CHAPTER02ARTICLE02

Fluid mechanics / Write conservation laws

FLUID MECHANICS · 02–02 / BASIC

Conservation of momentum

Formulate the relation between forces acting on a fluid, momentum change within a control volume, and momentum flux across its control surface.

14 min read2026-08-13Definitions, equations & units checkedJA version

Abstract

Momentum conservation is Newton's second law applied to a fluid. External force on a control volume equals momentum accumulation plus net momentum flux through the control surface, connecting directly to nozzles, bends, jets, wings and CFD.

What momentum means

Momentum describes how much motion an object has and in which direction. For an object of mass $m$ and velocity $\boldsymbol{v}$ , momentum $\boldsymbol{p}$ is defined by Equation (1). Greater mass or speed gives greater momentum, and the momentum vector points in the velocity direction.

A fluid also carries momentum because mass moves with the flow. For a small fluid mass $\mathrm d m$ moving at velocity $\boldsymbol{u}$ , $\mathrm d\boldsymbol{p}=\boldsymbol{u}\,\mathrm d m$ . Momentum per unit volume is therefore $\rho\boldsymbol{u}$ , and total momentum within a control volume is $\boldsymbol{P}$ = ∫ $\rho\boldsymbol{u}$ $\mathrm dV$ .

An external force changes momentum. Newton's second law expresses this relation in Equation (2). Force is required both to accelerate flow through a nozzle and to turn it through a pipe bend.

Equation (1)Definition of momentum

\boldsymbol{p}=m\boldsymbol{v}

$\boldsymbol{p}$: Momentum vector[ $\mathrm{kg{\cdot}m/s}$ ]
$m$: Mass[ $\mathrm{kg}$ ]
$\boldsymbol{v}$: Velocity vector[ $\mathrm{m/s}$ ]

Equation (2)Newton's second law

\sum\boldsymbol{F}=\frac{\mathrm d\boldsymbol{p}}{\mathrm d t}

For constant mass, this is equivalent to

\sum\boldsymbol{F}=m\boldsymbol{a}

When momentum is conserved

Treat several objects or a body of fluid as one system. If the resultant external force on the system is zero, Equation (2) shows that its total momentum does not change with time. Equation (3) states this basic meaning of momentum conservation.

Fluid enters and leaves piping and nozzles, so we choose a control volume fixed in space. For steady flow with one inlet and one outlet, when each section can be represented by a uniform velocity, the resultant force on the fluid follows Equation (4).

Equation (4) is a vector balance between the momentum flow entering and leaving. A nozzle changes speed, while a bend changes direction; either change requires a force.

Equation (3)Momentum conservation for an isolated system

\sum\boldsymbol{F}_{\mathrm{ext}}=\boldsymbol{0}\quad\Longrightarrow\quad\boldsymbol{P}_{\mathrm{tot}}=\text{constant}

The resultant external force on the system is zero. Internal action–reaction pairs do not change total momentum.

Equation (4)Steady one-dimensional momentum balance

\sum\boldsymbol{F}=\dot m\left(\boldsymbol{V}_2-\boldsymbol{V}_1\right)

Steady flow with one inlet and one outlet. Include every external force on the fluid, such as pressure, gravity and support forces.

Symbols & units

Begin with the symbols used in Equations (1)–(4). Keep vectors and scalars distinct and check SI dimensions. Momentum flow rate $\dot m\boldsymbol{u}$ has units kg· $\mathrm{m/s^2}$ = N, the same dimension as force.

Principal symbols in the basic equations
Symbol	Meaning	SI unit
$\boldsymbol{p}$	Momentum vector	$\mathrm{kg{\cdot}m/s}$
$m$	Mass	$\mathrm{kg}$
$\boldsymbol{v}$	Object velocity vector	$\mathrm{m/s}$
$\rho$	Fluid density	$\mathrm{kg/m^3}$
$\boldsymbol{u}$	Fluid velocity vector	$\mathrm{m/s}$
$\sum\boldsymbol{F}$	Resultant external force on the fluid	$\mathrm{N}$
$\dot m$	Mass flow rate	$\mathrm{kg/s}$

Extension to the general form (advanced)

Equation (4) is the practical form for steady flow when inlet and outlet velocities can be represented by section values. Unsteady flow and nonuniform velocity profiles require momentum inside the control volume and momentum crossing its surface to be integrated.

