Aircraft Performance & Propulsion PDF

6ENT1132 Aerospace Performance & Propulsion Introduction to aircraft performance Dr Zhengzhong Sun Introduction to aircraft performance Module learning aims and outcomes Aims (for this part of the module): Develop an in-depth understanding of the theory and practices associated with...

6ENT1132 Aerospace Performance & Propulsion Introduction to aircraft performance Dr Zhengzhong Sun Introduction to aircraft performance Module learning aims and outcomes Aims (for this part of the module): Develop an in-depth understanding of the theory and practices associated with the design and performance analysis of aircraft. Learning outcomes (for this part of the module): Successful students will typically be able to: 1. Identify the design and operational factors governing the performance of an aircraft; 2. Evaluate an aircraft's performance. 2 Introduction to aircraft performance Topics of aircraft performance Atmospheric modelling and effects Take-off performance Climbing flight Cruising (straight and level) flight Range and endurance 3 Introduction to aircraft performance Aircraft performance measures Performance: The action or process of performing a task or function (Oxford Dictionary, 2016) (Good) aircraft performance is fundamentally the need to maximise some/all of the following: How much can it move from A to B i.e. payload. How high can it go i.e. ceiling How long will it take i.e. speed How far can it go i.e. range. How long can it fly for i.e. endurance How much will it cost to run i.e. aerodynamic/propulsion efficiency 4 Introduction to aircraft performance Aircraft performance measures It may be impossible to maximise all of these performance measures together, so aircraft designers will need to optimise the design to focus on achieving the important performance criteria for the aircraft’s task. A minimum level of performance, as outlined by customer or certification requirements, defines the requirements for the aircraft from which the aircraft specification is drawn up. 5 Introduction to aircraft performance Aircraft flight profile Cruise Climb Descent Transition to Approach & climb landing Taxi & take-off Taxi Range Mission time and fuel Block time and fuel Typical aircraft mission profile 6 Introduction to aircraft performance Optimising performance Each flight phase has specific demands and requirements that need to be designed for in the aircraft specification. Flight phase (ICAO taxonomy) Considerations Taxi TXI Take-off run and distance available; Stall speed; Rotation Take-off TOF speed; Minimum speed for safe control; Lift-off speed (AEO and OEI); Field balance length. Initial climb ICL Best climb angle (1500-10000 ft); Best rate of climb (10000 ft – cruise altitude). En-route ENR Cruise speed; Minimum drag; Minimum power; Payload range; endurance. Approach APR Minimum speed for safe control. Landing LDG Flare; Ground roll. Each of these aircraft performance characteristics will depend on the environment it operates in, so a detailed understanding of the atmosphere, and ways of modelling it are necessary. 7 Introduction to aircraft performance Sessions (aircraft performance) Unit Lecture Lecture Topics Tutorials A1 Performance 1 Atmospheric modelling and effects Atmospheric modelling and effects A2 Performance 2 Take-off performance Take-off performance A3/4 Performance 3 Climbing flight / Cruising flight Climbing flight / Cruising flight A5 Performance 4 Range and endurance Range and endurance 8 Introduction to aircraft performance Summary Aircraft performance can be defined in a multitude of ways, some of which present design conflicts such that an optimised design need to be presented that satisfies the aircraft requirements. A typical flight profile has seven different phases, each having its own performance measures. Designing for one measure in one phase is likely to influence the performance of a characteristic in another phase. Each performance measure will be used as a constraint in the aircraft optimisation process and are among many constraints imposed by the certifications, regulations, or customer requirements. 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion Revision guide: Summary of useful equations (Aircraft performance) This note provides a list of important equations used in this part of the module (aircraft performance) you should ensure you can both remember but also make use of in a variety of contexts. Please note that this is not an exhaustive list of all the equations you may encounter in your exam. Nor is it the case that all of these will need to be used in your exam, depending on the questions. Also, some of the equations may be provided to you in a particular question. Regardless, it is worthwhile for you to take the time to become familiar with all of them. You may also be expected to recall and use equations from previous years that should be quite familiar to you by now. You may also, for example, be asked to express equations (e.g., the equation for Mach number) in words, and thus would not necessarily need to know the mathematical form but would need to know the individual components and a physical description of what the equation means. 1 Dynamic pressure 𝑞 = 𝜌 𝑉( 2 1 Lift force 𝐿 = 𝜌𝑉 ( 𝑆𝐶, = 𝑞𝑆𝐶, 2 1 Drag force 𝐷 = 𝜌𝑉 ( 𝑆𝐶. = 𝑞𝑆𝐶. 2 Perfect gas law 𝑝 = 𝜌𝑅𝑇 Speed of sound 𝑎 = 3𝛾𝑅𝑇 𝜌𝑉789 𝑐 Reynolds number Re = 𝜇 𝑉789 Mach number 𝑀= 𝑎 2(𝑝 − 𝑝@ ) 2𝑞 Airspeed 𝑉== == 𝜌 𝜌 𝑉B89 𝑉B89 𝑉789 = = True airspeed √𝜎 𝜌 E𝜌 F 2𝐿 2𝑊 Stall speed 𝑉G = = ≈= 𝜌𝑆𝐶,HIJ 𝜌𝑆𝐶,HIJ Page 1 of 3 Drag coefficient 𝐶. = 𝐶. M + 𝐶. O = 𝐶. F + 𝐾QR 𝐾𝐶,( 𝐶,( Induced drag coefficient 𝐶. O = 𝐾QR = 𝐾QR 𝐾𝐶,( 𝜋𝑒𝐴V note: at altitudes above wingspan 𝑏, 𝐾QR = 1 𝑏( Aspect ratio 𝐴V = 𝑆 Rolling friction 𝐹 = µ(𝑊 − 𝐿) where µ is the rolling friction coefficient 𝑇−𝐷 Climb angle (deg) 𝛾 = asin ^ _ 𝑊 𝑇−𝐷 𝑉b Climb angle (%) 𝛾=^ _ × 100 = 𝑊 𝑉 𝑇−𝐷 Climb rate 𝑉c = 𝑉 ^ _ = 𝑉 sin 𝛾 𝑊 Time to climb between ℎg − ℎh 𝑉b 𝑡=e i ln e g i heights a and b 𝑉b g − 𝑉b h 𝑉b h 𝐶. 