This document provides an overview of gas turbine engine design, focusing on compressor and turbine components. It discusses: 1) How gas turbine engines work by compressing air, mixing it with fuel, combusting the mixture to produce thrust or shaft power via Newton's third law. 2) The major components of compressors (axial, centrifugal) and turbines (axial, radial), how they operate to compress or expand the working fluid, and examples of each type. 3) Key design challenges like thermal issues, blade stalls, and dynamic surge; and methods to address them like various cooling techniques. 4) The basic process of axial compressor design which involves defining needs, determining rotational speed, estimating

1. Turbine and Compressor
Design

2. Major Topics
• Compressor and Turbine Design
• Cooling
• Dynamic Surge
• Stall Propagation

3. Background
History:
• First gas turbine was developed in 1872 by Dr. F. Stolze.
Gas Turbine Engine…What does it do?
• Generates thrust by mixing compressed ambient air with
fuel and combusting the mixture through a nozzle to
propel an object forward or to produce shaft work.

4. How Does it Work?
• Newton’s third law
For every action, there is an equal and
opposite reaction.
• As the working fluid is exhausted out the nozzle
of the gas turbine engine, the object that the
engine is attached to is pushed forward. In the
case of generating shaft work, the shaft turns a
generator which produces electrical power.

5. How Does it Work? Cont.
Exhaust
Gas
Ambient
Air In
Shaft

6. Operation
• Compressor is connected to the turbine via a
shaft. The turbine provides the turning moment
to turn the compressor.
• The turning turbine rotates the compressor fan
blades which compresses the incoming air.
• Compression occurs through rotors and stators
within the compression region.
– Rotors (Rotate with shaft)
– Stators (Stationary to shaft)

7. Types of Gas Turbines
• Centrifugal
– Compressed air output is around the outer perimeter
of engine
• Axial
– Compressed air output is directed along the centerline
of the engine
• Combination of Both
– Compressed air output is initially directed along
center shaft of engine and then is compressed
against the perimeter of engine by a later stage.

8. Example of Centrifugal Flow
Airflow being
forced around
body of engine
Centrifugal
Compressor
Intake airflow is being forced around the
outside perimeter of the engine.

9. Example of Axial Flow
Multistage
Axial
Compressor
Center
Shaft
Intake airflow is forced down the center shaft
of the engine.

10. Example of Combination Flow
Centrifugal
Compressor
Intake Air
Flow
Axial Compressor
Intake air flow is forced down the center
shaft initially by axially compressor stages,
and then forced against engine perimeter
by the centrifugal compressor.

11. Major Components of Interest
• Compressor
– Axial
– Centrifugal
Axial Compressor Centrifugal Compressor
• Turbine
– Axial
– Radial

12. Axial Compressor Operation
Axial compressors are designed in a divergent Average Velocity
shape which allows the air velocity to remain
almost constant, while pressure gradually
increases.
A&P Technician Powerplant Textbook published by Jeppesen Sanderson Inc., 1997

13. Axial Compressor Operation cont.
• The airflow comes in through the inlet and
first comes to the compressor rotor.
– Rotor is rotating and is what draws the airflow
into the engine.
– After the rotor is the stator which does not move
and it redirects the flow into the next stage of the
compressor.
• Air flows into second stage.
– Process continues and each stage gradually
increases the pressure throughout the
compressor.

14. Axial Compressor Staging
• An axial compressor stage consists of a rotor
and a stator.
• The rotor is installed in front of the stator and
air flows through accordingly. (See Fig.)
www.stanford.edu/ group/cits/simulation/

15. Centrifugal Compressor Operation
Centrifugal compressors rotate ambient air about an
impeller. The impeller blades guide the airflow toward the
outer perimeter of the compressor assembly. The air
velocity is then increased as the rotational speed of the
impeller increases.

