20 Momentum Equation

Introduction

The second physical principle used in deriving the governing equations that describe aerodynamic flows (or the flow of a fluid, in general) is the conservation of momentum, i.e., the application of Newton’s second law. This principle is one of the most important for solving aerodynamic motion because it allows solving problems involving the forces produced on or by flows, such as the lift and drag on a wing or the thrust produced by a jet or rocket engine.

From elementary physics, Newton’s second law is usually written as \vec{F} = m\, \vec{a}, where \vec{F} is the force on the system, m is the mass of the system, and \vec{a} is the resulting linear acceleration of the system. However, the most general form of Newton’s second law is when it is written in terms of the time rate of change of momentum, i.e.,

(1)   \begin{equation*} \vec{F} = \frac{d (m\vec{V})}{dt} \end{equation*}

where m\vec{V} is the total linear momentum in this case; naturally, there are analogous equations for angular momentum. Remember that in reference to “momentum,” the meaning is that it is the momentum of something, be it a mass, a fluid element, or something else. Also, note that the plural of momentum is “momenta” and never “momentums.”

Learning Objectives

  • Understand the principle of conservation of momentum when applied to a fluid.
  • Be able to set up the most general form of the momentum equation for a fluid, including all sources of forces.
  • Use the momentum and continuity equations to solve elementary flow problems.

Setting Up the Momentum Equation

  1. The objective is to apply the conservation of linear momentum principle to a flow to find a mathematical expression for the forces produced in terms of the familiar macroscopic flow field variables, such as density \varrho, velocity \vec{V} and pressure, p. Pressure comes into the problem of describing the behavior of a flow because any non-uniform pressure in a flow when it acts over an area, can produce a force. Such non-uniform pressures are called pressure gradients in that a pressure gradient will cause an imbalance of forces on the flow to accelerate or decelerate it and cause changes in its momentum.

Consider the previous control volume approach, as shown in the figure below. The volume is assumed to be fixed in space with the fluid moving through it, i.e., an Eulerian modeling approach. In general, the forces on the air, as it flows through the control volume, will manifest from two sources, namely:

  1. Body forces like gravity, inertial accelerations (maybe even electromagnetic, electrostatic forces), etc., may act on the fluid elements inside the control volume. The idea of a body force resulting from gravity was introduced when the hydrostatic equation was established.
  2. Surface forces: Pressure and shear stress acting on the area of control surface S, which will lead to forces.
A finite control volume, fixed in space, is used to set up the momentum equation in its most general form.

Pressure Forces

The pressure force acting on the elemental area dS is -p \, d\vec{S}, where the negative sign indicates that the force is inward and in the opposite direction to the unit normal vector area, which always points outward by convention. Therefore, the total pressure force on the fluid in the control volume is the summation of the elemental forces over the entire control surface, i.e., the pressure force is given by

(2)   \begin{equation*} \vec{F}_p = -\oiint_{S} p \, d\vec{S} \end{equation*}

Body Forces

Now consider the body forces, where \vec{f}_b is defined as the net body force per unit mass exerted on the fluid inside {\cal {V}}. The body force on the elemental fluid volume d{\cal {V}} is \varrho\, \vec{f}_b \, d{\cal {V}}, and the total body force is the summation over the volume {\cal {V}}, i.e.,

(3)   \begin{equation*} \vec{F}_b = \oiiint_{{\cal V}} \varrho \, \vec{f}_b \, d{\cal {V}} \end{equation*}

An example of a body force is gravity, so in this case, the body force per unit mass is simply the acceleration under gravity, i.e., g, so that in terms of the body force vector, then \vec{F}_b = \vec{f}_b = -g \, \vec{k}. By convention, the direction of \, \vec{k} is usually vertically upward, so the minus sign indicates that gravity is directed downward.

Viscous Forces

If the flow is viscous, and all fluids are viscous to a lesser or greater degree, then the viscous shear stresses \tau that arise over the control surface will exert a surface force. These forces will depend on the fluid’s viscosity and the velocity gradients in the flow, making evaluating such forces potentially difficult. Therefore, at least for now, the viscous effects can be recognized but not evaluated per se. The total viscous force on the flow as it passes through the control volume will be given by

(4)   \begin{equation*} \vec{F}_{\mu} = \iint_{S} \tau \, d{\vec{S}} \end{equation*}

It must be understood that the actual quantitative evaluation of this integral may be complicated and require some work, especially if turbulence is present.

