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Chapter 9
GAS POWER CYCLES(based on notes from Profs. Kaya & Gauthier)
Vinh Q. Tang, Ph.D., P.Eng.
Adjunct Recearch Professor
MAAE 2400
Thermodynamics and Heat Transfer
1
Carleton University
Department of Mechanical and Aerospace Engineering
Office: ME2186
Fall 2020
OUTLINE
9.1 Introduction (Review of ideal gas model)
9.2 Otto Cycle (For your interest)
9.3 Brayton Cycle
9.4 Summary
Reading Assignment: Related Sections in Chapter 9 in Fundamentals of
Engineering Thermodynamics, 7th Edition ; M. Moran and H. N. Shapiro, et al
2Fall 2020
9.1 INTRODUCTION
3
Equations of state :
ππ = πΉπ» ππ ππ½ = ππΉπ»
Fall 2020
Review of Ideal Gas Model (presented in Chapters 3 and 6)
4
Changes in βuβ and βhβ
ππ β ππ = ππ(π»π β π»π)
ππ β ππ = ππ·(π»π β π»π)
1) Constant specific heats
2) Variable specific heats
u(T) and h(T) are evaluated from tables:
βͺ Table A-22 for air (mass basis)
βͺ Table A-23 for other gases (molar basis)
Fall 2020
5
Changes in entropy
1) Constant specific heats
2) Variable specific heats :
ππ β ππ = πππππ»ππ»π
+ πΉππ(ππππ)
ππ β ππ = ππ·πππ»ππ»π
β πΉππ(π·π
π·π)
ππ β ππ = πππ β ππ
π β πΉππ(π·π
π·π)
s0 can be obtained from tables:
βͺ Table A-22 for air (mass basis)
βͺ Table A-23 for other gases (molar basis)
(See Tables A-20 & 21 for cv and cP data)
Fall 2020
6
For Isentropic Processes
1) Constant specific heats
π»ππ»π
=π·π
π·π
πβππ π»π
π»π=
ππππ
πβππ·π
π·π=
ππππ
π
Where π =ππ
πππππππ ππ π»ππππ π¨ β ππ
2) Variable specific heats
π·π
π·π=π·ππ
π·ππ
ππππ
=ππππππ
Where Pr and vr are provided for air π’π§ ππππ₯π π β ππ
for air only
Fall 2020
End of Review of Ideal Gas Model
β’ Starting section 9.2
in the next slide
Fall 2020 7
9.2 OTTO CYCLE (1)
β’ Used to model spark-
ignition engines
β’ In a spark-ignition
engine, the air-fuel
mixture is ignited by a
spark plug
8Fall 2020
9.2 OTTO CYCLE (2)
β’ Pressure-
displacement curve
for a four-stroke
reciprocating
internal combustion
engine
(TDC) (BDC)9Fall 2020
9.2 OTTO CYCLE (3)
1. A fixed mass of air is the working fluid throughout the entire cycle. Thus, there is no inlet and no exhaust process.
2. The internal combustion process is replaced by a heat transfer from an external source.
3. The cycle is completed by heat transfer to the surroundings.
4. All processes are internally reversible.
5. Air is an ideal gas with constant specific heats (Cold Air Standard Analysis).
AIR STANDARD ANALYSIS
10Fall 2020
9.2 OTTO CYCLE (4)
1-2 reversible adiabatic (isentropic) compression
2-3 constant volume heat transfer (heat added)
3-4 reversible adiabatic (isentropic) expansion
4-1 constant volume heat transfer (heat rejected)
2
3
4
v=c
v=c
1
11Fall 2020
9.2 OTTO CYCLE (5)
β’ Air Standard Otto Cycle (ASOC) is an ideal cycle
approximating a spark-ignition internal
combustion engine.
β’ Due to the simplifications in the ASOC, the
analysis is only valid on a QUALITATIVE BASIS.