16. Axial Turbine Operation
Hot combustion gases
expand, airflow
pressure and
temperature drops. This
drop over the turbine
blades creates shaft
work which rotates the
compressor assembly.
Airflow through stator
Axial Turbine with airflow Airflow around rotor

17. Radial Turbine Operation
• Same operation
characteristics as axial flow
turbine.
• Radial turbines are simpler
in design and less
expensive to manufacture.
• They are designed much
like centrifugal
compressors. Radial Flow Turbine
• Airflow is essentially
expanded outward from the
center of the turbine.

18. Gas Turbine Issues
• Gas Turbine Engines Suffer from a
number of problematic issues:
• Thermal Issues
• Blade (airfoil) Stalls
• Dynamic Surge
http://www.turbosolve.com/index.html

19. Thermal Issues
• Gas Turbines are limited
to lower operating
temperatures due to the
materials available for
the engine itself.
• Operating at the lower
temperature will
decrease the efficiency
of the gas turbine so a
means of cooling the
components is
necessary to increase
temperatures at which

20. Cooling Methods
• Spray (Liquid)
• Passage
• Transpiration

21. Spray Cooling
• The method of spraying a
liquid coolant onto the
turbine rotor blades and
nozzle.
• Prevents extreme turbine
inlet temperatures from
melting turbine blades by
direct convection between
the coolant and the
blades.

22. Passage Cooling
• Hollow turbine blades
such that a passage is
formed for the
movement of a cooling
fluid.
• DOE has relatively new
process in which
excess high-pressure
compressor airflow is
directed into turbine
passages.
http://www.eere.energy.gov/inventions/pdfs/fluidtherm.pdf

23. Transpiration Cooling
• Method of forcing air
through a porous turbine
blade.
– Ability to remove heat at a
more uniform rate.
– Result is an effusing layer
of air is produced around
the turbine blade.
– Thus there is a reduction
in the rate of heat transfer
to the turbine blade.

24. Blade (airflow) Stalls
• When airflow begins
separating from the
compressor blades over
which it is passing as the
angle of attack w.r.t. the
blades exceeds the
design parameters.
• The result of a blade stall
is that the blade(s) no
longer produce lift and
Separation Regions
thus no longer produces a
pressure rise through the
compressor.

25. Dynamic Surge
• Occurs when the static (inlet) air
pressure rises past the design
characteristics of the
compressor.
• When there is a reversal of
airflow from the compressor
causing a surge to propagate in
the engine.
• Essentially, the flow is exhausted
out of the compressor, or front,
of the engine.
• Result, is the compressor no
Compressor Inlet Turbine Exit
longer able to exhaust as quickly
as air is being drawn in and a
“bang” occurs.
http://www.turbosolve.com/index.html

26. Dynamic Surge Effects
• Cause: Inlet flow is reversed
– Effect: Mass flow rate is reduced into engine.
– Effect: Compressor stages lose pressure.
– Result: Pressure drop allows flow to reverse back into
engine.
– Result: Mass flow rate increases
• Cause: Increased mass flow causes high
pressure again.
– Effect: Surge occurs again and process continues.
– Result: Engine surges until corrective actions are taken.

27. Dynamic Surge Process
Compressor Surge Point,
P
Pressure Loss Flow Reverses
Occurs
No Surge
Condition
Flow reverses Corrective
back into engine Action Taken
mout
V
min
mout

28. Axial Compressor Design
• Assumption of Needs
• Determination of Rotational Speed
• Estimation of number of stages
• General Stage Design
• Variation of air angles

29. Assumption of Needs
• The first step in compressor design in the
determination of the needs of the system
• Assumptions:
– Standard Atmospheric Conditions
– Engine Thrust Required
– Pressure Ratio Required
– Air Mass Flow
– Turbine inlet temperature

30. Rotational Speed Determination
• First Step in Axial Compressor Design
– Process for this determination is based on
assumptions of the system as a whole
– Assumed: Blade tip speed, axial velocity, and
hub-tip ratio at inlet to first stage.
Rotational Speed Equation

31. Derivation of Rotational Speed
• First Make Assumptions:
– Standard atmospheric conditions
– Axial Velocity:
m
C a 150 − 200
s
– Tip Speed: U t 350 m
s
– No Intake Losses
– Hub-tip ratio 0.4 to 0.6

32. Compressor Rotational Speed
• Somewhat of an iterative process in
conjunction with the turbine design.
• Derivation Process:
– First Define the mass flow into the system
mdot = ρAU where U = C a1
C a1
– is the axial velocity range from the root
of the compressor blades to the tips of the
blades.

