aircraft engine technology - antonov an-225 mriya

Aircraft Engine Technology

Imperial College London

Mechanical Engineering Department

March 2010

Antonov An-925 Vira

Ambrose Tey

Nabilah Hamid

Ren Yin Tai

Ikwan Jamaludin

ABSTRACT

The new engine proposed was able to reduce sfc and increase range

whist reducing the number of engines aboard. Though having a

considerably larger size to its predecessor, the proposed engine’s

ability to reduce overall noise enables it to operate with less

interference to its surroundings therefore increasing its versatility in a

multitude of applications.

The size and thrust produced by this engine is comparable to that of

the largest civil aircraft engine known today, the GE-90 series. But

lower sfc for both design and off-design conditions makes it a more

suitable engine for carrying large payloads in long haul.


Coursework Task 3 | Introduction 1

TABLE OF CONTENTS

1 Introduction .................................................................................................................................... 1

1.1 Antonov An-225 Mriya ............................................................................................................ 1

1.2 Antonov An-925 Vira® ............................................................................................................. 1

2 Required Inputs Calculations .......................................................................................................... 2

2.1 Weight at start of cruise ......................................................................................................... 2

2.2 Net Thrust ............................................................................................................................... 2

2.2.1 Cruise .............................................................................................................................. 3

2.2.2 Top-of-climb .................................................................................................................... 3

2.2.3 Sea-level static (SLS) take-off .......................................................................................... 3

3 Engine Performance ........................................................................................................................ 4

3.1 Initial Conditions ..................................................................................................................... 4

3.2 Optimisation ........................................................................................................................... 4

3.3 Parametric Study ..................................................................................................................... 4

3.3.1 Outer Fan Pressure Ratio ................................................................................................ 4

3.3.2 Bypass Ratio .................................................................................................................... 5

3.3.3 Turbine Entry Temperature ............................................................................................ 6

3.3.4 Operating Line ................................................................................................................. 6

3.3.5 Performance Map ........................................................................................................... 6

3.4 Results ..................................................................................................................................... 8

4 Aircraft Performance ...................................................................................................................... 9

4.1 Flight Time ............................................................................................................................... 9

4.2 Range .................................................................................................................................... 10

4.3 Operating Cost of Fuel .......................................................................................................... 10

5 Improvements ............................................................................................................................... 10

6 Engine Dimensions ........................................................................................................................ 12

6.1 Compressors ......................................................................................................................... 12

6.1.1 Low Pressure Compressor ............................................................................................. 12

6.1.2 High Pressure Compressor ............................................................................................ 14

6.2 Turbines ................................................................................................................................ 15

6.2.1 High Pressure Turbine ................................................................................................... 15

6.2.2 Low Pressure Turbine .................................................................................................... 16

7 Velocity Triangles .......................................................................................................................... 17



7.1 Compressor ........................................................................................................................... 17

7.1.1 Mean Blade Angles........................................................................................................ 17

7.1.2 Number of blades at mean radius ................................................................................ 17

7.1.3 Rotor and stator blade angles at varying radii .............................................................. 18

7.2 Turbine .................................................................................................................................. 21

7.2.1 At mean radius .............................................................................................................. 21

7.2.2 Hub-mean-tip radiuses ................................................................................................. 22

8 HP Turbine Stresses ...................................................................................................................... 25

8.1.2 HPT Disc......................................................................................................................... 26

8.1.3 HPT Blades..................................................................................................................... 29

8.1.4 Temperature effects ..................................................................................................... 29

9 Engine Sketch ................................................................................................................................ 30

10 Future Developments ............................................................................................................... 33

11 Conclusion ................................................................................................................................. 33

12 Acknowledgements ................................................................................................................... 34

13 References ................................................................................................................................ 34

14 Appendix ................................................................................................................................... 35

14.1 Weight at start of cruise ....................................................................................................... 35

14.2 Initial Iteration ...................................................................................................................... 35

14.3 Initial GasTurb Printouts ....................................................................................................... 35

14.4 Design and Off-Design Engine Conditions ............................................................................ 36

14.5 Final GasTurb Printouts ......................................................................................................... 37

14.6 Calculation: Cruise time and range ....................................................................................... 38

14.7 Fuel Cost Calculations ........................................................................................................... 41

14.8 Fan Calculations .................................................................................................................... 42

14.9 Booster Calculations ............................................................................................................. 44

14.10 HPC Calculations ............................................................................................................... 47

14.11 Sample of Turbine Dimension Calculations ...................................................................... 51

14.12 Sample of Compressors’ velocity triangles and number of blades ................................... 53

14.13 Sample of Turbines’ verlocity triangles and number of blades ........................................ 54

14.14 HP turbine disc stresses .................................................................................................... 57

14.14.1 Stress Calculations .................................................................................................... 57



NOMENCLATURE

bpr Bypass Ratio

FN Net Thrust per Engine

h Blade Height

HPC High Pressure Compressor

HPCPR High Pressure Compressor Pressure Ratio

HPT High Pressure Turbine

IFPR Inner Fan Pressure Ratio

LPC Low Pressure Compressor

LPT Low Pressure Turbine

Ma Mach Number

NGV Nozzle Guide Vanes

OFPR Outer Fan Pressure Ratio

r Radius

rpm Revolutions per Minute

sfc Specific Fuel Consumption

SLS Sea-Level Static

TET, T041 Turbine Entry Temperature

TOC Top-Of-Climb

U Blade Speed

Vx Axial Velocity

Vx/Um Flow Coefficient

ω Rotational Speed

Δh0 Specific Work

Δh0/Um2 Work Coefficient

Subscripts

For r and U:

t Blade Tip

h Hub

m Mean

For T, P and ρ: Station Numbers



1 INTRODUCTION

1.1 Antonov An-225 Mriya

The Antonov An-225 Mriya is a strategic airlifter built by the Antonov Design Bureau primarily to

transport the Buran space shuttle. It is currently the world’s largest fixed-wing aircraft and is

commercially available for flying over-sized, heavy payloads. It was designed more than 2 decades

ago and first flew on 21st

December 1988. Only one aircraft is operating today while a second aircraft

is being built due to recent demands.

Although the An-225 is the most powerful heavy airlifter in the world, it inevitably comes with

several disadvantages. In contrast, the American counterpart, Lockheed C-5 Galaxy, is the alternate

option albeit a smaller payload. The C-5 is however restricted entirely to military and government

use. Table 1 below shows the general performance and capabilities of these 2 airlifters.

Antonov

An-225 Mriya

Lockheed

C-5 Galaxy

Engines 6 (D-18T) 4 (GE-TF39) -

Maximum Payload 250 122.4 tonnes

Specific Fuel Consumption (Installed) 17.04 Classified g/kN*s

Range @ Max payload 4000 4440 km

Max. take-off weight (mtow) 600 381 tonnes

Empty weight 285 172.37 tonnes

Cruise mach number (M) 0.75 0.77 -

Cruising Altitude 33000 30000 ft

Cruise Speed(V) 222.22 230.39 m/s

Take-off run @ Max. payload 3500 2600 m

Wing area 905 576 m2

Table 1: Comparison of Antonov An-225 with Lockheed C-5 Galaxy

To accommodate recent growing demands, an improved heavy airlifter combining the advantages of

both aircrafts will undoubtedly be well-received by the officials and public alike. Therefore, a set of

parameters and limitations have been defined for a new airlifter of the century, the Antonov An-925

Vira®.

1.2 Antonov An-925 Vira®

The Antonov An-925 Vira is an upgraded version of the An-225, and hence retains the same

aerodynamic features. The aim is to design a new engine to improve the performance of the aircraft.

Firstly, the An-925 will be fitted with 4 high bypass engines instead of 6 and the take-off run will be

reduced from 3500m to 2600m, which is similar to the C-5 Galaxy. However, it should still carry a

maximum payload of 250 tonnes with a range of around 4000km.

These improvements essentially mean that each engine on the An-925 will have to provide a

significantly higher thrust but still have a specific fuel consumption (sfc) lower than 17.04 g/kN*s.

Thus improving efficiencies and reducing operating cost. The targeted performance of the An-925 is

shown in Table 2.


Coursework Task 3 | Required Inputs Calculations 2

Antonov

An-999 Vira

Number of engines (n) 4 -

Maximum Payload 250 tonnes

Range @ Max payload 4000 km

Maximum fuel capacity 300 tonnes

Maximum range 15400 km

Max. take-off weight (mtow) 600 tonnes

Empty weight 285 tonnes

Cruising Altitude 35000 ft

Cruise Mach Number(M) 0.75 -

Lift/Drag Ratio 19 -

Maximum TET 1850 K

Cruise TET 1450 K

Take-off run @ Max. payload 2600 m

Table 2: Targeted performance of the An-925 Vira

2 REQUIRED INPUTS CALCULATIONS

Based on the specified requirements of the An-925, the net thrusts for different conditions were

calculated. Calculations are primarily based on the original lift/drag ratio of 19 for the An-225 since

the aerodynamics, cruising altitude and Mach number remains unchanged. [1]

2.1 Weight at start of cruise

The versatility of the An-925 (and An-225) meant that it has a wide range depending on its payload,

i.e. it can either fly a shorter distance with maximum payload or a maximum range with maximum

fuel load. To calculate the weight of the aircraft at the start of cruise, it was estimated that 4% the

fuel in a long haul flight (12hours) is used for take-off and climb to cruising altitude. The mass

available for both payload and fuel is:

= − = 315 Assuming that the aircraft takes off with the maximum weight of 600 tonnes regardless of the mass

of payload and fuel it carries, the reduction in total weight during the flight due to fuel consumption

is therefore the same for maximum range (maximum fuel capacity) and minimum range (maximum

payload).

So the weight at start of cruise is: = 9.81 × " − #0.04 × &', )*+ = 5768.28 /0

2.2 Net Thrust

Once the weight at start of cruise is obtained, the respective net thrusts were calculated.


Coursework Task 3 | Required Inputs Calculations 3

2.2.1 Cruise

The net thrust per engine during cruise is given by:

12 = 34 ÷

12 = 75.90 kN

Where L/D is the lift-to-drag ratio and n is the number of engines.

2.2.2 Top-of-climb

The net thrust per engine at top-of-climb with a minimum rate of climb of 1.5m/s is given by:

12 = 8DL + sin θ@ × n

12 = 84.20 kN

Where θ is the angle of climb at around 0.33°.

