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I'd like to know whether the Hagen-Poiseuille equation can be used to solve for the velocity of fluid when the Reynolds number (Re) is less than 1.
From textbooks, I understand that the Hagen-Poiseuille equation can be used for laminar flows. However, I am not sure if it can be used when Re <1.
I'd like to ask for clarification of the above.
Hagen-Poiseuille equation is just a very useful and informative approximation solution to Stokes equation in infinitely long pipe when there is no acceleration of the fluid motion (no inertial and no convectional accelerations). From full incompressible Navier-Stokes equation for a Newtonian fluid you have:
$$\rho_{f} \frac{\partial \mathbf{u}}{\partial t} + \rho_{f} \mathbf{u} \cdot \nabla \mathbf{u} = -\nabla P + \mu \nabla^{2} \mathbf{u}$$
and
$$\nabla \cdot \mathbf{u} = 0$$
As you mentioned, the Reynolds number is a way to find out what's the ratio of inertial forces over viscous forces, defined as: $Re = \frac{\rho_{f} u^{*} L}{\mu}$, where $u^{*}$ and $L$ are characteristic velocity and length in your system. It's not always trivial for complex systems how to choose $u^{*}$ and $L$, but for a pip, the characteristic length is its diameter and characteristic velocity is the maximum velocity presents in the system: $u^{*} = u_{max}$ and $L = D$, so: $Re = \frac{\rho_{f} u_{max} D}{\mu}$. In fact, it's really hard to tell when you can ignore for example inertial forces and just consider the viscous forces when the Reynolds number is so low. It needs to do some experiment to find the onset of turbulence which is a whole new area of expertise, which I'm not gonna touch it here. But, it is known that hypothetically, for an infinitely long pipe, no matter how you increase the Reynolds number, it is still in laminar form and you never see the onset of turbulence. But, remember it's just a theory and in reality when you do experiment, you don't have infinitely long pipe and your pipe's surface is not perfect. Even, really tiny imperfections at the surface will lead to turbulence in practice when you increase the Reynolds number. So, if you forgot about the left side of the Navier-Stokes equation, you will get Stokes equation:
$$\mu \nabla^{2} \mathbf{u} = \nabla P$$
and
$$\nabla \cdot \mathbf{u} = 0$$
If I use the continuity equation ($\nabla \cdot \mathbf{u} = 0$) explicitly on the first Stokes equation, I would get:
$$\nabla \cdot (\mu \nabla^{2} \mathbf{u}) = \nabla^{2} P$$
But divergence and Laplacian operators commute and I have:
$$\mu \nabla^{2} (\nabla \cdot \mathbf{u}) = \nabla^{2} P$$
or:
$$\nabla^{2} P = 0$$
or:
$$-\nabla P = G \hat{\mathbf{z}}$$
Where $G$ is the constant pressure gradient along the pipe. That minus is there to remind you that fluid is moving from a high pressure zone to low pressure zone. So, the Stokes equation finally will be transformed to a Poisson equation:
$$\mu \nabla^{2} \mathbf{u} = -G \hat{\mathbf{z}}$$
And if you solve this equation in cylindrical coordinate, you'll get Hagen-Poiseuille equation:
$$\mathbf{u}(\mathbf{r}) = u_{max} (1 - \frac{|\mathbf{r} - \mathbf{r}_{C}|^{2}}{R^{2}})\hat{\mathbf{z}}$$
Where $u_{max}$ is the maximum velocity, $\mathbf{r}_{C}$ is the coordinate of the center of pipe (remember it doesn't matter for its axial coordinate to where it is located cause it's an infinite pipe), and $R$ is the radius of the pipe and $\hat{\mathbf{z}}$ is the normal vector of the axis of the pipe.
Conclusion: So, as long as based on the knowledge from the physics of your system that you know the inertial forces are not that strong and your geometry is pretty close to an infinite pipe (a cylinder with really high aspect ratio of length over diameter), you could use Hagen-Poiseuille equation.
