Analytical and Simplified Solutions for Common Flow Problems

TL;DR

You'll learn how the full Navier-Stokes equations collapse into solvable ODEs once you assume steady, fully-developed, unidirectional flow. You'll derive Hagen-Poiseuille pipe flow and Couette-Poiseuille plate flow from first principles. You'll know exactly which assumptions justify each simplification, so you can pick the right formula on an exam or in the field instead of guessing.

1. The Mental Model

The full Navier-Stokes equations are three coupled nonlinear PDEs — nobody solves those by hand for real engineering flows. But if you kill the right terms (no time dependence, no acceleration, only one velocity component), the mess collapses into a single linear ODE you can integrate directly. Every "named" laminar flow solution — Poiseuille, Couette, thin-film — is just Navier-Stokes with specific terms zeroed out. Simplification isn't cheating; it's matching the math to what the flow is actually doing.

2. The Core Material

2.1 Killing terms: from Navier-Stokes to a solvable ODE

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Start with the x-momentum equation for incompressible, constant-viscosity flow:

$$\rho\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z}\right) = -\frac{\partial p}{\partial x} + \mu\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right) + \rho g_x$$

For most civil/water engineering pipe and channel problems, you apply four assumptions in sequence:

1. Steady flow: $\partial u/\partial t = 0$. No time dependence.
2. Fully developed flow: the velocity profile no longer changes shape downstream, so $\partial u/\partial x = 0$. This kills the streamwise convective term.
3. Unidirectional flow: $v = w = 0$ (no cross-flow). This kills the other two convective terms entirely — the whole left side of the equation vanishes.
4. No streamwise variation in the cross-section geometry: pressure gradient $\partial p/\partial x$ is constant along the pipe/channel (call it $-\Delta P/L$).

What survives is a balance between pressure force and viscous shear:

$$0 = -\frac{dp}{dx} + \mu\frac{d^2u}{dy^2}$$

This is no longer a nonlinear PDE — it's a second-order linear ODE in one variable. You solve it with two boundary conditions (usually no-slip at walls) and you're done. This is the pattern behind every classical solution below.

2.2 Hagen-Poiseuille flow (circular pipes)

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Switch to cylindrical coordinates for a pipe of radius $R$. The same term-killing logic applied to the axial momentum equation gives:

$$0 = -\frac{dp}{dz} + \frac{\mu}{r}\frac{d}{dr}\left(r\frac{du_z}{dr}\right)$$

Integrate once:

$$r\frac{du_z}{dr} = \frac{1}{\mu}\frac{dp}{dz}\frac{r^2}{2} + C_1$$

Since $du_z/dr$ must stay finite at $r=0$, $C_1 = 0$. Integrate again:

$$u_z(r) = \frac{1}{4\mu}\frac{dp}{dz}r^2 + C_2$$

Apply no-slip at the wall, $u_z(R) = 0$:

$$u_z(r) = -\frac{1}{4\mu}\frac{dp}{dz}\left(R^2 - r^2\right)$$

The velocity profile is a parabola — maximum at the centerline, zero at the wall. Integrate over the cross-section to get volumetric flow rate:

$$Q = \int_0^R u_z(r)\, 2\pi r\, dr = \frac{\pi R^4}{8\mu}\left(-\frac{dp}{dz}\right) = \frac{\pi R^4 \Delta P}{8\mu L}$$

This is the Hagen-Poiseuille equation. Two things worth memorizing: $Q$ scales with $R^4$ (double the pipe radius, flow rate goes up 16x), and it's only valid for laminar flow, $Re = \rho \bar u D/\mu < 2300$ roughly.

2.3 Couette and Couette-Poiseuille flow (parallel plates)

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Now consider flow between two flat plates separated by gap $h$, where the top plate moves at velocity $U$ and there may also be a pressure gradient driving flow (think: a sluice gate or thin lubricating film). The same reduced ODE applies in Cartesian form:

$$\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}$$

Integrate twice:

$$u(y) = \frac{1}{2\mu}\frac{dp}{dx}y^2 + C_1 y + C_2$$

Boundary conditions: $u(0) = 0$ (