7
$\begingroup$

Wikipedia gives the following equation for the efficiency $\eta_p$ of an engine here:

$$\eta_p= \frac {2\, (\frac {v} {v_e})} {1 + ( \frac {v} {v_e} )^2 }$$

where $v$ is the rocket speed and $v_e$ is the exhaust speed. It is backed by some reference to the book I have no access to.

I am still deeply puzzled, how it could be the rocket efficiency depends on the velocity at all? As much as I understand about rockets, I would expect the same amount of fuel to be converted into the same increase of velocity, regardless of the current speed. Unless we are approaching the speed of light that is this formula not likely about.

$\endgroup$
  • 1
    $\begingroup$ Rocket should accelerate a payload to given fixed velocity V. (For example V=7.8 km/s for launch to low Earth orbit). The variable is Ve. If you try to plot the formula as function of Ve with fixed V, you'll see that higher Ve yields higher efficiency. It's because a rocket with high Ve has lower fuel mass needed (google Rocket Equation to understand why). Lower fuel mass of rocket (with fixed payload mass) - smaller rocket - higher efficiency. $\endgroup$ – Heopps Apr 29 at 12:34
  • $\begingroup$ Sorry I misread the question. See Oberth effect ? $\endgroup$ – Antzi Apr 29 at 12:36
  • 1
    $\begingroup$ Similarly to jet engines, matching the exhaust speed and the vehicle speed gives optimum efficiency, in theory. However, in practice, this results in a very low specific impulse, causing much greater losses due to the need for exponentially larger masses of propellant. Unlike ducted engines, rockets give thrust even when the two speeds are equal. src. In short, this equation assumes fuel is energy source, and reaction mass is free, which works for aircraft (Air-fuel ratio ranges from 50:1 to 130:1) but is completely wrong for rockets. $\endgroup$ – SF. Apr 29 at 12:52
  • 1
    $\begingroup$ It's about how much energy put into reaction mass goes towards accelerating the aircraft, and how much goes towards uselessly stirring air behind the aircraft through propelling the exhaust backwards faster than necessary. Which is fine and dandy if you're picking the reaction mass (air) up in front of the aircraft and accelerating it backwards, and not carry all of it on board. $\endgroup$ – SF. Apr 29 at 12:58
3
$\begingroup$

This is one excellent question.

The answer is that fuel in a moving rocket have some kinetic energy. It is moving at great speed relatively to the Earth after all. Amount of this energy depend on how fast rocket is moving and, somewhat surprisingly, rocket engine can extract this energy and convert it into useful work. You might notice that a formula that you shown is achieving 100% efficiency when velocity of rocket is equal to velocity of rocket exhaust (v=ve). This is so because when v=ve the exhaust speed relatively to earth would be zero and all kinetic energy stored in fuel will be fully extracted by rocket engine and used to accelerate the rocket.

On the other hand, work made by rocket engine goes to increase kinetic energy of both the rocket and remaining fuel still moving with the rocket. So in a rocket you essentially push some energy in fuel first, then recover some of that energy back via rocket engine. You always put 100% of energy in fuel, but can recover only a fraction of that back and your formula is actually saying how much energy would be recovered.

There are some practical effects that come out of this observation. For example, consider a car with a 200 hp engine. Even if you drop all friction, air resistance, etc., the faster the car go, the slower would be its acceleration. This happens because energy of a car grows linearly with time and kinetic energy is proportional to speed squared, so you have to do more and more work to get one extra mile per hour. But thanks to that "extra-energy-in-fuel effect" this does not affect rockets. In fact rockets will accelerate faster and faster as the time goes, despite rocket engine nominally having the constant power much like an engine in a car. This happens because there's more and more kinetic energy available in rocket fuel and with decreasing amount of fuel in rocket, less and less energy is going there.

Normally you don't really need to think about all this, because there are equations that are much more practical for actual rocketry. However there's one nice trick called Oberth effect that really capitalizes on this idea and is actually used in space exploration. In Oberth maneuver one uses planet gravitation to accelerate the spacecraft and fuel inside it, getting some extra kinetic energy in that fuel. Once spacecraft "collected" as much kinetic energy as possible rocket engine can be fired to extract useful work from it. And it actually works.

| improve this answer | |
$\endgroup$
10
$\begingroup$

The energy efficiency isn't terribly useful when applied to rocket engines for space exploration. Using this definition of engine efficiency a cold gas thruster is more efficient than an ion engine$^{*}$ It's concerning the efficiency of with which the energy extracted from the fuel is converted to a force on the vehicle. It basically says if your exhaust is travelling "backwards" after it's accelerated your craft (your vehicle is travelling at a speed lower than the exhaust velocity), then too much of the energy was used to accelerate the exhaust, and it could have been used to accelerate the vehicle instead$^{**}$. (conversely, if you are already travelling faster than your exhaust velocity, the efficiency could be improved by accelerating your exhaust even more)

The factors which tend to matter for a launch vehicle or spacecraft are

  • ISP
  • thrust to weight ratio
  • mass fraction.

One way make a rocket engine "energy efficient" it would carry around a vast amount of inert propellant, to reduce the exhaust velocity at low speeds. This would give your vehicle a poor ISP and a poor mass fraction.

An "Air Augmented Rocket" improves it's engine efficiency and ISP, but as soon as the launch vehicle is out of the thicker atmosphere it has lost it's advantages.


$*$ Their fuel efficiency is a different matter entirely.

${**}$ because a rocket engine accelerates it's exhaust to accelerate the vehicle there's really not much one can do about this.

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ In short, rockets usually operate in situations where there's an abundance of energy, and the concern is about the efficiency of turning mass into rocket movement. Jet engines operate in situations where mass is abundant, so the concern is about efficiently turning energy into movement. $\endgroup$ – Mark Apr 29 at 20:30
  • 1
    $\begingroup$ An odd fact that follows from this: a rocket that varied its exhaust velocity to precisely match its instantaneous velocity would be 100% energy efficient...it would leave a trail of stationary exhaust and it and its payload would end up with all the kinetic energy. Or it would if it could start moving with an exhaust velocity of 0. Energy efficiency isn't a very useful figure with rockets, but it's noteworthy that staged launch vehicles often do approximate this in the end, with high-specific impulse upper stages on top of high-thrust lower stages. $\endgroup$ – Christopher James Huff Apr 29 at 23:28
3
$\begingroup$

I would expect the same amount of fuel to be converted into the same increase of velocity, regardless of the current speed.

Let's start with our friend $ E_k = \frac{1}{2}mv^2 $. Every increment of fuel burned gives you a certain amount of energy, but the faster you go, the less velocity increase you can get from that same energy. This applies to everyday objects, not just to rockets, so it shouldn't be a surprise that a velocity term could show up somewhere.

$ \frac{1}{2}mv^2 $ applies to the exhaust, as well, which gets us to what @JCRM explains about the efficiency. Efficiency in terms of accelerating the rocket means that the maximum amount of energy should be transferred to the rocket, and the minimum amount left in the exhaust. The exhaust's kinetic energy is minimized if it's left "at rest", i.e. when $ v_e = v $. If $ v_e $ is more or less, then the exhaust will have some leftover kinetic energy, which therefore, doesn't go into the rocket.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.