Continuity and Momentum Balance Equations

  • In the previous lecture, we discussed the concept of differential balances , and derived the general balance equation for a property $B$ \begin{align*} \frac{\partial c_B}{\partial t} = -\nabla\cdot \boldsymbol{J}_B + \sigma_B \end{align*}
  • In this lecture we're going to generate two important balance equations from this general equation, the continuity equation (mass) and the Cauchy momentum equation.
  • Using these two equations, we will solve complex flow problems over the following weeks.
\begin{align*} \frac{\partial c_B}{\partial t} = -\nabla\cdot \boldsymbol{J}_B + \sigma_B \end{align*}
  • The simplest property to study using the general balance equation is the transport of mass ($B=mass$ ).
  • The concentration of mass in a fluid is simply the density, $c_{mass}=\rho$ .
  • When a volume of fluid flows, it has a velocity $\boldsymbol{v}$ ; therefore, using the density we can calculate the convective mass flux, $\boldsymbol{J}_{mass}=\rho \boldsymbol{v}$ . 1
  • Excluding nuclear processes, there is no generation or loss of mass in most systems, ($\sigma_{mass} = 0$ ).
  • Thus, the differential balance for the mass is \begin{align*} \frac{\partial c_{mass}}{\partial t} &= -\nabla\cdot \boldsymbol{J}_{mass} + \cancelto{0}{\sigma_{mass}}\\% \frac{\partial \rho}{\partial t} &= -\nabla\cdot \rho \boldsymbol{v} \end{align*}
1 There is also a diffusive mass flux, but we can ignore it in pure fluids where diffusion processes do not apply. We will add this back in later when studying mixtures in diffusion.
\begin{align*} \frac{\partial \rho}{\partial t} + \nabla\cdot \rho \boldsymbol{v}= 0 \end{align*}
  • This is a well-known equation in fluid dynamics called the continuity equation .
  • On its own, the continuity equation is not immediately useful; however, it will help us develop the Cauchy momentum equation!
  • It does produce one simple result which has applications in many liquids…
\begin{align*} \frac{\partial \rho}{\partial t} = -\nabla\cdot \rho \boldsymbol{v} \end{align*}
  • If the fluid is incompressible (constant $\rho$ ) the equation reduces to \begin{align*} \nabla\cdot \boldsymbol{v} = 0 \end{align*} I'll prove this now using index notation.
  • This is an important result in fluid flow, the so-called divergence of the velocity field ( $\nabla\cdot \boldsymbol{v}$ ) is zero in incompressible fluids.
  • Many liquids (e.g., water) are incompressible to a close approximation, so we can use this result to simplify many flow problems.
  • We will now derive the Cauchy momentum equation to move towards solutions of flow problems.
  • Starting from the general balance equation for a property $B$ : \begin{align*} \frac{\partial c_B}{\partial t} = -\nabla\cdot \boldsymbol{J}_B + \sigma_B \end{align*}
  • Setting $B=momentum$, we must determine what the concentration ($c_{momentum}$), flux ($J_{momentum}$), and generation of momentum ($\sigma_{momentum}$) terms are.
  • You should note that, in general, the momentum is a vector quantity ($c_B\equiv\boldsymbol{c}_{momentum}$ ).
  • Therefore, the concentration and source terms must be vectors too, and the flux must be a matrix ($\nabla\cdot\boldsymbol{J}_{momentum}$ must give a vector result $\therefore$ $\boldsymbol{J}_{momentum}$ is a tensor).
  • Technically, we will be generating three differential balance equations, one for each component of the momentum. \begin{align*} \frac{\partial \boldsymbol{c}_{momentum}}{\partial t} = -\nabla\cdot \boldsymbol{J}_{momentum} + \boldsymbol{\sigma}_{momentum} \end{align*}
