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5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
1/44
Chalmers University of Technology
Lecture 3 Some more thermodynamics: Brief discussion of cycle efficiencies - continued
Ideal cycles II Heat exchanger cycle
Real cycles Stagnation properties, efficiencies,
pressure losses The Solar Mercury 50
Real cycles Mechanical efficiencies Specific heats (temperature variation) Fuel air ratio, combustion and cycle
efficiencies Bleeds
Jet engine nozzles Radial compressor I
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
2/44
Chalmers University of Technology
Generalization of the Carnot efficiency
etemperaturhotaverageT
etemperaturcoldaverageT
T
T
T
T
H
L
H
L
H
Lcarnotth
11,
=
=
==
Is generalization of Carnot efficiency
to Brayton cycle possible?
Define average temp. to value that would
give the same heat transfer, i.e.:
sTTdsq
sTTdsq
Lout
Hin
==
==
1
4
3
2
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
3/44
Chalmers University of Technology
Generalization of Carnot efficiency
But for the isobar we have,
2
3
0
2
3
2
3 lnlnlnT
Tc
P
PR
T
Tcs pp ==
=
Thus, the average temperature is obtained from (dp=0):
2
3
2323
3
2
3
22
3
ln
][ln
T
TTTTcdTcrelationscombineTds
TTcTsT HpppHH ======
Derive an expression for the lower average temperature in the same way.
PdvTdsdu =
Furthermore, we have Gibbs equation
(Cengel and Boles):
as well as: vdppdvdhpvhddudwdq === )(
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
4/44
Chalmers University of Technology
Generalization of Carnot efficiency
( ) ( )
Carnotth
H
L
H
L
Braytonth
T
T
T
T
TT
TTTT
,
2
3
32
1
4
41
1
4
2
3
4
3
1
2
32
41
23
1243,
11
T
Tln
T
T
Tln
T
-1
T
T
T
T
T
T
T
T
T
T-1
===
=
===
=
=
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
5/44
Chalmers University of Technology
When T4> T2a heat exchanger can be introduced.
Thi i true !hen" )1(21
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
6/44
Chalmers University of Technology
#ere !e obtain the e$$icienc%" &ot inde'endento$ T3 (i'le c%cle
i inde'endent o$ t3)
*o!er out'ut i una$$ected b% heat exchanger ince the
turbine and co'reor !or+ are the ae a in the i'le c%cle.
t
r
T
TT
T
T
T
T
TT
T
TT
T
T
T
T
TT
TT
TT
TTTT
TTc
TTcTTc
p
pp
1
2
13
1
4
1
1
24
1
21
4
3
1
2
43
12
43
1243
3
1243
111
)1(
)1(
1)(
)(1
)(
)()(
)(
)()(
===
=
==
=
=
=
Theory 3.1 Ideal heat exchanger cycle
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Heat exchange cycle
er% high e$$iciencie can be theoreticall% be obtained
#eat exchanger etallurgical liit !ill be releant.
T4 / 1000.0 /> 0>
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Heat exchange cycle
o! 'reure ratio
/> high e$$icienc%
What ha''en
!ith the aerage
te'erature at
!hich heat i
addedre5ected
!hen the
'reure ratio
change in heat
exchange c%cle6
qinqin
qout qout
TH TH
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
9/44
Chalmers University of Technology
Cycles with losses
a. Change in kinetic energy between inlet and outletmay not be negligible :
b. Fluid friction =>
- burners- combustion chambers- exhaust ducts
++
++= iinletsall
ii
ii
eexitsall
ee
ee gz
V
hmgz
V
hm!
2
2
22
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
10/44
Chalmers University of Technology
Cycles with losses
c. Heat exchangers.Economic size =>terminal temperature
difference, i.e. T5 < T4.d. Friction losses in shaft, i.e.
the transmission of turbine power tocompressor. Auxiliary power requirementsuch as oil and fuel pumps.
e. and c pvary with temperature and gas
composition.
