This post assumes that the reader has knowledge of basic vector calculus and physics.

In the study of fluid dynamics the term ‘momentum transport’ is thrown around quite often and often in conjunction with the idea of ‘momentum accumulation’; it is a rather difficult concept to understand because it is quite impossible to *see* momentum. Hopefully this post will help to clarify the concept.

We will start with an easy to visualize concept: ‘mass transport’. Let us assume that there is a laminar and steady fluid flow of density \(\rho\) and velocity \(\mathbf{v}\). We would like to know the rate \(K_T\) at which mass is transported in this flow. The answer—

\[ K_T = \rho |\mathbf{v}| \]Makes sense until you consider the units of \(K_T\) and realize that they are \(\frac{\text{g}}{\text{cm}^2 \cdot \text{s}}\) which definitely doesn't make much sense. In fact it is difficult to determine at this point what the correct units of the rate of mass transport should even be. To solve this problem we need to consider mass transport from the other side of the equation: ‘mass accumulation’.

If the total mass transport \(\sum K_T\) to a location is not equal to zero we should see accumulation or dispersion at that location. We can therefore characterize the rate at which mass is accumulating in terms of the mass transport rate such that—

\[ \sum K_T = \sum \frac{dm}{dt} \text{ so } K_T = \frac{dm}{dt} \] \[ \frac{dm}{dt} = \frac{dm}{dV} \frac{dV}{dt} = \rho\,Q\]Where \(m\) is mass, \(V\) is volume, and \(Q\) is the *volumetric flow rate* and is defined as—

Where \(A\) is the cross sectional surface through which the fluid flow occurs, \(\mathbf{n}\) is the unit outward vector on this surface, and \(dS\) is the surface element. If we then define the average velocity \(\left<v\right>\) as—

\[ \left<v\right> = \frac{1}{A} \int_A \mathbf{v} \cdot \mathbf{n}\,dS \]And assume that the cross sectional surface is constant we can the mass transport rate as—

\[ \sum K_T = A \sum \rho \left<v\right> \]Now we have the framework with which to address the issue of momentum transport. Let us assume the same fluid flow as above and describe the rate of momentum transport accumulation—

\[ \sum K_T = \sum \frac{d(m \mathbf{v})}{dt} \text{ so } K_T = \frac{d(m \mathbf{v})}{dt} \] \[ \frac{d(m \mathbf{v})}{dt} = \frac{d(m \mathbf{v})}{dV} \frac{dV}{dt} \] \begin{aligned} K_T &= \left(m \frac{d \mathbf{v}}{dV} + \mathbf{v} \frac{dm}{dV}\right) \frac{dV}{dt} \\ &= m \frac{d \mathbf{v}}{dt} + \mathbf{v} \frac{dm}{dV} \frac{dV}{dt} \end{aligned}We now make the same substitution that we made before and take into account that force \(\mathbf{F}\) and acceleration are \(\mathbf{a}\) then—

\[ \mathbf{F} = m \mathbf{a} \text{ where } \mathbf{a} = \frac{d \mathbf{v}}{dt} \] \[ \sum K_T = \sum \mathbf{F} + A \sum \rho \left<v\right> \mathbf{v} \]This result is typically interpreted as that the rate of accumulation of momentum is equal to the sum of the momentum transport rate and the net external force on the system.

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