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 cruises at a calibrated 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 EAS from a suitable diagram. d) Calculate the equivalent airspeed EAS. e) Calculate the TAS. f) Calculate the .

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

r × H h = earth = 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 curve slope of the total ¶ C 1 aircraft is estimated to be L = 0.8×2p . The angle of attack (AOA) is 12°. Wing ¶ a rad

area is 16 m². Use g = g0 and r = r0 .

What is the aircraft's 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. The combined effect reveals a steeper climb as well as a higher rate of climb for the single engine general aviation aircraft.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 10 Tutorial Questions with Solutions

5.2 Jane's all the Worlds Aircraft from 1993 lists data of the Fokker 50 Series 100: · wing span 29 m · wing area 70 m² · MTOW 19950 kg · max. operating altitude 25000 ft · power plant: two Pratt & Whitney Canada PW125B turboprops, each flat rated at 1864 kW (2500 shp) at S/L.

Abbreviations: · shp shaft horse power · S/L sea level.

For cruise conditions, it can be assumed:

· maximum lift-to-drag ratio, E = 16 max Fokker 50 · Oswald efficiency factor, e = 0.85

· propeller efficiency, hP = 0.8 .

a) Calculate the aspect ratio, A.

b) Calculate the drag coefficient at zero lift, CD0 .

c) Calculate the aircraft's equivalent power, Pe at seal level conditions. Assume the ratio of jet thrust to propeller thrust to be 0.15 . d) Calculate the equivalent power at maximum operating altitude. Assume the variation 0.5 of power with height is given by Pe µ s . s is the relative density. e) Derive an equation for level steady flight which can be used to calculate the maximum cruising speed. f) Simplify the equation from task e) and calculate an approximation of the maximum cruising speed at max. operating altitude and MTOW. g) Complete the table below and draw a graph of the data. Consider max. operating altitude and MTOW.

equivalent airspeed, VE trust, T drag, D 15 m/s 25 m/s 60 m/s 115 m/s

h) Calculate the maximum cruising speed at max. operating altitude and MTOW. i) Comment on the feasibility of flying on the "back side of the power curve" under given conditions.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 11

Solution b2 292 m 2 a) A = = =12.01 S 70 m 2

1 pAe pAe p ×12.01×0.85 b) Emax = Þ CD0 = 2 = 2 = 0.031319 2 CD0 4 Emax 4×16

c) Considering the aircraft's two engines ( nE = 2) at sea level conditions: æ T ö P = n × P ×ç1+ J ÷ = 2×1864 kW ×(1+ 0.15) = 4287.2 kW ( e ) S / L E S è TP ø d) with s(25000 ft) = 0.66942 from the ISA table:

P 25000 ft = P × s 25000 ft = 4287.2 kW ×0.66942 = 2869.9 kW e ( ) ( e ) S / L ( )

2 -2 e) D = A1 ×VE + B1 ×VE ( 1 )

C ×r ×S with A = D0 0 = 1.3428 kg/m 1 2 2× m2 × g 2 kg× m3 and B1 = = 2.786 4 p × A× e×r0 × S s

2 hP × Pe hP × Pe × s 6 kg× m 1 T = = = 1.5369×10 3 × ( 2 ) V VE s VE

hP × Pe × s 2 -2 T = D Þ = A1 ×VE + B1 ×VE VE

3 -1 hP × Pe × s = A1 ×VE + B1 ×VE ( 3 )

4 A1 ×VE - hP × Pe × s ×VE + B1 = 0 ( 4 ) f) Dropping the last term of Equation ( 3 ):

3 2 3 hP × Pe × s 2×hP × Pe × s 2×0.8×2869.9×10 kg × m ×m ×0.66942 m 3 3 3 VE = = = 3 2 = 104.6 A1 CD0 ×r0 ×S s ×0.031319×1.225 kg ×70 m s

V 203.3 kt V = 203.3 kt V = E = = 303.7 kt E s 0.66942

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 12 Tutorial Questions with Solutions g)

