I need help with question 4

Supplementary information 1 ft/s = 0.3048 m/s 1 knot = 0.5144 m/s R= 287 J/kg/K for air 1 1bf = 4.448 N 1 HP = 745.6 W 1 HP = 550 ft 1bf/s 1 litre = 61.02 in3 1 US Gallon = 231 in3 Specific gravity of Avgas = 0.72 1 km = 0.6214 miles 1 nautical mile = 1.852 km 1 slug = 14.5943 kg 1 1bm = 0.4536 kg 1 kg/m3 = 0.0019403 slug/ft3 V L Winitial R = TSFCD" Wfinal Ps =V. ( I -D) W dV T - D -F dt m 0.5 Vmindrag (2 W 0.5 Vminpower K PS V 3 C DO 1 )max = 2 VKCD. WInternational standard atmosphere Altitude Temperature Kinematic Speed m ft 5: Pressure Density Viscosity 7iscasity of Ratio Ratio Ratio Ratio Sound 0 0 15.2 1 0030 1.0900 1.0000 1 0000 340 3 152 500 14.2 0 9521 0.9555 0.9973 1 0121 339 7 304 1000 15.2 0 9644 0.9711 0.9947 1 0245 339 1 457 1500 12.2 0 9470 0.9565 0.9920 1 9367 335 5 609 2000 11.2 0.9295 0.9425 0.9593 1.0493 335.0 762 2500 10.2 0 9129 0.9259 0.9566 1 0622 337 4 914 3000 9.3 0 3962 0.9151 0.9539 1 0752 336 B 1066 3500 5.3 0.8795 0.9515 0.9512 1.5554 336.2 1219 4000 7.3 0 8657 0.3951 0.9755 1 1015 335 6 1371 4500 6.3 0 8477 0.3745 0.9755 1 1155 335 0 1524 5000 5.3 0 8320 0.3617 0.9731 1 1293 334 4 1676 5500 4.3 0 8166 0.3457 0.9704 1 1434 333 5 1525 6000 3.3 0 8014 0.3559 0.9677 1 1577 333 2 1981 6500 2.3 0 7864 0.3232 0.9649 1 1722 332 6 2133 7000 1.3 0 7716 0.3106 0.9622 1 1570 332 0 2256 7500 0.3 0 7571 0.7955 0.9595 1 2020 331 4 2435 5000 70.6 0 7425 0.7560 0.9567 1 2172 330 B 2590 5500 -1.6 0 7257 0.7739 0.9540 1 2327 330 2 2743 9000 -2.6 0 7145 0.7620 0.9512 1 2454 329 6 2595 9500 73.6 0 7012 0.7501 0.9455 1 2644 329 0 3045 10000 4.6 0 6377 0.7355 0.9457 1 2507 325 4 3200 10500 75.6 0 6745 0.7269 0.9450 1 2972 327 5 3352 11000 6.6 0 6614 0.7155 0.9402 1 3140 327 2 3505 11500 77.5 0.6456 0.7545 0.9374 1.5310 325.6 3657 12000 5.6 0 6360 0.6932 0.9347 1 3454 326 0 3510 12500 -9.6 0.6236 0.6522 0.9319 1.3660 325.4 3962 13000 15.6 0.6113 0.6715 0.9291 1.5540 324.7 4114 13500 -11.5 0 5993 0.6606 0.9265 1 4022 324 1 4267 14000 12.5 0 5875 0.6500 0.9235 1 4207 323 5 4419 14500 13.5 0 5755 0.6396 0.9207 1 4396 322 9 4572 15000 -14.5 0 5643 0.6292 0.9179 1 4555 322 3 6245 20500 -25.4 0.4530 0.5235 0.5567 1.6927 315.4 6400 21000 -26.4 0.4406 0.5150 0.5535 1.7163 314.5 6553 21500 -27.4 0 4314 0.5362 0.5509 1 7403 314 1 6705 22000 -25.4 0 4223 0.4976 0.5751 1 7647 313 5 6555 22500 -29.4 0 4154 0.4591 0.5752 1 7595 312 9 7010 23000 -30.4 0.4046 0.4506 0.5723 1.5145 312.2 7162 23500 -31.4 0 3960 0.4723 0.5694 1 5406 311 6 7315 24000 -52.3 0 3576 0.4642 0.5665 1 5665 311 0 7467 24500 -53.3 0 3795 0.4561 0.5636 1 5935 319 3 7620 25000 -54.3 0 3711 0.4451 0.5607 1 9207 309 7 7772 25500 -35.3 0 3631 0.4402 0.5575 1 9454 309 0 7924 26000 -56.3 0 3552 0.4325 0.5545 1 9766 305 4 5077 26500 -37.3 0 3474 0.4245 0.5519 2 0053 307 7 5229 27000 755.3 0 3395 0.4173 0.5490 2 3345 307 1 5352 27500 -39.3 0 3324 0.4995 0.5460 2 9643 306 4 3534 25000 -43.3 0.3250 0.4025 0.5431 2.0947 305.5 5656 25500 741.3 0.3175 0.3955 0.5402 2.1256 305.1 5539 29000 -42.3 0 3107 0.3551 0 5372 2 1571 304 5 5991 29500 -43.2 0.3035 0.3511 0.5342 2.1592 303.5 9144 30000 -44.2 0 2970 0.3741 0 5313 2 2219 303 2 . Fuel Capacity: 1160 1b Stall Speed (clean wing): 68 kts IAS CD. (clean) = 0.026 Calculate the maximum rate of turn and the minimum turn radius that this aircraft can achieve at sea level. Assume a level turn and a maximum structural load factor of n = 3.5. (Ans R = 131 m, w = 28.8 /sec) Question 4 A vintage, single engine, jet aircraft has the following properties: Mass = 5500 Kg CD = 0.02 + 0.06 . CL2 CLmax = 1.2 S = 30 m2 Static Thrust = 15500 N TSFC = 0.45 N/hr/N If the aircraft has a fuel capacity of 2000 Kg, determine: 1) the still-air maximum range at an altitude of 5000 m. (Ans R = 5510 km) 2) the reduction in range if the aircraft encounters a steady headwind of 20 m/s. (Ans 920 km) Question 5 The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. 1) If the aircraft is in flight at 5000 ft when the engine fails, how far can it glide in terms of distance measured along the ground? (Ans 729 miles) 2) Calculate the equilibrium glide velocity at 3,000 ft, corresponding to the minimum glide angle. The aspect ratio of the aircraft is 4.11, the Oswald efficiency factor is 0.7, the weight is 1,400 1b, and the wing area is 231 ft2. (Ans 97.2 ft/s) N