1. ‫خدا‬ ‫نام‬ ‫به‬

2. A BRIEF SEMINAR PRESENTATION ABOUT
TURBOMACHINES
• 2nd project for thermodynomics
• Presented by Shahin Tavakoli and Hasan Parvin

3. OVERLOOK
• Introduction – categorization – classiffication
• Types of turbomachines and the components
• Systems using in turbomachines
• Off-design (power transfer equation)
• On-design (ideal turbomachine analysis)

4. TURBOMACHINES

5. ‫توربوماشین‬
• ‫انتقال‬ ‫عامل‬ ‫و‬ ‫میکنند‬ ‫دریافت‬ ‫انرژی‬ ‫سیال‬ ‫از‬ ‫یا‬ ‫میدهند‬ ‫انرژی‬ ‫سیال‬ ‫به‬ ‫خود‬ ‫محور‬ ‫دوران‬ ‫با‬ ‫که‬ ‫ماشینهایی‬ ‫تمامی‬
‫است‬ ‫ماشین‬ ‫محور‬ ‫دوران‬ ‫قدرت‬.

6. ‫توربوماشینها‬ ‫تقسیمبندی‬
‫توربوماشینها‬
‫پذیر‬ ‫تراکم‬ ‫سیال‬
‫ماشین‬ ‫به‬ ‫سیال‬ ‫از‬ ‫انرژی‬ ‫انتقال‬
(‫گازی‬ ‫های‬ ‫توربین‬)
‫سیال‬ ‫به‬ ‫ماشین‬ ‫از‬ ‫انرژی‬ ‫انتقال‬
(‫ها‬ ‫کمپرسور‬)
‫ناپذیر‬ ‫تراکم‬ ‫سیال‬
‫ماشین‬ ‫به‬ ‫سیال‬ ‫از‬ ‫انرژی‬ ‫انتقال‬
(‫هیدرولیکی‬ ‫های‬ ‫توربین‬)
‫سیال‬ ‫به‬ ‫ماشین‬ ‫از‬ ‫انرژی‬ ‫انتقال‬
(‫ها‬ ‫فن‬ ‫و‬ ‫ها‬ ‫پمپ‬)

7. PUT TURBOMACHINES TOGETHER TO CREAT
“THRUST”
Propulsive turbomachines
Air breathing
Turbine powered
Ram powered
Non-continuous combustion
Non-air breathing
rocket
hybrid
waterjet

8. TURBINE POWERED AIR BREATHING
TURBOMACHINES
turbojet
turboprop
turbofan

9. TURBOJETS

10. TURBOPROPS

11. TURBOFANS

12. FOR EXAMPLE
differences between an aero-drivative and a heavy industrial turbomachine

13. TURBOSHAFT TO TURBOFAN
• Weight heavier
• Shaft speed slower
• Air flow higher
• Maintenance time longer
• Maintenance lay-down space longer

14. COMPRESSORS
Centrifugal Axial

15. COMPRESSORS
• Pressure ratio 5-1 world war 2
12-1 newer
30-1 recently
Better pressure ratio better thermal efficiiency
The best thermal efficiency is about 35 %
To achieve this thermal efficiency we should use some forms of “waste heat recovery”
Special material (Al , Fe and their alloys)

16. TURBINES
Turbofan high aspect ratio (long , thin blades) with tip shrouds
1.to dampen vibration
2.improve blade tip sealing characteristics
Turboshaft low aspect ratio (short , thick blades)
with no tip shrouds
Long thin airfoils need 1.lacing wire to dampen vibration
2.tip shrouds or mid-spam shrouds

17. TURBINES

18. BEARING DESIGN AND LUBRICATION
Turbofans anti friction roller anb ball bearings
25 – 400 gallons of oil
Turboshafts hydrodynamic bearings
1500 – 2500 gallons of oil