𝐶, for minimum drag 𝐶. M = 𝐶. O ⇒ 𝐶, lm = = F 𝐾 1 3𝐶. F 𝐶, for minimum power 𝐶. M = 𝐶. O ⇒ 𝐶, lM = = 3 𝐾 𝐶, for max. range (thrust- 𝐶. F rated a/c) 𝐶, = = 3𝐾 𝐶, for max. range (power- 𝐶. F rated a/c) 𝐶, = = 𝐾 𝐶, for max. endurance 𝐶. F (thrust-rated a/c) 𝐶, = = 𝐾 𝐶, for max. endurance 3𝐶. F (power-rated a/c) 𝐶, = = 𝐾 Page 2 of 3 You will NOT need to learn the following equations: 1 Ground run distance between 𝑊 (𝑇 − µ𝑊 ) + 𝜌𝑆(µ𝐶, − 𝐶. )𝑉g( 𝑠= ln q 2 r points a and b 𝑔𝜌𝑆(µ𝐶, − 𝐶. ) 1 (𝑇 − µ𝑊) + 𝜌𝑆(µ𝐶, − 𝐶. )𝑉h( 2 Ground run distance from rest 𝑊 𝜌𝑆(µ𝐶, − 𝐶. ) ( 𝑠= ln v1 + 𝑉 w (𝑉h = 0) to lift-off (𝑉g = 𝑉stu ) 𝑔𝜌𝑆(µ𝐶, − 𝐶. ) 2(𝑇 − µ𝑊) stu Ground run distance from speed V 𝑊 −µ𝑊 𝑠= ln q r to rest 𝑔𝜌𝑆(µ𝐶, − 𝐶. ) 1 −µ𝑊 + 2 𝜌𝑆(µ𝐶, − 𝐶. )𝑉 ( ( Ground distance covered whilst 𝑊 𝑉(( − 𝑉stu 𝑠= xe i + ℎ( y climbing to screen height 𝑇−𝐷 2𝑔 Brequet range 𝑉 𝐶, 𝑊{ 𝑅z = ln ^ _ (thrust-rated a/c, constant speed) 𝑐z 𝐶. 𝑊( Breguet range 2 2 3𝐶, 𝑅z = = |3𝑊{ − 3𝑊( } (thrust-rated a/c, constant altitude) 𝑐z 𝜌𝑆 𝐶. 𝜂M 𝐶, 𝑊{ Breguet range (power-rated a/c) 𝑅~ = ln ^ _ 𝑐~ 𝐶. 𝑊( 1 𝐶, 𝑊{ Endurance (thrust-rated a/c) 𝐸z = ln ^ _ 𝑐z 𝐶. 𝑊( Maximum endurance for a power- 𝜂M 𝜌𝑆 𝐶, /( 1 1 rated aircraft 𝐸~ = = e − i 𝑐z 2 𝐶. 3𝑊( 3𝑊{ However, you will be expected to know how to use them, and how to calculate the individual variables in each. You will also be expected to be able to perform linear interpolation in order to estimate various atmospheric properties. You will also be expected to be able to show (derive) the different conditions (i.e. 𝐶, value) for operating at minimum drag, and for maximum range and maximum endurance for both power and thrust-rated aircraft. For example, for minimum drag you should be able to evaluate d 𝐶. ^ _=0 d𝐶, 𝐶, and show the necessary computations (steps) to arrive at 𝐶. F 𝐶, lm = =. 𝐾 Page 3 of 3  INTERNATIONAL STANDARD ATMOSPHERE Dynamic Relative Speed of Relative Altitude Pressure Relative Temperature Relative Density Relative viscosity dynamic sound speed of (m) (kPa) pressure (K) temperature (kg/m3) density (kg/m-s) viscosity (m/s) sound x 10-5 0 101.3250 1 288.150 1 1.2250 1 1.7830 1 340.292 1 250 98.3575 0.9707 286.525 0.9944 1.1959 0.9762 1.7790 0.9977 339.368 0.9973 500 95.4068 0.9421 284.900 0.9887 1.1673 0.9529 1.7671 0.9911 338.368 0.9943 750 92.6336 0.9142 283.275 0.9831 1.1392 0.9300 1.7593 0.9867 337.401 0.9915 1000 89.8746 0.8870 281.650 0.9774 1.1117 0.9075 1.7513 0.9822 336.432 0.9887 1250 87.1824 0.8604 280.025 0.9718 1.0846 0.8854 1.7432 0.9777 335.460 0.9858 1500 84.5560 0.8345 278.400 0.9662 1.0581 0.8637 1.7354 0.9733 334.486 0.9829 1750 81.9940 0.8092 276.775 0.9605 1.0320 0.8425 1.7274 0.9688 333.508 0.9801 2000 79.4952 0.7846 275.150 0.9549 1.0065 0.8216 1.7192 0.9642 332.527 0.9772 2250 77.0584 0.7605 273.525 0.9492 0.9814 0.8012 1.7111 0.9597 331.544 0.9743 2500 74.6825 0.7371 271.900 0.9436 0.9569 0.7811 1.7031 0.9552 330.558 0.9714 2750 72.3663 0.7142 270.275 0.9380 0.9328 0.7614 1.6949 0.9506 329.569 0.9685 3000 70.1085 0.6919 268.650 0.9323 0.9091 0.7241 1.6869 0.9461 328.576 0.9656 3250 67.9082 0.6702 267.025 0.9267 0.8860 0.7232 1.6787 0.9415 327.581 0.9626 3500 65.7641 0.6490 265.400 0.9210 0.8632 0.7047 1.6705 0.9369 326.583 0.9597 3750 63.6751 0.6284 263.775 0.9154 0.8410 0.6865 1.6623 0.9323 325.581 0.9568 4000 61.6402 0.6083 262,150 0.9098 0.8191 0.6687 1.6541 0.9277 324.577 0.9538 4250 59.6583 0.5888 260.525 0.9041 0.7977 0.6512 1.6459 0.9231 323.569 0.9509 4500 57.7283 0.5697 258.900 0.8985 0.7768 0.6341 1.6375 0.9184 322.559 0.9479 4750 55.8492 0.5512 257.275 0.8929 0.7562 0.6173 1.6293 0.9138 321.545 0.9449 5000 54.0199 0.5331 255.650 0.8872 0.7361 0.6009 1.6209 0.9091 320.528 0.9419 5250 52.2394 0.5156 254.025 0.8816 0.7164 0.5848 1.6125 0.9044 319.507 0.9389 5500 50.5068 0.4985 252.400 0.8759 0.6971 0.5691 1.6042 0.8997 318.484 0.9359 5750 48.8210 0.4818 250.775 0.8703 0.6782 0.5536 1.5958 0.8950 317.457 0.9329 6000 47.1810 0.4656 249.150 0.8647 0.6597 0.5385 1.5874 0.8903 316.427 0.9299 6250 45.5859 0.4499 247.525 0.8590 0.6416 0.5237 1.5788 0.8855 315.393 0.9268 6500 44.0348 0.4346 245.900 0.8534 0.6238 0.5093 1.5705 0.8808 314.356 0.9238 6750 42.5267 0.4197 244.275 0.8477 0.6065 0.4951 1.5619 0.8760 313.316 0.9207 7000 41.0607 0.4052 424.650 0.8421 0.5895 0.4812 1.5535 0.8713 312.272 0.9177 7250 39.6359 0.3912 241.025 0.8365 0.5729 0.4677 1.5450 0.8665 311.225 0.9146 7500 38.2154 0.3775 239.400 0.8308 0.5566 0.4544 1.5364 0.8617 310.174 0.9115 7750 36.9063 0.3642 237.775 0.8252 0.5407 0.4414 1.5277 0.8568 309.119 0.9084 8000 35.5998 0.3513 236.150 0.8195 0.5252 0.4287 1.5191 0.8520 308.061 0.9053 8250 34.3310 0.3388 234.525 0.8139 0.5100 0.4163 1.5104 0.8471 306.999 0.9022 8500 33.0990 0.3267 232.900 0.8083 0.4951 0.4042 1.5018 0.8423 305.934 0.8990 8750 31.9031 0.3149 231.275 0.8026 0.4806 0.3923 1.4931 0.8374 304.865 0.8959 9000 30.7424 0.3034 229.650 0.7970 0.4664 0.3807 1.4843 0.8325 303.792 0.8927 9250 29.6162 0.2923 228.025 0.7913 0.4525 0.3694 1.4756 0.8276 302.715 0.8896 9500 28.5236 0.2815 226.400 0.7857 0.4389 0.3583 1.4669 0.8227 301.634 0.8864 9750 27.4639 0.2710 224.775 0.7801 0.4257 0.3475 1.4580 0.8177 300.550 0.8832 10000 26.4362 0.2609 223.150 0.7744 0.4127 0.3369 1.4492 0.8128 299.462 0.8800 10250 25.4400 0.2511 221.525 0.7688 0.4001 0.3266 1.4403 0.8078 298.369 0.8768 10500 24.4743 0.2415 219.900 0.7631 0.3877 0.3165 1.4314 0.8028 297.273 0.9736 10750 23.5386 0.2323 218.275 0.7575 0.3757 0.3067 1.4225 0.7978 296.173 0.8703 11000 22.6320 0.2234 216.650 0.7519 0.3639 0.2971 1.4136 0.7928 295.068 0.8671 PROPERTIES OF DRY AIR AT LOW PRESSURE At 1 atmos (101.325 kPa) 𝜇 𝑘 T Cp Cv 𝛾 x 10-5 x 10-5 Pr 𝜈 (K) (kJ/kg.K) (kJ/kg.K) 𝜌 (kg.m.s) (kW.m.K) x 10-5 (kg/m3) (m2/s) 175 1.0023 0.7152 1.401 1.182 1.593 0.744 2.017 0.586 200 1.0025 0.7154 1.401 1.329 1.809 0.736 1.765 0.753 225 1.0027 0.7156 1.401 1.467 2.020 0.728 1.569 0.935 250 1.0031 0.7160 1.401 1.599 2.227 0.720 1.412 1.132 275 1.0038 0.7167 1.401 1.725 2.428 0.713 1.284 1.343 300 1.0049 0.7178 1.400 1.846 2.624 0.707 1.177 1.568 325 1.0063 0.7192 1.400 1.962 2.816 0.701 1.086 1.807 350 1.0082 0.7211 1.398 2.075 3.003 0.697 1.009 2.056 375 1.0106 0.7235 1.397 2.181 3.186 0.692 0.9413 2.317 400 1.0135 0.7264 1.395 2.286 3.365 0.688 0.8824 2.591 450 1.0206 0.7335 1.391 2.485 3.710 0.684 0.7844 3.168 500 1.0295 0.7424 1.387 2.670 4.041 0.680 0.7060 3.782 550 1.0398 0.7527 1.381 2.849 4.357 0.680 0.6418 4.439 600 1.0511 0.7640 