Lecture 33
Lect 33
Matching of Engine Components
Lect 33
• Any variable in the engine is expressed as a function of ṁ f , P a , T a and V.
• Because the engine is self contained; fixing one set of engine non-dimensional parameters fixes all others.
V e
P e
V a
Lect 33
• Instantaneous cycle temperature ratio, T 03 /T 01 fixes the cycle pressure ratio π c , the instantaneous fuel
flow ṁ f and the instantaneous rotational speed, n.
• Similarly in a multi-shaft engine the ratio between the shaft speeds, n 2 /n 1 is also fixed by T 03 /T 01 .
• For most operating conditions of the engine the final propelling nozzle for the core flow and the bypass flow will be chocked, by design, for maximizing, thrust.
• Thus the engine responds to the inlet stagnation
conditions but is unaware of the forward speed.
Lect 33
Non-dimensional Variables of the Engine
1) Consider mass flow of air through the engine.
This can be written,
as a function of the rotational speed N of the shafts ṁ a = f (N, P 01 ,T 01 )
or, in terms of turbine inlet temperature ṁ a = f (T 03 , P 01 ,T 01 )
or, in terms of fuel flow rate
ṁ a = f (ṁ f , P 01 ,T 01 )
Lect 33
The non-dimensional mass flow rate is derived from dimensional analysis (e.g. Buckingham ∏ Theorem), as
01 2
01
.
) .
( .
P D
T c
m m a p
=
Where, D is a characteristic diameter of the engine,
typically the diameter of the inlet of the fan and D 2
denotes a representative area
Lect 33
Functional
dependence of
engine parameters is seen in engine characteristics
The lines of constant T
04and constant rpm are assumed parallel, so that once one
variable is chosen
the other variables
are also fixed .
Lect 33
Practical Normalizing of Parameters
For mass flow of air c p is constant and D 2 is constant for a given engine.
it has units.
Normalized fuel flow the abbreviated form is derived,
Non-dimensional speed is abbreviated : N/(T 02 ) 1/2 The corrected mass flow (kg/s) is
Where. Θ = T 02 / T 02ref , δ = P 02 /P 02ref, and T 02ref , P 02ref are STP
01
. 01
P T m m a
=
03 03 .P T
m f m f
=
δ
= m a . Θ m
Lect 33
Matching Procedure :
1) Select operating point (Altitude, Flight Condition) - P a , T a , and M
2) From the ambient condition obtain - P 01 , T 01 3) Select max Turbine entry temp. T 03
4) Select Rotational speed, N – then obtain 5) Obtain - N/√T 01 , and N/√T 03
6) Select a compressor pressure ratio , π 0C = P 02 /P 01 7) Obtain mass flow parameter,
8) The parameters in (5), (6) and (7) completely define the compressor operation point
01
P 01
T
m
Lect 33
01
P
01T m
π
Lect 33
9) Actual mass flow through the compressor is :
10) The turbine mass flow is :
where, f - is fuel/air ratio, b = air bleed
and Turbine entry Pressure, P 03 = P 02 (1 - ΔP cc ) 11) Based on the actual mass flows through the compressor and the turbine, the work done for these mass flows need to be equated (turbojet engine). The same may be done spool–wise.
01 01 01
c 01 T
P P
T
m = m .
b) f
.(1 m
m T = c + −
Lect 33
12) Work equivalence :
for a turbojet engine, or for a spool, i.e.
for LP spool
f or fan-turbine or
for HP spool
η mech
T .
T . W c m
W c .
m =
LP -
η mech
LPT .
LPT LPC
LPC . W m . W m =
HP -
η mech
HPT .
HPT HPC
HPC . W m . W m =
FT -
η mech
LPT2 .
LPT2 Fan
Fan . W m . W m =
Where η
mechis the
Mechanical efficiency of
the shaft
Lect 33
13) Compressor /Fan specific work (per unit mass flow)
−
=
= ∆
−
P 1 P η
T C
η
T
W C
airair
γ γ 1
01 02 c
01 air
- p c
012 air
- p c
−
=
∆
= −
gas gas
γ γ 1 03 03
gas - p T 034
gas - p T
T
P P 1 1
T η C
T η .C
W
14) Turbine Specific work (per unit mass) :
Lect 33
01 03 air
gas 03
02 02
01 01
air 01 03
gas 03
T T m
m P
P P
P P
T m
P
T m
. .
= .
