401
11.2 Thermodynamic Modelling
The thermal mixing equation determining T 2 reads:
(11.30)
This implies
(11.31)
The heat capacities in (11.30) do not exactly equal those in (11.29), but we neglect
the small differences. As a result thereof, calculation of the work and the final
temperature of the expansion becomes very simple.
Assuming that the cooling air does not contribute any momentum to the main
flow, the decrease of the momentum flow rate within the main flow, thus the
decelerating force, is v m c
, with v the average through-flow velocity just before
mixing. The corresponding pressure drop is −dp A g , with A g the through-flow area.
The gas flow rate is g
g
m
vA
r
=
.
So:
where M is the Mach number of the flow just before mixing. The mixing occurs
nearby the trailing edge, i.e. where the Mach number is close to 1. The pressure loss
thus ensues from
(11.32)
We may assume M ≈ 0.8, so that μ ≈ 0.85. In reality, the cooling air supplies some
momentum to the main flow. This moderates the pressure loss coefficient, with a
realistic value about μ ≈ 0.7 [4].
11.2.7 Compression with Extraction
The calculation of the cooled turbine cannot be separated from the calculation of
the compressor from which the cooling air is extracted. In practice, there are only
a few extractions (2–3). In order to become independent of the precise position of
these extractions, the extractions are distributed here over the compression. This is
achieved by dividing the entire pressure interval in the compressor and the turbine
into an equal number of subintervals. This is done according to a power series,
so that the pressure ratio of the fictitious stages always is the same. We choose
e.g. as many stages as the overall pressure ratio amounts to (40 stages at pressure
ratio 40). Due to the pressure drop in the combustion chamber, corresponding points
g pg 2a
2
c pc c
2
m c (T
T ) m c (T T ) 0.
−
+
−
=
2a
c
2
T
T
T
.
1
d
d
+
=
+
2
2
2
or
,
c
c
c
g
g
g
m
m
m
dp
v
dp
v
M
m
p m p
m
r
r
g
− =
−
=
=
with
pg
2
2
c
2'
2
2
g
pc
c
m
p
p
M
M .
p
m
c
g
d m
m
g
−
=
=
=
11.2 Thermodynamic Modelling
The thermal mixing equation determining T 2 reads:
(11.30)
This implies
(11.31)
The heat capacities in (11.30) do not exactly equal those in (11.29), but we neglect
the small differences. As a result thereof, calculation of the work and the final
temperature of the expansion becomes very simple.
Assuming that the cooling air does not contribute any momentum to the main
flow, the decrease of the momentum flow rate within the main flow, thus the
decelerating force, is v m c
, with v the average through-flow velocity just before
mixing. The corresponding pressure drop is −dp A g , with A g the through-flow area.
The gas flow rate is g
g
m
vA
r
=
.
So:
where M is the Mach number of the flow just before mixing. The mixing occurs
nearby the trailing edge, i.e. where the Mach number is close to 1. The pressure loss
thus ensues from
(11.32)
We may assume M ≈ 0.8, so that μ ≈ 0.85. In reality, the cooling air supplies some
momentum to the main flow. This moderates the pressure loss coefficient, with a
realistic value about μ ≈ 0.7 [4].
11.2.7 Compression with Extraction
The calculation of the cooled turbine cannot be separated from the calculation of
the compressor from which the cooling air is extracted. In practice, there are only
a few extractions (2–3). In order to become independent of the precise position of
these extractions, the extractions are distributed here over the compression. This is
achieved by dividing the entire pressure interval in the compressor and the turbine
into an equal number of subintervals. This is done according to a power series,
so that the pressure ratio of the fictitious stages always is the same. We choose
e.g. as many stages as the overall pressure ratio amounts to (40 stages at pressure
ratio 40). Due to the pressure drop in the combustion chamber, corresponding points
g pg 2a
2
c pc c
2
m c (T
T ) m c (T T ) 0.
−
+
−
=
2a
c
2
T
T
T
.
1
d
d
+
=
+
2
2
2
or
,
c
c
c
g
g
g
m
m
m
dp
v
dp
v
M
m
p m p
m
r
r
g
− =
−
=
=
with
pg
2
2
c
2'
2
2
g
pc
c
m
p
p
M
M .
p
m
c
g
d m
m
g
−
=
=
=