Total Forces

Therefore, the total forces acting on the flow in the control volume will be

(5)   \begin{equation*} \vec{F} = \vec{F}_p + \vec{F}_b + \vec{F}_{\mu} = \underbrace{ -\oiint_{S} p \, d\vec{S}}_{\begin{tabular}{c} \scriptsize Pressure \\[-4pt] \scriptsize forces \end{tabular}} + \underbrace{ \oiiint_{{\cal V}} \varrho\, \vec{f}_b \, d{\cal {V}}}_{\begin{tabular}{c} \scriptsize Body \\[-4pt] \scriptsize forces \end{tabular}} + \underbrace{ \iint_{S} \tau \, d{\vec{S}}}_{\begin{tabular}{c} \scriptsize Viscous \\[-4pt] \scriptsize forces \end{tabular}} \end{equation*}

which will constitute the net force that must appear on the left-hand side of the final momentum equation.

Momentum Terms

Now consider the right-hand side of the momentum equation, i.e., the term that must describe the fluid’s time rate of change of momentum. There will be two contributions to this, namely:

  1. The net momentum flows out of the control volume across surface S per unit of time.
  2. The time rate of momentum change inside {\cal {V}} will only occur if there are any unsteady fluctuations in the flow properties.

Consider the first term. At any point inside the control volume or on the control surface, the flow velocity is \vec{V}. The mass flow across the element dS is given by \varrho\vec{V}\bigcdot d\vec{S} so the flow of momentum across dS will be

(6)   \begin{equation*} \mbox{Flow of momentum over $dS$} = \left( \varrho \, \vec{V} \bigcdot d\vec{S} \right) \vec{V} \end{equation*}

The net flow of momentum out of the control volume over S is the sum of all the elemental contributions, i.e.,

(7)   \begin{equation*} \mbox{Total flow of momentum over $S$} = \iint_{S} \left( \varrho\vec{V}\bigcdot d\vec{S} \right) \vec{V} \end{equation*}

Now consider the second term. The momentum of the fluid inside the element of volume d{\cal {V}} will be (\varrho \, d{\cal {V}})\vec{V}, and so the total momentum contained inside the control volume at a given time is

(8)   \begin{equation*} \mbox{Total momentum inside ${\cal {V}}$} = \oiiint_{{\cal V}} \varrho\vec{V} d{\cal {V}} \end{equation*}

and its time rate of change (which can only result from unsteady flow fluctuations) will be

(9)   \begin{equation*} \mbox{Time rate of change of momentum inside ${\cal {V}}$} = \frac{\partial }{\partial t} \oiiint_{\cal V} \varrho\vec{V} d{\cal {V}} \end{equation*}

Final Form of the Momentum Equation

Combining all the above equations into the form \vec{F} = d(m\vec{V})/dt gives an expression for momentum conservation in terms of the total forces and time rate of change of momentum of the flow as it sweeps through the control volume, i.e.,

(10)   \begin{equation*} \vec{F} = \oiiint_{\cal{V}} \varrho \, \vec{f} _b \, d{\cal{V}} -\oiint_S p \, d\vec{S} + \vec{F}_{\mu} = \frac{\partial}{\partial t}\oiiint_{\cal{V}} \varrho \, \vec{V} \, d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V} \end{equation*}

In other words, this previous equation states that Total Forces = Body Forces + Pressure Forces + Viscous Forces = Time rate of change of momentum inside {\cal {V}} from any unsteadiness in the flow + Net flow of momentum out of {\cal {V}} across the control surface S per unit time. This general equation applies to any type of flow, compressible or incompressible, viscous or inviscid, etc.

Equation 10 is the momentum equation in its integral form. It will be apparent that it is a vector equation, so three scalar equations are rolled into one, i.e., one equation for each coordinate direction, i.e., if \vec{F} = ( F_x, F_y, F_z ), then

(11)   \begin{eqnarray*} F_x & = & \frac{\partial}{\partial t}\oiiint_{\cal{V}} \varrho \, u \, d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) u \\ F_y & = & \frac{\partial}{\partial t}\oiiint_{\cal{V}} \varrho \, v \, d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) v \\ F_z & = & \frac{\partial}{\partial t}\oiiint_{\cal{V}} \varrho \, w \, d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) w \end{eqnarray*}

What are the units of momentum?