VE [m/s] D(VE) [N] T(VE) [N] V [kt] 10 2,7869E+05 2,2959E+05 29 15 1,2410E+05 1,5306E+05 44 20 7,0175E+04 1,1480E+05 58 25 4,5407E+04 9,1838E+04 73 30 3,2158E+04 7,6532E+04 87 35 2,4383E+04 6,5599E+04 102 40 1,9557E+04 5,7399E+04 116 45 1,6473E+04 5,1021E+04 131 50 1,4497E+04 4,5919E+04 145 55 1,3268E+04 4,1745E+04 160 60 1,2569E+04 3,8266E+04 174 65 1,2263E+04 3,5322E+04 189 70 1,2261E+04 3,2799E+04 203 75 1,2501E+04 3,0613E+04 218 80 1,2941E+04 2,8699E+04 232 85 1,3551E+04 2,7011E+04 247 90 1,4309E+04 2,5511E+04 261 95 1,5198E+04 2,4168E+04 276 100 1,6206E+04 2,2959E+04 290 105 1,7322E+04 2,1866E+04 305 110 1,8540E+04 2,0872E+04 319 115 1,9854E+04 1,9965E+04 334 120 2,1259E+04 1,9133E+04 348

3,0E+05

2,5E+05

2,0E+05

D(V_E) 1,5E+05 T(V_E)

T und D [N] 1,0E+05

5,0E+04

0,0E+00 0 20 40 60 80 100 120 140 V_E [m/s]

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 13

f V (( E ) n ) h) Newton iteration of Equation (4): (VE ) = (VE ) - n+1 n f ¢ V (( E ) n )

VE [m/s] f (VE) [kg*m³/s4] f ' (VE) [kg*m²/s³] V [kt] 100,000 -6,75397E+07 3,07205E+06 290,4 121,985 4,49372E+07 7,44798E+06 354,2 115,952 4,21963E+06 6,07249E+06 336,7 115,257 5,20640E+04 5,92294E+06 334,7 115,248 8,26451E+00 5,92106E+06 334,7 115,248 1,93715E-07 5,92106E+06 334,7

Answer: The maximum cruising speed at max. operating altitude and MTOW is 334 kt.

Comment: Jane's all the Worlds Aircraft states: "Typical cruising speed 282 kt".

m i) The graph above shows a second solution of Equation (4) at about V = 15 . E s Flight at this speed would require a lift coefficient of

2 × m × g 2 ×19500kg × 9.81m × m 3 × s2 CL = 2 = 2 2 2 2 = 19.8 r 0 ×VE × S s ×1.225kg ×15 × m × 70m

Such lift coefficient is apparently not possible.

Answer: Under given conditions, this power setting results in (only) one aircraft speed of 334 kt.

5.3 A jet performs a steady climb. Aircraft data is given as follows:

· drag coefficient at zero lift, CD0 = 0.02 , · wing area, S = 100 m2 , m · wing loading, = 700 kg/m² , S · aspect ratio, A = 10 , · Oswald efficiency factor, e = 0.85 , T · thrust to weight ratio, = 0.2 . m× g

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 14 Tutorial Questions with Solutions

a) Calculate mass and thrust of the aircraft. b) For standard sea level conditions, plot rate of climb, ROC and climb gradient g versus equivalent airspeed. Assume a parabolic drag polar. c) Determine graphically for standard sea level conditions: · maximum rate of climb and equivalent airspeed for the maximum rate of climb, · maximum climb angle and equivalent airspeed for the maximum climb angle. d) Determine numerically for standard sea level conditions: · maximum rate of climb and equivalent airspeed for the maximum rate of climb, · maximum climb angle and equivalent airspeed for the maximum climb angle.

Solution m kg a) m = ×S = 700 × 100 m 2 = 70000 kg S m 2 T N T = ×m×g = 0.2 × 70000 kg× 9.81 = 137340 N . m× g kg

2 -2 b) D = A1 × V E + B1 ×VE T D T A B sing= - = - 1 × V 2 - 1 ×V -2 W W W W E W E æ T A B ö g= arcsinç - 1 × V 2 - 1 ×V -2 ÷ èW W E W E ø T A B V = V × sing = × V - 1 × V 3 - 1 ×V -1 v E E W E W E W E V VE vE ROC= V × sin g = × sin g = for sea level conditions: ROC= Vv s s E C × r × S 0.0 2× 1.225 kg ×100 m 2 kg A = D0 0 = = 1.225 1 2 m 3 ×2 m 2 2 2 2 2 2 3 2 2× m× g 2× 70000 kg × 9.81 N× m 8 N m B1 = = 2 2 = 2.8831×10 2 p ×A×e × r0 × S kg ×p × 10× 0.8 5× 1.225 kg ×100 m s

