CHAPTER
9
Gas Power Cycles
9-1 Basic Considerations in the Analysis
of Power Cycles
Air Standard Cycles
The Spark Ignition Engine
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FIGURE 9-1
Modeling is a
powerful
engineering
tool that
provides great
insight and
simplicity at
the expense of
some loss in
accuracy.
General classes of engines
• Reciprocating internal combustion engines.
• Reciprocating compressors
• Reciprocating steam engines
Single vs. double action engines
Single action reciprocating engine.
Double action reciprocating engine.
Indicator diagrams
An analog instrument that measures pressure vs.
Displacement in a reciprocating engine. The graph
is in terms of P and V.
p
Area is proportional to the
work done per cycle
V
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FIGURE 9-2
The analysis of
many complex
processes can be
reduced to a
manageable level
by utilizing some
idealizations.
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FIGURE 9-6
P-v and T-s diagrams of a
Carnot cycle.
9-2 The Caront Cycle and Its Value in
Engineering
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FIGURE 9-7
A steady-flow
Carnot engine.
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FIGURE 9-8
T-s diagram for
Example 9–1.
9-3 Air Standard Assumptions
• The working fluid is air, which continuously
circulates in a closed loop and always behaves as an
ideal gas.
• All the processes that make up the cycle are
internally reversible.
• The combustion process is replaced by a heataddition process from an external source.
• The exhaust process is replaced by a heat-rejection
process that restores the working fluid to its initial
state.
9-4 An Overview of Reciprocating
Engines
Air Standard Cycles
The Spark Ignition Engine
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FIGURE 9-10
Nomenclature for
reciprocating engines.
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FIGURE 9-11
Displacement
and clearance
volumes of a
reciprocating
engine.
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FIGURE 9-12
The net work
output of a cycle
is equivalent to
the product of the
mean effective
pressure and the
displacement
volume.
Indicator Diagrams
Work per cycle is represented in terms of a
mean effective pressure and the displacement.
p
MEP = Mean effective pressure
V
X = Displacement
Indicator Diagrams
To compute the work per cycle, use the MEP and
the displacement.
Work = MEP x A x Displacement.
MEP = kiY, where ki is a constant
Y = the average ordinate on the indicator diagram.
X = displacement
A = area of cylinder heat.
Work  kiYA
Ideal indicator diagram
Spark ignition engine - The Otto cycle
p
d
c
e
a
b
V
Displacement
Clearance volume
Processes:
a-b Intake
b-c Compression
c-d Combustion
(spark ignited)
d-e Power Stroke
e-f Gas Exhaust
b-a Exhaust Stroke
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FIGURE 9-13
Actual and ideal cycles in
spark-ignition engines
and their P-v diagrams.
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FIGURE 9-14
Schematic of a twostroke reciprocating
engine.
The spark ignition engine The Otto cycle
The Otto cycle
The air standard Otto
cycle approximates the
automotive internal
combustion engine and
aircraft engines.
Assume a quasistatic
process, isentropic
compression, and
isentropic exhaust.
p 4
s = Constant
3
5
1
6
2
Displacement
V
The Otto cycle
s = Constant
4
T
p 4
3
3
5
1
6
2
P-V diagram for the
actual variable mass
system.
5
V
2 6
V = Constant
This is the T-s diagram for
the fixed mass system.
s
The air-standard Otto cycle
Real Process
•
1-2 Intake of fuel/air mixture at P1
•
2-3 Compression to P3
•
3-4 Combustion, V = cons. and increasing P
•
4-5 Expansion
•
5-6 Exhaust at constant V and falling pressure
•
6-1Exhaust at P1
Air Cycle Approximation
(Constant mass, closed internally reversible, cycle)
•
2-3 Isentropic compression
•
3-4 Heat addition at constant volume
•
4-5 Isentropic expansion
•
5-6 Heat rejection at constant volume
The reversible Otto cycle
• Internally reversible processes
• Constant specific heats
– k = Cp/Cv = Constant
• Polytropic compression and expansion, PVk = Const.
• Treat air as an ideal gas.
Thermodynamic analysis for the
reversible Otto cycle
T
QH
4
QC
5
QH  C v (T4  T3 )
QL  C v (T5  T6 )
3
2 6
V = Constant
s
QH  QC
T5  T6