When we speak of “momentum,” we mean the momentum of something. Momentum is a vector quantity, so it has both magnitude and direction. If m denotes mass and \vec{V} denotes velocity, then the translational momentum is m \vec{V}. Therefore, in terms of base units

    \[ \left[ m \vec{V} \right] = \rm M \, L \, \rm T^{-1} = \left( \rm M \, L \, T^{-2} \right) \rm T \]

noting that the units inside the parenthesis represent a force. Therefore, in the SI system, the units of the translational momentum of a fluid will be kg m s^{-1}, which is also equivalent to a Newton-second, i.e., N s. In the USC system, the units of momentum will be slug ft s^{-1}, which is also equivalent to a pound-second, i.e., lb s.

Simplifications to the Momentum Equation

As in the use of the continuity equation for practical problem solving, the apparent complexity of the general form of the momentum equation can be simplified by making justifiable assumptions, the advantage being that the solutions of the resulting sets of governing equations (mass, momentum, and energy) most likely become commensurately easier. Remember that any assumptions must be justified, which may require information from an experiment or a judgment call, i.e., based on experience. Experience is acquired from repeated problem solving, i.e., starting from the exemplar problems in the classroom and doing many more on one’s own.

Consider, for example, a steady flow. In this case, \partial/\partial t \equiv 0, so the reduced form of the momentum equation will be

(12)   \begin{equation*} \vec{F} = \oiiint_{\cal{V}} \varrho \, \vec{f}_b \, d{\cal{V}} - \oiint_S p \, d\vec{S} + \vec{F}_{\mu} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V} \end{equation*}

Furthermore, if the flow is additionally assumed to be inviscid, then \vec{F}_{\mu} = 0. Therefore, in this case

(13)   \begin{equation*} \oiiint_{\cal{V}} \vec{f}_{b} \, \varrho \, d{\cal{V}} - \oiint_S p \, d\vec{S} = \varrho \oiint_S ( \vec{V} \bigcdot d\vec{S}) \vec{V} \end{equation*}

Proceeding even further, if the flow is assumed to be steady, inviscid, and incompressible, then

(14)   \begin{equation*} \varrho \oiint_S ( \vec{V} \bigcdot d\vec{S}) \vec{V} = \varrho \oiiint_{\cal{V}} \, \vec{f}_{b} \, d{\cal{V}} - \oiint_S p \, d\vec{S} \end{equation*}

and without any body forces, then

(15)   \begin{equation*} - \oiint_S p \, d\vec{S} = \varrho \oiint_S ( \vec{V} \bigcdot d\vec{S}) \vec{V} \end{equation*}

Proceeding even further to make two-dimensional or one-dimensional flow assumptions means volume integrals become area integrals per unit depth. The surface integrals become line integrals per unit depth that circumvent the path around the defined control surface. For two-dimensional (or three-dimensional) problems, the momentum equation must written appropriately so that the scalar components of the force on the fluid can be determined. For example, for a two-dimensional, steady flow without pressure forces, the momentum equation reduces to

(16)   \begin{eqnarray*} F_x & = & \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) u \\ F_y & = & \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) v \end{eqnarray*}

and for uniform properties, then

(17)   \begin{equation*} F_x = \overbigdot{m} \, \Delta u \quad \mbox{and} \quad F_y = \overbigdot{m} \, \Delta v \end{equation*}

where \Delta u and \Delta v are the changes in velocity of the fluid in the x and y direction, respectively.

Learning by Examples

How far the various simplifications to the momentum equation can be taken partly depends on the complexity of the problem itself and the strength of the justifications that can be made. Such steps are best learned by reviewing some examples. In learning fluid dynamics through examples, it is best to start with more straightforward problems and progressively work up to more complex scenarios. Besides the example below, practicing with real-world problems and using simulation software tools can be incredibly helpful in developing your understanding of problems in fluid dynamics. Additionally, there are many online resources to learn about fluid dynamics, including short courses and open resources from educational institutions, which can be used for more in-depth study and practice.

Worked Example #1 – Force on a converging duct or nozzle

Reduce the general form of the momentum equation to a form suitable for application to the steady, compressible, one-dimensional flow of air through a converging duct (often called a contraction) of circular cross-section, i.e., a nozzle, as shown below.

Proceed first by sketching the nozzle problem, showing the control surface, and annotating the inlet and outlet conditions appropriately, not only with what is known but also with what is unknown. In this case, there is a net pressure force on the fluid by the nozzle, which is to be determined.

The flow is steady, so \partial/\partial t \equiv 0. The fluid is air (which has a relatively low density), and there is no change in elevation from one side of the nozzle to the other, so it is possible to neglect all body forces, including the effects of gravity.

