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Tutorial Questions with Solutions Flight Mechanics Tutorial Questions with Solutions Flight Mechanics Prof. Dr.-Ing. Dieter Scholz, MSME 2 Tutorial Questions with Solutions Prof. Dr.-Ing. Dieter Scholz, MSME university of applied sciences - fachhochschule hamburg Department of Vehicle and Aeronautical Engineering Berliner Tor 5 D - 20099 Hamburg Tel.: +49 - 40 - 709 716 46 Fax: +49 - 40 - 709 716 47 E-mail: [email protected] WWW: http://www.fh-hamburg.de/pers/Scholz/ Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 3 1 Introduction to Flight Mechanics and the ISA 1.1 An aircraft cruises at a calibrated airspeed of 320 kt in FL 200. The outside air temperature is -23 °C. a) Calculate the air pressure p in FL 200. b) Calculate the air density r in FL 200 under given conditions. c) Determine the equivalent airspeed EAS from a suitable diagram. d) Calculate the equivalent airspeed EAS. e) Calculate the true airspeed TAS. f) Calculate the Mach number. Solution Given: hp = 20000 ft CAS = VC = 320 kt t = -23°C T = 250.15 K 5.25588 5.25588 æ L ö æ 1.9812×10-3 K ö a) p = p0 ç1- ×hp ÷ = 101325Paç1- ×20000ft÷ = 46563Pa è T0 ø è ft ×288.15K ø Answer: p = 466 hPa . p 46563N × kg × K kg b) r = = = 0.6485 R×T m 2 ×287.053Nm ×250.15K m3 kg Answer: r = 0.649 . m3 c) DVC = 9.5 kt VE = VC - DVC = 310.5 kt Answer: EAS = VE = 311 kt . d) d = p / p0 = 46563/101325 = 0.45954 ì 1 ü é ì 2 3.5 ü ù 3.5 ï 1 ïé æV ö ù ï ï V = a 5díê íê1+ 0.2ç C ÷ ú -1ý +1ú -1ý E 0 ê ú d ê è a0 ø ú ïë îïë û þï û ï îï þï Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 4 Tutorial Questions with Solutions ì 1 ü é 2 3.5 ù 3.5 ï 1 ïìé æ 320 ö ù ïü ï V = 661.48 kt 5×0.45954×íê íê1+ 0.2ç ÷ ú -1ý +1ú -1ý = 310.5 kt E ê0.45954 è 661.48ø ú ïë îïë û þï û ï î þ Answer: EAS = VE = 311 kt . e) s = r / r0 = 0.6485 /1.225 = 0.5294 V = VE / s = 310.5 kt / 0.5294 = 426.7 kt Answer: TAS = V = 427 kt . Nm f) a = gRT = 1.4×287.053 ×250.15 K = 317.06 m / s = 1141.4 km / h = 616.32 kt kg × K V 426.7 M = = = 0.6923 a 616.32 Answer: M = 0.69 . 1.2 The map shows the elevation of a mountain: 10000 ft. An airplane passes an airport near the mountain in FL 100. A manual indicates the elevation of the airport: 4000 ft. The Automatic Terminal Information Service (ATIS) of the airport reports a temperature of 0°C and QNH 993 hPa. How many feet is the airplane above or below the mountain top? Solution Given: mountain: h = 10000 ft airplane: hp = 10000 ft airport: h = 4000 ft , t = 0 °C , T = 273.15 K mean sea level: h = 0 ft , p = 993 hPa · Airport data used to check if ISA conditions exist: rearth × h 7 H = = 3999 ft with rearth = 2.0902×10 ft rearth + h (In the future we could save time. In most cases H » h will be of enough accuracy.) Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 5 -3 TISA (h = 4000 ft) = T0 - L× H = 288.15 K -1.9812×10 K / ft ×3999 ft = 280.22 K DT = T - TISA = 273.15 K - 280.22 K = -7.08 K The temperature at the present day is 7 °C below ISA standard conditions. · Calculating the geometric height of the airplane above mean sea level (MSL): T0 + DT 288.15 K - 7.08 K H = hp × = 10000 ft × = 9754.5 ft T0 288.15 K rearth × H h = = 9759 ft geometric height above a level with p = 1013.25 hPa rearth - H hMSL = h - Dh hMSL 1013 hPa 993 hPa mean sea level, MSL h Dh 1013 hPa ISA today On this present day: T + DT æ 1 ö Dh » DH = H = 0 ç1- d 5.25588 ÷ L è ø æ 1 ö 288.15 K - 7.08K æ 993 ö 5.25588 = ft ×ç1- ç ÷ ÷ = 544 ft 1.9812×10-3 K ç è1013.25ø ÷ è ø hMSL = h - Dh = 9759 ft - 544 ft = 9215 ft Answer: The airplane is flying 10000 ft - 9215 ft = 785 ft below the mountain top. 1.3 On 01.02.99 the news group sci.aeronautics.airliners lists an article in which