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
Chalmers University of Technology
Generalization of the Carnot efficiency
TL
TL
th,carnot  1 
 1
TH
TH
TL  average cold temperature
TH  average hot temperature
Is generalization of Carnot efficiency
to Brayton cycle possible?
Define average temp. to value that would
give the same heat transfer, i.e.:
3
qin 
 Tds T
H
s
2
1
qout 
 Tds T
L
4
s
Chalmers University of Technology
Generalization of Carnot efficiency
But for the isobar we have,
0


T3
P3
T3
s  c p ln  R ln
 c p ln
T2
P2
T2
Furthermore, we have Gibbs equation
(Cengel and Boles):
du  Tds  Pdv
as well as:
dq  dw  du  d (h  pv)  dh  pdv  vdp
Thus, the average temperature is obtained from (dp=0):
T
T
TH s  TH c p ln 3   Tds  [combine relations ]   c p dT  c p T23  TH  23
T
T2 2
2
ln 3
T2
3
3
Derive an expression for the lower average temperature in the same way.
Chalmers University of Technology
Generalization of Carnot efficiency
th, Brayton

T3  T4   T2  T1 

T3  T2
T3 T4 
T41  T2 T3
 1  
 
T32  T1 T4
T2 T1 
T41
T4
ln
TL
TL
T1
1 1
 1
 th,Carnot
T32
TH
TH
T3
ln
T2
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Heat exchange cycle
When T4 > T2 a heat exchanger can be introduced.

This is true when:
1  r  t 2( 1)
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Theory 3.1 – Ideal heat exchanger cycle
Here we obtain the efficiency:

c p (T3  T4 )  c p (T2  T1 )
c p (T3  T5 )

(T3  T4 )  (T2  T1 )

(T3  T4 )
Not independent
of T3!!! (simple cycle
is independent of t3)
T2
 1
 1)
(T2  T1 )  T2 T3 
T1
T1
r
T1
1


 1
 1  1
 1
T2
T
(T3  T4 )  T1 T4 
T
t
4
T4 (  1)
T3 1
T1
T2
T1 (
Power output is unaffected by heat exchangers since the
turbine and compressor work are the same as in the simple cycle.
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Heat exchange cycle
Very high efficiencies can be theoretically be obtained!
Heat exchanger metallurgical limits will be relevant.
T4 = 1000.0 K
=> 
 70%
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Heat exchange cycle
What happens
with the average
temperature at
which heat is
added/rejected
when the
pressure ratio
changes in heat
exchange cycle?
TH
qin
qout
Low pressure ratio
=> high efficiency
qin
qout
TH
Chalmers University of Technology
Cycles with losses
a.
Change in kinetic energy between inlet and outlet
may not be negligible :




Ve2
Vi 2
Q  W   m e  he 
 gze    m i  hi 
 gzi 
2
2
all exits e

 all inletsi 

b. Fluid friction =>
- burners
- combustion chambers
- exhaust ducts
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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 to
compressor. Auxiliary power requirement
such as oil and fuel pumps.
e. γ and cp vary with temperature and gas
composition.
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Cycles with losses
f.
g.
Efficiency is defined by SFC
(specific fuel consumption = fuel
consumption per unit net work
output). Cycle efficiency
obtained using fuel heating value.
Cooling of blade roots and
turbine disks often require
approximately the same mass
flow of gas as fuel flow => air
flow is approximated as constant
for preliminary calculations. This
is done in this course.
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!




Ve2
Vi 2
Q  W   m e  he 
 gze    m i  hi 
 gzi  
2
2
all exits e

 all inletsi 









V22
V12
[single input - single output ]  m 2  h2 

gz2   m 1  h1 

gz1 
2
2
 
 


0 
0 


h
h
 02

 01

• 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
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Stagnation properties
V22
V12
q  w  h2 
 h1 
 h02  h01
2 
2



h02
h01
For a perfect gas we get the stagnation temperature T0,
according to:
2
V
c pT0  c pT 
2
2
V
 T0  T 
2c p
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Stagnation pressure
• Defined in same manner as stagnation temperature (no
heat or work transfer) with added restriction
– retardation is thought to occur reversibly
• Thus we define the
stagnation pressure p0 by:
• Note that for an isentropic
process between 02 and 01
we get
P02 P02 P2 P1

