10. Airplane PerformancePart 3
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MEC ENG C162
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Dec 20, 2024
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C 162Flight MechanicsLecture 10:Aircraft Performance:Climb, Glide, CeilingOctober 2, 2024Dr Ir Thomas Lombaerts1/17
➢Rate of Climb ➔Vertical VelocityΣ 𝐹∥= 0 ⟹ ? cos 𝛼 − ? − ? sin𝛾 = 0Σ 𝐹⊥= 0 ⟹ 𝐿 + ? sin 𝛼 − ? cos𝛾 = 0steady flightrelative to flight pathor windoNote that the thrust T needs to overcome drag & component ofweight W!oThe required lift Lis smaller than weight W. Now, multiplying ?∞to Σ 𝐹∥= 0and assuming cos 𝛼 ≈ 1 ⟹? ?∞− ??∞?= ?∞sin𝛾power available“power required” forlevel flight; not climbing. A/C’s vertical velocityrate of climb⟹ Excess Power = ??∞− ??∞Good for 𝜸 < 𝟐𝟎°!horizonflight pathchord line?∞?ℎሶℎ = ?∞sin 𝛾rate of climb:(αsmall)2/17
oNote that the power required for level flight and climbing is different,because drag Dis smaller for climbing than for level flight @ the same velocity. WHY?❖To see this, let’s consider the following example:Example: An airplane with ? = 5,000 lb, ? = 100 ft2, ??,0= 0.015, 𝑒 = 0.6,and AR= 6. If the velocity ?∞= 500ft/s at sea-level, then compute: a) ??𝐚?? ?𝑳in level flight, b) ??𝐚?? ?𝑳for climbing flight.Solution: a) The lift coefficient ?𝐿=𝐿𝑞∞?⟹ ?𝐿=?12𝜌∞𝑉∞2?= 0.168⟹ ??= ??,0+?𝐿2? 𝑒 𝐴?= 0.0175b) For a 20-deg climb, i.e. 𝛾 = 20° and αsmall: sin 𝛼 ≈ 0?𝐿=?cos20°12𝜌∞𝑉∞2?= 0.145⟹ ??= ??,0+?𝐿2? 𝑒 𝐴?= 0.0169(Lift is reduced!)(Hence, lift-induceddrag is reduced!)oMaximum climb rate is given bymax. rateofclimb=?𝐚????? ?????? ?????𝑾3/17𝐿 = ? cos𝛾steady flight
HodographmaxV,maxoIllustration of Excess Power (for 𝜸 < 𝟐𝟎°) oMaximum climb angle, 𝜸?𝐚?Given ?∞, there are two components, ሶℎand ?ℎ, with respect tothe flight path, where ሶℎis the vertical velocity component (i.e.Rate of climb) and ?ℎhorizontal velocity component. ❖Note that maximum climb rate does not occur @maximum climb angle❖For large γ, the (full) equations must be solved algebraically, resulting in an exact solutionvalid for any γ. D is smaller in climb than in level flight!? ?∞− ??∞?= ?∞sin𝛾𝐿 = ?cos𝛾?∞Power𝑃req= ??∞𝑃𝐴= ??∞Jet EngineAirplane?∞Power𝑃𝐴Propeller-drivenAirplane(max)excesspower(max)excesspower?∞?ℎሶℎ = ?∞sin 𝛾rate of climb:4/17ሶℎ𝑃req= ??∞?max ???max ??
