Reynolds’ Transport Theorem

Section 3.1 Reynolds’ Transport Theorem

Subsection 3.1.1 Jacobian of Lagrangian to Eulerian

Consider a fluid volume, $V \subset D\text{,}$ that is initially dyed a certain colour. The packet of fluid is initially located at $V(0)\text{.}$ As the fluid evolves in time, it then occupies the volume $V(t)$ with $t > 0\text{.}$

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As an example, we may express the density of the fluid as

\begin{equation*} \rho = \rho(\bx, t), \qquad \textrm{$\bx \in V(t)$} \end{equation*}

i.e. at every point in space and moment in time, the above retrieves the density of the fluid within a designated region (which may be changing). This is a natural quantity to study in a fixed frame of the fluid.

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Alternatively, we can write

\begin{equation*} \rho = \rho(\bX, t), \qquad \textrm{$\bX \in V(0)$} \end{equation*}

which is the density of fluid for those particles making up an originally chosen volume, $V(0)\text{.}$ This is a natural quantity to study if we were to move with the fluid; for instance, if we were to colour the volume $V(0)$ with a dye and track its density in time.

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The correspondence between the Euclidean and Lagrangian coordinates is written as

\begin{equation*} \bx = \bx(\bX, t). \end{equation*}

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Theorem 3.1.1. Jacobian of Lagrangian to Eulerian.

Since we assume the medium is continuous, then we would generally require that the mapping from Lagrangian to Eulerian coordinates is continuous and one-to-one; then the map assigns every element (label) in the original reference configuration, denoted $\bX\text{,}$ a unique position, $\bx\text{,}$ in the deformed state.

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From Multivariable Calculus, a sufficient condition for this to be true is that the Jacobian of the transformation,

\begin{equation*} J = \pd{\bx}{\bX} = \pd{(x, y, z)}{(X, Y, Z)} = \begin{vmatrix} \pd{x}{X} & \pd{x}{Y} & \pd{x}{Z} \\ \pd{y}{X} & \pd{y}{Y} & \pd{y}{Z} \\ \pd{z}{X} & \pd{z}{Y} & \pd{z}{Z} \end{vmatrix} \end{equation*}

is finite and non-zero:

\begin{equation*} 0 \lt J \lt \infty. \end{equation*}

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The following requires a bit of algebra, and you are not required to prove the result.

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Theorem 3.1.2. Euler’s identity.

The material derivative of the Jacobian of the transformation $\bx = \bx(\bX, t)$ is given by

\begin{equation} \DD{J}{t} = J \nabla \cdot \bu,\tag{3.1.1} \end{equation}

where $\bu = \DD{\bx}{t}\text{.}$

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Proof.

The proof follows from direct differentiation on the determinant and use of the identity of the material derivative.

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Subsection 3.1.2 Reynolds’ Transport Theorem

We are now ready to derive a key result that eases our path towards developing the governing equations for a fluid. The result is as follows.

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Theorem 3.1.3. Reynolds’ Transport Theorem.

Consider a time-dependent volume, $V(t)\text{,}$ that is transported by the fluid so that it always consists of the same fluid particles. Then, for any function, $f(\bx, t)\text{,}$ continuously differentiable with respect to its arguments, Reynolds’ Transport Theorem is as follows:

\begin{equation} \dd{}{t} \iiint_{V(t)} f \, \de{x} \, \de{y} \, \de{z} = \iiint_{V(t)} \left[ \pd{f}{t} + \nabla \cdot (f\bu)\right] \, \de{x} \, \de{y} \, \de{z}.\tag{3.1.2} \end{equation}

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Proof.

We transform the integral in Euclidean coordinates to Lagrangian coordinates, integrating in the label space:

\begin{equation*} I(t) \equiv \iiint_{V(t)} f \, \de{x} \de{y} \de{z} = \iiint_{V(0)} f J \, \de{X} \de{Y} \de{Z}, \end{equation*}

and notice that we now only need to integrate over the fixed volume as defined in Lagrangian space, at the expense of adding the Jacobian factor. We now write

\begin{align*} \dd{I}{t} &= \iiint_{V(0)} \DD{(fJ)}{t} \, \de{X}\de{Y}\de{Z} \end{align*}