2.2.3 Sea-level static (SLS) take-off

To calculate the take-off thrust per engine, the kinematic equations are used as an approximation to

reduce the take-off run from 3500m to 2600m translating to an increase of about 1.5 times the

acceleration and twice the net thrust per engine.

The thrust required at take-off for a take-off run of 3500m is:

FB = thrust to weight ratio × L9.81 × mtowN

FB = 1377 kN

So the required acceleration (a1) is:

OB = 1B = 2.295 /Q

To estimate the required net thrust to achieve the same take-off speed (v) within 2600m, the

kinematic equation was used to relate the distance (s) with acceleration (a), assuming constant

acceleration:

RQ = SQ + 2O

But the initial speed (u) is zero and the take of speed remains the same, therefore:

OBB = OQQ

OQ = OB BQ = 3.089 /Q

∴ 12 = × OQ

12 = 463.41 kN


Coursework Task 3 | Engine Performance 4

3 ENGINE PERFORMANCE

The range of the An-225 with maximum payload and maximum fuel capacity is 4000km and

15400km respectively, indicating that the flight time is approximately 5 to 19 hours depending on

the payload. The design point was therefore chosen at the cruising altitude of 35000 feet and the

off-design point would be at take-off.

3.1 Initial Conditions

The on-design condition was first calculated using single cycle simulation in GasTurb to obtain rough

input values through iteration to produce the required cruising thrust, TET and velocities ratio. A

print screen showing the iterated targets, variables and outputs can be found in Appendix 14.2 and

14.3, where sfc was 13.78 g/kNs.

3.2 Optimisation

Optimisation of the engine performance was then performed to minimize the sfc while ensuring that

sufficient thrust is produced for off-design conditions during take-off and top-of-climb. The HPCPR

was set to 20.00. The NGV and HPT cooling rate were fixed to 10% and 8% respectively. Furthermore,

a polytropic efficiency of 0.9 was used for all engine components to ensure a consistent approach in

the design process.

Optimisation was first performed for on-design conditions by specifying the variables, constraints

where sfc was the figure of merit to minimize. Optimisation was initiated by varying the IFPR, OFPR,

bpr and corrected mass flow while constricting FN, T41 and the ideal jet velocity ratio. The sfc was

reduced to 13.60 g/kNs with design point optimisation. However, the off-design take-off conditions

had to be incorporated to ensure that the TET and T03 are acceptable.

The required SLS take-off thrust is 463.41kN with a maximum TET of 1850K while the cruising thrust

is 75.9kN with a maximum TET of 1450k. Optimisation of the engine was performed again to

accommodate the take-off conditions by creating an off-design point. The chosen figure of merit to

minimize was the on-design sfc while constraining the SLS take-off thrust and TET in addition to the

aforementioned design point constraints. The need to accommodate off-design constraints resulted

in a slight increase of the sfc to 13.66 g/kNs. It is however still well below that of the original engines.

3.3 Parametric Study

To validate the results from optimisation, parametric studies were performed in order to acquire the

performance and operating line of the engine. This is done by varying OFPR, TET and bpr. The sfc

values presented from in this section onwards are the calculated installed sfc after accounting for

nacelle drag. sfcinstalled is estimated using the following formula:

UVWX = Y1.04 + 0.01LZ[\ − 1N]UV

UVWX = 15.19 ^//0

3.3.1 Outer Fan Pressure Ratio

Figure 1 shows the variation in sfc and net thrust with OFPR. Net thrust peaks at 75.9kN and

corresponds with the minimum sfc, thus showing that the OFPR is optimised at 1.65. This confirms

that the engine is able to generate sufficient thrust while using minimum fuel at design point.

Figu

3.3.2 Bypass Ratio

Figure 2 depicts the graph of thrust and sfc against increasing bpr. As expected, net thrust decreases

as bpr increases. The minimum sfc corresponds to a

However, there will be insufficient thrust on design point. Hence the bpr is lower (sfc higher) to

accommodate the required cruise thrust of 75.9kN.

Figure

73

73.5

74

74.5

75

75.5

76

76.5

77

1.5 1.55

Th

rust

(k

N)

60

65

70

75

80

85

90

95

7 7.5

Th

rust

(k

N)

Thrust & sfc vs Bypass ratio


Coursework Task 3 | Engine Performance

Figure 1: Optimising Outer Fan Pressure Ratio.

depicts the graph of thrust and sfc against increasing bpr. As expected, net thrust decreases

as bpr increases. The minimum sfc corresponds to a bpr of around 8.75, which is the ideal bpr.


accommodate the required cruise thrust of 75.9kN.

Figure 2: Parametric study on varying bypass ratio.

1.6 1.65 1.7 1.75

Outer Fan Pressure Ratio

Thrust & sfc vs OFPR

8 8.5 9 9.5

BPR

Thrust & sfc vs Bypass ratio


Engine Performance 5

depicts the graph of thrust and sfc against increasing bpr. As expected, net thrust decreases

bpr of around 8.75, which is the ideal bpr.


15

15.1

15.2

15.3

15.4

15.5

15.6

15.7

15.8

15.9

16

1.8

sfc

(g/k

N*

s)

Thrust

sfc

15

15.1

15.2

15.3

15.4

15.5

15.6

15.7

15.8

15.9

16

10

sfc

(g/k

N*

s)

Thrust

sfc

3.3.3 Turbine Entry Temperature

Figure 3 depicts the graph of thrust and sfc against increasing TET. As expected, net thrust rises with

increasing TET. The minimum sfc corresponds to a TET of 1415K. However, there will be

thrust on design point. Hence the TET is higher (sfc higer) to accommodate the required cruise thrust

of 75.9kN. It can be seen that the design point TET is 1440.6K, which complies with the limit of

1450K.

3.3.4 Operating Line

The operating lines of the HP and LP compressors were evaluated on GasTurb to ensure that the

engine is not susceptible to surging. However, surge was found to occur in the HP compressor and

thus an automatic handling bleed was employed to rectify the problem. The new operating line with

handling bleed has a good surge margin and is shown in

respectively. This also provides

use either handling bleeds or variable stator blades to avoid HPC surge.

3.3.5 Performance Map

The optimisation of the engine can be clearly seen from the performance map (

outer fan pressure ratio and bpr, giving the lowest sfc for the required net thrust. The coloured

contours also display the range of propulsive efficiency at different o

propulsive efficiency increases with rising net thrust but also at a cost of larger sfc. Hence, the trade

off between efficiency and sfc is justified at the design point since the design aim is to design a

commercially viable engine for the aircraft: one that can provide the maximum range with the

lowest sfc.

50

55

60

65

70

75

80

85

90

95

100

1350 1400 1450

Th

rust

(k

N)

Thrust & sfc vs TET, T


Coursework Task 3 | Engine Performance

Turbine Entry Temperature

depicts the graph of thrust and sfc against increasing TET. As expected, net thrust rises with

increasing TET. The minimum sfc corresponds to a TET of 1415K. However, there will be



Figure 3: Parametric Study on varying TET.



c handling bleed was employed to rectify the problem. The new operating line with

handling bleed has a good surge margin and is shown in Figure 4 and Figure 5

respectively. This also provides a more realistic approach to the engine design since most engines

use either handling bleeds or variable stator blades to avoid HPC surge.

The optimisation of the engine can be clearly seen from the performance map (


contours also display the range of propulsive efficiency at different operating conditions. Ideally,

propulsive efficiency increases with rising net thrust but also at a cost of larger sfc. Hence, the trade


engine for the aircraft: one that can provide the maximum range with the

1500 1550 1600 1650 1700 1750

TET, T041 (K)

Thrust & sfc vs TET, T041


Engine Performance 6

depicts the graph of thrust and sfc against increasing TET. As expected, net thrust rises with

increasing TET. The minimum sfc corresponds to a TET of 1415K. However, there will be insufficient





c handling bleed was employed to rectify the problem. The new operating line with

for the LPC and HPC

a more realistic approach to the engine design since most engines

The optimisation of the engine can be clearly seen from the performance map (Figure 6) of varying


perating conditions. Ideally,

propulsive efficiency increases with rising net thrust but also at a cost of larger sfc. Hence, the trade-


engine for the aircraft: one that can provide the maximum range with the

14

14.5

15

15.5

16

16.5

17

17.5

18

1750

sfc

(g/k

N*

s)

Thrust

sfc



Figure 4: LPC Operating Line

Figure 5: HPC Operating Line

.8

1

1.2

1.4

1.6

1.8

2

0 200 400 600 800 1000 1200 1400 1600

Mass Flow W2RStd [kg/s]

LPC

0.3 0.4

0.5

0.6

0.7

0.8

0.9

0.9

5

1

1.1

0.93

0.92

0.91

0.90 0.88

0.85

0.80 0.70 0.60 0.50 0.40

Reference

With Bleeding

26/02/2010 GasTurb 10

0

4

8

12

16

20

24

20 40 60 80 100

Mass Flow W25RSTD [kg/s]

HPC

0.5

0.6

0.7

0.7

5 0

.8

0.8

5 0

.9

0.9

5 1

1.0

5

0.86 0.85

0.84

0.82

0.80

0.75

0.70

Reference

With Bleeding

Here the HPC

will surge.



Figure 6: Varying Outer Fan Pressure Ratio and Bypass Ratio

3.4 Results

The parametric study confirms that the required variables were optimised and the engine meets all

required constrains as specified, namely net thrust, T03 and TET. The engine data were extracted

from GasTurb for further analysis and interpretation.

A summary of the design and off-design engine conditions is shown in Table 3 below. The complete

data along with GasTurb print-outs can be found in Appendix 14.4 and 0.

Fundamentally, moving from on-design engine conditions to off-design engine conditions, the net

thrust and the overall pressure ratio should be higher while the bypass ratio and the ideal jet

velocity ratio should decrease. As for specific fuel consumption, take-off condition will require lower

sfc than that of cruise condition. However, top-of-climb condition requires a slightly higher sfc than

the sfc for cruise condition because during top-of-climb, the fuel flow is still high, but the sfc is

normalized by a much lower thrust than take-off, therefore a higher sfc. All of the mentioned trends

were observed during off-design conditions; take-off and top-of-climb. This is shown in the following

table.

Additionally, it is evident that the HPT and LPT pressure ratios are within acceptable range. T03, TET,

velocity ratios, overall and outer fan pressure ratios are also within acceptable range.