 1
QH
T4  T3
(1)
s3 4  s56
s3 4
Thermodynamic analysis
for the air-standard Otto Cycle
Qrev
Cv dt
 
 
T
T
path
3 4
 T4 
s3 4  Cv ln  ,
 T3 
T4 T5 (2)


T3 T6
s56
 T6 
 Cv ln  
 T5 
T4 T5

T3 T6
s 2  s3
Thermodynamic Analysis
for the air-standard Otto Cycle
T3  v 2 

 
T2  v3 
T4  v5 
 
T5  v 4 
k 1
k 1
Thermal efficiency
T4 T5


T3 T6
T6
  1
T3
v6
rv 
v3
T4  T3 T5  T6

T3
T6


 v3 
  1  
 v6 
  1 r
1 k
v
1 k
Thermal efficiency
v6
rv 
v3
   1  rv
1 k
Key assumptions: (1) Internally reversible processes
(2) Constant specific heats
Important consequence: (1) Efficiency is independent of
working fluid
(2) Efficiency independent of
temperatures
Efficiency of Air Standard Otto Cycle
k = 1.3 = Cp/Cv
QH

T
4
3
5
2,6
WNET
QC
s
rv
Air Standard Otto Cycle
0.8
0.7
0.6
Efficiency
0.5
k = 1.4
0.4
k = 1.3
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Pressure Ratio, Pv
Se
ri
es
1
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FIGURE 9-16
Thermal efficiency
of the ideal Otto
cycle as a
function of
compression ratio
(k = 1.4).
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FIGURE 9-18
The thermal
efficiency of the
Otto cycle
increases with the
specific heat ratio
k of the working
fluid.
The Otto cycle with real gases
• Variable specific heats.
• Efficiency based on internal energies obtained from the gas
tables which take into account variable specific heats.
u5  u6
  1
u 4  u3
Q
   (r , T , H
v
2
T
QH
4
QC
5
3
mTOT
)
2,6
V = Constant
s
9-6 Diesel Cycle:
Air Standard Cycles
The Compression Ignition Engine
The compression ignition engine - the
Diesel cycle
Air Standard Compression
Ignition Engine - The Diesel cycle
• An IC power cycle useful in many forms of automotive
transportation, railroad engines, and ship power plants.
• Key assumptions are,
– Constant specific heats, the ideal gas, and internally reversible processes.
Ideal Indicator Diagram
Compression Ignition Engine
p c
d
e
a
b
V
Displacement
Clearance volume
Processes:
a-b Intake
b-c Compression
c-d Combustion
d-e Power Stroke
e-f Gas Exhaust
b-a Exhaust Stroke
Ideal Indicator Diagram
Compression Ignition Engine
p
a
b
c
e
d
V
Displacement
Clearance volume
Processes:
a-b
Combustion
(P = Const.)
b-c
Expansion
(s = Const)
c-d
Exhaust
(V = Const)
d-e
Exhaust
(P = Const)
e-d
Intake
(P = Const)
d-a
Compression
(s = Const)
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FIGURE 9-21
T-s and P-v
diagrams for the
ideal Diesel cycle.
Advantages of the Diesel Cycle
• Eliminates pre-ignition of the fuel-air
mixture when compression ratio is high.
• All heat transfer to a constant mass system.
The air-standard compression ignition engine
T
p = Constant
b
a
QH
Win
QC
Wout
c
V = Constant
d
s
T
p = Constant
a
b
QH
QC
d
Thermal efficiency of the airstandard Diesel cycle
c
V = Constant
s
QC
  1
QH
QH  C p Tb  Ta 
QC  Cv (Tc  Td )
Cv (Tc  Td )
(Tc  Td )
  1
 1
C p (Tb  Ta )
k (Tb  Ta )
Key parameters for
the Diesel cycle
p
a
Compression
v Ratio
rv 
b
c
e
d
V
Displacement
Clearance volume
d
va
Expansion Ratio
vc
re 
vb
Cut Off Ratio
b
c
a
v
r 
v
Thermal efficiency
p
 1  r  1 