Apply $\sum\boldsymbol{F}=\frac{\mathrm d\boldsymbol{P}}{\mathrm dt}$ to a material system and write $\boldsymbol{P}$ = ∫ $\rho\boldsymbol{u}$ $\mathrm dV$ . The Reynolds transport theorem separates its change into accumulation inside the control volume and flux through the control surface, giving Equation (5) [1,2].

Equation (6) is the local form used for stress fields and computational fluid dynamics. A first reading need only proceed through Equation (4).

Equation (5)General integral form for a fixed control volume

\begin{aligned}\sum\boldsymbol{F}&=\frac{\mathrm d}{\mathrm d t}\int_{\mathrm{CV}}\rho\boldsymbol{u}\,\mathrm dV\\&\quad+\oint_{\mathrm{CS}}\rho\boldsymbol{u}(\boldsymbol{u}\cdot\boldsymbol{n})\,\mathrm dA\end{aligned}

\boldsymbol{n}

is the outward unit normal to

\mathrm{CS}

. Outflow has

\boldsymbol{u}\cdot\boldsymbol{n}>0

and inflow has

\boldsymbol{u}\cdot\boldsymbol{n}<0

Equation (6)Cauchy momentum equation (advanced)

\rho\frac{\mathrm D\boldsymbol{u}}{\mathrm D t}=\nabla\cdot\boldsymbol{\sigma}+\rho\boldsymbol{b}

Continuum approximation;

\boldsymbol{\sigma}

is the Cauchy stress tensor and

\boldsymbol{b}

is body force per unit mass.

01: Apply Newton's second law to the material system.
02: Write momentum as a volume integral.
03: Split the material rate into accumulation and flux.
04: Resolve all external forces consistently.

Validity conditions

State the control-volume choice, steadiness, velocity-profile approximation, force directions and sign convention.

Fixed control volume: Equation (5) is for a stationary control volume; moving boundaries require the general relative-velocity form.
Steady flow: Accumulation vanishes, but momentum flux remains.
1-D approximation: Mean velocity may require a momentum correction factor.
Vector nature: Resolve $x$ , $y$ and $z$ components for bends.

Worked example: horizontal nozzle

For steady water flow, let $\rho$ = 1000 $\mathrm{kg/m^3}$ , $A_1=0.010\ \mathrm{m^2}$ , $V$ ₁ = 2.0 $\mathrm{m/s}$ and $V$ ₂ = 6.0 $\mathrm{m/s}$ . Neglect pressure-force difference and $x$ -directed gravity only for this simplified example.

The mass flow rate is 20 $\mathrm{kg/s}$ and the required $x$ -directed force on the fluid is $80\ \mathrm{N}$ . The reaction exerted by the fluid on the nozzle is opposite in direction.

Equation (7)Mass flow rate

\dot m=\rho A_1V_1=20\ \mathrm{kg/s}

Equation (8)Net x-force

F_x=\dot m(V_2-V_1)=80\ \mathrm{N}

Calculator model

A calculator can restrict Equation (4) to collinear flow and replace mass flow rate by ρA₁ $V$ ₁, giving $F=\rho A_1V_1(V_2-V_1)$ . This does not mean pressure and gravity may always be ignored; it is a simplified calculation used only after confirming the relevant assumptions.

Common mistakes

First fix which body the force acts on and define the positive direction.

Forgetting pressure force: Include pressure surface forces unless both sections share the same ambient pressure.
Fluid vs hardware force: Solve force on the fluid first, then reverse direction for the hardware reaction.
Treating a bend as scalar: Momentum is a vector; balance each component.
Steady means no force: Steady flow can still have different inlet and outlet momentum fluxes.

Applications & related topics

Momentum conservation underpins nozzle thrust, pipe-bend reactions, jet impact, lift and drag, turbomachinery and finite-volume CFD. It is normally used together with mass conservation.

References

Sources accessed 13 August 2026. Independent expert review has not yet been completed.