03/03/2010 GasTurb 10

13.4

13.6

13.8

14

14.2

14.4

14.6

14.8

15

70 72 74 76 78 80 82 84 86

Net Thrust [kN]

Outer Fan Pressure Ratio = 1.5 ... 1.7

Design Bypass Ratio = 7 ... 8.8

7

7.2

7.4

7.6

7.8

8

8.2

8.4

8.6

8.8

Dotted Lines = Propulsive Efficiency

0.7

4

0.7

450

.75

0.7

55

0.7

6

0.7

65

7

7.2

7.4

7.6

7.8

8

8.2

8.4

8.6

8.8

1.5

1.54

1.58

1.62

1.66 1.7

1%


Coursework Task 3 | Aircraft Performance 9

Cruise Take-off Top-of-climb

Altitude m 10670.00 0.00 10670.00

Mach Number - 0.75 0.00 0.75

Inner Fan Pressure Ratio - 1.99 2.06 2.06

Outer Fan Pressure Ratio - 1.65 1.69 1.70

HP Compressor Pressure Ratio - 20.00 20.87 21.04

Design Bypass Ratio - 8.32 7.92 8.03

Burner Exit Temperature K 1514.78 1840.13 1599.19

Inlet Corr. Flow W2Rstd kg/s 1434.96 1434.96 1434.96

Net Thrust kN 75.90 463.41 84.20

Sp. Fuel Consumption (bare) g/(kN*s) 13.66 8.32 14.16

Sp. Fuel Consumption (installed) g/(kN*s) 15.20 9.23 15.73

Overall Pressure Ratio P3/P2 - 39.80 42.87 43.34

HPT Pressure Ratio - 5.05 5.03 5.05

LPT Pressure Ratio - 6.82 6.11 6.87

HPC Exit Temperature T3 K 763.49 918.97 795.42

HPC Exit Pressure P3 kPa 1377.98 4344.28 1500.62

HPT Stator Outlet Temp T41 K 1440.59 1750.60 1520.20

Engine Mass Flow W2 kg/s 533.44 1451.45 546.51

HPC Inlet Flow W25 kg/s 57.24 162.74 60.55

Bypass Inlet Flow W12 kg/s 476.20 1288.71 485.96

Ideal Jet Velocity Ratio V18/V8 - 0.84 0.80 0.78

Propulsive Efficiency - 0.76 0.00 0.74

Flight Velocity V0 m/s 222.48 0.00 222.48

Table 3: Summary of design and off-design engine conditions.

4 AIRCRAFT PERFORMANCE

Several specifications and performance, such as flight time, of the An-925 with the new engines

were calculated. The same values were also evaluated from the original An-225. The difference in

performance is largely attributed to a lower sfc for the new engines, effectively improving the range

and operating costs during flight. Furthermore 2 different flight configurations were considered for

thorough comparison: maximum payload and maximum fuel capacity.

The bare sfc for the original engines (D-18T) on the An-225 is 15.68 g/kNs [6] and the installed sfc is

17.04 g/kNs. Again, a realistic approach was adopted and the installed sfc values were used for all

aircraft performance calculations.

4.1 Flight Time

It was estimated that the aircraft consumes 4% of the maximum fuel capacity to takeoff and reach

the cruising altitude. Another 15% of the fuel capacity must be kept as reserves in the event of

emergencies.

The flight time was calculated according to the equation below:


Coursework Task 3 | Improvements 10

Time of Cruise = b mcdef dgeh ij kldigemmnelmoe × g × sfcq rLDs

Where ^ = 9.81 /Q and the average cruising mass:

mmnelmoe = t12u Y + WX] 4.2 Range

The expected range of the original aircraft as described in the specifications is 4000km at maximum

payload and 15400km with maximum fuel capacity. The range was calculated using the Breguet

range equation [5]:

s = − Vg × sfc 8LD@ × ln 8 wejhwgxmlx@

4.3 Operating Cost of Fuel

As for fuel consumption, the new An-925 consumed less fuel than the original aircraft, both for

maximum payload and maximum range. Assuming the jet fuel used is the standard kerosene fuel, Jet

A-1, the price of fuel for each aircraft was calculated. [2]

To calculate the volume of fuel (Vf) required per km for maximum payload and maximum range, the

density of Jet A-1 required is 804 kg/m3. [3]

Vc = Mass of fuel per kmFuel density × 1000 According to the International Airport Transport Association (IATA) [4], the current Jet Fuel price is

191.6 US cents per gallon which is £0.32 per litre. Therefore the cost of fuel for each aircraft, per km

is given by:

Cost of fuel per km = Vc × £0.32

5 IMPROVEMENTS

A comparison of the original and the new aircraft was done to verify that the improved An-925 Vira

is superior to the An-225 Mriya in terms of performance and operating costs.

The performance and specification of the two aircrafts are summarised in the following table. In

overall, the new An-925 has have a lower sfc and higher range at both maximum payload or

maximum fuel. It is able to fly further with the same amount of fuel and payload, which effectively

reduces the fuel cost for the same distance travelled. The take-off run has also been reduced to

2600m.

An-925 Vira An-225 Mriya Improvement

Number of engines 4 6 -2

Take-off run (m) 2600 3500 -25.71%

sfcinstalled (g/kNs) 15.20 17.04 -10.86%


Coursework Task 3 | Improvements 11

Flight time:

Max. Payload 2hrs 42mins 2hrs 24mins 12.50%

Max. Range 24hrs 57mins 22hrs 14mins 12.22%

Range (km):

Max. Payload 2165 1929 12.23%

Max. Range 15107 13467 12.18%

Fuel Cost (£/km):

Max. Payload 11.95 13.41 -10.89%

Max. Range 7.90 8.87 -10.94%

Fan tip diameter (m) 3.20 2.33 37.34%

Table 4: Summary of aircraft performance.


Coursework Task 3 | Engine Dimensions 12

6 ENGINE DIMENSIONS

For both compressor and turbine components, there are several constraints on each component‘s

characteristics which effectively determine the components’ design. The initial step of designing the

components’ dimensions and stages was to specify the different constraints on each component as

well as the assumptions made in the calculations. The criteria to be satisfied for each component are

shown in Table 5.

Although the guidelines given in the brief states that the maximum pressure ratio per stage should

not be more than 1.3, it was found that the General Electric engine, GE90-115B, has a pressure ratio

of 1.36 per stage. Therefore, according to recent standards, a pressure ratio of 1.36 is acceptable.

Several general assumptions were made to facilitate the calculation models, these are:

1. Uniform axial flow through each of the components.

2. Constant mean radius across each component.

3. Change in enthalpy per stage, Δh0,Stage, is equal across each component.

4. Constant pressure ratio per stage, PRStage.

Flow Coefficient

Vx/Um

Work Coefficient

Δh0/Um2

PR per Stage

PRStage

Blade Height

h (m)

Compressors 0.40 - 0.70 0.35 - 0.50 1.36 > 0.01

HPT 0.50 - 0.65 < 2.50 2.50 > 0.01

LPT 0.90 - 1.00 < 2.50 2.50 > 0.01

Table 5: Constraints on engine components.

6.1 Compressors

6.1.1 Low Pressure Compressor

The LPC includes the fan and booster components, which are powered by the LPT on the LP shaft. To

define an accurate model for the LPC calculations, it was approximated that the pressure ratio across

the fan is 1.65 while that for the booster is 1.206. Thereby giving an OFPR of 1.65 and an IFPR of 1.99.

The following table shows the input parameters for the fan and booster.

Input Parameters Booster Fan

Relative Mach @ Blade Tip Marel 0.736 1.6 -

Specific Work Δh0 17.436 42.171 kW/(kg/s)

Mass Flow ~ 57.24 533.44 kg/s

Hub-Tip Ratio rh/rt 0.70 0.35 -

Mach @ entry (Actual) Ma 0.55 0.60 -

Specific Heat Ratio γ 1.4 -

Booster Specific Heat Cp 1004.5 J/KgK

Normalised Mass Flow 1.022 1.078 -

Table 6: Input parameters for fan and booster.



Fan From the actual area, the tip radius was calculated and subsequently the hub radius, mean radius

and followed by the blade height. The chord length was calculated from the blade height and aspect

ratio, h/c. For fan, the blade aspect ratio is 2.5. For sketching purposes, the chord length was

approximated to represent the blade width. Table 26 and Table 7 show the dimensions and

characteristics of the fan respectively.

Characteristics

Work Coefficient Δh0/Um2 - 0.461

Flow Coefficient Vx/Um - 0.599

Rotational Speed ω rpm 2672.79

Axial Velocity Vx m/s 181.315

Fan mean Speed Um m/s 302.55

Table 7: Fan Characteristics.

The Mach number at entry was taken to be 0.6 and the static temperature was calculated using the

stagnation temperature from Gasturb. The axial velocity at the fan inlet was then computed. The

relative velocity of the blade tip was also calculated using the same method but with a relative Mach

number of 1.6. The tip speed was then calculated from the axial and relative velocities and

subsequently, the rotational speed and blade mean speed was computed.

Booster The actual annulus area was computed from the normalised mass flow rate. Assuming the hub to tip

ratio of booster to be 0.70, the tip radius was calculated from the area obtained earlier. Next, the

hub and mean radius, blade height and chord length was calculated.

The axial velocity and rotational speed of the booster is the same as that of the fan as both are on

the LP shaft. The mean blade speed was calculated based on the rotational speed and mean radius.

The enthalpy change per stage was obtained from the mean blade speed and work coefficient.

Finally, the number of stages in the booster can be determined from the total enthalpy change

across booster and per stage.

Characteristics

Work Coefficient Δh/Um2 - 0.47


Specific work per Stage Δh0,Stage kJ/kg 19.908

Number of Stages - - 2.0

Pressure ratio per Stage PRStage - 1.098

Table 8: Booster characteristics.

As the booster is on the LP shaft with a low rpm compared to the HP shaft, the mean blade speed,

Um, is therefore also relatively lower. This directly results in a significantly high flow coefficient of

1.33. Although the flow coefficient is outside the normal range of 0.4 – 0.7, it is worth noting that

the operating conditions of the booster is very much different from that of the HPC. However, a low

Um reduces the loading/compression capabilities of the booster stages, which results in a low PRStage

of only 1.098. In comparison to the booster stages of the GE90-115B, which has similar performance,

the PRStage is about 1.102. Therefore the number of stages and pressure ratio are justified.



6.1.2 High Pressure Compressor

Input Parameters The input parameters used to calculate the size of the HPC are extracted from GasTurb, such as inlet

and outlet conditions. The hub-tip ratio and work coefficient was set as input variables to allow

iteration of the calculations to meet the desired constraints on the HPC properties. The final input

parameters for the HPC, giving acceptable work and flow coefficient as well as number of stages, are

shown in Table 9.