  1   k 1 
rv  k rc  1 

c
k
c
a
b
e
d
V
Displacement
Clearance volume
p a
b
Air-standard Diesel cycle
c
d
V
e
Displacement
Thermal Efficiency
 1  rck  1 

  1   k 1 
 rv  k rc  1 
Clearance volume
vb
Cut Off Ratio: rc 
va
vd
Compression Ratio: rv 
va
Expansion Ratio:
vc
re 
vb
Important features of the Diesel cycle
• At rc = 1, the Diesel and Otto cycles have the same efficiency.
– Physical implication for the Diesel cycle: No change in volume when heat
is supplied.
– A high value of k compensates for this.
• For rc > 1, the Diesel cycle is less efficient than the Otto cycle.
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FIGURE 9-22
Thermal efficiency of
the ideal Diesel cycle
as a function of
compression and
cutoff ratios (k = 1.4).
Efficiency Comparisons
(Approximate)

rc > 1
rv
Effect of variable specific heats
• Efficiency is based on internal
energies and enthalpies obtained
from the gas tables which take into
account variable specific heats.
u a  ub
  1
ha  hb
Key terms and concepts
Air-standard Diesel Cycle
Expansion ratio
Pressure ratio
Cut of ratio
Compression ignition
Comparison of the Otto and Diesel
cycles
Comparison of the Otto and Diesel cycles
Comparison No. 1: (a) Same inlet state (P,V)
(b) Same compression
ratio, rv
(c) Same QH
Key factor: Constant volume heat addition of the Otto
cycle vs. constant pressure heat addition of
the Diesel cycle.
p
c
b
Otto cycle with a specified
inlet condition at “a” with
a given compression ratio
rv = va/vb
d
a
Displacement
V
p
c
b
Otto and Diesel cycles with
same compression inlet
conditions at “a” and the
same compression ratio, rv.
c*
d*
d
a
Displacement
V
p
c
b
c*
d*
d
a
Diesel Cycle
k

 1  rc  1 

  1   k 1 
 rv  k rc  1 
V
Otto Cycle
  1 r
1 k
v
First Law analysis of the heat addition process
dU  Q  W
Otto: Heat addition with V = 0 in process b  c.
W = 0, and P and T increase
Diesel: Heat addition with P = Constant
in process b  c*.
dW > 0, and P and T lower that
in the Otto cycle.
V = Const.
T
Tc
Tc*
c
Otto Cycle
c*
P = Const.
Diesel Cycle
b
d*
a
d
s
T-s diagrams for equal heat addition
T
Tc
Tc*
c
c*
QH
b
a
d*
d
s
QH,Otto = QH,Diesel
The areas under the
process paths b 
c and b  c* are
equal under the
assumption of
equal heat addition,
QH.
Efficiency comparisons
T
Tc
Tc*
c
c*
When QH and rv are
the same for both
d* cycles,
b
a
OTTO  DIESEL
d
s
QC,Otto < QC,Diesel
Comparison of the Otto and Diesel cycles
Comparison No. 2: (a) Same inlet state (P,V)
(b) Same maximum P
(c) Same QH
Comparison No. 2 is more practical when the “knocking”
effect is considered. In the Otto cycle, about 11 atm are
needed to achieve combustion with engine knock.
T
Diesel Cycle
c
c*
Otto Cycle
b*
b
a
d
d* V = Const.
P = Const.
s
DIESEL  OTTO
Gas Power Systems - 3
The Dual Cycle
The Dual cycle
• The dual cycle is designed to capture capture some of the
advantages of both the Otto and Diesel cycles.
• It it is a better approximation to the actual operation of the
compression ignition engine.
QH,P
p
b
QH,V
The Dual cycle
c
s = Constant
a
d
QC,V
e
V
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FIGURE 9-23
P-v diagram of
an ideal dual
cycle.
p
b
QH,P
c
QH,V
a
d
e
QC,V
V
C v (Td  Te )
  1
CV (Tb  Ta )  C P (TC  Tb )
(Td  Te )
1 
(Tb  Ta )  k (TC  Tb )
b
QH,P
c
QH,V
a
d
e
V
Ve
rv 
Va
Vc
rc 
Vb
Pb
rP 
Pa

