Navier Stokes Equation

Question

The general Navier Stokes Equation is $\dfrac{D\vec{v}}{D t}= \dfrac{d\vec{v}}{d t}+ \vec{v} .\nabla \vec{v} = \vec{g} - \dfrac{1}{\rho} \nabla p + \nu \nabla^2 \vec{v}$

The above equation can be reduced by the following assumptions

• Two dimensional flow

• Steady state

• u >> v

• $\mid \dfrac{\partial u}{\partial y}\mid \gg \mid\dfrac{\partial u}{\partial x}\mid$

• p = $\neq f(y)$

• $\nu =$ constant

The above equation reduces to In the x- direction

$u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial x} + \nu\left(\dfrac{\partial^2 u}{\partial y^2}\right)$

Then how can the above equation be reduced to the form $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial uv}{\partial y} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial x} + \nu\left(\dfrac{\partial^2 u}{\partial y^2}\right)$

The above equation was seen by me in A heat transfer textbook by John H Linehard around page 280, while deriving the Navier Stokes Equation.

Answer 1

Take the 2D, incompressible continuity equation $$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$$ and multiply it with $u$, leading to $$u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y}=0\,.$$ Now add the above equation to the reduced x-direction equation (the one you wrote correctly in your post); this will lead to $$2u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho}\frac{dp}{dx}+\nu\frac{\partial^2 u}{\partial y^2}$$ $$\Rightarrow \frac{\partial u^2}{\partial x}+\frac{\partial(uv)}{\partial y}=-\frac{1}{\rho}\frac{dp}{dx}+\nu\frac{\partial^2 u}{\partial y^2} \,.$$ Notice that the pressure derivative is a total one, since there is no dependence on other variables.

I think this is the equation you're looking for, check if you can that no typo was made while copying the equation from Lienhard's textbook. Let me know if I missed anything!

Answer 2

The reduction of the Navier-Stokes equation to the form ∂2u/∂x2 + ∂(uv)/∂y = -1/ρ * ∂p/∂x + ν * (∂2u/∂y2) can be obtained by combining the x-component and y-component equations and using the continuity equation.

First, let's recall the x-component equation obtained earlier: u∂u/∂x + v∂u/∂y = -1/ρ * ∂p/∂x + ν * (∂2u/∂y2).

Now, let's take the y-component equation: u∂v/∂x + v∂v/∂y = -1/ρ * ∂p/∂y + ν * (∂2v/∂y2).

We can eliminate the pressure term by taking the partial derivative of the x-component equation with respect to y and the partial derivative of the y-component equation with respect to x, and then adding the resulting equations. This gives:

u∂2u/∂x2 + v∂(∂u/∂y)/∂y = -1/ρ * (∂2p/∂x∂y) + ν * (∂3u/∂y2∂x)

Using the continuity equation, we know that ∂u/∂x + ∂v/∂y = 0. Rearranging this, we get ∂v/∂y = -∂u/∂x. Substituting this into the equation above and simplifying gives:

∂2u/∂x2 + ∂(uv)/∂y = -1/ρ * (∂p/∂x) + ν * (∂2u/∂y2)

which is the desired form.