equal heat addition, QQHH..

The areas under The areas under the process paths the process paths b b c and b c and b c* c* are equal under the are equal under the assumption ofassumption ofequal heat addition, equal heat addition, QQHH..

QQH,OttoH,Otto = Q = QH,DieselH,Diesel

QQHH

T-s diagrams for equal heat addition

DIESELOTTO

When QWhen QHH and r and rvv are are

the same for both the same for both cycles,cycles,

TT

ss

d*da

b

TTcc

TTc*c* c*c

QQC,OttoC,Otto < Q < QC,DieselC,Diesel

Efficiency comparisons

Comparison No. 2 is more practical when the “knocking”effect is considered. In the Otto cycle, about 11 atm are needed to achieve combustion with engine knock.

Comparison No. 2 is more practical when the “knocking”effect is considered. In the Otto cycle, about 11 atm are needed to achieve combustion with engine knock.

Comparison No. 2: (a) Same inlet state (P,V) (b) Same maximum P

(c) Same QH


DIESEL OTTO

TT

ss

d*

d

a

b

b*

c*

V = Const.V = Const.

P = Const.P = Const.

cDiesel CycleDiesel CycleOtto CycleOtto Cycle

Gas Power Systems - 3

The Dual Cycle

The Dual cycle

• The dual cycle is designed to capture capture some of the advantages of both the Otto and Diesel cycles.

• It it is a better approximation to the actual operation of the compression ignition engine.

pp

V

d

e

a

b cQQH,PH,P

QQH,VH,V

QQC,VC,V

The Dual cycleThe Dual cycle

s = Constants = Constant


FIGURE 9-23P-v diagram of an ideal dual cycle.

pp

QQC,VC,V

)()(

)(1

)()(

)(1

bCab

ed

bCPabV

edv

TTkTT

TT

TTCTTC

TTC

VV

d

e

a

b cQQH,PH,P

QQH,VH,V

)1(1

111

1cpp

kcp

kv rkrr

rr

r

a

bP

b

cc

a

ev

P

Pr

V

Vr

V

Vr

VV

d

e

a

b cQQH,PH,P

QQH,VH,V


The Gas Turbine

Overview

• Gas Turbines

– The Carnot cycle as an “air standard” cycle

• The Ericcson Cycle

• The Brayton Cycle

– The standard cycle

– The reheat cycle

– Inter-cooling

• Long-haul automotive power.• Large aircraft flight envelopes• Commercial aircraft

• Altitude (30,000 to 40,000 ft)

• Long range and low specific fuel consumption (sfc)

• Moderate speed and thrust

The modern gas turbine cycle

• Military aircraft• High altitude ( > 40,000 ft)

• Moderate range and high specific fuel consumption (sfc)

• High speed and large thrust

• Relevant cycles– The Ericsson and Brayton cycles

The modern gas turbine cycle

General features of the aircraft gas turbine engine

Intake AirIntake AirIntake AirIntake Air

FuelFuelFuelFuel

CompressorCompressorCompressorCompressorTurbineTurbine

CombustorCombustorCombustorCombustor

Exhaust SectionExhaust SectionExhaust SectionExhaust Section

Compressor Drive Shaft, WCompressor Drive Shaft, WcompcompCompressor Drive Shaft, WCompressor Drive Shaft, Wcompcomp

The air-standard turbine cycle

• Open system modeled as a closed system - fixed with fixed mass flow.

• Air is the working fluid.

• Ideal gas assumptions are applied.

• Approximate the combustor as the high temperature source.

• Internally reversible processes.

Air Standard Air Standard Brayton CycleBrayton Cycle

Air Standard Air Standard Carnot CycleCarnot Cycle

Air Standard Air Standard Ericcson CycleEriccson Cycle

Ideal gas power cycles

Air-standard Carnot cycle

Employ the same assumptions as for the air-standard turbine cycle.Employ the same assumptions as for the air-standard turbine cycle.

TT

ss

pp11 pp22

pp33pp44

QQHH

QQCC

1

311T

T

T

T

H

C

4321 , PPPP

)1(

2

3

1

1

4

1

3

2

1

4

1

11

11

kk

kkkk

V

V

V

V

P

P

P

P

Air-standard Carnot cycle

• Heat addition at constant temperature is difficult and costly.