r r 1
1 
  1  k 1 

rv  rp  1  krp (rc  1) 
k
p c
Gas Power Systems - 4
The Gas Turbine
Overview
• Gas Turbines
– The Carnot cycle as an “air standard” cycle
• The Ericcson Cycle
• The Brayton Cycle
– The standard cycle
– The reheat cycle
– Inter-cooling
The modern gas turbine cycle
• Long-haul automotive power.
• Large aircraft flight envelopes
• Commercial aircraft
• Altitude (30,000 to 40,000 ft)
• Long range and low specific fuel consumption (sfc)
• Moderate speed and thrust
The modern gas turbine cycle
• Military aircraft
• High altitude ( > 40,000 ft)
• Moderate range and high specific fuel consumption (sfc)
• High speed and large thrust
• Relevant cycles
– The Ericsson and Brayton cycles
General features of the aircraft gas turbine engine
Combustor
Fuel
Intake Air
Compressor
Turbine
Exhaust Section
Compressor Drive Shaft, Wcomp
The air-standard turbine cycle
• Open system modeled as a closed system - fixed with fixed
mass flow.
• Air is the working fluid.
• Ideal gas assumptions are applied.
• Approximate the combustor as the high temperature source.
• Internally reversible processes.
Ideal gas power cycles
Air Standard
Carnot Cycle
Air Standard
Ericcson Cycle
Air Standard
Brayton Cycle
Air-standard Carnot cycle
Air-standard Carnot cycle
Employ the same assumptions as for the air-standard turbine cycle.
T
QH
p1
p4
P1  P2 , P3  P4
TC
T3
  1
 1
TH
T1
p2
 P1 
  1  
 P4 
p3
QC
s
1 k 
V4 
 1  
 V1 
k
1 k 
 P2 
 1  
 P3 
 V3 
 1  
V2 
1 k 
(1 k )
k
Limitations of the air-standard Carnot cycle
• Heat addition at constant temperature is difficult and costly.
– Work is required because fluid expands.
• Heat addition is limited because a large change in volume would
imply a low mean pressure in the heat addition process.
– Frictional effects might become too great if the mean pressure is
too low.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-26
T-s and P-v diagrams of
Carnot, Stirling, and Ericsson
cycles.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-27
The execution of the
Stirling cycle.
Air-standard Ericsson cycle
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-28
A steady-flow Ericsson
engine.
The air-standard Ericsson cycle
• Constant pressure heat addition and rejection
• Constant temperature compression and expansion
Qb-c
p
T = Constant
c
b
Qa-b
Qc-d
a
Qd-a
d
v
Qb-c
p b
T = Const.
c
Qa-b
The Air Standard
Ericcson Cycle
Qc-d
a
Qd-a
d
v
P = Const.
T
Qc-d
c
d
Qb-c
Qd-a
b
Qa-b
a
s
p
Qb-c
T = Const.
b
Thermal Efficiency
Ericcson Cycle
c
Qa-b
Qc-d
Wa-b
a
Qd-a
d
Wc-d
v
WNET
Wc d  Wa b


Q Added Qb c  Qc d
The Brayton cycle
The Brayton Cycle
• Modern gas turbines operate on an open Brayton cycle.
– Ambient air is drawn at the inlet.
– Exhaust gases are released to the ambient environment.
• The air standard Brayton cycle is a closed cycle.
– All processes are internally reversible.
– Air is the working fluid and assumed an ideal gas.
General features of the aircraft gas turbine engine
Combustor
Exhaust Gases &
Work Output,
Wturb
Fuel
Intake Air
Compressor
Turbine
Compressor Drive Shaft, Wcomp
The open, actual
cycle for the gas
turbine.
QH
The closed, air
standard cycle
for the
gas turbine.
WCOMP
WTURB
QL
The gas turbine processes
•
•
•
•
Isentropic compression to TH
Constant pressure heat addition at TH
Isentropic expansion to TC
Constant pressure heat rejection at TC
QH
2
Air Standard
Brayton Cycle
3
WCOMP
WTURB
1
4
QL
p
QH
2
3
S = Constant
1
QC 4
Vv
p
QH
2
3
s = Constant
1
Air Standard
Brayton Cycle
QC 4
VV
Vv
T
3
QH
2
4
1
s1 = s2
p2 = p3
p1 = p4
QC
s3 = s4
s
3
T
QH
p2 = p3
WOUT
WIN
2
4
1
Thermal efficiency of
the ideal Brayton cycle
p1 = p4
QC
s
WOUT  Win (h3  h4 )  h2  h1 