HPC Input Parameters

Relative Mach @ Blade Tip Marel 1.1 -

Mach @ entry (Actual) Ma23 0.518 -

HPC Spec. Work Δh0HPC 479224.7 kW/(kg/s)

Specific Heat Ratio γ 1.38 -

Specific Heat Capacity Cp 1040.64 J/KgK

Hub-Tip Ratio rh/rt 0.8 -

HPC Inlet Mass Flow m23 57.24 kg/s

Normalised Mass Flow 0.992 -

Table 9: HPC input parameters.

Component Characteristics & Dimensions After obtaining the normalised mass flow, the remaining dimensions and characteristics of the HPC

were calculated accordingly. The dimensions and characteristics are shown in Table 26 and Table 10

respectively.

The number of stages is primarily dependant on the work coefficient and hub-tip ratio. From the

calculated values, it can be seen that the work coefficient is at the maximum allowable value of 0.5.

Therefore, the inlet hub-tip ratio was increased to 0.8 to reduce the number of stages required.

Although the weight of the engine is not a limiting factor in the design, it is prudent to keep the

number of stages (and weight) at a minimum. Increasing the hub-tip ratio also resulted in a decrease

in the rotational speed of the HP shaft, which was deemed acceptable after the HPT calculations in

the following section. Additionally, the flow coefficient, blade height and pressure ratio per stage are

well within acceptable ranges.

Characteristics

Rotational Speed ω 4856.66 rpm

Flow Coefficient Vx/Um 0.59 -

Work Coefficient Δh0/Um2 0.50

Axial Velocity Vx 175.08 m/s

Number of Stages 11.00 Stages

Specific work per Stage ΔhStage 43.566 kJ/kg

Pressure ratio per Stage PRStage 1.31 Pa

Table 10: HPC Characteristics.



6.2 Turbines

6.2.1 High Pressure Turbine

Input Parameters The main limitation of the design is that the maximum pressure ratio per stage should be less than

2.50. Therefore the minimum number of stages for the HPT is 2, which also corresponds to about 5

HPC stages per HPT stage. It is prudent to keep the number of stages at a minimum in order to

minimise the engine weight and size. From Gasturb, the total HPT specific work is 510.84kJ/kg.

Assuming equal contributions of each HPT stage, this translates to 255.42kJ/kg specific work per

stage.

Utilising the Smith chart, taking polytropic efficiency as 0.9, the corresponding flow and work

coefficients for the HPT were chosen. Since the engine does not use a gearbox, the rotational speed

of the HPT is the same as the HPC. The input parameters for HPT calculations are shown in Table 11.

Inputs


Work Coefficient Δh/Um2 - 1.7


Specific work per Stage Δhstage kJ/kg 255.42

Number of Stages - - 2


Table 11: HPT Input Parameters

Component Characteristics and Dimensions From the definitions of the work coefficient and flow coefficient, and the specific work per stage, Um

and Vx were then calculated. These velocities are assumed to be constant throughout HPT. From

these values, the dimensions of the HPT were calculated accordingly. A summary of the key

parameters for HPT is described below, and the detailed calculations for each parameter can be

found in Appendix 14.11.

Flow Characteristics

Axial Flow Velocity Vx m/s 251.95

Blade Mean Radius Velocity Um m/s 387.62

Blade Mean Radius rm m 0.76

Table 12 Summary of parameters for HPT

With the initial assumption of uniform axial flow, the flow directions for input and output, α2 and α4

were obtained using the Euler turbomachinery equation. In order to compute the static

temperatures and pressures, the entry and exit Mach numbers were calculated using:

= V = V 1 + − 12 Q



= V 2L2 − L − 1NQN

∴ = 2VQ2 − L − 1NVQ

The local density of the flow was then obtained through the static flow properties. With the local

mass flow rate obtained from Gasturb and the local density and axial velocity computed, the areas of

HPT inlet and outlet was then calculated. Using the equation of area for circular annulus, the blade

height for each stage was then obtained. Thus, tip and hub radius was then obtained through simple

geometry equation.

Detailed calculations can be found in Appendix 14.11 and the summary of the key dimensions for

HPT is given in Table 26.

6.2.2 Low Pressure Turbine

Input Parameters The total enthalpy change across the LPT obtained from Gasturb was 403.13 kJ/kg. For turbine, the

Smith chart was used to determine the suitable flow and work coefficient for a polytropic efficiency

of 0.90. The number of stages was estimated and the resulted conditions were evaluated whether

they are acceptable or not. The minimum number of stages that satisfies the constraints was found

to be 3. However it was later discovered that the resulting dimensions do not provide a smooth flow

between the HPT and LPT. Therefore the number of stages was increased to 4 to reduce losses in the

engine at the cost of extra weight.

Inputs


Work Coefficient Δh/Um2 - 1


Specific work per Stage Δhstage kJ/kg 100.78

Number of Stages - - 4


Table 13: LPT Input Parameters.

Component Characteristics and Dimensions Subsequently, Um was then calculated from the work coefficient and specific work per stage. Using

the flow coefficient, the corresponding axial velocity was found. The mean radius was then

calculated from the mean blade speed of LPT and the rotational speed of LP shaft. The calculated

values are shown in the following table.

Flow Characteristics

Axial Flow Velocity Vx m/s 241.48

Blade Mean Radius Velocity Um m/s 268.31

Blade Mean Radius rm m 0.96

Table 14: LPT flow characteristics.


Coursework Task 3 | Velocity Triangles 17

Next, the dimension of the low pressure turbine was determined using the same method as

mentioned above in the HPT section. The key dimensions of the LPT are show in the following table.

7 VELOCITY TRIANGLES

7.1 Compressor

7.1.1 Mean Blade Angles

A number of assumptions were made to ensure a consistent model for each component:

1. All stages in the compressor are repeating i.e. α1=α3=0

2. Zero incidence and deviation i.e. i=0=α1-β1 and δ=0=α2-β2

3. Constant mean radius through each component

4. Angles and velocities are taken as positive in the direction of blade rotation

Applying the Euler turbomachinery equation with α1=0 gives α2 which can be used to find the

necessary angles by simple trigonometry. From velocity triangles, the mean blade angles were

computed as shown in table 2.

Component α1 α2 α1rel α2rel α3

Fan 0.00 37.55 -59.07 -41.98 0.00

Booster 0.00 19.45 -36.91 -21.70 0.00

HPC 0.00 32.01 -60.75 -49.26 0.00

Table 15: Mean blade angles of compressor components in degrees

To ensure that this design is capable of operating within aerodynamic limits, De Haller’s criterion

was also shown to be not less than 0.72 in table 3. Since the compressor experiences an adverse

pressure gradient, deflection in the rotor is required not to exceed 45°.

Component deflection

ε (°)

De Haller's

(rotor)

De

Haller's

(stator)

Reaction

Λ

Lieblien

diffusion

factor

(rotor)

Lieblien

diffusion

factor

(stator)

Booster 15.21 0.86 0.94 0.76 0.45 0.45

HPC 11.50 0.75 0.85 0.83 0.45 0.45

Table 16: Mean blade characteristics

7.1.2 Number of blades at mean radius

To calculate the number of blades required evaluating the pitch to chord ratio (s/c). This was carried

out by using the Lieblien’s Diffusion Factor approximation. For all components, a diffusion factor of

0.45 was used to provide sufficient diffusion whilst avoiding extensive losses. Since the aspect ratio

for compressors can vary from 1-2.5, the blade pitch was calculated for these values. The number of

blades approximated for all the compressor components were based on the mean pitch hence for

each stage, the chord length of the rotor and stator will be the same. Blade thickness was taken to

be 10% of the chord in all cases; blade thickness should be between 10-20% of the chord.



Component Mean

location

Row Pitch/chord

ratio (s/c)

Aspect

ratio

(h/c)

Pitch

(m)

Chord

(m)

Blade

Thickness

(m)

Number

of blades

Fan Fan Rotor 0.716 2.500 0.298 0.416 0.042 23

Booster Front Rotor 2.200 1.750 0.216 0.098 0.010 14

Stator 2.892 1.750 0.284 0.098 0.010 11

Outlet Rotor 2.200 1.750 0.188 0.086 0.009 16

Stator 2.892 1.750 0.247 0.086 0.009 12

HPC Front Rotor 1.300 1.750 0.106 0.082 0.012 30

Stator 2.014 1.750 0.165 0.082 0.012 19

Middle Rotor 1.300 1.750 0.039 0.030 0.007 81

Stator 2.014 1.750 0.060 0.030 0.004 53

Rear Rotor 1.300 1.000 0.021 0.026 0.004 147

Stator 2.014 1.000 0.033 0.017 0.002 95

Table 17: Key component dimensions and number of blades at mean radius.

Since it is desirable to restrict the tip relative Mach number at HPC inlet to 1.1, increasing this value

whilst decreasing stage loading coefficient does have an effect on reducing the blade number. This

however, is a matter as compromise as a greater rotor deflection can be achieved by increasing the

load coefficient.

A suitable number of blades had to matched with the required number of stages to which the engine

is designed to have, which is the main reason why different aspect ratios were considered.

The process was quite iterative as an optimum compromise had to be achieved in to meet the fixed

design criteria.

7.1.3 Rotor and stator blade angles at varying radii

Table 4 shows the rotor and stator blade angles at the hub, mean and radius for the HP compressor.

The rotor deflection i.e. α1rel

- α2rel in absolute form is less than 45°, which is required in a compressor.

The degree of reaction implies that the rotor contributes most of the increase in static pressure

within the stage.

Radial

location

α1 α2 α1rel α2

rel α3 Deflection,

ε

Reaction

Hub 0 36.10 -56.84 -38.71 0 18.13 0.76

Mean 0 32.01 -60.75 -49.26 0 11.50 0.83

Tip 0 28.67 -63.90 -56.20 0 7.69 0.87

Table 18: HPC blade angles in degrees and rotor deflection

The radial variation of the absolute tangential velocities was obtained by conservation of angular

momentum; the tangential velocities at the radial locations can be derived from the mean values

from \ = VO.