– Work is required because fluid expands.

• Heat addition is limited because a large change in volume would imply a low mean pressure in the heat addition process.

– Frictional effects might become too great if the mean pressure is too low.

Limitations of the air-standard Carnot cycle


FIGURE 9-26T-s and P-v diagrams of Carnot, Stirling, and Ericsson cycles.


FIGURE 9-27The execution of the Stirling cycle.

Air-standard Ericsson cycle


FIGURE 9-28A steady-flow Ericsson engine.

• Constant pressure heat addition and rejection

• Constant temperature compression and expansion

Qb-c

The air-standard Ericsson cycle

vv

pp

a

b c

d

T = Constant

Qd-a

Qc-dQa-b

The Air Standard Ericcson Cycle

The Air Standard Ericcson Cycle

vv

pp

a

b c

dQQd-ad-a

QQc-dc-dQQa-ba-b

QQb-cb-c T = Const.

ss

TT

ab

c d

QQd-ad-a

QQc-dc-d

QQa-ba-b

QQb-cb-c

P = ConstP = Const.

Thermal EfficiencyEriccson Cycle

Thermal EfficiencyEriccson Cycle

QQb-cb-c

dccb

badc

Added

NET

QQ

WW

Q

W

pp

a

b c

d

T = Const.

QQd-ad-a

QQc-dc-dQQa-ba-b

WWc-dc-dWWa-ba-b

vv

The Brayton cycle

The Brayton Cycle

• Modern gas turbines operate on an open Brayton cycle.

– Ambient air is drawn at the inlet.

– Exhaust gases are released to the ambient environment.

• The air standard Brayton cycle is a closed cycle.

– All processes are internally reversible.

– Air is the working fluid and assumed an ideal gas.

General features of the aircraft gas turbine engine

Intake AirIntake AirIntake AirIntake Air

FuelFuelFuelFuel

CompressorCompressorCompressorCompressorTurbineTurbine

CombustorCombustorCombustorCombustor

Compressor Drive Shaft, WCompressor Drive Shaft, WcompcompCompressor Drive Shaft, WCompressor Drive Shaft, Wcompcomp

Exhaust Gases &Exhaust Gases &Work Output,Work Output,WWturbturb

Exhaust Gases &Exhaust Gases &Work Output,Work Output,WWturbturb

The open, actual The open, actual cycle for the gas cycle for the gas turbine.turbine.

The open, actual The open, actual cycle for the gas cycle for the gas turbine.turbine.

The closed, airThe closed, airstandard cycle standard cycle for the for the gas turbine.gas turbine.

The closed, airThe closed, airstandard cycle standard cycle for the for the gas turbine.gas turbine.

QQHH

QQLL

WWCOMPCOMP

WWTURBTURB

• Isentropic compression to TH

• Constant pressure heat addition at TH

• Isentropic expansion to TC

• Constant pressure heat rejection at TC

The gas turbine processes

1

2 3

4

Air Standard Air Standard Brayton CycleBrayton CycleAir Standard Air Standard Brayton CycleBrayton Cycle

QQHH

QQLL

WWCOMPCOMP

WWTURBTURB

pp

Vvv

1 4

32S = ConstantS = Constant

QQHH

QQCC

ss11 = s = s22 ss33 = s = s44

Air Standard Brayton CycleAir Standard Brayton Cycle

pp22 = p = p33

pp11 = p = p44

TT

ss

1

4

3

2

QQHH

QQCC

VV

pp

Vvv

1 4

32

QQHH

QQCC

s = Constants = Constant

Thermal efficiency of the ideal Brayton cycle

Thermal efficiency of the ideal Brayton cycle

23

1243 )(

hh

hhhh

Q

WW

H

inOUT

pp22 = p = p33

pp11 = p = p44

TT

ss

1

4

3

2

QQHH

QQCC

WWOUTOUT

WWININ

Thermal EfficiencyIdeal Brayton Cycle

Thermal EfficiencyIdeal Brayton Cycle

kk

P

P)1(

1

21

For an ideal gas, h-h0 = Cp(T - T0). The compression and expansion process are polytropic with constant k.

pp22 = p = p33

pp11 = p = p44

TT

ss

1

4

3

2

QQHH

QQCC

WWOUTOUT

WWININ

Cycle efficiency

• Ideal gas assumptions apply.