QH
h3  h2
3
T
QH
Thermal Efficiency
Ideal Brayton Cycle
p2 = p3
WOUT
WIN
2
4
1
p1 = p4
QC
For an ideal gas, h-h0 =
Cp(T - T0). The compression
and expansion process are
polytropic with constant k.
s
 P2 
  1  
 P1 
(1 k )
k
Cycle efficiency
• Ideal gas assumptions apply.
• All processes internally reversible
– Compressor and turbine efficiency are each 100%.
• Assume fuel added in the combustor is a small percent (mass or
moles) of total flow, and thus air properties provide a good
estimate of cycle performance.
Gas Power Systems - 5
Real Turbine Performance
Overview
• Internally irreversible Brayton cycles
– Case study
• Real turbine performance
Internally irreversible Brayton cycles
Real cycle analysis - Brayton cycle
• Internal irreversibility arises in each process of the open
cycle.
• Compression (compressor efficiency)
• Heat addition
• Expansion (turbine efficiency)
Internal irreversibility will lower
work output and decrease
thermal efficiency. Greater
entropy production will also
result.
2
3
1
4
p2 = p3
T,h
3
4
2
p1 = p4
1
s 2  s1
s3 = s4
s4
s
Case Study
Case study
Given: Turbine and compressor efficiencies of
80%, and the operating data as given below. No
internal irreversibility in the combustor.
Find: For the actual cycle, the thermal efficiency,
the ratio of Wcomp/Wturb and entropy production.
c
T
b’
b
b’
QH
c
d
d’
WIN
d’
a
a
s
QC
WOUT
State
a
b
b’
c
d
d’
T
P
h
S0
(R)
(p sia)
(Btu / lbm )
(Btu / lbm R)
530
881
967
1700
1059
1192
15
90
90
90
15
15
126.7
211.6
232.7
422.6
255.7
289.3
0.5963
0.7191
0.7421
0.8876
0.7647
0.7946
Note: Values of h and so are from the gas tables, and all
initial data are supplied here but this is not necessarily
the case in practice.
Process Calculations
Process
a-b
a-b’
b’-c
c-d
c-d’
d’-a
NET
Note:
h
Q
(Btu/lbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbmR)
84.9
106
189.9
-166.9
-133.3
-162.6
0
-84.9
-106
0
166.9
133.3
0
27.3
0
0
189.9
0
0
-162.6
0
0
0.0230
0.1455
0
0.0299
-.1983
0
si  j  s
o
i j
TURB
s
W
Pj 
 R ln  
Pi 
h
hs

,  COMP 
hs
h
The pressure ratios for
the isentropic and actual
processes are the same,
and therefore s = so.
Thermal Efficiency and Work Ratio
 CYCLE
Wcomp
Wturb
Wnet 133.3  106 27.3



 0.144
Qb ' c
189
189
106

 0.795
133.3
Notice that here we have a high ratio of the work of compression
to that of expansion, which is typical in gas turbine systems. The
key variable is the turbine inlet temperature, Tc, and this is to be
made as large a possible. Tc is limited by the material and turbine
blade cooling capability.
Entropy Production
Process
Q
(Btu/lbm)
a-b
a-b’
b’-c
c-d
c-d’
d’-a
NET
0
0
189.9
0
0
-162.6
27.3
s
Q/T
0
0.0230
0.1455
0
0.0299
-0.1983
0
0
0
0.1455
0
0
-0.1983
-0.0528