The blades are shown to operate within accepted aerodynamic limits at the various radii as shown in

table 5:



Row Radial

location

Pitch/chord

ratio (s/c)

Aspect

ratio

(h/c)

De

Haller's

Lieblien

Diffusion

Factor

Rotor Hub 0.76 1.75 0.72 0.45

Mean 1.30 1.75 0.75 0.45

Tip 2.00 1.75 0.79 0.45

Stator Hub 1.89 1.75 0.81 0.45

Mean 2.01 1.75 0.85 0.45

Tip 2.16 1.75 0.88 0.45

Table 19: Radial variation of blade characteristics

The angles computed are sufficient to sketch out a stage of the compressor. The sketch shown

below is that of the 1st

stage HPC and depicts the actual scale of the blades.



Fig

ure

7:

HP

C 1

st S

tag

e B

lad

e s

ha

pe

at

va

ryin

g R

ad

ius.



7.2 Turbine

Once the general dimensions for the HPT and LPT components has been fixed, a more detailed

dimension of the local area and blade height, between each stages is calculated as the middle stage

has to be taken into account.

7.2.1 At mean radius

Assuming repeating stages and axial inlet and outlet flow for each stage, the middle flow angle at

mean radius were obtained using Euler Turbomachinery equation. Once the flow angles and axial

velocity were known, the remaining velocity components and flow angles were calculated using

trigonometric relations. The recommended blade turning (α2rel

- α3rel

) is less than 120˚. The maximum

blade turning for HPT is 104.10˚ and for LPT is 71.98˚. An example of the calculations can be seen in

Appendix 14.13.

The reaction was calculated by applying conservation of energy to the fundamental definition of the

reaction (the ratio of the static enthalpy change in the rotor to the total enthalpy change across the

stage). The pitch-chord ratio (s/c) was obtained using tangential lift coefficient (Cl). The Zweifel’s

criterion suggests that the best compromise for Cl is 0.8 and this is the value used to calculate the s/c

ratio for both HPT and LPT blades. It is essential to ensure that the reaction and s/c for all stages and

radiuses are within the recommended range to optimise the performance of the engine. The

optimised range for s/c ratio and h/c ratio are shown below:

s/c h/c

HPT Stator 0.5-1.2

1.0-2.0

Rotor 2.0-3.0

LPT Stator 0.5-1.0

3.0-4.0

Rotor 3.0-4.0

Table 20: Recommended range for dimension ratios

The work coefficient and flow coefficient were changed until the reaction and s/c for all stages were

within the recommended range. Once the s/c is set and by using an appropriate aspect (h/c) ratio,

the number of blades for each stator and rotor stage was computed through simple algebra. The

summary of s/c, h/c, pitch, chord and no. of blades is shown below:

s/c h/c s c no. of blades

HPT Front

Stator 1.199 1.9 0.025 0.021 192

Rotor 0.515 2.5 0.016 0.030 307

Rear Stator 1.199 1.9 0.048 0.040 100

Rotor 0.515 2.5 0.030 0.058 160

LPT Front

Stator 0.879 3 0.025 0.028 244

Rotor 0.575 3 0.021 0.037 287

Middle Stator 0.879 4 0.024 0.027 250

Rotor 0.575 4 0.026 0.045 233

Rear Stator 0.879 4.5 0.060 0.068 101

Rotor 0.575 4.5 0.049 0.086 122

Table 21: Summary of s/c, h/c, pitch, chord and no. of blades



Once the values for the reaction and s/c are acceptable, the actual velocity triangles for each blade

can be obtained. A summary of the properties that define the velocity triangles is shown below:

Stage Station α Vx Vθ V αrel

Vxrel

Vθrel

Vrel

Units ms-1

ms-1

ms-1

ms-1

ms-1

ms-1

HPT Front 1 0.00 251.95 0.00 251.95

2 69.08 251.95 658.95 705.48 47.12 251.95 271.33 370.27

3 0.00 251.95 0.00 251.95 -56.98 251.95 -387.62 462.31

Rear 1 0.00 251.95 0.00 251.95

2 69.08 251.95 658.95 705.48 47.12 251.95 271.33 370.27

3 0.00 251.95 0.00 251.95 -56.98 251.95 -387.62 462.31

LPT Front 1 0.00 241.48 0.00 241.48

2 57.26 241.48 375.63 446.55 23.96 241.48 0.00 107.32

3 0.00 241.48 0.00 241.48 -48.01 241.48 0.00 -268.31

Middle 1 0.00 241.48 0.00 241.48

2 57.26 241.48 375.63 0.00 23.96 241.48 0.00 107.32

3 0.00 241.48 0.00 241.48 -48.01 241.48 0.00 -268.31

Rear 1 0.00 241.48 0.00 241.48

2 57.26 241.48 375.63 0.00 23.96 241.48 0.00 107.32

3 0.00 241.48 0.00 241.48 -48.01 241.48 0.00 -268.31

Table 22: Summary of values for velocity triangles

7.2.2 Hub-mean-tip radiuses

Using the radial equilibrium equation, and assuming that axial velocity is constant across radius and

stagnation enthalpy is constant with radius, it can be shown that,

\ = VO

From this relationship and the hub and tip radiuses computed previously, for both hub and tip

were obtained. The blade speed was obtained by computing the tangential velocity at the respective

radiuses as the rotational speed must be the same. With these 2 components and the axial velocity

known, the other velocity components and angles were easily obtained through trigonometric

relations similar to what was done at the mean radius. The summary of the velocity components and

flow angles for the HPT stage 1 are shown below:

Station α Vx Vθ V αrel Vxrel

Vθrel

Vrel

Units ms-1

ms-1

ms-1

ms-1

ms-1

ms-1

Hub 1 0.00 251.95 0.00 251.95

2 69.57 251.95 676.56 721.95 49.88 251.95 299.03 391.02

3 0.00 251.95 0.00 251.95 -56.28 251.95 -377.53 453.88

Mean 1 0.00 251.95 0.00 251.95

2 69.08 251.95 658.95 705.48 47.12 251.95 271.33 370.27

3 0.00 251.95 0.00 251.95 -56.98 251.95 -387.62 462.31

Tip 1 0.00 251.95 0.00 251.95

2 68.58 251.95 642.24 689.89 44.14 251.95 244.53 351.10

3 0.00 251.95 0.00 251.95 -57.65 251.95 -397.71 470.80



Table 23: Velocity components and flow angles at varying radii

The s/c at hub and tip radiuses were also computed through the tangential lift coefficient of 0.8.

Since the number of blades should be the same with the mean radius, and taking the pitch to be the

same as the mean radius because the spacing between the blades should be the same to produce

constant blade profiles, only the h/c ratio and c change across the radius.

s/c h/c s c no. of blades

HPT

Stage 1

Hub Stator 1.22 1.94 0.02 0.03 192

Rotor 0.48 2.35 0.03 0.02 307

Mean Stator 1.20 1.90 0.05 0.04 100

Rotor 0.51 2.50 0.03 0.06 160

Tip Stator 1.18 1.86 0.02 0.03 192

Rotor 0.55 2.66 0.03 0.02 307

Table 24: Dimensions of HPT blades at varying radii

It is reasonable to assume there is zero incidence and zero deviation for modern turbine blades.

Hence, the flow angles calculated resembles the blade angles in the sketch as shown below:



Figure 8: HPT 1st Stage Blade shape at varying radius


Coursework Task 3 | HP Turbine Stresses 25

8 HP TURBINE STRESSES

Stress calculations were carried out for HP turbine only. The proposed HP turbine will have 2 stages.

The following are the assumptions made to assist the stress calculations.

8.1.1.1 Assumptions

• The disc and blades were manufactured from Inconel 718 (IN718) [6].

• The disc is axially symmetric

• Uniform material properties

• Young’s Modulus, E and Poisson’s Ratio, ν are independent of radius

• Temperature, T independent of radius (uniform temperature)

• Thermal expansion coefficient, α is a function of temperature only

• Thin disc rotating at constant speed Ω rad/s

• The plane stress condition applies (σz=0)

• No through thickness variation of stresses

• The disc thickness is constant, h = axial chord

• Elastic stress distribution

• Centrifugal force acts in radial direction and no forces applied in the axial and hoop direction.

Properties and Dimensions Symbols Inlet Outlet Units

Density IN718 ρ 8220 kg/m3

Yield Stress IN718 σY 1000 Mpa

HP shaft rotational speed ω 502.65 rad/s

Poisson ratio IN718 ν 0.272 0.272

Disc

Inner radius Ri 0.2 0.2 m

Outer radius Ro 0.742 0.689 m

Thickness h 0.03 0.11 m

Blades

Number of blades attached to disc Nb 307 160

Radial force transmitted by each blade F 5.79 78.14 kN

Height hb 0.040 0.146 m

Thickness tb 0.005 0.009 m

Chord Cb 0.030 0.058 m

Volume of a blade Vb 3.66E-06 4.94E-05 m3

Mean radius rm 0.762 m

Table 25: Turbine properties and dimensions

The outer radius of the disc is taken as the hub radius of blades. The inner radius of the disc is also

the HP shaft radius and is assumed to be 0.20m. The inlet disc thickness was obtained from the

scaled blade sketch while the outlet disc thickness was computed from the pitch-to-chord ratio and

the inlet disc thickness. The blade was firstly assumed to behave like a rectangular cantilever.

However, the volume of the blade was taken to be 2/3 of the computed volume as the real blade is



not completely rectangular and also to take into consideration the cooling channel in the blade. The

blade thickness was assumed to be 15% of the blade chord.

8.1.2 HPT Disc

The radial force transmitted by each blade was calculated using the following equation:

1 = ΩQ\ = ΩQ\

1LN = 5.79/0

1LSN = 78.14/0

The radial and hoop stresses due to centrifugal loads along the radius of the disc were then

calculated using the following equation [7]:

= L3 + RN8 ΩQ Q + Q − QQ\Q − \Q + 012ℎLQ − QN 1 − 8\ @Q

= L3 + RN8 ΩQ Q + Q + QQ\Q − L1 + 3RNL3 + RN \Q + 012ℎLQ − QN 1 + 8\ @Q

These equations, which were obtained from the equilibrium and compatibility equations, are valid

by assuming uniform rectangular cross section of disc, no change in temperature and the disc is

subjected to a total force of Nb × FRIM due to the blades. The following figure shows the stresses

along the radius of the disc.

Figure 9: HPT inlet disc stress at varying radius

0

100

200

300

400

500

600

700

800

900

1000

0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75

Str

ess

(M

Pa

)

Radius (m)

σr and σθ along disc radius (HPT inlet)

Radial stress

Hoop stress



Figure 10: HPT outlet disc stress at varying radius

The hoop stress is at its maximum at the inner radius while the maximum radial stress is at mid

radius. Next, the safety factor against yielding at the most highly stressed location was computed.

