A pressure difference can drive a viscous fluid through a pipe. In a long straight circular pipe, sufficiently low-Reynolds-number flow approaches a particularly simple state called steady fully developed laminar flow.

Assume a Newtonian fluid with constant viscosity $\mu$ and density $\rho$ in a pipe of radius $R$. Far enough from the entrance, the axial velocity depends only on radial position:

$$u=u(r).$$

The profile no longer changes with axial position, and there is no mean radial flow.

Force balance on a cylindrical core

Consider a cylindrical fluid core of radius $r$ and length $L$. Let the pressure drop over that length be

$$\Delta p=p_{in}-p_{out}>0.$$

The net pressure force driving the core is

$$F_p=\Delta p,\pi r^2.$$

Viscous shear on its cylindrical surface resists the motion. If $\tau(r)$ is the shear-stress magnitude,

$$F_\tau=\tau(r)(2\pi rL).$$

Steady force balance gives

$$\Delta p,\pi r^2=\tau(r)2\pi rL,$$

so

$$\boxed{\tau(r)=\frac{\Delta p}{2L}r}.$$

The shear stress is zero at the centerline and increases linearly toward the wall.

Velocity profile

For a Newtonian fluid, shear stress is proportional to the velocity gradient. Because velocity decreases toward the wall,

$$\frac{du}{dr}<0,$$

and the signed relation is

$$\mu\frac{du}{dr}=-\frac{\Delta p}{2L}r.$$

Therefore

$$\frac{du}{dr}=-\frac{\Delta p}{2\mu L}r.$$

Integrating,

$$u(r)=-\frac{\Delta p}{4\mu L}r^2+C.$$

The no-slip boundary condition requires the fluid velocity to equal the stationary wall velocity:

$$u(R)=0.$$

Thus

$$C=\frac{\Delta p}{4\mu L}R^2,$$

and

$$\boxed{u(r)=\frac{\Delta p}{4\mu L}(R^2-r^2)}.$$

The laminar pipe-flow profile is parabolic.

The maximum speed occurs on the centerline:

$$\boxed{u_{max}=\frac{\Delta pR^2}{4\mu L}}.$$

Flow rate and mean velocity

The volumetric flow rate is

$$Q=\int_A u,dA.$$

For an annulus of radius $r$ and thickness $dr$,

$$dA=2\pi r,dr.$$

Therefore

$$Q=\int_0^R\frac{\Delta p}{4\mu L}(R^2-r^2)2\pi r,dr,$$

which gives

$$\boxed{Q=\frac{\pi R^4}{8\mu L}\Delta p}.$$

This is the Hagen-Poiseuille relation.

Since the pipe area is $A=\pi R^2$, the mean velocity is

$$\bar u=\frac QA =\boxed{\frac{\Delta pR^2}{8\mu L}}.$$

Hence

$$\boxed{u_{max}=2\bar u}.$$

Using diameter $D=2R$, the pressure drop can be written

$$\boxed{\Delta p=\frac{32\mu L\bar u}{D^2}}.$$

Worked example

Water-like fluid with

$$\mu=1.0\times10^{-3},\mathrm{Pa,s}$$

flows steadily through a pipe with

$$D=0.010,\mathrm m,\qquad L=2.0,\mathrm m$$

at mean speed

$$\bar u=0.10,\mathrm{m/s}.$$

The required pressure drop is

$$\Delta p =\frac{32(1.0\times10^{-3})(2.0)(0.10)}{(0.010)^2} =64,\mathrm{Pa}.$$

Thus

$$\boxed{\Delta p=64,\mathrm{Pa}}.$$

The centerline speed is

$$u_{max}=2\bar u=0.20,\mathrm{m/s}.$$

Scope of the result

Hagen-Poiseuille flow assumes steady, incompressible, Newtonian, axisymmetric, fully developed laminar flow in a straight circular pipe with no slip at the wall. Entrance regions, turbulent flow, non-Newtonian behavior, compressibility, or strong pipe curvature require different models.

The result is important not only as a pipe formula but as a direct demonstration of how a pressure gradient is balanced by viscous shear in internal flow.