33. Axial Velocity Relationship
 r 
2
 rr Radius to root of blade
C a1 = 1 −  r
r 
  * Ca
  t
  
 rt Radius to tip of blade
rt rr

34. Tip Radius Determination
• By rearranging the mass flow rate equation we can
obtain an iterative equation to determine the blade tip
radius required for the design.
mdot
rt 2
=
 r 
2

π 1Ca1  − r
ρ 1  
 
  rt
  

• Now Looking at the energy equation, we can determine the
entry temperature of the flow.
2
U2
U 2 C
c pT0 + 0
= c pT1 + 1 T1 = T0 − a1
2 2 2c p

35. Isentropic Relationships
• Now employing the isentropic relation between
the temperatures and pressures, then the
pressure at the inlet may be obtained.
γ
T1 ( γ − )
1
P =P  
1 0
T0 
• Now employ the ideal gas law to obtain the
density of the inlet air.
P1
ρ1 =
RT1

36. Finally Obtaining Rotational Speed
• Using the equation for tip speed.
U t = 2πrt N
• Rearranging to obtain rotational speed.
Ut
N =
2πrt
• Finally an iterative process is utilized to

37. Determining Number of Stages
• Make keen assumptions
– Polytropic efficiency of approximately 90%.
– Mean Radius of annulus is constant through
all stages.
• Use polytropic relation to determine the
exit temperature of compressor.
( n −1)
n = 1.4, Ratio of Specific Heats, Cp/Cv
 P02  n
T02 = T01   P02 is the pressure that the compressor outputs
 P01 
To1 is ambient temperature

38. Determine Temperature Change
• Assuming that Ca1=Ca
• λ is the work done factor
• Work done factor is estimate of stage efficiency
• Determine the mean blade speed.
U m = 2πrmean N
• Geometry allows for determining the rotor blade
angle at the inlet of the compressor.
Um
tan ( β 1 ) =
Ca

39. Temperature Rise in a Stage
• Determine the speed of the flow over the blade profile.
Ca
V1 = Velocity flow over
cos( β 1 ) blade V1.
• This will give an estimate of the maximum possible rotor
deflection. C
cos( β 2 ) = a
β 2 − β1 = Blade _ Deflection
V2
• Finally obtain the temperature rise through the stage.
λU m C a ( tan ( β1 ) − tan ( β 2 ) )
∆T0 s =
cp

40. Number of Stages Required
• The number of stages required is dependent
upon the ratio of temperature changes throughout
the compressor.
∆T
Stages = ∆T = T2 − Tamb
∆T0 s
∆T is the temperature change within a stage
∆T0 s is the average temperature change over all the stages

41. Designing a Stage
• Make assumptions
– Assume initial temperature change through
first stage.
– Assume the work-done factors through each
stage.
– Ideal Gas at standard conditions
• Determine the air angles in each stage.

42. Stages 1 to 2
• Determine the change in the whirl velocity.
– Whirl Velocity is the tangential component of
the flow velocity around the rotor.

43. Stage 1 to 2
• Change in whirl velocity through stage.
∆C w = C w 2 − C w1
c p ∆T
∆C w =
λU m
C w1 = Ca tan ( α1 ) Alpha 1 is zero at the first stage.
U m − Cw2
tan ( β 2 ) =
Ca
Cw2
tan ( α 2 ) =
Ca

44. Compressor Velocity Triangles

45. Pressure ratio of the Stage
• The pressure ratio in the stage can be determined through
the isentropic temperature relationship and the polytropic
efficiency assumed at 90%.
γ
P03  η s ∆T0 s  γ −1
η s = 0.9
Rs = = 1 + 
P01  Tamb 

46. Stage Attributes
• The analysis shows that the stage can be outlined by the
following attributes:
1.) Pressure at the onset of
the stage.
2.) Temperature at the onset
of the stage.
3.) The pressure ratio of the
stage.
4.) Pressure at the end of the
stage.
5.) Temperature at the end of
the stage.
6.) Change in pressure
through the stage.
Example of a single stage

47. Variation in Air Angles of Blade
• Assume the free vortex condition.
C w 2 r = const
• Determine stator exit angle.
Um
tan ( α 3 ) = − tan ( β1 )
Ca
• Then determine the flow velocity.
Um
C3 =
cos( α 3 )

48. Air Angle Triangle Alpha 1 is 0 at
the inlet stage
because there
are no IGV’s.
Thus,
Ca1=C1,
and Cw1 is 0
Note: This is
the whirl
velocity
component
and not a
blade spacing!

49. Velocity Triangle
Red is Ca
Ca
Green is β
Blue is α
Ca

50. Variation in Air Angles of Blade
• Determine the exit temp., pressure, and density of
stage 1 γ
C 2
 T3 
( γ −1) P3
T3 = T0 − a
P3 = P03   ρ3 =
2c p RT3
 T03 
• Determine the blade height at exit.
mdot A3
A3 = h=
ρ 3C a 2πrmean
• Finally determine the radii of the blade at stator exit.
h h
rts = rmean + rrs = rmean −
2 2

51. Variation in Air Angles of Blade
• Determine the radii at the rotor exit.
rtri + rts rrri + rrs
rtr = rrr =
2 2
Note: That rtri is the radius of the blade at the tip at rotor inlet.
Note: That
rrri is the radius of the blade at the root at rotor inlet.
• Determine the whirl velocities at therblade root and
rmean
tip. C w 2 r = C w 2 m r Cw 2t = C w 2 m
mean
rr rtr
Note: Cw2 m = Cw2 because there is no other whirl velocity component in the first stage.

52. Finally determine the Air Angles
Cw2r • Stator air angle at root of
tan ( α 2 r ) =
Ca blade
Cw2m • Stator air angle at middle of
tan ( α 2 m ) =
Ca blade
Cw 2t
tan ( α 2t ) = • Stator air angle at tip of blade
Ca
• Deflection air angle at root of
U rr − C w 2 r
tan ( β 2 r ) = blade
Ca
U m − Cw 2 m • Deflection air angle at middle
tan ( β 2 m ) = of blade
Ca
U tr − C w 2t • Deflection air angle at tip of
tan ( β 2t ) = blade
C a

53. Compressor Design Example
Design of a 5 stage axial compressor:
Givens:
rt = 0.2262m Use this and chart to get
Rotational speed of engine.
Ta = 288 K
T2 = 452.5 K
Ca = 150 m
s
λ = 0.98
Once rotational speed is found, determine mean blade tip speed.

54. Example
rt + rr
rmean = = 0.1697 m
2
m
U m = 2πrmean N = 266.6
s
Determine the total temperature rise through the first stage.
∆T = T2 − Tamb = 164.5 K
We are designing for more than just one stage, so we need
to define an average temperature rise per stage:
∆T
∆T0 s = = 32.9 K
# Stages

55. Example (Air Angle Determination)
Um
β1 = tan−1
= 60.64°
Ca
∆Cw = Cw 2 − C w1
m
C w1 = 0
s
c p ∆T0 s m
∆C w = = 126.55 = Cw 2
λU m s

56. Example (Air Angle Determination)
U m − Cw2
β 2 = tan −1 = 43.03°
Ca
Ca m
V2 = = 205.21
cos( β 2 ) s
Cw 2
α 2 = tan −1
= 40.15°
Ca

57. Questions???