As previously derived, the general form of the momentum equation is

    \[ \oiiint_{\cal{V}} \varrho \, \vec{f} _b \, d{\cal{V}} -\oiint_S p \, d\vec{S} + \vec{F}_{\mu} = \frac{\partial}{\partial t}\oiiint_{\cal{V}} \varrho \, \vec{V} \, d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V} \]

The reduced form for the assumptions made that the flow is steady without body forces leads to

    \[ - \oiint_S p \, d\vec{S} + \vec{F}_{\mu} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V} \]

The left-hand side can be written as

    \[ -\iint_{A_1} p \, d\vec{S} - \iint_{A_2} p \, d\vec{S} + \underbrace{\iint_{\rm Walls} \!\! \!-p \, d\vec{S} + \vec{F}_{\mu} }_{F_x} \]

In this case, the force on the fluid, F_x (to change its momentum), needs to be determined, i.e., this is the unknown. Notice that this force can include both pressure forces and viscous forces. There is also an equal and opposite force on the nozzle, say R_x, which ultimately is the force to be determined. Therefore, the momentum equation, in this case, is

    \[ F_x - \iint_{A_1} p \, d\vec{S} - \iint_{A_2} p \, d\vec{S} = \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V} \]

The flow can be assumed to be one-dimensional so that the flow properties can change only in one dimension, and this assumption makes the problem easier but realistic. With the one-dimensional assumption, then

    \[ F_{x} + p_1 A_1 - p_2 A_2 = \left( \varrho_2 A_2 V_2^2 - \varrho_1 A_1 V_1^2 \right) \]

noting the signs of the term on the first pressure integral over A_1 and remembering that d\vec{S} always points out of the control volume by convention. Also, the continuity of the flow means that

    \[ \varrho_1 A_1 V_1 = \varrho_2 A_2 V_2 \]

So, the final form of the momentum equation for this problem becomes

    \[ F_{x} = (p_2 A_2 - p_1 A_1) + \varrho_1 A_1 V_1 \left( V_2 - V_1 \right) \]

This force will be negative, i.e., to the left. The fact that this force is in the backward direction (opposite to the flow direction) is somewhat counterintuitive because momentum is being added to the fluid as it passes through the nozzle. But if a person is holding the nozzle, they must apply a forward force reaction while keeping the nozzle in place to prevent it from moving backward. This reaction force is

    \[ R_x = -F_x \]

which will be to the right. Notice also that this problem involves velocities and pressures. However, the Bernoulli equation(as discussed in the next chapter) cannot be applied to this problem because there are viscous losses as the fluid passes through the nozzle. However, the continuity equation is applicable for any type of fluid flow, viscous or inviscid, and so will apply in this case, as will the momentum equation.

Worked Example #2 – Force on an engine test stand

A static thrust stand is designed for testing a jet engine. The following conditions are expected for a typical test: intake air velocity = 200 m/s, exhaust gas velocity = 500 m/s, intake cross-section area = 1 m^2, intake static pressure = -22.5 kPa (= 78.5 kPa absolute), intake static temperature = 268K, exhaust static pressure = 0 kPa (101 kPa absolute). Estimate the thrust reaction force to design the test stand to carry if it must have a safety factor of 3.

Assume that the inlet is identified as station 1 and the outlet as station 2. Assume also that the exit area is the same as the inlet area. The density of the air at the inlet can be found using

    \[ \varrho_1 = \frac{p_1}{R T_1} = \frac{78.5 \times 10^3}{(287)(268)} = 1.0206 ~\mbox{kg m$^{-3}$} \]

From continuity considerations (assume 1-dimensional, steady flow), then

    \[ \varrho_1 A_1 V_1 = \varrho_2 A_2 V_2 = \overbigdot{m} = 204.12~\mbox{kg s$^{-1}$} \]

If the flow is additionally assumed to be inviscid and there are no gravitational effects, then \vec{F}_{\mu}, and the relevant form of the momentum equation is

    \[ - \oiint_S p \, d\vec{S} = \varrho \oiint_S ( \vec{V} \bigcdot d\vec{S}) \vec{V} \]

