Example 1: The Tomahawk is a long-range subsonic cruise missile powered by a solid-

fuel rocket booster and the small two-spool turbofan engine Williams Inter national
F107-WR-402 (Fig.1). It has the following data:
− Flight speed Vf = 247.22 m/s
− Ambient temperature Ta = 275 K
− Ambient pressure Pa = 0.79 bar
− Thrust force T= 3.1 kN
− Specific fuel consumption = 0.682 kg/kg-h
− Bypass ratio 𝛽 = 1
− Overall pressure ratio 𝜋𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 13.8
− Fan pressure ratio 𝜋𝑓 = 2.1
− Fan diameter Dfan = 0.305 m
− Fuel heating value QR = 43,000 kJ/kg

Fig. 1a Tomahawk missile and Williams F107-WR-402 turbofan engine

Fig. 1b Schematic diagram of Williams F107-WR-402 turbofan engine

Calculate
Air mass flow rate
Fuel-to-air ratio
Exhaust gas speed
Engine maximum temperature
Solution
1. Air mass flow rate
Flight speed
 𝑉𝑓 = 247,22 𝑚/𝑠
𝑑2 0,3052
A =𝜋 = 3,1416 × = 0,7306 𝑚2
4 4
𝑃𝑎 0,79 0,79 × 100
𝜌𝑎 = = = = 1,001 𝑘𝑔/𝑚3
𝑅𝑇𝑎 0,287 × 275 0.287 × 275
𝑚̇𝑎 = 𝜌𝑎 𝑉𝑓 𝐴 = 1,001 × 247,22 × 0.07306 = 18,08 𝑘𝑔/𝑠
𝑚̇𝑎 18,08
Khi Tỷ lệ bỏ qua β= 1 Thì 𝑚̇𝑐 = 𝑚̇ℎ = = = 9,04 𝑘𝑔/𝑠
2 2
𝑇 3.100
= = 171,7 𝑚/𝑠
𝑚̇𝑎 18,08
𝑘𝑔 0,628 0.628 𝑘𝑔 𝑔
𝑇𝑆𝐹𝐶 = 0,628 = = = 0,0000193 = 19.3
𝑘𝑔𝑓×ℎ 𝑔×3600 9,81×3600 𝑁.𝑠 𝑘𝑁.𝑠

Air mass flow rate are: 19.3 g/kN.s

2. Fuel-to-air ratio
𝑚̇𝑓 𝑚̇𝑓 /𝑚̇ℎ (𝑚̇𝑓 /𝑚̇ℎ ) 𝑓
𝑇𝑆𝐹𝐶 = = = =
𝑇 𝑇/𝑚̇ℎ (𝑇/𝑚̇𝑎 )×(1+𝛽) (𝑇/𝑚̇𝑎 )×(1+𝛽)

𝑇
𝑓 = 𝑇𝑆𝐹𝐶 × ( ) (1 + 𝛽 )
𝑚̇𝑎
Replacement value 𝑓 = 0.0000193 × 171,7 × (1 + 1) = 0.006618
Fuel-to-air ratio: 0.006618
3. Exhaust gas speed
We have:

