Lec0: Sizing a Rocket Engine from Scratch (Intro to Rocket Design)

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  • Опубликовано: 27 ноя 2024

Комментарии • 59

  • @liquidpropulsiongroup5017
    @liquidpropulsiongroup5017  3 года назад +12

    25:54 Forgot to add Bartz approximation equation.
    You can find it here, along with lecture notes about it:
    ocw.mit.edu/courses/aeronautics-and-astronautics/16-512-rocket-propulsion-fall-2005/lecture-notes/lecture_7.pdf (p. 9, fig. 30)

  • @wyattb3138
    @wyattb3138 Год назад +96

    Heard KSP music, so now I know this guy knows his stuff…

    • @nugs2727
      @nugs2727 Год назад +4

      I was thinking the same thing lmao

    • @robxcarlson
      @robxcarlson 8 месяцев назад

      yes ksp music

    • @dinklydorf9215
      @dinklydorf9215 7 месяцев назад

      Speaking the truths

    • @catt8822
      @catt8822 4 месяца назад

      this is actually so real

  • @gergelytolgyesi7326
    @gergelytolgyesi7326 2 года назад +37

    i love how the ksp music keeps playing xd

  • @muhammadhashim9238
    @muhammadhashim9238 Год назад +8

    cannot have enough words to thank you for this beautiful content you have delivered to me

  • @bnl1852
    @bnl1852 Год назад +22

    You just tied together my fluid mechanics, propulsion, and gas dynamics courses in a way that they actually make sense together. They were all so abstract to me before, thank you so much. Also notability is the best.

  • @pugglez4798
    @pugglez4798 4 месяца назад +4

    FOR ANYONE READING: There is a typo in the isentropic Mach relations, p and rho equations should be to the -y/(y-1) in the video there isn't a negative sign.
    Great video though, very imformative

    • @idris4587
      @idris4587 Месяц назад

      Thank you studying thermo at the moment and got stuck trying to figure this out

  • @alexconverse1122
    @alexconverse1122 Год назад +4

    awesome content thank you

  • @rverm1000
    @rverm1000 2 года назад +3

    very good content

  • @ryansmithers7143
    @ryansmithers7143 3 года назад +3

    Great video

  • @odedantebi8086
    @odedantebi8086 4 месяца назад

    OK, I will need to watch it like 8 times

  • @varunahlawat9013
    @varunahlawat9013 Год назад +2

    Pull up more such videos please!

  • @makarlock
    @makarlock 3 года назад +10

    100 views, pog?

  • @idris4587
    @idris4587 Месяц назад

    How does this change with monoprop or thermal engines with LOX variability? I assume then the sizing characteristic graph for optimumal ratios for fuel and oxider shift as it becomes less efficient to use oxider to raise temperature as you increase atomic mass.

  • @zerggod4190
    @zerggod4190 4 месяца назад

    Cool Vid

  • @harshauppuluri1181
    @harshauppuluri1181 2 года назад +1

    Chamber pressure is pressure created during combustion isn't right??

  • @marcovalentinoalvarado3290
    @marcovalentinoalvarado3290 Год назад

    This is so stupidly fascinating, but what are prior requirements to understand this subject? I have no clue about all the maths/physics behind it, but I want to know more and more!
    PD: Liked and subscribed!

  • @jeppelarsen6582
    @jeppelarsen6582 11 месяцев назад

    Hey mate, I have a question for you. What is more desirable regarding fuel oxidizer mixing in the chamber? That the fuel and oxidizer both comes into the chamber as gas phases, or both as liquid, or one as liquid and the other as a gas? Because I guess you would not, heat transfer wise, want the fuel to turn into fully gas(nuclear boiling ok) in the cooling jacket, but you would also not like to have a long liquid fuel penetration length in the chamber. So how would you design the engine so that you make sure the bulk average temperature of the fuel is just shy of boiling when entering the injector!? That is a hard problem I think! Also because I guess the amount of cooling mass flow of fuel is fixed, as well as the initial temperature of the fuel.

    • @carlgeorgbiermann2915
      @carlgeorgbiermann2915 6 месяцев назад

      Few thoughts. I think from a mixing perspective, gaseous injection is better, but typically fuels are present in liquid form so extra steps need to be taken in order to turn them into gases. IIRC SpaceX's Raptors do gaseous injection.
      But I gess, liquid-liquid injectors aim for mechanical atomization rather than boiling them, so having them right below boiling point is less of a concern

  • @MorningStar_BT
    @MorningStar_BT Год назад

    I have questions about the CC. With nasa cea, i can achieve stagnition conditions with infinite area nozzle. How it can be appliciable to reality ? How can I tide the CC to nozzle ?