The pressure integral on the left-hand side is

    \[ \oiint_S p \, d\vec{S} = \iint_{\rm Inlet} p \, d\vec{S} + \iint_{\rm Outlet} p \, d\vec{S} \iint_{\rm Engine} p \, d\vec{S} \]

or

    \[ -\oiint_S p \, d\vec{S} = - \iint_{A_1} p \, d\vec{S} - \iint_{A_2} p \, d\vec{S} - F \]

where F is the force of the engine on the fluid. Assuming one-dimensional flow gives

    \[ -F + p_1 A_1 - p_2 A_2 = \overbigdot{m} ( V_2 - V_1 ) \]

noting the signs of the terms, and remembering that d\vec{S} always points out of the control volume, or

    \[ F = -p_1 A_1 + p_2 A_2 + \overbigdot{m} ( V_2 - V_1 ) \]

Finally, substituting the known values gives

(18)   \begin{equation*} F = (-78.5 \times 10^3)(1.0) + (101.0 \times 10^3)(1.0) + 204.12 (500.0-200.0) = 83,736~\mbox{N} \end{equation*}

For the conditions stated, the residual force on the fluid must be 83,736 N in the direction of fluid flow, and the reaction force on the test stand will be in the opposite direction. Because there is a factor of safety requirement of 3, the force to which the stand should be designed is about 251 kN.

Worked Example #3 – Force on an angled plate

Water (\varrho_w = 1.94 slug ft^{-3}) flows at a mass flow rate of \overbigdot{m} = 6.0 slug s^{-1} through a nozzle of circular cross-section with an entrance diameter d_e = 10.0 in and an exit diameter of d_j = 4.0 in. The water jet then hits an angled plate with a circular groove and is deflected while also increasing in its circular cross-section, as shown in the figure below.

  1. Draw an appropriate control surface and volume with annotations to analyze this problem.
  2. Determine the velocity of the water entering and exiting the nozzle.
  3. Determine the static pressure change between the nozzle entrance and its exit.
  4. Determine the velocity of the water in the deflected jet.
  5. Determine the magnitude and direction of the force on the plate caused by the water jet.

1. The appropriate control surface and volume with annotations to analyze this problem are shown below.

 

2. This part of the problem requires the use of the continuity equation in its one-dimensional form, i.e.,

    \[ \overbigdot{m} = \varrho_w \ A_e \ V_e = \varrho_w \ A_j \ V_j \]

The entrance area, A_e, is

    \[ A_e = \frac{\pi d_e^2}{4} = \frac{\pi (10.0/12.0)^2}{4} = 0.5454~\mbox{ft$^2$} \]

Therefore, the entrance velocity is

    \[ V_e = \frac{\overbigdot{m}}{ \varrho_w \ A_j} \]

and substituting the numerical values gives

    \[ V_e = \frac{6.0}{1.94 \times 0.5454} = 5.67~\mbox{ft s$^{-1}$} \]

The exit jet area, A_j, is

    \[ A_j = \frac{\pi d^2}{4} = \frac{\pi (4.0/12.)^2}{4} = 0.0873~\mbox{ft$^2$} \]

Therefore, the exit jet velocity is

    \[ V_j = \frac{\overbigdot{m}}{ \varrho_w \ A_j} \]

and substituting the numerical values gives

    \[ V_j = \frac{6.0}{1.94 \times 0.0873} = 35.43~\mbox{m s$^{-1}$} \]

3. The change in pressure \Delta p between the nozzle inlet and its exit can be found using the Bernoulli equation, i.e.,

    \[ p + \frac{1}{2} \varrho V^2 = \mbox{constant} \]

In this case, then

    \[ p_e + \frac{1}{2} \varrho V_e^2 = p_j + \frac{1}{2} \varrho V_j^2 \]

We are asked for a change in pressure, so

    \[ p_e - p_j = \Delta p = \frac{1}{2} \varrho \left( V_j^2 - V_e^2 \right) \]

and substituting the numerical values gives

    \[ \Delta p = 0.5 \times 1.94 \times \left( 35.43^2 - 5.67^2 \right) = 1,186.44~\mbox{lb ft$^{-2}$} \]

4. The velocity V_d of the water in the deflected jet can be found using continuity, i.e.

    \[ \overbigdot{m} = \varrho_w \ A_d \ V_d \]

The deflected jet area A_d is

    \[ A_d = \frac{\pi d_d^2}{4} = \frac{\pi (5.0/12.0)^2}{4} = 0.1364~\mbox{ft$^2$} \]

Therefore, the deflected jet velocity is

    \[ V_d = \frac{\overbigdot{m}}{\varrho_w \ A_d} \]

and substituting the values gives

    \[ V_d = \frac{6.0}{1.94 \times 0.1364} = 22.67~\mbox{m s$^{-1}$} \]