𝑇 = [𝑚̇𝑐 + 𝑚̇ℎ × (1 + 𝑓)]𝑉𝑒 − 𝑚̇𝑎 𝑉𝑓

𝑇
= [𝑚̇𝑐 + 𝑚̇ℎ × (1 + 𝑓)]𝑉𝑒 − 𝑉𝑓
𝑚̇𝑎
𝑇 𝛽 1+𝑓
=[ + ] 𝑉 − 𝑉𝑓
𝑚̇𝑎 1+𝛽 1+𝛽 𝑒

(𝑇/𝑚̇𝑎 ) + 𝑉𝑓
𝑉𝑒 =
𝛽+1+𝑓
1+𝛽
(171,4622) + 247,22
𝑉𝑒 = = 417,3 𝑚/𝑠
2.006618
2
Exhaust gas speed: 417.3 m/s
4. Engine maximum temperature
Cycle analysis.
 − Intake
𝑉𝑓 247,22
𝑀= = = 0,7437
√𝛾𝑅𝑇𝑎 √1,4 × 0,287𝑥275
𝛾 1,4
𝛾 − 1 2 𝛾−1 1,4 − 1 1,4−1
𝑃02 = 𝑃𝑎 (1 + 𝑀 ) = 0,79 (1 + 0.74372 ) = 114,0522 𝑘𝑃𝑎
2 2
𝑃02 = 114,0522 𝑘𝑃𝑎
𝛾−1 2
𝑇02 = 𝑇𝑎 (1 + 𝑀 )
2
1.4 − 1
𝑇02 = 275 (1 + 0.74372 ) = 305,42
2
− Fan (𝛾 = 1.4)
𝑃03 = 𝑃02 𝜋𝑓
𝑃03 = 114,0522 × 2,1 = 239,51
𝛾−1
𝑇03 = 𝑇02 (𝜋𝑓 ) 𝛾
1,4−1
𝑇03 = 305,42(2,1) 1,4 = 377.62
− Compressor (𝛾 = 1.37)
𝑃04 = 𝑃03 𝜋𝑐
We have 𝜋𝑐 × 𝜋𝑓 = 𝜋𝑜𝑣𝑒𝑟𝑎𝑙𝑙
𝜋𝑜𝑣𝑒𝑟𝑎𝑙𝑙
𝑃04 = 𝑃03
𝜋𝑓
13.8
𝑃04 = 239,51 × = 1.574 𝑘𝑃𝑎
2.1
𝛾−1
𝑇04 = 𝑇03 (𝜋𝑐 ) 𝛾

1,37−1
𝑇04 = 305,42(2,1) 1,37 = 645 𝐾
− Combustion chamber
 𝑚̇𝑓 𝑄𝑅 + 𝑚̇ℎ 𝐶𝑝𝑐 𝑇04 = (𝑚̇ℎ + 𝑚̇𝑓 )𝐶𝑝ℎ 𝑇05
𝑓𝑄𝑅 + 𝐶𝑝𝑐 𝑇04
𝑇05 =
(1 + 𝑓)𝐶𝑝ℎ
0.006618 × 43000 + 1.005 × 645
𝑇05 = = 807.2 𝐾
(1 + 0.006618)1.148
Maximum temperature is then T0max = 807.20K.
Example 2: A two-spool forward unmixed turbofan engine is powering an airliner
flying at Mach number M = 0,9 at an altitude of H = 10.500 m (ambient pressure,
temperature, sonic speed, and air density ratio are Pa = 24.475 Pa, T = 220 K, V = 297,3
m/s, and 0,3165, respectively). The low-pressure spool is composed of a turbine driving
the fan and the low-pressure compressor. The high-pressure spool is composed of a
high-pressure compressor and a high-pressure turbine. Air is bled from the outlet of
high-pressure compressor. The total temperature difference (in K) and pressure ratios
for different modules are recorded and shown in Table 1:
Table 1

`
Diffuser, fan, and turbine nozzles have isentropic efficiency of Ma = 0,9. Maximum
cycle temperature is Tmax = 1.500 K. Inlet area is A = 3,14 m2. Assume the following
values for the different variables:
ηb = 0,96, QHV = 45.000 kJ/kg, γair = 1,4, and γgases = 1,33
It is required to calculate:
(a) Pressure recovery of diffuser rd
(b) Total air mass flow rate
(c) Isentropic efficiency of fan and low-pressure and high-pressure compressors
(d) The fuel-to-air ratio (f)
(e) The air bleed ratio (b)
(f) The bypass ratio (β)
(g) Area of cold and hot nozzles
(h) The thrust force

Solution
The modules of the engine and its different processes plotted on the T-s diagram are
shown in Figs. 1 and 2 below:
 Fig. 1 Layout of a double spool where fan and LPC driven by LPT