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 15

VE [m/s] T [N] VvE [m/s] g [deg] 40 137340 -2,6 -3,7 60 137340 4,6 4,4 80 137340 9,8 7,1 100 137340 14,0 8,1 120 137340 17,4 8,3 140 137340 20,1 8,3 160 137340 22,1 7,9 180 137340 23,3 7,4 200 137340 23,6 6,8 220 137340 23,1 6,0 240 137340 21,6 5,2 260 137340 19,0 4,2 280 137340 15,3 3,1 300 137340 10,4 2,0

Climb - Jet: VvE und Climb Angle

25,0

20,0

15,0

vVE [m/s] 10,0 g [deg]

5,0 VvE [m/s] and Climb Angle [deg]

0,0

-5,0 0 50 100 150 200 250 300 V_E [m/s]

c) · maximum rate of climb: 23.6 m/s = 4650 ft/min , · equivalent airspeed for the maximum rate of climb: 200 m/s = 389 kt , · maximum climb angle: 8.4 ° , · equivalent airspeed for the maximum climb angle: 125 m/s = 243 kt . d) For a jet, the equivalent airspeed for the maximum climb angle is the minimum drag speed

VE,md :

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 16 Tutorial Questions with Solutions

1/ 4 8 2 1/ 4 é B1 ù é2.8831×10 N m m ù m VE,md = ê ú = ê 2 ú = 123.86 ë A1 û ë s ×1.225kg û s At this speed with equations form b): m V = 17.99 vE s g = 8.35°

For a jet, the equivalent airspeed for the maximum rate of climb is calculated from: 1/ 2 é 1 2 ù VE = ê (T ± T +12 A1 B1 )ú ë6A1 û 1/ 2 é 1m æ kg Nm 2 öù m = ê ç137340 N + 1373402 N 2 +12×1.225 ×2.8831×108 ÷ú = 198.40 6×1.225kg ç m s2 ÷ s ëê è øûú At this speed with equations form b): m V = 23.63 vE s g = 6.84°

5.4 A piston-prop performs a steady climb. Aircraft data is given as follows:

· maximum lift-to-drag ratio, Emax = 10 , · wing area, S = 16.3 m2 , · aircraft mass, m = 1200 kg , · aspect ratio, A = 7.4 , · Oswald efficiency factor, e = 0.74 ,

· propeller efficiency, hP = 0.8 ,

· shaft power, PS = 150 hp = 111855 W .

a) Calculate the zero lift drag coefficient. b) For standard sea level conditions, plot rate of climb, ROC and climb gradient g versus equivalent airspeed. Assume a parabolic drag polar and constant propeller efficiency. c) Determine graphically for standard sea level conditions: · maximum rate of climb and equivalent airspeed for the maximum rate of climb, · maximum climb angle and equivalent airspeed for the maximum climb angle. d) Determine numerically for standard sea level conditions: · maximum rate of climb and equivalent airspeed for the maximum rate of climb, · maximum climb angle and equivalent airspeed for the maximum climb angle.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 17

Solution 1 p × A×e p × A×e p ×7.4×0.74 a) Emax = Þ CD0 = 2 = 2 = 0.043 2 CD0 4× Emax 4×10

2 -2 b) D = A1 ×VE + B1 ×VE T D T A B sin g = - = - 1 ×V 2 - 1 ×V -2 W W W W E W E T A B V = V ×sin g = ×V - 1 ×V 3 - 1 ×V -1 vE E W E W E W E

hP ×PS × s T = T ×VE = hP × PS × s VE h × P × s A B V = P S - 1 ×V 3 - 1 ×V -1 vE W W E W E æ h × P × s A B ö g = arcsinç P S ×V -1 - 1 ×V 2 - 1 ×V -2 ÷ è W E W E W E ø V VE vE ROC = V ×sin g = ×sin g = for sea level conditions: ROC = Vv s s E C ×r × S 0.043×1.225 kg×16.3 m 2 kg A = D0 0 = = 0.429 1 2 m 3 ×2 m 2 2 2 2 2 3 2 2×W 2×1200 kg ×9.81 N ×m 5 N m B1 = = 2 2 = 8.0685×10 2 p × A× e×r0 × S kg ×p ×7.4×0.74×1.225 kg ×16.3 m s