12Fall 2020
9.2 OTTO CYCLE (6)
CONSEQUENCES OF THE Air Standard Otto Cycle
ASSUMPTIONS
β’ The working fluid is a fixed amount of air.
β’ Heat rejection and addition (combustion) are
assumed to take place instantaneously.
β’ There are no pressure drops due to friction.
β’ Air is assumed to be an ideal gas.
β’ All processes are internally reversible.
13Fall 2020
9.2 OTTO CYCLE (7)
β’ USING 1st and 2nd Laws for a closed system with
DKE, DPE=0
121221For uuw β=β
433443For uuw β=β
232332For uuq β=β
144114For uuq β=β
WQU β=D
NOTE: wβs and qβs are all positive quantities. 14Fall 2020
( ) ( ) ( ) ( )
23
14
23
1423
23
1243
23
1234
1uu
uu
uu
uuuu
uu
uuuu
q
ww
q
w
cycle
cycle
cycle
added
net
cycle
β
ββ=
β
βββ=
β
βββ=
β=
=
9.2 OTTO CYCLE (8)
1212 uuw β=
4334 uuw β=2323 uuq β=
1441 uuq β=
We haveHence
15Fall 2020
9.2 OTTO CYCLE (9)
1
4
3
3
4
1
2
1
1
2 and
ββ
=
=
kk
V
V
T
T
V
V
T
T
For isentropic processes of 1-2 and 3-4 and air is an ideal gas:
2
1
3
4ButV
V
V
V=
4
3
1
2HenceT
T
T
T=
2
3
1
4OrT
T
T
T=
16Fall 2020
9.2 OTTO CYCLE (10)
1
2
1
1
2
2
3
1
4
23
141
β
=
=
=
β
ββ=
k
cycle
V
V
T
T
T
T
T
T
dTcdu
uu
uu
( ) 1
21
2
1
23
14
2
1
23
14
/
11
1
1/
1/11
ββ=
β=
β
ββ=
β
ββ=
kcycle
cycle
cycle
VV
T
T
TT
TT
T
T
TT
TT
We have
Hence
17Fall 2020
9.2 OTTO CYCLE (11)
min
max
1
rationcompressioc)(volumetri
where
Cycle Otto1
1
V
Vr
r
v
k
v
cycle
=
β=β
Thermal efficiency of CASOC:
Remember that this is an ideal cycle, and does not
account for variation of properties with temperature,
friction, heat losses, combustion losses, intake and
exhaust processes, etc. 18Fall 2020
9.2 OTTO CYCLE (12)
1
11
ββ=
k
v
thr
β’ As the rv increases, the efficiency also
increases. However, in practice, this may
lead to DETONATION due to auto-ignition
of the fuel-air mixture.
β’ Lead (tetraethyl lead) in gasoline
improves detonation characteristics, but
is not environmentally friendly.
β’ Non-leaded gasolines with good
detonation characteristics have been
developed to reduce atmospheric
contamination.
19Fall 2020
9.2 OTTO CYCLE (13)
We can plot the Otto cycle efficiency
1
11
ββ=
k
v
thr
5 10 15 200
0.2
0.4
0.6
0.8
k = 1.5k = 1.4k = 1.3
Compression Ratio
Th
erm
al E
ffic
ien
cy
Real Spark Ignition Cycle
20Fall 2020
9.3 BRAYTON CYCLE
OPEN CYCLE CLOSED CYCLE
21Fall 2020
22
βAir-standard analysisβ involves an idealization that
simplifies the study of open gas turbine power plants.
Two basic assumptions:
1. The working fluid is air that behaves as an
ideal gas.
2. The temperature rise that would be
brought about by combustion is
accomplished by a heat transfer from an
external source.