Fig. 2 T-s diagram for a double-spool turbofan engine

(a) Pressure recovery of diffuser rd
− Diffuser:
𝛾𝑐 − 1 2
𝑇02 = 𝑇𝑜𝑎 = 𝑇𝑎 (1 + 𝑀𝑎 )
2
1,4 − 1 2
𝑇02 = 220 × (1 + 0,9 ) = 255.6 𝐾
2
𝛾𝑐
𝛾𝑐 − 1 2 𝛾𝑐−1
𝑃𝑜𝑎 = 𝑃𝑎 (1 + 𝑀𝑎 )
2
1,4
1,4−1 2 1,4−1
𝑃𝑜𝑎 = 𝑃𝑎 (1 + 0.9 ) =1,69 𝑃𝑎
2
𝛾𝑐
𝛾𝑐 − 1 2 𝛾𝑐−1
𝑃02 = 𝑃𝑎 (1 + 𝜂𝑑 𝑀𝑎 )
2
1,4
1.4 − 1 2 1,4−1
𝑃02 = 𝑃𝑎 (1 + 0.9 0.9 ) = 1.61 𝑃𝑎
2
𝑃02 = 1.61 𝑃𝑎 = 1,61 × 24.475 Pa = 39.400 Pa = 39,4 kPa
𝑃02
𝑟𝑑 =
𝑃𝑜𝑎
 1.61 𝑃𝑎
𝑟𝑑 = = 0,953
1.69 𝑃𝑎
Pressure recovery of diffuser rd = 0,953
(b) Total air mass flow rate.
𝑃𝑎
𝜌𝑎 =
𝑅𝑇𝑎
24.475
𝜌𝑎 = = 0,3877 𝑘𝑔/𝑚3
287 × 220
𝑉 =𝑀×𝑎
𝑉 = 0,9 × 297,3 = 267.6
𝑚̇𝑡𝑜𝑡𝑎𝑙 = 𝜌𝑎 𝑉𝐴𝑖𝑛𝑙𝑒𝑡
𝑚̇𝑡𝑜𝑡𝑎𝑙 = 0,3877 × 267,6 × 3,14 = 325,7 𝑘𝑔/𝑠
Total air mass flow rate: 𝑚̇𝑡𝑜𝑡𝑎𝑙 = 325,7 𝑘𝑔/𝑠
(c) Isentropic efficiency of fan and low-pressure and high-pressure compressors
− Fan
𝑇02 = 255,6 𝐾, Following Table 1 ∆𝑇0𝑓 = 42 𝐾, 𝜋𝑓 = 1,58, 𝛾𝑐 = 1,4
𝑇010 = 𝑇02 + ∆𝑇0𝑓 = 255,6 + 42 = 297,6 𝐾
𝑃010 = 𝑃02 × 𝜋𝑓 = 39,4 × 1.58 = 62.27 𝑘𝑃𝑎
The fan isentropic efficiency may be expressed as:
𝛾𝑐 −1
𝛾
𝑇02 (𝜋𝑓 𝑐 − 1)
𝑇010𝑠 − 𝑇02
𝜂𝑓 = =
𝑇010 − 𝑇02 ∆𝑇𝑜𝑓
255,6 × (1,580,2857 − 1)
𝜂𝑓 = = 0.85.5
42
The fan isentropic efficiency: 85,5%
Efficience Low-pressure compressor
𝑇03 = 𝑇010 + ∆𝑇0𝐿𝑃𝐶 = 297,6 + 43 = 340,6 𝐾
𝑃03 = 𝑃010 + 𝜋𝑓𝐿𝑃𝐶 =62,27 × 1,5 = 93,04 𝑘𝑃𝑎
𝛾𝑐 −1
𝛾
𝑇010 (𝜋𝐿𝑃𝐶𝑐 − 1)
297,6
𝜂𝐿𝑃𝐶 = = (1,50,286 − 1) = 0,85095
∆𝑇𝑜𝐿𝑃𝐶 43
Efficience of Low-pressure compressor: 85,1%
Efficience high-pressure compressor
𝑇04 = 𝑇03 + ∆𝑇0𝐻𝑃𝐶 = 340,6 + 355 = 695,6 𝐾
𝑃04 = 𝑃03 × 𝜋𝑓𝐻𝑃𝐶 =93,40 × 9,75 = 910,65 𝑘𝑃𝑎
 𝛾𝑐 −1
𝛾𝑐
𝑇03 (𝜋𝐻𝑃𝐶 − 1)
340,6
𝜂𝐻𝑃𝐶 = = (9,750,286 − 1) = 0,88079
∆𝑇𝑜𝐻𝑃𝐶 355
Efficience high-pressure compressor:88,1%
(d) The fuel-to-air ratio (f)
− Combustion chamber

