Open Cycle
mf_dot fuel
QH_dot
combustor
3
2
ma_dot air compressor
4
turbine
W_dotnet=
Wt_dot+Wc_dot
1
(
turbine ...
(
)
mf_dot ⎞
⎛
Wt_dot = −ma_dot + mf_dot ⋅ h 3 − h 4 = −ma_dot⋅ ⎜ 1 +
⋅ T − T4
⎟ ⋅c
ma_dot p_prod 3
⎠
⎝
(
Jet engine
)
Wc_dot = −ma_dot⋅ h 2 − h 1 = −ma_dot⋅ cp_air⋅ T2 − T1
power ... compressor ...
)(
)
(
)
as a side note: if the net work were converted to velocity via a nozzle (jet engine) the relationships would be Wnet_dot = Wt_dot − Wc_dot
determines state 4 out of turbine at p 4 > p1 atmosphere is state 5
(
wnet = cp ⋅ T3 − T4
T 4 determined from equation for net work
)
γ−1
could determine p 4 from
nozzle anlysis:
First law, Q = W = 0
⎛ p4 ⎞
=⎜ ⎟
T3
⎝ p3 ⎠
T4
γ−1
γ
determine T 5 from
⎛ p5 ⎞
=⎜ ⎟
T3
⎝ p3 ⎠
T5
γ
or ...
γ−1
⎛ p5 ⎞
= ⎜ ⎟
T4
⎝ p4 ⎠
T5
γ
2
V
h4 = h5 +
2
determines V, thrust from momentum change
combustor ...
1 = atmosphere ...
0 = HR2 − HP3
rewrite using LHV ...
adiabatic combustion Q = W = 0
0 = Enthaply of reactants at combustor inlet, compressor outlet
- Enthalpy of products out of combustor - first law
(
)
0 = HR2 − HR0 − HP3 − HP0 + LHV
rewrite using specifi enthalpy and mass flows ... on a per unit mass flow of fuel ...
ma_dot
ma_dot ⎞
⎛
0 = h f2 − h f0 +
⋅ h a2 − h a0 − ⎜ 1 +
⎟ ⋅ h p3 − h p0 + LHV
mf_dot
mf_dot
⎝
⎠
(
)
(
to account for incomplete combustion introduce combustion efficiency ...
12/19/2005
1
)
only obtain
η comb⋅HV
Given
ma_dot
ma_dot ⎞
⎛
0 = h f2 − h f0 +
⋅ h a2 − h a0 − ⎜ 1 +
⎟ ⋅ h p3 − h p0 + η comb⋅LHV
mf_dot
mf_dot
⎝
⎠
(
can solve for
)
) (
(
ma_dot
introduce average
specific heat ...
cp_bar_air =
)
(
) (
(
)
h p3 − h p0 − h a2 − h a0
h a2 − h a0
cp_bar_prod =
T2 − T0
(
)
h p3 − h p0
T3 − T0
(
cp_bar_fuel =
η comb⋅LHV + cp_bar_fuel⋅ T2 − T0 − cp_bar_prod⋅ T3 − T0
=
(
)
(
cp_bar_prod⋅ T3 − T0 − cp_bar_air⋅ T2 − T0
mf_dot
or ... inverting
ma_dot
)
η comb⋅LHV + h f2 − h f0 − h p3 − h p0
=
mf_dot
mf_dot
)
mf_dot
Find ma_dot → h f2 − h f0 − h p3 + h p0 + η comb⋅LHV ⋅
−h a2 + h a0 + h p3 − h p0
ma_dot/mf_dot
ma_dot
(
=
(
)
)
T2 − T0
)
(
cp_bar_prod⋅ T3 − T0 − cp_bar_air⋅ T2 − T0
(
h f2 − h f0
)
(
)
η comb⋅LHV + cp_bar_fuel⋅ T2 − T0 − cp_bar_prod⋅ T3 − T0
gas turbine efficiency efficiency dividing by m a_dot
mf_dot ⎞
⎛
1
+
⋅ ( T − T4 ) − cp_bar_air⋅ ( T2 − T1 )
⎜
⎟ ⋅c
Wnet_dot
Wt_dot + Wc_dot
ma_dot p_bar_prod 3
⎝
⎠
η=
=
=
mf_dot ⋅LHV
mf_dot ⋅LHV
mf_dot
ma_dot
kg⋅
SFC =
hr
power = kW
SFC =
12/19/2005
fuel
mf_dot
Wnet_dot
=
=
kg
kW⋅ hr
mf_dot
=
⋅
lb
not equality ...
