Gas Turbine Performance

A Simple hand calculation method

Dr. Suresh Sampath

Head of Gas Turbine Systems & Operations

www.cranfield.ac.uk

1 © Cranfield University 2017

1-spool turbojet example

Performance analysis of a simple 1-spool Turbojet

1 – Intake 5 – Combustor Exit

2 – Compressor Inlet 6 – Turbine Inlet
3 – Compressor Exit 7 - Turbine Exit
4 – Combustor Inlet 8 – Nozzle Exit

Simulation through Turbomatch – Webengine Ver 3.0 Copyright

© 2022 Cranfield University, United Kingdom

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1-spool turbojet example – Design point hand calculations

Component Characteristics Value

Altitude 0

Mach number 0

Compressor design pressure ratio 8.8

Compressor and Turbine efficiency 89%

Bleed mass flow Nil

Combustor pressure loss Nil

Inlet mass flow 77.2 kg/s

TET 1141K

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Design point performance

Compressor Calculations
•
.
• ’ = 288.15*( .

• ’ = 536.38K
• - =
• = 567.1K

Compressor Work =

CW = 77.2*1005*(567.1-288.15)

CW = 21.64 MW

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Design point performance

Combustor Calculations
• = = 8.8 bar
• = = 567.1K
• =
= 77.2*1150*(1141-567.1)
= 50.96 MW
• = /LCV = 50.96/42.68
= 1.194 kg/s

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Design point performance

Turbine Calculations
• = = 1141K
• = 77.2 kg/s
• = 78.394 kg/s
• CW = TW
• 21.64 x 10^6 = 78.394*1150*(1141- )
• = 901K
•

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Design point performance

Turbine Calculations
• = 871.34K
• = 1.333 (hot section)
•

• = 2.9427
• = 2.99 bar

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Design point performance

Nozzle Calculations
Add the isentropic flow eq
• = 1.852;

• = 1.614 bar (Choked

nozzle)
• = 1.1665; (static) = 772.4K
• = = 543.6 m/s
• = 0.7281 kg/m^3

• A= = 0.1981 m^2

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Design point performance

Performance Calculations
• = +
• = 78.394*543.6 + 0.1981(1.614-1)*10^5
• = 54.78 kN

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Design performance – Simulation

Please use the WebEngine – 138.250.13.56 on your web browser

• Login using provided user ID
• Password is isabe&2022
• Use the Run Engine tab
• Please select Turbojet_PCN engine
• Scroll down and click Set Operating Conditions
• Click Run Engine
• Click Results and observe the output

• Note that there will be some discrepancy in calculations as the simulation

uses variable Cp and values
• Hand calculations have used constant Cp and values

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Off-Design performance – hand calculations

Off-design calculation steps

• Step – 1: To guess compressor pressure ratio
• Step – 2: To guess combustor temperature rise
• Step – 3: To guess turbine enthalpy drop
• Step – 4:
Determine compatibility of mass flow ( = )
Determine compatibility of rotation speed ( = )
Determine compatibility of work (CW = TW)

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Off-Design performance – 1st iteration

Step – 1
• Let us select an off-design handle for PCN = 0.7 (70% rotation speed)
• Guess the compressor pressure ratio,
• Obtain the Non-dimension mass flow and corresponding compressor
efficiency from the compressor map.
• Compute the mass flow rate and temperature at compressor exit.

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Off-Design performance – 1st iteration

Compressor characteristic map

DP

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Off-Design performance – 1st iteration

Step – 1
• Guessed pressure ratio, = 3.3171
/
• Corresponding NDMF = = 48.094
/
• = 48.094 kg/s
• Corresponding efficiency of compressor = 0.75
• - = ; = 157K
• = 445.15K

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Off-Design performance – 1st iteration

Combustor characteristic map

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Off-Design performance – 1st iteration

Step-2
• Guess on the combustor map
• = 333K
• Corresponding = 1.0
• = 778.15K
• = = 48.094*1150*(778.15-445.15) = 18.42 MW
• = /LCV = 0.436 kg/s
• = 48.53 kg/s

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Off-Design performance – 1st iteration

Turbine characteristic map

DP

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Off-Design performance – 1st iteration

Step-3
∆
• Guess the turbine
∆
• = 236.67 J/kg-K
• = = 778.15K
• Corresponding NDMF * CN = 213.29
• Corresponding turbine efficiency = 0.86
•

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Off-Design performance – 1st iteration

Step-3
∆ ( )
• where = 1150 kJ/kg-K; = 778.15K
• = 618K
. .
• = 1.3146
( ) . . ( . )

• = = 2.989 ( = 1.333)
• = 1.11 bar

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Off-Design performance – 1st iteration

Step-4
• Rotation Compatibility:
=
(N = 0.7; = 288.15K; = 778.15K)

LHS = 0.02509
RHS = 0.02509
Note: If the condition is not satisfied, an internal loop runs repeating Step-2
where Combustor Outlet Temperature is guessed again.

