categories: physics, automation & robotics, fluid mechanics
Last post in this mini-series on sizing real components. This one answers a question that shows up anywhere fluid moves through a narrow gap — cooling channels in a heatsink, lubricant in a bearing, a microfluidic channel — how fast does the fluid actually move at each point across the gap, not just on average? The Navier-Stokes equations look intimidating in full, but for this specific, very common setup, almost every term drops out and you're left with something you can integrate by hand.
Task: velocity profile for laminar flow between two plates
Setup: steady, fully-developed flow of an incompressible viscous fluid between two flat, stationary plates a distance $$h$$ apart. Flow runs along the $$X$$ axis, driven by a constant pressure gradient $$\frac{dp}{dx}=\text{const}$$.
Step 1 — simplify the Navier-Stokes equation. Full 3D Navier-Stokes has a lot of terms, but this setup kills almost all of them: no flow across the gap or out of the page ($$v=w=0$$), steady state ($$\partial u/\partial t=0$$), and fully developed flow means the profile doesn't change as you move downstream ($$\partial u/\partial x=0$$), so there's no advection term either. Starting from the general equation and crossing out everything that's zero for this case: $$\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}\right)$$ $$0=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2} \implies \frac{d^2u}{dy^2}=\frac{1}{\mu}\frac{dp}{dx}$$ What's left is a genuinely simple ODE — a constant on the right-hand side — because almost the entire physical complexity of the original equation was ruled out by the setup's symmetry, not by approximation.
Step 2 — integrate twice. A constant second derivative integrates the same way it would in any kinematics problem — twice, picking up one integration constant each time: $$\frac{du}{dy}=\frac{1}{\mu}\frac{dp}{dx}y+C_1$$ $$u(y)=\frac{1}{2\mu}\frac{dp}{dx}y^2+C_1y+C_2$$
Step 3 — apply the no-slip boundary conditions. A real viscous fluid doesn't slide freely against a solid wall — it sticks to it, so velocity must be exactly zero at both plates: $$u(0)=0$$ at the bottom, $$u(h)=0$$ at the top. $$u(0)=0 \implies C_2=0$$ $$u(h)=\frac{1}{2\mu}\frac{dp}{dx}h^2+C_1h=0 \implies C_1=-\frac{1}{2\mu}\frac{dp}{dx}h$$ Substituting both constants back in gives the final, parabolic profile — this is Poiseuille flow: $$u(y)=\frac{1}{2\mu}\left(-\frac{dp}{dx}\right)(hy-y^2)$$ The shape makes physical sense once you see it: zero at both walls (no-slip), peaking exactly in the middle at $$y=h/2$$ — this parabola is the reason engineers care about the difference between average flow velocity and peak centerline velocity when sizing anything that depends on local flow speed, like convective cooling.
Same three-step pattern as the very first differential-equation posts in this series: simplify the governing equation using the specific symmetry of your setup, integrate however many times the highest derivative demands, then pin down the integration constants with the actual physical boundary conditions. Thank you for reading, and thanks for following this whole series :)
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