• All processes internally reversible

– Compressor and turbine efficiency are each 100%.

• Assume fuel added in the combustor is a small percent (mass or moles) of total flow, and thus air properties provide a good estimate of cycle performance.


Real Turbine Performance

Overview

• Internally irreversible Brayton cycles– Case study

• Real turbine performance

Internally irreversible Brayton cycles

• Internal irreversibility arises in each process of the open cycle.

• Compression (compressor efficiency)• Heat addition • Expansion (turbine efficiency)

Real cycle analysis - Brayton cycle

Internal irreversibility will lowerInternal irreversibility will lowerwork output and decrease work output and decrease thermal efficiency. Greater thermal efficiency. Greater entropy production will also entropy production will also result.result.

Internal irreversibility will lowerInternal irreversibility will lowerwork output and decrease work output and decrease thermal efficiency. Greater thermal efficiency. Greater entropy production will also entropy production will also result.result.

23

1 4

pp11 = p = p44

T,hT,hpp22 = p = p33

ss

s3 = s4

2 4

1

3

4s

12 ss

Case Study

GivenGiven: Turbine and compressor efficiencies of : Turbine and compressor efficiencies of 80%, and the operating data as given below. No 80%, and the operating data as given below. No internal irreversibility in the combustor. internal irreversibility in the combustor. FindFind: For the actual cycle, the thermal efficiency, : For the actual cycle, the thermal efficiency, the ratio of Wthe ratio of Wcompcomp/W/Wturbturb and entropy production. and entropy production.

Case study

QQHH

QQCC

WWININ WWOUTOUT

aa

b’b’

d’d’

ccccTT

ssaa

bbb’b’

d’d’dd

State T(R)

P(psia)

h(Btu/ lbm)

S0

(Btu/ lbmR)

a 530 15 126.7 0.5963

b 881 90 211.6 0.7191

b’ 967 90 232.7 0.7421

c 1700 90 422.6 0.8876

d 1059 15 255.7 0.7647

d’ 1192 15 289.3 0.7946

Note: Values of h and sNote: Values of h and soo are from the gas tables, and all are from the gas tables, and all initial data are supplied here but this is not necessarily initial data are supplied here but this is not necessarily the case in practice.the case in practice.

Note: Values of h and sNote: Values of h and soo are from the gas tables, and all are from the gas tables, and all initial data are supplied here but this is not necessarily initial data are supplied here but this is not necessarily the case in practice.the case in practice.

Process h(Btu/lbm)

W(Btu/lbm)

Q(Btu/lbm)

s(Btu/lbmR)

a-b 84.9 -84.9 0 0a-b’ 106 -106 0 0.0230b’-c 189.9 0 189.9 0.1455c-d -166.9 166.9 0 0c-d’ -133.3 133.3 0 0.0299d’-a -162.6 0 -162.6 -.1983NET 0 27.3 0 0

Process CalculationsProcess Calculations

si j si jo R ln

Pj

Pi

Note:

TURB h

hs

, COMP hs

h

The pressure ratios for The pressure ratios for the isentropic and actual the isentropic and actual processes are the same, processes are the same, and therefore and therefore s = s = ssoo. .

The pressure ratios for The pressure ratios for the isentropic and actual the isentropic and actual processes are the same, processes are the same, and therefore and therefore s = s = ssoo. .

Thermal Efficiency and Work RatioThermal Efficiency and Work Ratio

795.03.133

106

144.0189

3.27

189

1063.133

'

turb

comp

cb

netCYCLE

W

W

Q

W

Notice that here we have a high ratio of the work of compressionto that of expansion, which is typical in gas turbine systems. Thekey variable is the turbine inlet temperature, Tc, and this is to bemade as large a possible. Tc is limited by the material and turbineblade cooling capability.

