Prerequisite - Thermodynamics/ Propulsion Review of Fundamentals/ Ideal Turbojet Performance/ Component Performance/ Non Ideal Cycle Analysis/ Ideal Turbofan Performance

1. Aeropropulsion
Unit
Non Ideal Turbofan Cycle Analysis
2005 - 2010
International School of Engineering, Chulalongkorn University
Regular Program and International Double Degree Program, Kasetsart University
Assist. Prof. Anurak Atthasit, Ph.D.

2. Aeropropulsion
Unit
Kasetsart University
2
A. ATTHASIT
Actual Turbofan Cycle In-class Practice: Ch07P01
Prove
•Obj: Able to use the fundamental equation under the correct assumptions
Analysis
•Obj: Understand the physical meaning of each parameters
Calculation
•Obj: Able to solve the relations under the constraints of corrected unit, constant, … etc.
Standard Sea Level:
T0=288.2 K
P0=101.33 kPa
gc=1.4, gt=1.3
Cpc=1.004 kJ/kg/K
Cpt=1.239 kJ/kg/K
A turbofan flies at sea level at a Mach number of 0.75. It ingests 74.83 kg/s of air to the core. The compressor operates with a pressure ratio of 15 and an efficiency of 88 percent. The engine has a bypass ratio of 3. Fan pressure ratio is 1.6. The efficiency of the fan is 90 percent. The fuel has a heating value of 41,400 kJ/kg, and the burner total temperature is 1389K. The burner has an efficiency of 91 percent and a total pressure ratio of 0.95, whereas the turbine has an efficiency of 85 percent. The total pressure recovery for the Inlet diffuser is 0.92 and the shaft efficiency is 99.5 percent. Fan and Primary nozzle efficiency are 90 and 96 percent respectively, Find:
1. Isentropic diffuser efficiency
2. Compressor exit total temperature and pressure
3. The fuel mass flow rate
4. Turbine exit total temperature and pressure
5. Check if the nozzle is choked and find the nozzle exit area
6. The developed thrust
7. TSFC

3. Actual Turbofan Cycle In-class Practice: Ch07P01
A turbofan flies at sea level at a Mach number of 0.75. It ingests 74.83 kg/s of air to the core. The
compressor operates with a pressure ratio of 15 and an efficiency of 88 percent. The engine has a
bypass ratio of 3. Fan pressure ratio is 1.6. The efficiency of the fan is 90 percent. The fuel has a
heating value of 41,400 kJ/kg, and the burner total temperature is 1389K. The burner has an efficiency
of 91 percent and a total pressure ratio of 0.95, whereas the turbine has an efficiency of 85 percent. The
total pressure recovery for the Inlet diffuser is 0.92 and the shaft efficiency is 99.5 percent. Fan and
Primary nozzle efficiency are 90 and 96 percent respectively, Find:
1. Isentropic diffuser efficiency
2. Compressor exit total temperature and pressure
3. The fuel mass flow rate
4. Turbine exit total temperature and pressure
5. Check if the nozzle is choked and find the nozzle exit area
6. The developed thrust
7. TSFC Constants Properties of Air and Hot gaz:
Cold Section: γc := 1.4 Cpc 1.004⋅103 J
kg K ⋅
:= Rc
γc − 1
γc
:= ⋅Cpc
Hot Section: γt 1.35 := Cpt 1.09610⋅ 3 J
kg K ⋅
:= Rt
γt − 1
γt
:= ⋅Cpt
Temp and Pressure at Standard Sea Level:
T0 288.2K :=
P0 101.3310:= ⋅ 3Pa
a0 γc Rc := ⋅ ⋅T0
Flight Condition and Engine Operations:
M1 0.75 := u1 M1a0 := ⋅
pi_c 15 := pi_f 1.6 :=
m_dot_C 74.83
kg
s
:= α := 3
Component Performance:
Diffuser pressure recovery factor:Fuel Characterisitc:
pi_d 0.92 := hpr 4140010⋅ 3 J
kg
:=
Compressor and fan isentropic efficiency:
ηc 0.88 := ηf 0.90 :=
Total pressure ratio of the burner:The mechanical shaft efficiency:
pi_b 0.95 := ηm 0.995 :=
Combustion efficiency:Isentropic turbine efficiency:
ηb 0.91 := ηt 0.85 :=
Maximum Total Temperature at Turbine inletFan and Primary Nozzle efficiency:
Tt4 1389K := ηn 0.9 := ηfn 0.9 :=