19. SYSTEMS
Starting
systems
Thrust
augmentation Ignition
systems
Lubrication
systems

20. STARTING SYSTEMS

21. IGNITION SYSTEMS

22. LUBRICATION SYSTEMS

23. GAS TURBINE INLET TREATMENT
• Air inlet cooling

24. GAS TURBINE EXHAUST TREATMENT
• 1.pollution
• 2.power augmentation

25. POLLUTION
*carbon mono-oxide(co) 1.function of combustion design
2.can be treated with a catalytic converter
*oxides of nitrogen(NOx) 1. (organic NO)
2.produced in hot regions (thermal NO)
ℎ𝑦𝑑𝑟𝑜𝑐𝑎𝑟𝑏𝑜𝑛. 𝑓𝑢𝑒𝑙 + 𝑎𝑖𝑟 →. . . . . . . +𝑁𝑂𝑥

26. POLLUTION

27. POLLUTION

28. THRUST (POWER) AUGMENTATION
• First way : cooling air entering combustor
a) Increasing air density
b) Increasing mass flow
c) More air and more cooled air to the combustor permits more fuel to be burned
before reaching the turbine inlet temperaturre limit
To cooling the air enterring combustor, water or steam injection is ok

29. WATER INJECTION
into diffuser, compressor or combustor
reducing combustion and turbine temperature
reducing oxides of nitrogen up to 80 %
Example in a heavy industrial gas turbine with 80 MW power output
ratio of water to fuel is 0.6
water is 1.15 % of total air intaking

30. WATER INJECTION
.

31. STEAM INJECTION
energy to vaporizing the water is conserved
reducing combustion and turbine temperature
steam is hot so it’s quenching capabilities are reduced
so more steam than the water is required to accomplish the same
amounts of nitrogen oxides
Example in an airo-drivative gas turbine with 25 MW power outpuut
ratio of steam to fuel is up to 2.4
steam is 3.3 % of total air

32. STEAM INJECTION

33. SELECTIVE CATALYTIC REDUCTION
(SCR)
• Injecting ammonia

34. AFTER BURNING OR REHEATING
Effect of after burner raising the exhaust temperature
raising the velocity of exhaust gases
Example with afterburner thrust = 17,900 lbf
TSFC = 1.956 ((lbm/hr)/lbf)/hr
without afterburner thrust = 11,870 lbf
TSFC = 0.86((lbm/hr)/lbf)/hr

35. OFF-DESIGN
(ENGINE PERFORMANCE ANALYSIS)
‫اویلر‬ ‫ی‬ ‫معادله‬(‫نیوتن‬ ‫دوم‬ ‫قانون‬ ‫از‬ ‫استفاده‬ ‫با‬)
‫محوری‬ ‫سرعت‬–Va‫محور‬ ‫با‬ ‫موازی‬o-o
‫شعاعی‬ ‫سرعت‬–Vm‫با‬ ‫موازی‬r
‫مماسی‬ ‫سرعت‬–Vt‫روتور‬ ‫گردش‬ ‫جهت‬ ‫با‬ ‫موازی‬

36. ‫سرعت‬ ‫های‬ ‫مولفه‬ ‫تغییرات‬
‫کفگرد‬ ‫های‬ ‫یاتاقان‬ ‫با‬ ‫شدن‬ ‫خنثی‬ ‫محوری‬ ‫نیروی‬ ‫ایجاد‬ ‫محوری‬ ‫سرعت‬ ‫تغییرات‬
‫شعاعی‬ ‫نیروی‬ ‫ایجاد‬ ‫روتور‬ ‫طول‬ ‫در‬ ‫مومنتوم‬ ‫تغییرات‬ ‫شعاعی‬ ‫سرعت‬ ‫تغییرات‬
‫روتور‬ ‫محور‬ ‫حول‬ ‫کوپل‬ ‫ایجاد‬ ‫گردشی‬ ‫ممان‬ ‫تغییر‬ ‫مماسی‬ ‫سرعت‬ ‫تغییرات‬

