Reynolds analogy and Chilton Colburn analogy

Поделиться
HTML-код
  • Опубликовано: 12 сен 2024

Комментарии • 36

  • @indianhuman356
    @indianhuman356 4 года назад +3

    Nice and simple explanation. My prof should learn from you. Thank you.

  • @libiaperezmontufar5208
    @libiaperezmontufar5208 5 лет назад +3

    Teacher,I loved your video, I'm from Mexico, and my teacher didn't teach us this correctly so i was lost, but you save me thank you

  • @Thestormrider21
    @Thestormrider21 6 лет назад +3

    Beautiful explanation. Thank you !

  • @xiaoweizhu995
    @xiaoweizhu995 5 лет назад +3

    Thank you for the nice interpretation. Recently I'm working on the CFD modelling of complex channel flow and heat transfer. The magnitude of j_h I caculated does not match with the measured j_D, but the trend of the two dataset perfectly agree with each other. The dataset can be matched by adding a modification constant to j_D. So my conclusion is that the analogy is widely applicable but a modification constant is generally needed for turbulent flows j_h roughly =0.7*j_d

    • @PLE_LU
      @PLE_LU  5 лет назад +1

      I would hesitate to draw _any general_ conclusions at all. As G.E.P. Box said, "All models are wrong. Some models are useful." I would for example guess that if you try with fluids with weird rheological properties, you will get other results than if you try Newtonian fluids.

    • @basil9019
      @basil9019 4 года назад +1

      Sir..i wish my professors had that much knowledge as you

  • @rajendrakumar-bn7ch
    @rajendrakumar-bn7ch Год назад

    Great Explanation, Sir!!

  • @justsomeguy2918
    @justsomeguy2918 6 месяцев назад

    Doing mechanical engineering, thank you for your video

  • @KazeReload
    @KazeReload 4 года назад

    Fantastic video, thank you! This is the first time I see mass transport described with a reference concentration. I can see why there needs to be a reference for temperature, as you can't define a definite amount of energy but only an energy difference, since no real zero exists (that's the reason, right? My thermodynamics course was so many years ago, and sorry if I didn't go back to read the books before writing). But I can't understand the reason behind the reference concentration needed in the convective mass transport along the direction in which the flow is directed. It would be really great to know something more about it. Maybe you talked about it in more detail in some other lesson?

    • @PLE_LU
      @PLE_LU  4 года назад +2

      This turned out to be a long rather philosophical "answer"
      On one level, I have chosen to talk about reference concentration as a pedagogical tool. An analogy is by its very nature a way to look at one phenomena (e.g. heat transfer) and how that phenomena works and for a moment pretend that another phenomena (e.g. mass transfer) works exactly the same way. When we talk about heat and enthalpies we define a reference state (aggregation state and temperature) at which the enthalpy is zero, so why not play the mind game that we treat mass the same way.
      At another level, however, what we are interested in here is the transfer of mass, heat and momentum. For all three of these we know that transfer happens spontaneously from high to low so we always compare the condition in one point with a condition in some reference point:
      Matter diffuses from high concentration to low concentration areas (areas here implying locality, not an area in e.g. square meter)
      Heat diffuses from high temperature to low temperature areas
      Momentum (in fluids) diffuses from high momentum to low momentum areas
      If the temperature in the center of the fluid is higher than the temperature at the wall, then heat is transported to the center of the stream
      If the concentration at the wall is higher (e.g. due to pipes being covered by some partly soluble substance) than in the center of the stream, mass is being transported to the center of the stream.
      For momentum, if we do have a pipe it is often reasonable to assume the no-slip condition, so the momentum is then always higher in the center of the stream and thus transport of momentum is out from the center of the stream.
      So, what do we do when we use transfer coefficients, e.g. mass transfer coefficients? We take the mass transfer coefficient and multiply that with the concentration in the point we are studying and the concentration in the surrounding: k*(C1-C2). We do the same for heat transfer h(T1-T2) and for transport of momentum.
      In this video I call the condition in the other point for the reference, the reference we need to compare with in other to calculate the transport using transfer coefficients. Note: The choice of reference point for temperature is not arbitrary in this video, i.e. not a reference point as the choice of reference point when calculating enthalpies. Instead, we look at the surrounding of a point in the flow and and try to decide: What would be the best reference condition to the condition in this point if we are to calculate mass, heat and momentum transfer to/from that point.
      I don't think I have heard anyone else talking about reference concentrations the way I do in this video. On the other hand, I don't think I've heard or seen anyone trying to explain Reynolds analogy using Lego figures ;-) So, in short, what I'm trying to do in this video is to come up with a way to describe the rather abstract idea of Reynolds analogy by making a simile with a rich filantropist (the Lego figure in the car). A filantropist doesn't give money to people that have more money than the filantropist has, a filantropist give money to people who have less. Transport of heat, mass and momentum works the same way.
      Note: We are used with comparing to movements we experience to a reference velocity, e.g. comparing movements of a car to the movement of the street (we _are_ hurdling through the cosmos), the movement of the International Space Station to the movement of a point on the surface of the earth, etc. There is nothing that prevents us from inventing a reference concentration since in most cases it is the _difference_ in concentration that determines transport, not the actual value of the concentration.
      The actual value of temperature, concentration and momentum do matter, however, in other situations.