VE [m/s] T [N] VvE [m/s] g [deg] 10 8948 0,71 4,1 11 8135 1,32 6,9 12 7457 1,83 8,8 13 6883 2,25 10,0 14 6392 2,61 10,7 15 5966 2,91 11,2 17 5264 3,39 11,5 20 4474 3,88 11,2 25 3579 4,29 9,9 30 2983 4,33 8,3 35 2557 4,08 6,7 40 2237 3,55 5,1 45 1989 2,75 3,5 50 1790 1,67 1,9

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 18 Tutorial Questions with Solutions

Climb - PistonProp: ROC und Climb Angle

12,00

10,00

8,00

vVE [m/s] 6,00 g [deg]

4,00 ROC [m/s] und Climb Angle [deg]

2,00

0,00 0 5 10 15 20 25 30 35 40 45 50 V_E [m/s]

c) · maximum rate of climb: 4.3 m/s = 850 ft/min , · equivalent airspeed for the maximum rate of climb: 30 m/s = 58 kt · maximum climb angle: 11.5 ° , · equivalent airspeed for the maximum climb angle: 17 m/s = 33 kt . At the very low speed associated with this flight condition (33 kt), the accuracy of both the drag and power idealisations, is reduced. The aircraft will be operating near to the stall, introducing discrepancies between the parabolic drag polar and the actual drag polar. Furthermore, the assumption that shaft power and propeller efficiency do not change with speed yields a very poor approximation at this low speed condition. d) For a piston-prop, the equivalent airspeed for the maximum rate of climb is the minimum

drag power speed VE,mp :

1/ 4 5 2 1/ 4 é B1 ù é8.0685×10 N m m ù m VE,mp = ê ú = ê 2 ú = 28.14 ë3A1 û ë s 3×0.429kg û s At this speed with equations form b): m V = 4.35 vE s g = 8.9°

For a piston-prop, the equivalent airspeed for the maximum climb angle is calculated from:

d sin g -hP PS s -2 2A1 2B1 -3 = VE - VE + VE = 0 dVE W W W

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 19

m This solved iteratively (e.g. with Newton Iteration) yields: V = 17.19 E s At this speed with equations form b): m V = 3.43 vE s g = 11.5°

5.5 A jet climbs with constant equivalent airspeed, VE = 150 m/s. Aircraft data is given as in Question 5.3:

· drag coefficient at zero lift, CD0 = 0.02 , · wing area, S = 100 m2 , · mass, m = 70000 kg , · aspect ratio, A = 10 , · Oswald efficiency factor, e = 0.85 ,

· thrust at sea level, T0 = 135900 N .

a) Calculate the variation of thrust with altitude in the troposphere. b) Calculate and plot the variation of true airspeed V with altitude in the troposphere. c) Calculate the variation of Mach number with altitude in the troposphere. d) Calculate and plot the variation of rate of climb, ROC and climb angle, g . e) Calculate the time to climb to an altitude of 8 km. f) Calculate the absolute ceiling.

Assume: · an ideal jet engine, · constant aircraft mass, · a parabolic drag polar, · a linear variation of the rate of climb within intervals Dh = 1000 m.

Solution a) In the troposphere, thrust is roughly proportional to density. Neglecting the difference between geopotential height and geometric height: 4.25588 æ 1 ö T = T0 ×s = T ×ç1- 0.022558 × h÷ è km ø Results are given in the Table below.

VE VE b) V = = 4.25588 / 2 s æ 1 ö ç1- 0.022558 × h÷ è km ø

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 20 Tutorial Questions with Solutions

Results are given in the Table below. V V c) M = = a 1 m K 20.0468 × × 288.15K -6.5 × h K s km Results are given in the Table below. d) Compare with solution to Question 5.3: æ T A B ö g = arcsinç - 1 ×V 2 - 1 ×V -2 ÷ èW W E W E ø T A B V = V ×sin g = ×V - 1 ×V 3 - 1 ×V -1 vE E W E W E W E V V ROC = V ×sin g = E ×sin g = vE s s C ×r × S 0.02×1.225 kg×100 m 2 kg A = D0 0 = = 1.225 1 2 m 3 ×2 m 2 2 2 2 2 2 3 2 2× m × g 2× 70000 kg ×9.81 N × m 8 N m B1 = = 2 2 = 2.8831×10 2 p × A×e×r0 × S kg × p ×10×0.85×1.225 kg×100 m s Results are given in the Table below.