From the graphs above, the inner radius is the point at which the difference between radial and

hoop stresses is the largest. The yield stress of IN718 was obtained from the following graph. There

is a temperature gradient between the blade and the combustion gas. However, the cooling

temperature, T3 best represents the condition near the disc. For T3 which is 763.5 K (470°C), the

corresponding yield stress is 1000 MPa.

Figure 11: Yield stress against Temperature for Inconel718

[6]

The Tresca yield criterion was assumed:

B = ; Q = ; = = 0 = O YLB − N; LB − QN; L − QN]

= B − =

0

100

200

300

400

500

600

700

800

900

0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7

Str

ess

(M

Pa

)

Radius (m)

σr and σθ along disc radius (HPT outlet)

Radial

stress



8.1.2.1 Inlet of HPT = 978.6¡O ∴ ¢1X = ) = 1.1

8.1.2.2 Outlet of HPT = 879.8¡O ∴ ¢1X = ) = 1.2

Next, the disc thickness profile was designed as such it has a uniform equivalent stress along the

radius.

= = £O =

For this purpose, the optimum safety factor was taken to be 1.25[8] and using the yield stress of

IN718, the constant was computed which was 800 MPa.

The thickness profile was obtained using the following equation:

ℎℎ = ¤Ω¦Q§¨ L©¦ª¦N

Where ℎ is the thickness at the inner radius.

Figure 12: Blade profiles at HPT inlet and outlet

It can be seen that the maximum thickness is at the bore and it reaches the minimum at the edge of

disc.

0

0.2

0.4

0.6

0.8

1

1.2

-0.04 -0.02 0 0.02 0.04

r/R

o

Thickness, h (m)

Thickness profile of disc (HPT Inlet)

0

0.2

0.4

0.6

0.8

1

1.2

-0.2 -0.1 0 0.1 0.2

r/R

o

Thickness, h (m)

Thickness profile of disc (HPT Outlet)



8.1.3 HPT Blades

The stresses due to the centrifugal effects at the root and mid-section of the blade were calculated

using the following equations:

L\N = ΩQ2 L«Q − \QN ℎ\ « = + ¬O ^ℎ

8.1.3.1 Inlet of HPT

At blade root, \ =

LN = 62.8¡O

At mid-section, \ = + ®X WQ = X

L XN = 31.8¡O

Safety factor:

¢1X = ) = 100062.8 = 15.9

8.1.3.2 Outlet of HPT

At blade root, \ =

LN = 230.4¡O

At mid-section, \ = + ®X WQ = X

L XN = 120.7¡O

Safety factor:

¢1X = ) = 1000230.4 = 4.3

8.1.4 Temperature effects

As temperature increases, stress increases and this leads to a decrease in the safety factor. For the

on-design cruise conditions, the assumptions made are acceptable. However, during take-off, the

temperature gradient effect will be greater and need to be considered appropriately when

computing the safety factor.


Coursework Task 3 | Engine Sketch 30

9 ENGINE SKETCH

Following the calculation of key dimensions of the compressors and turbines, the values were used

to sketch a technical drawing of the engine to evaluate and assess the flow path through the engine.

The primary objective was to ensure that there are minimum losses in the flow stream across the

engine. The compiled dimensions of the engine components are detailed overleaf in Table 26 and

the scaled drawing on squared paper is depicted in Figure 13 on the following page.

To maintain consistency in both axes, each major unit of the squared paper is 0.50m while each

minor unit is 0.05m. The technical drawing is evident that the calculated dimensions provide a

smooth flow stream across the engine and the approximate proportions of the engine components

can be perceived easily.



Ta

ble

26

: K

ey

dim

en

sio

ns

of

the

En

gin

e.



Fig

ure

13

: Sch

em

atic o

f the

en

gin

e.


Coursework Task 3 | Future Developments 33

10 FUTURE DEVELOPMENTS

There are several aspects of the engine design which could be improved in light of recent

developments in the aircraft engine industry. These developments include advance blade cooling

technologies, the use of improved thermal barrier coatings (TBC), improved component effciciencies,

improved aerodynamic design of turbomachinery bladings and new composite materials.

Firstly, the use of improved blade cooling coupled with TBC allows the HPT blades to attain a much

higher TET. Current trend shows that the maximum TET at cruise is as high as 1800K and increases

above 2000K during take-off. This effectively means that engines of the same size are able to

produce a much higher thrust due to a higher T04/T02.

Additionally, a conservative approach was adopted in the engine calculations where the polytropic

efficiencies for the engine components were taken as 0.90. However, contemporary engines are able

to operate at polytropic efficiencies as high as 0.93. Therefore the engine performance can be

improved further by using a polytropic efficiency of 0.93.

Improved aerodynamic design of turbomachinery bladings have seen the introduction of new 3D

aerodynamic blades design, allowing further increase in engine mass flow rates and and higher

overall compression ratios.

Developments in the use of composite materials have also expanded the boundaries of aircraft

engine design. New composite materials have allowed contemporary engines such as the GE90-115B

to achieve a bigger fan tip diameter (3.25m) without the risk of material failure.

11 CONCLUSION

In the design and calculation process of the engine size, it was found that there are unique critical

limiting constraints on each component which eventually determined the minimum number of

stages. The primary limiting factor for the fan is the work coefficient and the diameter (Fan diameter

should be less than 3.15m). The booster was limited by the flow coefficient and pressure ratio across

each stage. The HPC was limited by the work coefficient and the number of stages (More than 11

HPC stages are undesirable). The HPT was limited by the pressure ratio across each stage. Finally the

LPT stages had to be increased due to undesirable flow paths.

These limitations show that, although engineers seek to design the ideal engine that provides a high

thrust/weight ratio, there are physical and performance constraints on size (and weight) of the

engine. Therefore, new innovative technologies are constantly being developed to tackle and break

through these barriers. Just as what is happening in the industry today.


Coursework Task 3 | Acknowledgements 34

12 ACKNOWLEDGEMENTS

Special thanks to Dr. Andrew L. Heyes and Dr. Ricardo F. Martinez-Botas for guidance and advice

with calculations as well as the conduct of this project.

13 REFERENCES

(1) Tey A, Tai RY, Hamid N, Jamaludin I. Aircraft Engine Technology: Coursework Task 2. 1. ; 2010.

(2) BP Products Handbook [Online] Available from: http://www.bp.com/

(3) International Airport Transport Association (IATA)[Online] Available from: http://www.iata.org/

(4) Nicholas Cumpsty. A simple guide to the aerodynamic and thermodynamic design and

performance of jet engines. Jet Propulsion. 2.; 2009.

(5) Motrsich [Online] Available from: http://www.ukrainetrade.com/company/motrsich/d18t.htm

(6) Special Metals – Inconel Alloy 718

(7) AET Course: Stress Analysis of Blades and Discs

(8) http://www.mech.uwa.edu.au/DANotes/SSS/safety/safety.html

(9) AET Course: Stress Analysis, Material Design Issues and Failure Analysis


Coursework Task 3 | Appendix 35

14 APPENDIX

14.1 Weight at start of cruise = 600000 /^

&', ) = 300000/^

= 9.81 × " − #0.04 × &', )*+

= 9.81 × Y600000 − L0.04 × 300000N]

∴ = 5768.28 /0

14.2 Initial Iteration

14.3 Initial GasTurb Printouts



14.4 Design and Off-Design Engine Conditions

* Input * Cruise Take-off Top-of-climb

Altitude m 10670.00 0.00 10670.00

Mach Number 0.75 0.00 0.75

Inner Fan Pressure Ratio 1.99 2.06 2.06

Outer Fan Pressure Ratio 1.65 1.69 1.70

HP Compressor Pressure Ratio 20.00 20.87 21.04

Design Bypass Ratio 8.32 7.92 8.03

Burner Exit Temperature K 1514.78 1840.13 1599.19

Inlet Corr. Flow W2Rstd kg/s 1434.96 1434.96 1434.96

NGV Cooling Air W_Cl_NGV/W25 0.10 0.10 0.10

HPT Cooling Air W_Cl/W25 0.08 0.08 0.08

* Output *

Net Thrust kN 75.90 463.41 84.20

Sp. Fuel Consumption (bare) g/(kN*s) 13.66 8.32 14.16

Sp. Fuel Consumption (installed) g/(kN*s) 15.20 9.23 15.73

Overall Pressure Ratio P3/P2 39.80 42.87 43.34

HPT Pressure Ratio 5.05 5.03 5.05

LPT Pressure Ratio 6.82 6.11 6.87

Fan Inner Exit Temp T21 K 302.98 362.77 308.67

Fan Inner Exit Press P21 kPa 68.90 208.18 71.33

Fan Outer Exit Temp T13 K 285.49 341.01 289.68

Fan Outer Exit Press P13 kPa 57.13 171.50 58.73

HPC Exit Temperature T3 K 763.49 918.97 795.42

HPC Exit Pressure P3 kPa 1377.98 4344.28 1500.62

Burner Exit Pressure P4 kPa 1377.98 4344.28 1500.62

Burner Exit Temperature T4 K 1514.78 1840.13 1599.19

HPT Stator Outlet Temp T41 K 1440.59 1750.60 1520.20

HPT Exit Pressure P44 kPa 272.74 863.86 297.24

LPT Inlet Pressure P45 kPa 272.74 863.86 297.24

LPT Inlet Temperature T45 K 1001.76 1231.61 1059.63

LPT Exit Temperature T5 K 645.46 822.92 685.42

LPT Exit Pressure P5 kPa 40.02 141.35 43.24

Core Nozzle Vel. V8 m/s 423.61 388.44 465.85

Bypass Nozzle Vel. V18 m/s 309.24 309.38 311.49

Engine Mass Flow W2 kg/s 533.44 1451.45 546.51

HPC Inlet Flow W25 kg/s 57.24 162.74 60.55

Bypass Inlet Flow W12 kg/s 476.20 1288.71 485.96

Ideal Jet Velocity Ratio V18/V8 0.84 0.80 0.78

Propulsive Efficiency 0.76 0.00 0.74

Core Efficiency 0.57 0.52 0.57

Flight Velocity V0 m/s 222.48 0.00 222.48



14.5 Final GasTurb Printouts

Design Point

Off-Design (Take-Off)



Off-Design (Top-Of-Climb)

14.6 Calculation: Cruise time and range

For Maximum payload Original aircraft,

Cruising sfcbare = 15.68 g/(kN*s) = 0.54 kg/N.hr

Cruising sfcinstalled = (1.04+0.01(bpr-1) sfcbare = (1.04+0.01(5.7-1) 15.68

= 17.04 g/(kN*s)