37. ‫انرژی‬ ‫انتقال‬ ‫ی‬ ‫رابطه‬ ‫آوردن‬ ‫بدست‬
‫کنترل‬ ‫حجم‬ ‫عنوان‬ ‫به‬ ‫آن‬ ‫اطراف‬ ‫سیال‬ ‫و‬ ‫روتور‬ ‫گرفتن‬ ‫نظر‬ ‫در‬ ‫و‬ ‫نیوتن‬ ‫دوم‬ ‫قانون‬ ‫اعمال‬
‫ورودی‬:‫با‬ ‫سیال‬‫و‬‫در‬
‫خروجی‬:‫با‬ ‫سیال‬‫و‬‫در‬
‫پایا‬ ‫جریان‬ ‫فرض‬ ‫با‬:
‫پس‬:
‫بنابراین‬:
𝑉𝑡1𝑚1𝑟1
𝑉𝑡2𝑟2 𝑚2
𝑇 = 𝑚(𝑟1. 𝑉𝑡1 − 𝑟2. 𝑉𝑡2
𝑊 = 𝑇. 𝜔 = 𝑚(𝑟1. 𝑉𝑡1. 𝜔 − 𝑟2. 𝑉𝑡2. 𝜔
𝑢 = 𝑟𝜔
𝑊 = 𝑚(𝑢1. 𝑉𝑡1 − 𝑢2. 𝑉𝑡2

38. ‫توربوماشین‬ ‫هد‬
• H 1 1 2 2t tuV u VW
H
mg g

 
&
&
𝐻 =
𝑊
𝑚𝑔
=
𝑢1 𝑉𝑡1 − 𝑢2 𝑉𝑡2
𝑔

39. ‫واقعی‬ ‫و‬ ‫آل‬ ‫ایده‬ ‫های‬ ‫توربوماشین‬ ‫تفاوت‬
‫آل‬ ‫ایده‬
‫واقعی‬
‫سیال‬ ‫هد‬ ‫توربین‬>‫روتور‬ ‫هد‬
‫سیال‬ ‫هد‬ ‫کمپرسور‬<‫روتور‬ ‫هد‬
‫سیستم‬ ‫توسط‬ ‫شده‬ ‫انجام‬ ‫کار‬ ‫قرارداد‬+‫سیستم‬ ‫روی‬ ‫شده‬ ‫انجام‬ ‫کار‬–
‫پس‬:‫کمپرسور‬ ‫هد‬-‫پمپ‬ ‫هد‬-‫توربین‬ ‫هد‬+
𝑉𝑡‫سیال‬ = 𝑉𝑡‫روتور‬
𝑉𝑡‫سیال‬ ≠ 𝑉𝑡‫روتور‬

40. ‫روتور‬ ‫و‬ ‫سیال‬ ‫بین‬ ‫انرژی‬ ‫انتقال‬
‫سیال‬ ‫نسبی‬ ‫سرعت‬2 2 2
2 2 2 2 2
2 2 2
1 1 1 1 1
1
( )
2
1
( )
2
r
t r
t r
V V u
u V V u V
uV V u V
 
  
  
𝑉𝑟 = 𝑉 − 𝑢2 2 2 2 2 2
1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )
2 2 2
r rV V u u V V
H
g g g
  
  
𝑢1 𝑉𝑡1 =
𝑉1
2
+ 𝑢1
2
− 𝑉𝑟1
2
2
𝑢2 𝑉𝑡2 =
𝑉2
2 + 𝑢2
2 − 𝑉𝑟2
2
2
𝐻 =
𝑉1
2 − 𝑉2
2
2𝑔
+
𝑢1
2 − 𝑢2
2
2𝑔
+
𝑉𝑟1
2 − 𝑉𝑟2
2
2𝑔

41. DYNAMIC HEAD
‫سیال‬ ‫جنبشی‬ ‫انرژی‬ ‫تغییرات‬
2 2
1 2
2 2
1 2
2 2
1 2
( ) ( )
2
( ) ( )
2
( ) ( )
2
r r
V V
g
u u
g
V V
g