Air-standard analysis (For gas turbine open cycles)
More on assumptions
Fall 2020
1. The working fluid is air, which behaves as an ideal gas.
2. The mass and properties of the fuel are neglected.
3. The combustion process is replaced by a heat transfer
from an external source
[It regards heat as being added through a perfect heat
exchanger and removed by exhaust to an infinite
atmosphere (or another perfect heat exchanger)].
4. All processes are internally reversible.
23
For COLD AIR STANDARD : constant specific heat is assumed
MAIN ASSUMPTIONS FOR IDEAL AIR
STANDARD BRAYTON CYCLE ANALYSIS
Fall 2020
24
Air-standard Ideal Brayton Cycle
Fall 2020
1-2 Compression: reversible and adiabatic (isentropic)
2-3 Heat added at constant pressure
3-4 Expansion: reversible and adiabatic (isentropic)
4-1 Heat rejected at constant pressure
β
2
3
4p= cb
p=ca
1
25
P-v and T-s diagrams
Fall 2020
Applying the first law for an open system to
each CV shown in diagram
Recall the first law:
26Fall 2020
π π¬πͺπ½π π
= αΆπΈπͺπ½ β αΆπΎπͺπ½ + αΆπ ππ β ππ +π½ππ β π½π
π
π+ π(ππ β ππ)
Steady-state
ΞKE = ΞPE = 0
π = αΆπΈπͺπ½ β αΆπΎπͺπ½ + αΆπ ππ β ππ
27
Recall :
Fall 2020
π = αΆπΈπͺπ½ β αΆπΎπͺπ½ + αΆπ ππ β ππ
π =αΆπΈπͺπ½
αΆπβ
αΆπΎπͺπ½
αΆπ+ ππ β ππ
π = π βπ + ππ β ππTo be applied to each control volume
Applying to specific control volume
shown on the diagram, the energy
equation becomes :
1-2 Compression: isentropic
2-3 Heat added at constant pressure
3-4 Expansion: isentropic
4-1 Heat rejected at constant pressure
28Fall 2020
π = π βπ + ππ β ππRecall :
πππ ππππππππππ: ππ = ππ β ππ
πππ π―πΏπΈππ: πππ = ππ β ππ
πππ π―πΏπΈπππ: ππππ = ππ β ππ
πππ π»ππππππ: ππ = ππ β ππ
29
πΌπππππ =ππππ
ππππΌπππππ =
ππ βππ
πππ
πΌπππππ =(ππ β ππ) β (ππ β ππ)
(ππ β ππ)
πΌπππππ =(ππ β ππ) β (ππ β ππ)
(ππ β ππ)
πΌπππππ = π β(ππ β ππ)
(ππ β ππ)
Cycle Thermal Efficiency
Fall 2020
Evaluating cycle thermal efficiency
30
1) Air standard analysis: Using air table data
(not assuming constant specific heat)
2) Cold air standard analysis: assuming
constant specific heat
Two different types of analyses
Fall 2020
β
ββ
2
3
4
p=const
p=const
1
When Air Table Data are used, recall the
following relationships (from Eq. 6.41)
Note also that P2 = P3 and P1 = P431
isentropic
processes
π·ππ
π·ππ=π·π
π·π
π·ππ
π·ππ=π·π
π·π
Where the values of βPrβ can be found in the tables
1) For Air standard analysis, i.e.,
Using air table data (not assuming constant specific heat)
Fall 2020
32
2) For Cold Air Standard Brayton Cycle analysis
We will develop an expression for the cycle
thermal efficiency, assuming constant
specific heat (or using cold air standard
analysis)
But first we will review some useful
pressure-temperature relationships
i.e., Assuming constant specific heat
Fall 2020
4
3
1
2NoteP
P
P
P=
For isentropic processes of 1-2
(compressor) and 3-4 (turbine) and
air is an ideal gas:
β
ββ
2
3
4
p=c
p=c
1
Review P-T relationship
