This is a flow in a 2-D microchannel (h/L<<1). Low Reynolds number flow.

The mass - conservation equation states $$\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$ Where, $$\rho=\rho(x)$$ How to get this equation $$\frac{\partial p}{\partial x}=\mu \frac{\partial ^2u}{\partial y^2}$$ from the momentum balance equation i.e. $$\rho( {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u}) = - \nabla p + \mu\nabla^2{\bf u} + {\mu}{1\over3}\nabla (\nabla \cdot {\bf u})$$ Taking scales can eliminate the left- hand side as this a low Reynolds number flow. But I can't vanish the last term on the right- hand side.

Is it possible $$\rho{\bf u} \cdot \nabla {\bf u}={\bf u} \cdot \nabla (\rho{\bf u})=0 $$ even after considering $$\rho=\rho(x)$$

Or should I use this form of equation $$ {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u} = - {1\over\rho} \nabla p + \gamma\nabla^2{\bf u} + {g_{x}} $$ as $$\rho g_{x}=0$$ please help.

  • $\begingroup$ You say that you cant vanish tge last term on the right hand side, then go on to discuss the last term on the left hand side. Which term are you having problems with? $\endgroup$
    – Eddy
    Commented Jan 3, 2018 at 20:32
  • $\begingroup$ If I take length, velocity& pressure scale to make Re as it is low Re flow the left-hand side would vanish. $\rho=\rho(x)$density is the function of x. The mass conservation equation is $$\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$ So it looks like$$ \nabla (\rho{\bf u})=0 $$ May I write it as $$ \rho\nabla {\bf u}=0$$ That means, may I take density as constant? although density is a function of x. Or is there any other way to get $\frac{\partial p}{\partial x}=\mu \frac{\partial ^2u}{\partial y^2}$ from the momentum balance equation. $\endgroup$
    – Artisan
    Commented Jan 3, 2018 at 21:05
  • $\begingroup$ Why do you believe that you should be able to get that equation? I have only ever seen that equation derived for flows of constant density. $\endgroup$
    – Eddy
    Commented Jan 3, 2018 at 21:12
  • $\begingroup$ In the spatial coordinate frame has constant density at every value of x. and the density profile is stationary. In other way may I use this form of equation $ {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u} = - {1\over\rho} \nabla p + \gamma\nabla^2{\bf u} + {g_{x}} $ as $\rho g_{x}=0$. Is it possible ? $\endgroup$
    – Artisan
    Commented Jan 3, 2018 at 21:24
  • $\begingroup$ Including gravity won't help you. Why do you think that $\frac{\partial p}{\partial x}=\mu \frac{\partial ^2u}{\partial y^2}$ is true? $\endgroup$
    – Eddy
    Commented Jan 3, 2018 at 21:27

1 Answer 1


If we apply the same coordinate transformation to the momentum equation that we employed in the other forum (Physics Forums), we obtain for the X component of the momentum equation: $$U\frac{\partial U}{\partial X}+V\frac{\partial U}{\partial Y}=-\frac{1}{\rho}\frac{\partial p}{\partial X}+\frac{\mu}{\rho}\left[\frac{\partial^2U}{\partial X^2}+\frac{\partial^2U}{\partial Y^2}\right]+\frac{\mu}{3\rho}\left[\frac{\partial ^2U}{\partial X^2}+\frac{\partial^2V}{\partial X\partial Y}\right]\tag{1}$$ We next define the following dimensionless parameters: $$\bar{U}=\frac{U}{c}$$ $$\bar{V}=\frac{\lambda}{h}\frac{V}{c}$$$$\bar{X}=\frac{X}{\lambda}$$ $$\bar{Y}=\frac{Y}{h}$$$$\bar{\rho}=\frac{\rho}{\rho_0}$$ $$\bar{p}=\frac{h^2p}{\mu c \lambda}$$ If we substitute these dimensionless variables into Eqn. 1, we obtain: $$Re\left(\frac{h}{\lambda}\right)^2\bar{\rho}\left[\bar{U}\frac{\partial \bar{U}}{\partial \bar{X}}+\bar{V}\frac{\partial \bar{U}}{\partial \bar{Y}}\right]=-\frac{\partial \bar{p}}{\partial \bar{X}}+\left(\frac{h}{\lambda}\right)^2\frac{\partial^2\bar{U}}{\partial \bar{X}^2}+\frac{\partial^2\bar{U}}{\partial \bar{Y}^2}+\frac{1}{3}\left(\frac{h}{\lambda}\right)^2\left[\frac{\partial ^2\bar{U}}{\partial \bar{X}^2}+\frac{\partial^2\bar{V}}{\partial \bar{X}\partial \bar{Y}}\right]\tag{2}$$where the Reynolds number Re is given by: $$Re=\frac{\rho_0 c \lambda}{\mu}$$In the limit of $Re\rightarrow 0$ and $(h/\lambda)^2\rightarrow 0$, Eqn. 2 becomes: $$0=-\frac{\partial \bar{p}}{\partial \bar{X}}+\frac{\partial^2\bar{U}}{\partial \bar{Y}^2}\tag{3}$$In terms of the original dimensional variables, this now becomes:$$0=-\frac{\partial p}{\partial X}+\mu \frac{\partial^2U}{\partial Y^2}\tag{3}$$

  • $\begingroup$ +1 Dimensional scaling is always the best way to determine the most (in)significant terms in these situations. However, it may be unclear for others why the scale (in this case) for $V$ is $hc/\lambda$ and for $p$ is $uc\lambda/h^2$. Can you motivate your choice? $\endgroup$
    – nluigi
    Commented Jan 4, 2018 at 7:40
  • $\begingroup$ The choice was not random but, instead, was based on a specific rational methodology taught to us as first year graduate students at the Univ. of Michigan by S. W. Churchill in 1963. It is based on a paper published in 1962 by Hellums and Churchill, as follow-up to Hellums' doctoral thesis. The details are more lengthy than I'd be inclined to explain in the present format. See if you can locate the reference, and, if so, tell us what you think. $\endgroup$ Commented Jan 4, 2018 at 12:02

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