Viscous flow — Poiseuille flow, velocity profile in pipe

WHAT is viscosity? (first, the building block)

WHY this form? The faster one layer moves relative to its neighbour, the harder the molecules drag on each other → stress proportional to the gradient, not the speed itself.

Setting up Poiseuille flow

We consider steady, laminar, incompressible flow through a horizontal cylindrical pipe of radius $R$ and length $L$ , driven by a pressure difference $\Delta P = P_1 - P_2$ .

Derivation from scratch (force balance on a cylinder)

From profile to flow rate $Q$ (the practical payoff)

Worked Examples

Worked example Example 1 — Find the pressure drop

Water ( $\eta = 1.0\times10^{-3}\,\text{Pa·s}$ ) flows at $Q = 2.0\times10^{-5}\,\text{m}^3/\text{s}$ through a pipe of $R = 5.0\,\text{mm}$ , $L = 2.0\,\text{m}$ . Find $\Delta P$ .

Step 1: Rearrange Hagen–Poiseuille: $\Delta P = \dfrac{8\eta L Q}{\pi R^4}$ . Why? We want $\Delta P$ , so solve for it.

Step 2: Plug in: $R^4 = (5\times10^{-3})^4 = 6.25\times10^{-10}\,\text{m}^4$ .

\approx \frac{3.2\times10^{-7}}{1.963\times10^{-9}} \approx 163\,\text{Pa}$$ *Why this matters:* a modest flow needs only a small push because $R^4$ is in the denominator.

Common Mistakes (Steel-manned)

Active Recall

Recall Quick self-test (cover the answers!)

What boundary condition fixes the integration constant? → No-slip: $v(R)=0$ .
Shape of the velocity profile? → Parabolic, $v \propto R^2-r^2$ .
How does $Q$ depend on $R$ ? → $Q \propto R^4$ .
Mean speed vs max speed? → $\bar v = \tfrac12 v_{\max}$ .
What kind of flow is required? → Laminar, steady, incompressible, Newtonian.

Recall Feynman: explain to a 12-year-old

Imagine pushing a long line of people through a narrow hallway. The people brushing the walls get stuck and barely move, but the ones in the middle can run freely. So the middle goes fastest and the edges crawl — that's the curved "parabola" shape. To keep everyone moving you have to keep pushing from behind (that's the pressure). And a wider hallway helps SO much that doubling the width lets 16 times more people through — because both more room AND less wall-stickiness help.

Connections

Viscosity and Newton's law of viscosity — the $\tau = \eta\,dv/dr$ foundation used here.
Reynolds number and turbulence — tells you when Poiseuille's law breaks down.
Equation of continuity — relates $Q$ , area and speed in connected pipes.
Bernoulli's principle — the non-viscous counterpart; Poiseuille adds friction losses.
Stokes' law and terminal velocity — another viscous-drag application.
Blood flow and circulatory system — biological use of the $R^4$ scaling.

Newton's law of viscosity (formula)

\tau = \eta\,\dfrac{dv}{dr}

— shear stress = viscosity × velocity gradient.

What boundary condition is the no-slip condition?

Fluid velocity equals zero at the pipe wall,

v(R)=0

Velocity profile in Poiseuille flow

v(r) = \dfrac{\Delta P}{4\eta L}(R^2 - r^2)

— a parabola, max at centre, zero at wall.

Maximum (centre) velocity in a pipe

v_{\max} = \dfrac{\Delta P\,R^2}{4\eta L}

Hagen–Poiseuille flow rate

Q = \dfrac{\pi R^4 \Delta P}{8\eta L}

How does flow rate depend on pipe radius?

Q \propto R^4

— double radius → 16× flow.

Relation between mean and max velocity

\bar v = \tfrac{1}{2} v_{\max}

(parabolic profile).

Assumptions of Poiseuille's law

Steady, laminar, incompressible, Newtonian fluid in a long straight pipe.

How is the side-surface viscous force written for a coaxial cylinder of radius r, length L?

F = \eta\,\dfrac{dv}{dr}(2\pi r L)

Why does higher viscosity reduce flow?

Q \propto 1/\eta

; more internal friction resists motion for the same pressure.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Poiseuille flow ka pura funda yeh hai ki real fluids "sticky" hote hain — unme viscosity hoti hai. Jab fluid pipe me behta hai, toh wall ko touch karne wala layer bilkul ruka rehta hai (no-slip condition), aur beech wala fluid sabse tez bhaagta hai. Isi wajah se velocity ka shape ek parabola ban jaata hai: centre pe maximum, wall pe zero. Formula hai $v(r) = \frac{\Delta P}{4\eta L}(R^2 - r^2)$ .

Yeh formula hum bina ratte derive kar sakte hain. Ek imaginary coaxial cylinder (radius $r$ ) lo. Steady flow me acceleration zero hai, toh net force zero. Aage se pressure dhakka deta hai ( $\Delta P \cdot \pi r^2$ ), aur side surface pe viscous friction rokti hai ( $\eta \frac{dv}{dr}\cdot 2\pi r L$ ). Dono ko barabar karke integrate karo, no-slip condition ( $v=0$ wall pe) lagao — bas, parabola mil gaya. Phir patle rings ke upar integrate karke flow rate nikalta hai: $Q = \frac{\pi R^4 \Delta P}{8\eta L}$ .

Sabse important baat — yeh $R^4$ waala part. Radius double karo toh flow $2^4 = 16$ guna ho jaata hai! Isiliye blood vessel thoda sa patla ho jaye toh blood flow bahut gir jaata hai. Aur yaad rakho: $Q$ viscosity $\eta$ ke ulta proportional hai — gaadha fluid dhire behta hai. Mean velocity hamesha max velocity ka aadha hota hai. Exam me bas ek galti mat karna: yeh law sirf laminar (smooth, low Reynolds number) flow ke liye hai, turbulent flow me nahi chalta.

Viscous flow — Poiseuille flow, velocity profile in pipe

WHAT is viscosity? (first, the building block)

Setting up Poiseuille flow

Derivation from scratch (force balance on a cylinder)

From profile to flow rate $Q$ (the practical payoff)

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Fluid Mechanics

Connections