Note: Cold air standard analysis assumes
constant specific heat 33
v
p
c
ckand =
πππ·π
=π»ππ»π
ππβπ
πππ·π
=π»ππ»π
ππβπ
Fall 2020
34
β
ββ
2
3
4
p=c
p=c
1
Also note :
πππ·π
=π»ππ»π
ππβπ ππ
π·π=
π»ππ»π
ππβπRecall
π·π
π·π=π·π
π·π
π»ππ»π
=π»ππ»π
π»ππ»π
=π»ππ»π
π»ππ»π
=π·π
π·π
πβππ
Fall 2020
Substitute in
35
πΌπππππ = π β(ππ β ππ)
(ππ β ππ)
πΌπππππ = π β(π»π β π»π)
(π»π β π»π)
πΌπππππ = π βπ»π(
π»ππ»π
β π)
π»π(π»ππ»π
β π)
Ξh = cpΞT for constant cP
Fall 2020
Recall :
36
For Cold Air
Standard Brayton
Cycle analysis
(Previously found)
πΌπππππ = π βπ»π(
π»ππ»π
β π)
π»π(π»ππ»π
β π)
π4π1
=π3π2
πΌπππππ = π βπ»ππ»π
Fall 2020
37v
p
c
ckNote =:
Recall :
For Cold Air
Standard Brayton
Cycle analysis
(Previously found)
πΌπππππ = π βπ»ππ»π
π»ππ»π
=π·π
π·π
πβππ
πΌπππππ = π βπ
π·ππ·π
πβππ
Fall 2020
Thermal efficiency using Cold Air Standard Brayton
Cycle analysis
Note also that it is for an ideal cycle, which does not account for
β variation of properties with temperature
β components inefficiency
β friction
β heat losses
β combustion losses, etc. 38
πΌπππππ = π βπ
ππ
πβππ
Where rp is the pressure ratio
ππ=π·πππ
π·πππ
πΌπππππ = π βπ
π·ππ·π
πβππ
Recall :
Fall 2020
We can plot the Brayton cycle efficiency
5 10 15 200
0.2
0.4
0.6
0.8
k = 1.5k = 1.4k = 1.3
Pressure Ratio
Th
erm
al E
ffic
ien
cy
.
Real Gas Turbine Cycle
ββ=
k
k
p
cycle
r
1
11
39Fall 2020
β’ Ideal Cycle
Ξ·th
Pressure Ratio, rp
100%
0%1
Ideal- The efficiency
continually
increases with
increasing
pressure ratio
Ξ·th
Pressure Ratio, rp
Increasing Tmax
Real Cycle
β’ Real Cycle
- Efficiency and
work output peek as
the pressure ratio
increases.
- Pressure ratio for
maximum
efficiency and work
output increase
with Tmax.
40Fall 2020
β’ Due to the irreversibilities and losses, the actual T-s diagrams would be as illustrated
41
Effects of Irreversibilities
Fall 2020
42
πΌπ =ππβππ
ππβπ=πππ β ππππ β ππ
πΌπ =ππβπ
ππβππ=
ππ β ππππ β πππ
Isentropic efficiency
For compressor For turbine
Fall 2020
Compressor
Combustor
Turbine
Inlet Air Exhaust
Power
Fuel
Shaft
β’ Brayton cycle is used in several gas turbine engines.
43
Applications of Brayton Cycle
Fall 2020
OGT 2500 Turboshaft
(Electricity generation)
T56 Turboshaft (Turboprop)
(Thrust generation)
Turbojet 44Fall 2020
Turboprop Turbofan
Ramjet or Scramjet
Different engine types:
45Fall 2020
2
expansion)c(isentropiexit at
expansion)(real
2
exita
V
V t
nozzle
Brayton cycle for jet engines (turbojet)
46Fall 2020
F119-PW-100 Afterburning Turbojet
47Fall 2020
Examples from textbook
Example 9.4: Analyzing
the Ideal Brayton Cycle
Example 9.6 : Evaluating
Performance of a
Brayton Cycle with
Irreversibilities
Fall 2020 48
Example 9.4: Analyzing the Ideal Brayton
Cycle
49
Find: Determine the thermal
efficiency, the back work
ratio, and the net power
developed
Known: An ideal air-
standard Brayton cycle
operates with given
compressor inlet conditions,
given turbine inlet
temperature, and a known
compressor pressure ratio.