(Btu/lbm-R) (Btu/lbm-R) (Btu/lbm-R)
0
0.0230
0
0
0.0299
0
0.0529
The entropy production rates are based on the application of the entropy
balance for an open system to each process in the cycle. The heat addition
and rejection processes are at the high and low temperatures respectively.
Thus, these processes produce an entropy production from the external heat
transfer process that is not included in the above calculations.
The Carnot Efficiency
for the Cycle
TC
530
 Carnot  1 
 1
TH
1700
 0.68
  0.144
QH, TH
QC, TC
(General cycle representation)
Here, the difference in the efficiency
calculations indicates a high level
of external entropy production.
p2 = p 3
Efficiency Curves
T,h
4
2
TURB  COMP  1
3
p1 = p 4
0.9
0.25
1
0.8

0.7
0.1
2
4
Pressure Ratio, p2/p1
s’2
S1 = S2
S3 = S4
s’4
s
Brayton cycle efficiency is
greatly dependent on the
efficiencies of the turbine
and compressor. The
curves at the left are
approximate.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-29
An open-cycle gas-turbine engine.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-30
A closed-cycle
gas-turbine
engine.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-31
T-s and P-v
diagrams for the
ideal Brayton
cycle.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-32
Thermal
efficiency of the
ideal Brayton
cycle as a
function of the
pressure ratio.
Gas Power Systems - 6
Cycle Improvements
Cycle Improvements
• Regeneration
– Reduces heat input requirements and lowers heat rejected.
• Inter-cooling
– Lowers mean temperature of the compression process
• Reheat
– Raises mean temperature of the heat addition process
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-33
For fixed values
of Tmin and Tmax ,
the net work of
the Brayton cycle
first increases
with the pressure
ratio, then
reaches a
maximum at rp =
(Tmax /Tmin) k/[2(k –
1)], and finally
decreases.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-36
The deviation of
an actual gasturbine cycle
from the ideal
Brayton cycle as
a result of
irreversibilities.
The Brayton cycle with regeneration
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-38
A gas-turbine
engine with
regenerator.
QH
The standard Brayton cycle
shown as an open cycle.
2
3
1
Regeneration is
accomplished
by preheating the
combustor air
with the exhaust
gas from
the turbine.
WIN
4
WOUT
5
QH
2
1
3
WCOMP
4
WTURB
T,h
QH
3
x
4
2
y
1
Internal heat transfer,
States 2 - x.
QC
s
The maximum benefit of regeneration is obtained when
the exhaust temperature Ty is brought to temperature
T2 by the regenerative heat exchanger.
T,h
QH
3
The ideal
regenerator
x
4
The ideal regenerator
will heat the combustion
air up to T4
2
y
1
QC
s
 CYCLE
(h2  h1 )
 1
h3  h4
T,h
QH
3
The actual
regenerator
x
x
y
2
y
1
QC
4
Actual internal heat transfer takes
place between States 2 - x’.
The effectiveness of the regenerator is
the actual heat transfer divided by the
maximum possible heat transfer.
s
C p (Tx  T2' )
(Tx  T2 )


C p (T4  T2 ) (T4  T2 )
T,h
QH
3
Thermal efficiency
Regeneration reduces
net heat input at the
high temperature.
x
x
y
2
4
y
1
QC
s
 cycle
(h3  h4 )  (h2  h1 )

