chapter 9 gas power cycles - vinhqtang.com

Chapter 9

GAS POWER CYCLES(based on notes from Profs. Kaya & Gauthier)

Vinh Q. Tang, Ph.D., P.Eng.

Adjunct Recearch Professor

MAAE 2400

Thermodynamics and Heat Transfer

1

Carleton University

Department of Mechanical and Aerospace Engineering

[email protected]

Office: ME2186

Fall 2020

OUTLINE

9.1 Introduction (Review of ideal gas model)

9.2 Otto Cycle (For your interest)

9.3 Brayton Cycle

9.4 Summary

Reading Assignment: Related Sections in Chapter 9 in Fundamentals of

Engineering Thermodynamics, 7th Edition ; M. Moran and H. N. Shapiro, et al

2Fall 2020

9.1 INTRODUCTION

3

Equations of state :

𝒑𝒗 = 𝑹𝑻 𝒐𝒓 𝒑𝑽 = 𝒎𝑹𝑻

Fall 2020

Review of Ideal Gas Model (presented in Chapters 3 and 6)

4

Changes in “u” and “h”

𝒖𝟐 − 𝒖𝟏 = 𝒄𝒗(𝑻𝟐 − 𝑻𝟏)

𝒉𝟐 − 𝒉𝟏 = 𝒄𝑷(𝑻𝟐 − 𝑻𝟏)

1) Constant specific heats

2) Variable specific heats

u(T) and h(T) are evaluated from tables:

▪ Table A-22 for air (mass basis)

▪ Table A-23 for other gases (molar basis)

Fall 2020

5

Changes in entropy


2) Variable specific heats :

𝒔𝟐 − 𝒔𝟏 = 𝒄𝒗𝒍𝒏𝑻𝟐𝑻𝟏

+ 𝑹𝒍𝒏(𝒗𝟐𝒗𝟏)

𝒔𝟐 − 𝒔𝟏 = 𝒄𝑷𝒍𝒏𝑻𝟐𝑻𝟏

− 𝑹𝒍𝒏(𝑷𝟐

𝑷𝟏)

𝒔𝟐 − 𝒔𝟏 = 𝒔𝟐𝟎 − 𝒔𝟏

𝟎 − 𝑹𝒍𝒏(𝑷𝟐

𝑷𝟏)

s0 can be obtained from tables:

▪ Table A-22 for air (mass basis)

▪ Table A-23 for other gases (molar basis)

(See Tables A-20 & 21 for cv and cP data)

Fall 2020

6

For Isentropic Processes


𝑻𝟐𝑻𝟏

=𝑷𝟐

𝑷𝟏

𝒌−𝟏𝒌 𝑻𝟐

𝑻𝟏=

𝒗𝟏𝒗𝟐

𝒌−𝟏𝑷𝟐

𝑷𝟏=

𝒗𝟏𝒗𝟐

𝒌

Where 𝒌 =𝒄𝒑

𝒄𝒗𝒈𝒊𝒗𝒆𝒏 𝒊𝒏 𝑻𝒂𝒃𝒍𝒆 𝑨 − 𝟐𝟎

2) Variable specific heats

𝑷𝟐

𝑷𝟏=𝑷𝒓𝟐

𝑷𝒓𝟏

𝒗𝟐𝒗𝟏

=𝒗𝒓𝟐𝒗𝒓𝟏

Where Pr and vr are provided for air 𝐢𝐧 𝐓𝐚𝐛𝐥𝐞 𝐀 − 𝟐𝟐

for air only

Fall 2020

End of Review of Ideal Gas Model

• Starting section 9.2

in the next slide

Fall 2020 7

9.2 OTTO CYCLE (1)

• Used to model spark-

ignition engines

• In a spark-ignition

engine, the air-fuel

mixture is ignited by a

spark plug

8Fall 2020

9.2 OTTO CYCLE (2)

• Pressure-

displacement curve

for a four-stroke

reciprocating

internal combustion

engine

(TDC) (BDC)9Fall 2020

9.2 OTTO CYCLE (3)

1. A fixed mass of air is the working fluid throughout the entire cycle. Thus, there is no inlet and no exhaust process.

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

3. The cycle is completed by heat transfer to the surroundings.

4. All processes are internally reversible.

5. Air is an ideal gas with constant specific heats (Cold Air Standard Analysis).

AIR STANDARD ANALYSIS

10Fall 2020

9.2 OTTO CYCLE (4)

1-2 reversible adiabatic (isentropic) compression

2-3 constant volume heat transfer (heat added)

3-4 reversible adiabatic (isentropic) expansion

4-1 constant volume heat transfer (heat rejected)

2

3

4

v=c

v=c

1

11Fall 2020

9.2 OTTO CYCLE (5)

• Air Standard Otto Cycle (ASOC) is an ideal cycle

approximating a spark-ignition internal

combustion engine.

• Due to the simplifications in the ASOC, the

analysis is only valid on a QUALITATIVE BASIS.

12Fall 2020

9.2 OTTO CYCLE (6)

CONSEQUENCES OF THE Air Standard Otto Cycle

ASSUMPTIONS

• The working fluid is a fixed amount of air.

• Heat rejection and addition (combustion) are

assumed to take place instantaneously.

• There are no pressure drops due to friction.

• Air is assumed to be an ideal gas.