Fall 2020
Example 9.4 (continued)
50Find: Determine the thermal efficiency, the back work
ratio, and the net power developedFall 2020
Solution Approach
51Fall 2020
Fixing the states
State 1:
π1 = 300 πΎ β πππππ π΄. 22 β ππ = πππ. ππππ±
ππ; π·ππ = π. πππ
State 2:
ππ2
ππ1=
π2
π1(for isentropic process only)
ππ2 = ππ1π2
π1= 1.386 10 = 13.86
Pr2 = 13.86 β πππππ π΄. 22 β ππ = πππ. πππ±
ππ
52
State 3:
π3 = 1400 πΎ β πππππ π΄. 22 β ππ = ππππ. πππ±
ππ; π·ππ = πππ. π
State 4:
ππ4
ππ3=
π4
π3(for isentropic process only)
ππ4 = ππ3π4π3
= 450.51
10= 45.05
ππ4 = 45.05 β πππππ π΄. 22 β ππ = πππ. πππ±
ππ
Fixing the states (continued)
Fall 2020
Substituting values in solving equations
53
A) Find thermal efficiency
π€π‘ = β3 β β4 = 1515.4 β 808.5ππ½
ππ= 706.9
ππ½
ππ
π€π = β2 β β1 = 579.9 β 300.19ππ½
ππ= 279.7
ππ½
ππ
πππ = β3 β β2 = 1515.4 β 579.9ππ½
ππ= 935.5
ππ½
ππ
πππ¦πππ =π€π‘βπ€π
πππ=
πππ.πβπππ.πππ½
ππ
πππ.πππ½
ππ
πΌπππππ = ππ. π% β π¨π΅πΊπΎπ¬πΉ (π)
Fall 2020
πππ¦πππ =π€π‘ βπ€ππππ
B) Find Back Work
Ratio (bwr)
ππ€π =π€ππ€π‘
=πππ. ππ
ππ½ππ
πππ. πππ½ππ
πππ = ππ. π%β π¨π΅πΊπΎπ¬πΉ (π)
Fall 2020 54
55
C) Find Net Power Developed (in kW)
Fall 2020
αΆπΎπ΅ππ = αΆπΎπππππ = αΆπΎπ β αΆπΎπ = αΆπ(ππ βππ)
πΎπ ππππ ππππππ π πππππ "ππ" πππ "ππ"π΅ππππ ππππ ππ ππππ αΆπ
αΆπ = αΆππ = ππ π½π¨ π =π½π¨ π
ππ
56Fall 2020
π€βπππ π£πππ’πππππ‘πππ ππππ€ πππ‘π ππ΄ 1 ππ πππ£ππ
π£1 =π ππππ1
π1(ππππ ππ£ = π π)
Where π πππ =ΰ΄€π
ππππ=
8.314ππ½
ππππ.πΎ
28.97ππ
ππππ
= 0.287ππ½
ππ.πΎ
Recall : αΆπ = αΆππ = ππ π½π¨ π =π½π¨ π
ππ
π£1 =0.287
ππ½
ππ.πΎ300 πΎ
100 πππ= 0.861 (
ππ½
ππ.πππ)
ππ.π
ππ½
πππ
ππ/π2
π£1 = 0.861π3
ππ
πΎπ ππππ ππ ππππ πππ ππππππππ ππππππ "ππ"
57
αΆπ =π½π¨ π
ππ=
5π3
π
0.861π3
ππ
= 5.807ππ
π
Substituting all values in solving equation:
αΆππππ‘ = αΆπ(π€π‘ β π€π)
αΆππππ‘ = (5.807ππ
π )(706.9 β 279.7)
ππ½
ππ
αΆπΎπ΅ππ = ππππ ππΎ β π¨π΅πΊπΎπ¬πΉ (π)
Fall 2020
αΆπ =π½π¨ π
ππRecall :
Example 9.6 : Evaluating Performance of a
Brayton Cycle with Irreversibilities
58
pr
Known: An air-standard Brayton cycle
operates with given compressor inlet
conditions, given turbine inlet temperature,
and known compressor pressure ratio. The
compressor and turbine each have an
isentropic efficiency of 80%.
