# Derivation of non-dimensional Navier Stoke equation

• I
• ussername
In summary, the Navier-Stokes equation can be simplified by taking the first three terms and defining the length of the velocity vector field with a unit vector field. However, the derivation process can be affected if the velocity is not a constant reference speed. Additionally, in the non-dimensional Navier-Stokes equation, it is important to determine which terms are negligible in order to accurately understand and interpret the equation.
ussername
Take the first three terms of Navier Stoke equation:
$$\rho \cdot \left ( v_{x}\cdot \frac{\partial \vec{v}}{\partial x} + v_{y}\cdot \frac{\partial \vec{v}}{\partial y} + v_{z}\cdot \frac{\partial \vec{v}}{\partial z}\right )$$

Define the length ##v## of the velocity vector field:
$$\vec{v}=v\cdot \vec{v}^{0}$$
where ##\vec{v}^{0}## is the unit vector field with same direction as ##\vec{v}##. Thus it is:$$v_{x}=v\cdot v_{x}^{0}$$$$v_{y}=v\cdot v_{y}^{0}$$$$v_{z}=v\cdot v_{z}^{0}$$

Now both terms ##v\cdot \vec{v}^{0}## can change with ##x## coordinate, so their derivation is:
$$\frac{\partial (v\cdot \vec{v}^{0})}{\partial x}=\vec{v}^{0}\cdot \frac{\partial v}{\partial x}+v\cdot \frac{\partial \vec{v}^{0}}{\partial x}$$

When I substitute these derivations into the NS equation (1. equation), I get something like this:
$$\rho \cdot\vec{v}^{0}\cdot v\cdot \left ( \vec{v}^{0}\cdot \left ( \frac{\partial v}{\partial x} , \frac{\partial v}{\partial y} , \frac{\partial v}{\partial z}\right ) \right )+\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )$$

But according to the literature it should be just the second term:
$$\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )=\rho \cdot v^{2}\cdot \left ( \vec{v}^{0}\cdot \left ( \mathrm{div} \vec{v}^{0} \right ) \right )$$

I don't think you are understanding it right. If the point is to get rid of the dimensions then ##v## should be some constant reference speed, not the speed of the fluid at each point.

Edit: I also suggest you stop using ##\cdot## when you don't mean scalar product, it makes your post difficult to read.

Orodruin said:
If the point is to get rid of the dimensions then ##v## should be some constant reference speed, not the speed of the fluid at each point.
So what is the definition of the unit velocity vector?

Edit: If ##v## is the average constant speed, than the length of ##\vec{v}^0## is not generally unity.

Last edited:
ussername said:
So what is the definition of the unit velocity vector?

Edit: If ##v## is the average constant speed, than the length of ##\vec{v}^0## is not generally unity.
Indeed. It is just a dimensionless velocity. Note that ##v## can be any fixed non-zero speed. You would typically fix it to some speed that appears in your problem, such as the speed of the flow at infinity.

The non-dimensional Navier Stokes equation is:
$$\frac{L\cdot f}{v}\frac{\partial \vec{v}^*}{\partial \vec{\tau}^*}+ v_x^*\frac{\partial \vec{v}^*}{\partial x^*}+ v_y^*\frac{\partial \vec{v}^*}{\partial y^*}+v_z^*\frac{\partial \vec{v}^*}{\partial z^*}=\frac{g\cdot L}{v^2}\vec{g}^*+\frac{p^0}{\rho \cdot v^2} \bigtriangledown^*p^*+\frac{\eta }{L\cdot v\cdot \rho } \bigtriangledown^{*2}\vec{v}^*$$
I understand that by setting physical parameters (##v,\rho,L...##) I can achieve in my scaled flow e.g. identical ratio of friction force to net force as in the original flow.

What I don't understand is how can I determine which terms in this equation are negligible. I know the physical dimensional parameters (##v,\rho,L...##) but I generally don't know the non-dimensional vectors (##v_x^*\frac{\partial \vec{v}^*}{\partial x^*}+ v_y^*\frac{\partial \vec{v}^*}{\partial y^*}+v_z^*\frac{\partial \vec{v}^*}{\partial z^*}##, ##\bigtriangledown^{*2}\vec{v}^*##...).
So how can I know which values are standing behind every term in the equation?

## 1. What is the purpose of deriving the non-dimensional Navier-Stokes equation?

The non-dimensional Navier-Stokes equation is a simplified version of the original Navier-Stokes equation, which describes the motion of a fluid. This derivation is done to remove any physical units from the equation and make it easier to solve for a wide range of problems. It allows for a better understanding of the underlying physics and behavior of fluid flow.

## 2. What are the assumptions made during the derivation process?

The derivation process assumes that the fluid is incompressible, meaning its density remains constant, and that the flow is steady, meaning its properties do not change over time. Additionally, it assumes that the flow is laminar, meaning it is smooth and orderly, and that the fluid has constant viscosity and density.

## 3. How is the non-dimensional Navier-Stokes equation used in practical applications?

The non-dimensional Navier-Stokes equation is used in a wide range of practical applications, such as aerodynamics, hydrodynamics, and chemical engineering. It is used to model and predict fluid flow behavior in various systems, including pipes, pumps, turbines, and aircraft wings. It is also used in the design and optimization of these systems.

## 4. What are the main differences between the non-dimensional and dimensional Navier-Stokes equations?

The main difference between the two equations is that the non-dimensional Navier-Stokes equation removes all physical units, such as length, time, and mass, from the equation. This makes it easier to solve and compare results for different problems. The dimensional equation, on the other hand, includes physical units and is more complex.

## 5. Are there any limitations to using the non-dimensional Navier-Stokes equation?

Although the non-dimensional Navier-Stokes equation is a useful tool in fluid mechanics, it has some limitations. It is only applicable to incompressible and laminar flows, and it cannot account for certain phenomena, such as turbulence. It also assumes that the fluid properties are constant, which may not always be the case in real-world scenarios.

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