• All processes are internally reversible.

13Fall 2020

9.2 OTTO CYCLE (7)

• USING 1st and 2nd Laws for a closed system with

DKE, DPE=0

121221For uuw −=−

433443For uuw −=−

232332For uuq −=−

144114For uuq −=−

WQU −=D

NOTE: w’s and q’s are all positive quantities. 14Fall 2020

( ) ( ) ( ) ( )

23

14

23

1423

23

1243

23

1234

1uu

uu

uu

uuuu

uu

uuuu

q

ww

q

w

cycle

cycle

cycle

added

net

cycle

−

−−=

−

−−−=

−

−−−=

−=

=

9.2 OTTO CYCLE (8)

1212 uuw −=

4334 uuw −=2323 uuq −=

1441 uuq −=

We haveHence

15Fall 2020

9.2 OTTO CYCLE (9)

1

4

3

3

4

1

2

1

1

2 and

−−

=

=

kk

V

V

T

T

V

V

T

T

For isentropic processes of 1-2 and 3-4 and air is an ideal gas:

2

1

3

4ButV

V

V

V=

4

3

1

2HenceT

T

T

T=

2

3

1

4OrT

T

T

T=

16Fall 2020

9.2 OTTO CYCLE (10)

1

2

1

1

2

2

3

1

4

23

141

−

=

=

=

−

−−=

k

cycle

V

V

T

T

T

T

T

T

dTcdu

uu

uu

( ) 1

21

2

1

23

14

2

1

23

14

/

11

1

1/

1/11

−−=

−=

−

−−=

−

−−=

kcycle

cycle

cycle

VV

T

T

TT

TT

T

T

TT

TT

We have

Hence

17Fall 2020

9.2 OTTO CYCLE (11)

min

max

1

rationcompressioc)(volumetri

where

Cycle Otto1

1

V

Vr

r

v

k

v

cycle

=

−=−

Thermal efficiency of CASOC:

Remember that this is an ideal cycle, and does not

account for variation of properties with temperature,

friction, heat losses, combustion losses, intake and

exhaust processes, etc. 18Fall 2020

9.2 OTTO CYCLE (12)

1

11

−−=

k

v

thr

• As the rv increases, the efficiency also

increases. However, in practice, this may

lead to DETONATION due to auto-ignition

of the fuel-air mixture.

• Lead (tetraethyl lead) in gasoline

improves detonation characteristics, but

is not environmentally friendly.

• Non-leaded gasolines with good

detonation characteristics have been

developed to reduce atmospheric

contamination.

19Fall 2020

9.2 OTTO CYCLE (13)

We can plot the Otto cycle efficiency

1

11

−−=

k

v

thr

5 10 15 200

0.2

0.4

0.6

0.8

k = 1.5k = 1.4k = 1.3

Compression Ratio

Th

erm

al E

ffic

ien

cy

Real Spark Ignition Cycle

20Fall 2020

9.3 BRAYTON CYCLE

OPEN CYCLE CLOSED CYCLE

21Fall 2020

22

“Air-standard analysis” involves an idealization that

simplifies the study of open gas turbine power plants.

Two basic assumptions:

1. The working fluid is air that behaves as an

ideal gas.

2. The temperature rise that would be

brought about by combustion is

accomplished by a heat transfer from an

external source.

Air-standard analysis (For gas turbine open cycles)

More on assumptions

Fall 2020

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

2. The mass and properties of the fuel are neglected.

3. The combustion process is replaced by a heat transfer

from an external source

[It regards heat as being added through a perfect heat

exchanger and removed by exhaust to an infinite

atmosphere (or another perfect heat exchanger)].

4. All processes are internally reversible.

23

For COLD AIR STANDARD : constant specific heat is assumed

MAIN ASSUMPTIONS FOR IDEAL AIR

STANDARD BRAYTON CYCLE ANALYSIS

Fall 2020

24

Air-standard Ideal Brayton Cycle

Fall 2020

1-2 Compression: reversible and adiabatic (isentropic)

2-3 Heat added at constant pressure

3-4 Expansion: reversible and adiabatic (isentropic)

4-1 Heat rejected at constant pressure

‘

2

3

4p= cb

p=ca

1

25

P-v and T-s diagrams

Fall 2020

Applying the first law for an open system to

each CV shown in diagram

Recall the first law:

26Fall 2020

𝒅𝑬𝑪𝑽𝒅𝒕

= ሶ𝑸𝑪𝑽 − ሶ𝑾𝑪𝑽 + ሶ𝒎 𝒉𝒊 − 𝒉𝒆 +𝑽𝒊𝟐 − 𝑽𝒆

𝟐

𝟐+ 𝒈(𝒛𝒊 − 𝒛𝒆)

Steady-state

ΔKE = ΔPE = 0

𝟎 = ሶ𝑸𝑪𝑽 − ሶ𝑾𝑪𝑽 + ሶ𝒎 𝒉𝒊 − 𝒉𝒆

27

Recall :

Fall 2020

𝟎 = ሶ𝑸𝑪𝑽 − ሶ𝑾𝑪𝑽 + ሶ𝒎 𝒉𝒊 − 𝒉𝒆

𝟎 =ሶ𝑸𝑪𝑽

ሶ𝒎−

ሶ𝑾𝑪𝑽

ሶ𝒎+ 𝒉𝒊 − 𝒉𝒆

𝟎 = 𝒒 −𝒘 + 𝒉𝒊 − 𝒉𝒆To be applied to each control volume

Applying to specific control volume

shown on the diagram, the energy

equation becomes :