1.376 3.017 4.661 0.680 0.5883 5.128 650 1.0629 0.7758 1.370 3.178 4.954 0.682 0.5430 5.853 700 1.0750 0.7879 1.364 3.332 5.236 0.684 0.5043 6.607 750 1.0870 0.7999 1.359 3.482 5.509 0.687 0.4706 7.399 800 1.0987 0.8116 1.354 3.624 5.774 0.690 0.4412 8.214 850 1.1101 0.8230 1.349 3.763 6.030 0.693 0.4153 9.061 900 1.1209 0.8338 1.344 3.897 6.276 0.696 0.3922 9.936 950 1.1313 0.8442 1.340 4.026 6.520 0.699 0.3716 10.83 1000 1.1411 0.8540 1.336 4.153 6.754 0.702 0.3530 11.76 1050 1.1502 0.8631 1.333 4.276 6.985 0.704 0.3362 12.72 1100 1.1589 0.8718 1.329 4.396 7.209 0.707 0.3209 13.70 1150 1.1670 0.8799 1.326 4.511 7.427 0.709 0.3069 14.70 1200 1.1746 0.8875 1.323 4.626 7.640 0.711 0.2941 15.73 1250 1.1817 0.8946 1.321 4.736 7.849 0.713 0.2824 16.77 1300 1.1884 0.9013 1.319 4.846 8.054 0.715 0.2715 17.85 1350 1.1946 0.9075 1.316 4.952 8.253 0.717 0.2615 18.94 1400 1.2005 0.9134 1.314 5.057 8.450 0.719 0.2521 20.06 1500 1.2112 0.9241 1.311 5.264 8.831 0.722 0.2353 22.36 1600 1.2207 0.9336 1.308 5.457 9.199 0.724 0.2206 24.74 1700 1.2293 0.9422 1.305 5.646 9.554 0.726 0.2076 27.20 1800 1.2370 0.9499 1.302 5.829 9.899 0.728 0.1961 29.72 1900 1.2440 0.9569 1.300 6.008 10.233 0.730 0.1858 32.34 2000 1.2505 0.9634 1.298 - - - 0.1765 - Page 2 of 5 USEFUL EQUATIONS FOR AIRCRAFT PERFORMANCE ANALYSIS Take-off distance Ground distance travelled whilst accelerating on ground between speeds 𝑉' to 𝑉( : : 𝑊 (𝑇 − µ𝑊 ) + 𝜌𝑆(µ𝐶1 − 𝐶3 )𝑉(; ; 𝑠= ln 7 : < 𝑔𝜌𝑆(µ𝐶1 − 𝐶3 ) (𝑇 − µ𝑊) + 𝜌𝑆(µ𝐶1 − 𝐶3 )𝑉'; ; Ground distance travelled whilst climbing between speeds 𝑉' to 𝑉( : 𝑊 𝑉(; − 𝑉@; 𝑠= >? B + (ℎ( − ℎ' )D 𝑇−𝐷 2𝑔 Range equations Breguet range for thrust-rated aircraft at constant speed, constant angle of attack: 𝑉 𝐶1 𝑊: 𝑅F = ln H I 𝑐F 𝐶3 𝑊; Breguet range for thrust-rated aircraft at constant altitude, constant angle of attack: 2 2 K𝐶1 𝑅F = J LK𝑊: − K𝑊; M 𝑐F 𝜌𝑆 𝐶3 Breguet range for power-rated aircraft: 𝜂P 𝐶1 𝑊: 𝑅N = ln H I 𝑐N 𝐶3 𝑊; Endurance equations Endurance for thrust-rated aircraft: 1 𝐶1 𝑊: 𝐸F = ln H I 𝑐F 𝐶3 𝑊; Endurance for power-rated aircraft: S/; 𝜂P 𝜌𝑆 𝐶1 1 1 𝐸N = J ? − B 𝑐F 2 𝐶3 K𝑊; K𝑊: Page 3 of 5 USEFUL EQUATIONS FOR AIRCRAFT PROPULSION ANALYSIS Gas Turbine Performance UVWX Z[\ F ηthermal = ]^'\ _` = 1 − F abc defd Work = m(h2 - h1) = mCp(T2 - T1) Axial Compressors Page 4 of 5 UNIT CONVERSIONS Distances 1 metre (m) = 3.28084 feet (ft) = 0.000539957 nautical miles (nmi) 1 foot (ft) = 0.3048 metre (m) = 0.000164579 nautical miles (nmi) 1 kilometre (km) = 1000 metres (m) = 0.539957 nautical mile (nmi) 1 nautical mile (nmi) = 1.852 kilometre (km) = 1852 metres (m) Speeds 1 metre per second (m/s) = 3.28084 feet per second (ft/s) = 1.94384 knots (knt) 1 foot per second (ft/s) = 0.3048 metre per second (m/s) = 0.592484 knots (knt) 1 knot (knt) = 0.5144 metre/second (m/s) = 1.68781 feet per second (ft/s) Weights 1 kilogram (kg) = 0.001 tonnes (tn) = 2.20462 pounds (lb) = 0.0685218 slugs 1 tonne (tn) = 1000 kilograms (kg) = 2204.62 pounds (lb) = 68.5218 slugs 1 pound (lb) = 0.453592 kilograms (kg) = 0.000453592 tonnes (tn) = 0.031081 slugs 1 slug = 14.5939 kilograms (kg) = 0.0145939 tonnes (tn) = 32.174 pounds (lb) Forces 1 newton (N) = 0.001 kilo-newtons (kN) = 0.224809 pound-force (lbf) 1 pound-force (lbf) = 4.44822 newtons (N) = 0.00444822 kilo-newtons (kN) Fuel consumption 1 lb/lbf-hr = 28.3255 × 10−6 kg/N-s 1 lb/hp-hr = 1.6897 × 10−7 kg/J Page 5 of 5  AIRCRAFT PERFORMANCE The Atmosphere International Standard Atmosphere Atmospheric Characteristic Affects Aircraft Performance 1 2 1 2 Lift and Drag 𝐿= 𝝆𝑉 𝑆𝐶𝐿 𝐷= 𝝆𝑉 𝑆𝐶𝐷 2 2 Density 𝝆 Combustion efficiency (fuel-air mixture in engine) Propeller efficiency Air density (𝝆), pressure (𝒑) and temperature (𝑻) are inter-correlated through: Ideal gas law 𝒑 = 𝝆𝑹𝑻 𝑹: Gas constant for air 𝑹 = 𝟐𝟖𝟕 𝒎𝟐 Τ(𝒔𝟐 𝑲) R changes slightly depending on the molecular composition of air at a particular location. 𝝆: Air density 𝝆 changes depending on altitude. Addition of water vapour to air (increased humidity) reduces air density The Atmosphere Space Thermosphere (500-600 km) Mesosphere (80-85 km) Stratosphere upto ~164,000 ft (50 km) Troposphere upto ~23000 ft (7 km) at poles upto ~56000 ft (17 km) at equator Troposphere Tropopause: top of troposphere Tropopause altitude ~7 km at poles ~17 km at equator Temperature T decreases linearly at ~6.51K/km Pressure P decreases exponentially 50% reduction per 5.6km It features different weather conditions and streams. Stratosphere Stratopause: top of stratosphere Stratopause altitude ~50 km Temperature T Up to 25 km: Constant T 25-50km: T increases, due to UV absorption in ozone. Pressure P P decreases with altitude Stratosphere is free of gusts and turbulence. But it is characterised by high steady wind (jet stream). Mesosphere Mesopause: ~80-85 km Temperature T T decreases to ~-85°C at mesopause. Most meteors burn up in mesosphere. Thermosphere Thermopause: ~500-600 km This region is bombarded by extremely short wave radiation from the sun and populated by ionised atoms which align with the lines of force of the earth’s magnetic field. This ionisation of oxygen and nitrogen releases photons which may be visible as aurorae (so- called northern lights). International Standard Atmosphere (ISA) ISA: A standard atmosphere to define the conditions prevailing for most of the year for aircraft operating up to an altitude of ~100,000 ft (~30 km) in temperate latitudes (e.g. Europe and North America). 𝐿𝑜𝑐𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 Relative Parameter 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 @ 𝑠𝑒𝑎 𝑙𝑒𝑎𝑣𝑒𝑙  AIRCRAFT PERFORMANCE The Atmosphere International Standard Atmosphere International Standard Atmosphere (ISA) ISA: A standard atmosphere to define the conditions prevailing for most of the year for aircraft operating up to an altitude of ~100,000 ft (~30 km) in temperate latitudes (e.g. Europe and North America). 