Turbineand compressordesign.senatorlibya


Published on

Published in: Education
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Turbineand compressordesign.senatorlibya

  1. 1. Turbine and Compressor Design
  2. 2. Major Topics• Compressor and Turbine Design• Cooling• Dynamic Surge• Stall Propagation
  3. 3. BackgroundHistory:• 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. 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. 5. How Does it Work? Cont. Exhaust GasAmbientAir In Shaft
  6. 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. 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. 8. Example of Centrifugal Flow Airflow being forced around body of engineCentrifugalCompressor Intake airflow is being forced around the outside perimeter of the engine.
  9. 9. Example of Axial FlowMultistageAxialCompressor Center Shaft Intake airflow is forced down the center shaft of the engine.
  10. 10. Example of Combination Flow Centrifugal CompressorIntake AirFlow 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. 11. Major Components of Interest• Compressor – Axial – Centrifugal Axial Compressor Centrifugal Compressor• Turbine – Axial – Radial
  12. 12. Axial Compressor OperationAxial compressors are designed in a divergent Average Velocityshape which allows the air velocity to remainalmost constant, while pressure graduallyincreases. A&P Technician Powerplant Textbook published by Jeppesen Sanderson Inc., 1997
  13. 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. 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. 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. 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 statorAxial Turbine with airflow Airflow around rotor
  17. 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. 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. 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. 20. Cooling Methods• Spray (Liquid)• Passage• Transpiration
  21. 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. 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. 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. 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. 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. 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. 27. Dynamic Surge Process Compressor Surge Point, P Pressure Loss Flow Reverses Occurs No Surge ConditionFlow reverses Correctiveback into engine Action Taken mout V min mout
  28. 28. Axial Compressor Design• Assumption of Needs• Determination of Rotational Speed• Estimation of number of stages• General Stage Design• Variation of air angles
  29. 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. 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. 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. 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. 33. Axial Velocity Relationship  r  2  rr Radius to root of bladeC a1 = 1 −  r r    * Ca   t     rt Radius to tip of blade rt rr
  34. 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 theentry temperature of the flow. 2 U2 U 2 C c pT0 + 0 = c pT1 + 1 T1 = T0 − a1 2 2 2c p
  35. 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. 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. 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. 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. 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 rotordeflection. 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. 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. 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. 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. 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. 44. Compressor Velocity Triangles
  45. 45. Pressure ratio of the Stage• The pressure ratio in the stage can be determined throughthe isentropic temperature relationship and the polytropicefficiency assumed at 90%. γ P03  η s ∆T0 s  γ −1 η s = 0.9 Rs = = 1 +  P01  Tamb 
  46. 46. Stage Attributes• The analysis shows that the stage can be outlined by thefollowing 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. 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. 48. Air Angle Triangle Alpha 1 is 0 at the inlet stage because there are no IGV’s. Thus, Ca1=C1, and Cw1 is 0Note: This isthe whirlvelocitycomponentand not ablade spacing!
  49. 49. Velocity Triangle Red is Ca Ca Green is β Blue is α Ca
  50. 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. 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. 52. Finally determine the Air Angles Cw2r • Stator air angle at root oftan ( α 2 r ) = Ca blade Cw2m • Stator air angle at middle oftan ( α 2 m ) = Ca blade Cw 2ttan ( α 2t ) = • Stator air angle at tip of blade Ca • Deflection air angle at root of U rr − C w 2 rtan ( β 2 r ) = blade Ca U m − Cw 2 m • Deflection air angle at middletan ( β 2 m ) = of blade Ca U tr − C w 2t • Deflection air angle at tip oftan ( β 2t ) = blade C a
  53. 53. Compressor Design ExampleDesign of a 5 stage axial compressor:Givens:rt = 0.2262m Use this and chart to get Rotational speed of engine.Ta = 288 KT2 = 452.5 KCa = 150 m sλ = 0.98Once rotational speed is found, determine mean blade tip speed.
  54. 54. Example rt + rr rmean = = 0.1697 m 2 m U m = 2πrmean N = 266.6 sDetermine the total temperature rise through the first stage. ∆T = T2 − Tamb = 164.5 KWe are designing for more than just one stage, so we needto define an average temperature rise per stage: ∆T ∆T0 s = = 32.9 K # Stages
  55. 55. Example (Air Angle Determination) Umβ1 = tan−1 = 60.64° Ca∆Cw = Cw 2 − C w1 mC w1 = 0 s c p ∆T0 s m∆C w = = 126.55 = Cw 2 λU m s
  56. 56. Example (Air Angle Determination) U m − Cw2β 2 = tan −1 = 43.03° Ca Ca mV2 = = 205.21 cos( β 2 ) s Cw 2α 2 = tan −1 = 40.15° Ca
  57. 57. Questions???