  • This derivation is quite involved and long; however, it is invaluable as it will help you to understand the nature of all of the terms involved.
\begin{align*} \frac{\partial \boldsymbol{c}_{momentum}}{\partial t} = -\nabla\cdot \boldsymbol{J}_{momentum} + \boldsymbol{\sigma}_{momentum} \end{align*}
  • The momentum is simply mass$\times$velocity.
  • Velocity is an intensive property, it does not directly depend on the amount of fluid present.
  • Mass is an extensive property, it directly depends on the amount of fluid present. This means it has a concentration ($c_{mass}=\rho$).
  • Therefore, the concentration of momentum (or the momentum per unit volume) is the density$\times$velocity, or $\boldsymbol{c}_{momentum}=\rho \boldsymbol{v}$.
\begin{align*} \frac{\partial \rho \boldsymbol{v}}{\partial t} = -\nabla\cdot \boldsymbol{J}_{momentum} + \boldsymbol{\sigma}_{momentum} \end{align*}
  • For now we will ignore the flux of momentum and skip to the generation of momentum, $\boldsymbol{\sigma}_{momentum}$.
  • When momentum is “generated” in a system, it is actually due to external forces acting on the system, causing an acceleration1.
  • By comparing Newton's equation of motion ( $\boldsymbol{F}=m \boldsymbol{a}$ ) to the generation term: \begin{align*} \left(\frac{\partial \rho \boldsymbol{v}}{\partial t}\right)_{\sigma} = \boldsymbol{\sigma}_{momentum} \end{align*} we can see that $\boldsymbol{\sigma}_{momentum}$ has units of force per unit volume.
  • The simplest force to consider is gravity: \begin{align*} \boldsymbol{\sigma}_{momentum}^{grav.} = \rho \boldsymbol{g} \end{align*}
1 Momentum is never generated, but we occasionally ignore the equal and opposite force, such as the acceleration of the earth caused by smaller falling object on its surface.
  • The force due to pressure differences is derived here (it can be viewed as a source term, but over the volume momentum is conserved by this force)…
  • Performing a pressure balance over the volume, we can determine the pressure contribution:
    Pressure forces in the $z$-dimension omitted for clarity. Please notice that the pressure is a scalar! The force pressure applies is always to contract the volume (inward direction).
Each force acts on one face of the control volume, so the total force is the pressure multiplied by the area: \begin{align*} \Delta x \Delta y \Delta z \boldsymbol{\sigma}_{momentum}^{press.} &= \left[\begin{matrix}\left(p(x,y,z) - p(x+\Delta x,y,z)\right)\Delta y  \Delta z \\ \left(p(x,y,z) - p(x,y+\Delta y,z)\right) \Delta x  \Delta z \\ \left(p(x,y,z) - p(x,y,z+\Delta z)\right)\Delta x  \Delta y \end{matrix}\right] \end{align*} Dividing by $\Delta x \Delta y \Delta z$ and taking the limit as $\Delta x\to0$ , $\Delta y\to0$ ,and $\Delta z\to0$ we obtain: \begin{align*} \boldsymbol{\sigma}_{momentum}^{press.} = -\left[\begin{matrix} {\partial p}/{\partial x} \\ {\partial p}/{\partial y} \\ {\partial p}/{\partial z} \end{matrix}\right] = -\nabla p \end{align*}
\begin{align*} \frac{\partial \rho \boldsymbol{v}}{\partial t} = -\nabla\cdot \boldsymbol{J}_{momentum} - \nabla p + \rho \boldsymbol{g} \end{align*}
  • We've now determined the concentration and “generation” terms, now on to the more complicated flux terms.
  • As with the mass, the flux of momentum consists of a convective term and a diffusive term. \begin{align*} \boldsymbol{J}_{momentum} = \boldsymbol{J}^{conv.}_{momentum} + \boldsymbol{J}^{diff.}_{momentum} \end{align*}
  • The convective momentum flux is the flux of $B$ due to the bulk movement of the fluid (analogous to the convective mass flux). \begin{align*} \boldsymbol{J}^{conv.}_{momentum} = \left(\rho \boldsymbol{v}\right)\boldsymbol{v} \end{align*}