𝐿𝑜𝑐𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 Relative Parameter 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 @ 𝑠𝑒𝑎 𝑙𝑒𝑎𝑣𝑒𝑙 Sound Speed 𝑎 = 𝑘𝑅𝑇 ISA Temperature model vs. local temperature The ISA model is standardised set of reference parameters at different altitudes, but local weather and climate will influence the actual temperature and pressure. Mean temperature profiles at various latitudes. (Diston, 2009, Computational modelling and simulation of aircraft and the environment) Altitude Measurement Altitude can be measured by measuring pressure and use the ISA table Example: The pressure is measured to be 90kPa on the top of a mountain, what’s the height of mountain? 𝑝 − 𝑝1 ℎ − ℎ1 𝑝 − 𝑝1 = ℎ = ℎ2 − ℎ1 + ℎ1 𝑝2 − 𝑝1 ℎ2 − ℎ1 𝑝2 − 𝑝1 90 − 92.6336 = 1000 − 750 + 750 89.8746 − 92.6336 = 238.8547 + 750 = 988.85 m  AIRCRAFT PERFORMANCE The Atmosphere Airspeed Ground Speed The relative velocity between the aircraft and a fixed point on the ground. Airspeed The relative velocity between the aircraft and the moving air surrounding the aircraft. 𝑽𝒂𝒊𝒓𝒔𝒑𝒆𝒆𝒅 = 𝑽𝒈𝒓𝒐𝒖𝒏𝒅 − 𝑽𝒘𝒊𝒏𝒅 Vwind Vwind Vground Vground Airspeed = Vground — Vwind Airspeed = Vground + Vwind Measuring Airspeed Pitot Tube: Total pressure 𝑝𝑡 Static Port: Static pressure 𝑝 1 2 Dynamic Pressure (q): 𝑞 = 𝑝𝑡 − 𝑝 = 𝜌𝑉 2 2(𝑝𝑡 − 𝑝) Airspeed: 𝑉= 𝜌 Airspeed Indicator 1 knot = 0.5144 m/s Indicated Airspeed: 2(𝑝𝑡 − 𝑝) 𝑉𝐼𝐴𝑆 = 𝜌0 𝝆𝟎 Air density at sea level  2(𝑝𝑡 − 𝑝) Indicated Airspeed: 𝑉𝐼𝐴𝑆 = 𝜌0 : Air density at sea level 𝜌0 2(𝑝𝑡 − 𝑝) True Airspeed: 𝑉𝑇𝐴𝑆 = 𝜌 : Air density at flight altitude 𝜌 Dynamic pressure at flight altitude: 1 2 1 2 𝑞 = 𝑝𝑡 − 𝑝 = 𝜌𝑉𝑇𝐴𝑆 = 𝜌0 𝑉𝐸𝐴𝑆 2 2 2(𝑝𝑡 − 𝑝) Equivalent Airspeed: 𝑉𝐸𝐴𝑆 = 𝜌0 : Air density at sea level 𝜌0 EAS is the airspeed at sea level conditions which gives the same dynamic pressure as that for given by the TAS. 1 2 1 2 𝜌0 1 𝜌𝑉𝑇𝐴𝑆 = 𝜌0 𝑉𝐸𝐴𝑆 𝑉𝑇𝐴𝑆 = 𝑉 = 𝑉 𝜌 2 2 𝜌 𝐸𝐴𝑆 𝜌Τ𝜌0 𝐸𝐴𝑆 relative density 𝜌0 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion Performance tutorial 1: Atmospheric modelling & effects 1. Identify which of the following (one or more) statements are correct. (a) Density: i. increases with increasing temperature. ii. increases with increasing altitude. iii. reduces with increasing altitude. iv. is unaffected by temperature change. (b) A piston engine aircraft is likely to fly in the: i. troposphere. ii. stratosphere. iii. mesosphere. iv. thermosphere. (c) The properties of the Earth’s atmosphere that influence the performance of an aircraft are: i. humidity. ii. nitrogen content. iii. pressure. iv. temperature. v. atmospheric radiation. vi. air density. vii. geomagnetic field. (d) Dynamic pressure is: i. the pressure change due to the mass of air pressing down on the air beneath. ii. the pressure change caused by heating when a moving airflow is brought completely to rest. iii. the total pressure at a point where a moving airflow is brought completely to rest. iv. the amount by which the pressure rises at a point where a moving airflow is brought completely to rest. Solution: (a) Density: iii. reduces with increasing altitude. (b) A piston engine aircraft is likely to fly in the: i. troposphere. University of Hertfordshire Performance tutorial 1: Atmospheric modelling & effects 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion (c) The properties of the Earth’s atmosphere that influence the performance of an aircraft are: i. humidity, iii. pressure, iv. temperature, vi. air density. (d) Dynamic pressure is: iv. the amount by which the pressure rises at a point where a moving airflow is brought completely to rest.  2. Explain the differences between indicated, calibrated, equivalent, and true airspeed. Solution: Indicated airspeed (IAS) is the airspeed read directly from the airspeed indicator (ASI). The ASI is driven by the pitot-static tube which is used to calculate the dynamic pressure of the airflow. This is then referenced to the air density at international standard atmosphere (ISA) sea level to give the IAS. Since the aerodynamics of an aircraft, for a given weight, will be (mostly) the same for a given dynamic pressure the aircraft will behave the same at the same IAS, regardless of the density changes that occur at different altitudes. For example, the important stall speed of the aircraft will be the same IAS for all altitudes. Similarly, all aircraft V-speeds are published as IAS. Calibrated airspeed (CAS) is IAS corrected for instrument errors and position error. This is sometimes caused by upwash or other airflow disruptions at the pressure ports which give incorrect pressure readings. At high speeds and altitudes (typically above 10,000 ft and faster than 200-250 kn, or 100-120 m/s), CAS must be corrected for compressibility error to give equivalent airspeed (EAS). Compressibility error arises because the impact pressure of the airflow will cause the air to compress in the stagnation port of the pitot tube and indicate a higher pressure. For low speeds and altitudes no compressibility correction is practically needed. Equivalent airspeed can be thought of as the airspeed at sea level conditions of the ISA which gives a dynamic pressure that is the same as the dynamic pressure at the true airspeed. Through Bernoulli’s equation this then allows true airspeed (TAS) to be calculated, as a function of EAS and the relative air density as defined by the ISA. The TAS is the actual airspeed of the aircraft relative to mass of air in which it is flying. It can also be thought of as EAS that has been corrected for pressure altitude and temperature (which both define the air density). As the air density decreases with altitude, the TAS will increase with altitude and increasingly differ from the EAS.  2 of 5 University of Hertfordshire Performance tutorial 1: Atmospheric modelling & effects 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion 3. The indicated airspeed (IAS) for an aircraft at an altitude of 20,000 ft is measured as 293 kn. Assuming the instrumentation is free of errors and neglecting any compressibility errors calculate the true airspeed (TAS). (206.5 m/s) Solution: An IAS of 293 kn is 150.732 m/s. Therefore, neglecting position and compressibility errors, VIAS = VCAS = VEAS = 150.732 m/s. The relationship between EAS and TAS is VEAS VEAS VTAS = √ = r. σ ρ ρ0 The density at sea level, ρ0 , is 1.225 kg/m3. In order to find the air density at the operating altitude of 20,000 ft (or 6096 m) linear interpolation of the ISA tabulated data is required. The air density at 6000 m, ρ1 , is 0.6597 kg/m3 , and at 