5. The applicable form of the momentum equation is

    \[ \vec{F}= \overbigdot{m} \, \Delta \vec{V} \]

We can ignore all pressure forces because the plate is surrounded by free air at ambient conditions, so there is no pressure force on the control volume. In scalar form, because we have forces in two directions, then

    \[ F_x = \overbigdot{m} \ \Delta V_x \mbox{~~~~~and~~~~~~} F_y = \overbigdot{m} \ \Delta V_y \]

In the x direction, the force on the fluid is

    \[ F_x = \overbigdot{m} \ \Delta V_x \]

where \Delta V_x is the change in the flow velocity in the x direction. Therefore,

    \[ F_x = \overbigdot{m} \left( V_d \ \cos \theta - V_j \right) = 6.0 \left( 22.67 \ \cos 45^{\circ} - 35.43 \right) = -116.4~\mbox{lb} \]

In the y direction,  the force on the fluid is

    \[ F_y = \overbigdot{m} \ \Delta V_y \]

where \Delta V_y is the change in the flow velocity in the y direction. Therefore,

    \[ F_y = \overbigdot{m} \left( V_j \ \sin \theta - 0 \right) = 6.0 \left( 22.67 \ \sin 45^{\circ} - 0.0 \right) = 96.2~\mbox{lb} \]

If the force on the fluid is

    \[ \vec{F} = -116.4 \ \, \vec{i} + 96.2 \, \vec{j}~~\mbox{lb} \]

then, the reaction force on the plate is

    \[ \vec{R} = 116.4\ \, \vec{i} - 96.2 \ \, \vec{j}~~\mbox{lb} \]

which is to the right and downward.

Momentum Equation from the RTE

Recall that the Reynolds Transport Equation (RTE) can be expressed as

(19)   \begin{equation*} \underbrace{\frac{D}{Dt} \oiiint_{\mathrm{sys}} \beta \, \varrho \, d\mathcal{V}}_{\begin{tabular}{c} \scriptsize Time rate of \\[-3pt] \scriptsize change of $B$ \\[-3pt] \scriptsize inside the system.\end{tabular}} = \underbrace{\frac{\partial}{\partial t} \oiiint_{\mathcal{V}} \beta \, \varrho \, d\mathcal{V}}_{\begin{tabular}{c} \scriptsize Time rate \\[-3pt] \scriptsize of change of $B$ inside\\[-3pt] \scriptsize the control volume.\end{tabular}} + \underbrace{\oiint_{S} \beta \, \varrho ( \vec{V}_{\mathrm{rel}} \bigcdot d\vec{S}) }_{\begin{tabular}{c} \scriptsize Rate at which $B$ is \\[-3pt] \scriptsize leaving through the \\[-3pt] \scriptsize control surface.\end{tabular} } \end{equation*}

In the case of momentum, then \beta = \vec{V}, and for the system then, any change in momentum is equivalent to a force, so

(20)   \begin{equation*} \frac{D}{Dt} \oiiint_{\mathrm{sys}} \varrho \, \vec{V} \, d\mathcal{V} = \vec{F} \end{equation*}

where \vec{F} is the net force acting on the fluid in the C.V. Recall that the net force depends on body forces, pressure forces, and viscous forces, i.e.,

(21)   \begin{equation*} \vec{F} = \oiiint_{\cal{V}} \varrho \, \vec{f}_b \, d{\cal{V}} -\oiint_S p \, d\vec{S} + \vec{F}_{\mu} \end{equation*}

Therefore, the RTE becomes

(22)   \begin{equation*} \vec{F} = \frac{\partial }{\partial t} \oiiint_{{\cal{V}}} \varrho \, \vec{V} \, d {\cal{V}} + \oiint_{S} \varrho \vec{V} (\vec{V} \bigcdot d\vec{S}) \end{equation*}

which is the linear momentum equation. Remember that the momentum equation is a vector equation. For steady flow, then the momentum equation becomes

(23)   \begin{equation*} \vec{F} = \oiint_{S} \varrho (\vec{V} \bigcdot d\vec{S} ) \vec{V} \end{equation*}

and if the flow is incompressible where \varrho = constant, then the density term can be taken outside of the integral to get

(24)   \begin{equation*} \vec{F} = \varrho \oiint_{S} ( \vec{V} \bigcdot d\vec{S} ) \vec{V} \end{equation*}