hp⋅
hr
LHV
Wnet_dot LHV
=
1
η ⋅ LHV
2
⋅LHV
)
Open cycles have similar alternatives to closed
analysis would be similar as well so not repeated here
Open Cycle Regenerative (Recouperative)
6
regenerator
3
combustor
mf_dot fuel
mp_dot products
2
5
4
ma_dot air compressor
turbine
W_dotnet=
Wt_dot+Wc_dot
1
static data for plot
T-s diagram
temperature
1000
irreversible cycle
irreversible, heat exchanger maximum
regeneration inlet temperature irreversible
500
1
1.2
1.4
1.6
1.8
2
entropy
N.B. cycle is drawn closed from state 6 to 1 but is taking place in atmosphere
12/19/2005
3
2.2
Intercooled Regenerative (Recouperative) Cycle
stack
Rolls Royce WR-21 is an example see links
regenerator
8
5
combustor
mf_dot fuel
6 mp_dot products
QL_dot
1 compressor
2
ma_dot air
7
4
3
turbine
compressor
W_dotnet=
Wt_dot+Wc_dot
static data for plot
T-s diagram
1200
irreversible cycle
irreversible, heat exchanger maximum
regeneration inlet temperature irreversible
temperature
1000
800
600
400
0.8
1
1.2
1.4
entropy
12/19/2005
4
1.6
1.8
2
thermodynamic models for combustion
Various thermodynamic models can be used for analysis of products of combustion:
1. Single gas model
perfect gas, constant cp (1 kJ/kg*K close enough), γ = 1.4
2. Two gas model
a) perfect gas - air for compression, c p = 1.0035 kJ/kg*K, γa = 1.4
b) perfect gas combustion products; c pp = 1.13 kJ/kg*K, γp = 1.3
3. Tabulated data (e.g. Keenan & Kaye Gas Tables)
property data for air:
Table 1: Air at low pressure: T deg F abs, t deg F, h, pr, u, v r, φ
Table 2. Air at low pressures: T, t, c p , cv, k = cp /cv, a, Gmax/pi , μ, λ, Pr
Table 3: R Log N for air
Table 4: Products - 400% Theoretical Air (for One Pound Mole)
Table 5: Products - 400% Theoretical Air (for One Pound Mole) fuel data
Table 6: Products - R_bar Log e N +4.57263 n
Table 7: Products - 200% Theoretical Air (for One Pound Mole)
etc. data for oxygen, hydrogen, carbon monoxide, dioxide etc.
T = deg F abs
t = deg F
h = enthalpy per unit mass
pr = relative pressure
c p = specific heat at constant pressure
c v = specific heat at constant volume
G = flow per unit area or mass velocity
k = cp /cv
u = internal energy per unit mass
vr = relative volume
⌠
⎮
φ= ⎮
⎮
⌡
p = pressure
Pr = Prandtl number = cp*μ/λ
R = gas constant for air
a = velocity of sound
λ = thermal conductivity
μ = viscosity
T
cp
T
dT
T0
Notes: Appendix (Sources and methods
- " ...calculated for one particular composition of the hydrocarbon fuel, it has been shown that it represents
with high precision the properties of the productsof combustion of fuels of a wide range of composition - all
for 400% theoretical air." page 205 bottom
- problems involving intermediate mixtures to Table B:
can be solved by interpolation based on theoretical air
or ... extrapolated to 100% for products is valid except for effects of disassociation
.products
.
reactants
air_and
⎛ .
⎞
⎜ Table %theor %theor %theor water_vapor ⎟
⎜
⎟
Number
air
fuel
fuel
mass_%_water ⎟
⎜
Table_B =
⎜ 1
⎟
inf
0
0
0
⎜ 4
⎟
400
25
14
6.7
⎜
⎟
200
50
28
.
