What is the velocity distribution of the exhaust for a typical rocket engine?Temperature and pressure of rocket exhaustAre rocket exhaust flames ever opaque?What does the exhaust plume of a rocket look like in vacuum?What is the electromagnetic spectrum profile of rocket flame/exhaust?What is the 'Summerfield criterion' regarding rocket exhaust expansionDoes velocity of the exit gasses of a rocket affect visibility of the plume?Why are exhaust flames “jumping around” the bases of the Falcon-9 engine nozzles; NROL-76?What makes exhaust from aluminum-based SRB propellant so bright?Temperature and pressure of rocket exhaustHow does the camera make the exhaust of the Electron's RP-1/LOX exhaust transparent?Is it possible to create different colors in rocket exhaust?

Where can I find the list of all tendons in the human body?

How to detach yourself from a character you're going to kill?

What was this black-and-white film set in the Arctic or Antarctic where the monster/alien gets fried in the end?

Is it possible to change original filename of an exe?

If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?

What is game ban VS VAC ban in steam?

Uncommanded roll at high speed

How can I offer a test ride while selling a bike?

Thousands and thousands of words

The qvolume of an integer

Different PCB color ( is it different material? )

Adding strings in lists together

Can a non-EU citizen travel within the Schengen area without identity documents?

Biblical Basis for 400 years of silence between old and new testament

Why use water tanks from a retired Space Shuttle?

Why don't I have ground wiring on any of my outlets?

Points within polygons in different projections

How can I grammatically understand "Wir über uns"?

Self-Preservation: How to DM NPCs that Love Living?

Can a wire having a 610-670 THz (frequency of blue light) AC frequency supply, generate blue light?

What does "tea juice" mean in this context?

Mother abusing my finances

Humans meet a distant alien species. How do they standardize? - Units of Measure

What is the intuition behind uniform continuity?



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


Temperature and pressure of rocket exhaustAre rocket exhaust flames ever opaque?What does the exhaust plume of a rocket look like in vacuum?What is the electromagnetic spectrum profile of rocket flame/exhaust?What is the 'Summerfield criterion' regarding rocket exhaust expansionDoes velocity of the exit gasses of a rocket affect visibility of the plume?Why are exhaust flames “jumping around” the bases of the Falcon-9 engine nozzles; NROL-76?What makes exhaust from aluminum-based SRB propellant so bright?Temperature and pressure of rocket exhaustHow does the camera make the exhaust of the Electron's RP-1/LOX exhaust transparent?Is it possible to create different colors in rocket exhaust?













8












$begingroup$


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.
enter image description here



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










share|improve this question











$endgroup$











  • $begingroup$
    Where did your image come from? It seems like it answers your question.
    $endgroup$
    – Russell Borogove
    May 16 at 1:02






  • 3




    $begingroup$
    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.
    $endgroup$
    – Ingolifs
    May 16 at 1:17






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    May 16 at 1:32










  • $begingroup$
    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.
    $endgroup$
    – DrSheldon
    May 16 at 2:06






  • 1




    $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    May 16 at 2:29















8












$begingroup$


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.
enter image description here



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










share|improve this question











$endgroup$











  • $begingroup$
    Where did your image come from? It seems like it answers your question.
    $endgroup$
    – Russell Borogove
    May 16 at 1:02






  • 3




    $begingroup$
    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.
    $endgroup$
    – Ingolifs
    May 16 at 1:17






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    May 16 at 1:32










  • $begingroup$
    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.
    $endgroup$
    – DrSheldon
    May 16 at 2:06






  • 1




    $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    May 16 at 2:29













8












8








8


1



$begingroup$


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.
enter image description here



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










share|improve this question











$endgroup$




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.
enter image description here



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







exhaust velocity






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 16 at 2:27







Ingolifs

















asked May 15 at 23:07









IngolifsIngolifs

2,800831




2,800831











  • $begingroup$
    Where did your image come from? It seems like it answers your question.
    $endgroup$
    – Russell Borogove
    May 16 at 1:02






  • 3




    $begingroup$
    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.
    $endgroup$
    – Ingolifs
    May 16 at 1:17






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    May 16 at 1:32










  • $begingroup$
    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.
    $endgroup$
    – DrSheldon
    May 16 at 2:06






  • 1




    $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    May 16 at 2:29
















  • $begingroup$
    Where did your image come from? It seems like it answers your question.
    $endgroup$
    – Russell Borogove
    May 16 at 1:02