1-2 Compression: isentropic


3-4 Expansion: isentropic


28Fall 2020

𝟎 = 𝒒 −𝒘 + 𝒉𝒊 − 𝒉𝒆Recall :

𝑭𝒐𝒓 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒐𝒓: 𝒘𝒄 = 𝒉𝟐 − 𝒉𝟏

𝑭𝒐𝒓 𝑯𝑿𝑸𝒊𝒏: 𝒒𝒊𝒏 = 𝒉𝟑 − 𝒉𝟐

𝑭𝒐𝒓 𝑯𝑿𝑸𝒐𝒖𝒕: 𝒒𝒐𝒖𝒕 = 𝒉𝟒 − 𝒉𝟏

𝑭𝒐𝒓 𝑻𝒖𝒓𝒃𝒊𝒏𝒆: 𝒘𝒕 = 𝒉𝟑 − 𝒉𝟒

29

𝜼𝒄𝒚𝒄𝒍𝒆 =𝒘𝒏𝒆𝒕

𝒒𝒊𝒏𝜼𝒄𝒚𝒄𝒍𝒆 =

𝒘𝒕 −𝒘𝒄

𝒒𝒊𝒏

𝜼𝒄𝒚𝒄𝒍𝒆 =(𝒉𝟑 − 𝒉𝟒) − (𝒉𝟐 − 𝒉𝟏)

(𝒉𝟑 − 𝒉𝟐)

𝜼𝒄𝒚𝒄𝒍𝒆 =(𝒉𝟑 − 𝒉𝟐) − (𝒉𝟒 − 𝒉𝟏)

(𝒉𝟑 − 𝒉𝟐)

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −(𝒉𝟒 − 𝒉𝟏)

(𝒉𝟑 − 𝒉𝟐)

Cycle Thermal Efficiency

Fall 2020

Evaluating cycle thermal efficiency

30

1) Air standard analysis: Using air table data

(not assuming constant specific heat)

2) Cold air standard analysis: assuming

constant specific heat

Two different types of analyses

Fall 2020

‘

‘‘

2

3

4

p=const

p=const

1

When Air Table Data are used, recall the

following relationships (from Eq. 6.41)

Note also that P2 = P3 and P1 = P431

isentropic

processes

𝑷𝒓𝟐

𝑷𝒓𝟏=𝑷𝟐

𝑷𝟏

𝑷𝒓𝟒

𝑷𝒓𝟑=𝑷𝟒

𝑷𝟑

Where the values of “Pr” can be found in the tables

1) For Air standard analysis, i.e.,

Using air table data (not assuming constant specific heat)

Fall 2020

32

2) For Cold Air Standard Brayton Cycle analysis

We will develop an expression for the cycle

thermal efficiency, assuming constant

specific heat (or using cold air standard

analysis)

But first we will review some useful

pressure-temperature relationships

i.e., Assuming constant specific heat

Fall 2020

4

3

1

2NoteP

P

P

P=

For isentropic processes of 1-2

(compressor) and 3-4 (turbine) and

air is an ideal gas:

‘

‘‘

2

3

4

p=c

p=c

1

Review P-T relationship

Note: Cold air standard analysis assumes

constant specific heat 33

v

p

c

ckand =

𝒑𝟐𝑷𝟏

=𝑻𝟐𝑻𝟏

𝒌𝒌−𝟏

𝒑𝟑𝑷𝟒

=𝑻𝟑𝑻𝟒

𝒌𝒌−𝟏

Fall 2020

34

‘

‘‘

2

3

4

p=c

p=c

1

Also note :

𝒑𝟐𝑷𝟏

=𝑻𝟐𝑻𝟏

𝒌𝒌−𝟏 𝒑𝟑

𝑷𝟒=

𝑻𝟑𝑻𝟒

𝒌𝒌−𝟏Recall

𝑷𝟐

𝑷𝟏=𝑷𝟑

𝑷𝟒

𝑻𝟐𝑻𝟏

=𝑻𝟑𝑻𝟒

𝑻𝟒𝑻𝟏

=𝑻𝟑𝑻𝟐

𝑻𝟐𝑻𝟏

=𝑷𝟐

𝑷𝟏

𝒌−𝟏𝒌

Fall 2020

Substitute in

35

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −(𝒉𝟒 − 𝒉𝟏)

(𝒉𝟑 − 𝒉𝟐)

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −(𝑻𝟒 − 𝑻𝟏)

(𝑻𝟑 − 𝑻𝟐)

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −𝑻𝟏(

𝑻𝟒𝑻𝟏

− 𝟏)

𝑻𝟐(𝑻𝟑𝑻𝟐

− 𝟏)

Δh = cpΔT for constant cP

Fall 2020

Recall :

36

For Cold Air

Standard Brayton

Cycle analysis

(Previously found)

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −𝑻𝟏(

𝑻𝟒𝑻𝟏

− 𝟏)

𝑻𝟐(𝑻𝟑𝑻𝟐

− 𝟏)

