Explained: Area-Mach Number Relation

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  • @jaik195701
    @jaik195701 4 ปีที่แล้ว +3

    I am glad I can pause and rewind; I can’t imagine a 50 minute live lecture at this rate

    • @JoshTheEngineer
      @JoshTheEngineer  4 ปีที่แล้ว +2

      That's the beauty of TH-cam. When I lectured, I would never go this fast.

    • @navsquid32
      @navsquid32 4 ปีที่แล้ว

      @@JoshTheEngineer Are you trying to get in under a particular time cap? As a lecturer myself, I've found that if you take pauses that you perceive to be too long when lecturing, they actually are a good length and seem natural.

    • @JoshTheEngineer
      @JoshTheEngineer  4 ปีที่แล้ว +1

      @@navsquid32 Not necessarily; all my videos are pretty long, so I'm not worried about time. I was mainly trying to adapt my editing to comments on other videos saying that my breathing was distracting or something along those lines. I ended up over-editing, by cutting out any breathing point that I had that was noticeable to me, which makes it seem like I'm talking faster than the raw video actually shows. In my more recent videos, you can see the format that I like and that I'm sticking to now.

    • @navsquid32
      @navsquid32 4 ปีที่แล้ว +1

      JoshTheEngineer Thanks for that. I was curious about the edits, actually. I put your videos on 0.75, and it makes it “normal speed.” That makes sense with the edits, though, because the longer segments don’t seem particularly fast.

    • @JoshTheEngineer
      @JoshTheEngineer  4 ปีที่แล้ว +1

      @@navsquid32 Yep, no problem. There's a lot of videos I'd like to re-film to get them slower. The best example is my afterburners video. Nothing is wrong with any of the content, but most of the comments I have to sift through are repeated terrible jokes, which gets a little tedious. In the end, I just don't have the time or the patience to re-film videos that are perfectly fine aside from maybe being a little on the fast side. I recommend people do what you do, which is slow it down a little bit, and hopefully that's ok. If not, there's always other videos on the topic that might be more suitable for them.

  • @junuhunuproductions
    @junuhunuproductions 5 ปีที่แล้ว +6

    Wow... you are really Fast.. lol. But your unique teaching method makes it really easy to understand.

  • @joro8604
    @joro8604 5 ปีที่แล้ว +4

    Great explanation- I really like the fact that you mention it is counter-intuitive. The math is pretty perplexing too. You are hella smart. I wish I had youtube when I went to school instead of a class with a boring professor 1st thing in the morning (and me showing up with a hangover- that's an oblique shock).

  • @ubaidkhan8349
    @ubaidkhan8349 4 ปีที่แล้ว +1

    I had a lecture at supersonic speed, concepts were so fast one by one and things were made so easier...
    Incredible Sir !

  • @ngwakomakudubela8189
    @ngwakomakudubela8189 ปีที่แล้ว +1

    Clean and quick. Bravo!

  • @riturajrajput5012
    @riturajrajput5012 3 ปีที่แล้ว +1

    You are a legend. you are way better than our professors.

  • @pavanbharathi478
    @pavanbharathi478 ปีที่แล้ว +1

    Thanks for great lectures

  • @jamespessoa9386
    @jamespessoa9386 4 ปีที่แล้ว +1

    Great work! I would like to discuss my AirPower project with you, where I have a high-mass subsonic airflow at the exit of a heat insulated converging duct, tangentially impacting a radial turbine. The thrust of this flow rotates the turbine which is linked to an electric generator. The airflow was induced and created by an electric axial fan aspirating ambient air at the converging duct inlet. My point is that we can aspirate and accelerate ambient air, and generate more electric power than we injected at the fan, as we are converting internal energy from the air into kinetic energy. The air comes out of the system cooler than at the entrance, some 6, 7 degrees C. So, we have an open-cycle thermal machine which generates kinetic energy from the incoming internal energy, converts a portion of this kinetic energy into electric power, and cooler air leaves the system. In short, we can generate clean electric power from ambient air, at the same time obtaining cooler air and condensating part of the contained moisture. So, clean power, cool air and freshwater. From ambient air. And fully respecting the first and second laws. I need more theoretical expertise to prove it. If we could talk, would greatly appreciate your assistance. My email is james.pessoa@gmail.com
    Thank you very much.

