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Consider an expansion channel, where isothermal incompressible flow enters the domain from the smaller channel.enter image description here

Assume I inject infinitesimal particles (with same density as the flow) into the domain from the red area, and let them move with the streamlines. I think the scalar quantity representing the volume fraction of these particles is evolved by: $\partial v_f / \partial t + div (v_fu) = 0 $ where $u$ is the flow velocity. The boundary condition of $v_f$ at inlet (red area) is $1$.

Will the domain be filled with red points after sufficiently long time and the scalar field $v_f$ will be $1$ in the whole domain?

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  • $\begingroup$ please show the derivation of your equation. $\endgroup$
    – Chet Miller
    Commented Jan 6 at 12:25
  • $\begingroup$ It is the continuity equation. $\endgroup$
    – Denis
    Commented Jan 6 at 13:57
  • $\begingroup$ In the continuity equation, v is specific volume, not volume fraction, $\endgroup$
    – Chet Miller
    Commented Jan 6 at 14:32

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If the red is an inert tracer, there will be no movement of red across the streamlines (which you have indicated on your diagram). So the red will alway be sandwiched between the upper wall and the lower streamline of the injected red. The only way you are going to have the red spread is if there is diffusion of the tracer relative to the "solvent" carrier fluid.

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  • $\begingroup$ Thank you. If the red is the tracer, shouldn't the volume fraction of red be same in every unit volume around the domain? Assume we trace 10 particles between two streamlines which are injected "continuesly". When the streamlines get closer, does the concentration/volume fraction of these particles change? Because if the density is assumed to be constant and same as the fluid, in every unit volume the mass must remain constant. $\endgroup$
    – Denis
    Commented Jan 7 at 14:23
  • $\begingroup$ I agree. The volume fraction and mass fraction of tracer does not change within the streamline domain into which they were placed. $\endgroup$
    – Chet Miller
    Commented Jan 7 at 16:48

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