⎝
7
⎠
12/19/2005
5
4. Polynomial equations
- example in combustion example c p = f(θ
polytropic process
compressor
isentropic process
dT
dp
ds = cpo⋅
− R⋅
T
p
(7.21) in gas relationships
R⋅
dp
dT
= η pc⋅ cpo⋅
p
T
T
dp
p
=
dT
1
⋅c ⋅
R po T
T
⌠ 1s
⎛ p1 ⎞ 1
ln⎜ ⎟ = ⋅ ⎮
⎝ p 2 ⎠
R ⎮
⌡T
p1
cp
T
p2
dT
=e
1s
cp
η pc ⌠
⎮
⋅
dT
T
R ⎮
⌡T
2
turbine
2
T
p1
p2
=e
⌠ 1s c
p
1 ⎮
⋅
dT
T
R ⎮
⌡T
T
⌠ 1s c
p
⋅⎮
dT
⎮
T
η pt⋅R
⌡T
1
p1
2
p2
=e
High Temperature Gas Turbines
Advantages:
high efficiency - low specific fuel consumption
high specific horsepower - small size and weight
Disavantages:
materials strength problems (Creep) see separate notes re: creep
corrosion
Solutions:
better materials
blade and combustor cooling
ceramic materials
12/19/2005
6
2
blade cooling
mf_dot fuel
QH_dot
combustor
3
2
4
to blades
ma_dot air compressor
turbine
W_dotnet=
Wt_dot+Wc_dot
1
cooling flow
compressed air ducted into stationary AND rotor blades. Temperature reuduced by:
convective heat transfer
transpiration (evaporation of water from surface)
film
nominal data for plot
Cooling effectiveness (nominal)
effectiveness %
60
40
20
convective
film
transpiration
0
0
1
2
3
4
cooling flow/turbine flow %
cooling_effectiveness =
12/19/2005
Tblade_gas − Tblade_metal
Tblade_gas − Tcooling_air
7
5
6
1500 K
T3
nominal ΔT over stages defining where
cooling is required
1st nozzle @ stator
1st rotor
2nd nozzle @ stator
2nd rotor
metal temperature
@ critical value
800 K
T4
Ceramic materials
examples silicon nitride, silicon carbide
can be pressed, bonded and/or sintered to produce complete rotor system
25 _deg_C ( 75_deg_F)
⎛ .
⎜
.
MPa
tensile_strength = ⎜
552
⎜ Si3 ⋅ N4
⎜
193
⎝
Si⋅ C
12/19/2005
.
.
1400 _deg_C ( 2500_deg_F) ⎞
ksi MPa
ksi
80 172
25
28 138
20
8
⎟
⎟
⎟
⎟
⎠
Intercooled Regenerative Gas Turbine
typically two spool design
stack
regenerator
10
combustor
5
mf_dot fuel
6 mp_dot products
QL1_dot
3
2
7
4
1
9
lp
compressor
power
turbine
lp
turbine
hp
turbine
hp
compressor
ma_dot air
8
W_dotnet=
Wt_dot+Wc_dot
6
7
8
T
5
9
10
4
2
3
1
s
powers ... (review) reversible
(
)
HP_comp = −mair_dot⋅ ( h 4 − h 3 )
HP_turb = ( mair_dot + mfuel_dot) ⋅ ( h 6 − h 7 )
(
LP_comp = −mair_dot⋅ h 2 − h 1
(
)(
LP_turb = mair_dot + mfuel_dot ⋅ h 7 − h 8
(
)(
12/19/2005
)
w_dot LP_comp = −w_dot LP_turb
w_dot HP_comp = −w_dot HP_turb
mair_dot
)
Power_turb = mair_dot + mfuel_dot ⋅ h 8 − h 9
)(
QH_dot = mair_dot + mfuel_dot ⋅ h 6 − h 5
mfuel_dot
)
9
from combustion analysis
Marinization
Problems:
1. sea water droplets in air (inlet)
2. sea water in fuel
3. coupling to the propeller
4. long ducting
Solutions:
1. sea water in air
1. design of inlet - demisters to remove droplets
demisters
wire mesh
inertial separation
2. select corrosion resistant materials
3. surface treatment of components - plating to improve corrosion resistance
4. water washing and abrasive cleaning
2. sea water in fuel
1. treat to remove sodium
3. coupling to propeller (later)
4. long ducting
inlet and exit pressures reduce the pressure ratio across turbine
reduction in power
increase in fuel consumption
additional effect from inlet density
p ⋅ v = R⋅T =>
p
ρ
= R⋅ T
ρ =
p
R⋅ T static data for plot
12/19/2005
10
T-s diagram
irreversible cycle
nominal pressure losses inlet and exaust
temperature
1000
500
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
entropy
similar effect for T inlet > nominal
cycle will walk up p1 curve
normally cannot increase T H to account for these losses
other issues/topics
Materials
coatings
use of titanium
fuel treatment
sodium - bad - corrosion from products
remove by washing
add agents such as demulsifiers
water combines with sodium - remove by centrifuge
vanadium - in Bunker C combines with sulfur - creates corrosive combustion products
GE fro example has an additive to modify ash to prevent adhering to blades
problem 3 above: coupling to propeller
1. Controllable Reversible Pitch Propeller (CRP)
2. reversing gearbox
3. electric drive
4. reversing turbine
concentric opposite direction direction blade annuli
12/19/2005
11
Brayton cycle applied to turbocharging reciprocating engines
power ...