  • @slevinshafel9395
    @slevinshafel9395 Год назад +1

    4:50 Lightes gas dont make sense. We want kinetic energy wich before you mention is speed*mass so if reduce mas you lose thrust.
    What i see is: more HYDROGEN more combustion(chimic reaction) more ISP: H2>CH4(methane)>C2H6(Ether)....>C3H7NO2S(RP1(i think is the worst because less Hydrogen per Carbon and "NO2S" is just useless weight. That make me think in NH3(Amonio but its nasty reaction and Nitrogen weght more than Carbon))

    • @tristanlongangue5671
      @tristanlongangue5671 7 месяцев назад

      If you you double the speed of an object you quadruple the kinetic energy and if you double it’s mass you double its kinetic energy so by having a lighter gas you can have more kinetic energy because you can accelerate it faster. And you can carry more too I think

  • @shrinivasnymthabad7622
    @shrinivasnymthabad7622 Год назад

    Ok but there are like a million different specific impulse formulas on the internet which one is true can someone help

  • @therocketera
    @therocketera 2 года назад +1

    🙌🚀

  • @Turbo_autism
    @Turbo_autism 8 месяцев назад

    im confused how do you get mass flow rate when calculating a star and at what point in the engine is this measured

    • @quattrocity9620
      @quattrocity9620 20 дней назад

      I know this is an old comment and you might have an answer already but since there isn't a response here...
      when you say "a star" I assume you mean c*, which is the characteristic velocity.
      If you are asking which point in the engine you "measure" mass flow rate, it's where it's most practical to do so. If you are utilizing a gauge on a valve or a weight sensor to measure how much mass is being used in a certain amount of time that would be where you are "measuring". c* is calculated by taking the chamber pressure (pc) and multiplying it by the area of the throat (At) and dividing that by the mass flow rate (m). c*= (pc*At)/m. Keep in mind this isn't actually standard notation as I don't know how to input the actual symbols. Also keep in mind this is only really useful when disregarding transient (situations of extreme changes in values such as engine startup) conditions.

  • @bradleyjomard
    @bradleyjomard Год назад

    how do you find the combustion temperature T0?

  • @omargaber3122
    @omargaber3122 2 года назад

    wonderfull

  • @geforcekk6
    @geforcekk6 5 месяцев назад

    Is that KSP music LOL?

  • @MorningStar_BT
    @MorningStar_BT Год назад +1

    question : why i choose the o/f value that maximize T0/M , What about the highest spesific impulse?

    • @angl3_275
      @angl3_275 Год назад +2

      T0/M is proportional to exhaust velocity/specific impulse, so maximized T0/M is maximized specific impulse I believe.

    • @Prometheus7272
      @Prometheus7272 6 месяцев назад

      @@angl3_275Correct

  • @yungexperiment
    @yungexperiment Год назад

    No sound?

  • @emmetray9703
    @emmetray9703 2 года назад

    How did you get combustion chamber pressure Po?

    • @xxWender98xx
      @xxWender98xx 2 года назад +3

      you choose it, it's a design parameter. Typically is fixed due to technological limits (around 250bar, but spacex increased the limiti to 300bar)

    • @emmetray9703
      @emmetray9703 2 года назад +1

      @@xxWender98xx and according for what I have to choose it? according a chamber material ? I mean , for example, if someone chooses 1000 psi, it means that he did choose it according the material he bought?

    • @xxWender98xx
      @xxWender98xx 2 года назад +2

      @@emmetray9703 You must always seek the highest pressure in order to have the maximum efficiency and thrust, so you choose it based on the material you want to use. If you want a cheap simple engine without regenerative cooling maybe you choose a lower pressure.

    • @emmetray9703
      @emmetray9703 2 года назад +2

      @@xxWender98xx got it , so at the end , it depends on material I want to use and engine structure. Thank you,

    • @xxWender98xx
      @xxWender98xx 2 года назад +1

      @@emmetray9703 yes, you are welcome

  • @MrAlexey89
    @MrAlexey89 Год назад

    вот вроде нормальное объяснение с картинками, жаль только, что на импортном языке

    • @danielalasoo2930
      @danielalasoo2930 Год назад

      Мог бы английский подтянуть

  • @malwinabielecka4392
    @malwinabielecka4392 11 месяцев назад

    nice haircut

  • @flloyd
    @flloyd Год назад

    Its not rock.....hey......

  • @vacantpalesoul8982
    @vacantpalesoul8982 2 года назад +1

    U look like sykkuno lol

    • @makarlock
      @makarlock 2 года назад +1

      Lmfao I’ve cut my hair since then I swear