V - V 1 æVv ö e) A = vi+1 vi Dt = × lnç i+1 ÷ i Dh i A ç V ÷ i è vi ø

j

The time to climb to an altitude h = j ×Dh is t = å Dti . i=1

h [km] T [N] V [m/s] M [-] VvE [m/s] Vv [m/s] g [deg] Ai [1/s] Dt [s] t [s] 0 135900 150,0 0,44 20,87 20,87 8,0 -0,001846 50 0 1 123324 157,5 0,47 18,12 19,02 6,9 -0,001842 55 50 2 111658 165,5 0,50 15,57 17,18 6,0 -0,001842 62 105 3 100856 174,1 0,53 13,21 15,34 5,1 -0,001846 69 167 4 90873 183,4 0,57 11,03 13,49 4,2 -0,001855 80 237 5 81663 193,5 0,60 9,02 11,63 3,4 -0,001868 94 316 6 73185 204,4 0,65 7,17 9,77 2,7 -0,001887 114 410 7 65398 216,2 0,69 5,47 7,88 2,1 -0,001912 145 524 8 58261 229,1 0,74 3,91 5,97 1,5 -0,001945 203 669 9 51735 243,1 0,80 2,48 4,02 0,9 -0,001986 343 872 10 45784 258,4 0,86 1,18 2,04 0,5 -0,002037 11 40372 275,2 0,93 0,00 0,00 0,0 0,000002

Answer: The jet needs 669 s = 11 min 9 s (under given assumptions) as time to climb to an altitude of 8 km.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 21

TAS during Climb with Constant EAS

12

10

8

6 h [km] h [km] 4

2

0 0,0 50,0 100,0 150,0 200,0 250,0 300,0 TAS [m/s]

Climb with constant EAS - Jet: ROC and Climb Angle

25,00

20,00

15,00 V_v [m/s] 10,00

[deg] gamma [deg] 5,00

0,00 ROC [m/s] and Climb Angle -5,00 0 2 4 6 8 10 12 h [km]

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 22 Tutorial Questions with Solutions

time t to climb to altitude h Dt to climb Dh = 1000 m 800 700 600 500 Delta t [s] 400 t [s] time [s] 300 200 100 0 0 2 4 6 8 10 altitude [km]

f) The table above yields: vertical speed Vv = 0 m/s at altitude 11 km and vertical speed. The absolute ceiling is 11 km.

5.6 A jet with an absolute ceiling of 11000 m climbs to an altitude of 8000 m. At sea level, the

rate of climb ROC0 is 20.87 m/s. Calculate the time to climb. Assume a linear variation of rate of climb with altitude during the whole manouver.

Solution h æ h ö 11000m s æ 8000 abs ç ÷ ö tCLB = - ×lnç1- ÷ = - ×lnç1- ÷ = 685s ROC0 è habs ø 20.87m è 11000 ø

Compared to Question 5.5 (e), this approximate solution produces a relative error of 2.3%.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 23 6 Stall, Speed Stability and Turning Performance

6.1 During an IFR flight, a passenger looks out the window while relaxing in his seat. He observes a turn and estimates the bank angle to be 30°. At the same time, the passenger observes the free surface of the orange juice in his glass: it is parallel to the tray. a) The passenger assumes the turn being flown as a rate one turn. Explain the term rate one turn. Why is it correct to assume a rate one turn. b) The passenger assumes the turn being flown as a coordinated turn. Explain the term coordinated turn. Why is it correct to assume a coordinated turn. c) Calculate the aircraft's true airspeed.

Solution a) A rate one turn is a turn with a heading change of 180° in 60 seconds. During IFR flights turns are performed as rate one turns. b) In a coordinated turn (correctly banked turn) · the lift force lies in the aircraft plane of symmetry, · the ball in the turn and slip indicator is centered, · there is no along the y-axis of the aircraft. This phenomenon is also shown by the free surface of the orange juice in the glass which is parallel to the tray.

V ×W c) tan F == g g 9.81m ×60s m V = × tan F = ×tan 30°= 108.2 = 210kt W s2 ×p s

Answer: The aircraft's true airspeed is 210 kt.

6.2 An aerobatic airplane flies a perfect circular looping. The pilot reads an altitude of 5000 ft at the highes point of the looping and an altitude of 4000 ft a the lowest point of the looping. The looping is flown at such a speed that the lowest load factor in the looping is n = 0. Assume that aircraft speed is nearly constant in the looping. a) During which part of the looping does the airplane experience the highest load factor? b) What is the aircraft's speed in the looping? c) Calculate the highest load factor n during the looping.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 24 Tutorial Questions with Solutions

Solution

Load factor: B L n = m× g

In point A: r = 500ft V 2 (•) L - m× g = m× a = m× r A V m m g L V B n +1 A × g m× g L m× g

In point B: V •) - L - m× g = - m×a = - m× r = 500ft = 152.4 m r V 2 L = m× - m× g r V 2 n = -1 r × g a) Answer: The highest load factor is experienced in point A.