mf_dot fuel
in
(
comp = −mair_dot⋅ h 2 − h 1
out
2
ma_dot air compressor
(
) (
turb = mair_dot + mfuel_dot ⋅ h 3 − h 4
stack
3
)
4
)
turbine
1
(
) (
)
(
w_dot comp + w_dot turb = 0 = mair_dot + mfuel_dot ⋅ h 3 − h 4 − mair_dot⋅ h 2 − h 1
mfuel_dot ⎞
⎛
h2 − h1 = ⎜ 1 +
⎟ ⋅ h3 − h4
mair_dot
⎝
⎠
(
)
)
p3 may be > or < p2 depending on
what happens in engine
combined cycles - gas turbine and Rankine - or other
3
T
maximum available power from T 4 -> T5
4
2
5
1
s
⎛ w_dot rev ⎞
⎜
⎟
= ψ 4 − ψ 5 = h 4 − T0 ⋅ s4 − ( h 5 − T0 ⋅ s5 )
= h 4 − h 5 − T0 ⋅ (
s4 − s5 )
⎝
m_dot ⎠
max
second law ...
T⋅ ds = dh − v ⋅ dp
assuming cpp constant
ds =
dh
T
if ...
=
cpp⋅dT
p 4 = p 5 = p atmos
=>
T
⎛ w_dot rev ⎞
⎛
⎛ T4 ⎞ ⎞
⎜
⎟
= ψ 4 − ψ 5 = cpp⋅ ⎜ T4 − T5 − T0 ⋅ ln⎜
⎟⎟
⎝
m_dot ⎠
max
⎝
⎝ T5 ⎠ ⎠
12/19/2005
12
⎛ T4 ⎞
s4 − s5 = cpp⋅ ln⎜
⎟
⎝ T5 ⎠
dp = 0
ds =
dh
T
example LM 2500
T4 := 825
K
T0 := 300
K
GT_power := 330
⎛ 500 ⎞
T5 := ⎜ 400 ⎟
⎜ ⎟
⎝
300 ⎠
⎛ 325 ⎞
T4 − T5 = ⎜ 425 ⎟
⎜ ⎟
⎝
525 ⎠
kJ
kg⋅ K
⎛ w_dot rev ⎞
⎜
⎟
= W_m_dot_max
⎝ m_dot ⎠
max
⎛ T4 ⎞
T0 ⋅ ln⎜
⎟=
T5
⎝ ⎠
⎛ 150.233 ⎞
⎜ 217.176 ⎟
⎜
⎟
⎝
303.48 ⎠
⎛
⎛ T4 ⎞ ⎞
W_m_dot_max := cp_prod⋅ ⎜ T4 − T5 − T0 ⋅ ln⎜
⎟ ⎟ ⋅K
T5
⎝
⎝ ⎠⎠
⎛ 188.749 ⎞
kJ
W_m_dot_max = ⎜ 224.45 ⎟
⎜
⎟ kg
⎝
239.241 ⎠
W_m_dot_max
GT_power
12/19/2005
kJ := 1000J
kg
s
cp_prod := 1.08
1 < cp_prod < 1.33
kW
⎛ 0.572 ⎞
= ⎜ 0.68 ⎟
⎜
⎟
⎝
0.725 ⎠
13