𝑉1
2
− 𝑉2
2
2𝑔

42. STATIC HEAD
‫سیال‬ ‫نسبی‬ ‫سرعت‬ ‫تغییرات‬ ‫اثر‬ ‫بر‬ ‫سیال‬ ‫جنبشی‬ ‫انرژی‬ ‫تغییرات‬
‫است‬ ‫روتور‬ ‫سر‬ ‫دو‬ ‫در‬ ‫فشار‬ ‫تغییرات‬ ‫ایجاد‬ ‫آن‬ ‫ی‬ ‫نتیجه‬
𝑉𝑟1
2
− 𝑉𝑟2
2
2𝑔

43. CENTRIFUGAL ENERGY
‫انرژی‬ ‫تغییرات‬‫بر‬ ‫سیال‬‫شعاع‬ ‫تغییرات‬ ‫و‬ ‫چرخش‬ ‫اثر‬
𝑢1
2
− 𝑢2
2
2𝑔
1
2
( 𝑢2
2
− 𝑢1
2
=
𝑝1
𝑝2
𝜐. 𝑑𝑝

44. IMPULSE
DEGREE OF REACTION
‫پذیر‬ ‫تراکم‬ ‫سیال‬ ‫در‬ ‫آنتالپی‬ ‫تغییر‬=‫استاتیکی‬ ‫فشار‬ ‫تغییر‬ ‫اثر‬ ‫بر‬ ‫انرژی‬ ‫انتقال‬
𝑅 =
𝑢1
2 − 𝑢2
2 + 𝑉𝑟2
2 − 𝑉𝑟1
2 2 𝑔
𝐻

45. EFFICIENCY OF TURBINE
Overal efficiency
(‫سیال‬ ‫ی‬ ‫استفاده‬ ‫قابل‬ ‫هیدرودینامیکی‬ ‫انرژی‬(/)‫انرژی‬‫سیال‬ ‫سوی‬ ‫از‬ ‫توربین‬ ‫محور‬ ‫مکانیکی‬= )
Adiabatic or hydraulic efficiiency
(‫سیال‬ ‫ی‬ ‫استفاده‬ ‫قابل‬ ‫هیدرودینامیکی‬ ‫انرژی‬(/)‫انرژی‬‫روتور‬ ‫به‬ ‫شده‬ ‫داده‬ ‫مکانیکی‬)=
Mechanical efficiency
)‫انرژی‬‫روتور‬ ‫به‬ ‫شده‬ ‫داده‬ ‫مکانیکی‬(/)‫سیال‬ ‫سوی‬ ‫از‬ ‫توربین‬ ‫محور‬ ‫مکانیکی‬ ‫انرژی‬= )
𝜂 𝑜 𝑡
t
t
t
o
m
h




𝜂ℎ 𝑡
𝜂 𝑚 𝑡
=
𝜂 𝑜 𝑡
𝜂ℎ 𝑡

46. EFFICIENCY OF COMPESSOR
Overal efficiency
=(‫توربوماشین‬ ‫در‬ ‫سیال‬ ‫مفید‬ ‫هیدرودینامیکی‬ ‫انرژی‬ ‫توربوماشین(/)افزایش‬ ‫محور‬ ‫به‬ ‫شده‬ ‫داده‬ ‫مکانیکی‬ ‫)انرژی‬
Adiabatic or hydraulic efficiiency
=(‫توربوماشین‬ ‫در‬ ‫سیال‬ ‫مفید‬ ‫هیدرودینامیکی‬ ‫انرژی‬ ‫(/)افزایش‬ ‫مکانیکی‬ ‫انرژی‬‫توربوماشین‬ ‫روتور‬ )
Mechanical efficiency
=(‫توربوماشین‬ ‫روتور‬ ‫مکانیکی‬ ‫توربوماشین(/)انرژی‬ ‫محور‬ ‫به‬ ‫شده‬ ‫داده‬ ‫مکانیکی‬ ‫)انرژی‬
c
c
c
o
m
h