4. Solutions:
Diffuser
τr 1
γc − 1
2
:= + ⋅M12 τr 1.113 =
pi_r τr
γc
γc−1
:= pi_r 1.452 =
Tt1 τr T0 := ⋅ Tt1 320.623K =
Pt1 pi_rP0 := ⋅ Pt1 1.47210= × 5 Pa
Note: Since the diffuser is adiabatic, the total temperature at the diffuser exit is
Tt2 Tt1 :=
Pressure recovery factor for the diffuser is
0.92;
Pt2
Pt1
=0.92
Pt2 pi_dPt1 := ⋅ Pt2 1.35410= × 5 Pa
Isentropic Efficiency of Diffuser is
ηd
τr pi_d
γc−1
⋅ γc − 1
τr − 1
:= ηd 0.767 = <Ans>
Find air mass flow rate at Bypass (secondary) and Core (primary) stream
m_dot_F m_dot_C := ⋅α m_dot_F 224.49
kg
s
=
m_dot_0 m_dot_Fm_dot_C := + m_dot_0 299.32
kg
s
=
Fan
Pt13 Pt2pi_f := ⋅ Pt13 2.16610= × 5 Pa
τf
pi_f
γc−1
γc − 1
ηf
:= + 1 τf 1.16 =
Tt13 τf Tt2 := ⋅ Tt13 371.823K =

5. Fan Nozzle
P19_sonic Pt131
1 − γc
ηfn 1 ⋅( + γc)
+ ⎡⎢⎣
⎤⎥⎦
γc
γc−1
:= ⋅ P19_sonic 1.05810= × 5 Pa
P19 P19_sonic :=
Note that, the nozzle is choked when P19>P0 which gives the result of M19=1
Then the nozzle is choked, and gives M19=1.
T19 Tt13
1
1
γc − 1
2
+
⎛⎜⎜⎝
⎞⎟⎟⎠
:= ⋅ T19 309.852K =
u19 2Cpc ⋅ Tt13 T19 := ⋅( − ) u19 352.756
m
s
=
ρ19
P19_sonic
Rc T19 ⋅
:= ρ19 1.19
kg
m3
=
A19
m_dot_F
ρ19 u19 ⋅
:= A19 0.535m= 2 <Ans>
Compressor
Pt3 Pt2pi_c := ⋅ Pt3 2.03110= × 6 Pa
τc
pi_c
γc−1
γc − 1
ηc
:= + 1 τc 2.327 =
Tt3 τc Tt2 := ⋅ Tt3 746.116K = <Ans>
Combustor
C m
f m
C f m +m