  • 3




    $begingroup$
    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.
    $endgroup$
    – Ingolifs
    May 16 at 1:17






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    May 16 at 1:32










  • $begingroup$
    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.
    $endgroup$
    – DrSheldon
    May 16 at 2:06






  • 1




    $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    May 16 at 2:29















$begingroup$
Where did your image come from? It seems like it answers your question.
$endgroup$
– Russell Borogove
May 16 at 1:02




$begingroup$
Where did your image come from? It seems like it answers your question.
$endgroup$
– Russell Borogove
May 16 at 1:02




3




3




$begingroup$
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.
$endgroup$
– Ingolifs
May 16 at 1:17




$begingroup$
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.
$endgroup$
– Ingolifs
May 16 at 1:17




1




1




$begingroup$
Fine to have it as long as the question makes its origin clear.
$endgroup$
– Russell Borogove
May 16 at 1:32




$begingroup$
Fine to have it as long as the question makes its origin clear.
$endgroup$
– Russell Borogove
May 16 at 1:32












$begingroup$
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.
$endgroup$
– DrSheldon
May 16 at 2:06




$begingroup$
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.
$endgroup$
– DrSheldon
May 16 at 2:06




1




1




$begingroup$
Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
$endgroup$
– Ingolifs
May 16 at 2:29




$begingroup$
Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
$endgroup$
– Ingolifs
May 16 at 2:29










1 Answer
1






active

oldest

votes


















7












$begingroup$

@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



$$v_T = sqrtfrack_mathrm B Tm.$$



Testing example using air or nitrogen ($m=28 times 1.673times10^-27 mathrmkg$) at $293 mathrm K$ with Boltzmann constant $k_mathrm B = 1.381times 10^-23 mathrmJ/K$ gives $297 mathrmm/s$ 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



$$f(v),mathrm dv = left( fracm2 pi k_mathrm B Tright)^1/2 expleft(-fracmv^22 k_mathrm B Tright),mathrm dv$$



The directed exhaust velocity might be close to zero in the combustion chamber, and roughly $3600 mathrmm/s$ 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.



schematic of rocket combustion and exhaust velocities



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()


enter image description hereSource






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Organic Marble
    May 16 at 3:53






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:04






  • 1




    $begingroup$
    Thanks! That clears it up.
    $endgroup$
    – Organic Marble
    May 16 at 4:05






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:05











  • $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:10











Your Answer








StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "508"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f36181%2fwhat-is-the-velocity-distribution-of-the-exhaust-for-a-typical-rocket-engine%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

@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



$$v_T = sqrtfrack_mathrm B Tm.$$



Testing example using air or nitrogen ($m=28 times 1.673times10^-27 mathrmkg$) at $293 mathrm K$ with Boltzmann constant $k_mathrm B = 1.381times 10^-23 mathrmJ/K$ gives $297 mathrmm/s$ 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



$$f(v),mathrm dv = left( fracm2 pi k_mathrm B Tright)^1/2 expleft(-fracmv^22 k_mathrm B Tright),mathrm dv$$



The directed exhaust velocity might be close to zero in the combustion chamber, and roughly $3600 mathrmm/s$ 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.



schematic of rocket combustion and exhaust velocities



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()


enter image description hereSource






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Organic Marble
    May 16 at 3:53






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:04






  • 1




    $begingroup$
    Thanks! That clears it up.
    $endgroup$
    – Organic Marble
    May 16 at 4:05






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:05











  • $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:10















7












$begingroup$

@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



$$v_T = sqrtfrack_mathrm B Tm.$$



Testing example using air or nitrogen ($m=28 times 1.673times10^-27 mathrmkg$) at $293 mathrm K$ with Boltzmann constant $k_mathrm B = 1.381times 10^-23 mathrmJ/K$ gives $297 mathrmm/s$ 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



$$f(v),mathrm dv = left( fracm2 pi k_mathrm B Tright)^1/2 expleft(-fracmv^22 k_mathrm B Tright),mathrm dv$$



The directed exhaust velocity might be close to zero in the combustion chamber, and roughly $3600 mathrmm/s$ 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.



schematic of rocket combustion and exhaust velocities



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()


enter image description hereSource






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – Organic Marble
    May 16 at 3:53






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:04






  • 1




    $begingroup$
    Thanks! That clears it up.
    $endgroup$
    – Organic Marble
    May 16 at 4:05






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:05











  • $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:10













7












7








7





$begingroup$

@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



$$v_T = sqrtfrack_mathrm B Tm.$$



Testing example using air or nitrogen ($m=28 times 1.673times10^-27 mathrmkg$) at $293 mathrm K$ with Boltzmann constant $k_mathrm B = 1.381times 10^-23 mathrmJ/K$ gives $297 mathrmm/s$ 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



$$f(v),mathrm dv = left( fracm2 pi k_mathrm B Tright)^1/2 expleft(-fracmv^22 k_mathrm B Tright),mathrm dv$$



The directed exhaust velocity might be close to zero in the combustion chamber, and roughly $3600 mathrmm/s$ 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.



schematic of rocket combustion and exhaust velocities



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()


enter image description hereSource






share|improve this answer











$endgroup$



@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



$$v_T = sqrtfrack_mathrm B Tm.$$



Testing example using air or nitrogen ($m=28 times 1.673times10^-27 mathrmkg$) at $293 mathrm K$ with Boltzmann constant $k_mathrm B = 1.381times 10^-23 mathrmJ/K$ gives $297 mathrmm/s$ 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



$$f(v),mathrm dv = left( fracm2 pi k_mathrm B Tright)^1/2 expleft(-fracmv^22 k_mathrm B Tright),mathrm dv$$



The directed exhaust velocity might be close to zero in the combustion chamber, and roughly $3600 mathrmm/s$ 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.



schematic of rocket combustion and exhaust velocities



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()


enter image description hereSource







share|improve this answer














share|improve this answer



share|improve this answer








edited May 16 at 6:01









Loong

1214




1214










answered May 16 at 1:42









uhohuhoh

43.2k20169558




43.2k20169558







  • 1




    $begingroup$
    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.
    $endgroup$
    – Organic Marble
    May 16 at 3:53






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:04






  • 1




    $begingroup$
    Thanks! That clears it up.
    $endgroup$
    – Organic Marble
    May 16 at 4:05






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:05











  • $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:10












  • 1




    $begingroup$
    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.
    $endgroup$
    – Organic Marble
    May 16 at 3:53






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:04






  • 1




    $begingroup$
    Thanks! That clears it up.
    $endgroup$
    – Organic Marble
    May 16 at 4:05






  • 1




    $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:05











  • $begingroup$
    @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.
    $endgroup$
    – uhoh
    May 16 at 4:10







1




1




$begingroup$
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.
$endgroup$
– Organic Marble
May 16 at 3:53




$begingroup$
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.
$endgroup$
– Organic Marble
May 16 at 3:53




1




1




$begingroup$
@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.
$endgroup$
– uhoh
May 16 at 4:04




$begingroup$
@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.
$endgroup$
– uhoh
May 16 at 4:04




1




1




$begingroup$
Thanks! That clears it up.
$endgroup$
– Organic Marble
May 16 at 4:05




$begingroup$
Thanks! That clears it up.
$endgroup$
– Organic Marble
May 16 at 4:05




1




1




$begingroup$
@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.
$endgroup$
– uhoh
May 16 at 4:05





$begingroup$
@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.
$endgroup$
– uhoh
May 16 at 4:05













$begingroup$
@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.
$endgroup$
– uhoh
May 16 at 4:10




$begingroup$
@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.
$endgroup$
– uhoh
May 16 at 4:10

