𝑇4𝑇1

=𝑇3𝑇2

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −𝑻𝟏𝑻𝟐

Fall 2020

37v

p

c

ckNote =:

Recall :

For Cold Air

Standard Brayton

Cycle analysis

(Previously found)

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −𝑻𝟏𝑻𝟐

𝑻𝟐𝑻𝟏

=𝑷𝟐

𝑷𝟏

𝒌−𝟏𝒌

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟏 −𝟏

𝑷𝟐𝑷𝟏

𝒌−𝟏𝒌

Fall 2020

Thermal efficiency using Cold Air Standard Brayton

Cycle analysis

Note also that it is for an ideal cycle, which does not account for

❑ variation of properties with temperature

❑ components inefficiency

❑ friction

❑ heat losses

❑ combustion losses, etc. 38


𝒓𝒑

𝒌−𝟏𝒌

Where rp is the pressure ratio

𝒓𝒑=𝑷𝒎𝒂𝒙

𝑷𝒎𝒊𝒏


𝑷𝟐𝑷𝟏

𝒌−𝟏𝒌

Recall :

Fall 2020

We can plot the Brayton cycle efficiency

5 10 15 200

0.2

0.4

0.6

0.8

k = 1.5k = 1.4k = 1.3

Pressure Ratio

Th

erm

al E

ffic

ien

cy

.

Real Gas Turbine Cycle

−−=

k

k

p

cycle

r

1

11

39Fall 2020

• Ideal Cycle

ηth

Pressure Ratio, rp

100%

0%1

Ideal- The efficiency

continually

increases with

increasing

pressure ratio

ηth

Pressure Ratio, rp

Increasing Tmax

Real Cycle

• Real Cycle

- Efficiency and

work output peek as

the pressure ratio

increases.

- Pressure ratio for

maximum

efficiency and work

output increase

with Tmax.

40Fall 2020

• Due to the irreversibilities and losses, the actual T-s diagrams would be as illustrated

41

Effects of Irreversibilities

Fall 2020

42

𝜼𝒄 =𝒘𝟏−𝟐𝒔

𝒘𝟏−𝟐=𝒉𝟐𝒔 − 𝒉𝟏𝒉𝟐 − 𝒉𝟏

𝜼𝒕 =𝒘𝟑−𝟒

𝒘𝟑−𝟒𝒔=

𝒉𝟑 − 𝒉𝟒𝒉𝟑 − 𝒉𝟒𝒔

Isentropic efficiency

For compressor For turbine

Fall 2020

Compressor

Combustor

Turbine

Inlet Air Exhaust

Power

Fuel

Shaft

• Brayton cycle is used in several gas turbine engines.

43

Applications of Brayton Cycle

Fall 2020

OGT 2500 Turboshaft

(Electricity generation)

T56 Turboshaft (Turboprop)

(Thrust generation)

Turbojet 44Fall 2020

Turboprop Turbofan

Ramjet or Scramjet

Different engine types:

45Fall 2020

2

expansion)c(isentropiexit at

expansion)(real

2

exita

V

V t

nozzle

Brayton cycle for jet engines (turbojet)

46Fall 2020

F119-PW-100 Afterburning Turbojet

47Fall 2020

Examples from textbook

Example 9.4: Analyzing

the Ideal Brayton Cycle

Example 9.6 : Evaluating

Performance of a

Brayton Cycle with

Irreversibilities

Fall 2020 48

Example 9.4: Analyzing the Ideal Brayton

Cycle

49

Find: Determine the thermal

efficiency, the back work

ratio, and the net power

developed

Known: An ideal air-

standard Brayton cycle

operates with given

compressor inlet conditions,

given turbine inlet

temperature, and a known

compressor pressure ratio.

Fall 2020

Example 9.4 (continued)

50Find: Determine the thermal efficiency, the back work

ratio, and the net power developedFall 2020

Solution Approach

51Fall 2020

Fixing the states

State 1:

𝑇1 = 300 𝐾 → 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 → 𝒉𝟏 = 𝟑𝟎𝟎. 𝟏𝟗𝒌𝑱

𝒌𝒈; 𝑷𝒓𝟏 = 𝟏. 𝟑𝟖𝟔

State 2:

𝑃𝑟2

𝑃𝑟1=

𝑃2

𝑃1(for isentropic process only)

𝑃𝑟2 = 𝑃𝑟1𝑃2

𝑃1= 1.386 10 = 13.86

Pr2 = 13.86 → 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 → 𝒉𝟐 = 𝟓𝟕𝟗. 𝟗𝒌𝑱

𝒌𝒈

52

State 3:

𝑇3 = 1400 𝐾 → 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 → 𝒉𝟑 = 𝟏𝟓𝟏𝟓. 𝟒𝒌𝑱

𝒌𝒈; 𝑷𝒓𝟑 = 𝟒𝟓𝟎. 𝟓

State 4:

𝑃𝑟4

𝑃𝑟3=

𝑃4


𝑃𝑟4 = 𝑃𝑟3𝑃4𝑃3

= 450.51

10= 45.05

𝑃𝑟4 = 45.05 → 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 → 𝒉𝟒 = 𝟖𝟎𝟖. 𝟓𝒌𝑱

𝒌𝒈

Fixing the states (continued)