h3  hx
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-39
T-s diagram of a
Brayton cycle
with
regeneration.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-40
Thermal efficiency
of the ideal
Brayton cycle with
and without
regeneration.
The Brayton cycle with inter-cooling
The standard Brayton cycle with on stage of intercooling
7
QH
WCOMP
1
5
6
3
WTURB
4
3
T
7’
7
6
5 5’
4
4’
1
s
One stage of inter-cooling with turbine
and compressor inefficiencies.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-42
Comparison of
work inputs to a
single-stage
compressor
(1AC) and a twostage
compressor with
intercooling
(1ABD).
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-43
A gas-turbine
engine with twostage
compression with
intercooling, twostage expansion
with reheating,
and regeneration.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-44
T-s diagram of an
ideal gas-turbine
cycle with
intercooling,
reheating, and
regeneration.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-45
As the number of
compression and
expansion stages
increases, the
gas-turbine cycle
with intercooling,
reheating, and
regeneration
approaches the
Ericsson cycle.
The Brayton cycle with reheat
3
T,h
QH  Q23  Q45
4
5
6
2
1
Qc  Q61
s
The re-heating of the fluid at an intermediate
pressure raises the enthalpy input to the cycle
without raising the temperature of the external
thermal reservoir.
Example
This example compares several modifications to the basic
Brayton cycle: multi-stage compression and expansion
with reheat, inter-cooling and regeneration.
Description:
In an air-standard gas turbine engine, air at 60o F
and 1 atm enters the compressor with a pressure
ratio of 5:1. The turbine inlet temperature is
1500o F, and the exhaust pressure is 1 atm.
Determine the ratio of turbine work to compressor
work and the thermal efficiency under the
following operating conditions.
Operating conditions:
(a) The engine operates on an ideal Brayton
cycle.
(b) The adiabatic efficiency of the turbine
and compressor are 0.83 and 0.93,
respectively.
(c ) The conditions of (b) hold and a
regenerator with an effectiveness of 0.65 is
introduced.
Operating conditions:
(d) The conditions of (b) and (c ) hold, and an
inter-cooler that cools the air to 60o F at a
constant pressure of 35 psia is added.
(e) The inter-cooler is removed and a re-heater
the heats the fluid to 1500o F at a constant
pressure of 35 psia is added.
(f) The conditions of (e) hold and the intercooler of (d) is put back.
Assumptions:
(1) All cycles operate on the air-standard
basis.
(2) Ideal gas
(3) Constant specific heats; k = constant,
Cp = 0.24 BTU/lbm-R, k = 1.40 = constant.
(4) Internally reversible processes except
where isentropic efficiency (adiabatic
efficiency) is less than one.
(5) 1 atm = 14.7 psia.
Solution:
(a) The engine operates as an ideal Brayton cycle.
 k 1 


 k 
3
T
T4  P4 
  
T3  P3 
h3 4  C p T3 4
4
2
1
s
State D ata
State
1
2
3
4
h1 2  C p T1 2
P
(psia)
14.7
73.5
73.5
14.7
T
(R)
520
825
1960
1237
Process quantities
Process Quantities
h
BTU/Lbm
Q
BTU/lbm
1-2
2-3
3-4
4-1
Total
73.1
272.4
-173
-172
0.5 (~0)
0
272.4
0
-172
100.32
Process quantities
W3 4  h3 4  C p T3  T4   173 BTU lbm
W1 2  h1 2  73.1 BTU lbm
Wturbine
W3 4 173


 2.37
Wcompressor W1 2 73.1
Wnet W3 4  W1 2 h23  h1 2
 cycle 


Q2  3
Q2  3
h23
173  73.1

 0.367
(0.24)(1960  825)
(b) The adiabatic efficiency of the compressor and turbine
are 0.83 and 0.92 respectively.
T
 k 1 


 k 
T4  P4 
  
T3  P3 
h3 4  C p T3 4
3
2’
2
4 4’
1
s
h1 2  C p T1 2
T  0.92, comp  0.83
State D ata
State
1
2
3
4
P
(psia)
14.7
73.5
73.5
14.7
Ws ,1 2 T2  T1
 comp 

W1 2
T2  T1
T2  T1
T2  T1 
 887 R
T
(R)
520
825
1960
1237
 comp
 turb
W3 4

 T4  1295 R
W s , 3 4
State D ata
State
1
2’
3
4’
P
(psia)
14.7
73.5
73.5
14.7
T
(R)
520
887
1960
1295
W3 4  h3 4  C p T3  T4   159.5 BTU lbm
W1 2  h1 2  88.8 BTU lbm
Wturbine
W3 4 159.5


 1.81
Wcompressor W1 2
88.8
Wnet W3 4  W1 2 h3 4  h1 2
 cycle 


Q23
Q23
h23
159.5  88.8

 0.28
(0.24)(1960  887)
(c) The conditions of (b) hold, and a regenerator with
an effectiveness of 0.65 is introduced.
 k 1 


 k 
3
T
2’
2
x
4’
y
1
s
T4  P4 
  
T3  P3 
h3 4  C p T3 4
h1 2  C p T1 2
(Tx  T2 ' )