Fuel-to-air ratio
𝑚̇𝑓 (1 − 𝑏)(𝐶𝑝ℎ 𝑇05 − 𝐶𝑝𝑐 𝑇04)
𝑓= =
𝑚̇ℎ 𝜂𝑐𝑐 𝑄𝐻𝑉 − 𝐶𝑝ℎ 𝑇05
𝑘𝐽
We have: 𝐶𝑝ℎ = 1,148; 𝐶𝑝ℎ = 1,005, 𝑄𝐻𝑉 = 45.000 ; 𝑇05 = 𝑇𝑀𝑎𝑥 = 1500 𝐾,
𝑘𝑔
𝜂𝑐𝑐 = 𝜂𝑏 = 0,96
Add value to aboce equation, we get:
(1−𝑏)(1,148×1500−1,005×695,6
𝑓= = 0,02466(1 − 𝑏)
0.96×4.500−1,005×1500

𝑓 = 0,02466(1 − 𝑏) (1)
Energy balance for high-pressure spool
𝐶𝑝𝑐 (𝑇04 − 𝑇03 ) = (1 + 𝑓 − 𝑏)𝐶𝑝ℎ (𝑇05 − 𝑇06 )
Substitute from (1)
1,005 × (695,6 − 340,6) = [1 + 0,02466(1 − 𝑏) − 𝑏]𝑥1,148 × 380
Bleed b = 0,202 = 20,2%
Comment: Though this bleed ratio looks like a great percentage, however, it will be
described in the turbine cooling sections that each blade row may need some 2 % of air
mass flow rate for cooling, and the two turbines here have a total of five stages and so
ten blade rows. In addition bleeding has its other applications for cabin air conditioning
as well as other anti-icing applications.
From equation (1)
Fuel-to-air ratio f = 0.01968
This fuel-to-air ratio is a reasonable figure which assures the value previously obtained
for the bleed ratio.
(e) The air bleed ratio (b)
Bleed b = 20.2 %
(f) The bypass ratio (β)
Energy balance for low-pressure spool:
(1 + 𝛽)𝐶𝑝𝑐 ∆𝑇𝑜𝑓 + 𝐶𝑝𝑐 ∆𝑇𝑜𝐿𝑃𝑇 = (1 + 𝑓 − 𝑏)𝐶𝑝ℎ ∆𝑇𝑜𝐿𝑃𝑇
Apply Value, we get:
(1 + 𝛽)1,005 × +1,005 × 43 = (1 + 0,01968 − 0,202) × 1,148 × 390
Or 𝛽 = 6,649
Thus, the bypass ratio is 𝛽 = 6,649
(g) Area of cold and hot nozzles
Area of cold nozzles
𝛽
Fan airflow rate is then 𝑚̇𝑓𝑎𝑛 = 𝑚̇𝑡𝑜𝑡𝑎𝑙
𝛽+1
6,649
Add value we get: 𝑚̇𝑓𝑎𝑛 = × 325,7 = 283,12 𝑘𝑔/𝑠
7,649
1
Core airflow rate is then 𝑚̇𝑐𝑜𝑟𝑒 = 𝑚̇𝑡𝑜𝑡𝑎𝑙
𝛽+1
1
Add value we get: 𝑚̇𝑐𝑜𝑟𝑒 = × 325,7 = 42,58 𝑘𝑔/𝑠
7,649

Fan nozzle
The first step in nozzle analysis is to check whether it is choked or not by calculating
the pressure ratio:
𝑃010 1
= 𝛾𝑐
𝑃𝑐 1 𝛾𝑐 − 1 𝛾𝑐−1
(1 − )
𝜂𝑓𝑛 𝛾𝑐 + 1
Add value we get
𝑃010 1
= 1,4 = 2,0478
𝑃𝑐 1 1,4 − 1 1,4−1
(1 − )
0.9 1,4 + 1
𝑃010 62,27
And = = 2,5447
𝑃𝑎 24,47
𝑃010 𝑃010
Sine < then nozzle choked:
𝑃𝑐 𝑃𝑎