Differential Form of the Momentum Equation

The differential form of the continuity equation can also be derived from the RTE. Using the divergence theorem, the RTE for a fixed control volume becomes

(25)   \begin{equation*} \vec{F} = \oiiint_{{\cal{V}}} \bigg[ \frac{\partial (\beta \, \varrho)}{\partial t} + \beta \, \varrho \,\left( \nabla \bigcdot \vec{V} \right) \bigg] d{\cal{V}} \end{equation*}

where in this case \beta = \vec{V}, so in differential form, then

(26)   \begin{equation*} d\vec{F} =\left[ \frac{\partial (\varrho \vec{V}) }{\partial t} + \varrho \, \vec{V} \left( \nabla \bigcdot \vec{V} \right) \right] d{\cal{V}} \end{equation*}

which will apply at every point in the flow. Notice that

(27)   \begin{equation*} \frac{\partial (\varrho \vec{V}) }{\partial t} = \vec{V} \, \frac{\partial \varrho }{\partial t} + \varrho \, \frac{\partial \vec{V} }{\partial t} \end{equation*}

Furthermore, the mass of a differential fluid element is \varrho \, d{\cal{V}}, so in terms of per unit mass, then

(28)   \begin{equation*} \varrho \vec{f} = \vec{V} \, \frac{\partial \varrho }{\partial t} + \varrho \, \frac{\partial \vec{V} }{\partial t} + \varrho \, \vec{V} \left( \nabla \bigcdot \vec{V} \right) = \vec{V} \, \frac{\partial \varrho }{\partial t} + \varrho \frac{D \vec{V}}{Dt} \end{equation*}

where D /D t is the substantial derivative. Remember that this equation is a vector equation. Finally, if the density is assumed constant, i.e., an incompressible flow, then

(29)   \begin{equation*} \varrho \vec{f} = \varrho \frac{D \vec{V}}{Dt} \end{equation*}

The force per unit mass \vec{f} will comprise the sum of pressure forces, body forces, gravitational forces, and viscous forces so that the momentum equation can be written as

(30)   \begin{equation*} \underbrace{ -\nabla p}_{\rm pressure} + \underbrace{ \varrho \vec{f_b}}_{\rm body} \underbrace{ -\varrho g\, \vec{k}}_{\rm gravity} + \underbrace{ \vec{f_{\mu}} }_{\rm viscous} = \varrho \frac{D \vec{V}}{Dt} \end{equation*}

which is just a statement of Newton’s second law, or

(31)   \begin{equation*} \varrho \frac{D \vec{V}}{Dt} = -\nabla p + \varrho \vec{f_b} + \varrho \vec{g} + \vec{f}_{\mu} \end{equation*}

recognizing that \vec{g} is pointed downward. Notice the pressure gradient term (pressure force per unit mass) arose previously in the derivation of the hydrostatic equation.

Navier-Stokes Equation

The viscous term, \vec{f}_{\mu}, requires considerable mathematics work to evaluate. However, it can be shown that

(32)   \begin{equation*} \vec{f}_{\mu} = \mu \, \nabla^2 \,\vec{V} \end{equation*}

where \nabla^2 is the vector Laplacian. Remember that

(33)   \begin{equation*} \nabla^2 \,\vec{V} = \nabla ( \nabla \bigcdot \vec{V} ) \end{equation*}

which is the gradient of a divergence field. In matrix form, then

(34)   \begin{equation*} \mu \, \nabla \, \vec{V} = \mu \begin{bmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} & \dfrac{\partial u}{\partial z} \\[12pt] \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} & \dfrac{\partial v}{\partial z} \\[12pt] \dfrac{\partial w}{\partial x} & \dfrac{\partial w}{\partial y} & \dfrac{\partial w}{\partial z} \end{bmatrix} \end{equation*}

which are the viscous stresses per unit mass. Notice that the viscous terms depend on the velocity gradients in the flow, as they should be, i.e., based on Newton’s formula of viscosity. Therefore,

(35)   \begin{equation*} \mu \, \nabla^2 \,\vec{V} = \mu \, \nabla ( \nabla \bigcdot \vec{V} ) = \mu \begin{bmatrix} \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} + \dfrac{\partial^2 u}{\partial z^2} \\[12pt] \dfrac{\partial^2 v}{\partial x^2} + \dfrac{\partial^2 v}{\partial y^2} + \dfrac{\partial^2 v}{\partial z^2} \\[12pt] \dfrac{\partial^2 w}{\partial x^2} + \dfrac{\partial^2 w}{\partial y^2} + \dfrac{\partial^2 w}{\partial z^2} \end{bmatrix} \end{equation*}

which represents a viscous force per unit mass of the fluid. Therefore, the incompressible form of the momentum equations in differential form for a laminar flow (no turbulence) and without body forces can be written as

