Fluids

Fluids

Calculating heat transfer rate in fluid cross section regions

    • Karthikkurinji
      Subscriber

      Hello all,

      I have this setup shown below

    • Karthik R
      Administrator
      Hi If you estimate the surface integral of the Total Surface Heat Flux, you should be able to obtain the net heat crossing these surfaces.
      Karthik
    • Karthikkurinji
      Subscriber
      Dear Karthik I tried that. But it shows it as zero. Moreover, I could not create the planes in the FLUENT tab. I can only create these in CFD-Post, but there again, I don't know how to find it because there is no Total heat flux in the function calculator. I have attached images of the panels here for a better understanding.

      Thanks again Karthik
    • Rob
      Ansys Employee
      In Fluent use iso-surfaces of mesh and select the fluid region when you create the surface. Repeat for the solid surface. You can then see the flux (suspect it's not available) but you will get the temperature values (mass weighted mean) along with flow rate to estimate the energy in the fluid and solid zones at each position.
    • Karthikkurinji
      Subscriber
      Hi Rob Thanks for introducing iso-surfaces. I did try that. I created an Iso-surface at the inlet itself. I extracted the mass-weighted average of the static temperature and also the mass flow rate. The values were :
      T_inlet = 300.09743 K
      m_dot = 0.003181 Kg/s
      Cp_water = 4000 J/KgK ( i have manually input it as 4000 in the material property)
      Q_dot = m_dot x Cp x Delta_T ( In my case, 12.8 J/Ks x Delta_T)
      Delta_T is the problem area.
      As you can see in the below-attached image(also attached the iso-surface values attained), I have the heat transfer at the inlet is 23.35 W. I do not understand how it can be that value.


      Thanks Karthik
    • Rob
      Ansys Employee
      The fluxes on the outer flow boundaries are relative to a reference value, if you use Q = m cp dT you should find the reference is around 298.15K.
    • Karthikkurinji
      Subscriber
      Oh Right. Thanks !
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