Heat Transfer (24) - Flat plate convection heat transfer coefficients

Поделиться
HTML-код
  • Опубликовано: 7 янв 2025

Комментарии •

  • @emt4921
    @emt4921 2 года назад +10

    Thanks, CPP. I really hope that Dr Biddle finished this playlist before he retired. I'll be waiting for more videos.

    • @CPPMechEngTutorials
      @CPPMechEngTutorials  2 года назад +9

      He did. The remaining videos will be released in the next couple weeks.

  • @vicentereiman368
    @vicentereiman368 10 месяцев назад

    This professor is amazing! Thank you so much sir!!!

  • @chichorro1000
    @chichorro1000 2 года назад +3

    24/37 let´s go! thank for the content guys!

  • @cathysunny1804
    @cathysunny1804 2 года назад

    Thank you for keep updating this series!

    • @CPPMechEngTutorials
      @CPPMechEngTutorials  2 года назад +3

      You're welcome. Not even a pandemic could stop us from completing the series.

  • @ZalanK77
    @ZalanK77 Год назад +4

    One step is very painfully missing from this otherwise excellent presentation. To calculate the turbulent heat transfer over the plate h(x)=k*Nu(x)/x is to be used and that needs to be integrated and divided by the length. So, it becomes the mysterious h_avg=0.037*Pr^(1/3)*k*(rho/mu*v_inf)^4/5*L^(-1/5). In this the Reynolds number was already substituted by Re(x)=v_inf/nu*x=rho*v_inf/mu*x as it is a function of x. nu is the greek letter for kinematic viscosity, rho is the density and mu is the dinamic viscosity. Nu(x) is the Nusselt number at location x, h(x) is the heat transfer coefficient the location x.

    • @rzeeonthemic
      @rzeeonthemic Год назад +1

      What is v infinite?

    • @ZalanK77
      @ZalanK77 Год назад +1

      @@rzeeonthemic The velocity far away from the plate, so basically the velocity of the undisturbed fluid.

  • @gajendrasingh_384
    @gajendrasingh_384 Год назад +2

    Great lecture watching from India ❤❤

  • @CarlMichaelAAdame
    @CarlMichaelAAdame 2 года назад

    Thank you very much, cpp

  • @mohammadaiman4710
    @mohammadaiman4710 Год назад

    what is eqn for hx for turbulent?? The lecturer didn't show in this video. Can somebody help me.

    • @habibbolgiba8095
      @habibbolgiba8095 Год назад

      I believe he derived it from the Nusselt number, by making hx the subject, therefore multiplying Nu by K and dividing by x. Not sure thou

  • @willcolletti4357
    @willcolletti4357 Год назад

    he kept referring to NU bar as h bar, is there a reason for that

  • @isk3804
    @isk3804 2 года назад +3

    Dr. Biddle is an excellent instructor, however cameraman ruins it a lot!!! :(

    • @CPPMechEngTutorials
      @CPPMechEngTutorials  2 года назад +1

      The issue got fixed in future lectures.

    • @jeffsmith7247
      @jeffsmith7247 Год назад +2

      @@CPPMechEngTutorials you always say that. lol, Here I am in future lectures..... :/

    • @CPPMechEngTutorials
      @CPPMechEngTutorials  Год назад +1

      @@jeffsmith7247 Hang in there... I swear it gets better.

    • @jeffsmith7247
      @jeffsmith7247 Год назад +1

      @@CPPMechEngTutorials haha the last video better be a beauty

    • @CPPMechEngTutorials
      @CPPMechEngTutorials  Год назад +8

      @@jeffsmith7247 If it's not, you'll get a full refund.

  • @dilannemutlu3835
    @dilannemutlu3835 23 дня назад

    Bad