𝑃11 = 𝑃𝑐 = 30,41 𝑘𝑃𝑎

2
𝑇11 = 𝑇𝑐 = 𝑇
𝛾𝑐 + 1 010
2
Or 𝑇11 = × 297,6 = 248 𝐾
1,4+1

𝑉11 = 𝑉𝑒𝑓𝑎𝑛 = √𝛾𝑐 𝑅𝑇11

Aad value 𝑉11 = 𝑉𝑒𝑓𝑎𝑛 = √1,4 × 287 × 248 = 315,67
𝑃11
𝜌11 =
𝑅𝑇11
30,41
Or 𝜌11 = = 0,4273 𝑘𝑔/𝑚3
0,287×248
𝑚̇𝑓𝑎𝑛
Therefore 𝐴11 =
𝜌11 𝑉11

283,12
𝐴11 = = 2,099 𝑚2
0,4273 × 315,67
Area of cold nozzles are 2,099 m2
Area of hot nozzles
Turbine nozzle
Here also a check for the choking of turbine nozzle is followed:
𝑃08 1
= 𝛾ℎ
𝑃𝑐 1 𝛾ℎ−1 𝛾ℎ −1
(1 − )
𝜂𝑡ℎ 𝛾ℎ + 1
With 𝛾ℎ = 1,33; 𝜂𝑡ℎ = 0,9
We get
𝑃08 1
= 1,33 = 1,9853
𝑃𝑐 1 1,33 − 1 1,33−1
(1 − )
0,9 1,33 + 1
𝑃08 84,91
= = 3,469
𝑃𝑎 24,47
𝑃08 𝑃08
Sine < then nozzle checked
𝑃𝑐 𝑃𝑎

𝑃9 = 𝑃𝑐 = 42,81 𝑘𝑃𝑎
2
𝑇9 = 𝑇𝑐 = 𝑇
𝛾ℎ + 1 08
2
𝑇9 = × 730 = 625,53 𝐾
1,33 + 1
𝑉9 = 𝑉𝑒ℎ = √𝛾ℎ 𝑅𝑇9
𝑉9 = 𝑉𝑒ℎ = √1,33 × 287 × 625,53 = 488,64 𝑚/𝑠
𝑃9 42,81
𝜌9 = = = 0,23846 𝑘𝑔/𝑚3
𝑅𝑇9 0,287 × 625,53
𝑚̇𝑐𝑜𝑟𝑒 (1+𝑓−𝑏)
We have: 𝐴9 =
𝜌9 𝑉9

Replace by value, we get:

 42,58[1 + 0,02466(1 − 0,202) − 0,202]
𝐴9 = = 0,298 𝑚2
0,23846 × 488,64
Area of hot nozzles are 0,298 m2
(a) Thrust force (T)
We have:
𝑇 = 𝑚̇𝑓𝑎𝑛 𝑉𝑒𝑓𝑎𝑛 + (1 + 𝑓 − 𝑏)𝑚̇ℎ 𝑉𝑒ℎ − 𝑚̇𝑡𝑜𝑡𝑎𝑙 𝑉𝑓𝑙𝑖𝑔ℎ𝑡 + 𝑃11 𝐴11 + 𝑃9 𝐴9 − 𝑃𝑎 𝐴𝑖
Replace by above value:
𝑇 = 283,12 × 315,67 + 0,81769 × 42,58 × 489,26 − 325,7 × 267,6
+ 103 (30,4 × 2,099 + 42,81 × 0,298 − 24,457 × 3,14)
𝑇 = 18,97 𝑘𝑁
Thrust force (T) =18,97 kN
Example 3 A. Triple-spool unmixed turbofan engine (Trent 700) is shown in Fig. 1.
Bleed air having a percentage of 8 % is taken from HP compressor to cool HP and IP
turbines as shown in Fig. 2. Engine has the following data in Table 1 below:
For takeoff operation where Mach number M = 0.2 at sea level, calculate:
1. The total temperature at the outlet of the fan and intermediate- and high-pressure
compressors
2. The total temperature at the outlet of the high-, intermediate-, and low-pressure
Turbines
3. The pressure and temperature at the outlet of cold and hot nozzles
4. Specific thrust, thrust specific fuel consumption, and propulsive, thermal, and
overall efficiencies

