[aero_head]

Hello Everyone,

I was looking for an explanation for the meaning of mass flow density. There is only one post on this website back from 2004 that is unanswered. I would like to re-ask it, this time with my hypothesis of what it would be at the bottom.

For reference, the Function Calculator in the Post-processor calculates the following:

Function: MassFlowAve Location: Inlet Variable: Particle.MassFlowDensity

What is it actually calculating? I looked in the CFX-Post User's Guide, but while it mentions it can calculate it, it does not provide a description.

The units of mass flow density are [kg m^-2 s^-1], mass flow rate is in [kg s^-1], and energy density is [kg m^-1 s^-2], so would it be something related to the energy the particles carry with them? From the energy density units, I see I would need another m^-1 s^1. If I were to use the mass flow rate, I would need a [m^-2].

[flotus1]

I don't see how energy density comes into this.
From your description, Particle.MassFlowDensity with the unit of is just that: the mass of particles passing through an area per second.
Plugging this quantity into a "mass flow average" function for a surface should not change the units. It calculates the average of the Particle.MassFlowDensity, weighted by mass flow rate.

[aero_head]

Originally Posted by flotus1

I don't see how energy density comes into this.
From your description, Particle.MassFlowDensity with the unit of is just that: the mass of particles passing through an area per second.
Plugging this quantity into a "mass flow average" function for a surface should not change the units. It calculates the average of the Particle.MassFlowDensity, weighted by mass flow rate.

Hello Alex,

Thank you very much for your answer, makes sense to me now. I was just trying to determine the physical meaning of mass flow density and was probably over-thinking it a bit.

[FMDenaro]

kg/(m^2*s) =(kg/m^3)*(m/s)





If I am right, the quantity could be seen as rho*u, that is a momentum quantity

[aero_head]

Originally Posted by FMDenaro

kg/(m^2*s) =(kg/m^3)*(m/s)





If I am right, the quantity could be seen as rho*u, that is a momentum quantity

Hello Professor,

Ah, right! This was more like what I was trying to figure out. Thank you for handing me this piece of the puzzle, I think it was what I was looking for.