6. Pt4 := pi_b⋅Pt3 Pt4 1.92910= × 6 Pa
m_dot_fuel
m_dot_C Cpt ⋅ Tt4 Tt3 ⋅( − )
ηb hpr ⋅ Cpt Tt4 − ⋅
:= m_dot_fuel 1.458
kg
s
= Ans
Note: The burner specific heat is evaluated at the exit burner condition. One
may evaluate the specific heat by averaging burner temperature from inlet
and exit.
Turbine
1
1
1
t
t
t
γ
γ
τ
η
π
−
−
=
−
Power required to drive Compressor and Fan:
Pow_Comp m_dot_CCpc ⋅ Tt3 Tt2 := ⋅( − )
Pow_Fan m_dot_FCpc ⋅ Tt13 Tt2 := ⋅( − )
From the shaft power balance:
Tt5 Tt4
1
m_dot_C m_dot_fuel +
⎛⎜⎝
⎞⎟⎠
1
Cpt⋅ηm
⎛⎜⎝
⎞⎟⎠
⋅ Pow_Comp Pow_Fan := − ⋅( + )
Tt5 866.043K = Ans
Note: The turbine specific heat is evaluated at the exit burner condition. One
may evaluate the specific heat by averaging turbine temperature from inlet
and exit.
To obtain Pt5, we must calculate Tt5i (Isentropic Total Temp.) and then use the
isentropic relation between temperature and pressure ratio.
pi_t 1
1
Tt5
Tt4
−
ηt
−
⎛⎜⎜⎝
⎞⎟⎟⎠
γt
γt−1
:= pi_t 0.105 =
Pt5 Pt4pi_t := ⋅ Pt5 2.0210= × 5 Pa Ans
Nozzle
Adiabatic at Nozzle:
Tt9 Tt5 := Tt9 866.043K =
Checking chok condition, assuming M9=1

7. T9_sonic
Tt9
1
γt − 1
2
+ ⎛⎜⎝
⎞⎟⎠
:= T9_sonic 737.058K =
T9i_sonic Tt9
Tt9 T9_sonic −
ηn
⎛⎜⎝
⎞⎟⎠
:= − T9i_sonic 722.726K =
P9_sonic
Pt5
Tt5
T9i_sonic
⎛⎜⎝
⎞⎟⎠
γt
γt−1
:= P9_sonic 1.00510= × 5 Pa
Which P9_sonicP0, then the nozzle is not choked
Resulting to P9 P0 :=
Recalculation M9=? and any others parameters
T9i Tt9
P0
Pt5
⎛⎜⎝
⎞⎟⎠
γt−1
γt
:= ⋅ T9i 724.228K =
T9 Tt9 ηn Tt9T9i := − ⋅( − ) T9 738.41K =
M9
2
γt − 1
Tt9
T9
1 − ⎛⎜⎝
⎞⎟⎠
:= ⋅ M9 0.994 =
Pt9 P9
Tt9
T9
⎛⎜⎝
⎞⎟⎠
γt
γt−1
:= ⋅ Pt9 1.87410= × 5 Pa
Note, the pressure drop in primary nozzle can define by πn (using in Mattingly Textbook)
pi_n
Pt9
Pt5
:= pi_n 0.928 =
u9 2Cpt ⋅ Tt9 T9 := ⋅( − ) u9 528.935
m
s
=
ρ9
P9
Rt T9 ⋅
:= ρ9 0.483
kg
m3
=
A9
m_dot_fuel m_dot_C +
ρ9 u9 ⋅
:= A9 0.299m= 2 Ans
Total Thrust:
Thrust_Fan m_dot_F ( )u19u1 ⋅( − ) A19 P19P0 := + ⋅( − )
Thrust_Core m_dot_Cm_dot_fuel ( + ) u9 ⋅ m_dot_C u1 − ⋅ A9 P9P0 := + ⋅( − )
Thrust Thrust_FanThrust_Core := +
Thrust 4.55510= × 4 N
Ans
TSFC:
TSFC
m_dot_fuel
Thrust
:= TSFC 3.20210 − 5 × s
m
= Ans

8. Fuel/Air Ratio:
f
m_dot_fuel
m_dot_C
:= f 0.019 =
Thrust Ratio (FR):
FR
Thrust_Core
m_dot_C
Thrust_Fan
m_dot_F
:= FR 2.626 =
Note: Parametric Performance from Mattingly (Pg. 397)
Spec_Thrust_Core
1 a0 ⋅
1 + α
(1 + f)
u9
a0
⋅ − M1 (1 + f)
Rt
T9
T0
⋅
Rc
u9
a0
⋅
⋅
1
P0
P9
−
γc
+ ⋅
⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
:= ⋅
Spec_Thrust_Fan
α⋅a0
α + 1
u19
a0
− M1
T19
T0
u19
a0
1
P0
P19
−
γc
+ ⋅
⎛⎜⎜⎜⎝
⎞⎟⎟⎟⎠
:= ⋅
Spec_Thrust Spec_Thrust_CoreSpec_Thrust_Fan := +
Thrust_ m_dot_0Spec_Thrust := ⋅
Thrust_ 4.55510= × 4 N