6250 m is ρ2 = 0.6416 kg/m3. Thus ρ2 − ρ1 ρ − ρ1 =. h2 − h1 h − h1 ρ2 − ρ1 ∴ ρ= (h − h1 ) + ρ1 h2 − h1 0.6416 − 0.6597 = (6096 − 6000) + 0.6597 6250 − 6000 = 0.6527 kg/m3. Hence 150.732 150.732 VTAS = √ = √ = 206.463 m/s. 0.6527/1.225 0.533  4. The Reynolds number is an important dimensionless quantity that is used to help predict similar flow patterns in different fluid flow situations. For an aircraft in moving air the associated Reynolds number is typically defined as ρVc Re = , µ where ρ is the air density, V is the airspeed, c is the chord length, µ is the dynamic viscosity. A trainer aircraft with a rectangular wing of span 6 m is flying at a true airspeed of 25 m/s. 3 of 5 University of Hertfordshire Performance tutorial 1: Atmospheric modelling & effects 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion If the wing area is 7.4 m2 determine the Reynolds number of the trainer at: (a) sea level. (2.11 × 106 ) (b) 1,000 ft above sea level. (2.06 × 106 ) Solution: Given the wingspan, b, and the wing planform area, S , and that the wing is rectangular, then 7.4 bc = S ∴ c= = 1.23 m. 6 (a) At sea level ρ = ρ0 = 1.225 kg/m3 and µ = µ0 = 1.783 × 10−5. Thus 1.225 × 25 × 1.23 Re = −5 ≈ 2.113 × 106. 1.783 × 10 (b) At 1,000 ft (or 304.8 m) both ρ and µ can be found by linear interpolation from the tabulated data: At h1 = 250 m, ρ1 = 1.1959 kg/m3 , µ1 = 1.7752 × 10−5 kg/m-s. At h2 = 500 m, ρ2 = 1.1673 kg/m3 , µ2 = 1.7671 × 10−5 kg/m-s. ρ2 − ρ1 ∴ ρ= (h − h1 ) + ρ1 h2 − h1 1.1673 − 1.1959 = (304.8 − 250) + 1.1959 500 − 250 = 1.1896 kg/m3. µ2 − µ1 µ= (h − h1 ) + µ1 h2 − h1 (1.7671 − 1.7752) × 10−5 = (304.8 − 250) + 1.7752 × 10−5 500 − 250 = 1.7734 × 10−5 kg/m-s. Thus 1.1896 × 25 × 1.23 Re = ≈ 2.063 × 106. 1.7734 × 10−5  5. The Mach number for subsonic speeds can be related to the relative difference between total and ambient static pressure of the air by the relationship !  P 3.5 = 1 + 0.2M 2. Ps Using this expression, show that the equivalent airspeed can be calculated with the expression s  !1/3.5  q VEAS = a0 5  1 + − 1.   Ps 4 of 5 University of Hertfordshire Performance tutorial 1: Atmospheric modelling & effects 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion where a0 and P0 are the speed of sound and static pressure at sea level conditions, respectively. Solution: Given the expression !  P 3.5 = 1 + 0.2M 2 , Ps rearranging with respect to M yields !1/3.5   P 1 + 0.2M 2 = , Ps !1/3.5  P 1/3.5    !  1  P M = − 1 = 5  − 1 , 2    0.2 P s Ps s   P 1/3.5 !  M = 5  − 1.   Ps Now, the total pressure is the sum of the static and dynamic pressures: P = P s + q. Hence s   s   {P s + q} 1/3.5 ! !1/3.5  q M = 5  − 1 = 5  1 + − 1.    Ps Ps The equivalent airspeed, VEAS , determines the dynamic pressure based on sea level conditions. Since VTAS = aM then VEAS = a0 M, thus s  !1/3.5  q VEAS = a0 5  1 + − 1.   Ps  5 of 5  AIRCRAFT PERFORMANCE TAKE-OFF PERFORMANCE Ground Run & V-Speeds Ground Run The aircraft accelerates from stationary to achieve a speed so that the lift produced is larger than the weight (L>W). During the ground run, the thrust should also overcome the drag (T>D). Directional Nose Wheel @ low speed Control during Ground Run Rudder @ high speed Rotation Speed: VR Lift-Off Speed: VLOF Stall Speed: Vs Lift Drag Thrust Weight Rotation Speed: VR As the aircraft approaches the airspeed where the lift exceeds the weight, the elevators are used to rotate the aircraft nose-up. Lift-Off Speed: VLOF The aircraft will continue to accelerate after this rotation speed VR and a short while later ends its ground run and starts its flight at the lift-off speed. Stall Speed: Vs The airspeed at which the lift generated by a wing operating at CL,max equals the weight of the aircraft is, by definition, the stall speed for that particular configuration (e.g. flap and undercarriage settings): 2W 𝑉𝑠 = For safety: 𝑉𝐿𝑂𝐹 > 1.2𝑉𝑠 𝜌𝑆𝐶𝐿,𝑚𝑎𝑥 VR VLOF Decision Speed: V1 When an aircraft has OEI (One Engine Inoperative) in ground run, the pilot can safely bring the aircraft to stop with the remaining runway distance, the speed at this point is the decision speed. V>V1 Take-Off be Continued V VMCG surfaces after lift-off Take-Off Speed V2 at Completion of take-off once screen height is achieved. Screen Height: 35ft or 10.5m VR VLOF 35ft 10.5m Screen Height SUMMARY VMCG Minimum control speed on the ground VS Stall speed V1 Decision speed VR Rotation speed VLOF Lift-off speed VLOF > 1.2VS V2 Take-off speed 35ft 10.5m 0 Screen Height VMCG VS V1 VR VLOF V2  AIRCRAFT PERFORMANCE TAKE-OFF PERFORMANCE Drag：Induced Drag & Friction Drag  Lift Drag: D Thrust: T Friction Drag: F Weight 𝑑𝑉 In the direction parallel with the ground: 𝑚 =𝑇−𝑫−𝐹 𝑑𝑡 Drag (D) = Parasitic Drag (Dp) + Induced Drag (Di) Parasitic drag Dp Drag that results when an object is moved through a fluid. (aerodynamic friction) V increases Dp becomes larger Induced drag Di A result of the increasing amount of lift generated during the take-off ground run INDUCED DRAG (Drag due to lift generation) 𝐶𝐿2 1 Induced Drag Coefficient 𝐶𝐷𝑖 = = 𝐾𝐶𝐿2 𝐾= 𝜋𝑒𝐴𝑅 𝜋𝑒𝐴𝑅 𝑒 Oswald efficiency number, a correction factor that represents the change in drag with lift of a 3D wing. 𝑏2 𝐴𝑅 Wing aspect ratio 𝐴𝑅 = 𝑏 – wing span 𝑆 – wing area 𝑆 𝑏 For rectangular wing 𝐴𝑅 = 𝑐 – wing chord 𝑐 INDUCED DRAG (Drag due to lift generation) ~ Wing Span (b) Ground Effect Increase of lift and reduction of induced drag when an aeroplane is flown at slow speed very near the surface. Reason: The ground surface obstructs the creation of wingtip vortices and interrupts downwash behind the wing. ℎ increases to 1.0 → 𝐾𝐺𝐸 approaches to 1.0 𝑏 FRICTION DRAG (from rolling undercarriage) 𝐹 = 𝜇(𝑊 − 𝐿) 𝑑𝑉 𝜇: coefficient of rolling friction 𝑚 =𝑇−𝐷−𝑭 𝑑𝑡 lift L ↑ Friction Drag F ↓ Induced drag ↑ Coefficient of rolling friction can be affected by runway surface and wheel slip ratio. Runway surface effects on 𝝁 Wheel slip ratio  AIRCRAFT PERFORMANCE TAKE-OFF PERFORMANCE Take-Off Distance/Ground Run Distance  Take-Off Distance = Ground Run Distance + Airborne Distance to Screen Height Ground Run Distance Let’s consider a small time frame ∆𝑡, we can have: ∆𝑠 = 𝑉∆𝑡 ∆𝑡 ∆𝑉 ∆𝑉 𝑑𝑉 𝑑𝑉 ∆𝑠 = 𝑉∆𝑉 =𝑉 =𝑉 (1) ∆𝑉 ∆𝑡 ∆𝑠 𝑑𝑡 𝑑𝑠 In differential form: From the forces acting on aircraft: 𝑑𝑉 𝑚 = 𝑇 − 𝐷 − 𝜇(𝑊 − 𝐿) (2) 𝑑𝑡 Combine Eq.