 0.65
(T4 '  T2 ' )
Tx  T2   T4  T2   1052 R
T  0.92
 comp  0.83
State D ata
State
1
x’
3
4’
P
(psia)
14.7
73.5
73.5
14.7
T
(R)
520
1052
1960
1295
W3 4  h3 4  C p T3  T4   159.5 BTU lbm
W1 2  h1 2  88.8 BTU lbm
Wturbine
W
159.5
 3 4 
 1.81
Wcompressor W1 2
88.8
Wnet W3 4  W1 2 h3 4  h1 2
 cycle 


Q x 3
Q x 3
hx3
159.5  88.8

 0.37
(0.24)(1960  1052)
(d) The conditions of (b) and (c ) hold, and an inter-cooler
that cools the air to 60o F at 35 psia is added.
3
T
P = 35 psia
x
7’
520 R
6
5’
y
4’
1
s
3
T
P = 35 psia
x
7’
520 R
6
5’
y
1
s
4’
Compute T5’ ,T7’,
and the actual
s
temperature between
state 7’ and x due
to regeneration (7”).
State D ata
State
P
(psia)
14.7
35
35
14.7
73.5
73.5
73.5
73.5
14.7
73.5
14.7
1
5
5’
6
7
7’
7”
3
4’
x
y
 k 1 


k


T5  P5 
  
T1  P1 
T5  667 R
T5  T1 
T
(R)
520
667
697
520
643
668
616
1960
520
1295
888
1

T5  T1 
T
3
P = 35 psia
7’
6
x
5’
4’
W3 4  159.5 BTU / lbm
W15  W67  78 BTU / lbm
y
1
s
Wsturbine
 2.04
Wcompressor
Wnet

 0.38
C p (T3  T7 )
(e) Eliminate inter-cooling but add one stage of reheat
to 1500o F.
  0.38
(f) Combine parts (d) and (e).
  0.40
Wturbine
 2.28
WCompressor
Summary of results:
(a) Ideal Cycle
(b) Actual Basic
Cycle
(c) Actual Basic
Cylce and
Regenerator ( =
0.65)
(d) Actual Cycle
w ith Intercooling at
o
35 psia to 60 F and
part (c)
(e) Actual Basic
Cycle w ith Reheat at
35 psia and part (c)
(f) Parts (c ) ,(d), and
(e) combined

Work Ratio
0.37
0.28
2.37
1.81
173
160
0.37
1.18
160
0.38
2.04
160
0.38
2.02
178
0.40
2.28
178
Wt/Wc
Wt
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-48
Basic components of
a turbojet engine and
the T-s diagram for the
ideal turbojet cycle.
[Source: The Aircraft Gas Turbine Engine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-51
Energy supplied to an
aircraft (from the
burning of a fuel)
manifests itself in
various forms.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-52
A turbofan engine.
[Source: The Aircraft Gas Turbine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951,
FIGURE 9-53Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A modern jet
engine used to
power Boeing
777 aircraft.
This is a Pratt &
Whitney
PW4084
turbofan
capable of
producing
84,000 pounds
of thrust. It is
4.87 m (192 in.)
long, has a 2.84
m (112 in.)
diameter fan,
and it weighs
6800 kg (15,000
lbm).
Photo Courtesy of Pratt&Whitney Corp.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-54
A turboprop engine.
[Source: The Aircraft Gas Turbine Engine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-55
A ramjet engine.
[Source: The Aircraft Gas Turbine Engine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.]
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-57
Under average
driving
conditions, the
owner of a 30-mpg
vehicle will spend
$300 less each
year on gasoline
than the owner of
a 20-mpg vehicle
(assuming
$1.50/gal and
12,000 miles/yr).
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-62
Aerodynamic
drag increases
and thus fuel
economy
decreases
rapidly at speeds
above
55 mph.
(Source: EPA and U.S. Dept. of Energy.)