Fig. 1 Layout of unmixed three-spool engine (Trent 700)

 Fig. 2 Layout of threespool engine and air bleed details
Table 1 A triple-spool unmixed turbofan engine operating conditions (Trent 700)

Solution
1. The total temperature at the outlet of the fan and intermediate and high-pressure
compressors
At takeoff (M = 0.2) at sea level
Cycle analysis from Figs. 1 and 2:
Diffuser (a–2)
𝛾𝑐
𝛾𝑐 − 1 2 𝛾𝑐−1
𝑃02 = 𝑃𝑎 (1 + 𝜂𝑑 𝑀𝑎 )
2
𝛾𝑐 −1
And 𝑇02 = 𝑇𝑎 (1 + 𝑀𝑎2 )
2

At sea level, the ambient temperature and pressure are Ta = 288,158 K and Pa = 1,01325.
105 Pa, and papametters give in table 1 we get as:
1,4
1,4 − 1 2 1,4−1
𝑃02 = 1,01325. 105 (1 + 0,88 0,2 ) = 1,0384 × 105 Pa
2
 1,4 − 1 2
𝑇02 = 288,16 (1 + 0.2 ) = 290,4653 K
2
Fan (2–3)
𝑃03 = 𝑃02 × 𝜋𝑓
𝛾𝑐 −1
𝛾
𝜋𝑓 𝑐 −1
𝑇03 = 𝑇02 (1 + )
𝜂𝑓

We have: 𝜂𝑓 = 0,9; 𝜋𝑓 = 1,45; 𝛾𝑐 = 1,4

Replace above vulue we get:
𝑃03 = 1,0384 × 105 × 1,45 = 1,5057 × 105 Pa
1,4−1
1,45 1,4 −1
𝑇03 = 290,4653 (1 + ) = 326,6124 K
0,9

Temperature at the outlet of the fan: 𝑇03 = 326,6124 K

Intermediate-pressure compressor (IPC) (3–4)
𝑃04 = 𝑃03 × 𝜋𝐼𝑃𝐶
𝛾𝑐 −1
𝛾
𝜋𝐼𝑃𝐶𝑐 −1
𝑇04 = 𝑇03 (1 + )
𝜂𝐼𝑃𝐶

We have: 𝜂𝐼𝑃𝐶 = 0,89; 𝜋𝐼𝑃𝐶 = 5,8; 𝛾𝑐 = 1,4

𝑃04 = 1,5057 × 105 × 5,8 = 8,7333 × 105 Pa
1,4−1
5,8 1,4 −1
𝑇04 = 326,6124 (1 + ) = 566,0401 K
0,89

Temperature of intermediate pressure compressors: 𝑇04 = 566,0401 K

High-pressure compressor (HPC) (4–5)
We have: 𝜂𝐻𝑃𝐶 = 0,98; 𝜋𝐻𝑃𝐶 = 4,2 ; 𝛾𝑐 = 1,4
Bleed air is extracted at an intermediate state defined as:
𝑃04𝑏 = 𝑃04 × √𝜋𝐻𝑃𝐶 = 8,7333 × 105 × √4,2 = 1,7898 × 106
𝛾𝑐 −1
1,4−1
𝛾𝑐
𝜋𝐻𝑃𝐶 −1 4,2 1,4 −1
𝑇04𝑏 = 𝑇04 × (1 + ) = 566,0401 (1 + ) = 710,7556 K
𝜂𝐻𝑃𝐶 0,98