Fall 2020

Substituting values in solving equations

53

A) Find thermal efficiency

𝑤𝑡 = ℎ3 − ℎ4 = 1515.4 − 808.5𝑘𝐽

𝑘𝑔= 706.9

𝑘𝐽

𝑘𝑔

𝑤𝑐 = ℎ2 − ℎ1 = 579.9 − 300.19𝑘𝐽

𝑘𝑔= 279.7

𝑘𝐽

𝑘𝑔

𝑞𝑖𝑛 = ℎ3 − ℎ2 = 1515.4 − 579.9𝑘𝐽

𝑘𝑔= 935.5

𝑘𝐽

𝑘𝑔

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑡−𝑤𝑐

𝑞𝑖𝑛=

𝟕𝟎𝟔.𝟗−𝟐𝟕𝟗.𝟕𝑘𝐽

𝑘𝑔

𝟗𝟑𝟓.𝟓𝑘𝐽

𝑘𝑔

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟒𝟓. 𝟕% → 𝑨𝑵𝑺𝑾𝑬𝑹 (𝒂)

Fall 2020

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑡 −𝑤𝑐𝑞𝑖𝑛

B) Find Back Work

Ratio (bwr)

𝑏𝑤𝑟 =𝑤𝑐𝑤𝑡

=𝟐𝟕𝟗. 𝟕𝟏

𝑘𝐽𝑘𝑔

𝟕𝟎𝟔. 𝟗𝑘𝐽𝑘𝑔

𝒃𝒘𝒓 = 𝟑𝟗. 𝟔%→ 𝑨𝑵𝑺𝑾𝑬𝑹 (𝒃)

Fall 2020 54

55

C) Find Net Power Developed (in kW)

Fall 2020

ሶ𝑾𝑵𝒆𝒕 = ሶ𝑾𝒄𝒚𝒄𝒍𝒆 = ሶ𝑾𝒕 − ሶ𝑾𝒄 = ሶ𝒎(𝒘𝒕 −𝒘𝒄)

𝑾𝒆 𝒉𝒂𝒗𝒆 𝒂𝒍𝒓𝒆𝒂𝒅𝒚 𝒇𝒐𝒖𝒏𝒅 "𝒘𝒕" 𝒂𝒏𝒅 "𝒘𝒄"𝑵𝒐𝒘𝒘𝒆 𝒏𝒆𝒆𝒅 𝒕𝒐 𝒇𝒊𝒏𝒅 ሶ𝒎

ሶ𝒎 = ሶ𝒎𝟏 = 𝝆𝟏 𝑽𝑨 𝟏 =𝑽𝑨 𝟏

𝒗𝟏

56Fall 2020

𝑤ℎ𝑒𝑟𝑒 𝑣𝑜𝑙𝑢𝑚𝑒𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑉𝐴 1 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛

𝑣1 =𝑅𝑎𝑖𝑟𝑇1

𝑃1(𝑓𝑟𝑜𝑚 𝑝𝑣 = 𝑅𝑇)

Where 𝑅𝑎𝑖𝑟 =ത𝑅

𝑀𝑎𝑖𝑟=

8.314𝑘𝐽

𝑘𝑚𝑜𝑙.𝐾

28.97𝑘𝑔

𝑘𝑚𝑜𝑙

= 0.287𝑘𝐽

𝑘𝑔.𝐾

Recall : ሶ𝒎 = ሶ𝒎𝟏 = 𝝆𝟏 𝑽𝑨 𝟏 =𝑽𝑨 𝟏

𝒗𝟏

𝑣1 =0.287

𝑘𝐽

𝑘𝑔.𝐾300 𝐾

100 𝑘𝑃𝑎= 0.861 (

𝑘𝐽

𝑘𝑔.𝑘𝑃𝑎)

𝑘𝑁.𝑚

𝑘𝐽

𝑘𝑃𝑎

𝑘𝑁/𝑚2

𝑣1 = 0.861𝑚3

𝑘𝑔

𝑾𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒗𝒐𝒍𝒖𝒎𝒆 "𝒗𝟏"

57

ሶ𝑚 =𝑽𝑨 𝟏

𝒗𝟏=

5𝑚3

𝑠

0.861𝑚3

𝑘𝑔

= 5.807𝑘𝑔

𝑠

Substituting all values in solving equation:

ሶ𝑊𝑁𝑒𝑡 = ሶ𝑚(𝑤𝑡 − 𝑤𝑐)

ሶ𝑊𝑁𝑒𝑡 = (5.807𝑘𝑔

𝑠)(706.9 − 279.7)

𝑘𝐽

𝑘𝑔

ሶ𝑾𝑵𝒆𝒕 = 𝟐𝟒𝟖𝟏 𝒌𝑾 → 𝑨𝑵𝑺𝑾𝑬𝑹 (𝒄)

Fall 2020


𝒗𝟏Recall :

Example 9.6 : Evaluating Performance of a

Brayton Cycle with Irreversibilities

58

pr

Known: An air-standard Brayton cycle

operates with given compressor inlet

conditions, given turbine inlet temperature,

and known compressor pressure ratio. The

compressor and turbine each have an

isentropic efficiency of 80%.