P01 P2 P1 P01
 T02 
  
 T2 

 1
 T2 
 
 T1 

 1
 T1 
 
 T01 

P0  T0   1
 
P T 

 1
 T02 T2 T1 

 
 T2 T1 T01 

 1
 T02 
  
 T01 

 1
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Compressor and turbine efficiencies
Isentropic efficiency (compressors and turbines are approximately
adiabatic => if expansion is reversible it is isentropic).
The isentropic
efficiency is for the compressor is:
h c T
c 

h0 c p T0
'
0
Where
'
p
c , cp
'
p
'
0
are the averaged specific heats of the temperature
intervals 01-02´ and 01-02 respectively.
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:
c  cp
'
p
• We therefore define:
T02  T01
c 
T02  T01
• Similarly for the turbine:
T03  T04
t 
T03  T04
Chalmers University of Technology
Compressor and turbine efficiencies
T02  P02 

 
T01  P01 
 1

T03  P03 

and
 
T04  P04 
Using
frequently used expressions:
 1




T01   P02 
  1
T02  T01 
 
c   P01 



T03  T04





 1
 T03 t 1  

 P03



 P

  04

 1
 







 1

produces the
(2.11)








(2.12)
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Turbine efficiency options
• If the turbine exhausts directly
to atmosphere the kinetic
energy is lost and a more
proper definition of efficiency
would be:
 1



 
 

 

 1  
T03  T04  T03t 1  
 



P
   03   
   Pa   


• In practice some of the kinetic energy is recovered in an
exhaust diffuser => turbine pressure ratio increases.
• Here we put p04=pa for gas turbines exhausting into
atmosphere and think of ηt as taking both turbine and
exhaust duct losses into account
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Turbine diffusers
Recovered
energy
Chalmers University of Technology
Heat-exchanger efficiency
Conservation of energy (neglecting energy transfer to
surrounding):
c p , 46 T04  T06   c p , 25 T05  T02 
In a real heat-exchanger T05 will no longer equal T04 (T05 <T04). We
introduce heat exchanger effectiveness as:
• Modern heat exchangers are
designed to for effectiveness values
above 90%. Use of stainless steel
requires T04 around 900 K (or less).
More advanced steal alloys can be
used up to 1025 K.
T05  T02
Effectiveness 
T04  T02
Chalmers University of Technology
Pressure losses – burners & heat-exchangers
• Burner pressure losses
– Flame stabilizing & mixing
 pb pha
P  P02 1 

– Fundamental loss (Chapter 7 + Rayleigh- 03
p02
P02

line appendix A.4)
• Heat exchanger pressure loss
– Air passage pressure loss ΔPha
– Gas passage pressure loss ΔPhg
– Losses depend on heat exchanger
effectiveness. A 4% pressure loss is a
reasonable starting point for design.



pb
 2  3% (industria l gas turbine)
p02
pb
 3  6% (aircraft engine)
p02
Chalmers University of Technology
The Solar Mercury 50
• 4.3 MW output
• η = 40.5 %
• System was designed from
scratch to allow high performance
integration of heat-exchanger
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 ηm:
Wturbine 
1
m
c p ,12 T02  T01 
Usually power to drive fuel and oil pumps are transmitted from the
shaft. We will assume ηm=0.99 for calculations.
Chalmers University of Technology
Temperature variation of specific heat
We have already established:
cp=f1(T)
cv=f2(T)
Since γ =cp/cv we
have γ=f3(T)
The combustion product thermodynamic
properties will depend on T and f (fuel air
ratio)
Chalmers University of Technology
Pressure dependency?
At 1500 K dissociation begins
to have an impact on cp and γ.
K
 PH 2O

 P
 PO2

 P



1
2



 PH 2

 P



Detailed gas tables for afterburners
may include pressure effects. We
exclude them in this course.
1
H 2  O2  H 2O
2
1
CO  O2  CO2
2
Chalmers University of Technology
Temperature variation of specific heat
In this course we use:
c pa  1005 J/kg ,  a  1.400
c pg  1148 J/kg ,  g  1.333
Since gamma and cp vary in opposing senses some of the
error introduced by this approximation is cancelled.
Chalmers University of Technology
Determining the fuel air ratio
Calculate f that gives T03 for given T02? Use first law for
control volumes (q=w=0) and that enthalpy is a point
function (any path will produce the same result)
1  f c pg T03  298  f  H 25   c pa T02  298  fcpf T f
 298