Find: Determine the thermal efficiency, the
back work ratio, and the net power
developed, in kW
Fall 2020
Example 9.6 (continued)
59
pr
Find: Determine the thermal
efficiency, the back work
ratio, and the net power
developed, in kWFall 2020
Fixing the states
60
State 1:
π1 = 300 πΎ β πππππ π΄. 22 β β1 = 300.19ππ½
ππππ1 = 1.386
State 2s:ππ2π
ππ1=
π2π
π1(for isentropic process only)
ππ2π = ππ1π2π
π1= 1.386 10 = 13.86
ππ2π = 13.86 β πππππ π΄. 22 β β2π = 579.9ππ½
ππ
Fall 2020
61
State 2:
ππ =π€1β2π
π€1β2=
β2π ββ1
β2ββ1
β2 = β1 +β2π ββ1
ππ
β2 = 300.19ππ½
ππ+
579.9 β 300.19ππ½ππ
0.8
β2 = 649.8ππ½
ππ
Fall 2020
β2π = 579.9ππ½
ππRecall :
Fall 2020 62
State 3:
π3 = 1400 πΎ β πππππ π΄. 22 β β3 = 1515.4ππ½
ππππ3 = 450.5
State 4s:ππ4π
ππ3=
π4π
π3(for isentropic process only)
ππ4π = ππ3π4π π3
= 450.51
10= 45.05
ππ4π = 45.05 β πππππ π΄. 22 β β4π = 808.5ππ½
ππ
Fall 2020 63
State 4:
ππ‘ =π€3β4
π€3β4π =
β3ββ4
β3ββ4π
β4 = β3 β ππ‘(β3 β β4π )
β4 = 1515.4ππ½
ππβ 0.8 1515.4 β 808.5
ππ½
ππ
β4 = 949.9ππ½
ππ
β4π = 808.5ππ½
ππRecall :
Substituting values in solving equations
64
A) Find thermal efficiency
π€π‘ = β3 β β4 = 1515.4 β 949.9ππ½
ππ= 565.5
ππ½
ππ
π€π = β2 β β1 = 649.8 β 300.19ππ½
ππ= 349.6
ππ½
ππ
πππ = β3 β β2 = 1515.4 β 649.8ππ½
ππ= 865.6
ππ½
ππ
πππ¦πππ =π€π‘βπ€π
πππ=
565.5β349.6ππ½
ππ
865.6ππ½
ππ
πΌπππππ = ππ. π% β π¨π΅πΊπΎπ¬πΉ (π)
Fall 2020
65
B) Find Back Work Ratio (bwr)
ππ€π =π€ππ€π‘
=349.6
ππ½ππ
565.5ππ½ππ
= 61.8% β π¨π΅πΊπΎπ¬πΉ (π)
Fall 2020
66
C) Find Net Power Developed (kW)
αΆππππ‘ = αΆπππ¦πππ = αΆππ‘ β αΆππ = αΆπ(π€π‘ β π€π)
αΆπ = αΆπ1 = π1 ππ΄ 1 =ππ΄ 1
π£1ππππ’πππππ‘πππ ππππ€ πππ‘π ππ΄ 1 ππ πππ£ππ.
