Visual walkthrough — Viscous flow — Poiseuille flow, velocity profile in pipe

We only need one idea from before: fluids are sticky. That stickiness is called viscosity (see Viscosity and Newton's law of viscosity). Everything else we grow here.

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Step 1 — Picture the pipe and one hidden cylinder inside it

WHAT. Draw a straight horizontal pipe of radius $R$ (the widest the fluid can reach) and length $L$ (how far along we look). Now imagine a smaller invisible tube of fluid inside it, sharing the same centre-line, with some radius $r$ where $0\le r\le R$. This inner tube is our free body — the chunk of fluid we will do physics on.

WHY. In physics, when you cannot see forces directly, you isolate a piece and demand its forces balance. A cylinder is the natural piece here because the pipe's symmetry means everything depends only on how far you are from the centre — the distance $r$.

PICTURE. The outer grey wall is the pipe. The lavender inner cylinder is our chosen chunk. Two things will push and pull on it: pressure on its two flat ends, and friction on its curved side.

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Step 2 — The push: pressure on the two flat ends

WHAT. Fluid enters at pressure $P_1$ and leaves at lower pressure $P_2$. The difference $\Delta P=P_1-P_2$ is what shoves our cylinder forward. Pressure is force per unit area, so to get force we multiply by the area it acts on — the flat circular end, area $\pi r^2$.

$$F_{\text{push}}=\underset{\begin{array}{c}\text{pressure drop}\\ \text{(the shove per area)}\end{array}}{\underbrace{\Delta P}}\times \underset{\begin{array}{c}\text{area of the}\\ \text{cylinder’s end}\end{array}}{\underbrace{\pi r^2}}$$

WHY this and not something else? Why multiply pressure by area at all? Because pressure alone cannot move anything — a huge pressure on a pinhole is a tiny force. Force is what accelerates matter, and force $=$ pressure $\times$ area. We use the difference $P_1-P_2$ because the back end pushes forward while the front end pushes back; only the net counts.

PICTURE. Two coral arrows: a big one pushing the back face forward ($P_1$), a smaller one pushing the front face backward ($P_2$). Their net is the forward drive.

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Step 3 — The brake: viscous friction on the curved side

WHAT. The fluid outside our cylinder (the shell between radius $r$ and the wall) moves slower, because it is closer to the stationary wall. So it drags backward on our cylinder's curved skin. This drag per unit area is the shear stress $\tau$, and Newton's law of viscosity says it equals viscosity times how quickly speed changes as you move outward:

$$\tau =\underset{\begin{array}{c}\text{viscosity}\\ \text{(stickiness)}\end{array}}{\underbrace{\eta }}\underset{\begin{array}{c}\text{velocity gradient:}\\ \text{how fast }v\text{ changes with }r\end{array}}{\underbrace{\frac{dv}{dr}}}$$

To turn a stress into a force we again multiply by the area it acts on — but this time the curved side of the cylinder, whose area is circumference $2\pi r$ times length $L$:

$$F_{\text{visc}}=\tau \times \underset{\begin{array}{c}\text{curved side area}\\ =\text{(circumference)}\times L\end{array}}{\underbrace{2\pi rL}}=\eta \frac{dv}{dr}(2\pi rL)$$

WHY the gradient $dv/dr$ and not the speed $v$? A layer moving fast alongside a neighbour moving equally fast feels no friction — they don't slide. Friction comes from the difference between neighbours, i.e. how speed changes across a tiny step in $r$. That "change per step" is exactly what the derivative $dv/dr$ measures. We need calculus here precisely because the speed is different at every radius.

PICTURE. Arrows of different lengths at different radii — long in the middle, short near the wall. Where arrow lengths differ most, the sliding (and friction) is greatest.

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Step 4 — Balance the forces (steady flow means no net push)

WHAT. Our cylinder moves at constant speed — the fluid isn't speeding up or slowing down over time (that's what steady flow means). No acceleration means, by Newton's second law, the total force is zero. So the forward push must exactly cancel the backward drag:

$$\underset{\text{forward push}}{\underbrace{\Delta P(\pi r^2)}}=\underset{\text{backward drag}}{\underbrace{-\eta \frac{dv}{dr}(2\pi rL)}}$$

WHY the minus sign? Move outward (increase $r$) and the speed drops, so $dv/dr$ is a negative number. But friction physically pulls backward, a real resisting force. The minus sign flips that negative gradient into a positive resisting magnitude, so both sides are genuine forward-vs-backward balances. Now solve for the gradient:

$$\frac{dv}{dr}=-\frac{\Delta P}{2\eta L}r$$

Read this: the rate at which speed falls off gets steeper as $r$ grows. Near the centre the speed barely changes; near the wall it plunges. That single fact is the seed of the parabola.

PICTURE. A balance beam: pressure force on one pan, viscous force on the other, level because they're equal.

