Momentum equation in fluid dynamics

Abstract

The momentum equation in fluid dynamics is a direct extension of Newton’s Second Law, establishing that the change in momentum of a fluid is equal to the forces applied on it. This principle links the physical reality of motion with the mathematical description of control volumes and flow fields. Momentum can be altered through both body forces, such as gravity or electromagnetism, and surface forces, such as pressure and viscous stresses. By considering inflows, outflows, and unsteady changes within a volume, the momentum equation captures the dynamic balance between forces and the time rate of change of momentum. Simplified cases, such as steady or inviscid flows, lead to forms like the Euler equations, while including viscosity yields the Navier–Stokes equations. The concept is not merely theoretical but underpins phenomena ranging from the swaying of tree branches in the wind to the lift generated on an aircraft wing. It explains how forces and momentum interact, providing a foundation for understanding both everyday experiences and advanced engineering systems.

Key words

Momentum

The product of mass and velocity of a body or fluid element, representing the quantity of motion carried.

Control volume

A fixed or moving region in space through which fluid flows, used to analyze balance laws such as mass, momentum, and energy.

Body forces

Forces that act throughout the volume of a fluid, such as gravity or electromagnetic forces, rather than only on surfaces.

Surface forces

Forces exerted on the boundary of a control volume, including pressure and viscous stresses.

Inviscid flow

An idealized flow where viscosity is neglected, simplifying the momentum equation to the Euler equations.

Navier-Stokes equations

The governing equations of viscous fluid flow, derived from the momentum equation when viscosity is accounted for.

Euler equations

A simplified form of the momentum equation for inviscid flows, often applied to high-Reynolds-number problems.

Divergence theorem

A mathematical tool that relates a surface integral to a volume integral, used in fluid dynamics to convert surface fluxes into volume terms.

Unsteady flow

Flow in which velocity or other properties vary with time within the control volume.

Steady flow

Flow in which velocity and other properties remain constant with time at any given point.

Momentum equation

Let’s think for a minute about forces. Imagine that person A is moving on the street, but not just moving, that person is running because he is on a rush!; suddenly, person B appears on the scene, and due to the speed person A had, he is unable to evade the collision, so they crash. You can imagine this situation, of course that they are going to feel the impact, it is going to be painful, especially because person A had a great force, but how can we know this? According to Newton’s Second Law of Motion, we are sure that:

$F=m\vec a$

Since we assume that person A had a certain mass and we know that he was running, we can then say that he had a certain force accumulated, and this is how this equation works. However, since it was person A who had the force, how is person B going to feel it? Well, here comes into place the Momentum Equation, the force will be applied, transmitted, into person B.

In a different situation, imagine a UFC fight competition, when the fighter throws a punch, that punch has certain mass and acceleration, therefore, certain force; the other fighter receives all the force into his face due to the Momentum Equation and then he gets knocked out.

Now, we can think about something different. Newton’s First Law of Motion establishes that a body will remain at rest unless an external force acts on it, but what do you think? What is the natural state of things—rest or movement? Think about a beautiful day in the park, you are seated next to a magnificent tree, you see how the sun barely passes through the branches, so you look up and see how the branches and their leaves are moving, but wait a minute, how are they moving? There must be a force acting on them! Yes, the air particles. You are a leaf, you are hanging from the branch when suddenly air particles start hitting you, inducing a force, making you move; the Momentum Equation describes which forces are acting on you and how your velocity will change.

Understanding momentum in fluid dynamics

Let’s go with the basics, Newton’s second law is frequently written as [1]:

$F=m\vec a$

Where F is the force exerted to a body of mass m and a is the acceleration; however, a more general form would be:

$F=\frac{d}{dt}(m\vec V)$

In this new equation mV is the momentum of a body of mass m; not only that, this also represents a fundamental principle upon which theoretical fluid dynamics is based:

$Force=Time~Rate~of~Change~of~Momentum$

Now, let’s limit our analysis to a control volume fixed in space, now, the goal is to obtain expressions in terms of the flow-field variables:

$p, \rho, V,...$

From the last equation that we have, we can see, of course, two expressions, one in the left and the other on the right. The left side expresses the forces exerted on the fluid as it flows through the control volume (the change of momentum), while the right side is the sum of all these forces. These forces come up from:

Body forces: gravity, electromagnetic forces, or any other that acts at the distance on the fluid inside the control volume (V).
Surface forces: Pressure and shear stress acting on the control surface (S).

