Hello. The Intergate Variables filter just calculate the area integral of
each variable. If you need the mass flow then your workflow is correct. I
use the same approach:
1. Slice + coordinate thresholds if it's needed to bound the sloce.
2. Calculator filter with the function like
MassFlux=Density*VelocityVector*SurfaceNormalVector
3. Intergate Variables. The MassFlux integral is what we need.
Regarding the time dependence I cannot suggest you something because I
didn't work with transient cases in ParaView (I use CFD Post for
transients).
I also don't know the simple way to filter just one variable to integrate.
If you have lots of variables like in CFD results it's very inconvenient to
dig through all variables in Intagrate results.

2018-02-01 15:56 GMT+03:00 Andrew Roberts <and...@thevisualroom.com>:

> Hello,
>
> I have a function for mass flowrate that I want to plot over a time period
> DT. The velocity is time averaged per timestep dt, not the whole time
> period, DT. In otherwords the number of timesteps is DT/dt.
>
> mdot(T) = int_A (rho . u(T) . n . alpha(T)) dA
>
> mdot(T) is the massflow rate over time (kg/s)
> int_A () dA is the integral over the area of a slice through the domain
> (m^2)
> rho is the density of the particle phase and this is constant (kg/m^3)
> u(T) is the time averaged velocity of the particle phase for a timestep of
> dt and varies with time, T (m/s)
> n is the unit vector normal to the area (-)
> alpha is the time averaged volume fraction of the particle phase and also
> varies with time T (-)
>
> I think I am pretty sure that I can compute rho . u . n . alpha
>
> My process in paraview:
>
> Slice > Origin = 0, 0, 0, Normal = 0, 1, 0, Slice Type = Plane
>
> This is the part in the brackets of the massflow rate equation. Mean just
> means time averaged:
>
> Calculator > abs(U1Mean_Y)*2900*alpha1Mean
>
> Now I must integrate this over the area of the slice.
>
> *** Is it correct to use the Filter > IntegrateVariables? ***
>
> I then use PlotSelectionOverTime (which I'm pretty sure is right).
>
> Kind regards,
>
> Andrew
>
>
>
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