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Step 5 — Add up the slopes: integrate to get the speed

WHAT. Step 4 gave us the slope of the velocity at every radius. To recover the actual speed $v(r)$ we must sum all those slopes — that's what integration is: adding infinitely many tiny changes. We integrate from our radius $r$ outward to the wall $R$, because at the wall we know the answer for sure.

$${\int }_{v(r)}^{0}dv=-\frac{\Delta P}{2\eta L}{\int }_r^R{r}^{\prime }d{r}^{\prime }$$

WHY integrate from $r$ to $R$? Because of the no-slip condition: fluid touching the wall cannot move, so $v(R)=0$. That known endpoint is the anchor that pins down the otherwise-unknown constant. Doing the integral (the right side is $\frac{1}{2}(R^2-r^2)$):

$$0-v(r)=-\frac{\Delta P}{2\eta L}\cdot \frac{R^2-r^2}{2}⟹\boxed{v(r)=\frac{\Delta P}{4\eta L}(R^2-r^2)}$$

Term by term: $\Delta P$ up top (more push $\to$ faster), $\eta L$ underneath (stickier or longer $\to$ slower), and the shape-factor $R^2-r^2$ — biggest at the centre ($r=0$), exactly zero at the wall ($r=R$).

PICTURE. The filled-in parabola: a bullet-nose profile, fat and fast in the middle, pinned to zero at both walls.

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Step 6 — Slice the flow into rings and add up the volume

WHAT. We want $Q$, the volume of fluid passing per second. Different radii move at different speeds, so we can't just multiply one speed by one area. Instead cut the cross-section into thin rings: a ring at radius $r$, thickness $dr$, has area $2\pi rdr$ (unroll it — a strip of length $2\pi r$ and width $dr$). Everything on that ring moves at nearly the same $v(r)$.

$$Q={\int }_0^R\underset{\text{ring speed}}{\underbrace{v(r)}}\underset{\text{ring area}}{\underbrace{(2\pi rdr)}}={\int }_0^R\frac{\Delta P}{4\eta L}(R^2-r^2)2\pi rdr$$

WHY rings and not squares? The flow is circularly symmetric — speed depends only on distance from centre. Rings are the shape along which speed is constant, so each ring contributes cleanly. Working the integral:

$$Q=\frac{\pi \Delta P}{2\eta L}{\left[\frac{R^2{r}^2}{2}-\frac{r^4}{4}\right]}_0^R=\frac{\pi \Delta P}{2\eta L}\cdot \frac{R^4}{4}⟹\boxed{Q=\frac{\pi R^4\Delta P}{8\eta L}}$$

PICTURE. The disc cross-section broken into coloured concentric rings, each labelled with its own speed arrow — tall in the middle, tiny at the rim.

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Step 7 — The edge cases (never leave the reader guessing)

WHAT. Check every corner so no scenario surprises you.

• At the exact centre, $r=0$: $v=\frac{\Delta P{R}^2}{4\eta L}=v_{\max }$. The slope $dv/dr=0$ there — the parabola's flat top. Speed is largest and changing slowest.
• At the wall, $r=R$: $v=0$. This is no-slip, the boundary condition itself. The slope is steepest here, so friction is fiercest at the wall.
• No pressure drop, $\Delta P=0$: then $v(r)=0$ everywhere and $Q=0$. Without a push, sticky fluid just sits still — exactly what experience says.
• Zero viscosity, $\eta \to 0$: $v_{\max }\to \infty$. A frictionless fluid would need no push to race — which is why the ideal, non-viscous world of Bernoulli's principle never has this parabola; it's a flat profile.
• Too fast (turbulence): if you push hard enough the smooth layers break into chaotic swirls. Then Poiseuille's whole picture collapses; the Reynolds number and turbulence tells you when.

PICTURE. Two profiles side by side: the viscous parabola (pinned to zero at the walls) versus the idealised flat "plug" profile of a frictionless fluid.

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The one-picture summary

Everything at once: force balance on the cylinder $\to$ slope of $v$ $\to$ integrate to the parabola $\to$ sum rings to get $Q$.

Recall Feynman retelling — the whole walkthrough in plain words

I drew a fat pipe and imagined a skinny tube of fluid hiding inside it. Pressure at the back shoves that tube forward; the slower fluid outside it drags it back like sticky honey. I said "you're not speeding up, so these two must be equal" — that gave me a rule for how fast the speed drops as I walk outward. Adding up all those little drops (that's integrating), and using the fact that fluid glued to the wall can't move at all, I got a parabola: fastest bullet-nose in the middle, zero at the edges. Then to find how much fluid gushes out per second, I sliced the circle into thin rings, multiplied each ring's speed by its area, and stacked them up. Out popped $R^4$ — because both the room and the middle-speed grow with the radius. Wider pipe, WAY more flow. Stickier fluid, less flow. No push, no flow. That's the whole story.

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Connections

• Viscosity and Newton's law of viscosity — the $\tau =\eta dv/dr$ used in Step 3.
• Reynolds number and turbulence — when this whole derivation stops being valid (Step 7).
• Bernoulli's principle — the frictionless flat-profile counterpart.
• Equation of continuity — how $Q$ and area link across connected pipes.
• Blood flow and circulatory system — the $R^4$ law in your arteries.