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.


In the final form of the equation, the first terms is incorrect. It should read $\frac{\partial u^2}{\partial x}$. The correct form of the final equation is obtained by first taking the continuity equation and multiplying it by u to obtain: $$u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y}=0$$You then add this to your next-to-last equation.


Not the answer you're looking for? Browse other questions tagged or ask your own question.