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Introduction to High Speed Airbreathing Propulsion Systems

Introduction to High Speed Airbreathing Propulsion Systems. • Oxidizer … “liquid air” Have to take our own along So that the engine can breathe At high altitudes. • How Much Oxidizer? … depending op chosen propellants … 4-9 times as much As the fuel we carry!.

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Introduction to High Speed Airbreathing Propulsion Systems

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  1. Introduction to High Speed Airbreathing PropulsionSystems

  2. • Oxidizer … “liquid air” Have to take our own along So that the engine can breathe At high altitudes • How Much Oxidizer? … depending op chosen propellants … 4-9 times as much As the fuel we carry!

  3. Flight Without Oxidizer, I What if we could get enough oxygen from ambient air? X-43 airbreathing SCRAMjet engine What happens to our Isp? OK .. Lets loose the Oxidizer … and our Isp goes up by a factor Of 7! … but where do we the the Oxidizer for combustion? ….

  4. … from the atmosphere

  5. Flight Without Oxidizer, II Do Not Need Oxidizer here For a large portion of the Launch trajectory there is Plenty Enough air For combustion if you are going fast enough Need Oxidizer here

  6. Fuel Efficiencies of Various High Speed Propulsion Systems Isp Minimal turbo-machinery

  7. Operational Flight Envelope for Various Flight Vehicles

  8. Operations Payoffs for Airbreathing Launch • Decreased gross lift-off weight, resulting in smaller facilities and easier handling • Wider range of emergency landing sites for intact abort • Powered flyback/go-around & more margin at reduced power • Self-ferry & taxi capabilities • Greatly expanded launch windows (double or triple) • Rapid orbital rendezvous (up to three times faster than rockets) • Wider array of landing sites from orbit, with 2,000-mile cross range and increased range • Reduced sensitivity to weight growth

  9. Applications

  10. Our total launch mass goes from 10,000 kg/kg down to <1400 kg/kg wow!

  11. Squeeze Bang Blow Suck Airbreathing Propulsion Basics

  12. Brayton Cycle for Airbreathing Combustion Step Process 1) Intake (suck) Isentropic Compression 2) Compress the Air (squeeze) Adiabatic Compression 3) Add heat (bang) Constant Pressure Combustion 4) Extract work (blow) Isentropic Expansion in Nozzle 5) Exhaust Heat extraction by surroundings … step 5 above happens In the exhaust plume and has minimal Effect on engine performance Step 3 Step 1-2 Step 4 Step 5 (Credit Narayanan Komerath, Georgia Tech)

  13. Brayton Cycle for Airbreathing Combustion (cont’d) • Ramjet Engine Combustion Cycle Steps • Compression and Power extraction steps Performed passively (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  14. Turbine Combustor Nozzle Compressor Turbojet Afterburner Turbine Compressor Afterburner Turbofan Combustor Nozzle Fan Brayton Cycle for Airbreathing Combustion (cont’d) • Turbojet/Turbo fan engines Combustion Cycle steps Compression and Power extraction steps use Turbo-machinery To augment cycle Inlet Inlet

  15. Ideal Ramjet Cycle Analysis Step 1-2 Step 3 Step 4

  16. Ideal Ramjet Cycle Analysis T-s Diagram Step 1-2 Step 3 Step 4 T s

  17. Thermodynamic Efficiency of Ideal Ramjet • Net Work Available --> work perform by system in step 4 minus work required for step 1-2 • Net heat input --> heat input during step 3 (combustion) - heat lost in exhaust plume (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  18. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • Ideal Cycle Efficiency() = (Net work output)/(Net heat input} (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  19. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • Ideal Cycle Efficiency = (Net work output)/(Net heat input} (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  20. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • Assume … Cpair ~ Cpproducts (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  21. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • For simplicity … let … Cpair ~ Cpproducts

  22. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • From C-->D flow is isentropic … (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  23. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • Adiabatic compression across diffuser (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  24. Thermodynamic Efficiency of Ideal Ramjet (cont’d) • Sub into efficiency equation (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  25. Thermodynamic Efficiency of Ideal Ramjet (cont’d) (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4

