# What is the velocity distribution of the exhaust for a typical rocket engine?

The exhaust velocity figure of a given engine is an average value. I'm curious to know how the distribution of velocities of the exhaust particles look.

From a qualitative perspective, how does the distribution typically look? Is it Normal? Lognormal? Some variation on the Boltzmann distribution? Is it multimodal due to the presence of light molecules (H2) and heavier molecules (CO2)? Is the standard deviation small, meaning the velocities are clustered tightly around the mean, or is it large with particle velocities spanning many orders of magnitude? Does it differ substantially between different engines or fuel types?

An example velocity distribution is below. This is what I suspect what a velocity distribution might look like - it doesn't represent any physical reality.

I know this question may be interpreted as broad, which is why I'm asking for a handwavey general answer.

• Where did your image come from? It seems like it answers your question. – Russell Borogove May 16 at 1:02
• I made it myself to illustrate what I'm after. It is completely made up and not based in reality at all. If this is confusing, maybe I could get rid of it. – Ingolifs May 16 at 1:17
• Fine to have it as long as the question makes its origin clear. – Russell Borogove May 16 at 1:32
• Your question is not too broad, and I think our upvotes confirm that. However, I would take RussellBorogove's advice, and edit the question to explain that the graph represents the type of answer you are looking for. – DrSheldon May 16 at 2:06
• Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited. – Ingolifs May 16 at 2:29

@RussellBorogove's answer mentions temperatures of roughly 3700 K and 1800 K in the combustion chamber and exhaust of a big rocket engine.

The canonical Wikipedia plot of velocity and temperature for a de Laval nozzle is shown below as a schematic representation only.

Ignoring some aspects of gas theory, we can estimate the thermal velocity of a molecule using

$$v_T = \sqrt{\frac{k_\mathrm B T}{m}}.$$

Testing example using air or nitrogen ($$m=28 \times 1.673\times10^{-27}\ \mathrm{kg}$$) at $$293\ \mathrm K$$ with Boltzmann constant $$k_\mathrm B = 1.381\times 10^{-23}\ \mathrm{J/K}$$ gives $$297\ \mathrm{m/s}$$ which does agree with the speed of sound (a good rough indicator of average thermal velocity).

species      mass (kg)      293 K     1800 K     3700 K
-------      ---------     ------     ------     ------
H2        3.346E-27      1100       2700       3900
CO2        7.361E-26       230        580        830


The thermal velocity will be isotropic (all directions) but the exhaust velocity is directed mostly in one direction.

The Maxwell-Boltzman distribution projected in one dimension is given as

$$f(v)\,\mathrm dv = \left( \frac{m}{2 \pi k_\mathrm B T}\right)^{1/2} \exp\left(-\frac{mv^2}{2 k_\mathrm B T}\right)\,\mathrm dv$$

The directed exhaust velocity might be close to zero in the combustion chamber, and roughly $$3600\ \mathrm{m/s}$$ in the nozzle.

Assuming that the engine burns methane and a little bit of H2 is formed in the exhaust in order to fulfill the terms of the question, this plot shows the resulting estimated velocity distributions. For each species, the wide curve centered at zero represents the condition in the combustion chamber, and the narrower curve offset to the right represents the axial velocity exiting the nozzle.

The transverse velocity distribution would look similar except that the the offset for the exhaust curve would be closer to zero and depend on distance from the axis and details of under/over expansion. That's a more complicated calculation and has too many special cases for to address for this level of approximation.

def f(v, v0, m, T):
term_1 = np.sqrt((m)/(twopi * kB * T))
term_2 = (m * (v-v0)**2)/(2*kB*T)
return term_1 * np.exp(-term_2)

import numpy as np
import matplotlib.pyplot as plt

twopi = 2 * np.pi
kB    = 1.381E-23
mp    = 1.673E-27

temps       = np.array([3700, 1800])
v0s         = np.array([0,    3600])
m_H2, m_CO2 = mp * np.array([2, 44])
v           = np.linspace(-15000, 15000, 301)

f_H2   = [f(v, v0, m_H2,  T)  for (T, v0) in zip(temps, v0s)]
f_CO2  = [f(v, v0, m_CO2, T)  for (T, v0) in zip(temps, v0s)]

if True:
fig = plt.figure()
for i, (f, name) in enumerate(zip((f_H2, f_CO2), ('H2', 'CO2'))):
ax = fig.add_subplot(2, 1, i+1)
# plt.subplot(2, 1, i+1)
for thing in f:
ax.plot(v, thing)
ax.set_title(name, fontsize=18)
# ax.set_yticklabels([]) # no labels
plt.gca().axes.get_yaxis().set_visible(False) # no labels or ticks
ax.set_xlabel('velocity (m/s)', fontsize=16)
plt.show()

• I'm not familiar with this sort of analysis. What do the negative velocities mean physically? The magnitudes seem huge compared with the exhaust velocity. – Organic Marble May 16 at 3:53
• @OrganicMarble the OP is "curious to know how the distribution of velocities of the exhaust particles look." and lists H2 and CO2 molecules as examples. The plot shows the distribution of velocities of individual molecules projected along the nozzle axis. Even though the average flow is 3600 m/s "to the right", at any moment a fraction of the molecules will be moving to the left, or into the nozzle until they collide with another molecule. Mean free paths are like tens of nanometers,or hundreds of diameters of a given molecule. – uhoh May 16 at 4:04
• Thanks! That clears it up. – Organic Marble May 16 at 4:05
• @OrganicMarble the plot shown in the OP's question might be *velocity squared*(energy) or the square root of that; something like absolute velocity integrated over all directions on a sphere, and so it is zero at zero. This is velocity along a given direction so it can be both positive and negative. – uhoh May 16 at 4:05
• @OrganicMarble but I have no way to address the exhaust velocity profile across the opening of the nozzle. It must be faster in the center and slower near the edges, but I haven't a clue how to approach that. So I think that there is still room for more answers here. – uhoh May 16 at 4:10