Fluids

Fluids

Nusselt Number in Triple Concentric Tube Heat Exchangers

    • Christiantyo Moeprodjo
      Subscriber

      I am currently conducting a research study on optimising Triple Concentric Tube Heat Exchangers. To further validate my research, I am comparing the Nusselt number variations I obtain from my Fluent simulations to the experimental data obtained by a researcher named Gomaa. 

      My method of calculating the Nusselt number is by extracting the inlet and outlet temperatures of the 3 pipes directly from Fluent, and using heat transfer equations to manually calculate the Nusselt number.

      My main issue is that my Nusselt number variation is rather minimal compared to Gomaa's values. I am not too sure why this is the case.

       

      Nusselt Number variation I am currently getting:

      Nusselt number variation Gomaa obtained: 

      Any suggestions on different approaches that I could use to accurately extract Nusselt number values? Perhaps directly extracting it from Fluent or perhaps other alternative approaches?

      Many thanks,

      Chris

       

       

    • Rob
      Ansys Employee

      How does the heat flux compare between the publication and Fluent? How well refined is the near wall mesh? How did the experiment measure the outlet temperature? 

      The most common cause of disagreement with experimental data is monitoring/measuring different things. The second most is misusing the reference values in Fluent, but as you're taking inlet & outlet temperatures that shouldn't be possible. 

    • Christiantyo Moeprodjo
      Subscriber

      Thank you very much for your response!

      How does the heat flux compare between the publication and Fluent?

      Heat Flux is not necessarily measured in the experiment, but rather calculated using Q_dot=m_dot * Cp * deltaT. However, the paper does not outline the obtained experimental outlet temperatures. So, I am not too sure how I would approach this. Now with regards to my heat flux measured in Fluent, I followed the link below which was issued by Ansys to measure my heat flux.

      https://courses.ansys.com/index.php/courses/how-heat-exchangers-work/lessons/simulation-examples-homework-and-quizzes-how-heat-exchangers-work/topic/heat-transfer-in-a-shell-and-tube-heat-exchanger-simulation-example/

      This then gives me the Total Heat Transfer Rate at the inlet and outlet of the intermediate tube. If I was to use Q_dot=m_dot * Cp * deltaT, and measure the outlet temperature directly in Fluent I would get similar results to the Total Heat Transfer Rate value.

      Total Heat Transfer Rate:

      Q_dot=m_dot * Cp * deltaT:

      Q_dot = 0.09 [kg/s] * 4190 * (343 - 326.5) = 6278.25

       

      How well refined is the near wall mesh?

      I used edge sizing on the pipe edge at the inlet and outlet. This is the only specific mesh sizing method I used.

      How did the experiment measure the outlet temperature?

      The inlet and outlet temperatures were measured using six shelled precalibrated K-type thermocouples. All thermocouples were connected via a pre-calibrated data acquisition unit (accuracy ±0.01 degC) through computer software.

    • Rob
      Ansys Employee

      OK, for mesh resolution near the walls you want to look at inflation. Read up on combining inflation with swept meshes; whilst it's easy it's not immediately obvious how it works. 

      Inlet and outlet temperatures in Fluent mass weighted average on the surface? Or a point value more akin to the experiment? How do you differentiate from heat coming from a fluid zone of smaller annular diameter against from a higher annular diameter? 

    • Christiantyo Moeprodjo
      Subscriber

       

      I will try and look into your suggestion on inflation and swept mesh configurations! Thank you for this.

       

      Inlet and outlet temperatures in Fluent mass-weighted average on the surface? Or a point value more akin to the experiment?

      I have been trying to obtain the bulk temperature values at the inlet and outlet; however, I never managed to get this to work. So instead I placed a line probe which spans from the inlet to the outlet at the centre of the annulus I am interested in.

      I just tried using the Fluent mass-weighted average of temperature on the outlet surface, and it was giving me 54degC. The outlet temperature I obtained using the line probe method was 53degC.

       

      How do you differentiate from heat coming from a fluid zone of smaller annular diameter against from a higher annular diameter? 

      Massive apologies in advance, but I am unsure of what you meant here. Could you please elaborate?

       

    • Rob
      Ansys Employee

      OK. The line will give a "length" average: this assumes the whole area is moving at the same speed. A mass average accounts for the flow rate of material, ie a faster moving core of the flow may be warmer/cooler than the rest so correctly accounts for the amount of thermal energy in the flow. 