draft saved

draft discarded
















































Thanks for contributing an answer to Space Exploration Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f36181%2fwhat-is-the-velocity-distribution-of-the-exhaust-for-a-typical-rocket-engine%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Club Baloncesto Breogán Índice Historia | Pavillón | Nome | O Breogán na cultura popular | Xogadores | Adestradores | Presidentes | Palmarés | Historial | Líderes | Notas | Véxase tamén | Menú de navegacióncbbreogan.galCadroGuía oficial da ACB 2009-10, páxina 201Guía oficial ACB 1992, páxina 183. Editorial DB.É de 6.500 espectadores sentados axeitándose á última normativa"Estudiantes Junior, entre as mellores canteiras"o orixinalHemeroteca El Mundo Deportivo, 16 setembro de 1970, páxina 12Historia do BreogánAlfredo Pérez, o último canoneiroHistoria C.B. BreogánHemeroteca de El Mundo DeportivoJimmy Wright, norteamericano do Breogán deixará Lugo por ameazas de morteResultados de Breogán en 1986-87Resultados de Breogán en 1990-91Ficha de Velimir Perasović en acb.comResultados de Breogán en 1994-95Breogán arrasa al Barça. "El Mundo Deportivo", 27 de setembro de 1999, páxina 58CB Breogán - FC BarcelonaA FEB invita a participar nunha nova Liga EuropeaCharlie Bell na prensa estatalMáximos anotadores 2005Tempada 2005-06 : Tódolos Xogadores da Xornada""Non quero pensar nunha man negra, mais pregúntome que está a pasar""o orixinalRaúl López, orgulloso dos xogadores, presume da boa saúde económica do BreogánJulio González confirma que cesa como presidente del BreogánHomenaxe a Lisardo GómezA tempada do rexurdimento celesteEntrevista a Lisardo GómezEl COB dinamita el Pazo para forzar el quinto (69-73)Cafés Candelas, patrocinador del CB Breogán"Suso Lázare, novo presidente do Breogán"o orixinalCafés Candelas Breogán firma el mayor triunfo de la historiaEl Breogán realizará 17 homenajes por su cincuenta aniversario"O Breogán honra ao seu fundador e primeiro presidente"o orixinalMiguel Giao recibiu a homenaxe do PazoHomenaxe aos primeiros gladiadores celestesO home que nos amosa como ver o Breo co corazónTita Franco será homenaxeada polos #50anosdeBreoJulio Vila recibirá unha homenaxe in memoriam polos #50anosdeBreo"O Breogán homenaxeará aos seus aboados máis veteráns"Pechada ovación a «Capi» Sanmartín e Ricardo «Corazón de González»Homenaxe por décadas de informaciónPaco García volve ao Pazo con motivo do 50 aniversario"Resultados y clasificaciones""O Cafés Candelas Breogán, campión da Copa Princesa""O Cafés Candelas Breogán, equipo ACB"C.B. Breogán"Proxecto social"o orixinal"Centros asociados"o orixinalFicha en imdb.comMario Camus trata la recuperación del amor en 'La vieja música', su última película"Páxina web oficial""Club Baloncesto Breogán""C. B. Breogán S.A.D."eehttp://www.fegaba.com

Vilaño, A Laracha Índice Patrimonio | Lugares e parroquias | Véxase tamén | Menú de navegación43°14′52″N 8°36′03″O / 43.24775, -8.60070

Cegueira Índice Epidemioloxía | Deficiencia visual | Tipos de cegueira | Principais causas de cegueira | Tratamento | Técnicas de adaptación e axudas | Vida dos cegos | Primeiros auxilios | Crenzas respecto das persoas cegas | Crenzas das persoas cegas | O neno deficiente visual | Aspectos psicolóxicos da cegueira | Notas | Véxase tamén | Menú de navegación54.054.154.436928256blindnessDicionario da Real Academia GalegaPortal das Palabras"International Standards: Visual Standards — Aspects and Ranges of Vision Loss with Emphasis on Population Surveys.""Visual impairment and blindness""Presentan un plan para previr a cegueira"o orixinalACCDV Associació Catalana de Cecs i Disminuïts Visuals - PMFTrachoma"Effect of gene therapy on visual function in Leber's congenital amaurosis"1844137110.1056/NEJMoa0802268Cans guía - os mellores amigos dos cegosArquivadoEscola de cans guía para cegos en Mortágua, PortugalArquivado"Tecnología para ciegos y deficientes visuales. Recopilación de recursos gratuitos en la Red""Colorino""‘COL.diesis’, escuchar los sonidos del color""COL.diesis: Transforming Colour into Melody and Implementing the Result in a Colour Sensor Device"o orixinal"Sistema de desarrollo de sinestesia color-sonido para invidentes utilizando un protocolo de audio""Enseñanza táctil - geometría y color. Juegos didácticos para niños ciegos y videntes""Sistema Constanz"L'ocupació laboral dels cecs a l'Estat espanyol està pràcticament equiparada a la de les persones amb visió, entrevista amb Pedro ZuritaONCE (Organización Nacional de Cegos de España)Prevención da cegueiraDescrición de deficiencias visuais (Disc@pnet)Braillín, un boneco atractivo para calquera neno, con ou sen discapacidade, que permite familiarizarse co sistema de escritura e lectura brailleAxudas Técnicas36838ID00897494007150-90057129528256DOID:1432HP:0000618D001766C10.597.751.941.162C97109C0155020