Find: Determine the thermal efficiency, the

back work ratio, and the net power

developed, in kW

Fall 2020


59

pr

Find: Determine the thermal

efficiency, the back work

ratio, and the net power

developed, in kWFall 2020

Fixing the states

60

State 1:

𝑇1 = 300 𝐾 ⇒ 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 ⇒ ℎ1 = 300.19𝑘𝐽

𝑘𝑔𝑃𝑟1 = 1.386

State 2s:𝑃𝑟2𝑠

𝑃𝑟1=

𝑃2𝑠


𝑃𝑟2𝑠 = 𝑃𝑟1𝑃2𝑠

𝑃1= 1.386 10 = 13.86

𝑃𝑟2𝑠 = 13.86 ⇒ 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 ⇒ ℎ2𝑠 = 579.9𝑘𝐽

𝑘𝑔

Fall 2020

61

State 2:

𝜂𝑐 =𝑤1−2𝑠

𝑤1−2=

ℎ2𝑠−ℎ1

ℎ2−ℎ1

ℎ2 = ℎ1 +ℎ2𝑠−ℎ1

𝜂𝑐

ℎ2 = 300.19𝑘𝐽

𝑘𝑔+

579.9 − 300.19𝑘𝐽𝑘𝑔

0.8

ℎ2 = 649.8𝑘𝐽

𝑘𝑔

Fall 2020

ℎ2𝑠 = 579.9𝑘𝐽

𝑘𝑔Recall :

Fall 2020 62

State 3:

𝑇3 = 1400 𝐾 ⇒ 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 ⇒ ℎ3 = 1515.4𝑘𝐽

𝑘𝑔𝑃𝑟3 = 450.5

State 4s:𝑃𝑟4𝑠

𝑃𝑟3=

𝑃4𝑠


𝑃𝑟4𝑠 = 𝑃𝑟3𝑃4𝑠𝑃3

= 450.51

10= 45.05

𝑃𝑟4𝑠 = 45.05 ⇒ 𝑇𝑎𝑏𝑙𝑒 𝐴. 22 ⇒ ℎ4𝑠 = 808.5𝑘𝐽

𝑘𝑔

Fall 2020 63

State 4:

𝜂𝑡 =𝑤3−4

𝑤3−4𝑠=

ℎ3−ℎ4

ℎ3−ℎ4𝑠

ℎ4 = ℎ3 − 𝜂𝑡(ℎ3 − ℎ4𝑠)

ℎ4 = 1515.4𝑘𝐽

𝑘𝑔− 0.8 1515.4 − 808.5

𝑘𝐽

𝑘𝑔

ℎ4 = 949.9𝑘𝐽

𝑘𝑔

ℎ4𝑠 = 808.5𝑘𝐽

𝑘𝑔Recall :

Substituting values in solving equations

64

A) Find thermal efficiency

𝑤𝑡 = ℎ3 − ℎ4 = 1515.4 − 949.9𝑘𝐽

𝑘𝑔= 565.5

𝑘𝐽

𝑘𝑔

𝑤𝑐 = ℎ2 − ℎ1 = 649.8 − 300.19𝑘𝐽

𝑘𝑔= 349.6

𝑘𝐽

𝑘𝑔

𝑞𝑖𝑛 = ℎ3 − ℎ2 = 1515.4 − 649.8𝑘𝐽

𝑘𝑔= 865.6

𝑘𝐽

𝑘𝑔

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑡−𝑤𝑐

𝑞𝑖𝑛=

565.5−349.6𝑘𝐽

𝑘𝑔

865.6𝑘𝐽

𝑘𝑔

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟐𝟒. 𝟗% ⇒ 𝑨𝑵𝑺𝑾𝑬𝑹 (𝒂)

Fall 2020

65

B) Find Back Work Ratio (bwr)

𝑏𝑤𝑟 =𝑤𝑐𝑤𝑡

=349.6

𝑘𝐽𝑘𝑔

565.5𝑘𝐽𝑘𝑔

= 61.8% ⇒ 𝑨𝑵𝑺𝑾𝑬𝑹 (𝒃)

Fall 2020

66

C) Find Net Power Developed (kW)

ሶ𝑊𝑁𝑒𝑡 = ሶ𝑊𝑐𝑦𝑐𝑙𝑒 = ሶ𝑊𝑡 − ሶ𝑊𝑐 = ሶ𝑚(𝑤𝑡 − 𝑤𝑐)

ሶ𝑚 = ሶ𝑚1 = 𝜌1 𝑉𝐴 1 =𝑉𝐴 1

𝑣1𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑉𝐴 1 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛.

𝑣1 =𝑅𝑎𝑖𝑟𝑇1

𝑃1(𝑓𝑟𝑜𝑚 𝑝𝑣 = 𝑅𝑇)

with 𝑅𝑎𝑖𝑟 =ത𝑅

𝑀𝑎𝑖𝑟=

8.314𝑘𝐽

𝑘𝑚𝑜𝑙.𝐾

28.97𝑘𝑔

𝑘𝑚𝑜𝑙

= 0.287𝑘𝐽

𝑘𝑔.𝐾

Fall 2020

Where:

67

Substituting values:

𝑣1 =0.287

𝑘𝐽

𝑘𝑔.𝐾300 𝐾

100 𝑘𝑃𝑎= 0.861 (

𝑘𝐽

𝑘𝑔.𝑘𝑃𝑎)