9. Aeropropulsion
Unit
Kasetsart University
3
A. ATTHASIT
Actual Turbofan Cycle In-class Practice: Ch07P02
Prove
•Obj: Able to use the fundamental equation under the correct assumptions
Analysis
•Obj: Understand the physical meaning of each parameters
Calculation
•Obj: Able to solve the relations under the constraints of corrected unit, constant, … etc.
As a complement to Ch07P01, the parametric variation of the bypass ratio was studied. Write a computer program or use the spreadsheet to calculate and graph the performance of the separate-exhaust turbofan engine described in Ch07P01 for a range of bypass ratios from 1.5 to 5. Explain the results that you graphically depict for TSFC, Specific thrust.
Standard Sea Level: T0=288.2 K P0=101.33 kPa
gc=1.4, gt=1.3
Cpc=1.004 kJ/kg/K
Cpt=1.239 kJ/kg/K

10. Actual Turbofan Cycle In-class Practice: Ch07P02
As a complement to Ch07P01, the parametric variation of the bypass ratio was studied. Write a
computer program or use the spreadsheet to calculate and graph the performance of the
separate-exhaust turbofan engine described in Ch07P01 for a range of bypass ratios from 1.5 to
5. Explain the results that you graphically depict for TSFC, Specific thrust.
Constants Properties of Air and Hot gaz:
Cold Section: γc := 1.4 Cpc 1.004⋅103 J
kg K ⋅
:= Rc
γc − 1
γc
:= ⋅Cpc
Hot Section: γt 1.35 := Cpt 1.09610⋅ 3 J
kg K ⋅
:= Rt
γt − 1
γt
:= ⋅Cpt
Temp and Pressure at Standard Sea Level:
T0 288.2K :=
P0 101.3310:= ⋅ 3Pa
a0 γc Rc := ⋅ ⋅T0
Flight Condition and Engine Operations:
M1 0.75 := u1 M1a0 := ⋅
pi_c 15 := pi_f 1.6 :=
m_dot_C 74.83
kg
s
:=
rang 030 := .. αrang
rang 15 +
10
:=
Component Performance:
Diffuser pressure recovery factor:Fuel Characterisitc:
pi_d 0.92 := hpr 4140010⋅ 3 J
kg
:=
Compressor and fan isentropic efficiency:
ηc 0.88 := ηf 0.90 :=
Total pressure ratio of the burner:The mechanical shaft efficiency:
pi_b 0.95 := ηm 0.995 :=
Combustion efficiency:Isentropic turbine efficiency:
ηb 0.91 := ηt 0.85 :=
Maximum Total Temperature at Turbine inletFan and Primary Nozzle efficiency:
Tt4 1389K := ηn 0.9 := ηfn 0.9 :=

11. Solutions:
Diffuser
τr 1
γc − 1
2
:= + ⋅M12 τr 1.113 =
pi_r τr
γc
γc−1
:= pi_r 1.452 =
Tt1 τr T0 := ⋅ Tt1 320.623K =
Pt1 pi_rP0 := ⋅ Pt1 1.47210= × 5 Pa
Note: Since the diffuser is adiabatic, the total temperature at the diffuser exit is
Tt2 Tt1 :=
Pressure recovery factor for the diffuser is
0.92;
Pt2
Pt1
=0.92
Pt2 pi_dPt1 := ⋅ Pt2 1.35410= × 5 Pa
Isentropic Efficiency of Diffuser is
ηd
τr pi_d
γc−1
⋅ γc − 1
τr − 1
:= ηd 0.767 = Ans
Find air mass flow rate at Bypass (secondary) and Core (primary) stream
m_dot_Frang := m_dot_C⋅αrang
m_dot_0 m_dot_Fm_dot_C := +
Fan
Pt13 Pt2pi_f := ⋅ Pt13 2.16610= × 5 Pa
τf
pi_f
γc−1
γc − 1
ηf
:= + 1 τf 1.16 =
Tt13 τf Tt2 := ⋅ Tt13 371.823K =

