chalmers university of technology elementary axial turbine theory –velocity triangles –degree of...
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• Elementary axial turbine theory– Velocity triangles– Degree of reaction– Blade loading coefficient, flow coefficient
• Problem 7.1
• Civil jet aircraft performance
Lecture 7 – Axial flow turbines
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Axial flow turbines
• Working fluid is accelerated by the stator and decelerated by the rotor
– Expansion occurs in stator and in relative frame of rotor
• Boundary layer growth and separation does not limit stage loading as in axial compressor
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Elementary theory• Energy equation for control
volumes (again):
0103
0103
21
1
23
3
00103
22TTchh
Vh
Vhwq p
gasPerfect
hh
• Adiabatic expansion process (work extracted from system - sign convention for added work = +w)– Rotor => -w = cp(T03-T02) <=>
w = cp(T02-T03)– Stator => 0 = cp(T02-T01)
=> T02= T01
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How is the temperature drop related to the blade angles ?
• We study change of angular momentum at mid of blade (as approximation)
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Governing equations and assumptions• Relative and absolute refererence
frames are related by:
23
velocityrelativefor direction
of change Assume
23
22332233
2233
radiusconstant at Flow
wwww
wwww
ww
CCUCCU
UCUCrCrCorkspecific wlTheoretica
torquespecificlTheoreticarCrC
momentumangularspecificofchangeofRate
UCV • We only study designs where:
– Ca2=Ca3
– C1=C3
• We repeat the derivation of theoretical work used for radial and axial compressors:
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Principle of angular momentum
Stage work output w:
3322
32
tantan aa
ww
CCU
CCUw
Ca constant:
32
3322
tantan
tantan
a
aa
UC
CCUw
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(1)
(2)
(1)+(2) =>
3232 tantantantan
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Combine derived equations =>
32 tantan aUCw
stagep Tcw ,0
Exercise: derive the correct expression when 3 is small enough to allow 3 to be pointing in the direction of rotation.
(7.3) tantan 32,0 astagep UCTc
Energy equation
Energy equation:
We have a relation between temperature drop and blade angles!!! :
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Dimensionless parametersBlade loading coefficient
(7.6) tantan2
7.3Equation
21 32
2
,0
U
C
U
Tcastagep
Degree of reaction31
32
TT
TT
Exercise: show that this expression is equal to =>when C3= C1 0301
32
TT
TT
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can be related to the blade angles!
C3 = C1 =>
32,0 tantan UCTcTc stagepstagep
Relative to the rotor the flow does no work (in the relative frame the blade is fixed). Thus T0,relative is constant =>
22
3222
22
3 tantan2
1
2
1 arotorp CVVTc
Exercise: Verify this by using the definition of the relative total temperature: p
relative c
VTT
2
2
,0
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can be related to the blade angles!
Plugging in results in definition of =>
(7.7) tantan2 23
31
32
U
C
TT
TT a
The parameter quantifies relative amount of ”expansion” in rotor. Thus, equation 7.7 relates blade angles to the relative amount of expansion. Aircraft turbine designs are typically 50% degree of reaction designs.
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Dimensionless parameters Finally, the flow coefficient:
5.0
0.50.3
0.18.0
Current aircraft practice (according to C.R.S):
U
Ca
Aircraft practice => relatively high values on flow and stage loading coefficients limit efficiencies
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Dimensionless parameters Using the flow coefficient in 7.6 and 7.7 we obtain:
(7.8) tantan2 32
The above equations and 7.1 can be used to obtain the gas and blade angles as a function of the dimensionless parameters
22
1
2
1 tan 2
22
1
2
1 tan 3
1
tan tan 22
1
tan tan 33
(7.9) tantan2
23
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• Exercise: show that the velocity triangles become symmetric for = 0.5. Hint combine 7.1 and 7.9
• Exercise: use the “current aircraft practice” rules to derive bounds for what would be considered conventional aircraft turbine designs. What will be the range for 3? Assume = 0.5.
Two suggested exercises
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Turbine loss coefficients:Nozzle (stator) loss coefficients:
202
0201
22
22
2
pp
ppY
cC
TT
N
p
N
3,03
,03,02
22
33
2
pp
ppY
cV
TT
rel
relrelR
p
R
Nozzle (rotor) loss coefficients:
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Problem 7.1
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Civil jet aircraft performance
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Four forces of flight
L ift
W eight = m g
D rag
Thrust
V
x
y
Resulting force perpendicular
to the flight path
Net thrust from the engines
resulting force parallell to the flight path
α angle of attack
V velocity
sincos gmDFdt
dVm
Newton’s second law
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Aerodynamic equations
• L=Lift = q·S·CL [N]
• D=Drag = q·S·CD [N]
• q = dynamic pressure [N/m²]
• S = reference wing area [m²]
• CL = coefficient of lift CL = f(α,Re,M)
• CD = coefficient of drag CD = f(α,Re,M)
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Reference wing areaThe area is considered to extend without interruption
through the fuselage and is usually denoted S.
