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I have a tiny reactor, and the volume I need is 425ml. If there is no/few other reactants, then the volume in hydraulic retention time is clear: reactor/container's volume. and hrt is V/u(V is 425ml, u is flow rate) But if I fill 80% of container with something like lime mud, the real volume of water is only 20% left -- 85ml. How is my V should be? 425 or 85ml? IOW, the volume in hydraulic retention time is of reactor/container or liquid?

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  • $\begingroup$ The usable volume.. ie that not filled by mud. $\endgroup$ – Solar Mike Apr 17 at 10:11
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Water volume is 85 ml now, so the hrt is 85 ml/u. The liquid retention time because the mud doesn't react, right?

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  • $\begingroup$ because the mud doesn't react IMO, it's not right, because the mud will release Ca2+ $\endgroup$ – 陳 力 Apr 18 at 1:22
  • $\begingroup$ Then you should consider the mud to calculate the hrt. $\endgroup$ – Dibujo de Croquis Apr 19 at 1:10

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