0
f is small (typically around 0.02) and cpf is also small =>
last term is negligible. The equation determines f.
Chalmers University of Technology
Combustion temperature rise
Hypothetic fuel:
86.08% carbon
13.92% hydrogen
ΔH25 = - 43100 kj/kg
Curves ok for kerosene
burned in dry air. Not ok
in afterburner (fin≠0).
Chalmers University of Technology
Shaft cycle performance parameters
f
SFC  specific fuel consumption 
wN
wnet
  cycle efficiency 
fQnet , p
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Bleeds
• Combustor and turbine
regions require most of the
cooling air.
• Anti-ice
• Rule of thumb: take air as
early as possible (less work
put in)
• Accessory unit cooling (oil
system, aircraft power
supply (generator), fuel
pumps)
• Air entering before rotor
contributes to work!
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Aircraft propulsion – thrust generation
Rate of change of momentum 
Net Thrust 
mC j

mCa


gross momentum thrust
intake momentum drag
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Jet engine – principles of thrust generation
Net Thrust 




m C j  Ca   A j  p j  pa 
  

Aircraft
pressure thrust
velocity


Chalmers University of Technology
Jet engine – principles of thrust generation
No heat or work transfer in the jet engine nozzle
V52
V42
q  w
 h4 
 h05  h04  0
  h5 
2 
2
0





0
h0  c pT0
h05
Stagnation temperature
is constant
T05  T04
h04
Chalmers University of Technology
Mach number relations for stagnation properties
2
V
We have already introduced the
T0  T 
stagnation temperature as:
2c p
and shown that (revision task): c p  cv  R
The specific heat ratio γ is defined:

cp
cv
The Mach number is defined as:
V
V
V
M *

a
speed of sound
RT
Chalmers University of Technology
Mach number relations for stagnation properties
Thus:
M 2  RT  c p
 
  1 M 2  T
T0  T 
 


 T 
2c p

2
 R   1
T0
 1
 1
2
T
M T

 1
M 2
2
T
2

P0  T0   1
but we defined:
which directly gives:
 
P T 

P0    1 2   1
 1 
M 
P 
2

Chalmers University of Technology
Nozzle efficiencies
Nozzle may operate choked
or unchoked:
T04  T5
j 
T4  T5
C52
Temperature equivalent 
2c p
Chalmers University of Technology
Nozzle efficiencies
Critical pressure for irreversible nozzle is
obtained from:
T04 T05
 1 2  1

 1
M 
T5
T5
2
2
T04  Tc
j 
T04  Tc
which gives:
P04

Pc

P04  T04   1
  
Pc  Tc 
1

1
1 
 j

   1   1


   1 
Chalmers University of Technology
Basic operation of
radial compressor
• Impeller - work
is transferred to
accelerate flow
and increase
pressure
• Diffuser recover high
speed generated
in impeller as
pressure
Chalmers University of Technology
Radial compressor
operation
• Typical design takes 50 % of
increase in static pressure in
diffuser
• Conservation of angular momentum
governs performance:
Theoretical torque 
Rate of change of angular momentum 
Cw 2 r2  Cw1r1


prewhirl term
Theoretical work   Cw 2 r2  Cw1r1  
for cases without
prewhirl    Cw 2 r2  Cw 2U 2
Chalmers University of Technology
Slip factor
Due to inertia of flow Cw2 < U:
Theoretical work   U 22
Stanitz formula for estimating σ
0.63
n
n  number of impeller vanes
  1
Chalmers University of Technology
Power input factor - 
• Power is put into overcoming additional friction
not related to the flow in the impeller channels
• Converts energy to heat => additional loss =>
Work done  U
2
Chalmers University of Technology
Overall pressure rise:
p03   cU
 1
p01 
c pT01

2
  1



• P03 is here used to denote the pressure at
compressor exit. P02 is reserved for the
stagnation pressure between the impeller and
the diffuser vanes
Chalmers University of Technology
Example 4.1a
•
•
•
•
•
•
•
•
ψ =1.04, σ = 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
ηc = 0.78
• Compute pressure ratio and
power required
Chalmers University of Technology
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 gas
turbine components (turbine and compressor
performance are given in Lecture 4) and include
them in cycle analysis
• Know how to compute cycle efficiencies for the
heat exchanger cycle