Why does this formula say rocket efficiency depends on velocity?

Question

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.

Answer 1

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.

Answer 2

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.

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$*$ 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.

Answer 3

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.

Answer 4

Here is the math that leads to that efficiency formula:

Define efficiency $\eta$ to be the ratio of the power going to accelerate the payload, $P_{pld}$, to the power in the payload plus the power in the exhaust,$P_e$, as seen from the stationary observer.

$P_{pld} = Fv = \dot{m} v_e v$, where $v$ is velocity of payload, $\dot{m}$ is the rate of exhaust mass (assumed positive), and $v_e$ is exhaust speed.

$P_e = \frac{\dot{m}}{2}(v_e-v)^2$

So, $$ \eta = \frac{\dot{m}v_ev}{\dot{m} v_e v + \frac{\dot{m}}{2}(v_e-v)^2}=\frac{2\frac{v}{v_e}}{1+(\frac{v}{v_e})^2}$$.

Note that the change of momentum or thrust $\dot{m}v_e$ is independent of $v$ but the energy change per unit time of that exhaust, $(v_e - v)^2$, does depend on $v$.

Some problems to note are that efficiency is 0 at $v = 0$. But, I suppose that this formula is saying that one should gradually increase the exhaust velocity as the velocity grows. The change of energy per unit time in the load is actually $\frac{dE}{dt} = Fv - \frac{\dot{m}v^2}{2}$, but it was not counted because, well, it gets dropped. But, this formula is a quick and dirty way to see that efficiency is best somewhere around $v = v_e$.

Here is a way to find the constant exhaust velocity that minimizes the fuel required to attain $\Delta v = v$. Use the rocket equation to get the total energy expended by fuel to attain the velocity change.

$$E = \frac{m_i - m_f}{2}v_e^2 = \frac{m_f}{2}(e^{v/v_e} - 1)v_e^2$$

The minimum energy per final payload mass required is where the derivative goes to zero. $$ \frac{1}{m_f}\frac{dE}{dv_e} = \frac{1}{2}[(2v_e - v)e^{v/v_e} - 2v_e] = 0. $$ The minimum occurs at $v_e = 0.6275 v$, where $E = 1.54\frac{m_fv^2}{2}$.

Of course, the downside of using $v_e < v$ is that the mass of the fuel greatly exceeds the mass of the load. With chemical propellants, the energy density is so low that one has to accept whatever efficiency you get with the exhaust velocity. But, with the more energetic ion drives, RTG (radioisotope thermoelectric generators), or fusion engines a trade-off can be made between exhaust speed and propellant mass. The efficiency formula says that there is a benefit to adding inert mass to the exhaust to reduce its velocity and increase flow rate even though it increases the rocket's mass. The energy spent remains the same, but the momentum change is greater, i.e. higher efficiency.

Answer 5

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

Of course you are right and also the formula that you have cited appears to be correct. Unlike an automobile, a space rocket accelerates the same way for the same amount of fuel regardless how fast it is going. Also it may have a velocity in the sense of whatever arbitrary reference frame you would like.

In that case it will pick up next to zero kinetic energy per unit jet energy at times when it has a velocity that is next to zero. At those times almost all of the engine work will be spent on accelerating the propellant. When the rocket is moving right up close to the exhaust velocity then the engine work adds next to nothing to the kinetic energy of the jet.

The formula does not work in the case where the rocket is slowing down in the reference frame.