Thermodynamics 1 - ocw.tudelft.nl · Thermodynamics 1 Lecture 13: ... Thijs Vlugt. Theo Woudstra....

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March 22, 2010 1 Thermodynamics 1 Lecture 13: Kringprocessen Gasturbines Joule-Brayton Bendiks Jan Boersma Wiebren de Jong Thijs Vlugt Theo Woudstra Energy Technology

Transcript of Thermodynamics 1 - ocw.tudelft.nl · Thermodynamics 1 Lecture 13: ... Thijs Vlugt. Theo Woudstra....

March 22, 2010

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Thermodynamics 1Lecture 13: KringprocessenGasturbinesJoule-Brayton

Bendiks Jan BoersmaWiebren de JongThijs VlugtTheo Woudstra

Energy Technology

Energy Technology

Lecture 13, March 22, 2010 2

College 12• Diesel process• dual cycle

Energy Technology

Lecture 13, March 22, 2010 3

Recapitulation• Diesel cycle (s, p, s, v). ->Compression ratio

-> cut-off ratio

• Relation work per cycle and power.

• Dual cycle (s, v, p, s, v)

1

2

VrV

=

3

2c

Vr

V=

Energy Technology

Lecture 13, March 22, 2010 4

New• Joule-Brayton-process (gas-turbine process)

Energy Technology

Lecture 13, March 22, 2010 5

Simple gas-turbine cycle: configurations

Open Closed

Energy Technology

Lecture 13, March 22, 2010 6

Advanced gas-turbine cycle: combined cycle

• Higher thermal efficiencies as hot flue gas from gas turbine is used as input heat for a steam cycle

Energy Technology

Lecture 13, March 22, 2010 7

A biomass gasification integrated combined cycle (IGCC)

Varnamo (South Sweden)18 MWth

Energy Technology

Lecture 13, March 22, 2010 8

Gas-turbines: equipment

Marine application, propulsion:

Energy Technology

Lecture 13, March 22, 2010 9

Gas-turbines: equipment

Energy Technology

Lecture 13, March 22, 2010 10

Aviation, important:

High power to mass ratioCompactLow emissions (CO,UHC)

Gas-turbines: equipment

Energy Technology

Lecture 13, March 22, 2010 11

Joule – Brayton cycle (p,s)

( ) ( ) ( )2 212i o i o i oW Q m h h V V g z z⎧ ⎫= + − + − + −⎨ ⎬

⎩ ⎭

( )12

1 2

W

m h h

=

41 0W =1

2 3

4

v

p T

s

1

2

3

4p1

p2

p1

p2

( )34 3 4W m h h= −

23 0W =

34 0Q =

12 0Q =

( )41 1 4Q m h h= −

( )23

3 2

Q

m h h

=

o

i

algemeen:

d

mits reversibel

W m v p= ∫

Energy Technology

Lecture 13, March 22, 2010 12

Irreversibilities in the cycle

.

.

.

.

43

43t

t

ts

W

m

W

m

h hh h

s

η

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

−=

−=

.

.

.

.

12

12c

c

c

S

W

m

W

m

h hh h

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

−=

−=

Energy Technology

Lecture 13, March 22, 2010 13

Aircraft gas-turbine

General Electric F110-GE-100 Augmented Turbofan Engine -> T2 ~ 600 °C

1

2 3

5

v

T

s

1

2

3

4

54

p

compressor

combustor

turbine

jetV

1

2 34

5

Energy Technology

Lecture 13, March 22, 2010 14

Brayton-cycle 23Q

W

41Q

compressor

turbine

1

2 3

4

23 41W Q Q= +

4

141 1 4 1

3 2 223 23 3

2

11 1 1

1

TTQ T T TW

T T TQ Q TT

η

⎛ ⎞−⎜ ⎟

− ⎝ ⎠= = + = + = −− ⎛ ⎞

−⎜ ⎟⎝ ⎠1 1

4 4 3 3

1 1

1 1 2 2

T p T p

T p T p

κ κκ κ

κ κκ κ

− −− −

− −− −

=

=

Poisson3 34 2

1 2 4 1

en ookT TT T

T T T T= =

Cold air standard

Energy Technology

Lecture 13, March 22, 2010 15

Brayton cycle

1

21

TT

η = −1

212

1

pT T p

κκ⎛ ⎞

⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

=

( )1

1

2

2

1

TT

pp

κκ

− −

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

=

-> derived in working class 4!!

( )1

2

1

11

pp

κκ

η−

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

⇒ = −

0

10

20

30

40

50

60

0 2 4 6 8 10 12

Compressor pressure ratio (-)

η (

%)

Practical limit:Turbine inlet T3

Energy Technology

Lecture 13, March 22, 2010 16

System design considerations

Energy Technology

Lecture 13, March 22, 2010 17

Regenerative Gas Turbines

2

4 2reg

xh hh h

η −=

−Regenerator effectiveness:

60% 80%regη< <Typically,

Energy Technology

Lecture 13, March 22, 2010 18

Example problem 1

An internally reversible air-standard Brayton cycle receives air at 27 oC and 103 kPa. The upper limits of pressure and temperature of the cycle are 517 kPa and 316 oC. Determine the thermal efficiency of the cycle, assuming constant specific heat, Cp=1,0035 kJ/(kg.K).

P1 =103 kPaT1 =27 oC=300 K

P2 =517 kPaT2 =?

P3 =P2T3 =316 oC=589 K

P4 =P1T4 =?

Energy Technology

Lecture 13, March 22, 2010 19

Example problem 2

A regenerator of 75% effectiveness is used in an air standard Brayton cycle working between pressures of 1030 hPa and 5,17 bar.

Determine the work per kg of air and the thermal efficiency of the cycle if the maximum and minimum temperatures of the cycle are 671,3 oC and 32,4 oC, respectively. Assume constant Cp of 1,004 kJ/(kg.K) and κ=1,4.

Energy Technology

Lecture 13, March 22, 2010 20

Sensitivity study on influence of regenerator effectiveness

(example problem 2)

0

10

20

30

40

50

60

0 20 40 60 80 100 120Regenerator effectiveness (%)

Ther

mal

effi

cien

cy (%

)

Energy Technology

Lecture 13, March 22, 2010 21

Study-suggestions

• Study Chapter 9.1 - 9.8 and 9.11• Do some of the exercises 9.28 - 9.35 (Brayton-cycles)

Especially: finish 9.28, do 9.29• Exercises 9.36-9.39 regenerator• Problems 9.40 - 9.44 reheat• Problems 9.45 - 9.50 intercooling• 9.51 intercooling and reheat• 9.52 intercooling, reheat and regenerator