  • @dhonors999
    @dhonors999 7 ปีที่แล้ว +3

    that's really interesting! is there an intuitive explanation why the relation flips?

    • @JoshTheEngineer
      @JoshTheEngineer  7 ปีที่แล้ว +6

      I would say that the most intuitive way to think about it is that when you include the effects of changing density, this result naturally pops out of the conservation equations. What do I mean by that? I showed how we can use only the conservation of mass equation to see how velocity changes with changes in area when the flow is incompressible. The reason that was possible was because we assumed density was constant.
      When we relax the incompressible assumption and allow density to change, you can still relate the area change to the velocity change from the mass conservation equation, but you'll also have that density term in there, and it's not clear how it affects the velocity without further analysis. So we brought in the conservation of momentum equation to find a relationship between changes in velocity and changes in pressure. We were able to eliminate that density term by using the definition of the speed of sound, which relates the change in pressure to the isentropic change in density. So in the end, we can end up with an expression that again relates the area to the velocity, but we now have a term that modifies the velocity change that depends on whether the flow velocity is less than, equal to, or greater than the speed of sound.
      I know that discussion basically just rehashed the mathematics of the video, but it was harder than I thought to come up with a more intuitive way of thinking about this other than it happens to fall out of the combination of the mass and momentum conservation equations after rearranging some terms and introducing the convenient dimensionless Mach number.

    • @Cray25
      @Cray25 6 ปีที่แล้ว +1

      You're a great explaining stuff, better than anyone I know in engineering

    • @yyyyyyyyxxx334
      @yyyyyyyyxxx334 5 ปีที่แล้ว +3

      i was thinking about it and I came up with the following explanation. Look at the conservation of mass equation. Increasing the area requires some combination of lowering the velocity and lowering the density. This makes sense. If you give the gas more space it will have two choices. Either it can spread out- decreasing density, or it can slow down, so that more gas accumulates in the the area as the area increases.
      However, as derived from the conservation of momentum equation, the velocity and the density change in opposite directions. This also makes sense. In order for the density to decrease, the fluid must spread out. So imagine a small section of the flow. In order for it to spread out, the front of it needs to speed away from the back, so the section is now spread among more total area. Thus, decreasing density creates an increase in speed. The opposite is also true. If the speed decreases, The gas in the front slows down and the gas in the back comes up behind it, and the total space decreases, increasing the density.
      So, when the area is expanded, the flow has two coping mechanisms to choose from. It can decrease its density, or it can decrease its velocity., However, whichever it chooses, the other will work against it. That is where the final equation comes in. The mach number determines how strong the correlation is between the change in density and the change in velocity. If the mach number is low, then decreasing the velocity does not effect the density very much, so expanding the flow will slow it down without significantly changing the density. On the other hand, for a supersonic flow, decreasing the velocity would have a tremendous effect on the density, and the mass equation could not be satisfied. Rather, the density decreases, thereby increasing the velocity. And, in the middle case where the flow is exactly sonic, the changes in velocity and density exactly balance each other out, so neither one will happen just because the area changes.
      It makes sense that this depends on mach number. When the mach number is low, a disturbance in one place does not immediately effect the gas farther upstream. So decreasing the velocity of the front gas in the section will not instantaneously increase the density in the gas behind it. It can accelerate smoothly without compressing. However, in a supersonic flow, any gas that slows down will cause all of the gas behind it to pile up because the deceleration reverberated back through the flow.
      Thoughts?

    • @govindchari4586
      @govindchari4586 4 ปีที่แล้ว

      @@yyyyyyyyxxx334 Wow! That explanation is amazing. It finally gave me a very good intuition for such a counter-intuitive concept. Thank you so much for taking the time to write that out to help others like myself understand this idea! A million thanks!