Math Notes: Dyadic product

\begin{align*} \boldsymbol{a} \boldsymbol{b} = \begin{Bmatrix} a_x b_x & a_x b_y & a_x b_z \\ a_y b_x & a_y b_y & a_y b_z \\ a_z b_x & a_z b_y & a_z b_z \end{Bmatrix} = a_i b_j \end{align*}
  • If we take all of the terms that we've derived so far, we have: \begin{align*} \frac{\partial \rho \boldsymbol{v}}{\partial t} = -\nabla\cdot \left(\rho \boldsymbol{v} \boldsymbol{v}\right) -\nabla\cdot\boldsymbol{J}^{diff.}_{momentum} - \nabla p + \rho \boldsymbol{g} \end{align*}
  • But what is the diffusive flux? It didn't appear in the continuity equation as we're not considering mixtures (yet) but diffusion in mixtures will be introduced later.
  • The diffusion of momentum in fluids arises from viscosity. Viscosity is the fluid analogue of friction, so imagine “blocks” of fluid sliding past each other:
  • Block $C$ has a stress (force per unit area) pulling it forward from block $B$, and a stress pulling it back from block $D$.
  • The net effect of these stresses is $\tau_{xy}$.
  • $\tau_{xy}$ is the stress in the $x$-direction caused by a difference in velocities in the $y$-direction.
  • The diffusive momentum flux $\boldsymbol{J}_{momentum}^{diff.}$ contains the stresses on the fluid due to viscous forces.
  • We can make a definition and state that the diffusive momentum flux is equal to the stress tensor, $\boldsymbol{\tau}$. \begin{align*} \boldsymbol{J}_{momentum}^{diff.} \equiv \boldsymbol{\tau}= \left[\begin{matrix} \tau_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \tau_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \tau_{zz} \end{matrix}\right] = \tau_{ij} \end{align*}
  • If an element of fluid is moving in the $x$-direction, it will:
    • Drag the fluid behind/ahead of it, $\tau_{xx}$.
    • Drag the fluid above/below it (y-direction), $\tau_{xy}$.
    • And drag the fluid either side of it (z-direction), $\tau_{xz}$.
  • You should know from previous courses, that for a Newtonian fluid under 1D shear, the off-diagonal stress is related to the velocity difference by Newton's “law” \begin{align*} \tau_{xy} = -\mu \frac{\partial v_x}{\partial y} \end{align*}
  • But we must generalise to 3D systems, where shear can occur in any direction to give \begin{align*} \tau_{xy} = -\mu \left(\frac{\partial v_x}{\partial y} + \frac{\partial v_y}{\partial x}\right) \end{align*}
  • But what about the bulk terms $\tau_{xx}$, $\tau_{yy}$, $\tau_{zz}$?
  • The bulk terms account for viscous effects when the fluid is decelerating/compressing or accelerating/expanding.
  • Here there is another coefficient to consider known as the volume , bulk or second viscosity, $\mu^B$ .
  • Our final expression for the stress in a Newtonian fluid is (in index notation) \begin{align*} \boldsymbol{\tau}=\tau_{ij} = -\mu \left(\frac{\partial v_i}{\partial r_j} + \frac{\partial v_j}{\partial r_i}\right) + \delta_{ij} \mu^B \nabla\cdot\boldsymbol{v} \end{align*}