12. Fan Nozzle
P19_sonic Pt131
1 − γc
ηfn 1 ⋅( + γc)
+ ⎡⎢⎣
⎤⎥⎦
γc
γc−1
:= ⋅ P19_sonic 1.05810= × 5 Pa
P19 P19_sonic :=
Note that, the nozzle is choked when P19P0 which gives the result of M19=1
Then the nozzle is choked, and gives M19=1.
T19 Tt13
1
1
γc − 1
2
+
⎛⎜⎜⎝
⎞⎟⎟⎠
:= ⋅ T19 309.852K =
u19 2Cpc ⋅ Tt13 T19 := ⋅( − ) u19 352.756
m
s
=
ρ19
P19_sonic
Rc T19 ⋅
:= ρ19 1.19
kg
m3
=
A19rang
m_dot_Frang
ρ19 u19 ⋅
:= Ans
Compressor
Pt3 Pt2pi_c := ⋅ Pt3 2.03110= × 6 Pa
τc
pi_c
γc−1
γc − 1
ηc
:= + 1 τc 2.327 =
Tt3 τc Tt2 := ⋅ Tt3 746.116K = Ans
Combustor
C m
f m
C f m +m

13. Pt4 := pi_b⋅Pt3 Pt4 1.92910= × 6 Pa
m_dot_fuel
m_dot_C Cpt ⋅ Tt4 Tt3 ⋅( − )
ηb hpr ⋅ Cpt Tt4 − ⋅
:= m_dot_fuel 1.458
kg
s
= Ans
Note: The burner specific heat is evaluated at the exit burner condition. One
may evaluate the specific heat by averaging burner temperature from inlet
and exit.
Turbine
1
1
1
t
t
t
γ
γ
τ
η
π
−
−
=
−
Power required to drive Compressor and Fan:
Pow_Comp m_dot_CCpc ⋅ Tt3 Tt2 := ⋅( − )
Pow_Fan m_dot_FCpc ⋅ Tt13 Tt2 := ⋅( − )
From the shaft power balance:
Tt5 Tt4
1
m_dot_C m_dot_fuel +
⎛⎜⎝
⎞⎟⎠
1
Cpt⋅ηm
⎛⎜⎝
⎞⎟⎠
⋅ Pow_Comp Pow_Fan := − ⋅( + )
Ans
Note: The turbine specific heat is evaluated at the exit burner condition. One
may evaluate the specific heat by averaging turbine temperature from inlet
and exit.
To obtain Pt5, we must calculate Tt5i (Isentropic Total Temp.) and then use the
isentropic relation between temperature and pressure ratio.
pi_t 1
1
Tt5
Tt4
−
ηt
−
⎛⎜⎜⎝
⎞⎟⎟⎠
γt
γt−1
:=
Pt5 Pt4pi_t := ⋅ Ans
Nozzle
Adiabatic at Nozzle:
Tt9 Tt5 :=
Checking chok condition, assuming M9=1