  • @ramansinghthakur7085
    @ramansinghthakur7085 3 года назад

    Thank you Sir!

  • @shreusshinde
    @shreusshinde 4 года назад

    Thanks you sir very nicely explained this qstn came in isro Jan 2020 exam . Quite helpful . Great

    • @PLE_LU
      @PLE_LU  4 года назад +2

      Glad to hear it helped. I can't reveal if it appears on our January exam, since our students have exam this Friday ;-)

  • @aykikard
    @aykikard 2 года назад +1

    a brisa de que momentum, calor e massa se comportam igual é mto daora mano

    • @PLE_LU
      @PLE_LU  2 года назад +1

      Since your question(?) isn't in English (or Swedish), I'm not sure that I understand what you say, but transport of mass, heat and momentum are related and Reynolds analogy (gases only) and Chilton Colburn analogy (liquids and non-ideal gases) _might_ give you a good estimate of _how_ they are related. Since there are fundamental differences between the transport of heat, mass and momentum their diffusivities and their transfer coefficients may, however, be rather different.

    • @aykikard
      @aykikard 2 года назад +1

      @@PLE_LU It wasn't a question, I was just saying how good this class was, the analogies you made opened to me a whole new door of ideas that now can flow through my mind, thank you professor! :D

  • @mohammedzubairkhan3004
    @mohammedzubairkhan3004 2 года назад

    Sir, I was curious I learned that the Reynolds analogies are applicable only for turbulent flows where the eddies diffusivity is dominant over the other diffusivities of Momentum Heat and mass? But in your video, u mentioned it's usually applicable for laminar flows. Could u please throw some light upon this it will be really helpful!!

    • @PLE_LU
      @PLE_LU  2 года назад

      It is an analogy, not a truth about the world. If the analogy gives you enough accuracy for your system you use it, otherwise you don't. In some text books, the analogy is derived assuming laminar flow and in other text books it is derived assuming turbulent flow.
      Note:
      1. There is both mass, heat and momentum in an eddie
      2. In fact, if there is no mass there is no heat transfer (apart from radiation) and no transfer of momentum
      3. The assumption behind Reynolds analogy and Chilton Colburn analogy is that there is some relation between transfer of these three entities in two orthogonal directions (see 4.13) . The derivations I have seen has not gone into depth as to why that relation exist (but it does seem reasonable, doesn't it?)
      If radiation is important, I would say all bets are off. Heat transfer through radiation is fundamentally different than transfer of mass and momentum.

    • @mohammedzubairkhan3004
      @mohammedzubairkhan3004 2 года назад

      ​@@PLE_LU Sir it's an honor to learn from you and your videos really give a physical idea and imagination of the concepts. I was really happy when I learned the Reynolds analogy relation that takes turbulence into consideration whereas in all the other derivations we assume it to be laminar flow 😅.
      But nevertheless, my last question is whether the Chilton Colburn analogy is an experimental relation developed from a series of experiments?

    • @PLE_LU
      @PLE_LU  2 года назад +1

      As far as I understand, Chilton Colburn is based on experiments (see e.g. facstaff.cbu.edu/rprice/lectures/mtcoeff.html). I haven't read the original research papers behind Chilton-Colburn analogy, but I wonder a bit about the 2/3 power. I have difficulties seeing a mathematical derivation that would give you exactly e.g. Sc^(2/3), but how good was the fit?
      Anyway, I teach basic courses and my students later go onto a rather wide range of jobs so my main messages to my students on this topic are
      1. There are similarities between how mass, heat and momentum are transported
      2. Hence, there exists analogies that may make it possible to estimate one transfer coefficient from information on the transfer coefficient for another entity (e.g. mass from heat)
      3. Chilton Colburn simplifies to Reynolds analogy if Sc=Pr=1 (e.g. if you have an ideal gas)
      4. "All models are wrong, some models are useful" (as George E.P. Box phrased it) Thus, if a theory/correlation/model/… works with good enough accuracy for your application use it

    • @mohammedzubairkhan3004
      @mohammedzubairkhan3004 2 года назад

      @@PLE_LU i will surely look into this paper

  • @mohammadkamran1926
    @mohammadkamran1926 3 года назад

    Dear Sir, the values of density and heat capacity used in chilton coulburn analogy are material's property or air properties ?