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Off-Design performance – 1st iteration

Step-4
• Mass Flow Compatibility:
/ /
=
/ /
( = 48.53 kg/s; = 778.15K; = 3.3171 bar; = 48.094 kg/s; = 288.15K; = 1 bar)

LHS = 24.042
RHS = 24.042
Note: If the condition is not satisfied, an internal loop runs repeating Step-3
where turbine enthalpy drop is guessed again.

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Off-Design performance – 1st iteration

Step-4
Is the Compressor Work = Turbine Work?
CW = 48.094*1005*(445.15-288.15)
CW = 7.589 MW
TW = 48.53*1150*(778.15-618)
TW = 8.938 MW
• The Compressor work is not equal to the Turbine work.
• Step – 1 is repeated where a new pressure ratio on 0.7 speed line is
guessed.

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Off-Design performance – 2nd iteration

Compressor characteristic map

DP

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Off-Design performance – 2nd iteration

Step – 1
• Guessed pressure ratio, = 4.514
/
• Corresponding NDMF = = 46.307
/
• = 47.307 kg/s
• Corresponding efficiency of compressor = 0.791
• - = ; = 194.31K
• = 482.46K

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Off-Design performance – 2nd iteration

Step-2
• Guess on the combustor map
• = 359.34K
• Corresponding = 0.9996
• = 841.8K
• = = 46.307*1150*(841.8-482.46) = 19.136 MW
• = /LCV = 0.419 kg/s
• = 476.726 kg/s

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Off-Design performance – 2nd iteration

Turbine characteristic map

DP

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Off-Design performance – 2nd iteration

Step-3
∆
• Guess the turbine
∆
• = 244.234 J/kg-K
• = = 841.8K
• Corresponding NDMF * CN = 230.35
• Corresponding turbine efficiency = 0.872
•

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Off-Design performance – 2nd iteration

Step-4
• Rotation Compatibility
=
(N = 0.7; = 288.15K; = 841.8K)

LHS = 0.02413
RHS = 0.02413
Note: If the condition is not satisfied, an internal loop runs repeating Step-2
where Combustor Outlet Temperature is guessed again.

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Off-Design performance – 2nd iteration

Step-4
• Mass Flow Compatibility
/ /
=
/ /
( = 46.726 kg/s; = 841.8K; = 4.514 bar; = 46.307 kg/s; = 288.15K; = 1 bar)

LHS = 17.6926
RHS = 17.6926
Note: If the condition is not satisfied, an internal loop runs repeating Step-3
where turbine enthalpy drop is guessed again.

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Off-Design performance – 2nd iteration

Step-4
• Work Compatibility
Is the Compressor Work = Turbine Work?
CW = 9.113 MW
TW = 9.113 MW
• The Compressor work is equal to the Turbine work. And the solution is
now converged.
• Engine performance parameters can now be calculated
• The next step is to follow the calculation to the nozzle from the turbine exit
to check the choking condition

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Off-Design performance – 2nd iteration

Nozzle Calculations
• = 1.852; = 1.564

• = 1 bar as nozzle is unchoked ( > )

• = 663.02K; = 586.4K
•
• = 402.9 m/s
• = = 18.55kN

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Off-Design performance – Simulation

Please use the WebEngine – 138.250.13.56 on your web browser

• Use the Run Engine tab
• Please select Turbojet_PCN engine
• Select PCN as Power Setting and provide a value of 0.7
• Scroll down and click Set Operating Conditions
• Click Run Engine
• Click Results and observe the output

• Note that there will be some discrepancy in calculations as the simulation

uses variable Cp and values
• Hand calculations have used constant Cp and values

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Off-Design performance – Conclusions

• The engine simulation is an iterative process to calculate off-design

performance.