oHow to find the maximum climb angle 𝛾max?Return to the equations of motion: Σ 𝐹∥= 0 ⟹ ? cos 𝛼 − ? − ? sin𝛾 = 0? − ??= sin𝛾steady flightGood for 𝜸 < 𝟐𝟎°!oMaximum climb angle is given bymax. angleofclimb= arcsin?𝐚????? ?????? ?????? ???? ??𝐚?𝑾steady flight?∞?req= ?Thrust?𝐴max?max(max) excessThrust over Drag?γ maxJet EngineAirplane5/17?𝐿=?cosγ12𝜌∞𝑉∞2?⟹ ??= ??,0+?𝐿2? 𝑒 𝐴?= ??,0+1? 𝑒 𝐴??cos𝜸12𝜌∞𝑉∞2?2For 𝜸 > 𝟐𝟎°:??∞− ??12?∞?∞3??= ?∞sin𝜸
oNote: this was all for steady flight! (constant speed)oWhat about unsteadyflight? Revisiting the equations of motion, now with acceleration term:Σ 𝐹∥= 𝑚ሶ? ⟹ ? cos 𝛼 − ? − 𝑚𝑔 sin𝛾 = 𝑚ሶ?? ?∞− ??∞?=?∞𝑔ሶ? + ?∞sin𝛾power available“power required” for steadylevel flight; not climbing. A/C’s vertical velocity? cos 𝛼 − ? = 𝑚ሶ? + 𝑚𝑔 sin𝛾specific excess powerA/C’s accel/decel termObservation: Net excess power can be used to:•Accelerate in level flight ⇒increase kinetic energy only•Climb at steady airspeed ⇒increase potential energy only•Accelerated climb ⇒increase kinetic and potential energyReservoir analogy:“Zoom climb”: fly level at high speed, then climb by trading kinetic energy (speed) for potential energy (altitude).6/17
oIllustration: Altitude and speed records of the X-15:oAltitude record on Aug 22, 1963: 354,200ft with 3,794mph (M4.98)oSpeed record four years later: 4,520mph (M6.70) at 102,700ftprioritized potential energy over kinetic energyprioritized kinetic energy over potential energy7/17
➢Gliding Flight –Power-offFor an equilibrium, unaccelerated glide flight, the sum of all forcesmust be zero. That is,? = ? sin𝛾𝐿 = ?cos𝛾(? = 0for power-off)⟹ tan 𝛾 =1𝐿/?Glide angle is a function of thelift-to-drag ratiooThe smallest glide path angle occurs at 𝐿?max, which corresponds to the maximum range Rfor the glide.oGiven 𝛾and h, the range Rcan be derived as ? =ℎtan 𝛾= ℎ𝐿?oNote that𝐿?=?𝐿??Appendix D is useful here! WLDhRhorizonflight path8/17
❖Example: The max. lift-to-drag ratio for CP-1 is 13.6. Calculatethe minimum glide angle and the maximum range in a power-off glide starting at an altitude of 10,000 ft.❖Example: Consider the same CP-1 airplane as above. If the weight and wing area are 2,950 lb and 174 ft2, respectively. Calculate the equilibrium glide velocities at altitudes of 10,000 ft and 2,000 ft,each corresponding to the minimum glide angle.Solution: The minimum glide angle 𝛾minis obtained bytan(𝛾min) =1ൗ𝐿?max=113.6⟹ 𝛾min= 4.2°The maximum range ?maxis then given by?max= ℎ𝐿?max= 136,000 ft⟹ ?max= 25.6 miles Solution: Note that, from earlier, 𝐿?max= 13.6corresponds to the lift coefficient ?𝐿= 0.634. Both ?𝐿and 𝐿?are functions of AoA,and they are independent of flight conditions (i.e. level flight, climbing, etc…) 9/17
It’s given that?cos 𝛾 = 𝐿 =12?∞?∞2??𝐿⟹ ?∞=2cos 𝛾?∞?𝐿??Wing Loading!From Appendix B, at 10,000 ft ➔?∞= 0.001756 slug/ft3at 2,000 ft ➔?∞= 0.002241 slug/ft3Since 𝛾min= 4.2°@ 10,000 ft, the glide velocity is?∞=2cos 4.2°0.001756 ∙ 0.6342950174= 174.3 ft/s@ 2,000 ft, the glide velocity is?∞=2cos 4.2°0.002241 ∙ 0.6342950174= 154.3 ft/sThe equilibrium glide velocity decreases as altitude decreases!10/17