a rule of thumb is presented to calculate the true air speed: Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 6 Tutorial Questions with Solutions Example conditions: Altitude 35,000', IAS 280 knots TAT -15°C Altitude (1,000's) x 6 35 x 6 = 210 +IAS 210 + 280 = 490 + TAT 490 + (-15) = 475 KTAS a) Calculate the true air speed for the above conditions from physical principles. b) Compare your results with the true airspeed given by the rule of thumb. Calculate absolute and relative error. c) Several more checks of the same nature would be necessary for different altitudes, air speeds and total air temperatures (TAT) to show the overall accuracy of the rule of thumb. Use the computer with a suitable software tool for this task and present your results in tabular and graphical form. Solution Given: h = 35000 ft TT = 258.15 K VI = 280 kt » VC T 258.15K a) T = T = = 227.46K 1+ 0.2× M 2 1+ 0.2×0.821352 K T (35000ft) = T - LH = 288.15K -1.9812×10-3 ×35000ft = 218.81ft ISA 0 ft DT = T - TISA = 277.46K - 218.81K = 8.65K For non-ISA conditions and approximating h » H : 5.25588 5.25588 p æ L ö æ 1.9812×10-3 K / ft ö d = = ç1- × H÷ = ç1- ×35000ft÷ = 0.24696 p0 è T0 + DT ø è 288.15K + 8.65K ø ì 1 ü é ì 2 3.5 ü ù 3.5 ï 1 ïé æV ö ù ï ï M = 5íê íê1+ 0.2ç C ÷ ú -1ý +1ú -1ý êd ê è a0 ø ú ú ïë îïë û þï û ï îï þï Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 7 ì 1 ü é 2 3.5 ù 3.5 ï 1 ïìé æ 280 ö ù ïü ï M = 5× íê íê1+ 0.2ç ÷ ú -1ý +1ú -1ý = 0.80441 ê0.24696 è 661.48ø ú ïë îïë û þï û ï î þ Nm a = gRT = 1.4×287.053 ×227.46 K = 302.34 m / s = 587.70 kt kg ×K V = M ×a = 0.80441×587.70 kt = 472.7 kt Answer: TAS = V = 473 kt . b) Answer: The absolute error is 475 kt - 473 kt = + 2 kt. The relative error is +0.4 % This is an amazingly good result for a rule of thumb. c) The solution is left to the student. Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 8 Tutorial Questions with Solutions 2 Definitions and Aerodynamic Fundamentals 2.1 An aircraft is equipped with a wing of symmetrical airfoils. The lift curve slope of the total ¶ C 1 aircraft is estimated to be L = 0.8×2p . The stall angle of attack (AOA) is 12°. Wing ¶ a rad area is 16 m². Use g = g0 and r = r0 . What is the aircraft's mass during a flight on which a stall speed of 50 kt was observed. Solution ¶ C 1 Given: L = 5.0265 a = 12°= 0.20944 rad ¶ a rad max V = 50 kt = 92.6 km / h = 25.72 m / s 2 SW = 16 m ¶ C 1 C = L ×a = 5.0265 ×0.20944 rad = 1.05275 L,max ¶ a max rad 1 1 L = rV 2 ×C × S = r V 2 ×C ×S = W = m× g = m× g 2 L,max W 2 0 L,max W 0 2 2 2 2 r0 ×V 1.225 kg ×s ×25.72 ×m m = ×CL,max ×SW = 3 2 ×1.05275×16 m = 696 kg 2g0 m ×2×9.80665× m ×s Answer: The aircraft's mass is 696 kg. Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 9 5 Level, Climbing and Descending Flight 5.1 We consider two types of planes: a motor glider and a single engine piston prop general aviation aircraft. For each type, generic data is given as presented below. motor glider single engine aircraft PS / MTOW 7 W/N 12 W/N E 30 10 Consider a climb directly after take-off with MTOW at a speed of 100 km/h. Assume a propeller efficiency hP = 0.8. Calculate and compare the climb gradient, g and the rate of climb, ROC of both airplane categories! Assume small angle approximation. Solution Small angle approximation: L = W h × P P = h × P = T ×V Þ T = P S T P S V T D T D T 1 h × P 1 sin g = - = - = - = P - W W W L W E V ×W E ROC = V ×sin g motor glider single engine aircraft g 9.7° 14.2° ROC 4.7 m/s = 920 ft/min 6.8 m/s = 1340 ft/min Answer: The single engine aircraft shows a lift-to-drag ratio which is much less than that of the motor glider, however its power-to-weight ratio is higher.
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