(1) and Eq.(2): 𝑑𝑉 𝑇 − 𝐷 − 𝜇(𝑊 − 𝐿) 𝑑𝑉 𝑇 − 𝐷 − 𝜇(𝑊 − 𝐿) 𝑉 = 𝑉 = 𝑑𝑠 𝑚 𝑑𝑠 𝑊/𝑔 ∆𝑠 Ground Run Distance 𝑑𝑉 𝑇 − 𝐷 − 𝜇(𝑊 − 𝐿) 𝑉 = 𝑑𝑠 𝑊/𝑔 1 1 Assume: 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐿 = 𝜌𝑉 2 𝑆𝐶𝐿 2 𝐷 = 𝜌𝑉 2 𝑆𝐶𝐷 2 1 2 1 2 1 𝑑𝑉 𝑇 − 2 𝜌𝑉 𝑆𝐶𝐷 − 𝜇(𝑊 − 2 𝜌𝑉 𝑆𝐶𝐿 ) 𝑇 − 𝜇𝑊 + 𝜌𝑉 2 𝑆 (𝜇𝐶𝐿 − 𝐶𝐷 ) = 2 𝑉 = 𝑑𝑠 𝑊/𝑔 𝑊/𝑔 𝑊 Differential form: 𝑑𝑠 = 1 𝑉𝑑𝑉 𝑔 𝑇 − 𝜇𝑊 + 𝜌𝑉 2 𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 The distance required from rest to speed V can be integrated: 𝑠 𝑉 𝑊 𝑠 𝑉 1 𝑠 = න 𝑑𝑠 = න 𝑉 𝑑𝑉 𝑠 = න 𝑑𝑠 = න 𝑉𝑑𝑉 1 𝑔 𝑔1 0 0 𝑔 𝑇 − 𝜇𝑊 + 𝜌𝑉 2 𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 0 0 𝑇 − 𝜇𝑊 + 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 )𝑉 2 2 𝑊 𝑊2 𝑔 𝑔1 𝑐1 = 𝑇 − 𝜇𝑊 𝑐2 = 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 𝑊 𝑊2 𝑽 𝟏 Finally, 𝒔=න 𝑽𝒅𝑽 𝟎 𝒄𝟏 + 𝒄𝟐 𝑽𝟐 Now, the task is to do the integration: 𝑽 𝟏 𝟏 𝑽 𝟏 𝟐 𝒔=න 𝟐 𝑽𝒅𝑽 𝒔= න 𝒅𝑽 𝒄 𝟎 𝟏 + 𝒄 𝟐 𝑽 𝟐 𝟎 𝒄𝟏 + 𝒄𝟐 𝑽𝟐 𝒃 𝟏 𝟏 𝒃 𝟏 𝒄𝟏 + 𝒄𝟐 𝒃𝟐 We know from Calculus: න 𝟐 𝒅𝒙𝟐 = 𝐥𝐧 𝒄𝟏 + 𝒄𝟐 𝒙𝟐 = 𝐥𝐧 𝒄 𝒂 𝟏 + 𝒄𝟐 𝒙 𝒄𝟐 𝒂 𝒄𝟐 𝒄𝟏 + 𝒄𝟐 𝒂𝟐 𝑔 1 𝑉 1 2 1 𝑐1 + 𝑐2 𝑉 2 1 𝑐2 2 𝑐1 = 𝑊 𝑇 − 𝜇𝑊 So 𝑠= න 𝑑𝑉 = ln = ln 1 + 𝑉 2 0 𝑐1 + 𝑐2 𝑉 2 2𝑐2 𝑐1 2𝑐2 𝑐1 𝑔1 𝑐2 = 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 𝑊2 Ground Run Distance: 𝑊 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑠= ln 1 + 𝑉 𝑔𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑇 − 𝜇𝑊 Under constant thrust 𝑊 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑠= ln 1 + 𝑉 𝑔𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑇 − 𝜇𝑊  𝑊 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 Ground Run Distance: 𝑠= ln 1 + 𝑉 𝑔𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑇 − 𝜇𝑊 Example: Consider a commercial, twin turbofan, transport aircraft with the following specifications. Estimate the distance required for the aircraft to reach the lift off speed, sLOF. Lift-off speed VLOF 145 kn (or 74.594 m/s) Max take-off gross weight mTOW 175.7 tonnes W Wing planform area S 286.5 m2 𝐶𝐿2 Wing lift coefficient (with take-off flaps) CL 1.0 𝐶𝐷,𝑖 = 𝐾𝐺𝐸 𝜋𝑒𝐴𝑅 Wing span b 47.57 m Oswald efficiency number e 0.7 Ground effect factor KGE 0.4 CD for fuselage, wing, tailplane, and nacelles CD,ac 0.01378 CD for undercarriage CD,u 0.02108 CD,p CD for flaps at take-off setting CD,f 0.00649 Coefficient of rolling friction μ 0.02 Static engine thrust (for one engine) T 55600 lb.f (or 247.33 kN)  𝑊 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 Ground Run Distance: 𝑠= ln 1 + 𝑉 𝑔𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑇 − 𝜇𝑊 Example: Consider a commercial, twin turbofan, transport aircraft with the following specifications. Estimate the distance required for the aircraft to reach the lift off speed, sLOF. Weight: 𝑊 = 𝑚𝑔 = 175.7 × 1000 × 9.8 = 1,721,860 𝑁 Parasitic Drag: 𝐶𝐷,𝑝 = 𝐶𝐷,𝑎𝑐 + 𝐶𝐷,𝑢 + 𝐶𝐷,𝑓 = 0.01378 + 0.02108 + 0.00649 = 0.04135 𝑏2 47.572 Aspect Ratio: 𝐴𝑅 = = = 7.90 𝑆 286.5 Induced Drag including 𝐶𝐿2 1.02 𝐶𝐷,𝑖 = 𝐾𝐺𝐸 = 0.4 × = 0.023 the group effect: 𝜋𝑒𝐴𝑅 𝜋 × 0.7 × 7.90 Total Drag Coefficient: 𝐶𝐷 = 𝐶𝐷,𝑝 + 𝐶𝐷,𝑖 = 0.04135 + 0.023 = 0.06435  𝑊 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 Ground Run Distance: 𝑠= ln 1 + 𝑉 𝑔𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑇 − 𝜇𝑊 Example: Consider a commercial, twin turbofan, transport aircraft with the following specifications. Estimate the distance required for the aircraft to reach the lift off speed, sLOF. 1721860 𝑠𝐿𝑂𝐹 = 9.8 × 1.225 × 286.5(0.02 × 1.0 − 0.06435) 1.225 × 286.5 × (0.02 × 1.0 − 0.06435) 2 × ln 1 + × 74.594 2 × 2 × 247.33 × 103 − 0.02 × 1721860 𝑠𝐿𝑂𝐹 = 1115 𝑚  𝑊 𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 Ground Run Distance: 𝑠= ln 1 + 𝑉 𝑔𝜌𝑆(𝜇𝐶𝐿 − 𝐶𝐷 ) 2 𝑇 − 𝜇𝑊 Thrust Effect on SLOF. The above example has the assumption: 𝑇 = 55600 𝑙𝑏 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) A more realistic model of the engine thrust has a quadratic variation of thrust with speed. Unit: V – m/s T-N Take-off thrust variation 60 55600 lb-f (247.33 kN) 50 Engine thrust (1000 lb-f) 46437 lb-f (206.77 kN) 40 30 145 kn 20 (74.59 m/s) 10 0 20 40 60 80 100 120 140 160 180 200 Speed (knots) Improved Solution by Considering Thrust Variation Calculate SLOF using Thrust determined at VLOF. 𝑉𝐿𝑂𝐹 = 74.594 𝑚/𝑠 𝑇 = 206.765 𝑘𝑁 Use the new thrust and apply into the equation of SLOF: 𝑠𝐿𝑂𝐹 = 1370 𝑚 Thrust T=247.33 kN 𝑠𝐿𝑂𝐹 = 1115 𝑚 1115 + 1370 𝑠𝐿𝑂𝐹 = = 1243 𝑚 𝑠𝐿𝑂𝐹 = 1370 𝑚 2 Thrust T=206.765 kN Thrust output from engines affects the take-off distance! The example here is a simplified case. In reality, more parameters should be considered: Runway Altitude Ambient Air Temperature Runway Slope Head/Tail Wind during Ground Run  AIRCRAFT PERFORMANCE TAKE-OFF PERFORMANCE Take-Off Distance/Airborne Distance  Take-Off Distance = Ground Run Distance + Airborne Distance to Screen Height Excessive Power L Vc Assumption: Small AOA V 𝑾 𝒅𝑽 T = 𝑻 − 𝑫 − 𝑾 sin 𝜽 𝜃 𝒈 𝒅𝒕 𝑾 𝒅𝜽 D 𝑽 = 𝑳 − 𝑾 cos 𝜽 𝑾 cos 𝜽 𝒈 𝒅𝒕 𝑾 sin 𝜽 W 𝒅𝒉 𝒅𝒉 Rate of Climb: 𝑽𝒄 = 𝑽𝒄 = 𝑽 sin 𝜽 = 𝒅𝒕 𝒅𝒕 𝒅𝒉 𝑻 − 𝑫 𝟏 𝒅𝑽 𝑽𝒄 = =𝑽 − 𝒅𝒕 𝑾 𝒈 𝒅𝒕 𝑾 𝒅𝑽 𝑻 − 𝑫 𝟏 𝒅𝑽 = 𝑻 − 𝑫 − 𝑾 sin 𝜽 sin 𝜽 = − 𝒈 𝒅𝒕 𝑾 𝒈 𝒅𝒕 𝒅𝒉 𝑾 𝒅𝑽 𝒅𝒉 𝟏 𝑾 𝒅𝑽𝟐 𝒅 𝟏 Excessive Power: 𝑽 𝑻−𝑫 = 𝑾 + 𝑽 =𝑾 + = 𝑾𝒉 + 𝑾𝑽𝟐 𝒅𝒕 𝒈 𝒅𝒕 𝒅𝒕 𝟐 𝒈 𝒅𝒕 𝒅𝒕 𝟐𝒈 𝑽 𝑻−𝑫 𝒅 𝟏 𝟐 Weight-Specific Excessive Power: 𝑷𝒔 = 𝑾 = 𝒅𝒕 𝒉+ 𝟐𝒈 𝑽 Excessive Energy and Airborne Distance sA Thrust being greater than drag is used to increase the kinetic energy and potential energy (in the form of greater airspeed and altitude, respectively). ∆𝐸 = 𝐸𝐾 + 𝐸𝑃 𝑠𝑐𝑟𝑒𝑒𝑛 − 𝐸𝐾 + 𝐸𝑃 𝐿𝑂𝐹 ∆𝐸 = 𝑇 − 𝐷 𝑠𝐴 1 2 *This is an estimation = 𝑚 𝑉22 − 𝑉𝐿𝑂𝐹 + 𝑚𝑔 ℎ2 − ℎ𝐿𝑂𝐹 2 1𝑊 2 2 = 𝑉2 − 𝑉𝐿𝑂𝐹 + 𝑊 ℎ2 − ℎ𝐿𝑂𝐹 2𝑔 1𝑊 2 2 Equating both sides: 𝑇 − 𝐷 𝑠𝐴 = 𝑉2 − 𝑉𝐿𝑂𝐹 + 𝑊 ℎ2 − ℎ𝐿𝑂𝐹 2𝑔 2 𝑊 𝑉22 − 𝑉𝐿𝑂𝐹 𝑠𝐴 = + ℎ2 − ℎ𝐿𝑂𝐹 𝑇−𝐷 2𝑔 𝐿𝑂𝐹 ℎ𝑠𝑐𝑟𝑒𝑒𝑛 𝑠𝐴 Example: The same commercial, twin turbofan transport aircraft with following parameters. Estimate the required distance sA from lift-off to screen height. 2 𝑊 𝑉22 − 𝑉𝐿𝑂𝐹 𝑠𝐴 = + ℎ2 − ℎ𝐿𝑂𝐹 𝑇−𝐷 2𝑔 Note: T and D at LOF and screen height has different values. We can use an averaged value to estimate sA. Example: The same commercial, twin turbofan transport aircraft with following parameters. Estimate the required distance sA from lift- off to screen height. 2 𝑊 𝑉22 − 𝑉𝐿𝑂𝐹 𝑠𝐴 = + ℎ2 − ℎ𝐿𝑂𝐹 𝑇−𝐷 2𝑔 ℎ2 = 35 𝑓𝑡 = 10.7 𝑚 1 1 Avg. Thrust 𝑇= 𝑇 − 𝑇 = × 2 × 206.75 + 203.42 = 𝟒𝟏𝟎. 