𝜂 𝑚 𝑐
=
𝜂 𝑜 𝑐
𝜂ℎ 𝑐
𝜂 𝑜 𝑐
𝜂ℎ 𝑐

47. ON-DESIGN
(PARAMETRIC CYCLE ANALYSIS)
• Total or stagnation temprature
• Static or thermodynamic temperature
• Flow velocity
• First law of thermodynamic
• Mach number
𝑇𝑡
𝑇
𝑣
𝑇𝑡 = 𝑇 +
𝑣2
2𝑔 𝑐 𝑐 𝑝
𝑀 =
𝑣
𝑎
=
𝑣
𝛾𝑔 𝑐 𝑅𝑇
𝑇𝑡 = 𝑇(1 +
𝛾 − 1
2
. 𝑀2 𝑝𝑡
𝑝
=
𝑇𝑡
𝑇
𝛾−1
𝛾
𝑝𝑡 = 𝑝 1 +
𝛾 − 1
2
. 𝑀2
𝛾
𝛾−1

48. TOTAL
PRESSURE AND TEMPRETURE RATIO
diffuser (d) compressor (c) burner (b) turbine (t)
Nozzle (n) fan (f) free stream (r)
= ( total pressure leaving a / total pressure entring a )
= ( total temperature leaving a / total temperature entring a )
𝜋 𝑎
𝜏 𝑎
𝜏 𝑟 =
𝑇𝑡0
𝑇0
= 1 +
𝛾 − 1
2
. 𝑀2
𝜋 𝑟 =
𝑝𝑡0
𝑝0
= 1 +
𝛾 − 1
2
. 𝑀2
𝛾
𝛾−1
𝑇𝑡0 = 𝑇0. 𝜏 𝑟𝑝𝑡0 = 𝑝0. 𝜋 𝑟
𝜏 𝜆 =
ℎ 𝑡 𝑏𝑢𝑟𝑛𝑒𝑟
ℎ 𝑜
=
𝑐 𝑝 𝑇𝑡 𝑏𝑢𝑟𝑛𝑒𝑟
𝑐 𝑝 𝑇0

49. STEPS OF ENGINE PARAMETRIC
CYCLE ANALYSIS

50. DESIGN INPUT
• 1. flight conditions
• 2. design limits
• 3. component performance
• 4. design choise
𝑝0, 𝑇0, 𝑀0, 𝑐 𝑝, 𝜋 𝑟, 𝜏 𝑟
𝑐 𝑝. 𝑇𝑡 𝑏𝑢𝑟𝑛𝑒𝑟
𝜋 𝑑, 𝜋 𝑛, 𝜋 𝑏, 𝑒𝑡𝑐.
𝜋 𝑐, 𝜋 𝑓, 𝑒𝑡𝑐.

51. RELATIONS FOR ALL T AND P

52. STEP 1
• 1. equation for thrust
𝐹 =
1
𝑔 𝑐
𝑚9. 𝑣𝑒 − 𝑚0. 𝑣0 + 𝐴9 𝑝9 − 𝑝0
→
𝐹
𝑚0
=
𝑎0
𝑔 𝑐
𝑚9
𝑚0
𝑣9
𝑎0
− 𝑀0 +
𝐴9 𝑝9
𝑚0
1 −
𝑝0
𝑝9

53. STEP 2
• 2. velocity ratio(s) in terms of mach numbers , temperatures and …
𝑣9
𝑎0
2
=
𝑎9
2
𝑀9
2
𝑎0
2
=
𝛾9 𝑅9 𝑔 𝑐 𝑇9
𝛾0 𝑅0 𝑔 𝑐 𝑇0
. 𝑀9
2