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Lift versus angle of attack
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numberMachoffunctionasCD
DC
1M Mach
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The ISA AtmosphereFrom lecture 5
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Equations
222
2222
2
1
2
12
1
2
1
2
1
2
SLSL
a
SL
Sl
SLSLSL
aMT
TTRM
TRMaMVq
TRp
a
V
a
VM
TRap
p
T
T
SL
2
2
1Mp
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Lift equation
LL
LLSL
LSLSLLSL
LL
CSMCSMconstL
CSMconstCSMp
CSMaCSaM
CSaMCSVL
242
22
constant
2222
222
100928,7
2
12
1
2
12
1
2
1
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Drag equation
DD
DDSL
DSLSLDSL
DD
CSMCSMconstD
CSMconstCSMp
CSMaCSaM
CSaMCSVD
242
22
2222
222
100928,7
2
12
1
2
12
1
2
1
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Drag polarL
D
R
R
C L
C D
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High speed drag polar
LC
0.02 0.04 0.06 0.08 0.1 0.12 0.140.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Mach 0,63Mach 0,73Mach 0,80Mach 0,82Mach 0,83Mach 0,85
DC
LC
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A flight consists of:
• Taxi
• Take off
• Climb
• Cruise
• Descent
• Approach and landing
• Diversion to alternate airport?
Sector Distance
Flight Time & Fuel
Block Time & Fuel
En route Climb
Descent
Approach &Landing
1500 ft
Initial Cruise
Step Cruise
Takeoff &Initial ClimbStart-up
&Taxi-out
Taxi-in
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CruiseFor an airplane to be in level, unaccelerated flight, thrust and drag
must be equal and opposite, and the lift and weight must be equal andopposite according to the laws of motion, i.e.
Lift = Weight = mgThrust = Drag
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Range
m
s
sN
kgnconsumptiofuelspecific
F
mSFC
s
kgflowfuelm
s
mvelocityV
mrangeR
f
f
a
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Range
START
END
END
START
END
START
END
START
W
W
SLa
W
W
SL
W
W
W
W
a
a
af
a
ff
a
W
dW
D
LM
SFCg
aR
W
dW
D
LM
SFCg
a
W
dW
D
LM
SFCg
a
W
dW
D
LV
SFCgR
W
dW
D
LV
SFCgDFWL
WFSFCg
dWWVdR
FSFCg
dWV
FSFC
dmVdR
FSFC
dm
m
dm
V
dRdt
F
mSFC
dt
dmmV
dt
dR
1
1&
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Breguet range equation
END
STARTSL
W
W
SLa W
W
D
LM
SFCg
a
W
dW
D
LM
SFCg
aR
START
END
ln
For a preliminary performance analysis is the range equation usually simplified. If we assume flight at constant altitude, M, SFC and L/D the range equation becomes
which is frequently called the Breguet range equation
END
STARTa W
W
D
LV
SFCgR ln
1
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Breguet range equationThe Breuget range equation is written directly in terms of SFC. Clearly maximum range for a jetaircraft is not
dictated by maximum L/D, but rather the maximum value of the product M(L/D) or V(L/D).
ENDSTARTD
La
W
WD
L
D
La
D
La
D
L
D
L
L
LL
WWC
C
SSFCgR
W
dW
C
C
SSFCgW
dW
C
C
SSFCgdR
W
dW
C
C
S
W
SFCgW
dW
D
LV
SFCgdR
C
C
S
W
C
C
CS
W
D
LV
CS
WVCSVWL
END
START
22
2121
211
22
2
2
1 2
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Breuget range equationFrom the simplified range equation, maximum range is obtained from
• Flight at maximum
• Low SFC
• High altitude, low ρ
• Carrying a lot of fuel
D
L
C
C
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Range
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.82
4
6
8
10
12
14
16
Mach 0,63Mach 0,73Mach 0,80Mach 0,82Mach 0,83Mach 0,85
LC
D
LM
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Endurance
END
START
W
W
W
W
W
W
f
ff
W
W
D
L
SFCgt
simplifiedusuallyaboveequationtheisionapproximatfirstafor
W
dW
D
L
SFCgW
dW
D
L
SFCgW
dW
D
L
SFCgt
W
dW
D
L
SFCgDFWL
FSFCg
dW
FSFC
dm
m
dmdt
F
mSFC
dt
dmm
START
END
END
START
END
START
ln1
:
111
1&
Endurance is the amout of time that an aircraft can stay in the air on one given load of fuel
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Endurance
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.94
6
8
10
12
14
16
18
20
Mach 0,63Mach 0,73Mach 0,80Mach 0,82Mach 0,83Mach 0,85
D
L
LC
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Learning goals
• Understand how turbine blade angles relate to work output
• Be familiar with the three non-dimensional parameters: – Blade loading coefficient, degree of reaction,
flow coefficient
• Understand why civil aircraft are– operated at high altitude– at flight Mach numbers less than 1.0– How engine performance influences aircraft range