Sometimes the defining equation for mass flux in this article is used interchangeably with the defining equation in mass flow rate. For example, Fluid Mechanics, Schaum's et al uses the definition of mass flux as the equation in the mass flow rate article.
is the mass current (flow of mass m per unit time t) and A is the area through which the mass flows.
For mass flux as a vector jm, the surface integral of it over a surfaceS, followed by an integral over the time duration t1 to t2, gives the total amount of mass flowing through the surface in that time (t2 − t1):
The area required to calculate the flux is real or imaginary, flat or curved, either as a cross-sectional area or a surface.
For example, for substances passing through a filter or a membrane, the real surface is the (generally curved) surface area of the filter, macroscopically - ignoring the area spanned by the holes in the filter/membrane. The spaces would be cross-sectional areas. For liquids passing through a pipe, the area is the cross-section of the pipe, at the section considered.
The vector area is a combination of the magnitude of the area through which the mass passes through, A, and a unit vector normal to the area, . The relation is .
If the mass flux jm passes through the area at an angle θ to the area normal , then
where · is the dot product of the unit vectors. That is, the component of mass flux passing through the surface (i.e. normal to it) is jm cos θ, while the component of mass flux passing tangential to the area is jm sin θ, but there is no mass flux actually passing through the area in the tangential direction. The only component of mass flux passing normal to the area is the cosine component.
Consider a pipe of flowing water. Suppose the pipe has a constant cross section and we consider a straight section of it (not at any bends/junctions), and the water is flowing steadily at a constant rate, under standard conditions. The area A is the cross-sectional area of the pipe. Suppose the pipe has radius r = 2 cm = 2 × 10−2 m. The area is then
To calculate the mass flux jm (magnitude), we also need the amount of mass of water transferred through the area and the time taken. Suppose a volume V = 1.5 L = 1.5 × 10−3 m3 passes through in time t = 2 s. Assuming the density of water is ρ = 1000 kg m−3, we have:
(since initial volume passing through the area was zero, final is V, so corresponding mass is m), so the mass flux is
Substituting the numbers gives:
which is approximately 596.8 kg s−1 m−2.
Equations for fluidsEdit
Using the vector definition, mass flux is also equal to:
ρ = mass density,
u = velocity field of mass elements flowing (i.e. at each point in space the velocity of an element of matter is some velocity vector u).
Sometimes this equation may be used to define jm as a vector.
Mass and molar fluxes for composite fluidsEdit
In the case fluid is not pure, i.e. is a mixture of substances (technically contains a number of component substances), the mass fluxes must be considered separately for each component of the mixture.
When describing fluid flow (i.e. flow of matter), mass flux is appropriate. When describing particle transport (movement of a large number of particles), it is useful to use an analogous quantity, called the molar flux.
Using mass, the mass flux of component i is
The barycentric mass flux of component i is
where is the averagemass velocity of all the components in the mixture, given by
ρ = mass density of the entire mixture,
ρi = mass density of component i,
ui = velocity of component i.
The average is taken over the velocities of the components.
If we replace density ρ by the "molar density", concentrationc, we have the molar flux analogues.
The molar flux is the number of moles per unit time per unit area, generally:
So the molar flux of component i is (number of moles per unit time per unit area):
and the barycentric molar flux of component i is
where this time is the averagemolar velocity of all the components in the mixture, given by: