## Tuesday, July 31, 2018

### 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: