Thermodynamics 1
Lecture 13:
Kringprocessen
Gasturbines
Joule-Brayton
Bendiks Jan Boersma
Wiebren de Jong
Thijs Vlugt
Theo Woudstra
March 22, 2010
1
Energy Technology
College 12
• Diesel process
• dual cycle
Lecture 13, March 22, 2010
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Recapitulation
• Diesel cycle (s, p, s, v).
->Compression ratio
V1
r =
V2
-> cut-off ratio
V3
rc =
V2
• Relation work per cycle and power.
• Dual cycle (s, v, p, s, v)
Lecture 13, March 22, 2010
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New
• Joule-Brayton-process (gas-turbine process)
Lecture 13, March 22, 2010
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Simple gas-turbine cycle: configurations
Open
Lecture 13, March 22, 2010
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Closed
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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
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A biomass gasification integrated
combined cycle (IGCC)
Varnamo (South Sweden)
18 MWth
Lecture 13, March 22, 2010
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Gas-turbines: equipment
Marine application, propulsion:
Lecture 13, March 22, 2010
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Gas-turbines: equipment
Lecture 13, March 22, 2010
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Gas-turbines: equipment
Aviation, important:
High power to mass ratio
Compact
Low emissions (CO,UHC)
Lecture 13, March 22, 2010
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Joule – Brayton cycle (p,s)
p
W23 = 0
p2
p1
2
T
algemeen:
Q23 =
o
W = m ∫v dp
3
m ( h3 − h2 )
i
mits reversibel
W
34
W12 =
m ( h1 − h2 )
1
= m ( h3 − h4 )
4
W41 = 0
v
3
2
p2
Q12 = 0
1
p1
Lecture 13, March 22, 2010
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Q41 = m ( h1 − h4 )
1 2
⎧
⎫
2
W = Q + m ⎨( hi − ho ) + V i −Vo + g ( z i − z o ) ⎬
2
⎩
⎭
(
Q34 = 0
s
)
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Irreversibilities in the cycle
.
.
ηt
⎛W
⎞
t
⎜
. ⎟
⎜
⎟
h −h
m
⎝
= . ⎠ = 3 4
h3 − h4s
⎛W
⎞
⎜ t . ⎟
⎜
⎟
⎝ m ⎠s
Lecture 13, March 22, 2010
Energy Technology
ηc
⎛W
⎞
c
⎜
. ⎟
⎜
⎟
h −h
m
⎝
= . ⎠s = 2S 1
h2 − h1
⎛W
⎞
⎜ c . ⎟
⎜
⎟
⎝ m⎠
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Aircraft gas-turbine
General Electric F110-GE-100 Augmented Turbofan Engine -> T2~ 600 °C
combustor
compressor
2
3
4
jet
5
V
p
2
T
3
3
2
4
4
turbine
1
Lecture 13, March 22, 2010
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1
5
5
1
v
s
13
Brayton-cycle
compressor
W = Q23 + Q41
2
W
1
Q41
W
η = =1+ Q 23
Q 23
Poisson T 4 p 4
−
T 1 p1
κ −1
κ
κ −1
κ
Lecture 13, March 22, 2010
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turbine
Cold air standard
−
Q23
−
= T3p3
−
= T 2 p2
κ −1
κ
κ −1
κ
Q41
4
⎛T 4
⎞
⎜ − 1⎟
T1 − T 4
T1 ⎝ T1
⎠
=1+
=1−
T3 −T2
T 2 ⎛T 3
⎞
⎜ − 1⎟
⎝T 2
⎠
T3 T2
T4 T3
=
en ook
=
T1 T 2
T 4 T1
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Brayton cycle
T
η =1− 1
T
p2 ⎟
T 2 =T p ⎟ -> derived in working class 4!!
1⎟
⎛
⎜
1 ⎜⎜
⎝
2
T1 ⎛⎜
=
T 2 ⎜⎜
⎝
(κ −1)
− κ −1
⎞ κ
2⎟
⎟
1 ⎟⎠
p
p
50
40
η (%)
⎛p ⎞ κ
⎜ 2⎟
⎜p ⎟
⎝ 1⎠
( )
⎠
60
1
⇒η =1−
κ −1
⎞ κ
30
20
Practical limit:
Turbine inlet T3
10
0
0
2
4
6
8
10
12
Compressor pressure ratio (-)
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System design considerations
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Regenerative Gas Turbines
Regenerator effectiveness:
Typically,
60% <
Lecture 13, March 22, 2010
Energy Technology
ηreg
ηreg < 80%
hx − h2
=
h4 − h2
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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).
P2=517 kPa
T2=?
P1=103 kPa
T1=27 oC=300 K
Lecture 13, March 22, 2010
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P3=P2
T3=316 oC=589 K
P4=P1
T4=?
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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.
Lecture 13, March 22, 2010
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Sensitivity study on influence of
regenerator effectiveness
(example problem 2)
60
Thermal efficiency (%)
50
40
30
20
10
0
0
20
40
60
80
100
120
Regenerator effectiveness (%)
Lecture 13, March 22, 2010
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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
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