Question

Q7. Consider the twin-jet airplane described as follow: wing area = 47 m2, aspect ratio = 6.5, Oswald efficiency factor = 0.87, weight = 103,047N, and zero-lift drag coefficient = 0.032. The airplane is equipped with two jet engines with 40,298 N of static thrust each at sea level. The thrust-specific fuel consumption is 1.0 N of fuel per newton of thrust per hour, the fuel capacity is 1900 gal, and the maximum gross weight is 136,960 N. Calculate the range and endurance at a standard altitude of 8 km. Assume a paved runway. Also, during the ground roll, the angle of attack is restricted by the requirement that the tail not drag the ground. Hence, assume that Clmax during the ground roll is limited to 0.8. When the airplane is on the ground, the wings are 5 ft above the ground. Estimate the sea-level liftoff distance. Assume that the airplane is landing at full gross weight. The maximum lift coefficient with flaps fully employed at touchdown is 2.8. After touchdown, assume zero lift. Estimate the sea-level landing ground roll distance.

Answer 1

Gliven, Weight, w= 103047N Wing asea, se 47m² Aspect Ratio, AR= 6.5 er 0.87 Coo= 0.032 Throust specific fuel consumption (TSFC), Ç I Nof fixel = Iš 3600 N - hร Ç= 2.77X1045-7 D Gooss weight of aircoaft, W. = 136960 N. Weight of aiscraft without fuel, W,= Wo-Wf Wp = 1900gal = 1900X6-7 1b = 12730 lb = 12730x 4.45N Wp = 56,648.5 N Then, W = 136960 -56648.5 W, = 80311:5N at altitude of 8 km, So= 0.5258 kg/m3 Cele mai (COOTEAR)?? (0-032 XIT 20-87XG-5)'2 2Coo 2 X0.032 = 11.78 14 14 دے ( 2 Cootie AR) X0-032 XTT X0-8786-5) Co max 4.600 X0-032 = 15:46

For jet aircoaft, Range, R = 2 2 sos Chia [ wi CE Wo 42-61" ,, X 15:46 [136960)+(003155] J1 2 2 Х 0.5258X47 2:77x104 R = 2.752 X10 Endurance, E = 늘 뚱 In Wo w I 2:77x104 x 11.78 x In 136960 80311-5 = 22699.935 22699.93 ho 3600 E 6.305 ho 2 Lift off distance 0.8 W= 103047 N , 9-9.81 m/s2 , Cemax= = (-525 m for pared dunway, U=0.02. hese given that height, h=5 ft = 5x0.305m b=vs.AR Then, goound effect fackoo, 0 - (164) b = AR, 47X6.5 17.478m 116x1.525 17.478 2 1 + (15x1520) p= = 0.66

be Lift off velocity, Vo 1:2 Ystall = 1-2 2w SUSCLmax at sea level, So = 1-225 kg/m3 Vio 1.2 2X103047 80.26 6 m/s 1.225X47X0.8 Voo= 0.7VLO = 0.7 X 80.26 = 56.188 8 m/s Lift, L - L so veo SCL * x 1-225x(56-186)) x47x0-8 L = 72707.81 N Drag, D = 1 / 2 Lovers [coot Ø 6,² TEAR L { x 1.225 x (56.188){ x 47 0.032+ 0.66 x 0.32 TT X0.87X 6.5 5069.20 N For tunjet flight Thoust, T= 2x Tindividual = 2 X 40298 T = 80596N Then, Lift off distance, SLO 1:44 W2 9 Pos Cumax (T-0+ Mg (W-2) 1.44x (103047) 9.81x 1-225X47X0-8[80596-15069 270-02 [103047-72707-81] So = 451•69 m = Lift off distance

3 Goound soll distance W= 103047N Cmax" 2.8 assume U-20.4 Vy= 1.3 Vstau = 1.3 2W - 1.3 2x103047 SAS Climax 1.225X47X2:8 V = 46:48 m/s Vo = 0·7 V7 - 0:7X46:48 = 32:53 m/s Here after touchdown, L= 0 →4=0 Doag = 1 / 2 ove Scoo = { x 1:225 x(32:53)x 47X0032 974.81 N Groound soll distance, ş 1.69 w2 9 PAS CLman [D+uW] 169 X(1030472 9.81X1-225X47X 2.8 [9474-81 ton x 103047] 0.4 S = L 268.9 m Ground roll distance :