• The accuracy of the solution depends on the level of detail and accuracy of
component maps available for performing the calculations.

• The engine simulation did not involve bleeds, combustor pressure loss,
flow mixing. The level of complexity increases with double and triple spool
as compatibility has to be performed on all shafts.

• The alternate method for the obtaining solution is – Matrix method

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 Off-Design performance – Simulation exercise 1

Please use the WebEngine – 138.250.13.56 on your web browser

• Use the Run Engine tab
• Please select Turbojet_TET engine
• Please complete the table below

Compressor Turbine Performance

Net Fuel
SL Operating condition Pressure ratio Exit temperature Efficiency Work Pressure ratio Exit temperature Efficiency Work thrust SFC Mass flow flow
Design point
1 TET = 1141K
OP1
2 TET = 900K
OP2
3 TET = 1000K
OP3
4 TET = 1200K
OP4
5 TET = 1300K

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 Off-Design performance – Simulation exercise 2

• Please select Turbojet_TET engine

• Please complete the table below (TET is constant = 1141K)
• Operating condition – (Altitude, Temperature deviation, Mach no.)
Operating condition Compressor Turbine Performance
(Altitude, Temp dev, Net
SL Mach no) Pressure ratio Exit temperature Efficiency Work Pressure ratio Exit temperature Efficiency Work thrust SFC Mass flow Fuel flow
Design point
1 0, 0, 0
OP1
2 0, -15, 0
OP2
3 2000, -10, 0.5
OP3
4 5000, 0, 0.6
OP4
5 8000, 10, 0.7
OP5
6 10000, 0, 0.8
OP6
7 6000, 20, 0.8
OP7
8 0, 0, 0.6

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Off-Design performance – Degradation

Component degradation
• Engine components undergo degradation through its lifetime
• Common faults – fouling, erosion, shroud clearance
• Causes reduction in engine output

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Turboshaft example

Performance analysis of helicopter turboshaft engine

1 – Intake
2 – Compressor
3 – Combustor
4 – Compressor turbine
5 – Power turbine
6 – Exhaust

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Degradation – Simulation exercise 3

Please use the WebEngine – 138.250.13.56 on your web browser

• Use the Run Engine tab
• Please select Turboshaft engine

Compressor faults Turbine faults

Operating condition PR Deg FC Deg ETA Deg DH deg TF deg ETA deg

OP1 0% -1% -2% 0% 0% 0%

OP2 -2% 0% 0% 0% 0% -2%

OP3 0% 0% 0% -1% 0% -3%

OP4 -1% -1% -3% 0% 0% 0%

OP5 0% 0% -4% 0% 0% -2%

OP6 0% 0% 0% -2% -3% 0%

OP7 -3% 2% -3% 0% -1% -1%

OP8 0% 1% 0% 0% -2% -4%

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 Degradation – Simulation exercise 3

Please use the WebEngine – 138.250.13.56 on your web browser

• Use the Run Engine tab
• Please select Turboshaft engine and impose the degradation conditions
Compressor Turbine Power Turbine Performance

Exit Pressure Exit Pressure Exit

SL Operating condition Pressure ratio temperature Efficiency Work ratio temperature Efficiency Work ratio temperature Efficiency Work Power SFC Mass flow Fuel flow

1 Design point

2 OP1

3 OP2

4 OP3

5 OP5

6 OP6

7 OP7

8 OP8

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 2-spool turbofan example

Performance analysis of 2-spool turbofan engine

1 – Intake 10 – Core and SAS mix 17 – Bypass nozzle

2 – Fan 11 – HP turbine 22 – Turbine cooling bleed
3 – Bypass flow 12 – LP turbine 20 – LP accessory bleed
4 – LP compressor 13 – Core nozzle 24 – HP accessory bleed
5 – LP Bleed
6 – HP compressor
7 – HP bleed 1
8 – HP bleed 2
9 – Combustor

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 2-spool turbofan example – Design point hand calculations

Component Value Component Characteristics Value Component Characteristics Value

Characteristics
Altitude 10668m LP compressor pressure ratio 2.36 HP turbine auxiliary work 500 W