      No worries. I couldn't work out if you had two or three fluid annular flow paths. With three, how do you account for heat entering the middle one from the outer as compared to the inner? 

      • Christiantyo Moeprodjo
        Subscriber

        No worries, that means I'll just use the mass-weighted average method when extracting the outlet temperature from now on. Thanks for this!

         

        Apologies if my problem definition is unclear. I have three fluid annular flow paths as there are three channels within the triple concentric tube heat exchanger. Hopefully, the figure below would provide a better description of the system.

        The system setup is as follows:

        • Hot Fluid flows through the middle tube (70degC) --> main heat transfer fluid
        • Cold fluid flows through the inner tube (10degC)
        • Normal fluid flows through the outer tube (18degC)

        The aim of this project is to show that a triple tube heat exchanger has better heat transfer properties compared to the conventional double tube heat exchanger due to its larger heat transfer area. In a double tube heat exchanger, heat is transferred from the hot fluid to the cold fluid via convection and conduction through the pipe wall. On the other hand, in a triple tube heat exchanger, the hot fluid flows through the intermediate tube. This allows the heat to be transferred to the normal temperature fluid in the outer tube, and to the cold temperature fluid in the inner tube. This way, the heat transfer rate would be increased due to the increased heat transfer area. 

        Hence, to answer your question of how do you account for heat entering the middle one from the outer as compared to the inner? I am not too sure if this would be an area of concern of mine. Theoretically, heat would only be transferred from the middle tube to the outer and inner tubes. If there is heat transferred from the lower temperature medium to the higher temperature medium, I think this would be relatively minimal and hence could be neglected. Please do correct me if my understanding is incorrect.

         

    • Rob
      Ansys Employee

      Heat will transfer from middle to inner & middle to outer. Outer may also lose heat to outside depending on the boundary condition you set. Heat flux through the wall(s) will tell you what goes where. I assume you don't reach a point where the flow temperature is equalised between any of the tubes? 

      • Christiantyo Moeprodjo
        Subscriber

        What boundary condition should I apply to the outer tube so that the outer surface of the heat exchanger is insulated (to minimise convective heat loss to the surroundings)?

        No, the flow temperatures do not reach a point where they are equalised.

    • Rob
      Ansys Employee

      Heat flux = 0 W/m2   is the adiabatic condition, ie perfectly insulated. 

      • Christiantyo Moeprodjo
        Subscriber

         

        I will make sure to input this boundary condition!

        Coming back to the heat transfer rate issue, is my method of extracting the Total Heat Transfer Rate accurate enough?

        I have been reading this forum thread https://forum.ansys.com/forums/topic/calculating-heat-transfer-rate-in-fluid-cross-section-regions/ where a suggestion of using iso-surfaces was brought up, but I never knew how to get this to work in Fluent. Would this be applicable to my application, and if so how should I go about implementing this in my model? Are there perhaps any other alternative methods of extracting the heat transfer rate?

         

    • Rob
      Ansys Employee

      I'd use the heat flux through the wall: it's in the Reports section. You already have a surface! 

      • Christiantyo Moeprodjo
        Subscriber

         

        Hi, 

         

        Apologies for getting back to you this late. I have just been working away with the simulation.

        I am now measuring the Area-Weighted Average of the Total Surface Heat Flux from the pipe walls of interest; however, I am uncertain as to what this total surface heat flux indicates. Does the total surface heat flux in Fluent indicate the convective heat transfer from the fluid to the pipe wall, or does it indicate the total heat transfer (convective + conductive heat transfer) generated from the fluid to the pipe wall?

        Furthermore, I am also measuring the Area-Weighted Average of the Surface Heat Transfer Coefficient on the pipe wall that is in contact with the hot fluid. Does that surface heat transfer coefficient indicate the convective or conductive heat transfer coefficient of the pipe wall? Apologies if this question is rather trivial.

         

        Many thanks, 

        Chris

         

    • Rob
      Ansys Employee

      The flux is ALL heat passing through that facet, note it should have a sign and it is possible for postive and negative numbers to exist on the same surface. Surface HTC is defined in the manual, https://ansyshelp.ansys.com/account/Secured?returnurl=/Views/Secured/corp/v222/en/flu_ug/flu_ug_fvdefs.html  and given it's based on "q" I would expect it to contain the convective and conductive parts. 

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