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Aero Engineering 315

Aero Engineering 315. Lesson 28 Cruise Range and Endurance. T-38 Example. Given: W = 12,000 lbs h = Sea Level Find: g MAX. T-38 Example. Given: W = 12,000 lbs h = Sea Level Find: ROC MAX. T-38 Ceiling. What happens to T A - T R as we go higher?. Ceilings.

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Aero Engineering 315

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  1. Aero Engineering 315 Lesson 28Cruise Range and Endurance

  2. T-38 Example Given: W = 12,000 lbs h = Sea Level Find: gMAX

  3. T-38 Example Given: W = 12,000 lbs h = Sea Level Find: ROCMAX

  4. T-38 Ceiling What happens to TA - TR as we go higher?

  5. Ceilings Based on maximum climb rates • Absolute Ceiling = 0 ft/min ROC • Service Ceiling = 100 ft/min ROC • Cruise Ceiling = 300 ft/min ROC • Combat Ceiling = 500 ft/min ROC

  6. Cruise performance overview • Thrust Specific Fuel Consumption • Average Value Method • Endurance • Range • Breguet Equations (conceptual only) • Endurance • Range • Know points to fly for max range and endurance • Find velocities for max range and endurance from T-38 charts or drag polar

  7. Speaking of Range and Endurance…

  8. Thrust output in lb Wf T Thrust Specific Fuel Consumption Fuel use rate in lb/hr TSFC = TSFC = ct = Adjust for altitude or ct ALT = ct SL (aALT /aSL)

  9. TSFC - Typical Values • Engine type TSFC (1/hr) • Recip Prop 0.25 to 0.60 • Turboprop 0.35 to 0.60 • Turbofan 0.35 to 0.60 (high bypass) • 0.39 to 0.70 (medium bypass) • 0.80 to 1.00 (low bypass) • Turbojet 1.00 to 1.30 • 1.80 to 2.50 (with afterburner)

  10. Low and High Bypass Low Bypass Ratio with AfterburnerHigh Bypass Ratio Bypass Ratio = 0.2 - 1.0 Bypass Ratio = 2.0 - 8.0 TSFCDry = 0.8 - 1.3 TSFC = 0.5 - 0.7 TSFCWET = 2.2 - 2.7 BPR = Mf / Mc

  11. 17.3% loss T-38 Powerplant Ratings (see Note 1) Power Setting Normal Military Maximum PowerPowerPower Augmentation None None Afterburner Engine Speed (Note 2) 96.4 100 100 Thrust per engine - lb No losses 2140 2455 3660 Installed 1770 1935 2840 Specific fuel consumption (Note 3) Installed 1.09 1.14 2.64 Notes (1) Sea level static ICAO standard conditions with a fuel specific weight of 6.5 lb/gal. (2) Units are % RPM where 100% = 16,500 RPM. (3) Units are lb/hr per lb thrust. Supplemental Data

  12. Fuel use rate in lb/hr DWf DWf for SLUF E = = = ctTR Wf DWf DWf ctDavg ctD E = Endurance—Average Value Method How long will an airplane fly? Total fuel used in lb Endurance = But weight changes cause drag changes, so use the average drag over the segment

  13. To maximize endurance… …minimize drag Maximum Endurance Using our average endurance equation: DWf E = ctDavg and ctare constant for a mission segment and altitude DWf

  14. Example:T-38 Given: W = 11,000 lb hT = 20,000 ft Ct = 1.09 (sea level) Fuel burned = 2,000 lbFind: E for M=.7 EMAX

  15. Breguet Equations:Endurance For a complete endurance solution, integrate over weight changes Max E? E = ct min (high altitude) Wfuel max L/D)max where W0=initial weight and W1=final weight For our drag polar this means?

  16. DWf Vavg DWf R = R = ctDavg ct(D/V) avg Range—Average Value Method • Starting with Endurance • For our average situation • Max Range? Range = Endurance x Velocity R = EV or Minimize drag/velocity

  17. DWf R = ct(D/V1 ) = D for (D/V1) For a given velocity, say V1 Range TR T TA (WET) TA (DRY) Slope= (D/V1) V

  18. Slope = (D/V)MIN, avg = D for (D/V)MIN, avg = V for (D/V)MIN, avg Max Range TR T TA (WET) TA (DRY) V

  19. Example:T-38 again Slope tangent at Mavg = 0.63 Davg = 960 Given: W = 11,000 lb h = 20,000 ft Fuel Wt = 2,000 lbFind: RMAX

  20. CL1/2 CD )max Breguet Equations:Range For a complete range solution, integrate over weight changes Max R? R = Vdt = ct min & rmin (High altitude) Wfuel max where W0=initial weight and W1=final weight

  21. CL = (CD,0 /3k)1/2 CL1/2 CD )max Maximum Range using the Drag Polar Parasite Drag = Drag due to Lift occurs when: 3x CD,0 = 3CD,i or CD,0= 3k CL2 so: CD = CD,0 + CD,i = 4 CD,0 /3 = 4 kCL2 solving for CL:

  22. Range and Endurance -Aerodynamic Summary MAX RANGE: MAX ENDURANCE: Graphical Analytical Minimum of Thrust Curve Tangent to Thrust Curve CL = (CD,0 /k)1/2 CL = (CD,0 /3k)1/2

  23. Performance Summary(text p. 173) *for typical non-afterburning turbojet aircraft

  24. Performance Summary Climbing Cruise Glides

  25. 4 5 3 7 6 8 2 1 7 7

  26. Next Lesson (T29)… • Prior to class • Read text 5.10 • Complete problems #32, 33 and 34 • Complete FDP parts a, b, c, d, e, f, i, j, k, l, r • In class • Discuss takeoffs and landings

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