# Derivation of Momentum Equation (Eulers Equation)

• aeroboyo
In summary: The second force in the x direction is due to the right face at x+ \delta x and is given by the pressure at (x+\delta x,y,z) times the area of this face. This force is in the negative x direction. Hence the contribution is -p(x+\delta x,y,z)\,\delta y\,\delta z. The contribution from the left face at x\,\delta x is the pressure at (x,y,z) times the area of this face and is directed in the positive x direction. Again, using a Taylor series we have -p(x,y,z) \approx -
aeroboyo
hey, I'm having difficulty understanding how Eulers Equation is derived... its because my grasp of calculus is preety bad.

Eulers equation is:

dp = -qVdV, where p is pressure, q is density and V is velocity. And that's clearly a differential equation. But i'd like to know how its derived. I know i have to start with Newtons 2nd law:

F=ma

my textbook says that

The force in the x direction acting on a fluid particle is

F = p dydz - (p + dp/dx dx) dydz (1)

Hence F = dp/dx dxdydz = force on fluid element due to pressure. (d)

This is where my confusion begins... I don't understand why the Force would be pressure multiplied by dydz and then minus pressure plus dp/dx multiplied by dydz. In other words i really don't understand what's going on with equation one, or how it becomes equation (2).

Can any1 help? I think once i understand this example i'll be able to work out others. Thanks.

aeroboyo said:
hey, I'm having difficulty understanding how Eulers Equation is derived... its because my grasp of calculus is preety bad.

Eulers equation is:

dp = -qVdV, where p is pressure, q is density and V is velocity. And that's clearly a differential equation. But i'd like to know how its derived. I know i have to start with Newtons 2nd law:

F=ma

my textbook says that

The force in the x direction acting on a fluid particle is

F = p dydz - (p + dp/dx dx) dydz (1)

Hence F = dp/dx dxdydz = force on fluid element due to pressure. (d)

This is where my confusion begins... I don't understand why the Force would be pressure multiplied by dydz and then minus pressure plus dp/dx multiplied by dydz. In other words i really don't understand what's going on with equation one, or how it becomes equation (2).

Can any1 help? I think once i understand this example i'll be able to work out others. Thanks.

Imagine a cuboid with sides $\delta x$, $\delta y$ and $\delta z$. There are two contributions to the force in the $x$ direction on the fluid in this cuboid. The first is due to the left face at $x$, and is given by the pressure at $(x,y,z)$ multiplied by the area of this face, which is $\delta y\,\delta z$. This force is in the positive $x$ direction. Hence the contribution is $p(x,y,z)\,\delta y\,\delta z$. The contribution from the right face at $x+ \delta x$ is the pressure at $(x + \delta x ,y,z)$ times the area of this face and is directed in the negative $x$ direction. Using a Taylor series we have to first order $$p(x + \delta x,y,x) \approx p(x,y,z) + \frac{\partial p}{\partial x}\,\delta x.$$ Thus to first order the net force in the $x$ direction is $$-\frac{\partial p}{\partial z}\,\delta x\,\delta y\,\delta z.$$

fresh_42

## 1. What is the derivation of the momentum equation?

The momentum equation, also known as Euler's equation, is derived from Newton's second law of motion, which states that the rate of change of an object's momentum is equal to the net force acting on the object. This equation is used to describe the relationship between the forces acting on a fluid and its resulting acceleration.

## 2. What are the assumptions made in the derivation of the momentum equation?

The derivation of the momentum equation assumes that the fluid is inviscid (no internal friction), incompressible (constant density), and the flow is steady and irrotational (no vorticity). Additionally, it assumes that the fluid is flowing in a straight line and that the external forces acting on the fluid are conservative.

## 3. How is the momentum equation applied in fluid mechanics?

The momentum equation is applied in fluid mechanics to predict the motion of a fluid based on the forces acting on it. It is used to solve problems involving fluid flow, such as calculating the velocity or pressure of a fluid at a specific point, or determining the forces acting on a fluid in a given system.

## 4. What are the different forms of the momentum equation?

The momentum equation can be written in different forms, depending on the specific situation and variables involved. The most commonly used forms are the 1D, 2D, and 3D forms, which represent the equation in one, two, and three dimensions, respectively. There is also a form known as the integral form, which is used for systems with complex geometries.

## 5. What is the significance of the momentum equation in fluid dynamics?

The momentum equation is a fundamental equation in fluid dynamics, as it allows us to understand and predict the behavior of fluids in motion. It is used in various applications, such as aerodynamics, hydrodynamics, and weather forecasting, to analyze and design systems involving fluid flow. It is also the basis for other important equations in fluid mechanics, such as the Bernoulli equation and the Navier-Stokes equations.

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