54. STEP 3
• 3. find the exit mach number
Then
Where
r d c b t AB n      
𝑀9
𝑝𝑡9 = 𝑝9 1 +
𝛾 − 1
2
. 𝑀9
2
𝛾
𝛾−1
𝑀9
2 =
2
𝛾 − 1
𝑝𝑡9
𝑝9
𝛾−1
𝛾
− 1
𝑝𝑡9
𝑝9
=
𝑝0
𝑝9
𝑝𝑡0
𝑝0
𝑝𝑡2
𝑝𝑡0
𝑝𝑡3
𝑝𝑡2
𝑝𝑡4
𝑝𝑡3
𝑝𝑡5
𝑝𝑡4
𝑝𝑡7
𝑝𝑡5
𝑝𝑡9
𝑝𝑡7
=
𝑝0
𝑝9
𝜋 𝑟 𝜋 𝑑 𝜋 𝑐 𝜋 𝑏 𝜋 𝑡 𝜋 𝐴𝐵 𝜋 𝑛

55. STEP 4
• 4. find the temperature ratio
Where
𝑇9 𝑇0
𝑇9
𝑇0
=
𝑇𝑡9 𝑇0
𝑇𝑡9 𝑇9
=
𝑇𝑡9 𝑇0
𝑝𝑡9 𝑝9
𝛾−1
𝛾
𝑇𝑡9
𝑇0
=
𝑇𝑡0
𝑇0
𝑇𝑡2
𝑇𝑡0
𝑇𝑡3
𝑇𝑡2
𝑇𝑡4
𝑇𝑡3
𝑇𝑡5
𝑇𝑡4
𝑇𝑡7
𝑇𝑡5
𝑇𝑡9
𝑇𝑡7
= 𝜏 𝑟 𝜏 𝑑 𝜏 𝑐 𝜏 𝑏 𝜏 𝑡 𝜏 𝐴𝐵 𝜏 𝑛

56. STEP 5
• 5. apply the 1st law of thermodynamics to the burner and find an expression for the
fuel/air ratio
0 3 0 4p t f PR p tm c T m h m c T & & &
𝑚0 𝑐 𝑝 𝑇𝑡3 + 𝑚 𝑓ℎ 𝑃𝑅 = 𝑚0 𝑐 𝑝 𝑇𝑡4

57. STEP 6,7
• 6. when applicable, find an expression for the total temperature ratio across the
turbine by relating the turbine power output to the compressor, fan, and/or
propeller power requirements. This allows us to find in terms of other variables.
• 7. evaluate the specific thrust using the above results
t
𝜏 𝑡
𝜏 𝑡

58. STEP 8,9
• 8. evaluate the thrust specific fuel consumption , using the results for specific
thrust and fuel/air ratio
• 9. develop expressions for the thermal and propulsive efficiencies
1
1
d n
d n
 
 
 
 
𝑠
𝑠 =
𝑓
𝐹 𝑚0

59. ASSUMPTIONS
OF IDEAL CYCLE ANALYSIS
• 1.the working fluid is air and acts as a perfect gas
• 2.isentropic ( reversible and adiabatic ) processes
𝜏 𝑑 = 𝜏 𝑛 = 1 𝜋 𝑑 = 𝜋 𝑛 = 1
𝜏 𝑐 = 𝜋 𝑐
𝛾−1
𝛾 𝜏 𝑡 = 𝜋 𝑡
𝛾−1
𝛾

60. ASSUMPTIONS
OF IDEAL CYCLE ANALYSIS
• 3.the exhaust nozzle expand the gas to the ambient pressure
• 4.constant pressure combustion
; ;
𝑝 𝑒 = 𝑝0
𝜋 𝑏 = 1
𝑚 𝑓 <<< 𝑚 𝑎𝑖𝑟
𝑚 𝑓
𝑚 𝑐
<< 1 𝑚 𝑐 + 𝑚 𝑓 ≅ 𝑚 𝑐