𝑘𝑁.𝑚

𝑘𝐽

𝑘𝑃𝑎

𝑘𝑁/𝑚2

𝑣1 = 0.861𝑚3

𝑘𝑔

Fall 2020

𝑣1 =𝑅𝑎𝑖𝑟𝑇1𝑃1

(𝑓𝑟𝑜𝑚 𝑝𝑣 = 𝑅𝑇)Recall:

Fall 2020 68

Mass flow rate


𝒗𝟏=

5𝑚3

𝑠

0.861𝑚3

𝑘𝑔

= 5.8𝑘𝑔

𝑠

Substituting values in solving equation:ሶ𝑊𝑁𝑒𝑡 = ሶ𝑚(𝑤𝑡 −𝑤𝑐)

ሶ𝑊𝑁𝑒𝑡 = (5.8𝑘𝑔

𝑠)(565.5 − 349.6)

𝑘𝐽

𝑘𝑔

ሶ𝑾𝑵𝒆𝒕 = 𝟏𝟐𝟓𝟐 𝒌𝑾 ⇒ 𝑨𝑵𝑺𝑾𝑬𝑹 (𝒄)

End of Examples 9.4 & 9.6

Fall 2020 69

Regenerative Gas Turbines

70

Regenerator Effectivenessሶ𝑸𝒊𝒏

ሶ𝒎= 𝒒𝒊𝒏 = 𝒉𝟑 − 𝒉𝒙 𝜼𝒓𝒆𝒈. =

𝒉𝒙 − 𝒉𝟐𝒉𝟒 − 𝒉𝟐

ሶ𝑸𝒊𝒏 𝒇𝒓𝒐𝒎 𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍 𝒔𝒐𝒖𝒓𝒄𝒆𝒔 (𝒊. 𝒆. , 𝒉𝒐𝒕 𝒓𝒆𝒔𝒆𝒓𝒗𝒐𝒊𝒓)

Regenerator

(a heat exchanger)

Fall 2020

71

Ideal Gas Turbines with Reheat

Fall 2020

72

Ideal Gas Turbines with Reheat (continued)

(From referenced textbook)

For metallurgical reasons, the temperature of the

gaseous combustion products entering the turbine

must be limited. This temperature can be controlled

by providing air in excess of the amount required to

burn the fuel in the combustor (see Chap. 13).

As a consequence, the gases exiting the combustor

contain sufficient air to support the combustion of

additional fuel.

Some gas turbine power plants take advantage of

the excess air by means of a multistage turbine

with a reheat combustor between the stages.Fall 2020

Ideal Gas Turbines with Reheat (continued)

(From referenced textbook)

73

Despite the increase in net work with

reheat, the cycle thermal efficiency would

not necessarily increase because a greater

total heat addition would be required

Fall 2020

Two-stage compression with intercooling

74

Fig. 9.18 Two-stage compression with intercoolingFall 2020

75

Two-stage compression with

intercooling (continued)

The compressor work input can

be reduced by multistage

compression with intercooling

Fall 2020

EXAMPLE FROM TEXTBOOK

Example 9.7 : Evaluating Thermal Efficiency of a Brayton

Cycle with Regeneration

Fall 2020 76

Example 9.7 : Evaluating Thermal Efficiency

of a Brayton Cycle with Regeneration

77

pr

Known: A regenerative gas turbine

operates with air as the working

fluid. The compressor inlet state,

turbine inlet temperature, and

compressor pressure ratio are

known.

Find: For a regenerator

effectiveness of 80%, determine the

thermal efficiency. Also plot the

thermal efficiency versus the

regenerator effectiveness ranging

from 0 to 80%.

Fall 2020


78

pr

Find: For a regenerator effectiveness of 80%, determine

the thermal efficiency Fall 2020

Fall 2020 79

𝜼𝒄𝒚𝒄𝒍𝒆 =𝒘𝒕 −𝒘𝒄

𝒒𝒊𝒏

Find Thermal Efficiency

𝜼𝒄𝒚𝒄𝒍𝒆 =𝒉𝟑 − 𝒉𝟒 − (𝒉𝟐 − 𝒉𝟏)

(𝒉𝟑−𝒉𝒙)

Note: h1, h2, h3 and h4 are

the same as in Example 9.4

We need to find hx.

Note 1: On the Regenerator

(or Heat Exchanger)

80

𝜂𝑟𝑒𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝒐𝒓 =ℎ𝑥 − ℎ2ℎ4 − ℎ2

ℎ𝑥 − ℎ2 :Heat energy actually received by a unitmass of coolant in the “cooler stream"

ℎ4 − ℎ2 : Hypothetical maximum of heat energy that could be

received by a unit mass of coolant in the “cooler stream"

2

4

x

Hotter

stream

Cooler

stream

(4 to y)

(2 to x)

Efficiency of regenerator:

Where:

Fall 2020

81

൯𝒉𝒙 = 𝒉𝟐 + 𝜼𝒓𝒆𝒈(𝒉𝟒 − 𝒉𝟐

𝜂𝑟𝑒𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 =ℎ𝑥 − ℎ2ℎ4 − ℎ2

Recall:

𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝒉𝒙

Fall 2020

𝒉𝒙 = 𝟓𝟕𝟗. 𝟗𝒌𝑱

𝒌𝒈+ 𝟎. 𝟖 𝟖𝟎𝟖. 𝟓 − 𝟓𝟕𝟗. 𝟗

𝒌𝑱

𝒌𝒈

𝒉𝒙 = 𝟕𝟔𝟐. 𝟖𝒌𝑱

𝒌𝒈

Note: h1, h2, h3 and h4 are the same as in Example 9.4

2

4

x

Hotter

stream

Cooler

stream

(4 to y)

(2 to x)

Note 2: "𝒒𝒊𝒏"

82

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑡 − 𝑤𝑐𝑞𝑖𝑛

Where 𝒒𝒊𝒏 =ሶ𝑸𝒊𝒏

ሶ𝒎= 𝒉𝟑 − 𝒉𝒙

ሶ𝑸𝒊𝒏 𝒇𝒓𝒐𝒎 𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍 𝒔𝒐𝒖𝒓𝒄𝒆𝒔 (𝒊. 𝒆. , 𝒇𝒓𝒐𝒎 𝒂 𝒉𝒐𝒕 𝒓𝒆𝒔𝒆𝒓𝒗𝒐𝒊𝒓)

For Brayton Cycle with Regeneration

The heat energy that the working fluid receives in the

regenerator (𝐟𝐫𝐨𝐦 𝐬𝐭𝐚𝐭𝐢𝐨𝐧 𝟐 𝒕𝒐 𝒙) is not counted in the

cycle efficiency expression

It’s a free energy obtained from the exhaust gas !!!

Fall 2020

Fall 2020 83

𝜂𝑐𝑦𝑐𝑙𝑒 =𝑤𝑡 −𝑤𝑐𝑞𝑖𝑛

Substituting values

𝜼𝒄𝒚𝒄𝒍𝒆 =𝒉𝟑 − 𝒉𝟒 − (𝒉𝟐 − 𝒉𝟏)

(𝒉𝟑−𝒉𝒙)

𝜼𝒄𝒚𝒄𝒍𝒆 =𝟏𝟓𝟏𝟓. 𝟒 − 𝟖𝟎𝟖. 𝟓

𝒌𝑱𝒌𝒈

− 𝟓𝟕𝟗. 𝟗 − 𝟑𝟎𝟎. 𝟏𝟗𝒌𝑱𝒌𝒈

𝟏𝟓𝟏𝟓. 𝟒 − 𝟕𝟔𝟐. 𝟖𝒌𝑱𝒌𝒈

𝜼𝒄𝒚𝒄𝒍𝒆 = 𝟎. 𝟓𝟔𝟖 = 𝟓𝟔. 𝟖% ⇒ 𝑨𝒏𝒔𝒘𝒆𝒓

Related Notes on Regenerator

84

ሶ𝒎𝑯𝒐𝒕(𝒉𝟒 − 𝒉𝒚) ሶ𝒎𝑪𝒐𝒐𝒍(𝒉𝒙 − 𝒉𝟐)

Assuming the same mass flow rate in both hot and cool streams, then

The rate of heat

transfer from the

hotter stream

The rate of heat

received by the

cooler stream

(𝒉𝟒 − 𝒉𝒚) (𝒉𝒙 − 𝒉𝟐)

Ideally

Ideally

Ideally

Reality% 𝒐𝒇 (𝒉𝟒 − 𝒉𝒚) (𝒉𝒙 − 𝒉𝟐)

Fall 2020

2

4

x

Hotter

stream

Cooler

stream

9.4 SUMMARY (1)

• Cold Air Standard Otto Cycle

1

11

−−=

k

v

Ottor

- Used to model spark-ignition internal combustion

engines

rv: Volumetric

Compression Ratio

3

2

4

v=c

v=c

1

85Fall 2020

9.4 SUMMARY (2a)

121221For hhw −=−

433443For hhw −=−

232332For hhq −=−

144114For hhq −=−

NOTE: w’s and q’s are written such that all have positive values

Air Standard Brayton Cycle

1-2 Compression: isentropic


3-4 Expansion: isentropic


‘

‘‘

2

3

4p=c

p=c

1

- Used to model gas turbine engines

86Fall 2020

9.4 SUMMARY (2b)

87

)(

)(1

23

14

hh

hhcycle

−

−−=

Brayton’s Cycle thermal efficiency

Fall 2020

9.4 SUMMARY (3)

• Cold Air Standard Brayton Cycle

- Used to model gas turbine engines

−−=

k

k

p

Brayton

r

1

11

Pressure Ratio rp = pmax / pmin

‘

‘

3

2

4

p=c

p=c

1

88Fall 2020

Regenerative Gas Turbines

89

Regenerator Effectiveness

𝜼𝒓𝒆𝒈. =𝒉𝒙 − 𝒉𝟐𝒉𝟒 − 𝒉𝟐

9.4 SUMMARY (4)

Regenerator

(a heat exchanger)

ሶ𝑸𝒊𝒏

ሶ𝒎= 𝒒𝒊𝒏 = 𝒉𝟑 − 𝒉𝒙

ሶ𝑸𝒊𝒏 𝒇𝒓𝒐𝒎 𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍 𝒔𝒐𝒖𝒓𝒄𝒆𝒔 (𝒊. 𝒆. , 𝒉𝒐𝒕 𝒓𝒆𝒔𝒆𝒓𝒗𝒐𝒊𝒓)Fall 2020

chapter 9 gas power cycles - vinhqtang.com

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