Consider one-dimensional Poiseuille flow between two flat plates placed at $2h$ from each other, with a known pressure gradient $\partial p/\partial x=G$. The Navier-Stokes equation for this flow is $$G=\mu\frac{\mathrm{d}^2u}{\mathrm{d}y^2}$$ with the boundary conditions $u=0$ at $y=\pm h$. If we solve this equation we get $$u=-\frac{G}{2\mu}(h^2-y^2)$$ and the average velocity $$U=\frac{1}{2h}\int_{-h}^{h}u\mathrm{d}y=-\frac{Gh^2}{3\mu};$$ then the non-dimensional velocity is $$u^*=\frac{u}{U}=\frac{3}{2}\left(1-\left(\frac{y}{h}\right)^2\right)=\frac{3}{2}\left(1-y^{*2}\right)$$ which does not depend on any flow parameter.

On the other hand, if we make the equation dimensionless from scratch we get $$-Re=\frac{\mathrm{d}^2u^*}{\mathrm{d}y^{*2}}$$ where $Re=\frac{-Gh^2}{U\mu}$, and its solution is $$u^*=\frac{Re}{2}(1-y^{*2})$$ which is dependent on $Re$!

This means that we cannot freely choose $Re$ and it must be equal to $3$ in any case!

Why does this paradox occur?

  • $\begingroup$ Where did you get that definition of Re from? It should be $\frac{\rho Uh}{\mu}$ $\endgroup$ Sep 14, 2022 at 11:20

1 Answer 1


By restricting the solution to steady, fully-developed unidirectional flow , the velocity does not depend on time or the $x$-coordinate. The only relevant length scale in the $y-$dimension is $Y_c=h$. Leaving the characteristic scale for velocity $U_c$, pressure $p_c$, and $x$-dimension $X_c$ unspecified for the moment, the reduced Navier-Stokes equation is

$$\frac{\mu U_c}{h^2}\frac{d^2 u^*}{d y^{*2}}= \frac{p_c}{X_c}\frac{\partial p^*}{\partial x^*}$$

where the dimensionless variables are $u^* = u/U_c$, $y^* = y/h$, $p^* = p/p_c$, and $x^* = x/X_c$.

Rearranging we get,

$$\tag{1}\frac{d^2 u^*}{d y^{*2}}= \frac{h^2p_c}{\mu U_cX_c}\frac{\partial p^*}{\partial x^*}= \underbrace{\frac{\rho U_c h}{\mu}}_{Re} \frac{h}{X_c} \frac{p_c}{\rho U_c^2}\frac{\partial p^*}{\partial x^*}$$

At this point the usual Reynolds number $Re = \frac{\rho U_c h}{\mu}$ appears. Ostensibly there are three independent remaining scaling parameters $U_c$, $p_c$ and $X_c$. However for this fully developed flow, the pressure gradient is assumed to be a constant $G$ independent of the spatial variables. Consequently we must have

$$\frac{\partial p}{\partial x} = \frac{p_c}{X_c}\frac{\partial p^*}{\partial x^*}=G,$$

and (1) reduces to

$$\tag{2}\frac{d^2 u^*}{d y^{*2}}= \underbrace{\frac{\rho U_c h}{\mu}}_{Re} \frac{hG}{\rho U_c^2}$$

Note that the factor $\frac{hG}{\rho U_c^2}$ is a dimensionless pressure gradient where $\rho U_c^2$, the stagnation pressure for an incompressible fluid with velocity $U_c$, is the characteristic pressure scale. Regardless there is only one remaining degree-of-freedom in the choice for $U_c$.

Depending how this velocity scale is selected, we can obtain any value we like for the Reynolds number. However, in solving the differential equation we find that the maximum velocity is attained at $y=0$ and this value is an appropriate velocity scale

$$U_c = \frac{-Gh^2}{2\mu}$$

(Alternatively, we can use the average velocity as you suggest but this only differs by a factor of $2/3$.)

Upon substituting into (2) we get

$$\frac{d^2 u^*}{d y^{*2}}= -2$$

There is no paradox or inconsistency here. We expect the RHS to involve only dimensionless parameters and in this case we get a constant because the constant pressure gradient has been incorporated into the velocity scale.

  • $\begingroup$ In your dimensionless groups definition the same problem has rose, we can not choose arbitararty $U_c$, in fact I have partially figured it out by White's "Viscous Fluid Flow" where it is said there is no characteristic velocity (including $U_c$) for Poiseuille flow, so the Reynolds number will not appear and the equation is like yours but with $-1$ instead of $-2$, in fact the characterics velocity for this flow is $-GH^2/\mu$, and you can not find Reynolds number in dimensionlesss Poiseuille flow equation in any credible book, but It is used repeatedly by article writers! $\endgroup$ Sep 14, 2022 at 7:45
  • $\begingroup$ @DanialRezai: I think you are missing the point. For a given characteristic velocity $U_c$ and length $l_c$ the Reynolds number is $\frac{\rho U_c l_c}{\mu}$ and represents the ratio of inertial to viscous forces. It arises in a meaningful way when you nondimensionalize the full Navier-Stokes equations with inertial terms present. Those steps are: $\endgroup$
    – RRL
    Sep 14, 2022 at 20:26
  • $\begingroup$ $\rho \mathbf{u} \cdot \nabla \mathbf{u} = -\nabla p + \mu \nabla^2 \mathbf{u} \implies \frac{\rho U_c}{l_c} \mathbf{\tilde{u}} \cdot \tilde{\nabla} \mathbf{\tilde{u}} = -\frac{p_c}{l_c}\tilde{\nabla}\tilde{ p} + \frac{\mu U_c}{l_c^2}\tilde{\nabla}^2 \mathbf{\tilde{u}} \implies \frac{\rho U_c l_c}{\mu} \mathbf{\tilde{u}} \cdot \tilde{\nabla} \mathbf{\tilde{u}} = -\frac{p_cl_c}{\mu U_c}\tilde{\nabla}\tilde{ p} + \tilde{\nabla}^2 \mathbf{\tilde{u}}$ $\endgroup$
    – RRL
    Sep 14, 2022 at 20:27
  • $\begingroup$ So the Reynolds number appears in front of the inertial terms on the LHS. That means we can generally neglect inertial effects when the Reynolds number is small. When it is large the flow transitions to turbulence and the simple Poiseulle flow field is not even realized physically. For laminar Poiseuille flow the exact solution is obtained with the inertial terms identically zero and the Reynolds number in its usual role does not even enter into the problem. $\endgroup$
    – RRL
    Sep 14, 2022 at 20:32
  • $\begingroup$ Yes as you said It does not appear, but It is being used by authors! particularly journal stuff, not books. $\endgroup$ Sep 15, 2022 at 20:24

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