Notice that here we have a high ratio of the work of compressionto that of expansion, which is typical in gas turbine systems. Thekey variable is the turbine inlet temperature, Tc, and this is to bemade as large a possible. Tc is limited by the material and turbineblade cooling capability.

The entropy production rates are based on the application of the The entropy production rates are based on the application of the entropy entropy balance for an open systembalance for an open system to each process to each process inin the cycle. The heat addition the cycle. The heat addition and rejection processes are at the high and low temperatures respectively. and rejection processes are at the high and low temperatures respectively. Thus, these processes produce an Thus, these processes produce an entropy production from the external heat entropy production from the external heat transfer process that is not included in the above calculationstransfer process that is not included in the above calculations..

Entropy ProductionEntropy Production

The Carnot Efficiency The Carnot Efficiency for the Cyclefor the CycleThe Carnot Efficiency The Carnot Efficiency for the Cyclefor the Cycle

68.0

1700

53011

H

CCarnot

T

T

144.0

QH, TH

QC, TC

(General cycle representation)

Here, the difference in the efficiencyHere, the difference in the efficiencycalculations indicates a high levelcalculations indicates a high levelof external entropy production.of external entropy production.

Here, the difference in the efficiencyHere, the difference in the efficiencycalculations indicates a high levelcalculations indicates a high levelof external entropy production.of external entropy production.

Efficiency CurvesEfficiency Curves T,hT,h

p2 = p3

sS1 = S2 S3 = S4

2 4

1

3

s’2 s’4

p1 = p4

Pressure Ratio, p2/p1

22 44

TURB COMP 1

0.9

0.8

0.7

0.10.1

0.250.25

Brayton cycle efficiency is Brayton cycle efficiency is greatly dependent on the greatly dependent on the efficiencies of the turbineefficiencies of the turbineand compressor. The and compressor. The curves at the left are curves at the left are approximate.approximate.

Brayton cycle efficiency is Brayton cycle efficiency is greatly dependent on the greatly dependent on the efficiencies of the turbineefficiencies of the turbineand compressor. The and compressor. The curves at the left are curves at the left are approximate.approximate.


FIGURE 9-29An open-cycle gas-turbine engine.


FIGURE 9-30A closed-cycle gas-turbine engine.


FIGURE 9-31T-s and P-v diagrams for the ideal Brayton cycle.

FIGURE 9-32Thermal efficiency of the ideal Brayton cycle as a function of the pressure ratio.



Cycle Improvements

Cycle Improvements

• Regeneration

– Reduces heat input requirements and lowers heat rejected.

• Inter-cooling

– Lowers mean temperature of the compression process

• Reheat

– Raises mean temperature of the heat addition process


FIGURE 9-33For fixed values of Tmin and Tmax , the net work of the Brayton cycle first increases with the pressure ratio, then reaches a maximum at rp = (Tmax /Tmin) k/[2(k – 1)], and finally decreases.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-36The deviation of an actual gas-turbine cycle from the ideal Brayton cycle as a result of irreversibilities.

The Brayton cycle with regeneration


FIGURE 9-38A gas-turbine engine with regenerator.

WWOUTOUT

The standard Brayton cycle The standard Brayton cycle shown as an open cycle.shown as an open cycle.

WWININ11

3322

44

QQHH

Regeneration is Regeneration is accomplishedaccomplishedby preheating the by preheating the combustor aircombustor airwith the exhaustwith the exhaustgas from gas from the turbine.the turbine. WCOMP

443322

55

WTURB

11

QQHH

xx44

11

33

22yy Internal heat transfer,Internal heat transfer,

States 2 - x.States 2 - x.

Internal heat transfer,Internal heat transfer,States 2 - x.States 2 - x.

QQCC

QQHH

ss

T,hT,h

The maximum benefit of regeneration is obtained whenThe maximum benefit of regeneration is obtained whenthe exhaust temperature Tthe exhaust temperature Tyy is brought to temperature is brought to temperature

TT2 2 by the regenerative heat exchanger.by the regenerative heat exchanger.

