Momentum Transport?

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—

$$Q = \int_A \mathbf{v} \cdot \mathbf{n}\,dS$$

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.