  26. Thermodynamic Efficiency of Ideal Ramjet (cont’d) (Credit Narayanan Komerath, Georgia Tech) Step 1-2 Step 3 Step 4 As engine pressure ratio, PB/PA, goes up … h goes up As combustor temperature difference TC-TB goes up … h goes up iii) As inlet total pressure ratio (P0B/P0A) goes down … (stagnation pressure loss goes up) … h goes down

  27. Why is total pressure recovery important? • Choked Nozzle Throat massflow • Thrust is proportional to engine massflow

  28. Ideal Ramjet: Inlet and Diffuser • Take a Rocket motor and “lop the top off” • Works Ok for subsonic, but for supersonic flow … can’t cram enough air down the tube • Result is a normal shock wave at the inlet lip

  29. Ideal Ramjet: Inlet and Diffuser (cont’d) M M • Mechanical Energy is Dissipated into Heat • Huge Loss in Momentum

  30. Ideal Ramjet: Inlet and Diffuser (cont’d) • So … we put a spike in front of the inlet • How does this spike Help? • By forming an Oblique Shock wave ahead of the inlet

  31. Oblique Shockwave Theory Background Anderson, Compressible Flow, Chapter 4 pp.127-145 • In Supersonic flow, pressure disturbances cannot outrun “point-mass” generating object • Result is an infinitesimally weak “mach wave”

  32. Oblique Shock Wave (cont’d) • When generating object is larger than a “point”, shockwave is stronger than mach wave …. Oblique shock wave b>m • b -- shock angle • q -- turning or “wedge angle” q

  33. Oblique Shock Wave Geometry • Must satisfy i) continuity ii) momentum iii) energy Tangential Normal Ahead w1, Mt1 u1, Mn1 Of Shock Behind w2, Mt2 u2, Mn2 Shock

  34. Collected Oblique Shock Equations • Continuity • Momentum • Energy

  35. Oblique Shock Waves:Collected Algorithm • Properties across Oblique Shock wave ~ f(M1, b) • q is the geometric angle that “forces” the flow

  36. Oblique Shock Waves:Collected Algorithm (cont’d) • Can be re-written as third order polynomial in tan(q) • “Very Easy” numerical solution • Cubic equation has three solutions i) weak shock ii) Strong shock iii) non-physical solution

  37. Oblique Shock Waves:Collected Algorithm (cont’d) • “Less Obvious” explicit solution = 0 ---> Strong Shock = 1 ---> Weak Shock

  38. Oblique Shock Waves:Collected Algorithm (cont’d) • ... and the rest of the story …

  39. Oblique Shock Waves:Collected Algorithm (concluded) • ... and the rest of the story …

  40. Basic Oblique Shock Example: •M1 = 3.0, p1=1atm, T1=288K, q=20 g=1.4, • Compute shock wave angle (weak) • Compute P02, T02, p2, T2, M2 … Behind Shockwave

  41. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 • Explicit Solver for b =7.13226 =0.93825

  42. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 • d= 1 (weak shock) = 37.764

  43. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 •  Normal Component of Free stream mach Number =1.837 • Compute Pressure ratio across shock Normal Shock Solver p2 = 3.771(1 atm) = 3.771 atm • Flow is compressed

  44. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 •  Temperature ratio Across Shock Normal Shock Solver T2 = 1.5596(288 K) = 449.2 K

  45. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 •  Stagnation Pressure ratio across shock =1.837 Normal Shock Solver 0.7961

  46. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 •  Stagnation Pressure ratio (alternate method) =0.7961

  47. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 •  Stagnation Pressure =29.24 atm

  48. Basic Oblique Shock Example : (cont’d) •M1 = 3.0, p1=1atm, g=1.4, T1=288K, q=20 •  Stagnation Temperature behind shock =806.4 oK

  49. 2-D Ramjet Inlet Example P0B M1=4.0 B • Compare MB and P0 behind normal shockwaves b= 40 B P0B Assume g=1.4

  50. 2-D Ramjet Inlet Example (cont’d) P0B M1=4.0 B • From Normal Shock wave solver

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