and the material derivative passes through since the domain is fixed. The Theorem 2.1.6 now allows us to convert the material derivative to regular partial derivatives. By the chain rule:

\begin{align*} \DD{(fJ)}{t} &= \DD{f}{t}J + f\DD{J}{t} \\ &= \left[ \pd{f}{t} + \bu \cdot \nabla f\right]J + f \left[J \nabla \cdot \bu\right] \\ &= \left[ \pd{f}{t} + \nabla \cdot (f\bu) \right]J \end{align*}

and we have differentiated the Jacobian via Euler’s identity Theorem 3.1.2 in the second line. The last line follows from the chain rule applied to the vector identity:

\begin{equation*} \nabla \cdot (f\bu) = f \nabla \cdot \bu + \nabla f \cdot \bu. \end{equation*}

We can now revert from Lagrangian to Eulerian integration, and this thus completes the proof of the Reynolds’ Transport Theorem.

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Note 3.1.4.

It is helpful for you to convince yourself that the vector identity used in the proof is the only possible arrangement of operations that makes sense, i.e. in order for $\nabla \cdot (f\bu)$ to return a scalar.

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In summary: the Reynolds’ Transport Theorem thus gives an identity for how time differentiation can pass through the integral when the domain of integration is changing in time!

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Remark 3.1.5. A 1D version of the Reynolds’ Transport Theorem.

In 1D, Reynolds’ Transport Theorem reduces to an identity known as Leibniz’s rule. This is presented as an exercise in Exercise 3.6.3.

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Remark 3.1.6. RTT and conserved quantities.

Reynolds’ Transport Theorem is particularly useful when the integral quantity

\begin{equation*} B = \iiint_{V(t)} f \, \de{x} \, \de{y} \, \de{z} \end{equation*}

is a conserved quantity. A conserved quantity is something for which we can apply a principle of conservation. In practice for this course, this is usually mass or linear momentum:

Mass: $f=\rho$ and the left-hand side of (3.1.2) is zero because mass is conserved.
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Linear momentum: $\bf=\rho\bu$ is now a vector quantity and, by Newton’s second law, the left-hand side of (3.1.2) equals the sum of the all the forces acting on the fluid in the control volume $V\text{,}$ which are:
- Surface forces: Forces acting on the surfaces of the fluid, which could be reaction forces from the walls of a container, or forces due to pressure or stress from surrounding fluid. (Note that we will cover stress in Chapter 7.)
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- Body forces: Forces acting over the interior of the fluid, and in this course you will usually only need to consider the force of gravity, although in principle other body forces could act such as an electromagnetic force. In addition, if working in a noninertial frame of reference, other apparent forces need to be added (forces due to linear acceleration, angular acceleration and the Coriolis and centrifugal forces).
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Remark 3.1.7. General statement of RTT.

Non-examinable:

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A more general statement of the Reynolds’ Transport Theorem is as follows: We define a control volume $V(t)\text{.}$ This is a region of the fluid that we specify; it could be any fluid-containing volume, and in particular it doesn’t need to follow the fluid particles (unlike (3.1.2)). Then the rate of change of $B$ in the system of particles equals the integral of the rate of change of $f$ over the control volume plus the net flux of the $f$ through the surface of the control volume:

\begin{equation} \dd{B_{\textrm{syst}}}{t} = \dd{B_{CV}}{t}+\dot{B}_{CS}.\tag{3.1.3} \end{equation}

Here, $B_{\textrm{syst}}$ is the amount of $B$ in the system of particles, that is the Lagrangian derivative of $B\text{,}$ $B_{CV}$ is the amount of $B$ in the control volume and $\dot{B}_{CS}$ is the flux of $B$ across the surface of the control volume. Note that, if $B$ is conserved then we can write the left-hand side in terms of other things (zero if $B$ is mass and the sum of the forces if $B$ is linear momentum).

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This is in a form that can be used to solve many problems in engineering. The key point is that these problems are about quantities on the scale of the control volume and not about the details of the flow at particular points.

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Thus, for example we could consider a tank with inlets and outlets and, using the fluid in the tank as the control volume, we can apply the conservation of mass to find the volume flow rates in the inlets and outlets and the level of fluid in the tank.

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We can use momentum conservation to find the force on a section of a pipe or the dynamics of a piston moving a fluid to which a known force is applied.

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