Focusing in the left side, let’s use f to represent the net body force per mass exerted on the fluid inside V (in the same way, we want to know the force per unit volume, so we multiply f by the density at that specific point). So then, the body force on the elemental volume dV (which is a tiny piece of the control volume) is:

$\rho f d\mathcal{V}$

And the total body force exerted on the fluid of the volume control is the summation (integration in all three spatial dimensions) of the above over the volume V.

$\iiint\limits_{\mathcal{V}} \rho f \, d\mathcal{V}$

So this expression is telling us that we are integrating the body force at every tiny volume element (dV) throughout the whole control volume.

Following the same line of thought, as mentioned above, we have two types of forces, the ones in the body and the ones on the surface. Now, going to the surface forces, we can add the pressure that is acting on the element of area dS.

$-p dS$

Where the negative sign indicates that the pressure is directed into the control volume (due to the pressure of its surroundings). Therefore, the complete pressure force is the summation of the elemental surface over the entire control surface (remembering that the surface is in a 2D dimension):

$-\iint \limits_{\mathcal{S}}pdS$

We know that shear and normal viscous stresses also act on the surface, so we will recognize them as: $F_{viscous}$

At this point, the left side in the equation, therefore:

$F=\iiint\limits_{\mathcal{V}} \rho f \, d\mathcal{V}-\iint \limits_{\mathcal{S}}pdS+F_{viscous}$

So let’s go with the right side of the equation and remembering the physical principle of the force, the time rate of change of momentum of the fluid as it sweeps through the fixed control volume is the sum of two terms [2]:

$Net~flow~of~momentum~out~of~control~volume~across~surface~S\equiv G$

and

$Time~rate~of~change~of~momentum~due~to~unsteady~fluctuations~of~flow~properties~inside~\mathcal V\equiv H$

Focusing on term G, we can imagine that the flow enters the control volume with a certain momentum and it leaves it with a different momentum. With this said, we can note that the net flow of momentum out of the control volume across the surface S is simply this outflow minus the inflow of momentum across the control surface. To obtain the expression for G, we need to remember that the mass flow equation across the elemental area dS is:

$\dot m = \rho \vec V dS$

So, the flow momentum per second across dS is:

$(\rho \vec V\cdot dS)\vec V$

So, what this formula is telling is how much mass of fluid passes through it per second times its velocity; in other words, how much momentum is passing (or leaving). And just as we did earlier, to get the net flow of momentum out of the control volume through S is the summation of the elemental contributions:

$G=\iint \limits_{S}(\rho \vec V\cdot dS)\vec V$

It is important to denote that positive values of G represent that the mass flow rate if going out of the control volume, and viceversa. If G has a positive value, there is more momentum flowing out of the control volume per second than flowing in, and viceversa.

Now, it is time to consider H. The momentum of the fluid in the elemental volume dV is:

$(\rho d\mathcal V)\vec V$

So now, doing the summation (since we are talking about the volume, we are looking for a 3D sum):

$H=\iiint \limits_{\mathcal V}\rho \vec V d \mathcal V$

And, its time rate of change is:

$H=\frac{\partial}{\partial t}\iiint \limits_{\mathcal V}\rho \vec V d \mathcal V$

Now that we have both terms, G and H, we can combine them to have the total time rate of change of momentum as it sweeps through the fixed control volume, which in turn represents the right side of the equation:

$\frac{d}{dt}(mV)=G+H=\iint \limits_{S}(\rho \vec V\cdot dS)\vec V+\frac{\partial}{\partial t}\iiint \limits_{\mathcal V}\rho \vec V d \mathcal V$

Therefore, from Newton’s second law we got its application into fluids. Now, we recall both left (forces applied) and right sides (changes in the momentum) of the equation and organize a little bit:

$\frac{d}{dt}(mV)=F$

$\frac{\partial}{\partial t}\iiint \limits_{\mathcal V}\rho \vec V d \mathcal V ~+~\iint \limits_{S}(\rho \vec V\cdot dS)\vec V~=-\iint \limits_{\mathcal{S}}pdS~+~~\iiint\limits_{\mathcal{V}} \rho f \, d\mathcal{V}~+~F_{viscous}$

Now, this is a powerful equation that tells us how properties change for an entire region, but what about if we want to know about a specific point? So, we need to proceed to a partial differential equation which relates flow-field properties at a point in space. We can see how we have properties in terms of volume and others in terms of surfaces, for this, we need to recall the Gradient Theorem which relates integrals over surfaces and integrals over volumes:

$Gradient~Theorem: \iint \limits_{S} pdS=\iiint\limits_{\mathcal{V}}\nabla p d\mathcal{V}$

Now, if we apply it to the first term on the right hand side of the equation (pressure on the surface):

$-\iint \limits_{S} pdS=-\iiint\limits_{\mathcal{V}}\nabla p d\mathcal{V}$

We know that the volume control is fixed, constant, for this reason, we can place the time derivative from the momentum equation inside the integral, so then the equation would be:

$\iiint \limits_{\mathcal V}\frac{\partial (\rho \vec V)}{\partial t} d \mathcal V ~+~\iint \limits_{S}(\rho \vec V\cdot dS)\vec V~=-\iiint\limits_{\mathcal{V}}\nabla p d\mathcal{V}~+~~\iiint\limits_{\mathcal{V}} \rho f \, d\mathcal{V}~+~F_{viscous}$

We can see how this is a vector equation, so, in order to gain clarity and to see clearly how the fluid behaves for each component separately, we will use:

$\vec V=ui+vj+wk$

Starting with the x component, we will have the next equation:

$\iiint \limits_{\mathcal V} \frac{\partial (\rho u)}{\partial t}d\mathcal{V}+\iint \limits_{S}(\rho \vec V\cdot dS)u=-\iiint\limits_{\mathcal{V}}\frac{\partial p}{\partial x} d\mathcal{V}~+~~\iiint\limits_{\mathcal{V}} \rho f_x \, d\mathcal{V}~+~(F_x)_{viscous}$

Remember that the gradient of the pressure is expressed as:

$\nabla p=\frac{\partial p}{\partial x}i+\frac{\partial p}{\partial y}j+\frac{\partial p}{\partial z}k$

At this time, we can note that $(\rho \vec V\cdot dS)$ is a scalar (hence, does not have components), in the same line, in order to have properties in terms of volumes, we will use the divergence theorem, which relates a surface scalar to a volume vector:

$Divergence~Theorem:~\iint \limits_{S}A \cdot dS=\iiint \limits_{\mathcal V}(\nabla \cdot A) d \mathcal V$

Now, we apply it to the second component of the left side of the equation:

$\iint \limits_{S}(\rho \vec V \cdot dS) u =\iint \limits_{\mathcal S}(\rho u V)\cdot dS = \iiint \limits_{\mathcal V}\nabla \cdot (\rho u V)d\mathcal V$

What is happening in this theorem? We recognize that u is the component of x inside the vector V, so u is a scalar, for this reason, we can put it inside of the parenthesis which becomes in A according to the divergence theorem, the only left part is to apply the theorem.

