What is the velocity distribution of the exhaust for a typical rocket engine?Temperature and pressure of...

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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?













4












$begingroup$


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
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    2 hours ago










  • $begingroup$
    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.
    $endgroup$
    – Dr Sheldon
    1 hour ago










  • $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    1 hour ago
















4












$begingroup$


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
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    2 hours ago










  • $begingroup$
    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.
    $endgroup$
    – Dr Sheldon
    1 hour ago










  • $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    1 hour ago














4












4








4





$begingroup$


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 1 hour ago







Ingolifs

















asked 4 hours ago









IngolifsIngolifs

2,770730




2,770730












  • $begingroup$
    Where did your image come from? It seems like it answers your question.
    $endgroup$
    – Russell Borogove
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    2 hours ago










  • $begingroup$
    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.
    $endgroup$
    – Dr Sheldon
    1 hour ago










  • $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    1 hour ago


















  • $begingroup$
    Where did your image come from? It seems like it answers your question.
    $endgroup$
    – Russell Borogove
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago






  • 1




    $begingroup$
    Fine to have it as long as the question makes its origin clear.
    $endgroup$
    – Russell Borogove
    2 hours ago










  • $begingroup$
    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.
    $endgroup$
    – Dr Sheldon
    1 hour ago










  • $begingroup$
    Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
    $endgroup$
    – Ingolifs
    1 hour ago
















$begingroup$
Where did your image come from? It seems like it answers your question.
$endgroup$
– Russell Borogove
2 hours ago




$begingroup$
Where did your image come from? It seems like it answers your question.
$endgroup$
– Russell Borogove
2 hours ago




2




2




$begingroup$
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
2 hours ago




$begingroup$
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
2 hours ago




1




1




$begingroup$
Fine to have it as long as the question makes its origin clear.
$endgroup$
– Russell Borogove
2 hours ago




$begingroup$
Fine to have it as long as the question makes its origin clear.
$endgroup$
– Russell Borogove
2 hours ago












$begingroup$
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.
$endgroup$
– Dr Sheldon
1 hour ago




$begingroup$
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.
$endgroup$
– Dr Sheldon
1 hour ago












$begingroup$
Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
$endgroup$
– Ingolifs
1 hour ago




$begingroup$
Yeah you're right. The unreferenced picture was a bit 'Muze-like' and needed explanation. Edited.
$endgroup$
– Ingolifs
1 hour ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

@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



$$v_T = sqrt{frac{kB T}{m}}.$$



Testing example using air or nitrogen ($m=28 times 1.673times10^{-27}$ kg) at 293 K with Boltzmann constant $k_B = 1.381times 10^{-23}$ J/K gives 297 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



$$f(v) dv = left( frac{m}{2 pi k_B T}right)^{1/2} exp(-frac{mv^2}{2 k_B T}) dv$$



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



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











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    4












    $begingroup$

    @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



    $$v_T = sqrt{frac{kB T}{m}}.$$



    Testing example using air or nitrogen ($m=28 times 1.673times10^{-27}$ kg) at 293 K with Boltzmann constant $k_B = 1.381times 10^{-23}$ J/K gives 297 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



    $$f(v) dv = left( frac{m}{2 pi k_B T}right)^{1/2} exp(-frac{mv^2}{2 k_B T}) dv$$



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



    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











    $endgroup$


















      4












      $begingroup$

      @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



      $$v_T = sqrt{frac{kB T}{m}}.$$



      Testing example using air or nitrogen ($m=28 times 1.673times10^{-27}$ kg) at 293 K with Boltzmann constant $k_B = 1.381times 10^{-23}$ J/K gives 297 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



      $$f(v) dv = left( frac{m}{2 pi k_B T}right)^{1/2} exp(-frac{mv^2}{2 k_B T}) dv$$



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



      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











      $endgroup$
















        4












        4








        4





        $begingroup$

        @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



        $$v_T = sqrt{frac{kB T}{m}}.$$



        Testing example using air or nitrogen ($m=28 times 1.673times10^{-27}$ kg) at 293 K with Boltzmann constant $k_B = 1.381times 10^{-23}$ J/K gives 297 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



        $$f(v) dv = left( frac{m}{2 pi k_B T}right)^{1/2} exp(-frac{mv^2}{2 k_B T}) dv$$



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



        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



        $$v_T = sqrt{frac{kB T}{m}}.$$



        Testing example using air or nitrogen ($m=28 times 1.673times10^{-27}$ kg) at 293 K with Boltzmann constant $k_B = 1.381times 10^{-23}$ J/K gives 297 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



        $$f(v) dv = left( frac{m}{2 pi k_B T}right)^{1/2} exp(-frac{mv^2}{2 k_B T}) dv$$



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



        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







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        edited 1 hour ago

























        answered 2 hours ago









        uhohuhoh

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