The maximum benefit of regeneration is obtained whenThe maximum benefit of regeneration is obtained whenthe exhaust temperature Tthe exhaust temperature Tyy is brought to temperature is brought to temperature

TT2 2 by the regenerative heat exchanger.by the regenerative heat exchanger.

The ideal regenerator

The ideal regeneratorThe ideal regeneratorwill heat the combustionwill heat the combustionair up to Tair up to T44

43

12 )(1

hh

hhCYCLE

xx44

11

33

22yy

QQCC

QQHH

ss

T,hT,h

33

The actual regenerator

Actual internal heat transfer takesActual internal heat transfer takesplace between States 2 - x’.place between States 2 - x’.

The effectiveness of the regenerator is The effectiveness of the regenerator is the actual heat transfer divided by the the actual heat transfer divided by the maximum possible heat transfer.maximum possible heat transfer.

)(

)(

)(

)(

24

2

24

'2

TT

TT

TTC

TTC x

p

xp

xx44

11

22yy

QQCC

QQHH

ss

T,hT,h

xy

Thermal efficiency

xcycle

hh

hhhh

3

1243 )()(

Regeneration reducesRegeneration reducesnet heat inputnet heat input at the at thehigh temperature.high temperature.

xx44

11

22yy

QQCC

QQHH

ss

T,hT,h

xy

33


FIGURE 9-39T-s diagram of a Brayton cycle with regeneration.


FIGURE 9-40Thermal efficiency of the ideal Brayton cycle with and without regeneration.

The Brayton cycle with inter-cooling

The standard Brayton cycle with on stage of inter-cooling

44

5511

WWCOMPCOMPWWTURBTURB

QQHH 3377

66

TT

ss11

7’7’77

66

5’5’55 444’4’

33

One stage of inter-cooling with turbine One stage of inter-cooling with turbine and compressor inefficiencies. and compressor inefficiencies. One stage of inter-cooling with turbine One stage of inter-cooling with turbine and compressor inefficiencies. and compressor inefficiencies.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-42Comparison of work inputs to a single-stage compressor (1AC) and a two-stage compressor with intercooling (1ABD).

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-43A gas-turbine engine with two-stage compression with intercooling, two-stage expansion with reheating, and regeneration.


FIGURE 9-44T-s diagram of an ideal gas-turbine cycle with intercooling, reheating, and regeneration.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-45As the number of compression and expansion stages increases, the gas-turbine cycle with intercooling, reheating, and regeneration approaches the Ericsson cycle.

The Brayton cycle with reheat

The re-heating of the fluid at an intermediateThe re-heating of the fluid at an intermediatepressure raises the enthalpy input to the cycle pressure raises the enthalpy input to the cycle without raising the temperature of the external without raising the temperature of the external thermal reservoir. thermal reservoir.

The re-heating of the fluid at an intermediateThe re-heating of the fluid at an intermediatepressure raises the enthalpy input to the cycle pressure raises the enthalpy input to the cycle without raising the temperature of the external without raising the temperature of the external thermal reservoir. thermal reservoir.

44

11

33

22

55

ss

T,hT,h

66

16QQc

5432 QQQH

Example

DescriptionDescription::In an air-standard gas turbine engine, air at 60In an air-standard gas turbine engine, air at 60oo F F and 1 atm enters the compressor with a pressure and 1 atm enters the compressor with a pressure ratio of 5:1. The turbine inlet temperature is ratio of 5:1. The turbine inlet temperature is 15001500oo F, and the exhaust pressure is 1 atm. F, and the exhaust pressure is 1 atm.

Determine the ratio of turbine work to Determine the ratio of turbine work to compressor compressor work and the thermal efficiency under the work and the thermal efficiency under the following operating conditions.following operating conditions.

This example compares several modifications to the basicThis example compares several modifications to the basicBrayton cycleBrayton cycle: multi-stage compression and expansion: multi-stage compression and expansionwith reheat, inter-cooling and regeneration.with reheat, inter-cooling and regeneration.

Operating conditionsOperating conditions::(a) The engine operates on an ideal Brayton (a) The engine operates on an ideal Brayton cycle.cycle.

(b) The adiabatic efficiency of the turbine (b) The adiabatic efficiency of the turbine and compressor are 0.83 and 0.93, and compressor are 0.83 and 0.93, respectively.respectively.

(c ) The conditions of (b) hold and a (c ) The conditions of (b) hold and a regenerator with an effectiveness of 0.65 is regenerator with an effectiveness of 0.65 is introduced.introduced.

Operating conditionsOperating conditions::(d) The conditions of (b) and (c ) hold, and an(d) The conditions of (b) and (c ) hold, and an inter-cooler that cools the air to 60inter-cooler that cools the air to 60oo F at a F at a constant pressure of 35 psia is added.constant pressure of 35 psia is added.

(e) The inter-cooler is removed and a re-heater(e) The inter-cooler is removed and a re-heaterthe heats the fluid to 1500the heats the fluid to 1500oo F at a constant F at a constant pressure of 35 psia is added.pressure of 35 psia is added.

(f) The conditions of (e) hold and the inter-(f) The conditions of (e) hold and the inter-cooler of (d) is put back.cooler of (d) is put back.

AssumptionsAssumptions::

(1) All cycles operate on the air-standard (1) All cycles operate on the air-standard basis.basis.(2) Ideal gas(2) Ideal gas(3) Constant specific heats; k = constant, (3) Constant specific heats; k = constant, CCpp = 0.24 BTU/lbm-R, k = 1.40 = constant. = 0.24 BTU/lbm-R, k = 1.40 = constant.

(4) Internally reversible processes except (4) Internally reversible processes except where isentropic efficiency (adiabatic where isentropic efficiency (adiabatic efficiency) is less than one.efficiency) is less than one.(5) 1 atm = 14.7 psia. (5) 1 atm = 14.7 psia.

SolutionSolution::(a) The engine operates as an ideal Brayton cycle.(a) The engine operates as an ideal Brayton cycle.

TT

ss11

22

33

44

2121

4343

1

3

4

3

4

TCh

TCh

P

P

T

T

p

p

k

k

State DataState P

(psia)T

(R)1 14.7 5202 73.5 8253 73.5 19604 14.7 1237

Process Quantities hBTU/Lbm

QBTU/lbm

1-2 73.1 02-3 272.4 272.43-4 -173 04-1 -172 -172

Total 0.5 (~0) 100.32

Process quantitiesProcess quantities

367.0)8251960)(24.0(

1.73173

37.21.73

173

1.73

173

32

2132

32

2143

32

21

43

2121

434343

h

hh

Q

WW

Q

W

W

W

W

W

lbmBTUhW

lbmBTUTTChW

netcycle

compressor

turbine

p

Process quantitiesProcess quantities

(b) The adiabatic efficiency of the compressor and turbine(b) The adiabatic efficiency of the compressor and turbineare 0.83 and 0.92 respectively.are 0.83 and 0.92 respectively.

2121

4343

1

3

4

3

4

TCh

TCh

P

P

T

T

p

p

k

k

TT

ss11

22

33

442’2’

4’4’

83.0,92.0 compT

RTW

W

RTT

TT

TT

TT

W

W

sturb

comp

scomp

1295

887

443,

43

1212

12

12

21

21,

State DataState P

(psia)T

(R)1 14.7 5202 73.5 8253 73.5 19604 14.7 1237

State DataState P

(psia)T

(R)1 14.7 5202’ 73.5 8873 73.5 19604’ 14.7 1295

28.0)8871960)(24.0(

8.885.159

81.18.88

5.159

8.88

5.159

32

2143

32

2143

32

21

43

2121

434343

h

hh

Q

WW

Q

W

W

W

W

W

lbmBTUhW

lbmBTUTTChW

netcycle

compressor

turbine

p

(c) The conditions of (b) hold, and a regenerator with(c) The conditions of (b) hold, and a regenerator withan effectiveness of 0.65 is introduced. an effectiveness of 0.65 is introduced.