    • @navsquid32
      @navsquid32 4 ปีที่แล้ว

      @@yyyyyyyyxxx334 It's not just gas. This works for incompressible fluids as well. The flow of water through pipes, for example. Transitioning to a larger diameter pipe will reduce the velocity of the fluid by necessity. If this didn't occur, then mass would not be conserved, since more mass would be flowing upstream of the pipe transition than is flowing into it. That cannot occur. The mass of fluid flowing into the control volume needs to equal the mass of fluid flowing out.
      Whenever flow conditions are such that there is an abrupt change in conditions (like density in a compressible fluid like air), then a shock front develops, in which there is a prompt change in fluid conditions, which is a highly non-isentropic phenomenon.

  • @jurajkolesar1894
    @jurajkolesar1894 3 ปีที่แล้ว +1

    the most valuable 7 minutes 42 seconds i lived

  • @manelsouguir8126
    @manelsouguir8126 4 ปีที่แล้ว

    please, can you tell me :
    is always at the throat M=1 or this depends on the ratio P/P0?. I mean that: the throat is sonic (M=1) only if P/P0=0.5283.

  • @arrow9323
    @arrow9323 3 ปีที่แล้ว +1

    Thank you bro

  • @joro8604
    @joro8604 5 ปีที่แล้ว

    In the conservation of momentum equation, is the rho, u, and A a constant in the denominator? I’m struggling with the notation otherwise.

  • @jraimond
    @jraimond 4 ปีที่แล้ว +1

    Can you speak a little bit more about how it is that you can write pressure as a function of just density and entropy? I see the convenience to your derivation, but it isn't clear to me why you can do that.

    • @JoshTheEngineer
      @JoshTheEngineer  4 ปีที่แล้ว +3

      In general (there are cases when this is not true, but you don't need to worry about it), you can write any state variable (density, pressure, temperature, entropy, etc.) as a function of just two other state variables. Sometimes it's more convenient to write them as a function of something other than density and entropy. If you want to see this in action, here's the link to my video on specific heats (th-cam.com/video/vmrXXN5d4iI/w-d-xo.html).

  • @santhoshgowda5727
    @santhoshgowda5727 5 ปีที่แล้ว +1

    Energetic!!!

  • @thewetman736
    @thewetman736 3 ปีที่แล้ว

    Can someone please help me out. Is the notation dV the "change in velocity", therefore acceleration. Or is this not the case?

    • @JoshTheEngineer
      @JoshTheEngineer  3 ปีที่แล้ว +1

      The dV term just means "change in velocity", which can be related to changes in other variables, such as density, area, etc. It is not acceleration; acceleration is a change in velocity over a change in time.

  • @ahmetkaangok2925
    @ahmetkaangok2925 5 ปีที่แล้ว

    Can't we find the relation M= M(A/A*) for supersonic flows?

  • @dylanlundberg9609
    @dylanlundberg9609 4 ปีที่แล้ว +1

    This is the area velocity variation and not the area Mach relation!.... two different relations

    • @dylanlundberg9609
      @dylanlundberg9609 4 ปีที่แล้ว

      nikander.github.io/compflow/Anderson/Chapter5/#:~:text=The%20area%2DMach%20number%20relation%20gives%20the%20ratio%20of%20local,allowed)%20and%20calorically%20perfect%20gases.

    • @JoshTheEngineer
      @JoshTheEngineer  4 ปีที่แล้ว +1

      Yep, you're correct that this is actually the area-velocity relation. I guess the reason I had labeled it like this was because I then used the CPG assumption to simplify the expression to the area-Mach number relation in the next video. But yes, you're right.

  • @NizaanthNelson
    @NizaanthNelson 6 ปีที่แล้ว +2

    Is it just me or if anyone else feels like the video is fast forwarding?

    • @JoshTheEngineer
      @JoshTheEngineer  5 ปีที่แล้ว +3

      I use jump-cuts so you don't have to listen to 10 minutes of me breathing.

    • @junuhunuproductions
      @junuhunuproductions 5 ปีที่แล้ว

      @@JoshTheEngineer Lol