# Integral form of Navier-Stokes Equation

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• FluidStu
In summary: Thus, the total flux of momentum across ##V## is$$\rho\vec{u}=\partial_t \rho+\vec{\nabla}(\rho \vec{v}),$$and the total pressure across ##V## is$$P\rho=\vec{\nabla}(\rho \vec{v}).$$In summary, the Navier-Stokes equation may be written as: if we have a fixed volume (a so-called control volume) then the integral of throughout V yields, with the help of Gauss' theorem: the definition of Gauss' theorem: could someone show me how to go from the integral
FluidStu
The Navier-Stokes equation may be written as:

If we have a fixed volume (a so-called control volume) then the integral of throughout V yields, with the help of Gauss' theorem:

(from 'Turbulence' by Davidson).

The definition of Gauss' theorem:

Could someone show me how to go from the integral over the control volume to the closed surface integrals on the left of the second equation? Do we need to use Leibniz's integral rule or Reynolds Transport theorem?

FluidStu said:
The Navier-Stokes equation may be written as:
View attachment 101838

If we have a fixed volume (a so-called control volume) then the integral of throughout V yields, with the help of Gauss' theorem:
View attachment 101837
(from 'Turbulence' by Davidson).

The definition of Gauss' theorem:
View attachment 101836

Could someone show me how to go from the integral over the control volume to the closed surface integrals on the left of the second equation? Do we need to use Leibniz's integral rule or Reynolds Transport theorem?
I think you meant "right", not "left". The first equation assumes that the density is constant, so you can multiply both sides by ##\rho##. You have the identity:
$$\rho\vec{u}\cdot \nabla \vec{u}+\rho \vec{u}\nabla \cdot \vec{u}=\nabla \cdot (\rho \vec{u} \vec{u})$$
But, for an incompressible fluid, the second term on the left is equal to zero. Therefore, $$\rho\vec{u}\cdot \nabla \vec{u}=\nabla \cdot (\rho \vec{u} \vec{u})$$
Does this help?

Chestermiller said:
I think you meant "right", not "left". The first equation assumes that the density is constant, so you can multiply both sides by ##\rho##. You have the identity:
$$\rho\vec{u}\cdot \nabla \vec{u}+\rho \vec{u}\nabla \cdot \vec{u}=\nabla \cdot (\rho \vec{u} \vec{u})$$
But, for an incompressible fluid, the second term on the left is equal to zero. Therefore, $$\rho\vec{u}\cdot \nabla \vec{u}=\nabla \cdot (\rho \vec{u} \vec{u})$$
Does this help?

I have not encountered this identity. Secondly, the RHS of the identity has two vectors multiplied, which I don't understand since vector multiplication is either dot product, cross product or combinations thereof?

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FluidStu said:
Firstly, I can't see why the author chose to represent the material derivative as (U.∇)U - I've previously always encountered it as U.(∇U), which makes more intuitive sense to me. Clearly these must be the same, but I can't see how?
Write it out in component form for Cartesian coordinates and you will see how.
Secondly, I have not encountered this identity.
Again, write it out in component form for Cartesian coordinates.
Thirdly, the RHS of the identity has two vectors multiplied, which I don't understand since vector multiplication is either dot product, cross product or combinations thereof?
I guess you are not familiar with dyadic tensor notation. A dyad consists of two vectors placed in juxtaposition, without any operation such as dot product or cross product implied between the two. Such an entity is a 2nd order tensor. In this case, the tensor ##\rho \vec{u} \vec{u}## represents the momentum flux per unit volume of fluid. To learn more about the powerful use of dyadic tensor notation and the implementation of mathematical operations using dyadics, see Transport Phenomena by Bird, Stewart, and Lightfoot (Appendices). In my judgment, it is something very worthwhile to learn by those learning and using Fluid Mechanics.

By the way, when you wrote ##(\vec{\nabla} \vec{u})##, you were already employing a dyadic of the vector operator ##\vec{\nabla}## and the vector ##\vec{u}##.

FluidStu
Chestermiller is obviously using a tensor notation with the abbreviation ##\vec{u} \vec{u}=\vec{u} \otimes \vec{u}##. It's easier in components. The term in the NS equation we are discussing we have (in Cartesian coordinates)
$$(\vec{u} \cdot \vec{\nabla} \vec{u})_j=u_k \partial_k u_j=\partial_k(u_k u_j)-u_j \partial_k u_k.$$
Now obviously, it's tacidly assumed that the fluid is incompressible and thus
$$(\vec{u} \cdot \vec{\nabla} \vec{u})_j = \partial_k(u_k u_j),$$
i.e., a total divergence that can be transformed to a surface term when integrating over ##V##.

What's also used is the continuity equation
$$\partial_t \rho+\vec{\nabla}(\rho \vec{v})=0.$$

FluidStu

## 1. What is the integral form of Navier-Stokes Equation?

The integral form of Navier-Stokes Equation is a mathematical equation that describes the motion of fluid substances. It is an important tool in the field of fluid dynamics, used to model and predict the behavior of fluids in various scenarios.

## 2. How is the integral form of Navier-Stokes Equation derived?

The integral form of Navier-Stokes Equation is derived from the original differential form by applying the Reynolds Transport Theorem, which converts the partial derivatives into total derivatives. This allows for the equation to be written in terms of volume integrals instead of partial derivatives.

## 3. What are the applications of the integral form of Navier-Stokes Equation?

The integral form of Navier-Stokes Equation is used in many fields, including engineering, meteorology, and oceanography. It can be used to analyze and design fluid systems, such as pumps and turbines, as well as to predict weather patterns and ocean currents.

## 4. What are the limitations of the integral form of Navier-Stokes Equation?

One major limitation of the integral form of Navier-Stokes Equation is that it assumes the fluid being studied is continuous and homogeneous, which is not always the case in real-world scenarios. Additionally, it only applies to Newtonian fluids, which have constant viscosity and do not exhibit non-Newtonian behavior.

## 5. What are some current research topics related to the integral form of Navier-Stokes Equation?

Some current research topics related to the integral form of Navier-Stokes Equation include the development of numerical methods for solving the equation, the incorporation of turbulence models, and the extension of the equation to non-Newtonian fluids. Additionally, there is ongoing research on simplifying and approximating the equation for more efficient computational analysis.

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