𝟏𝟕 𝒌𝑵 2 𝐿𝑂𝐹 2 2 1 2 1 Drag at LOF 𝐷 = 𝜌𝑉𝐿𝑂𝐹 𝑆 𝐶𝐷,𝑝 + 𝐶 𝐷,𝑖 𝐿𝑂𝐹 = × 1.225 × 74.5942 × 286.5 × 0.04135 + 0.023027 2 2 = 62.911 𝑘𝑁 1 2 1 Drag at h2 𝐷= 𝜌𝑉 𝑆 𝐶𝐷,𝑝 + 𝐶𝐷,𝑖 = × 1.225 × 82.82 × 286.5 × 0.04135 + 0.17771 2 2 2 2 = 263.8 𝑘𝑁 1 Avg. Drag 𝐷= × 62.911 + 263.8 = 𝟏𝟔𝟑. 𝟑𝟔 𝒌𝑵 2 175.7 × 9.8 × 103 82.82 − 74.5942 Airborne Dis. 𝑠𝐴 = 410.17 × 103 − 163.36 × 103 2 × 9.8 + 10.7 − 0 = 𝟓𝟑𝟒 𝒎  AIRCRAFT PERFORMANCE TAKE-OFF PERFORMANCE Balanced Field Length/Stop Distance In the case of engine failure, there is a Decision Speed (V1) during ground run: If V > V1, Take-Off Continued If V < V1, Take-Off Aborted In the case of engine failure, there is a Balanced Field Length: Distance needed to Distance needed to complete take-off stop Typical Aircraft Field Lengths: Aircraft Field length (m) Airbus A300-600B 2190 Airbus A310-300 2408 Airbus A320-200 2336 Bae (Avro) RJ85 1385 Boeing 737-600 1877 Boeing 747-400 3322 Boeing 777 2469 Distance to Stop Lift L Thrust is zero: 𝑇 = 0 Drag D Friction: 𝐹 = 𝜇(𝑊 − 𝐿) Thrust T Friction F 𝐹 = 𝜇(𝑊 − 𝐿) Weight W From the force diagram: 𝑑𝑉 𝑑𝑉 𝑑𝑉 −𝐷 − 𝜇(𝑊 − 𝐿) 𝑚 = −𝐷 − 𝐹 𝑚 = −𝐷 − 𝜇(𝑊 − 𝐿) = 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑊/𝑔 𝑑𝑠 𝑑𝑠 𝑉= 𝑑𝑡 = 𝑑𝑡 𝑉 𝑑𝑉 −𝐷 − 𝜇(𝑊 − 𝐿) 1 𝑊 𝑉 = 𝑑𝑠 = 𝑉𝑑𝑉 𝑑𝑠 𝑊/𝑔 −𝐷 − 𝜇(𝑊 − 𝐿) 𝑔 𝑠0 0 1 𝑊 1 𝑊 𝑉12 𝑠𝑆𝑇𝑂𝑃 = න 𝑑𝑠 = න 𝑉𝑑𝑉 𝑠𝑆𝑇𝑂𝑃 = 𝑠1 𝑉1 −𝐷 − 𝜇(𝑊 − 𝐿) 𝑔 𝐷 + 𝜇(𝑊 − 𝐿) 𝑔 2 *Assume L & D does not vary with V  1 𝑊 𝑑𝑠 = 𝑉𝑑𝑉 What happens when L & D does vary with V? −𝐷 − 𝜇(𝑊 − 𝐿) 𝑔 1 1 2 We know: 𝐿 = 𝜌𝑉 2 𝑆𝐶𝐿 𝐷= 𝜌𝑉 𝑆𝐶𝐷 2 2 1 𝑊 𝑑𝑠 = 𝑉𝑑𝑉 1 2 𝑔 −𝜇W + 2 𝜌𝑉 𝑆 (𝜇𝐶𝐿 − 𝐶𝐷 ) The distance required for the aircraft to decelerate from V1 to stop: 𝑠0 0 1 𝑊 𝑠𝑆𝑇𝑂𝑃 = න 𝑑𝑠 = න 𝑉𝑑𝑉 𝑠1 1 𝑉1 −𝜇W + 𝜌𝑉 2 𝑆 (𝜇𝐶 − 𝐶 ) 𝑔 2 𝐿 𝐷 1𝑊 0 1 𝑠𝑆𝑇𝑂𝑃 = න 𝑑𝑉 2 1 2 𝑔 𝑉1 −𝜇W + 𝜌𝑆 (𝜇𝐶 − 𝐶 )𝑉 2 2 𝐿 𝐷 𝒃 𝟏 𝟏 𝒃 𝟏 𝒄𝟏 + 𝒄𝟐 𝒃𝟐 We know from Calculus: න 𝟐 𝒅𝒙𝟐 = 𝐥𝐧 𝒄𝟏 + 𝒄𝟐 𝒙𝟐 = 𝐥𝐧 𝒂 𝒄𝟏 + 𝒄𝟐 𝒙 𝒄𝟐 𝒂 𝒄𝟐 𝒄𝟏 + 𝒄𝟐 𝒂𝟐 1 𝑊 −𝜇W 𝑠𝑆𝑇𝑂𝑃 = ln 𝜌𝑆 (𝜇𝐶𝐿 − 𝐶𝐷 ) 𝑔 1 −𝜇W + 𝜌𝑆 (𝜇𝐶𝐿 − 𝐶𝐷 )𝑉12 2 Example of BFL for an aircraft The decision speed V1 = 130 kn BFL = 1500 m 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion Performance tutorial 2: Take-off performance 1. A twin turbofan-powered aircraft, with a take-off mass of 59 tonnes, is to take-off from a level tarmac runway at sea level. The following specification has been provided: Wing planform area S 115 m2 Wing span b 32 m Lift coefficient during ground run with take-off flaps selected CL 1.1 Lift coefficient at V2 with take-off flaps selected C L2 2.1 Drag coefficient for wings, fuselage, tailplane, and nacelles C Dac 0.021 Drag coefficient for flaps at take-off setting CD f 0.007 Drag coefficient for undercarriage C Du 0.015 Oswald efficiency factor for the aircraft e 0.75 Ground effect factor for ground run KGE 0.4 Ground effect factor at screen height KGE2 0.7 Runway friction factor µ 0.02 Aircraft lift-off speed VLOF 135 kn Aircraft take-off speed V2 148 kn Static thrust per engine Ts 82 kN Thrust per engine at lift-off speed T LOF 70 kN Thrust per engine at take-off speed T2 67 kN Note: 1 ft ≡ 0.3048 m, 1 kn ≡ 0.5144 m/s, 1 tonne ≡ 1000 kg. The ground distance before lift-off required for an aircraft to accelerate to a speed, V, is given by ρS (µC L − C D ) 2 " # W sG = ln 1 + V , (1) gρS (µC L − C D ) 2(T − µW) whilst the ground distance required for an aircraft to climb from height ha to height hb is " 2 Vb − Va2 ! # W sA = + (hb − ha ). (2) T −D 2g Calculate, approximately, the total take-off distance required for the aircraft to reach take-off speed, V2 , at a screen height of 35 ft. (Ground run ≈ 1080 m, Climb run ≈ 401 m, Total take-off run ≈ 1482 m) Solution: The total take-off distance is defined as the ground distance required for the aircraft to reach the take-off speed at the specified screen height. Thus, in (2) let hb = 35 ft (or 10.7 m), Vb = V2 = 148 kn (or 76.13 m/s) and ha = 0 m, Va = VLOF = 135 kn (or 69.44 m/s). In (1) let V = VLOF. The sum of (1) and (2) then gives the (approximate) total take-off distance required. For sG , the weight is W = mTOW g = (59 × 103 ) × 9.81 = 578790 N. University of Hertfordshire Performance tutorial 2: Take-off performance 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion The drag coefficient, C D , can be constructed by the sum of the total parasitic drag and induced drag from the aircraft:   C2 C D = C D p + C Di = C Dac + C Du + C D f + KGE L. πeAR The aspect ratio, AR = b2 /S = 322 /115 = 8.9. Hence the total drag coefficient is (at lift-off)   1.12 C D = 0.021 + 0.007 + 0.015 + 0.4 × π × 0.75 × 8.9 = 0.043 + 0.0231 = 0.0661. The total thrust provided by the twin turbofans at rest is T = 2T s = 164 kN. However at lift-off this reduces to T = 2T LOF = 140 kN. Therefore, we may obtain two values of S G for each thrust setting, and the average of these two will provide a more accurate value1. These data, along with the remaining specification, provides all the necessary data to solve (1): For T = 2T s = 164 kN: 578790 sG1 = 1.225 × 9.81 × 115(0.02 × 1.1 − 0.0661) " # 1.225 × 115(0.02 × 1.1 − 0.0661) × ln 1 + × 69.44 2 2(164000 − 0.02 × 578790) " # 578790 −6.2126 = ln 1 + × 69.44 2 −60.945 304848 = −9496.923 ln [1 − 0.0983] = 982.679 m. For T = 2T LOF = 140 kN: 578790 sG2 = 1.225 × 9.81 × 115(0.02 × 1.1 − 0.0661) " # 1.225 × 115(0.02 × 1.1 − 0.0661) × ln 1 + × 69.44 2 2(140000 − 0.02 × 578790) " # 578790 −6.2126 = ln 1 + × 69.44 2 −60.945 256848 = −9496.923 ln [1 − 0.1166] = 1177.402 m. 1 You could instead take the average of T s and T LOF and use this to calculate sG. Whilst this will give an answer somewhere between the two values of sG with either thrust values, it will not generate the same solution as calculating the average of sG with each thrust value. That approach gives a more conservative answer and is more accurate in this case. 2 of 5 University of Hertfordshire Performance tutorial 2: Take-off performance 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion Therefore sG1 + sG2 982.679 + 1177.402 sG = = = 1080.04 m. 2 2 For sA , the reciprocal of the weight-specific excess thrust, W/(T − D), between the two conditions (lift-off and screen height) can be obtained using average values of the thrust and drag forces: At lift-off: T = 2T LOF = 140000 N, C L2 1.12 C DLOF = C D p + KGE = 0.043 + 0.4 × πeAR π × 0.75 × 8.9 = 0.0661, 1 2 1 DLOF = ρVLOF S C DLOF = × 1.225 × 69.442 × 115 × 0.0661 2 2 = 22450.438 N. At screen height: T = 2T 2 = 134000 N, C L2 2 2.12 C D2 = C D p + KGE2 = 0.043 + 0.7 × πeAR π × 0.75 × 8.9 = 0.1902, 1 1 D2 = ρV22 S C D2 = × 1.225 × 76.132 × 115 × 0.1902 2 2 = 77647.255 N. Thus, the average values for T and D are: T av = 137000 N, Dav = 50048.847 N. Therefore " 2 2 ! # W V2 − VLOF sA = + (h2 − hLOF ) T av − Dav 2g 76.132 − 69.442 " ! # 578790 = + (10.7 − 0) 137000 − 50048.847 2 × 9.81 = 6.656 [49.636 + 10.7] = 401.596 m. Then the total take-off distance required is s = sG + sA = 1080.04 + 401.596 = 1481.636 m.  3 of 5 University of Hertfordshire Performance tutorial 2: Take-off performance 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion 2. The ground run needed to decelerate an aircraft from a speed V to zero can be found from the equation     W −µW sSTOP = .   ln  (3) gρS (µC L − C D )  1 −µW + ρS (µC L − C D )V  2 2 When the brakes are applied the rolling friction coefficient rises to 0.73. For the aircraft in Q1 if the decision speed, V1 , is 124 kn calculate the balanced field length needed for the aircraft. Assume the thrust from any operating engines is constant at the static thrust value. (sBFL ≈ 1215 m) Solution: The balanced field length comprises the ground roll distance to V1 AEO, followed by a braking roll with OEI. Thus we may use (1), with AEO (T = 2T s ) and nominal rolling friction (µ = 0.02), to calculate the distance required to the decision speed followed by the (3), with all engines off (T = 0) and braking (µ = 0.7), to calculate the stopping distance. The ground run to the decision speed is approximately 578790 s1 = 1.225 × 9.81 × 115(0.02 × 1.1 − 0.0661) " # 1.225 × 115(0.02 × 1.1 − 0.0661) × ln 1 + × 63.791 2 2(164000 − 0.02 × 578790) " # 578790 −6.2126 = ln 1 + × 63.791 2 −60.945 304848 = −9496.923 ln [1 − 0.0829] = 821.8520 m. Now the distance required to stop with µ = 0.73: 578790 sSTOP = 9.81 × 1.225 × 115(0.73 × 1.1 − 0.0661)      −0.73 × 578790  × ln   1 −0.73 × 578790 + × 1.225 × 115(0.73 × 1.1 − 0.0661) × 63.791    2 " 2 # 578790 −423674 = ln 1021.4668 −423674 + 52.0625 × 4069.299 = 566.96264 ln [2.0002] = 392.7554 m. 4 of 5 University of Hertfordshire Performance tutorial 2: Take-off performance 6ENT1131 & 6ENT1132 Aerospace Performance & Manufacture / Propulsion Thus the balanced field length is sBFL = s1 + sSTOP = 821.8520 + 392.7554 = 1214.61 m.  5 of 5  AIRCRAFT PERFORMANCE CLIMB PERFORMANCE Climb Angle  After Take-Off (achieving screen height), two performance parameters are important: Climb Angle Crucial to avoid high terrain or buildings close to the airfield. To minimise noise disturbance on the ground. Climb Rate To reach the optimum cruise altitude and speed in an efficient manner. Change in Height Climb Gradient (%) = Horizontal Distance Cruise Altitude Climb angle in percentage Climb Rate Climb Angle Climb Angle (𝜸) 𝑾 𝒅𝑽 = 𝑻 − 𝑫 − 𝑾 sin 𝜽 Along flight path 𝒈 𝒅𝒕 𝑾 𝒅𝜽 𝑽 = 𝑳 − 𝑾 cos 𝜽 Normal to flight path 𝒈 𝒅𝒕 𝒅𝑽 Assumption: 𝒅𝑽 Aircraft Acceleration along flight path Steady Climb =𝟎 𝒅𝒕 𝒅𝒕 𝑾 𝒅𝑽 𝑻−𝑫 = 𝑻 − 𝑫 − 𝑾 sin 𝜽 𝑾 sin 𝜽 = 𝑻 − 𝑫 𝜽 = 𝒂𝒔𝒊𝒏 𝒈 𝒅𝒕 𝑾 Assumption: Small Angle of Attack: 𝜸=𝜽 𝑻−𝑫 𝑻−𝑫 Climb Angle (𝜸) 𝜸 = 𝒂𝒔𝒊𝒏 Excess Thrust-To-Weight Ratio 𝑾 𝑾 𝑻−𝑫 =𝟏 𝜸 = 𝟗𝟎° The aircraft can climb vertically. 𝑾  𝑻−𝑫 Climb Angle (𝜸) 𝜸 = 𝒂𝒔𝒊𝒏 𝑾 Is there a maximum climb angle? 𝜸𝒎𝒂𝒙  𝑻−𝑫 𝑻−𝑫 Climb Angle 𝜸 = 𝒂𝒔𝒊𝒏 Excess thrust-to weight ratio 𝑾 𝑾 Minimum Drag Velocity - 𝑽𝒎𝒊𝒏 happens when Parasitic Drag = Induced Drag Speed for max angle of climb- 𝑽𝒙 maximises altitude gain per ground distance covered. Question: Compare 𝑉𝑥 and 𝑉𝑚𝑖𝑛 , if thrust does not vary with airspeed? 𝑉𝑚𝑖𝑛 = 𝑉𝑥 𝑉𝑥 𝑉𝑚𝑖𝑛 Example: An aircraft is in a steady climb: Aircraft mass m 175.7 tonnes Wing plan form area S 286.5 m2 Drag polar CD 0.015+0.0576CL2 Total climb thrust from all engines at 315kn is 260kN. Through the use of small throttle pushes, it may be assumed that this thrust remains constant for the range of speeds and altitudes. If the aircraft is flying with a TAS of 315kn at an altitude where ambient pressure of 89.9 kPa and temperature of 282 K. Calculate: (1) Climb angle (2) Climb gradient (3) the best-angle-of-climb speed – Vx (4) the minimum drag for the aircraft (5) the maximum climb angle for the aircraft Example: p = 89.9 kPa T = 282 K We need to determine the altitude and air density from ISA 𝜌2 − 𝜌1 𝜌 − 𝜌1 1.1392 − 1.1117 𝜌 − 1.1117 = = 𝜌 = 1.1176 𝑘𝑔/𝑚3 𝑇2 − 𝑇1 𝑇 − 𝑇1 283.275 − 281.650 282 − 281.650 𝑻−𝑫 𝟐𝟔𝟎 − 𝟏𝟎𝟑. 𝟖 (1) Climb angle 𝜸 = 𝒂𝒔𝒊𝒏 𝑾 = 𝒂𝒔𝒊𝒏 𝟏𝟕𝟐𝟏. 𝟖𝟔 = 𝒂𝒔𝒊𝒏 𝟎. 𝟎𝟗𝟏 = 𝟓. 𝟐° 𝑇 = 260 𝑘𝑁 𝑊 = 𝑚𝑔 = 175.7 × 103 × 9.8 = 1721.86 𝑘𝑁 𝐿 2W 2 × 1721.86 × 103 𝐶𝐿 = = = = 0.41 1 2 𝜌𝑉 2𝑆 1.1176 × 315 × 0.5144 2 × 286.5 𝜌𝑉 𝑆 2 Drag Polar 𝐶𝐷 = 0.015 + 0.0576𝐶𝐿2 = 0.015 + 0.0576 × 0.412 = 0.0247 1 𝐷 = 𝜌𝑉 2 𝑆𝐶𝐷 = 0.5 × 1.1176 × 315 × 0.5144 2 × 286.5 × 0.0247 = 103.8 𝑘𝑁 2 Example: (2) Climb Gradient [Climb angle in percentage] 𝑻−𝑫 𝟐𝟔𝟎 − 𝟏𝟎𝟑. 𝟖 𝜸 = 𝒂𝒔𝒊𝒏 = 𝒂𝒔𝒊𝒏 = 𝒂𝒔𝒊𝒏 𝟎. 𝟎𝟗𝟏 = 𝟓. 𝟐° 𝜸 % = 𝟗. 𝟏% 𝑾 𝟏𝟕𝟐𝟏. 𝟖𝟔 (3) Best angle of climb speed Parasitic Drag Coef. 𝐶𝐷,𝑝 = 0.015 Drag Polar 𝐶𝐷 = 0.015 + 0.0576𝐶𝐿2 Induced Drag Coef. 𝐶𝐷,𝑖 = 0.0576𝐶𝐿2 Best climb angle happens when total drag is minimum, namely Dp = Di 0.015 = 0.0576𝐶𝐿2 𝐶𝐿 =0.51 𝑉𝑥 = 145.29 𝑚/𝑠 1 2 𝜌𝑉 𝑆𝐶 = 𝐿 = 𝑊 2 𝑥 𝐿 2𝑊 2 × 1723617 𝑉𝑥 = = = 𝟏𝟒𝟓. 𝟐𝟗 𝐦/𝐬 𝜌𝑆𝐶𝐿 1.1176 × 286.5 × 0.51 Example: (4) The minimum drag during climb 𝐶𝐷,𝑚𝑖𝑛 = 2 × 0.015 = 0.03 1 𝐷𝑚𝑖𝑛 = 𝜌𝑉𝑥2 𝑆𝐶𝐷,𝑚𝑖𝑛 = 0.5 × 1.1176 × 145.292 × 286.5 × 0.03 = 101.385 𝑘𝑁 2 (5) The maximum climb angle 𝑇 − 𝐷𝑚𝑖𝑛 𝟐𝟔𝟎 − 𝟏𝟎𝟏. 𝟑𝟖𝟓 𝛾𝑚𝑎𝑥 = 𝑎𝑠𝑖𝑛 = 𝑎𝑠𝑖𝑛 = 5.3° 𝑊 𝟏𝟕𝟐𝟏. 𝟖𝟔  AIRCRAFT PERFORMANCE CLIMB PERFORMANCE Climb Rate After Take-Off (achieving screen height), two performance parameters are important: Climb Angle Crucial to avoid high terrain or buildings close to the airfield. To minimise noise disturbance on the ground. Climb Rate To reach the optimum cruise altitude and speed in an efficient manner. Vertical speed of an aircraft 𝑽𝒄 Cruise Altitude Climb Rate Climb Angle Climb Rate (Vertical Speed) 𝒅𝒉 𝑽𝒄 = = 𝑽 sin 𝜽 𝒅𝒕 𝑾 𝒅𝑽 Along flight path 𝒈 𝒅𝒕 = 𝑻 − 𝑫 − 𝑾 sin 𝜽 𝒅𝑽 𝑻−𝑫 Steady climb 𝒅𝒕 =𝟎 ⇒ 𝑻 − 𝑫 = 𝑾 sin 𝜽 ⇒ 𝒔𝒊𝒏 𝜽 = 𝑾 𝑻−𝑫 𝑽𝑻 Power available from engines 𝑽𝒄 = ?