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In a diverging pipe section like the following,

enter image description here

the pipe of radius $r$ splits into two pipes of radius $r/2$.

Consider a solute transported by convection from node 1. $$\frac{\partial C}{\partial t} = - v\frac{\partial C}{\partial x}$$

According to Gauss divergence theorem,

"The flux out of each region is equal to the sum of the flux through its two faces"

Therefore, would it be right to interpret $$\frac{\partial C}{\partial t}_\text{at point 2} = - (1/delx)[v_{12}C_{12}- v_{24}C_{24} - v_{23}C_{23}] \hspace{0.5cm} (2)$$

where $v$ and $c$ are the velocity and average concentration in the respective pipe section, respectively.

I am confused because, if the fluid is incompressible $$\nabla .v = 0; $$

This implies incoming flow is equal to the outgoing flow.

The continuity equation is in terms of flow and in the solute transport equation the products, $vC$, give flux in and flux out (in terms of moles/area/s).

In some references that I looked up, the fluxes in the solute transport equation are written in terms of moles/s. So, I am confused whether flux should be expressed in unit area i.e moles/area/time or moles/time while modeling flows through varying volumes.

Should the transport equation itself be expressed in terms of moles/time instead of moles/time/volume?

EDIT:

The condition for continuity is the following, $ A_{12}v_{12}C_{12} = A_{24}v_{24}C_{24} + A_{23}v_{23}C_{23} $

Could someone explain how the above condition can be incorporated in the discretized version of the solute transport equation?

Multiplying equation (2) by area changes the units of RHS to moles/length/time, whereas the units of LHS is in moles/volume/time. I am not sure how this has to be handled so that the units are the same on both sides.

Also, I couldn't completely understand the implication of Gauss Divergence theorem.In the sense, from what I understand the gauss-divergence talks about flux (per unit moles/area/time) that is entering and leaving a control volume element. However, here we are looking at the flux ( in terms of moles/time) entering and leaving the control volume (node 2).

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The condition that has to be satisfied at the node point 2 is that the influx equals the outflux. Here, this will then be $$ v_{12}C_{12} = v_{24}C_{24}+v_{23}C_{23}, $$ but $C_{ij}$ is not the average concentration in each pipe: It is the concentration at end point 2 of the pipes.

If the velocities are constant in time, then you can of course rewrite this as follows in time increment form: $$ v_{12}\frac{dC_{12}}{dt} = v_{24}\frac{dC_{24}}{dt}+v_{23}\frac{dC_{23}}{dt}. $$ Furthermore, since there are no sources or sinks in your system, it makes sense that each time, $C_{12}=C_{23}=C_{24}$, and then the mass conservation equation is simply $$ v_{12} = v_{24}+v_{23}. $$

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  • $\begingroup$ Thanks a lot. My confusion here is, since the pipes are of different volumes, should the influx = outflux condition be $v_{12}C_{12}A_{12} = v_{24}C_{24}A_{24} + v_{23}C_{23}A_{23}$ where C is concentration per unit volume. And could you please explain why $C_{ij}$ has to be $C_{j}$( could it be the concentration at end point 1 of each point instead of end point 2)? $\endgroup$
    – Natasha
    Nov 22 '19 at 2:54
  • $\begingroup$ For instance, while solving for the flow through a diverging pipe section, we always write $A_{12}v_{12} = A_{24}v_{24} + A_{23}v_{23} $ if the fluid is incompressible. $\endgroup$
    – Natasha
    Nov 22 '19 at 3:21
  • $\begingroup$ Ah, yes, I interpreted $v_{ij}$ as a flux rate, not a velocity. The flux rate is the average velocity times the cross section. $\endgroup$
    – Wolfgang Bangerth
    Nov 22 '19 at 16:03
  • $\begingroup$ I am sorry, $v_{ij}$ is the velocity. Could you please explain why $C_{ij}$ has to be $C_{j}$( could it be the concentration at endpoint 1 of each edge instead of endpoint 2) ? Also, could you please have a look at the edit made in the original post? $\endgroup$
    – Natasha
    Nov 23 '19 at 3:04
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    $\begingroup$ What I'm saying is that at a junction, the concentration at the end of one incoming pipe must equal the concentration at the start of each outgoing pipe. You would not preserve mass if that were not the case. But it's also the case because your concentration profile must be continuous (except, possibly, for individual points in time, but let's ignore this complication for a moment). $\endgroup$
    – Wolfgang Bangerth
    Nov 23 '19 at 3:56

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