Momentum balance

Section 3.3 Momentum balance

With mass conservation now handled, we turn our eyes towards a law that governs the conservation of momentum. This is Newton’s second law the rate of change of momentum of a body is equal to the sum of all forces acting on the body.

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Our task from this section is to prove the following equation for the conservation of mass of a fluid:

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Theorem 3.3.1. Momentum equation.

The differential form of the law of conservation of momentum, is

\begin{equation} \rho \DD{\bu}{t} = \rho \left(\pd{\bu}{t} + \bu \cdot \nabla\bu\right) = - \nabla p + \rho \bg\tag{3.3.1} \end{equation}

where \(\rho = \rho(\bx,t)\) is the density of the fluid, \(\bu = \bu(\bx, t)\) is the velocity of the fluid, \(p = p(\bx, t)\) is the pressure, and \(\bg\) is the acceleration due to a body force (typically gravity).

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Note that above, we have used the notation \(\nabla \bu\) for the gradient of a vector. This is explained in Remark 2.1.7.

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Subsection 3.3.1 Proof of the momentum equation for an inviscid fluid

Consider again a material volume \(V(t)\) of the fluid. Then the rate of change of net momentum of this volume is equal to

\begin{equation} \dd{}{t} \iiint_{V(t)} \rho \bu(\bx, t) \, \de{V}.\tag{3.3.2} \end{equation}

It is helpful to remember that the mass of a volume element is \(\rho \de{V}\) and the acceleration is \(\dd{\bu}{t}\text{,}\) so the above is similar to mass times acceleration. However, we work with the more general form above to allow for changes in the density throughout the fluid, and also for variable fluid elements, \(V(t)\text{.}\)

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We must equate the above to the sum of all surface and body forces applied to the fluid element.

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An example of a body force is the force of gravity. For a small volume element of mass \(\rho \de{V}\text{,}\) the force of gravity is equal to \((\rho \de{V})\bg\text{.}\) Therefore, the total external force due to gravity on the volume is equal to

\begin{equation} \iiint_{V} \rho \bg \, \de{V}.\tag{3.3.3} \end{equation}

There may be other external body forces. For example, if your fluid is electrically conductive (like a plasma), there may be electromagnetic forces that must be considered. In any case, \(\bg\) can be considered as the analogous body force.

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The final type of forces we should consider are surface forces, which are applied to the boundary of the fluid element, denoted \(\partial V\text{.}\) Let us assume that at each point on the boundary, there is a per-surface area surface force, \(\bF\text{,}\) that decomposes into component normal to the boundary, \(F_n \bn\text{,}\) and a component tangential to the boundary, \(F_t \bt\text{.}\) So the surface forces, summed over the boundary, will be

\begin{equation} \iint_{\partial V} \left[ F_n \bn + F_t \bt\right] \, \de{S}.\tag{3.3.4} \end{equation}

In the above, the interpretation is that the force on a small patch of surface area \(\de{S}\) is equal to the per-area force, say \(F_n\text{,}\) multiplied by the area, then directed into the normal direction, \(\bn\text{.}\)

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At this point, we make a key assumption that is applied in the particular case of inviscid fluids.

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Theorem 3.3.2. Forces in inviscid fluids.

In the case of inviscid fluids, we assume that the surface pressure exerted on (interior) volume elements is accounted solely by a pressure, \(p\text{,}\) which acts in the inward normal direction at each point, with

\begin{equation*} F_n \bn = -p \bn. \end{equation*}

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Consequently, the surface force, given by (3.3.4), is

\begin{equation} \iint_{\partial V} (-p) \bn \, \de{S} = \iiint_V (-\nabla p) \, \de{V}.\tag{3.3.5} \end{equation}

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In particular, for the case of inviscid fluids, we ignore tangential internal forces.

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Proof.

The result follows by applying the divergence theorem to each component of,

\begin{equation*} F_i = -p \be_i, \qquad i = 1, 2, 3. \end{equation*}

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Applying it once, we have

\begin{equation*} \iiint_V \nabla \cdot F_i \, \de{V} = \iint_{\partial V} (-p)\be_i \cdot \bn \, \de{S}. \end{equation*}

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Note that the LHS is:

\begin{equation*} \iiint_V (- \nabla p)_i \, \de{V} \end{equation*}

the ith element of \(-\nabla p\text{.}\)

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Now add up the above for the three elements of \(i = 1, 2, 3\text{,}\) and use the shorthand notation for the divergence of a vector field:

\begin{equation*} \nabla \cdot \bF = \sum_{i=1}^3 (\nabla \cdot F_i)\be_i = (-\nabla p) \end{equation*}

hence establishing the theorem.

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Finally, it follows by summation of the forces above that Newton’s law states that

\begin{equation*} \dd{}{t} \iiint_{V(t)} \rho \bu(\bx, t) \, \de{V} = -\iiint_V (\nabla p) \, \de{V} + \iiint \rho \bg \, \de{V}. \end{equation*}

We can now use the corollary to the transport theorem Corollary 3.2.5, with the assignment of \(f = \rho u_x, \rho u_y, \rho u_z\) and this allows us to the conclude that

\begin{equation*} \iiint_{V} \left( \rho \DD{\bu}{t} + \nabla p - \rho \bg\right) \, \de{V} = 0. \end{equation*}

Again, the above holds for all material volumes \(V\) and therefore it must follow that the integrand is zero. We conclude thus with the following result as stated in Theorem 3.3.1:

\begin{equation*} \rho \DD{\bu}{t} = - \nabla p + \rho \bg. \end{equation*}

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In Exercise 3.6.4, you will practice the derivation of the mass and momentum equations.

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