Temperature of high pressure compressors: 𝑇04𝑏 = 710,7556 K

 𝑃05 = 𝑃04𝑏 × √𝜋𝐻𝑃𝐶 = 1,7898 × 106 × √4,2 = 3,668 × 106
𝛾𝑐 −1
1,4−1
𝛾𝑐
𝜋𝐻𝑃𝐶 −1 4,2 1,4 −1
𝑇05 = 𝑇04𝑏 × (1 + ) = 710,7556 (1 + ) = 892,4693 K
𝜂𝐻𝑃𝐶 0,98

2. The total temperature at the outlet of the high-, intermediate-, and low-pressure
Turbines
Combustion chamber (CC) (5–6)
𝑇06 = 𝑇𝐼𝑇, 𝑃06 = 𝑃05 × (1 − ∆𝑃𝑐𝑐 )
𝐶𝑝ℎ × 𝑇06 − 𝐶𝑝𝑐 × 𝑇05
𝑓 = (1 − 𝑏 ) ( )
𝜂𝑏 × 𝑄𝑅 − 𝐶𝑝ℎ × 𝑇06
We have: TIT = 1543 K,
𝐽 𝐽
∆𝑃𝑐𝑐 = 0.03; 𝐶𝑝ℎ = 1184 𝑘𝑔.𝐾 ; 𝐶𝑝𝑐 = 1005 ; 𝑄𝑅 = 45.000kj/kg
𝑘𝑔.𝐾

𝑇06 = 1543 𝐾 , 𝑃06 = 3,668 × 106 × (1 − 0.03) = 3,5579 × 106 Pa

1184×1543−1005×892,4693
𝑓 = (1 − 0.08) ( )= 0.0213
0.98×45000−1005×3,5579×106

High-pressure turbine (HPT) (6–7)

𝐶𝑝𝑐
𝑇07 = 𝑇06 − [( ) × ([𝑇04𝑏 − 𝑇04 ] + (1 − 𝑏) × [𝑇04 − 𝑇04𝑏 ])]
𝜂𝑚1 × 𝜆1 × (1 + 𝑓 − 𝑏) × 𝐶𝑝ℎ
𝑇07 = 1206.8 𝐾
𝛾ℎ −1
(𝑇06 − 𝑇07 ) 𝛾ℎ
𝑃07 = 𝑃06 (1 − ) = 1,2134 × 106 Pa
𝜂𝐻𝑃𝑇 × 𝑇06
Intermediate-pressure turbine (IPT) (7–8)

𝐶𝑝𝑐 (𝑇04 − 𝑇03 )

𝑇08 = 𝑇07 − [( ) = 995,3980 𝐾]
𝑏
𝜆2 × 𝜂𝑚2 × (1 + 𝑓 − 2) × 𝐶𝑝ℎ
𝛾ℎ −1
(𝑇07 − 𝑇08 ) 𝛾ℎ
𝑃08 = 𝑃07 (1 − ) = 5,2318 × 105 Pa
𝜂𝐼𝑃𝑇 × 7
Low-pressure turbine (LPT) (8–9)
 3. The pressure and temperature at the outlet of cold and hot nozzles
Hot nozzle (10–11)

Cold nozzle (3–12)

4. Specific thrust, thrust specific fuel consumption, and propulsive, thermal, and overall
efficiencies
Specific thrust

Where:
T is the thrust force, m∘a is the air mass flow rate.
V is the flight speed, V11 is the exhaust hot gas speed.
V12 is the exhaust cooled gas speed.
Since the cold nozzle is unchoked, then P12 = Pa, and the corresponding pressure thrust
is zero.
Thrust specific fuel consumption
Propulsive efficiency

Thermal efficiency

Overall efficiency

Problem 1. The following data apply to a twin-spool turbofan engine, with the fan
driven by the LP turbine and the compressor by the HP turbine. Separate hot and cold
nozzles are used:
• Overall pressure ratio: 20.0
• Fan pressure ratio: 1.6
• Bypass ratio: 3.5
• Turbine inlet temperature: 1300 0K
• Air mass flow: 120 kg/s
• Find the sea-level static thrust and TSFC if the ambient pressure and
temperature are 1 bar and 288 K. Heat value of the fuel: 43 MJ/kg