    • @PLE_LU
      @PLE_LU  3 года назад

      When you're calculating dimensionless numbers that describe the flow characteristics of a fluid it is _always_ physical data for the fluid you should be using. (you write air, but air can usually be assumed to be an ideal gas and if you have an ideal gas you might be able to use Reynolds analogy rather than Chilton Colburn. If your fluid is a liquid, however, Chilton Colburns analogy is needed)
      (Note: The surface roughness of the wall may be important for the flow characteristics of a fluid)

    • @mohammadkamran1926
      @mohammadkamran1926 3 года назад

      @@PLE_LU thankyou sir for the information. But i have read in a text that in case of low mass flux, chilton colburn is applicable. And the material i am working on are brick, mortar and concrete which facilitates very low constant rate flux value as compared to biological products. Now its bit confusing which analogy to choose.
      Will be glad if you clarify

    • @PLE_LU
      @PLE_LU  3 года назад

      @@mohammadkamran1926 Chilton Colburn's analogy is more advanced than Reynolds analogy. In fact, Chilton Colburn's analogy simplifies to Reynolds Analogy for perfectly ideal gases. Thus, for gases it might be good enough to use Reynolds analogy, while for liquids Reynolds analogy is (at least to my knowledge) almost guaranteed to fit rather badly. Neither of the two analogies is the reality so if an analogy has proven to work in your particular application, just smile and be happy (Or as G E P Box said "All models are wrong. Some models are useful")
      I'm not sure I understand what you mean when you say that you are working with brick, mortar and concrete. Is that the _wall_ material or the fluid? (I have a hard time thinking of bricks as a fluid, but pyroclastic flows exist and they are probably even stranger than a fluid with bricks…) _If_ the fluid consists of _air_ (through or in contact with brick, mortar, concrete) and your uncertainties are "large" (or your demands on accuracy are "low") then Reynolds analogy _might_ be good enough, as it can be seen as a simplification of Chilton Colburn. But don't take my word for it. I come at this from a rather philosophical and general point of view and the devil usually lies in the details (as the saying goes)…
      (Interesting question by the way)

    • @mohammadkamran1926
      @mohammadkamran1926 3 года назад

      @@PLE_LU Got it Sir. Thank you very much for your valuable suggestions. I consider it as a learning session and I apologize for asking questions without clarity. I will try to describe my area of interest in few lines.
      As the standard textbooks and literature talks about the laminar flow over a flat plate, in my case, the flat plat has been replaced by a specimen of fully saturated brick. And like other materials, the drying kinetics of brick material also shows constant rate and falling rate periods. Now, I am able to estimate external mass transfer coeff. using constant rate data and interested to determine convective heat transfer coeff by using Chilton analogy.
      I hope you find it more interesting.

    • @PLE_LU
      @PLE_LU  3 года назад

      A, ok, now I understand better, thanks and good luck!

  • @jyotirdiptadas3989
    @jyotirdiptadas3989 4 года назад

    Sir, can you please tell me what are the assumptions of these two analogies

    • @PLE_LU
      @PLE_LU  4 года назад +2

      I answer that @5.46 in video and @6.56
      Remember "All models are wrong. Some models are useful" (G E P Box). If it proves to describe your system well, use it. If it doesn't, try something else

  • @J-Zone
    @J-Zone 6 лет назад

    Sir, can you please elaborate why Reynold's Analogy works better if laminar.

    • @PLE_LU
      @PLE_LU  6 лет назад +4

      The simple answer is: No, unfortunately I can't.
      The reason for my answer is that different literature sources disagree on a fundamental level: They disagree whether Reynolds Analogy is derived, valid and better suited for laminar flow or derived, valid and better suited for turbulent flow. (Yes, this is a bit surprising, especially that they disagree for what situation Reynolds Anology is _derived_)
      So (assuming you're taking a course), depending on what course material is used in a course, your teacher may want you to explain why Reynolds analogy is better for laminar flow or (if another text book is used) to explain why Reynolds analogy is better for turbulent flow. If you're in this unfortunate situation I can't really help you. Perhaps the course you're taking is given in a context where Reynolds analogy clearly is better suited for laminar flow or perhaps your teacher is unaware that different sources disagree or perhaps your teacher is more knowledgeable than me and knows for sure that some of these sources are incorrect…
      As G.E.P. box famously said: "All models are wrong. Some models are useful" Reynolds Analogy is not the truth about a system, it's a model. Whether or not that model is useful for your system will depend on your system. In the end. all you can do is do some measurements on your system and see if the model performance is adequate or not.

    • @KazeReload
      @KazeReload 4 года назад

      @@PLE_LU such a beautiful answer

  • @abhimaanpratap978
    @abhimaanpratap978 6 лет назад

    where is Q gone?

    • @PLE_LU
      @PLE_LU  6 лет назад +1

      It disappears since the volumetric flux is linked to the mass flux through
      Q=w/rho
      m3/s = kg/s * m3/kg

  • @shah-jahan198
    @shah-jahan198 2 года назад

    🤣 difficult

  • @chemengph2546
    @chemengph2546 6 лет назад

    Thank you!