15) Turbine mass flow parameter can be determined from :
16) The net turbine – compressor excess power (if any) may now be decided by:
mech c air
T gas
engine
η
w . w m
. m
P
−
=
For pure turbojet P engine = 0,
For multi-spool turbofan P spool =0
Where η
mechis the
Mechanical efficiency of
the shaft
Lect 33
• The power or work matching between compressor and turbine has to be exact.
• If they are unequal, either a new speed of the engine or new values of the compressor pressure ratio & mass flow are selected to try and arrive at perfect matching.
• For turboprop or turboshaft the excess shaft power P engine = P propeller/ rotor and requires exact matching.
• If they are unequal – either the engine speed setting (N) or mass flow setting (ṁ) , or
compression ratio π c , or TET, T 03 is to be selected --- or -- a new propeller/ rotor pitch setting needs to be selected.
• Logic of this selection has to be built in to the
control system logic of the engine (FADEC)
Lect 33
In case of a choked nozzle, the actual mass flow is invariant with change of other parameters:
const
=
=
01 03 air
gas 03
02 02
01 01
air 01 03
gas 03
T T m
m P
P P
P P
T m
P
T m
. .
.
Assume that pressure ratio across the combustion chamber, P 02 /P 03 = constant and = m air m gas
constant a
1 is K
,
. where
01 03 01
air 01 01
02 T
T P
T m
K 1 P
P
=
Lect 33
Choked flow
Unchoked
flow
Lect 33
assume that cycle temperature ratio is constant, then
constant another
2 is K
,
. where
01 air 01 01
02 P
T m
K 2 P
P
=
For a straight and level cruise flight
constant another
3 is K
,
. where
01 03 01
02 T
T K 3
P
P =
constant another
4 is K
, where K 4
P P
01 02 =
If T 01 / T 03 is held constant for a cruise flight
Lect 33
Off-Design Matching of Turbojet Engine
2 1
01 .
2 . 1
1
−
+ −
=
airair
a air
I a
P M
P γ
γ γ
η
−
+
= 2
01 .
2
1 air 1 a
a
T M
T γ
The Intake delivery Pressure and Temperature can be found from
• Higher the flight Mach number, M a higher would be P 01 and T 01 at any constant altitude
• For constant flight Mach number, M a - the P 01 and T 01
decreases with increase of altitude & vice versa
Lect 33
• The Ram pressure development in the intake
increases/ decreases the compressor inlet and then compressor outlet pressure.
• It then increases / decreases the turbine inlet / outlet pressure
• Thus the pressure ratio across the nozzle increases / decreases
• At high nozzle pressure ratio, the flow is choked – it is then independent of nozzle pressure ratio – and hence from flight speed
• This fixes the turbine operating point w.r.t nozzle
choked condition .
Lect 33
An aero engine is choked most of the time of flight except during – Taxiing, Approaching and Landing The nozzle pressure ratio may be computed from
a
a P
P P
P P
P P
P P
P 01
01 02 02
03 03
04
04 = . . .
Where, P 01 /P a is obtained from Intake analysis
Lect 33
a a
flight 5
5 5
5
R.T .
M V
by, given
is speed flight
where
), (
) (
γ
=
− +
−
= m V V flight A P P a
F
Thrust of the engine may be written as:
The gas exhaust speed V 5 depends on the flow
condition at nozzle exit, and with chocking reaches the maximum for that condition. For example,
some time during the climb operation if the nozzle
is choked, it will remain so during the climb , with
continuous fall in ambient pressure P a .
Lect 33
1 . . R.T 2
. V
by, given
is locity exhaust ve
where
), (
) (
04 5
5
5 5
5 5
= +
=
− +
−
=
gas gas gas
a flight
T R
P P
A V
V m
F
γ γ γ
For choked nozzle:
For un-choked nozzle :
−
=
−
=
− γ γ
γ
1
04 04
5 04
gas - p
5 . C .( ) 2 . 1
V P
T P C
T
T a
p gas
gas
Lect 33
For choked nozzle:
1 04
5 1
. 1 1 1
−
+
− −
=
=
gasgas
gas gas nozzle
c P
P P
γ γ
γ γ η
For un-choked nozzle :
P 5 = P a
Lect 33
• The engine performance seems to be decided by engine normalized speed N/√T 01 , but the maximum performance is capped by the engine speed, N max .
• This max speed is decided by stress limits of rotating components.
• At N max /√T 01 as the ambient temperature increases thrust will decrease, but engine speed cannot be
increased much more.
• The engines are often designed for 15 0 C, 288K. They are bound to give lower performance in tropical
atmospheres & higher performance in colder climates.
Lect 33
Next class :
Engine Component Matching and Sizing
