Mass flow rate

Alternative equations

Mass flow rate can also be calculated by

$${\displaystyle \dot{m}=\rho \cdot \dot{V}=\rho \cdot \mathbf{v}\cdot \mathbf{A}={\mathbf{j}}_{\text{m}}\cdot \mathbf{A},}$$

where

• ${\displaystyle \dot{V}}$ or ${\displaystyle Q}$ = volume flow rate,
• ${\displaystyle \rho }$ = mass density of the fluid,
• ${\displaystyle \mathbf{v}}$ = flow velocity of the mass elements,
• ${\displaystyle \mathbf{A}}$ = cross-sectional vector area/surface,
• ${\displaystyle {\mathbf{j}}_{\text{m}}}$ = mass flux.

The above equation is only true for a flat, plane area. In general, including cases where the area is curved, the equation becomes a surface integral: $${\displaystyle \dot{m}={\iint }_A\rho \mathbf{v}\cdot d\mathbf{A}={\iint }_A{\mathbf{j}}_{\text{m}}\cdot d\mathbf{A}.}$$

The area required to calculate the mass flow rate is real or imaginary, flat or curved, either as a cross-sectional area or a surface, e.g. 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, ${\displaystyle A}$, and a unit vector normal to the area, ${\displaystyle \hat{\mathbf{n}}}$. The relation is ${\displaystyle \mathbf{A}=A\hat{\mathbf{n}}}$.

The reason for the dot product is as follows. The only mass flowing through the cross-section is the amount normal to the area, i.e. parallel to the unit normal. This amount is

${\displaystyle \dot{m}=\rho vA\cos \theta ,}$

where ${\displaystyle \theta }$ is the angle between the unit normal ${\displaystyle \hat{\mathbf{n}}}$ and the velocity of mass elements. The amount passing through the cross-section is reduced by the factor ${\displaystyle \cos \theta }$, as ${\displaystyle \theta }$ increases less mass passes through. All mass which passes in tangential directions to the area, that is perpendicular to the unit normal, doesn't actually pass through the area, so the mass passing through the area is zero. This occurs when ${\displaystyle \theta =\pi /2}$: $${\displaystyle \dot{m}=\rho vA\cos (\pi /2)=0.}$$ These results are equivalent to the equation containing the dot product. Sometimes these equations are used to define the mass flow rate.

Considering flow through porous media, a special quantity, superficial mass flow rate, can be introduced. It is related with superficial velocity, ${\displaystyle v_s}$, with the following relationship: $${\displaystyle {\dot{m}}_s=v_s\cdot \rho =\dot{m}/A}$$ The quantity can be used in particle Reynolds number or mass transfer coefficient calculation for fixed and fluidized bed systems.

Usage

In the elementary form of the continuity equation for mass, in hydrodynamics: $${\displaystyle {\rho }_1{\mathbf{v}}_1\cdot {\mathbf{A}}_1={\rho }_2{\mathbf{v}}_2\cdot {\mathbf{A}}_2.}$$

In elementary classical mechanics, mass flow rate is encountered when dealing with objects of variable mass, such as a rocket ejecting spent fuel. Often, descriptions of such objects erroneously invoke Newton's second law ${\displaystyle \mathbf{F}=d(m\mathbf{v})/dt}$ by treating both the mass ${\displaystyle m}$ and the velocity ${\displaystyle \mathbf{v}}$ as time-dependent and then applying the derivative product rule. A correct description of such an object requires the application of Newton's second law to the entire, constant-mass system consisting of both the object and its ejected mass.

Mass flow rate can be used to calculate the energy flow rate of a fluid: $${\displaystyle \dot{E}=\dot{m}e,}$$ where ${\displaystyle e}$ is the unit mass energy of a system.

Energy flow rate has SI units of kilojoule per second or kilowatt.