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Cycles with losses
f. Efficiency is defined by SFC(specific fuel consumption = fuel
consumption per unit net work
output). Cycle efficiency
obtained using fuel heating value.
g. Cooling of blade roots and
turbine disks often require
approximately the same mass
flow of gas as fuel flow => air
flow is approximated as constantfor preliminary calculations. This
is done in this course.
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Stagnation properties For high-speed flows, the potential energy of the fluid can still
be neglected but the kinetic energy can not!
++
++=
=
++
++=
0
1
2
111
0
2
2
222
2
2
0102
22]out'utingle-in'utingle[
22
gzV
hmgzV
hm
gzV
hmgzV
hm!
hh
iinletsall
ii
ii
eexitsall
ee
ee
It is convenient to combine the static temperature and the kinetic
energy into a single term called the stagnation (or total) enthalpy,
h0=h+V2/2, i.e. the energy obtained when a gas is brought to rest
without heat or work transfer
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Stagnation properties
0102
21
1
22
2
0102
22hh
Vh
Vhwq
hh
=++=
For a perfect gas we get the stagnation temperature T0,
according to:
p
ppc
VTT
VTcTc 22
2
0
2
0 +=+=
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Stagnation pressure
Defined in same manner as stagnation temperature (noheat or work transfer) with added restriction
retardation is thought to occur reversibly
Thus we define the
stagnation pressure p0by:
Note that for an isentropic
process between 02 and 01
we get
100
=
TT
PP
1
01
021
01
1
1
2
2
021
01
11
1
21
2
02
01
1
1
2
2
02
01
02
=
=
==
T
T
T
T
T
T
T
T
T
T
T
T
T
T
P
P
P
P
P
P
P
P
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Compressor and turbine efficiencies
Isentropic efficiency (compressors and turbines are approximatelyadiabatic => if expansion is reversible it is isentropic).The isentropicefficiency is for the compressor is:
0
7
0
7
0
7
0
TcTc
hh
p
pc ==
pp cc ,7
Where are the averaged specific heats of the temperature
intervals 01-02and 01-02 respectively.
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Compressor and turbine efficiencies
Ideal and mean temperature
differences are not very
different. Thus it is a good
approximation to assume:
We therefore define:
pp cc 7
0102
0102
TT
TTc
=
Similarly for the turbine:
0403
0403
TT
TTt
=
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Compressor and turbine efficiencies
Using produces the
frequently used expressions:
1
04
03
04
03
1
01
02
01
02 and
== PP
TT
PP
TT
(2.11)1
1
01
02010102
=
P
PTTT
c
(2.12)1
1
1
04
03
030403
=
P
PTTT
t
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Turbine efficiency options
If the turbine exhausts directlyto atmosphere the kinetic energyis lost and a more properdefinition of efficiency would
be:
1
1
1
03
030403
=
a
t
P
PTTT
In practice some of the kinetic energy is recovered in an
exhaust diffuser => turbine pressure ratio increases.
Here we put p04=pafor gas turbines exhausting intoatmosphere and think oftas taking both turbine and
exhaust duct losses into account
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Turbine diffusers
Recovered
energy
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Heat-exchanger efficiency
( ) ( )0202,080448, TTcTTc pp =
Conservation of energy (neglecting energy transfer to surrounding):
In a real heat-exchanger T05will no longer equal T04(T05
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Pressure losses burners & heat-exchangers
Burner pressure losses
Flame stabilizing & mixing
Fundamental loss (Chapter 7 + Rayleigh-line
appendix A.4)
Heat exchanger pressure loss
Air passage pressure loss
Pha Gas passage pressure loss Phg
Losses depend on heat exchanger effectiveness. A
4% pressure loss is a reasonable starting point for
design.
engine)(aircra$t83
turbine)gal(indutria32
1
02
02
0202
0203
=
p
p
p
p
P
p
p
pPP
b
b
hab
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
The Solar Mercury 504.3 MW output= 40.5 %System was designed from
scratch to allow high performance
integration of heat-exchanger
C f
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Mechanical losses
Turbine power is transmitted directly from the turbine
without intermediate gearing => (only bearing and windage
losses). We define the transmission efficiency m:
( )010212,1
TTc pm
turbine =
Usually power to drive fuel and oil pumps are transmitted from the
shaft. We will assumem=0.99 for calculations.
Ch l U i it f T h l
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Temperature variation of specific heat
We have already established:
cp=f1(T)
cv=f2(T)
Since=cp/cvwe
have =f3(T)
The combustion product thermodynamic
properties will depend on T and f (fuel air
ratio)
Ch l U i it f T h l
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Pressure dependency?
$H$H 22221 +
=
P
P
P
P
P
P
%
H$
$H
22
2
2
1
At 1500 K dissociation begins to have animpact on cp and.
22
2
1C$$C$ +
Detailed gas tables for afterburnersmay include pressure effects. We exclude them in thiscourse.
Ch l U i it f T h l
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Temperature variation of specific heat
333.1,9+g114:
400.1,9+g100
==
==
gpg
apa
c
c
In this course we use:
Since gamma and cpvary in opposing senses some of the
error introduced by this approximation is cancelled.
Ch l U i it f T h l
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Calculate fthat gives T03for given T02? Use first law forcontrol volumes (q=w=0) and that enthalpy is a pointfunction(any path will produce the same result)
Determining the fuel air ratio
( ) ( ) ( ) ( ) ( ) 0
02203 2;:2;:2;:1++=+ #p#papg T#cTcH#Tc#
fis small (typically around 0.02) and cpfis also small => last term
is negligible. The equation determines f.
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Combustion temperature rise
Hypothetic fuel:
86.08% carbon
13.92% hydrogen
H25= - 43100 kj/kg
Curves ok for kerosene
burned in dry air. Not ok
in afterburner (fin0).
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Shaft cycle performance parameters
&w
#nconsumptio#uelspeci#ic'(C ==
pnet
net
#!
we##iciencycycle
,
==
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Bleeds
Combustor and turbineregions require most of thecooling air.
Anti-ice
Rule of thumb: take air asearly as possible (less workput in)
Accessory unit cooling (oilsystem, aircraft powersupply (generator), fuel
pumps) Air entering before rotor
contributes to work!
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
dragoentuinta+ethrutoentugro
a) mCmC
Thrust&et
momentumo#changeo#Rate
=
=
Aircraft propulsion thrust generation
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
( )
thrustpressure
a))
velocity*ircra#t
a) pp*CCm
Thrust&et
+
=
Jet engine principles of thrust generation
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Jet engine principles of thrust generation
0
22040
2
44
2
00
040
==++===
hhV
hV
hwq
hh
00 Tch p=
040 TT =
No heat or work transfer in the jet engine nozzle
Stagnation temperature
is constant
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Chalmers University of Technology
Mach number relations for stagnation properties
We have already introduced the
stagnation temperature as:pc
VTT
2
2
0 +=
and shown that (revision task): Rccvp
+=
The Mach number is defined as:
RT
V
soundo#speed
V
a
V+
===
additional loss =>
2 -doneor. =
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Overall pressure rise:
1
01
2
01
03 1
+=
Tc
-
p
p
p
c
P03is here used to denote the pressure at
compressor exit. P02is reserved for the stagnation
pressure between the impeller and the diffuser
vanes
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Example 4.1a
=1.04,= 0.90
N = 290.0 rev/s,
D = 0.5 m
Deye,tip = 0.3, Deye,root= 0.15 m = 9.0 kg/s
T01= 295 K
P01= 1.1 bar
c= 0.78
Compute pressure ratio and
power required
Chalmers University of Technology
5/19/2018 Chalmers University of Technology - Gas Turbine Technology Lecture 3
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Learning goals
Understand why the Carnot cycle can be used
for qualitative arguments also for the
Joule/Brayton cycle
Be able to state reasonable loss levels for gasturbine components turbine and compressor
performance are given in Lecture !" and include
them in cycle analysis
#now how to compute cycle efficiencies for the
heat e$changer cycle