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enter image description hereI am doing some simulation for natural circulation loop and have a heat rate q in the heating leg and a fixed temperature for the cooling leg. I know the heat rate for the heating section (128 W) and I know the fixed temperature for the cooler (9.87 °C). The issue that I am facing is, how to computed wall heat flux for the cooler from the fixed temperature.

I do not know the thickness of the wall or the initial temperature for $\Delta T$ for the cooler. the usual equation for heat flux is $q = h \cdot \Delta T$ or $q = -k \frac{\Delta T }{dx}$ the heat transfer coefficient is also unknown.

can someone help with this please.

I am trying to mimic the experiment in this article using 1-d full conservation equations https://www.sciencedirect.com/science/article/abs/pii/S0017931097000070

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    $\begingroup$ Please include a drawing of your simulation geometry. $\endgroup$
    – Algo
    Oct 12, 2021 at 9:44
  • $\begingroup$ unless the system is heating up forever, the heat out must equal the heat in $\endgroup$
    – Tiger Guy
    Oct 12, 2021 at 13:19
  • $\begingroup$ @TigerGuy you seem to not understand what I meant. I added the natural circulation loop schematic for you. cooling and heating do not necessarily equal each other so the principle you mentioned is not what I am talking about. $\endgroup$
    – yamifm0f
    Oct 12, 2021 at 13:44
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    $\begingroup$ And thus the steady state heat out must equal the heat in. $\endgroup$
    – Tiger Guy
    Oct 12, 2021 at 17:02
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    $\begingroup$ If the system is not steady state then there is more information we need, like the starting temp of the fluid, the entry and exit temps of cooling, flow rates for both primary and cooling fluids. The only way for heat out to not equal heat in is for the system to be heating up or cooling down. With the data given, the best assumption is steady state, for which heat out must equal heat in. So the heat out through the cooling system needs to equal 128W. $\endgroup$
    – Tiger Guy
    Oct 12, 2021 at 18:13

2 Answers 2

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If you are looking just for heat flux (aka heat flux density, heat-flow density or heat flow rate intensity) and you know that the heat rate is 128 W, then the heat flux is just:

$$\phi_q = \frac{\dot{q}}{A}$$

where:

  • $\phi_q$ is the heat flux
  • $\dot{q}$: the heat rate in W
  • ${A}$ the surface area.
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  • $\begingroup$ yeah that would be good if I knew the heat rate W. I know the heat rate for the heating leg but I only know the cooling temperature for the cooler. in the equation that you mentioned only area is known. $\endgroup$
    – yamifm0f
    Oct 12, 2021 at 11:00
  • $\begingroup$ if the cool surface keeps a constant temperature 9.87 C, then the $\Delta T$ (which will be different for convection and conductive) will adjust itself so that it accommodates heat rate of 128W. Unless, I am not understanding something in your setup and your terms "cooling leg" and "heating leg". $\endgroup$
    – NMech
    Oct 12, 2021 at 11:03
  • $\begingroup$ so a natural circulation system contains one heating leg where you apply a heat flux or a heat rate and parallel to that leg or pipe if you will you have a cooling system (heat exchanger) this can be with a constant temperature where it is lower than the temperature in the heating section or a negative heat flux (-128 W). sciencedirect.com/science/article/abs/pii/S0017931097000070 I am trying to mimic this article experiment in 1-d unsteady state with full conservation equations. $\endgroup$
    – yamifm0f
    Oct 12, 2021 at 12:07
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The paper you linked states clearly that this is a steady state investigation of natural circulation of water near density extremes (density inversion), so as @Tiger Guy said in the comments, the heat in (heater) must equal the heat out (cooler).

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