➢Absolute Ceiling & Service CeilingoRecall the effect of altitude on power-required, 𝑃req@sea level@h1@h2As altitude increases,+ move upward+ move to right+ rotate clockwisePower?∞oAs altitude increases, the maximum excess power decreases,hence the maximum rate of climb decreases!?∞Power𝑷????𝐚????? 𝑷𝑨𝑷𝑨,𝐚??𝑷???,𝐚??❖At some high enough altitude, the 𝑷𝑨- curve becomes tangent to the 𝑷???- curve: 1)The velocity at that tangent point is the ONLY velocityat which steady & level flight is possible!2)Obviously, excess power is zero, so is max. climb rate3)Absolute Ceiling: Altitude at which max. R/C = 0!4)Service Ceiling: Altitude at which max. R/C = 100 ft/min Power?∞𝑷???𝑷𝑨practical upper limit!sea level < h1< h211/17
oHow to determine absolute and service ceilings?1) Note that ??∞− ??∞≡ Excess Power andRate of Climb (R/C)=Excess Power?Compute maximum R/C for a number of different altitudes!2) Plot/Tabulate max. R/C (x-axis) vs. altitude (y-axis) 3) Extrapolate the curve to 100 ft/min and 0 ft/min to find Service Ceiling and Absolute Ceiling!10040080012001600max. R/C (ft/min)Altitude, ft0@sea-levelservice ceilingabsolute ceiling12/17
➢Summary - Overview✓Rate of Climb ➔Vertical Velocity? ?∞− ??∞?= ?∞sin𝛾power available“power required” forlevel flight; not climbing. A/C’s vertical velocityrate of climb⟹ Excess Power = ??∞− ??∞Good for 𝜸 < 𝟐𝟎°!HodographmaxV,maxIllustration of Excess Power (for 𝜸 < 𝟐𝟎°) oMaximum climb angle, 𝜸?𝐚??∞Power𝑃req= ??∞𝑃𝐴= ??∞Jet EngineAirplane?∞Power𝑃𝐴Propeller-drivenAirplane(max)excesspower(max)excesspower?∞?ℎሶℎ = ?∞sin 𝛾rate of climb:13/17ሶℎ𝑃req= ??∞steady flight
✓Gliding Flight –Power-off⟹ tan 𝛾 =1𝐿/?Glide angle is a function of thelift-to-drag ratio? =ℎtan 𝛾= ℎ𝐿?✓Absolute Ceiling & Service CeilingAt some high enough altitude, the 𝑷𝑨- curve becomes tangent to the 𝑷???- curve: 1)The velocity at that tangent point is the ONLY velocityat which steady & level flight is possible!2)Obviously, excess power is zero, so is max. climb rate3)Absolute Ceiling: Altitude at which max. R/C = 0!4)Service Ceiling: Altitude at which max. R/C = 100 ft/min Power?∞𝑷???𝑷𝑨practical upper limit!14/17?∞=2cos 𝛾?∞?𝐿??Wing Loading!
15/17Mid-Term Test: Tuesday October 10 Review Session: Thursday October 5 Thursday’s lecture: virtual / online
➢Air Transat Flight 236Date: August 24, 2001Aircraft: Airbus A330-200Itinerary: Toronto to LisbonCause accident: fuel leakdiverted to Lajes (Azores)Both engines ran out of fuel, last 120 km was glide!Everyone on-board survived⟹ 𝐿/? =1tan 𝛾arctanhR=16/17avgcruiseavg11.4185LDLD
➢US Airways Flight 1549 Date: January 15, 2009Aircraft: Airbus A320Itinerary: NY La Guardia to Charlotte (NC)Cause accident: bird strike with flock of geese, lost both engines, glide and water landing into Hudson River.Everyone on-board survived.17/17avgcruvisea g18.1616.33.15LDLD