Why does power go up with the cube of the airspeed?

Question

If i have an airplane with over 100 kN of thrust, and i want to accelerate from a velocity at which drag force is 25kN to a velocity twice higher, is it possible that i wont be able to do this because my engine doesn't have enough power, even though it has enough thrust to counter the drag force near this higher speed, so there should be a net forward acceleration on my plane?

Answer 1

W = F*d

Where W is the energy needed to apply a force F over a given distance d.

Power is energy divided by time and also equivalent to Force applied for a given velocity.

F x v = F x d/t = W/t

For an airplane and straight and level flight at constant true airspeed, the force of drag is equal to the force of thrust.

Fd = Ft

Since the force of drag is proportional to the square of the velocity by

Fd = 0.5 * CdpA*v^2

Where p is equal to the density of the air, Cd is a coefficient of drag, A is equal to the flat plate area of the object being pushed through the air, and v is the true airspeed of the air, this gives the power consumed by drag applied at a given airspeed as

Pd = Pt = Fdv = 0.5CdpA*v^3

Pt is proportional to the cube of v, where Pt is engine power.

Answer 2

Firstly, your hypothetical situation ("is it possible...") would represent an over-constrained problem. It would never be possible that an a/c has enough thrust to overcome drag at speed Y, but doesn't have enough power to overcome drag at the same speed Y. That's a fundamental contradiction, indicating a misunderstanding about the basic relationship between force, work, and power, possibly better suited to exploring on an engineering or physics site.

(This is assuming that by "has enough thrust" and "has enough" power, we mean at whatever particular airspeed we are talking about-- not the max thrust or power that we'd get at the optimum airspeed for maximizing that particular parameter. It's not completely clear from the question which you meant-- if you meant the latter case, then we should note that it certainly does often happen that there are airspeeds we can't reach (in level flight) due to "lack of thrust" even if though we can produce that same amount of thrust at some lower airspeed, particularly with piston engines that tend to have a very roughly constant power output, and thus experience a dramatic loss of thrust as airspeed is increased.)

Second, you haven't given enough us information to know or even guess the drag force at your second velocity Y(=2X). (Obviously I'm calling your second velocity "Y" and your first velocity "X".)

Third, you haven't told us whether the 100 kN thrust force is constant independent of airspeed, or what.

The answer to the title itself, "Why does power go up with the cube of the airspeed?", is "It doesn't-- power required is not directly proportional to airspeed cubed across the entire flight envelope, because the drag coefficient is not constant". Of course this is assuming that by "power" in the title, you meant the power required, not the maximum power that the propulsion system could produce at that airspeed. The latter quantity, naturally, also is not directly proportional to airspeed cubed!

There are situations where an aircraft has enough thrust and power to overcome drag in horizontal flight at some given airspeed, but doesn't have enough thrust and power to overcome drag in horizontal flight at some lower airspeed, because that lower airspeed is on the "back side of the thrust-required curve" where drag is very high, so the thrust and power required (for horizontal flight) are also very high. In such a case the aircraft could not accelerate, without giving up altitude, from the lower airspeed to the higher airspeed. Nor could the aircraft maintain altitude at the lower airspeed. But that doesn't seem to be what you are asking about here, as evidenced by your comment "so there should be a net forward acceleration on my plane".

To better understand how the power required for horizontal flight varies with airspeed, in any given actual aircraft (assuming fixed constant weight), where the drag coefficient is not constant because angle-of-attack must vary as airspeed varies, see these sections from John Denker's excellent See How It Flies website-- the graphs included here are much better than any verbal description, and you'll see that the required power does not simply vary according to the cube of airspeed--

Drag and the power curve-- introduction

More about the power curve

Related ASE questions that deal with how the thrust and power available from the propulsion system vary with airspeed

Why is thrust said to be constant over speed for a jet engine?

Why is thrust inverse to speed in piston engines?

How (and why) does engine thrust change with airspeed?

How do power and thrust curves compare?

What´s the logic behind the Power Curve??

Answer 3

$$F=m.a$$

At constant mass, we only need to consider the difference in thrust and drag forces to determine if the aircraft can accelerate to a higher velocity.

if i have an airplane with over 100 kN of thrust, and i want to accelerate from a velocity at which drag force is 25kN to a velocity twice higher

For incompressible flow, at standard aviation Reynolds number, drag force D is $$ D = C_D. \frac{1}{2} \rho V^2. S$$

At constant altitude and attitude, $D = C. V^2$ with C being a constant, so at twice the speed the drag is 4 times as high. Thrust T must be equal to this higher D, power does not appear in this consideration.

During engine design, required engine power is determined by required aerodynamic power = $T. V$ divided by transmission and thrust mechanism efficiencies. It is the aerodynamc required power that goes up by the cube of velocity, engine power must be able to handle that.