(36)   \begin{equation*} \varrho \frac{D \vec{V}}{Dt} = -\nabla p + \mu \, \nabla^2 \,\vec{V} \end{equation*}

which is the Navier-Stokes equation. Although one equation, they are normally referred to as the Navier-Stokes equations (plural) because they are a set of three coupled scalar equations rolled into one. These equations can also be written in the form

(37)   \begin{equation*} \frac{D \vec{V}}{Dt} = -\frac{1}{\varrho} \, \nabla p + \left( \frac{\mu}{\varrho} \right) \nabla^2 \,\vec{V} = -\frac{1}{\varrho} \, \nabla p + \nu \, \nabla^2 \,\vec{V} \end{equation*}

where \nu ( = \mu / \varrho ) is the kinematic viscosity. Finally, their most elegant form, they are written as

(38)   \begin{equation*} \frac{D \vec{V}}{Dt} = -\frac{1}{\varrho} \, \nabla p + \nu \, \nabla^2 \,\vec{V} \end{equation*}

The unknowns in this equation are the velocity field \vec{V}= (u, v, w, t) and the pressure p(x, y, z, t). Notice that there are three equations and four unknowns, i.e., the three scalar velocities, u, v, w, respectively, and the pressure, p, so they must be solved in conjunction with the continuity equation, i.e.,

(39)   \begin{equation*} \nabla \bigcdot \vec{V} = 0 \end{equation*}

Without the viscous terms, they are often referred to as the Euler equations, i.e.,

(40)   \begin{equation*} \frac{D \vec{V}}{Dt} = -\frac{1}{\varrho} \, \nabla p \end{equation*}

The Navier-Stokes equations are the most famous of all equations in fluid dynamics. These equations, a statement of Newton’s second law for a fluid, are fundamental to understanding fluid behavior. Despite the apparent simplicity of the Navier-Stokes equation, they represent a set of coupled nonlinear partial differential equations, so solving them poses a significant challenge. The equations remain a cornerstone in the study of fluid dynamics. Finding closed-form solutions remains one of the Clay Mathematics Institute’s Millennium Prizes recognizing the challenges in their solution.

However, the challenges of solving the Navier-Stokes equations extend well beyond their mathematical novelty because engineers must apply them to real-world scenarios. The Navier-Stokes equations are frequently employed to model turbulent flows, but predicting turbulence is also inherently problematic. Engineers now employ computational fluid dynamics (CFD) methods to find approximate numerical solutions to the Navier-Stokes equations and gain insights into the development of turbulent flows about flight vehicles. The Euler equations are often used as an initial approximation in CFD simulations.

Summary & Closure

Applying the principle of the conservation of momentum to a fluid is needed whenever forces are involved, i.e., the application of Newton’s second law of motion. The conservation of momentum is a fundamental principle that applies to fluid flows and is expressed mathematically through the momentum equation. The equation considers various forces that act on a fluid, such as body, pressure, and viscous forces. Depending on the specific problem, the momentum equation can be simplified by making appropriate assumptions, such as steady or inviscid flow, but only if these assumptions are justified. To solve for unknown quantities in fluid flow problems, the momentum equation can be combined with the continuity equation. In cases where pressure is involved, the principles of the conservation of energy must also inevitably be taken into account.

5-Question Self-Assessment Quiz

For Further Thought or Discussion

  • Consider the flow through a reducing pipe with a 180^{\circ} bend. Show how to set up the equations to calculate the force of the fluid on the bend.
  • If the value of a force calculated using the momentum equation is negative, then what does this mean?
  • Think of some flow problems where viscosity effects are dominant, and an inviscid assumption for modeling the flow would likely be invalid.

Additional Online Resources

To learn more about the conservation of momentum applied to flow problems, check out some of these additional online resources:

  • See an excellent, detailed explanation of the momentum equation in this video.
  • A video explaining how to find the force from a flow through a nozzle.
  • A (long!) video by an instructor explaining the conservation of momentum.

License

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Introduction to Aerospace Flight Vehicles Copyright © 2022, 2023, 2024 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n17