    Math Notes

    $\delta_{ij}$ is the Kronecker delta function and is defined as follows \begin{align*} \delta_{ij} = \begin{cases} 1 & \text{if $i=j$} \\ 0 & \text{if $i\neq j$} \end{cases} \end{align*}
  • Note: If the flow is incompressible, we have $\nabla\cdot\boldsymbol{v}=0$ and the bulk viscosity is not needed (a common assumption).
  • Including the diffusive flux for momentum, the complete momentum equation is: \begin{align*} \frac{\partial \rho \boldsymbol{v}}{\partial t} = -\nabla \cdot \left(\rho \boldsymbol{v} \boldsymbol{v}\right) - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g} \end{align*}
  • But we can simplify this equation a little using the continuity equation.
  • Using the normal product rule on the left side and the vector product rule, $\nabla \cdot \boldsymbol{a}\boldsymbol{b} = \boldsymbol{b}\nabla \cdot \boldsymbol{a} + \boldsymbol{a}\cdot\nabla \boldsymbol{b}$ (see tutorial 1) on the first right hand term: \begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} +\boldsymbol{v}\frac{\partial \rho}{\partial t} &= -\boldsymbol{v} \nabla\cdot \left(\rho \boldsymbol{v} \right) -\rho\boldsymbol{v}\cdot \nabla \boldsymbol{v} - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g} \end{align*}
  • We can then rearranging the equation to express and cancel the continuity equation: \begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} + \boldsymbol{v}\cancelto{0}{\left(\frac{\partial \rho}{\partial t} + \nabla\cdot \left(\rho \boldsymbol{v} \right)\right)} &= -\rho\boldsymbol{v}\cdot \nabla\boldsymbol{v} - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g}\\ \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\rho\boldsymbol{v}\cdot \nabla\boldsymbol{v} - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g} \end{align*}
\begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\rho\boldsymbol{v}\cdot \nabla\boldsymbol{v} - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g} \end{align*}
  • At last! We've finally derived the Cauchy momentum equation.
  • Although this equation is new to you, it should be at least slightly familiar from your use of a version of it.
  • If we assume $\boldsymbol{\tau}=0$, the equation becomes the equation of motion for inviscid flow.
  • What other equation do you know that applies to fluids without viscous losses?
\begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\rho\boldsymbol{v}\cdot \nabla\boldsymbol{v} - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g} \end{align*}
  • In brief, if we take the following vector calculus identity (proof here): \begin{align*} \boldsymbol{a}\cdot\nabla\boldsymbol{a} &= \frac{1}{2} \nabla \boldsymbol{a}^2 - \boldsymbol{a} \times \left[\nabla\times\boldsymbol{a}\right] \end{align*} and assume inviscid flow (no viscosity so $\boldsymbol{\tau}=\boldsymbol{0}$), we can obtain the following from the Cauchy momentum equation: \begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\frac{1}{2}\rho\,\nabla\,\boldsymbol{v}^2 -\rho\,\boldsymbol{v}\times\left[\nabla\times\boldsymbol{v}\right] - \nabla p + \rho \boldsymbol{g} \end{align*}
\begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\frac{1}{2}\rho\,\nabla\,\boldsymbol{v}^2 -\rho\,\boldsymbol{v}\times\left[\nabla\times\boldsymbol{v}\right] - \nabla p + \rho \boldsymbol{g} \end{align*}
  • If we assume that the fluid is irrotational, which also follows from assuming its inviscid, then $\nabla \times \boldsymbol{v}=\boldsymbol{0}$ everywhere.
  • This is something that you can prove with the maths you have now but we don't assess in this course, I'm just using it to show you what's hidden inside the cauchy momentum equation...
  • Assuming irrotational ($\nabla \times \boldsymbol{v}=\boldsymbol{0}$) and incompressible flow ($\rho=\mbox{constant}$), and that gravity is a constant, we can write \begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\nabla\left(\frac{1}{2}\rho \boldsymbol{v}^2 + p - \rho \boldsymbol{g}\cdot\boldsymbol{r}\right) \end{align*}
\begin{align*} \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\nabla\left(\frac{1}{2}\rho \boldsymbol{v}^2 + p - \rho \boldsymbol{g}\cdot\boldsymbol{r}\right) \end{align*}
  • A side note for those of you who know/remember physics, this function resembles the definition of a conservative force: $F=-\nabla E$ (A conservative force is the gradient of an energetic potential).
  • Assume the flow is at steady state $\partial/\partial t = 0$ and integrating the right hand side we have \begin{align*} \frac{1}{2}\rho \boldsymbol{v}^2 + p - \rho \boldsymbol{g}\cdot\boldsymbol{r} = \text{constant} \end{align*}
  • So, if we neglect viscous forces and assume inviscid (and thus irrotational), incompressible flow, the differential momentum balance reduces to Bernoulli's equation.
  • This is hugely more complicated than just taking a simple energy balance over the fluid, but it highlights that the continuity and Cauchy momentum equation contain all of the simple results you've already encountered.

Learning Objectives

  • Today, we have learnt to derive the equations for mass and momentum transport: \begin{align*} \frac{\partial \rho}{\partial t} &= -\nabla\cdot \rho\boldsymbol{v} & \text{(Continuity equation)}\\ \rho\frac{\partial\boldsymbol{v}}{\partial t} &= -\rho \boldsymbol{v}\cdot \nabla\boldsymbol{v} - \nabla\cdot\boldsymbol{\tau} - \nabla p + \rho \boldsymbol{g} & \text{(Cauchy momentum equation)} \end{align*}
  • We also learnt that for Newtonian fluids, the stress tensor is \begin{align*} \boldsymbol{\tau}=\tau_{ij} = -\mu \left(\frac{\partial v_i}{\partial r_j} + \frac{\partial v_j}{\partial r_i}\right) + \delta_{ij} \mu^B \nabla\cdot\boldsymbol{v} \end{align*}
  • If the fluid is incompressible, inviscid (and thus irrotational), the momentum balance reduces to Bernoulli's equation. We don't need to derive it, just know that it exists inside the Cauchy momentum equation. \begin{align*} \frac{1}{2}\rho \boldsymbol{v}^2 + p - \rho \boldsymbol{g}\cdot\boldsymbol{r} = \text{constant} \end{align*}
  • Finally, we're ready to solve a problem!