Exhaust Velocity and Mass of Exhaust Products

In my last post I went over specific impulse, and how it is essentially the same as exhaust velocity. However, I later started wondering why exhaust velocity was so important. Following that, I ran into the fact that if exhaust velocity was important, so was the mass of the exhaust products, as a lighter exhaust product would equal a higher exhaust velocity for the same energy. I was confused by this, as I mistakenly thought that the momentum would remain the same.

So, why does the mass of exhaust products matter? To understand that, we need to take a look at how rocket engines work. Rocket engines are a type of reaction engine, and reaction engines are engines that work according to Newton's 3rd law of motion, that for every action there is an equal and opposite reaction. Many types of propulsion other than rocket engines are reaction engines, for example: Propellers (on both boats and planes) or jet engines. However, internal combustion engines are not reaction engines.

Reaction engines work according to conservation of momentum. That means that the momentum imparted to the propellant is equal and opposite to the momentum imparted to the ship. We can simplify the system to number of discrete particles being pushed backwards, ignoring the source of energy, and any chemical reactions. The momentum imparted to each particle is $p = mv$ where $p$ = momentum, $m$ = mass, and $v$ = velocity.

Here, we run into a simple reason why lighter propellants are better: propellant mass = exhaust mass, and our ship gets more velocity per unit of momentum if its mass is lower, and since all exhausted mass must be carried as propellant mass, it is better to use a lighter propellant with a higher velocity than vice versa.

But what if we didn't have to carry our propellant with us? In a purely hypothetical situation where we have infinite, massless propellant available, would it matter whether we used a lighter particle as reaction mass? The answer is yes, because of the relationship between kinetic energy and momentum.

Momentum is $p = mv$, and the mass is what we're varying. The question is what the velocity term is. Kinetic energy is related to velocity by the equation $E = \frac{1}{2}mv^2$. Note that this is not how it works, rocket engines do not provide a perfect equal amount of kinetic energy to their exhaust mass, even in nuclear thermal engines. What this most closely models is a mass driver, but it is still quite a rough approximation. If we rearrange the equation to solve for velocity, we get: $$v = \sqrt{\frac{E}{\frac{1}{2}m}}$$
Then put that into the momentum equation: $$p = m\sqrt{\frac{E}{\frac{1}{2}m}}$$
Which we can simplify to $$p = \sqrt{2Em}$$
Note that momentum has diminishing returns as either energy or mass increase, nonetheless, clearly two particles with a mass of $0.5n$ have a higher momentum than one of $n$ mass for the same energy.
$$2 \dot{} \sqrt{2\dot{}10\dot{}0.5} \approx 6.3$$
$$\sqrt{2\dot{}10\dot{}1} \approx 4.5$$
This is what a graph of that looks like:
Backup image in case the embedded graph doesn't work:

Specific Impulse

I've made a previous post about Specific Impulse, but I've wanted to improve it for a while, since it's not very clear. This post is using a new system called MathJax to display equations, which should make them easier to read.

Specific impulse is essentially a measure of how efficiently a rocket engine converts the chemical potential energy of its fuel into the kinetic energy driving the rocket forward. It is defined as the thrust of a rocket engine $T$ divided by the weight flow rate $\dot{W}$: $$I_{sp}=\frac{T}{\dot{W}}$$ First, we need to distinguish between weight and mass. Mass is how much matter an object consists of, while weight is the force exerted on an object by the local gravitational field. The local gravity on Earth exerts 9.81 m/s^2 of acceleration on everything, so if force = mass times acceleration, the force, or weight, exerted by an object with 1 kg of mass is 1 kg * 9.81 m/s^2. Since one Newton is the amount of force needed to accelerate one kg at 1 m/s^2, the object has a weight of 9.81 N. Weight is normally measured in kilograms (or pounds, but they're the same dimensionally), which seems odd, since we know weight can change independently of mass on different planets with different gravities. This is because the unit commonly referred to by a kilogram is actually kilogram-force, or a kilopond, which is exactly the force of one kilogram of mass in standard Earth gravity, or about 9.81 N.  Therefore, in common use, both the definition and unit of mass and weight are the same.

Specific impulse $I_{sp}$ is defined as the thrust $T$ produced by a rocket engine, divided by the weight flow rate of propellant $\dot{W}$, or: $${I_{sp}}=\frac{T}{\dot{W}}$$ It seems like it should be thrust per unit weight of propellant consumed, but thrust is instantaneous, so it wouldn't make sense without converting to some weird unit like newton-seconds.

Since $\dot{W}$ is weight, and weight is mass multiplied by standard gravity, the weight flow rate is the mass flow rate $\dot{m}$ multiplied by gravity: $$\dot{W}=g\dot{m}$$ The mass flow rate is the first derivative of the propellant mass $m_p$ with respect to time: $$\dot{m}=\frac{dm_p}{dt}$$ Therefore: $$\dot{W}=g\frac{dm_p}{dt}$$
Thrust is simply the exhaust velocity of the propellant from the engine $v_e$ multiplied by the mass flow rate $\dot{m}$ which is exactly the same as the mass flow rate already defined: $$T=v_e\frac{dm_p}{dt}$$
If we substitute these back into the original $I_{sp}$ formula, we get: $$I_{sp}=\frac{v_e\frac{dm_p}{dt}}{g\frac{dm_p}{dt}}$$ Since the mass flow rate is on both the top and bottom of the equation, we can get rid of them, and get: $$I_{sp}=\frac{v_e}{g}$$ The important thing to remember here is that $g$ is a constant in this equation. It is quite literally the acceleration due to gravity at the earth's surface, regardless of where the rocket engine is, or what kind of gravity it is in. In fact, any constant with the unit of $m/s^2$ would work. Not only that, but when calculating $\Delta{V}$ using a specific impulse term, as is frequently seen, the specific impulse term is multiplied by $g$, turning the term back into $v_e$. Another weird effect of this is that specific impulse has a unit of seconds.

This weird definition was originally used, as far as I can tell, because the unit of seconds is the same whether you're using metric or imperial units to calculate it, which is less confusing than m/s and ft/s. Using metric units, exhaust velocity and specific impulse can be calculated approximately by multiplying or dividing by ten. In short, specific impulse is a way of making exhaust velocity, which is the real measure of rocket engine efficiency, into a number which is the same whatever units you use to calculate it.

Interestingly, $v_e$ is effective exhaust velocity, which takes into account any propellant not exhausted at the full velocity of the engine, such as with turbine exhaust or film cooling. Effective exhaust velocity can be calculated on an existing engine with the specific impulse formula by using: $$v_e=\frac{T}{\dot{m}}$$ Except $T$ and $\dot{m}$ are based on measurements of a engine rather than calculated.