Question

What is the maximum angle of climb (steady velocity climb) for a turbojet powered aircraft with an (L?D)max equal to 10.5 and a gross weight of 8000 lb. if it is producing a thrust of 2000 lb.? If the aircraft using the same available thrust, wishes to establish a rate of climb equal to 1800fpm, what airspeed will it need to establish? (Assume the required thrust equals 764pounds)

Answer 1

We consider that the velocity is steady (and there is no acceleration). We will use T for Thrust, W for weight( m*g); L for lift force, D for Drag and θ for angle of climb.

T = D + W sinθ (equation 1)

L = W cos θ (equation 2)

As, L/D = 10.5 or D = L/10.5

Equation 1 will become

T = L/10.5 + W sinθ

T = W cosθ /10.5 + W sinθ

cosθ + 10.5 sinθ = (T/W) *10.5 = (2000/8000)*10.5 = 2.625 (as T = 2000 lb & W = 8000 lb)

Solving this trigonometric equation, we will find θ = 8.97^o

There are two ways to solve above equation, I will explain the simpler one (approximation) valid for our case. In our case Angle of climb is very small and we can assume sinθ = θ and cos θ =1.

cosθ + 10.5 sinθ = 2.625

1 + 10.5*θ = 2.625;

θ = (2.625 -1)/10.5 = 0.156 rad = 8.94^o degree

As we require Rate of climb of 1800 fpm; which is equal to v sinθ ; where v is speed of plane.

v sinθ = 1800

v = 1800 / sinθ

If thrust of 2000 lb is used with maximum L/D and at maximum Angle of climb, then we will have sinθ = 8.97^o = 0.156 rad

v = 1800 / 0.156 = 11543 fpm = 192.4 fps