14. T9_sonic
Tt9
1
γt − 1
2
+
:=
T9i_sonic Tt9
Tt9 T9_sonic −
ηn
⎛⎜⎝
⎞⎟⎠
:= −
P9_sonic
Pt5
Tt5
T9i_sonic
⎛⎜⎝
⎞⎟⎠
γt
γt−1
:=
Which P9_sonicP0, then the nozzle is not choked
Resulting to P9 P0 :=
Recalculation M9=? and any others parameters
T9irang Tt9rang
P0
Pt5rang
⎛⎜⎝
⎞⎟⎠
γt−1
γt
:= ⋅
T9 Tt9 ηn Tt9T9i := − ⋅( − )
M9
2
γt − 1
Tt9
T9
1 − ⎛⎜⎝
⎞⎟⎠
:= ⋅
Pt9 P9
Tt9
T9
⎛⎜⎝
⎞⎟⎠
γt
γt−1
:= ⋅
Note, the pressure drop in primary nozzle can define by πn (using in Mattingly Textbook)
pi_n
Pt9
Pt5
:=
u9 2Cpt ⋅ Tt9 T9 := ⋅( − )
ρ9
P9
Rt T9 ⋅
:=
A9
m_dot_fuel m_dot_C +
ρ9 u9 ⋅
:= A9 9.83710 − 3 = × m2 Ans
Total Thrust:
Thrust_Fan m_dot_F ( )u19u1 ⋅( − ) A19 P19P0 := + ⋅( − )
Thrust_Core m_dot_Cm_dot_fuel ( + ) u9 ⋅ m_dot_C u1 − ⋅ A9 P9P0 := + ⋅( − )
Thrust Thrust_FanThrust_Core := +
Ans
TSFC:
TSFC
m_dot_fuel
Thrust
:= Ans

15. Fuel/Air Ratio:
f
m_dot_fuel
m_dot_C
:= f 0.019 =
Thrust Ratio (FR):
FR
Thrust_Core
m_dot_C
Thrust_Fan
m_dot_F
:=
Specific Thrust:
Spec_Thrust
Thrust
m_dot_0
:=
1.5 22.533.544.5
3.2 .10 5
3.3 .10 5
3.4 .10 5
100
150
200
250
TSFCrang Spec_Thrustrang
α rang

16. Aeropropulsion
Unit Kasetsart University A. ATTHASIT 4
Actual Turbofan Cycle
In-class Practice: Ch07P02
Prove
• Obj: Able to
use the
fundamental
equation
under the
correct
assumptions
Analysis
• Obj:
Understand
the physical
meaning of
each
parameters
Calculation
• Obj: Able to
solve the
relations
under the
constraints of
corrected unit,
constant, …
etc.
As a complement to Ch07P01, the parametric variation of the bypass ratio was studied.
Write a computer program or use the spreadsheet to calculate and graph the
performance of the separate-exhaust turbofan engine described in Ch07P01 for a range
of bypass ratios from 1.5 to 5. Explain the results that you graphically depict for TSFC,
Specific thrust.
Spec_Thrust
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
2 3 6 2 2 8 2 2 0 2 13 2 0 6 2 0 0 19 4 18 18 3 17 8 17 16 8 16 4 16 0 15 15 
1.5 2 2.5 3 3.5 4 4.5
3.2 10
5
3.3 10
5
3.4 10
5
100
150
200
250
3.493 10
 5

3.201 10
 5

TSFCrang
236.228
101.459
Spec_Thrustrang
1.5  rang 4.5
Optimum Bypass Ratio
See
Mattingly
Pg. 405-
407

17. Aeropropulsion
Unit Kasetsart University A. ATTHASIT 5
Conclusion
*
2
1
*
2
1
1
*
*
2
*
1
2( 1)
2
*
1
2
1
1
2
1
2
1
1
2
1
2
1
1
2
1
1
1 2
1
2
T
T
M
P
P
M
P
P
T
M
T
P
m AV AM
R T
M
A
A M
g
g
g
g
g
g
g
g
g
g

 g
g

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g
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 
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  
           
  
 
   
           
 
    
    
    
  
 
 
2
0
0 t
dA d du
A u
udu dP
dh dh udu
dP d dT
P T
a
P






g
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P dP
T dT
d
A dA
u du
 





P
T
A
u

dx
2
dP
P 
See You
Next Class!