Fluids

Fluids

Heat transfer coefficient

    • hugo CFD
      Subscriber

      Hello.


      I have some doubts that i would like to clarify. Consider a fixed plate heat exchanger air-air with quasi counter flow. The heat exchange is through forced convection.


      How should I calculate the coefficient of heat transfer? Directly from the fluent across the surfaces integrals --> area weighted average or through the formula h = q / at, where do I take the values of twall and tref and q from the fluent?


      The next question is about boundary conditions. when defining the boundary conditions, is it necessary to consider heat transfer by convection between the air and the plate, by selecting the CONVECTION point in the THERMAL conditions?


       


      Thanks

    • Raef.Kobeissi
      Subscriber

      Hello Hugo,


      Question 2 first: It depends on how you're simulating your case, if it is forced convection I would assume air flow is involved around the fixed plate HE and in this case you don't need to use convection. 


      Question1: I am assuming that you are using a solid wall(meshed) to simulate the HE, in that case you can calculate it through fluent as you mentioned. Can you attach the case file and I can show you how to do it through a small video or series of pictures.


      Regards

    • hugo CFD
      Subscriber

      Hello


      I am sending the study case. I have read that in certain situations, a UDF is required to calculate h. Am i correct? What are these situations?


       


      Regards

    • raul.raghav
      Subscriber

      For laminar flow in Ansys Fluent, the convective heat transfer coefficient (h) is calculated based on the equation:


      h = q / (T_wall - T_ref); where T_ref is the reference temperature you provide. You need to be careful providing the T_ref value, as that can result in the underestimation or overestimation of convective heat transfer coefficient (h) value. I'd highly recommend you to go through the following paper which compares the convective heat transfer coefficient obtained with T_ref value evaluated in different ways:


      Determination of surface convective heat transfer coefficients by CFD


      You can plot contours of the local heat transfer coefficients. And you can use "Surface integrals" to plot the average heat transfer coefficient (see image below):





      PS: I haven't checked any of the inputs or outputs in your case file, so I won't be commenting on the results.

    • Raef.Kobeissi
      Subscriber

      Hi Hugo, have you been able to resolve your issue?

    • hugo CFD
      Subscriber

      No. I am looking for another way to get heat transfer coefficient, instead to calculate by T ref method. 


      Hugo

    • raul.raghav
      Subscriber

      Hugo, as far as I'm aware of, heat transfer coefficient can either be calculated using the standard definition as I mentioned my previous post or by the Reynolds Analogy as turbulent heat transfer is modeled using the analogy.


      From the Reynolds analogy: h = (Cf/2) * density * U_ref * Cp


      where Cf is the skin friction coefficient, and U_ref is the reference velocity, which is difficult to be estimated. For turbulent flows, you can use wall functions to evaluate h:





       


       

    • hugo CFD
      Subscriber

      yes. you are right. maybe i didn´t express well.


      i considered follow expression to calculte heat transfer coefficient: h = q / ( T _ wall - T _ adj)


      q is heat flux and I get this value from fluent (surface integrals---wall flux--- total surface heat flux) referring to the boundary condition "heat transfer 1".


      T_wall is a temperature of the wall and i get this value from fluent referring to the boundary condition ""heat transfer 1-shadow".


      T_adj is the temperature of the first fluid layer following the wall referring to the boundary condition "heat transfer 1".


      My problem is how i get the values of the expression from fluent. I don t know if what i have considered is right.


       


      Hugo

    • raul.raghav
      Subscriber

      You can define a custom field function with the formula shown in the image below [Total Surface Heat Flux / (Wall Temperature - Static Temperature)]. Once you've created the function, use a contour plot to display the function. Make sure when you create the contour plot, you have the "Node Values" unchecked.






    • hugo CFD
      Subscriber

      Hi.


      I did what you explained me. Why i obtain two diferents values to HTC to the same fluid?  And one is negative, that is not possible.


      See images.



       



      hugo

    • raul.raghav
      Subscriber

      The heat transfer coefficient that is area-weighted average is the average heat transfer coefficient. The contour plot shows the local heat transfer coefficient. And the heat transfer coefficient depends on the fluid, hydrodynamic and thermal conditions and also on the geometry under consideration.


      I'm not sure why the coefficient is negative though. Again as I mentioned in my earlier posts, the way you calculate the coefficient makes a whole lot of difference.

    • hugo CFD
      Subscriber

      Hi.


      How can I obtain/calculate the local heat transfer coefficient along the length of the heat exchanger? And make a chart of this behavior?


       


      hugo

    • raul.raghav
      Subscriber
    • atulsingh92
      Subscriber

      @Raul.raghav Could you please explain in more detail, as to how, can one define this reference temperature? The thing is, this value, is good enough to make the nusselt number negative.

      So, lets say I am simulating for internal flow in a corrugated tube with periodic bc. (yes i have segregated solver and const heat flux bc as mentioned in Ansys manual). The reference temperature I have chosen, is by reverse calculating the nusselt number mentioned in an experimental paper. i.e the paper mentiones the fluid -water entering at 26 degrees C and then showed a nusselt of 65 for example. The q applied is 21100.


      This would leave me by htc= 21100 / (26 - x); Nu = 60 = htc * 0.00575 / 0.61; I enter this x as my reference temperature.


      But the question is , here, I knew my nusselt from an experimental data, what to give when I dont know this before hand
      Also, the way I mentioned, is it the correct way?

    • irineupetri
      Subscriber

      Hi.

      Follow the advice from colleagues above, but don't forget to consider the outer walls of your adiabatic system, setting the heat flux equal to zero (Q = 0) on the external walls.

      Good luck

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