V 2 V 2 b) Point B: 0 = -1 Þ =1 r × g r × g m m V = r × g = 152.4 m×9.81 = 38.67 = 75.2 kt s2 s Answer: The aircraft speed is 75 kt.

V 2 38.672 m 2 s2 c) Point A: n = +1 = +1 = 2 r × g s2 152.4m×9.81m Answer: Under given assumptions (which are not conservative!), the highest load factor during the looping is n = 2 .

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 25 7 Range and Endurance

7.1 A jet cruises with constant lift coefficient and constant Mach number of 0.82 in the stratosphere. The take-off mass with full fuel tanks is 271000 kg, the maximum fuel volume is 135000 liters, fuel density is 800 kg/m³. The specific fuel consumption is 16 mg/(Ns). The aircraft has a specified range of 7200 NM. a) Write down the appropriate equation for these flight conditions from which the range can be calculated. How is this equation named? b) Calculate the aircraft's mass without fuel. c) On its fuel reserves, the aircraft could fly additional 920 NM plus further 30 min. How far could the aircraft fly in these 30 min. if we assume that the 30 min. are also flown at cruise speed? How far could the aircraft fly on its fuel reserves? d) Calculate the aircraft's lift-to-drag ratio.

Solution a) From the information given we know: 1.) jet aircraft 2.)

CL = const. M = const. Þ V = const. (even considering a shallow climb; flight in stratosphere) Þ 2. flight schedule

V × E m R = × ln 1 this equation is known as Breguet Range Equation. c× g m2

b) m1 = 271000 kg 3 3 3 V f = 135000 l = 135 m m f = V f ×r f = 135 m ×800 kg / m = 108000 kg

m2 = m1 - m f = 271000 kg -108000 kg = 163000 kg c) In the stratosphere: a = 295.07 m / s (from ISA-Table) V = M ×a = 0.82×295.07 m / s = 241.96 m / s

s30min = 241.96 m / s×1800 s = 435.5 km = 235.2 NM

sreserve = 920 NM + 235.2 NM = 1155.2 NM = 2139.3 km

d) Rtheory = R = 7200 NM ×1.852 km / NM + 2139.3 km =15474 NM

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK 26 Tutorial Questions with Solutions

R× c× g 15474×106 m ×16×10-6 kg ×9.81 m ×s E = = = 19.7 æ m ö æ 271000 kgö V × lnç 1 ÷ N× s× s2 ×241.96 m × lnç ÷ è m2 ø è 163000 kg ø Answer: From given data with given assumptions, a lift-to-drag ratio of 19.7 for cruise flight was calculated.

7.2 Derive an equation for the range of a jet aircraft in a (shallow) steady climb with constant airspeed and constant lift coefficient.

Solution Q = c×T T = D + mg×sin g æ D ö Q = c ×ç × mg + mg ×sin g÷ è L ø L æ 1 ö Q = c× mg ×ç + sin g÷ è E ø g m2 V R = - dm ò Q m1 g V m2 1 R = - dm m× g L = m× g × cos g æ 1 ö ò m c× g ×ç + sin g÷ m1 è E ø æ m ö m× g ×sin g V × lnç 1 ÷ è m2 ø R = aircraft with lift and weight vector æ 1 ö c × g ×ç + sin g÷ è E ø

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK Tutorial Questions with Solutions 27 18 Miscellaneous Questions

18.1 The category of effect of a failure is judged to be hazardous. Following ACJ No. 1. to JAR 25.1309 a) What is the largest permissible failure probability? b) What is the mean time to failure MTTF ?

Solution

F(t) probability of failure, l failure rate MTTF mean time to failure FH flight hour a) hazardous : F(t = 1 FH) £10-7

1 b) For small probabilities of failure: l » F / t = 10-7 × FH

1 MTTF = 1/ l = FH = 10 000 000 FH 10-7

Answer: If a failure has a hazardous effect, the mean time to this failure may not be less than 10 000 000 FH.

Prof. Dr.-Ing. Dieter Scholz fachhochschule hamburg FACHBEREICH FAHRZEUGTECHNIK