Problem 2 Atwin-spoolmixed turbofan engine operates with an overall pressure ratio

of 18. The fan operates with a pressure ratio of 1.45 and the bypass ratio of 5.0. The
turbine inlet temperature is 1400K. The engine is operating at aMach number of 0.85 at
an altitude where the ambient temperature and pressure are 223.2 K and 0.2645 bar.
Calculate
• The thrust
 • TSFC
• Propulsive efficiency
Problem 3. The shown block diagram illustrated in Fig. 1 represents a single-spool
mixed afterburning turbofan engine flying at altitude 24,000 ft and Mach number 1.8.
The fan pressure ratio and mass flow rate are 1.8 and 90 kg/s, respectively. Compressor
pressure ratio and mass flow rate are 8.0 and 45 kg/s. Tubine inlet temperature and
maximum engine temperature are 1500 and 26000K. Fuel heating value is 45,000 kJ/kg.
Nozzle is of the convergent– divergent type.

Figure 1. Block diagram for a single-spool mixed afterburning turbofan engine

No pressure drop is assumed in combustor, mixer, or afterburner. All other processes
are assumed ideal. It is required to:
1. Draw a schematic diagram for engine showing different states
2. Plot T-s diagram for the cycle
3. Calculate bypass ratio
4. Calculate fuel-to-air ratios in combustor and afterburner
5. Calculate the thrust and TSFC
6. Calculate the propulsive, thermal and overall efficiencies
Problem 4 Figure 1 shows a fighter airplane propelled by a low-bypass ratio mixed
afterburning turbofan engine during takeoff from an air carrier. Temperature and
pressure of ambient air are 290 K and 101 kPa. Air is ingested into the engine intake at
a Mach number of 0.2. Turbofan engine has the following data:
• Fan pressure ratio πf = 1:8.
• Compressor pressure ratio πc = 5.
• Fuel-to-air ratio in combustion chamber f = 0.02.
• Afterburner fuel-to-air ratio fab = 0.04.
 • Fuel heating value QR = 45 MJ=kg.
• Percentages of power extracted by fan and compressor from the driving
• turbines are respectively λ1 = λ2 = 0:8.
• Assume ideal processes and variable properties: (γ, Cp).

Figure 1. Fighter airplane powered by a low-bypass ratio mixed afterburning turbofan

engine
Calculate:
1. Bypass ratio
2. Jet speed
3. TSFC
Problem 5 A double-spool turbofan engine; Fig. 1, is used to power an aircraft flying
at speed of 250 m/s at an altitude of 11,000 m. As shown in the figure below, the low-
pressure turbine drives the fan and low-pressure compressor, while the high-pressure
turbine drives the high-pressure compressor. The engine has the following data:
• Bypass ratio = 8.
• Total ingested air flow rate = 180 kg/s.
• Overall pressure ratio OPR = 35.
• Fan pressure ratio = 1.6.
• Pressure ratio of high-pressure compressor is four times that of the
low-pressure compressor; πHPC = 4 πLPC.
• Turbine inlet temperature = 1650 K.
• Fuel heating value = 43 MJ/kg.
Assuming all processes are ideal and neglecting any pressure drop, it’s required to:
1. Find the thrust, TSFC, and efficiencies of the engine
2. Plot the velocity and temperature distribution over the engine cross section (rear end)
 Figure 1. Double spool high bypass ratio turbofan engine
Problem 6. A double-spool turbofan engine is used to power an aircraft flying at speed
of 250 m/s at an altitude of 11,000 m. As shown in Fig. 1, the low-pressure turbine
drives the fan and low-pressure compressor, while the high-pressure turbine drives the
high-pressure compressor. Inlet and outlet temperature and velocity of engine are
plotted beside the engine layout. The engine has the following data:
• Bypass ratio = 8.
• Total ingested air flow rate = 180 kg/s.
• Overall pressure ratio OPR = 35.
• Pressure ratio of high-pressure compressor is four times that of the low-pressure
compressor; πHPC = 4 πLPC.
• Fuel heating value = 43 MJ/kg.
Assuming all processes are ideal and neglecting any pressure drop, it’s required to find:
1. Whether the nozzles are choked or not
2. Fan pressure ratio
3. TIT
4. The thrust force

Figure 1.