18 Conservation of Momentum: 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 problems involving the forces produced by flows to be solved, 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 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.

Learning Objectives

  • Understand the principle of conservation of momentum applied to a fluid.
  • Be able to set up the most general form of the momentum equation, including all sources of forces.
  • Use the momentum equation in conjunction with the continuity equation to solve some simple flow problems.
Here is a short video lesson on Newton’s laws of motion from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Setting Up the Momentum Equation

The objective is to apply the conservation of 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, then it 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.

A finite control volume, fixed in space, as used to set up the momentum equation in its most general form.

Consider the control volume approach used previously, which 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, such as gravity, electromagnetic, electrostatic, etc., may act on the fluid elements inside the control volume. The idea of a body force resulting from gravity has already been introduced when establishing the hydrostatic equation.
  2. Surface forces: Pressure and shear stress acting on the area of control surface S, which will lead to forces.

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 so 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 = \iiint_{{\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. Herefore, 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} = -\oiint_{S} p \, d\vec{S} + \iiint_{{\cal V}} \varrho\vec{f}_b d{\cal V} + \iint_{S} \tau \, d{\vec{S}} \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 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 ${\cal V}$} = \iiint_{{\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} \iiint_{\cal V} \varrho\vec{V} d{\cal V} \end{equation*}

Final Form of 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} = \iiint_{\cal{V}} \varrho \vec{F} _b d{\cal{V}} -\oiint_S p \, d\vec{S} + \vec{F}_{\mu} = \frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, \vec{V} d {\cal{V}} + \oiint_S (\varrho \, \vec{V} \bigcdot d\vec{S}) \vec{V} \end{equation*}

In 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} per unit time.

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. This general equation applies to any type of flow, compressible or incompressible, viscous or inviscid, etc.

What are the units of momentum?

Momentum is a vector quantity, so it has both magnitude and direction. If m denotes mass and \vec{V} denotes velocity, then the momentum \vec{p}p is

    \[ \vec{p} = m \vec{V} \]

Therefore, in terms of base units

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

noting that the units inside the parenthesis represent a force. Therefore, in the SI system, then the units of momentum 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.

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

(11)   \begin{equation*} \vec{F} = \iiint_{\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

(12)   \begin{equation*} \iiint_{\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

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

and without any body forces then

(14)   \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 that 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. How far the various simplifications 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.

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, inviscid, 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 and by showing the control surface and annotating the inlet and outlet conditions appropriately, not only with what is known but also 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, and the flow is inviscid, so \vec{F}_{\mu} = 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

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

and the reduced form for the assumptions made that the flow is steady, compressible and inviscid, leads to

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

The pressure integral on the left-hand side can be written as

    \[ \oiint_S p \, d\vec{S} = \iint_{A_1} p \, d\vec{S} + \iint_{A_2} p \, d\vec{S} + \iint_{\rm Walls} p \, d\vec{S} \]

In this case, it is the force of the walls on the fluid, say F_x (to change its momentum), is what needs to be determined, i.e., this is the unknown. There is also equal and opposite force on the nozzle, say R_x, which ultimately is the force to be determined.

The flow can also be assumed to be one-dimensional, so the flow properties can change only in one dimension and this assumption makes the problem easier. Therefore, the left-hand side of the equation can be written as

    \[ -\oiint_S p \, d\vec{S} = - F_x - \int_{A_1} p \, d\vec{S} - \int_{A_2} p \, d\vec{S} \]

so the momentum equation in this case is

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

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 terms and remembering that d\vec{S} always points out of the control volume.

Also, continuity of the flow means that

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

so the momentum equation becomes

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

and so, finally, the reaction force is

    \[ R_x = -F_x \]

Worked Example #2 – Force on an Engine Test Stand

A static thrust stand is being 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 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 on the fluid by the engine. 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

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

The residual force on the fluid for the conditions stated 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, then the force to which the stand should be designed is about 251 kN.

Worked Example #3 – Water Jet Striking a Vertical Plate

A water jet with velocity V that exits from a nozzle strikes a vertical plate. The exit area of the nozzle is A. Find an expression for the horizontal force on the plate. If the volume flow rate through the nozzle is 0.5 m^{3} s^{-1} and the diameter of the jet is 5 cm, then find the force on the plate in units of Newtons. The density of water can be assumed to be 1,000 kg m^{-3}.

The force on the plate is obviously related to the force of the fluid and so to the exit velocity from the nozzle. Using the momentum equation and assuming that the flow is steady and inviscid then the force on the plate from the water jet is equal to the time rate of change in horizontal momentum, i.e.,

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

Using a one-dimensional assumption, then

    \[ F = \overbigdot{m} V - 0 \]

The use of the continuity equation gives

    \[ \varrho V A = \overbigdot{m} \]

and because we are dealing with water then it is incompressible, so

    \[ V A = \overbigdot{Q} = 0.5~\mbox{m$^{3}$ s$^{-1}$} \]

The value of V is

    \[ V = \frac{\overbigdot{Q}}{A} = \frac{0.5}{\pi d^2/4} = \frac{0.5}{\pi (0.05)^2/4} = 254.65~\mbox{m s$^{-1}$} \]

Therefore, the force on the plate is

    \[ F = \overbigdot{m} V = \left( \varrho \overbigdot{Q} \right) V = 1,000 \times 0.5 \times 254.65 =127.33~\mbox{kN} \]

Worked Example #4 – 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 control 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 control volume with annotations to analyze this problem is shown below.

 

2. This part of the problem requires the use of the continuity equation is 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 the 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, then  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.

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 it is expressed mathematically through the momentum equation. The equation takes into account various forces that act on a fluid, such as body forces, pressure forces, and viscous forces. Depending on the specific problem, the momentum equation can be simplified by making appropriate assumptions, such as steady flow 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 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 from the use of the momentum equation is negative, then what does this mean?
  • Think of some flow problems where the effects of viscosity are dominant and where 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 a good 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 conservation of momentum.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles 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