Mass of fuel = 600,000 – 285,000 – 250,000

= 65,000kg (with max load)

mgxmlx ¯c kldige = 600,000 − 0.04 × 300,000 = 588,000 kg

Assuming 15% fuel reserve,

mejh ¯c kldige = me°±x² + m±m²f¯mh + 0.15mcdef = 285,000 + 250,000 + 0.15 × 65,000

= 544,750kg

mkldige mnelmoe = t12u Ymgxmlx ¯c kldige + mejh ¯c kldige] = 0.5 × L588,000 + 544,750N



= 566,375 kg

mcdef dgeh ij kldige = mcdef − mxm³e ¯cc cdef − mcdef legelne

= 65,000 − 0.04 × 300,000 − 0.15 × 65,000

= 43,250kg

-Flight range, s = − ´µ¶o gck × ln ·¸¹º·»¼½¾¼

= ªQQQ.QQ×B¿¿.ÀB×BÁ.Â×BÃÄ × ln LÅÂÂ,ÁÅÅÀÀ,N

= 1,929.71 km

Time of Cruise = b mcdef dgeh ij kldigemkldige mnelmoegsfcq rLDs

= Æ Â,QÅÅÇÇ,ÁÅ×¿.ÀB×BÁ.Â×BÃÄÈ É19Ê = 2 hours 24minutes

For new aircraft,

Cruising sfcbare = 13.66 g/(kN*s)


= 15.19 g/(kN*s)

Flight range, s = − VLDg sfc × ln 8 wejhwgxmlx@

= − QQQ.QQ×B¿¿.ÀB×BÅ.B¿×BÃÄ × ln ÅÂÂ,ÁÅÅÀÀ,

= 2,165km


= Æ Â,QÅÅÇÇ,ÁÅ×¿.ÀB×BÅ.B¿×BÃÄÈ É19Ê

= 2 hours 42 minutes

For maximum range Original aircraft,

Cruising sfcbare = 15.68 g/(kN*s) = 0.54 kg/N.hr




= 17.04 g/(kN*s)

Mass of fuel = 300,000 kg (max)

mgxmlx ¯c kldige = 588,000kg

Assuming 15% fuel reserve,

mejh ¯c kldige = me°±x² + m±m²f¯mh + 0.15mcdef = 345,000kg

mkldige mnelmoe = ÆBQÈ Ymgxmlx ¯c kldige + mejh ¯c kldige] = 466,500 kg

mcdef dgeh ij kldige = 0.81mcdef = 243000 kg


= ªQQQ.QQ×B¿¿.ÀB×BÁ.Â×BÃÄ × ln LÂÅ,ÅÀÀ,N

= 13,467.1km


= Æ QÂ,ÂÅ,×¿.ÀB×BÁ.Â×BÃÄÈ É19Ê = 22 hours 14 minutes

For new aircraft,

Cruising sfcbare = 13.66 g/(kN*s)


= 15.19 g/(kN*s)


= − QQQ.QQ×B¿¿.ÀB×BÅ.B¿×BÃÄ × ln ÂÅ,ÅÀÀ,

= 15,107km


= Æ QÂ,ÂÅ,×¿.ÀB×BÅ.B¿×BÃÄÈ É19Ê = 24 hours 57 minutes



14.7 Fuel Cost Calculations

Calculation: Volume of fuel per km An-225 Mriya, maximum payload:

Vc = 33.7 kg/km804 kg/km × 1000 = 41.92

An-225 Mriya, maximum range:

Vc = 22.28 kg/km804 kg/km × 1000 = 27.71

An-925 Vira, maximum payload:

Vc = 30.02 kg/km804 kg/km × 1000 = 37.34 An-925 Vira, maximum range:

Vc = 19.86 kg/km804 kg/km × 1000 = 24.7 Calculation: Cost of fuel per km An-225 Mriya, maximum payload:

Cost of fuel = 41.92 × £0.32 = £13.41

An-225 Mriya, maximum range:

Cost of fuel = 27.71 × £0.32 = £8.87

An-925 Vira, maximum payload:

Cost of fuel = 37.34 × £0.32 = £11.95

An-925 Vira, maximum range:

Cost of fuel = 24.7 × £0.32 = £7.90

From the computed costs of fuel, the reduction in cost of fuel is given by:

Reduction in fuel cost = Cost of fuel¯lioijmf − Cost of fuelje·Cost of fuel¯lioijmf %



14.8 Fan Calculations

The normalised mass flow for fan was given by:

= − 1 81 + − 12 Q@ªLÎÏBN/QLÎªBN

= 1.4√1.4 − 1 0.5865 81 + 1.4 − 12 0.5865Q@ªLB.ÂÏBN/QLB.ÂªBN

= 1.063

The actual annulus area of fan was given by

ÑQ = ~ £Q × ¡Q

ÑQ = 476.2 √1004.5 × 243.641.063 × 34600

ÑQ = 6.40 Q

The tip radius was given by:

\ = Ò ÑQL1 − L\'\ NQN ×

\ = 6.40L1 − 0.35QN ×

\ = 1.523

While the hub radius:

\' = \ × \'\

\' = 1.523 × 0.35

\' = 0.533

And the mean radius:

\ W = L\ + \'N2

\ W = L1.523 + 0.533N2

\ W = 1.028



The blade height:

ℎX = \ − \'

ℎX = 1.523 − 0.533

ℎX = 0.990

The chord length:

£ℎ\, V = ℎXℎ/V

£ℎ\, V = 0.9902.5

£ℎ\, V = 0.396 ≈ Ôℎ U ZO,

The static temperature was calculated using the following equation:

Õ = WW1 + − 12 Q

Q = 243.641 + 1.4 − 12 0.587Q

Q = 227.96 Ö

While the static pressure is:

¡Õ = ¡WW1 + − 12 QÎ LÎªBN×

¡Q = 346241 + 1.4 − 12 0.587QB.Â LB.ÂªBN×

¡Q = 27431.2 ¡O

The axial velocity was given by:

) = &W W × Q

) = 0.587 × √1.4 × 287 × 227.96

) = 177.5 ªB

While the relative velocity:

= &W × Q



= 1.6 × √1.4 × 287 × 227.96

= 484.23 ªB

Next, the tip speed was given by:

Ø = Q − )Q

Ø = 484.23Q − 177.5Q

Ø = 450.52 ªB

The rotational speed was found from the following equation:

Ù = Ø × 602 × \ \[

Ù = 450.52 × 602 × 1.523 \[

Ù = 2823.9 \[

The blade mean speed was the computed:

Ø W = Ù × \ W × 30

Ø W = 2823.9 × 1.028 × 30

Ø W = 304.10 ªB

14.9 Booster Calculations

Inlet of Booster The actual cross-sectional area into the core can be computed.

ÑQ,Õ' = Ñ∗0.85

ÑQ = 0.6380.85

ÑQ = 0.751 Q

The tip radius of booster was given by:

\ = ÑÕ'L1 − 0.80QN × = 0.751L1 − 0.80QN ×

\Q, = 0.815



While the hub radius:

\' = \ × 0.80

\Q,' = 0.815 × 0.80

\Q,' = 0.652

And the mean radius:

\ W = L\ + \'N2

\ W = L0.815 + 0.652N2

\ W = 0.733

The blade height:

ℎX = ÑÕ'2 × \ W

ℎX = 0.7512 × 0.733

ℎX = 0.163

The chord length:


£ℎ\, V = 0.1631.75

£ℎ\, V = 0.093 ≈ Ôℎ U ZO,

The enthalpy change per stage for booster was given by: ∆ℎØ Q = 0.4

∆ℎ = 0.4 × L253.48QN

∆ℎ = 25.70 /Ü /^ªB

Next, the number of stages in a booster can be calculated using the following equation:

0 U O^ = ∆ℎ∆ℎ

0 U O^ = 59.7225.70



0 U O^ = 2.32 ≈ 3 O^

The pressure ratio per stage was computed using the following equation:

¡Ý = L¡N2 &

¡ = fjLÞ¨ßàáâããN2 &

¡ = fj B.¿¿

¡ = 1.26

Outlet of Booster The static temperature and pressure were given by:

Q = 302.981 + 1.4 − 12 0.522Q

Q = 287.30 Ö

¡Õ = ¡WW1 + − 12 QÎ LÎªBN×

¡Q = 688991 + 1.4 − 12 0.522QB.Â LB.ÂªBN×

¡Q = 57203.78 ¡O

Therefore, the outlet to inlet static density ratio is:

ρQρQ = PQRTQ PQRTQæ

ρQρQ = 0.6940.419

ρQρQ = 1.65

Therefore the annulus area of the booster outlet:

ÑQ = ÑQ1.65

ÑQ = 0.454 Q

Thus, blade height at the outlet is:



ℎX = 0.4542 × 0.733

ℎX = 0.099

While the tip and hub radii are:

\, = \ W + ℎ2

\, = 0.783

\,' = \ W − ℎ2

\, = 0.684

Finally, the chord length:


£ℎ\, V = 0.0991.75

£ℎ\, V = 0.056 ≈ Ôℎ U ZO,

14.10 HPC Calculations

Inlet of HPC The annulus area of HPC inlet was calculated from the normalised mass flow per unit area for flow at

Mach number 0.518 and = 1.381:

Q = 1.381√1.381 − 1 0.518 81 + 1.381 − 12 0.518Q@ªLB.ÀBÏBN/QLB.ÀBªBN

Q = 0.992

∴ ÑQ = 57.244√1040.639 × 302.9868899 × Q

ÑQ = 0.470Q

It is sufficiently accurate to assume at this stage that the hub/tip ratio is 0.8.