4’4’

TT

ss

11

22

33

yy2’2’ xx

2121

4343

1

3

4

3

4

TCh

TCh

P

P

T

T

p

p

k

k

83.0

92.0

comp

T

RTTTT

TT

TT

x

x

1052

65.0)(

)(

242

'2'4

'2

State DataState P

(psia)T

(R)1 14.7 520x’ 73.5 10523 73.5 19604’ 14.7 1295

37.0)10521960)(24.0(

8.885.159

81.18.88

5.159

8.88

5.159

3

2143

3

2143

3

21

43

2121

434343

xxx

netcycle

compressor

turbine

p

h

hh

Q

WW

Q

W

W

W

W

W

lbmBTUhW

lbmBTUTTChW

(d) The conditions of (b) and (c ) hold, and an inter-cooler(d) The conditions of (b) and (c ) hold, and an inter-coolerthat cools the air to 60that cools the air to 60oo F at 35 psia is added. F at 35 psia is added.

520 R520 R

ss

3

TT

1

7’

6

5’ 4’

P = 35 psiaP = 35 psia

xx

yy

520 R520 R

ss

3

TT

1

7’

6

5’ 4’


xx

yy

Compute TCompute T5’ 5’ ,T,T7’7’,,

and the actual and the actual temperature betweentemperature betweenstate 7’ and x duestate 7’ and x dueto regeneration (7”).to regeneration (7”).

Compute TCompute T5’ 5’ ,T,T7’7’,,

and the actual and the actual temperature betweentemperature betweenstate 7’ and x duestate 7’ and x dueto regeneration (7”).to regeneration (7”).

ss

State DataState P

(psia)T

(R)1 14.7 5205 35 6675’ 35 6976 14.7 5207 73.5 6437’ 73.5 6687” 73.5 6163 73.5 19604’ 14.7 520x 73.5 1295y 14.7 888

RT

P

P

T

T k

k

6675

1

1

5

1

5

1515

1TTTT

3T

1

7’

6

5’ 4’


xx

yy

ss

38.0)(

04.2

/78

/5.159

73

7651

43

TTC

W

W

W

lbmBTUWW

lbmBTUW

p

net

compressor

turbine

ss

38.0(e) Eliminate inter-cooling but add one stage of reheat(e) Eliminate inter-cooling but add one stage of reheatto 1500to 1500oo F. F.

(f) Combine parts (d) and (e).(f) Combine parts (d) and (e).

28.2

40.0

Compressor

turbine

W

W

Work RatioWt/Wc

Wt

(a) Ideal Cycle 0.37 2.37 173(b) Actual BasicCycle

0.28 1.81 160

(c) Actual BasicCylce andRegenerator ( =0.65)

0.37 1.18 160

(d) Actual Cyclewith Intercooling at35 psia to 60o F andpart (c)

0.38 2.04 160

(e) Actual BasicCycle with Reheat at35 psia and part (c)

0.38 2.02 178

(f) Parts (c ) ,(d), and(e) combined

0.40 2.28 178

Summary of resultsSummary of results::


FIGURE 9-48Basic components of a turbojet engine and the T-s diagram for the ideal turbojet cycle.

[Source: The Aircraft Gas Turbine Engine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]


FIGURE 9-51Energy supplied to an aircraft (from the burning of a fuel) manifests itself in various forms.


FIGURE 9-52A turbofan engine.

[Source: The Aircraft Gas Turbine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]


FIGURE 9-53A modern jet engine used to power Boeing 777 aircraft. This is a Pratt & Whitney PW4084 turbofan capable of producing 84,000 pounds of thrust. It is 4.87 m (192 in.) long, has a 2.84 m (112 in.) diameter fan, and it weighs 6800 kg (15,000 lbm).

Photo Courtesy of Pratt&Whitney Corp.


FIGURE 9-54A turboprop engine.



FIGURE 9-55A ramjet engine.


Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-57Under average driving conditions, the owner of a 30-mpg vehicle will spend $300 less each year on gasoline than the owner of a 20-mpg vehicle (assuming $1.50/gal and 12,000 miles/yr).

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 9-62Aerodynamic drag increases and thus fuel economy decreases rapidly at speeds above 55 mph.

(Source: EPA and U.S. Dept. of Energy.)

chapter 9 gas power cycles. 9-1 basic considerations in the analysis of power cycles air standard...

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