Finally, if we substitute again, we will have:

$\iiint \limits_{\mathcal V}[\frac{\partial (\rho u)}{\partial t}+\nabla \cdot(\rho uV)+\frac{\partial p}{\partial x}-\rho f_x-(f_x)_{viscous}]d\mathcal V=0$

Now, since the control volume is arbitrary and we are the ones that can select its values and also due to this last equation, we can know that the integrand at all points will be zero, therefore, we can rewrite the equation for all its components:

$For~x:~~\frac{\partial (\rho u)}{\partial t}+\nabla \cdot (\rho uV)=-\frac{\partial p}{\partial x}+\rho f_x+(f_x)_{viscous}$

$For~y:~~\frac{\partial (\rho v)}{\partial t}+\nabla \cdot (\rho vV)=-\frac{\partial p}{\partial y}+\rho f_y+(f_y)_{viscous}$

$For~z:~~\frac{\partial (\rho w)}{\partial t}+\nabla \cdot (\rho wV)=-\frac{\partial p}{\partial z}+\rho f_z+(f_z)_{viscous}$

It is worth to remember that these equations represent the scalar values, they are partial differential equations that relate flow-field properties at any point in the flow.

But for now, let’s go back to Newton’s Second Law applied to fluids:

There are some things that we can notice:

$\frac{\partial}{\partial t}$ → this tells us that the physical properties may change in time.
$\rho$ → is a variable, so the fluid could be compressible.
$f$ → body forces are present.
$F_{viscous}$ → Viscous forces are being taken into account.

Therefore, we can notice how this equation acts for steady or unsteady flows. If we want them specialized for steady flows, then we set:

$\frac{\partial}{\partial t}=0$
$F_{viscous}=0$
$f=0$

But why we would want it to be specialized for steady flows? This helps to simplify equations, so they become algebraic or simpler integrals; also because it matches many real world situations as many engineering systems are designed to operate at this conditions. With this said, our equation would then become:

$\iint \limits_{\mathcal S}(\rho V\cdot dS)V=-\iint \limits_S pdS$

And their components turn to:

$For~x:~~\nabla \cdot (\rho uV)=-\frac{\partial p}{\partial x}$

$For~y:~~\nabla \cdot (\rho vV)=-\frac{\partial p}{\partial y}$

$For~z:~~\nabla \cdot (\rho wV)=-\frac{\partial p}{\partial z}$

Finally we can just mention that the equations for inviscid flows are called as Euler equations, while the ones for viscous flows are the Navier-Stokes equations.

Visually thinking

Our momentum equation is a direct consequence from the super famous Newton’s Second Law. Let’s remember that an equation is a balance, something is balanced/equal with other thing, in the case of momentum, the change in momentum is equal to the forces applied:

$\iint \limits_S (\rho V\cdot dS)V=-\iint \limits_SpdS$

So then, we can see how the net rate at which momentum leaves or enters the control surface must be equal to the total force due to pressure acting on the surface.

Now, let’s return to the situation at the beginning of this mini-essay. Imagine you are in a beautiful sunny day, seated under a magnificent tree, you look up and see how the branches of the tree are moving. It is clear for us now to see how the air carries out a mass and it has certain velocity, then it impacts the branch changing redirecting the air and changing its momentum (changing its velocity); this change is balanced by the pressure that acts in the branch, making it to move.

In the same way, think about an airplane, think about the wing that is the object that allows an airplane to flight. The airplane starts accelerating to take off, the air is impacting the wing, the air has a velocity and a mass (a momentum), when this air impacts the wing, the momentum changes and it is balanced through the pressure that acts on the wing, creating lift and allowing the airplane to fly.

Conclusions

The momentum equation is not just an abstract thing with abstract shapes on its equation, it is a fundamental term that allows us to explain and try to understand the world. It is present in fun things like playing soccer, beautiful nature like leaves moving, and more fantastic things like allowing humans to do the impossible, to fly.

Understanding this equation, is understanding movement in the world.

Sources of information

[1] Fundamentals of Aerodynamics (p. 130) – John D. Anderson

[2] Fluid Mechanics Fundamentals (p. 256) – Yunus A. Cengel, John M. Cimbala