π£1 =π ππππ1
π1(ππππ ππ£ = π π)
with π πππ =ΰ΄€π
ππππ=
8.314ππ½
ππππ.πΎ
28.97ππ
ππππ
= 0.287ππ½
ππ.πΎ
Fall 2020
Where:
67
Substituting values:
π£1 =0.287
ππ½
ππ.πΎ300 πΎ
100 πππ= 0.861 (
ππ½
ππ.πππ)
ππ.π
ππ½
πππ
ππ/π2
π£1 = 0.861π3
ππ
Fall 2020
π£1 =π ππππ1π1
(ππππ ππ£ = π π)Recall:
Fall 2020 68
Mass flow rate
αΆπ =π½π¨ π
ππ=
5π3
π
0.861π3
ππ
= 5.8ππ
π
Substituting values in solving equation:αΆππππ‘ = αΆπ(π€π‘ βπ€π)
αΆππππ‘ = (5.8ππ
π )(565.5 β 349.6)
ππ½
ππ
αΆπΎπ΅ππ = ππππ ππΎ β π¨π΅πΊπΎπ¬πΉ (π)
End of Examples 9.4 & 9.6
Fall 2020 69
Regenerative Gas Turbines
70
Regenerator EffectivenessαΆπΈππ
αΆπ= πππ = ππ β ππ πΌπππ. =
ππ β ππππ β ππ
αΆπΈππ ππππ ππππππππ πππππππ (π. π. , πππ πππππππππ)
Regenerator
(a heat exchanger)
Fall 2020
71
Ideal Gas Turbines with Reheat
Fall 2020
72
Ideal Gas Turbines with Reheat (continued)
(From referenced textbook)
For metallurgical reasons, the temperature of the
gaseous combustion products entering the turbine
must be limited. This temperature can be controlled
by providing air in excess of the amount required to
burn the fuel in the combustor (see Chap. 13).
As a consequence, the gases exiting the combustor
contain sufficient air to support the combustion of
additional fuel.
Some gas turbine power plants take advantage of
the excess air by means of a multistage turbine
with a reheat combustor between the stages.Fall 2020
Ideal Gas Turbines with Reheat (continued)
(From referenced textbook)
73
Despite the increase in net work with
reheat, the cycle thermal efficiency would
not necessarily increase because a greater
total heat addition would be required
Fall 2020
Two-stage compression with intercooling
74
Fig. 9.18 Two-stage compression with intercoolingFall 2020
75
Two-stage compression with
intercooling (continued)
The compressor work input can
be reduced by multistage
compression with intercooling
Fall 2020
EXAMPLE FROM TEXTBOOK
Example 9.7 : Evaluating Thermal Efficiency of a Brayton
Cycle with Regeneration
Fall 2020 76
Example 9.7 : Evaluating Thermal Efficiency
of a Brayton Cycle with Regeneration
77
pr
Known: A regenerative gas turbine
operates with air as the working
fluid. The compressor inlet state,
turbine inlet temperature, and
compressor pressure ratio are
known.
Find: For a regenerator
effectiveness of 80%, determine the
thermal efficiency. Also plot the
thermal efficiency versus the
regenerator effectiveness ranging
from 0 to 80%.
Fall 2020
Example 9.7 (continued)
78
pr
Find: For a regenerator effectiveness of 80%, determine
the thermal efficiency Fall 2020
Fall 2020 79
πΌπππππ =ππ βππ
πππ
Find Thermal Efficiency
πΌπππππ =ππ β ππ β (ππ β ππ)
(ππβππ)
Note: h1, h2, h3 and h4 are
the same as in Example 9.4
We need to find hx.