61. ANALYZING AN IDEAL TURBOJET

62. IDEAL CYCLE

63. IDEAL
TURBOJET CYCLE ANALYSIS
Assumptions
𝜂 𝑡ℎ = 1 −
𝑇0
𝑇𝑡3
= 1 −
1
𝜏 𝑟 𝜏 𝑐
t cw w& &
𝑤𝑡 = 𝑤𝑐 𝑇𝑡4 > 𝑇𝑡3 𝑝𝑡4 > 𝑝𝑡3

64. STEP 1
F
𝐹
𝑚0
=
1
𝑔 𝑐
(𝑣9 − 𝑣0 =
𝑎0
𝑔 𝑐
(
𝑣9
𝑎0
− 𝛭0

65. STEP 2
v , a
𝑣9
𝑎0
2
=
𝑎9
2
𝑀9
2
𝑎0
2 =
𝛾9 𝑅9 𝑔 𝑐 𝑇9 𝑀9
2
𝛾0 𝑅0 𝑔 𝑐 𝑇0
=
𝑇9
𝑇0
𝑀2

66. STEP 3
Total exit pressure
;
However thus
So and so
𝑝𝑡9 = 𝑝9 1 −
𝛾 − 1
2
. 𝑀9
2
𝛾
𝛾−1
𝑃𝑡9 = 𝑃0
𝑝𝑡0
𝑝0
𝑝𝑡2
𝑝𝑡1
𝑝𝑡3
𝑝𝑡2
𝑝𝑡4
𝑝𝑡3
𝑝𝑡5
𝑝𝑡4
𝑝𝑡9
𝑝𝑡5
0 r d c b t np      
= 𝑝0 𝜋 𝑟 𝜋 𝑑 𝜋 𝑐 𝜋 𝑏 𝜋 𝑡 𝜋 𝑛
𝜋 𝑑 = 𝜋 𝑏 = 𝜋 𝑛 = 1 𝑝𝑡9 = 𝑝0 𝜋 𝑟 𝜋 𝑐 𝜋 𝑡
𝑀9
2
=
2
𝛾 − 1
[
𝑝𝑡9
𝑝9
𝛾−1
𝛾
− 1 𝑀9
2
=
2
𝛾 − 1
(𝜏 𝑟 𝜏 𝑐 𝜏 𝑡 − 1

67. STEP 4
Total exit temperature
Then
Thus
𝛵𝑡9 = 𝑇0
𝑇𝑡0
𝑇0
𝑇𝑡2
𝑇𝑡0
𝑇𝑡3
𝑇𝑡1
𝑇𝑡4
𝑇𝑡2
𝑇𝑡5
𝑇𝑡3
𝑇𝑡9
𝑇𝑡5
= 𝑇0 𝜏0 𝜏 𝑟 𝜏 𝑑 𝜏 𝑐 𝜏 𝑏 𝜏 𝑡 𝜏 𝑛 = 𝑇0 𝜏 𝑟 𝜏 𝑐 𝜏 𝑏 𝜏 𝑡
𝑇9
𝑇0
= 𝜏 𝑏
𝑇9
𝑇0
=
𝑇𝑡9
𝑇0
𝑇𝑡9
𝑇9
=
𝜏 𝑟 𝜏 𝑐 𝜏 𝑏 𝜏 𝑡
𝑝𝑡9
𝑝9
𝛾−1
𝛾
=
𝜏 𝑟 𝜏 𝑐 𝜏 𝑏 𝜏 𝑡
− 𝜋 𝑟 𝜋 𝑐 𝜋 𝑡
𝛾−1
𝛾
=
𝜏 𝑟 𝜏 𝑐 𝜏 𝑏 𝜏 𝑡
𝜏 𝑟 𝜏 𝑐 𝜏 𝑡

68. STEP 5
Steady flow energy equation
𝑚0ℎ 𝑡3 + 𝑚 𝑓ℎ 𝑃𝑅 = 𝑚0 + 𝑚 𝑓 ℎ 𝑡4
𝑓 =
𝑚 𝑓
𝑚0
=
𝑐 𝑝 𝑇0
ℎ 𝑃𝑅
𝜏 𝜆 − 𝜏 𝑟 𝜏 𝑐
𝑓 =
𝑚 𝑓
𝑚0
=
𝑐 𝑝 𝑇0
ℎ 𝑃𝑅
𝑇𝑡4
𝑇0
−
𝑇𝑡3
𝑇0