Mach number 0.8 HP compressor pressure ratio 8.45 HP compressor efficiency 90%

Fan pressure ratio 1.65 LP compressor auxiliary work 500 W

LP and HP compressor efficiency 88%
Fan efficiency 90%
LP compressor efficiency 93%
LP accessory bleed Nil
Bypass ratio 5.5
HP cooling bleed 13% Engine handle TET/PCN
Bypass pressure loss 1%
HP accessory bleed 1%
DP net thrust 25.7 kN
Inlet mass flow 170 kg/s
Combustor pressure loss 5%
TET 1360K DP fuel flow 0.4 kg/s
Combustion efficiency 99.9%
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 Off-Design performance – Simulation exercise 4

Please use the WebEngine – 138.250.13.56 on your web browser

• Use the Run Engine tab
• Please select Turbofan_PCN engine

Input

Operating condition Time (mins) Altitude (m) Mach number PCN

Taxi 5 0 0 0.7

Takeoff 1 0 0 1.1

Climb 1 5 3000 0.5 0.9

Climb 2 10 7000 0.7 0.8

Cruise 123 10000 0.8 1

Low Rating 30 5000 0.75 0.7

Reverse Thrust 1 0 0.4 1.1

Taxi 5 0 0 0.7

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Off-Design performance – Simulation exercise 4

• The temperature-stress and stress-LMP relations are provided below:

TET (K) Stress (MPa) Stress (MPa) P
1570 350 100 37.0
1525 300 150 36.0
1290 265 200 35.0
1215 250 250 34.5
1070 200 265 34.1
1040 150 300 34.0
900 100 350 33.9

Assuming that we have been able to determine the operating temperatures and
stress levels, from the Larson-Miller chart we can provide the parameter P in
the equation:
• = ( + 20)

• Where,
T = Temperature K (TET)
tf = Time to failure

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 Off-Design performance – Simulation exercise 4

• Calculate the number of cycles the engine could operate in this mode until failure. Also
apply a factor of safety of 1.5 and compute the hours of operation the engine can sustain
based on LMP. the engine can sustain based on LMP.

Input Output
Net Turbine tf
Operating condition Time (mins) Altitude (m) Mach number PCN Thrust Fuel flow inlet T11 Stress (Mpa) P (hours)
Taxi 5 0 0 0.7
Takeoff 1 0 0 1.1
Climb 1 5 3000 0.5 0.9
Climb 2 10 7000 0.7 0.8
Cruise 123 10000 0.8 1
Low Rating 30 5000 0.75 0.7
Reverse Thrust 1 0 0.4 1.1
Taxi 5 0 0 0.7

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Off-Design performance – Simulation exercise 4

Operating condition Time (mins) tf (hours) T/tf

Taxi 5
Takeoff 1
Climb 1 5
Climb 2 10
Cruise 123
Low Rating 30
Reverse Thrust 1
Taxi 5
Total 180

 Total number of cycles using Miner’s Law, N = 1/(Total T/tf)

 Applying FOS = 1.5, the number of cycles, Nfos = N/1.5

 Total number of hours before failure criteria, T = Nfos x 180/60

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 Off-Design performance – Simulation exercise 5

• Consider Fan efficiency degradation of 3%. Estimate the reduction in the operating hours due
to the faulty component for the same mission profile

• Enter -3 in Compressor efficiency degradation index (Corresponds to 97% fan efficiency

scaling)
e engine can sustain based on LMP.
Input Output
Net Turbine tf
Operating condition Time (mins) Altitude (m) Mach number PCN Thrust Fuel flow inlet T11 Stress (Mpa) P (hours)
Taxi 5 0 0 0.7
Takeoff 1 0 0 1.1
Climb 1 5 3000 0.5 0.9
Climb 2 10 7000 0.7 0.8
Cruise 123 10000 0.8 1
Low Rating 30 5000 0.75 0.7
Reverse Thrust 1 0 0.4 1.1
Taxi 5 0 0 0.7

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Off-Design performance – Simulation exercise 5

Operating condition Time (mins) tf (hours) T/tf

Taxi 5
Takeoff 1
Climb 1 5
Climb 2 10
Cruise 123
Low Rating 30
Reverse Thrust 1
Taxi 5
Total 180

 Total number of cycles using Miner’s Law, N = 1/(Total T/tf)

 Applying FOS = 1.5, the number of cycles, Nfos = N/1.5

 Total number of hours before failure criteria, T = Nfos x 180/60

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 Thank You
Email: s.sampath@cranfield.ac.uk
Phone: +44-1234-754712

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