Note 1: On the Regenerator
(or Heat Exchanger)
80
ππππππππππ‘ππ =βπ₯ β β2β4 β β2
βπ₯ β β2 :Heat energy actually received by a unitmass of coolant in the βcooler stream"
β4 β β2 : Hypothetical maximum of heat energy that could be
received by a unit mass of coolant in the βcooler stream"
2
4
x
Hotter
stream
Cooler
stream
(4 to y)
(2 to x)
Efficiency of regenerator:
Where:
Fall 2020
81
ΰ΅―ππ = ππ + πΌπππ(ππ β ππ
ππππππππππ‘ππ =βπ₯ β β2β4 β β2
Recall:
ππππ πππ ππ
Fall 2020
ππ = πππ. πππ±
ππ+ π. π πππ. π β πππ. π
ππ±
ππ
ππ = πππ. πππ±
ππ
Note: h1, h2, h3 and h4 are the same as in Example 9.4
2
4
x
Hotter
stream
Cooler
stream
(4 to y)
(2 to x)
Note 2: "πππ"
82
πππ¦πππ =π€π‘ β π€ππππ
Where πππ =αΆπΈππ
αΆπ= ππ β ππ
αΆπΈππ ππππ ππππππππ πππππππ (π. π. , ππππ π πππ πππππππππ)
For Brayton Cycle with Regeneration
The heat energy that the working fluid receives in the
regenerator (ππ«π¨π¦ π¬ππππ’π¨π§ π ππ π) is not counted in the
cycle efficiency expression
Itβs a free energy obtained from the exhaust gas !!!
Fall 2020
Fall 2020 83
πππ¦πππ =π€π‘ βπ€ππππ
Substituting values
πΌπππππ =ππ β ππ β (ππ β ππ)
(ππβππ)
πΌπππππ =ππππ. π β πππ. π
ππ±ππ
β πππ. π β πππ. ππππ±ππ
ππππ. π β πππ. πππ±ππ
πΌπππππ = π. πππ = ππ. π% β π¨πππππ
Related Notes on Regenerator
84
αΆππ―ππ(ππ β ππ) αΆππͺπππ(ππ β ππ)
Assuming the same mass flow rate in both hot and cool streams, then
The rate of heat
transfer from the
hotter stream
The rate of heat
received by the
cooler stream
(ππ β ππ) (ππ β ππ)
Ideally
Ideally
Ideally
Reality% ππ (ππ β ππ) (ππ β ππ)
Fall 2020
2
4
x
Hotter
stream
Cooler
stream
9.4 SUMMARY (1)
β’ Cold Air Standard Otto Cycle
1
11
ββ=
k
v
Ottor
- Used to model spark-ignition internal combustion
engines
rv: Volumetric
Compression Ratio
3
2
4
v=c
v=c
1
85Fall 2020
9.4 SUMMARY (2a)
121221For hhw β=β
433443For hhw β=β
232332For hhq β=β
144114For hhq β=β
NOTE: wβs and qβs are written such that all have positive values
Air Standard Brayton Cycle
1-2 Compression: isentropic
2-3 Heat added at constant pressure
3-4 Expansion: isentropic
4-1 Heat rejected at constant pressure
β
ββ
2
3
4p=c
p=c
1
- Used to model gas turbine engines
86Fall 2020
9.4 SUMMARY (2b)
87
)(
)(1
23
14
hh
hhcycle
β
ββ=
Braytonβs Cycle thermal efficiency
Fall 2020
9.4 SUMMARY (3)
β’ Cold Air Standard Brayton Cycle
- Used to model gas turbine engines
ββ=
k
k
p
Brayton
r
1
11
Pressure Ratio rp = pmax / pmin
β
β
3
2
4
p=c
p=c
1
88Fall 2020
Regenerative Gas Turbines
89
Regenerator Effectiveness
πΌπππ. =ππ β ππππ β ππ
9.4 SUMMARY (4)
Regenerator
(a heat exchanger)
αΆπΈππ
αΆπ= πππ = ππ β ππ
αΆπΈππ ππππ ππππππππ πππππππ (π. π. , πππ πππππππππ)Fall 2020