69. STEP 6
The power output of the turbine
The power required to drive the compressor
0 4 1p t
T
m c T
 
  
 
& 𝑤𝑡 = 𝑚0 + 𝑚 𝑓 ℎ 𝑡4 − ℎ 𝑡5 = 𝑚0 𝑐 𝑝 ℎ 𝑡4 − ℎ 𝑡5
= 𝑚0 𝑐 𝑝 𝑇𝑡4 1 −
𝑇𝑡5
𝑇𝑡4
= 𝑚0 𝑐 𝑝 𝑇𝑡4 1 − 𝜏 𝑡
𝑤𝑐 = 𝑚0 ℎ 𝑡3 − ℎ 𝑡2 = 𝑚0 𝑐 𝑝 𝑇𝑡3 − 𝑇𝑡2 = 𝑚0 𝑐 𝑝 𝑇𝑡2
𝑇𝑡3
𝑇𝑡2
− 1
= 𝑚0 𝑐 𝑝 𝑇𝑡2 𝜏 𝑐 − 1

70. STEP 6
Since for the ideal turbojet, then
or
thus
c tw w& &
𝑤𝑐 = 𝑤𝑡 𝑚0 𝑐 𝑝 𝑇𝑡2 𝜏 𝑐 − 1 = 𝑚0 𝑐 𝑝 𝑇𝑡4 1 − 𝜏 𝑡
𝜏 𝑡 = 1 −
𝑇𝑡2
𝑇𝑡4
𝜏 𝑐 − 1
𝜏 𝑡 = 1 −
𝜏 𝑟
𝜏 𝜆
𝜏 𝑐 − 1

71. STEP 7
Specific thrust
But
thus
   
2
29 9
9
0 0
2
1
1
r c t
r c
v T
M
a T

  
  
 
   
 
𝑣9
𝑎0
2
=
𝑇9
𝑇0
𝑀9
2
=
2
𝛾 − 1
𝜏 𝜆
𝜏 𝑟 𝜏 𝑐
𝜏 𝑟 𝜏 𝑐 𝜏 𝑡 − 1
𝐹
𝑚0
=
𝑎0
𝑔 𝑐
[
2
𝛾 − 1
𝜏 𝜆
𝜏 𝑟 𝜏 𝑐
𝜏 𝑟 𝜏 𝑐 𝜏 𝑡 − 1 − 𝑀0
𝐹
𝑚0
=
𝑎0
𝑔 𝑐
(
𝑣9
𝑎0
− 𝑀0

72. STEP 8
Specific thrust fuel consumption
So
𝑠 =
𝑓
𝐹 𝑚0
pc
s 
𝑠 =
𝑐 𝑝 𝑇0 𝑔 𝑐 𝜏 𝜆 − 𝜏 𝑟 𝜏 𝑐
𝑎0ℎ 𝑃𝑅
2
𝛾 − 1
𝜏 𝜆
𝜏 𝑟 𝜏 𝑐
𝜏 𝑟 𝜏 𝑐 𝜏 𝑡 − 1 − 𝑀0

73. STEP 9
Efficiencies
Thermal efficiency
Propulsive efficiency
Overal efficiency
.o th p  
𝜂 𝑡ℎ = 1 −
1
𝜏 𝑟 𝜏 𝑐
𝜂 𝑝 =
2𝑀0
𝑣9 𝑎0 + 𝑀0
𝜂 𝑜 = 𝜂 𝑡ℎ. 𝜂 𝑝

83. REFERENCES
1. Elements of gas turbine propulsion

84. REFERENCES
2. Gas turbine handbook

85. REFERENCES
3. Gas turbine theory

86. REFERENCES
4. Turbomachines