A “kulbit” (Russian for “summersault”) is an aerial maneuver were a friendly air

Question

A “kulbit” (Russian for “summersault”) is an aerial maneuver were a friendly aircraft performs a tight loop to get directly behind an enemy aircraft. Immediately before beginning the maneuver, the pilot reduces the aircraft’s thrust to zero. The aircraft as two engines capable of 75.22 kN of thrust each. If the 16,380 kg aircraft’s initial speed is 650 km/h at the beginning of the loop of radius 27.4 m, what is the aircraft’s speed at

(a) the top of the loop and

(b) at the bottom of the loop after the maneuver?

(c) How much specific energy is lost performing the maneuver?

(d) How much specific power is lost during the maneuver?

(there is no more information available)

Explanation / Answer

a) At top of loop

final kinetic energy = 0.5 * 16380 * 180.55 * 180.55 + 2 * 75220 * 54.8 - 16380 * 9.8 * 54.8

                                 = 266427494.3 J

=>   1/2 * 16380 * v2 = 266427494.3

=> v =   180.36 m/sec

            = 649.3068   km/hr      -----------> top of the loop

b)   At bottom of loop

=> final kinetic energy = 0.5 * 16380 * 180.36 * 180.36 + 2 * 75220 * 54.8 + 16380 * 9.8 * 54.8

                                       =   283459312.6 J

=>      1/2 * 16380 * v2 = 283459312.6

=> v =    186.038 m/sec

           =   669.7368 km/hr     -------------> bottom of loop

c)   specific energy is lost performing the maneuver = 2 * 75220 * 54.8 * 2

                                                                                    =   16488224 J

d)    specific power is lost during the maneuver = 2 * 75220 * ( 186.038 - 180.55 )

                                                                             =   825614.72 W

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