\ = 0.470L1 − ç0.8èQN

\ = 0.645

Correspondingly;



\' = 0.8\ = 0.8 × 0.644

\' = 0.516

and

\O = L0.645 + 0.516N2

\ W = 0.580

Hence the blade height at the first stage of the HPC:

ℎX = 0.4702 × 0.580

ℎQ = 0.129

The static density at inlet which will be used when computing the HPC outlet area is therefore:

Q = ¡Q Q× = 57510.38 L287 × 288.253N×

Q = 0.69517 /^/

The axial was calculated using the equation described earlier:

) = 0.518 × √1.381 × 287 × 288.25

) = 175.08 ªB

And

= 1.1 × √1.381 × 287 × 288.253

= 371.78 ªB

By assuming a purely axial flow into the first stage of the HPC inlet:

Ø = 371.78Q − 175.08Q

Ø = 327.9796/

The calculated flow coefficient is then: )Ø = 0.534

The rotational speed based on HPC conditions is therefore:

Ù = 327.97960.645 = 508.59 \O/

Ù = 4856.66 \[



Superimposing the rotational speed on the mean plane gives the mean blade speed:

Ø W = Ù\ W = 508.59 × 0.580 Ø W = 295.18 /

The cord length is expressed by:

£ℎ\Q = 0.1291.75

£ℎ\Q = 0.074

Outlet of HPC In order to convert the stagnation quantities to static at HPC outlet, the Mach number at that

location needs to be evaluated:

O = V = V 1 + − 12 OQ

O = 2VQ2 − L − 1NVQ

O = 2L175.08NQL2 × 1.381 × 287 × 763.490N − L1.381 − 1NL175.08NQ

O = 0.321

Using this Mach number, it is possible to calculate the static temperature and pressure at the outlet: = 763.49 81 + L1.381 − 1N2 0.321Q@æ

= 748.763Ö

And

¡ = 1377983 1 + 1.381 − 12 0.321QB.ÀB LB.ÀBªBN×æ

¡ = 1284,020 ¡O

Therefore, the outlet to inlet static density ratio is:



ρρQ = PRT PQRTQæ

ρρQ = 5.9750.694

ρρQ = 8.595

The annulus area at HPC exit is then:

Ñ = Q ÑQ = 0.4708.595

Ñ = 0.055 Q

The outlet blade height is given by:

ℎ = 0.0552 × 0.580

ℎ = 0.0150

Accordingly:

\, = 0.580 + 0.01502

\, = 0.5879

\,' = 0.580 − 0.01502

\,' = 0.5729

Chord length is therefore:

£ℎ\ = 0.01501.75

£ℎ\ = 0.009

From Gasturb, the overall enthalpy change within the HPC is 479.224 kJ/kg. Given the chosen load

coefficient: ∆ℎ,Ø Q = 0.5

∴ ∆ℎ, = 0.5 × 295.18Q = 43,566 Ü//^

. U O^ = 479,224.743,566



. U O^ = 10.9999 ≈ 11 O^

With a design HPC pressure ratio of 20, each stage will have:

¡ = fj QBB

¡ = 1.3

14.11 Sample of Turbine Dimension Calculations

The 1st

stage of HPT is used

The mean velocity was calculated from the work coefficient obtained:

U°Q = ∆hφ

= 225.42 × 102

U° = 357.37 m/s

And the axial velocity was then obtained from the flow coefficient:

V = ∅ × U°

V = 0.55 × 357.37

V = 196.55 m/s

Hence the mean radius was computed by:

\ = VÙ

\ = 196.55508.58

\ = 0.703

Initially flow angles, α2 and α4 were calculated using Euler turbomachinery equation and degree of

reaction equation:

∆ℎ = Ø )Ltan ∝Q− tan ∝N

í = 0°

í = tanªB ïℎ/ Ø2)/Ø

íQ = 74.62°



Property Units Stage 41 Stage 45

Mach no 0.3456 0.4121

P kPa 1275.2 244.4

T K 1414.5 994.7

ρ Kg/m3

3.14 0.856

The areas of the inlet and outlet of HPT were then obtained using the following equation:

~ = Ñ)

ÂB~ = ÂBÑÂB)

ÑÂB = ÂB~ÂB) = 0.068Q

Once areas were obtained, the blade heights of the 2 stages of HPT were then obtained using:

Ñ = 2\ ℎX cos í

ℎX,ÂB = ÑÂB2\ cos íQ

= 0.040

The tip and hub radii were obtained using:

\ÂB, = 0.762 + 0.0402

\ÂB, = 0.782

And:

\ÂB,' = 0.762 − 0.0402

\ÂB,' = 0.742

The chord length:

£ℎ\, VLO\N = ℎX1.9

£ℎ\, ÂB, = 0.02088



14.12 Sample of Compressors’ velocity triangles and number of blades

For stage 1 HPC:

Flow into the rotor:

) = B = 185.32 /

B = − Ø = 0 /

The expression is similar for flow out of rotor.

Euler turbomahinery equation:

Δℎ = Ø)Ltan íQ − OíBN

íQ = OªB 8ïℎØ)@ = 32.0°

Δℎ is enthalpy change per stage and is obtained by multiplying the overall enthalpy change from

Gasturb with the load coefficient.

The remaining angles can be obtained from the trigonometric relations ad the axial and calculated

lade velocities.

Reaction:

Λ = )2Ø × #tan íB + íQ* = 0.825

Lieblien Diffusion Factor approximation:

DF is the diffusion factor, taken a 0.45 I all cases here for HPC.

Rotor:



V = 2 b41 − 1 − QBq b BQ − Bq = 2.01

Stator: V = 2 t41 − 81 − QB@u t BQ − Bu = 1.30

Pitch: = Vℎ × V × ℎ = 0.165 LU\ \\N

No of blades: ¬O . = 2\ = 20 LU\ \\N

The blades number is computed by taking the aspect ratio (h/c) =1.75

Inlet (mean)

Stator

h/c s/c blade height s mean circumference no of blades

1 2.013551 0.143303145 0.288548259 3.151400756 10.92157258

1.25 2.013551 0.143303145 0.230838607 3.151400756 13.65196572

1.5 2.013551 0.143303145 0.192365506 3.151400756 16.38235887

1.75 2.013551 0.143303145 0.16488472 3.151400756 19.11275201

2 2.013551 0.143303145 0.14427413 3.151400756 21.84314515

2.5 2.013551 0.143303145 0.14427413 3.151400756 21.84314515

Rotor

h/c s/c blade height s mean circumference no of blades

1 1.300488 0.143303145 0.186363952 3.151400756 16.90992661

1.25 1.300488 0.143303145 0.149091162 3.151400756 21.13740826

1.5 1.300488 0.143303145 0.124242635 3.151400756 25.36488991

1.75 1.300488 0.143303145 0.106493687 3.151400756 29.59237156

2 1.300488 0.143303145 0.093181976 3.151400756 33.81985322

Where mean pitch is assumed the same throughout the radius.

14.13 Sample of Turbines’ verlocity triangles and number of blades

Sample calculations for HPT stage 1:

From Euler Turbomachinery equation:

Δℎ = Ø)LOíQ − OíN

Since íB = í = 0, hence

íQ = tanªB ΔℎØ) = 69.08U\ ò¡ O^ 1



Vx = 251.95ms-1

Um = 387.62 ms-1

The velocity triangles were obtained using the following method:

Stage 1:

óôõ = óõ ö÷ø ∝õ = óô = ùúõ. ûúmsªB

óüõ = óõ øýþ ∝õ =

Stage 2:

Assuming axial velocity remains constant throughout the component,

óô = óù ö÷ø ∝ù

óù = óôö÷ø ∝ù = ú.msªB

óüù = óù øýþ ∝ù = ú. ûúmsªB

óüù = óüù − = ùõ.msªB

∝ù = þªõ óüùóô =. õù

óù = óôö÷ø ∝ù = ùúõ. ûúmsªB

Stage 3:

= óô = ùúõ. ûúmsªB



óü = ó øýþ ∝ =

óü = óü − = −.ùmsªB

∝ = tanªB óüóô = −ú. û

Calculation of the reaction, ⋀

öý÷þ,∧ = ÷÷ ÷,ø = ù −

÷

= #*ù − #ù*ùù ÷ = . õú þ ýø

Calculation of pitch-chord ratio (s/c)

For stator: V = 0.8 × £2cosQ íQLtan íB − tan íQN = 1.199

For rotor: V = 0.8 × £2cosQ íLtan íQ − tan íN

= 0.515

Pitch, s was calculated using:

= V × Vℎ × ℎ

Hence, the number of blades was obtained using:

. = ù!

For the hub and tip velocity components,

óü," = #$óü#$ = óü,#$

The remaining velocity components were calculated using trigonometric relations, what differs at

hub and tip radius is that the aspect ratio is calculated instead of being an input. ℎV = V × 1 × ℎ



14.14 HP turbine disc stresses

14.14.1 Stress Calculations

The radial force transmitted by each blade is given by:

1 = ΩQ\ = ΩQ\

1 = 8220 × 3.65 × 10ªÇ × 502.7Q × 0.762 = 5.79/0

The radial stress is given by:

= L3 + RN8 ΩQ Q + Q − QQ\Q − \Q + 012ℎLQ − QN 1 − 8\ @Q

LN = L3 + 0.272N8 8220 × 502.7Q 0.2Q + 0.742Q − 0.2Q0.742Q0.742Q − 0.742Q + 307 × 5790.3 × 0.7422 × 0.03L0.742Q − 0.2QN 1 − 8 0.20.742@Q

LN = 13.1 ¡O

The hoop stress is calculated using the following equation:

= L3 + RN8 ΩQ Q + Q + QQ\Q − L1 + 3RNL3 + RN \Q + 012ℎLQ − QN 1 + 8\ @Q

LN = L3 + 0.272N8 8220 × 502.7Q 0.2Q + 0.742Q + 0.2Q0.742Q0.2Q − L1 + 3 × 0.272NL3 + 0.272N 0.2Q + 307 × 5790.3 × 0.7422 × 0.03L0.742Q − 0.2QN 1 + 80.20.2@Q

LN = 978.6 ¡O

Safety factor at inlet calculation:

¢1X = ) = 1000978.6 = 1.1

Safety factor at outlet calculation:

¢1X = ) = 1000879.8 = 1.2

Stress at optimum safety factor:

= )¢1 = 10001.25 = 800¡O

Disc thickness was given by:

ℎ = ℎ × ¤Ω¦Q§¨ L©¦ª¦N

ℎLN = 0.03 × 8220×502.72¦×%&& L.Q¦ª.ÁÂQ¦N=0.0154 m

Stress in blade at the root was given by:

L\N = ΩQ2 L«Q − \QN ℎ\ « = + ¬O ^ℎ



LN = 8220 × 502.7Q2 LL0.742 + 0.04NQ − 0.742QN = 62.8 ¡O

Stress in blade at mid-section, \ = + ®X WQ = X

L XN = 8220 × 502.7Q2 8L0.742 + 0.04NQ − L0.742 + 0.042 NQ@ = 31.8 ¡O

Safety factor:

¢1X = ) = 100062.8 = 15.9

aircraft engine technology - antonov an-225 mriya

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