Viscous flow — Poiseuille flow, velocity profile in pipe: Question bank

Before we start, three symbols we keep reusing, restated in plain words so nothing here is a mystery:

$R$ = pipe radius (distance from centre to wall); $r$ = distance of a point from the centre line, so $0 \leq r \leq R$ .
$Δ P = P_{1} - P_{2}$ = the pressure drop along the pipe — the push.
$Q$ = volume flow rate (cubic metres of fluid crossing any cross-section per second).
$η$ = viscosity (the fluid's internal stickiness); $L$ = pipe length.

The two master results we test against: $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2}), Q = \frac{π R ^{4} Δ P}{8 η L} .$

True or false — justify

The fluid at the very centre of the pipe experiences the largest shear stress.

False. Shear stress is

τ = η d v / d r

; the gradient

d v / d r = - \frac{Δ P}{2 η L} r

is zero at the centre and largest at the wall, so the wall — not the centre — feels the most shear.

Doubling the pipe length halves the flow rate at fixed pressure drop.

True.

Q \propto 1/ L

, so twice the length means half the flow — a longer pipe has more wall area dragging over more distance, so the same push moves less fluid.

If you double the pressure drop, the maximum speed at the centre doubles.

True.

v_{m a x} = \frac{Δ P R ^{2}}{4 η L}

is linear in

Δ P

, so twice the push gives twice the centre speed (as long as flow stays laminar).

The mean speed equals the centre speed.

False. For a parabola the average is exactly half the peak:

\overset{v}{ˉ} = \frac{1}{2} v_{m a x}

, because slow fluid near the walls drags the average down.

Poiseuille's law applies equally well to a fast-moving river.

False. It requires laminar flow (low Reynolds number); a river is turbulent and obeys a completely different, non-linear relationship — see Reynolds number and turbulence.

Making the fluid twice as viscous, keeping everything else fixed, halves the flow rate.

True.

Q \propto 1/ η

, so doubling the stickiness halves the flow — more internal friction resists the same push.

At the pipe wall the fluid is stationary because friction with the wall material stops it.

True in effect, subtle in cause. It's the no-slip condition: fluid molecules touching the wall have zero bulk velocity, which is what pins

v (R) = 0

and fixes the whole profile.

The velocity profile would still be parabolic if the fluid were non-viscous (

η = 0

False. With no viscosity there is no shear stress, no wall drag, and no reason for the wall fluid to stop — you'd get uniform "plug" flow, not a parabola. The parabola is the fingerprint of viscosity.

For a fixed flow rate

Q

, the required pressure drop is independent of viscosity.

False. From

Δ P = \frac{8 η L Q}{π R ^{4}}

, the needed push is directly proportional to

η

: a stickier fluid demands more pressure to force the same

Q

through.

Spot the error

"Since

Q = v \times area

and area

\propto R^{2}

, flow rate scales as

R^{2}

The error: the velocity itself also grows as

R^{2}

(centre speed

\propto R^{2}

), so

Q \propto R^{2} \cdot R^{2} = R^{4}

. Both a bigger cross-section and a flatter gradient multiply together.

"The pressure inside the pipe is highest at the walls and lowest at the centre."

The error: pressure varies along the pipe (high at inlet, low at outlet), not across it. At a given cross-section pressure is essentially uniform; it's the velocity that varies across the radius.

"Because velocity decreases outward,

d v / d r

is positive."

The error: as

r

increases (moving outward)

v

decreases, so the slope

d v / d r

is negative. That minus sign is exactly why the derivation carries

\frac{d v}{d r} = - \frac{Δ P}{2 η L} r

"A thicker (more viscous) fluid is heavier, so gravity pushes it through faster."

The error confuses two things: viscosity is internal friction, not density/weight, and Poiseuille flow here is horizontal so gravity does no driving work anyway. Higher

η

means slower flow.

"The shear force acts on the flat circular ends of the coaxial cylinder in the derivation."

The error: pressure acts on the flat ends (area

π r^{2}

); the viscous shear acts on the curved side surface (area

2 π r L

). Mixing up which force acts on which area breaks the whole force balance.

"Integrating the profile, we can start from

v (0) = 0

since the centre is a natural reference."

The error: the centre is where speed is maximum, not zero. The correct boundary condition is

v (R) = 0

at the wall — that's the physical no-slip fact that fixes the integration constant.

"If we halve the radius, flow drops to a quarter."

The error:

Q \propto R^{4}

, so halving

R

multiplies

Q

(1/2)^{4} = 1/16

— flow drops to a sixteenth, not a quarter.

Why questions

Why does flow rate depend on

R^{4}

rather than

R^{2}

Because a wider pipe helps twice: it offers more cross-sectional area (

\propto R^{2}

) and allows a higher centre speed with a gentler gradient (

v_{m a x} \propto R^{2}

). Multiplying the two effects gives

R^{4}

Why is the velocity profile a parabola and not, say, a straight cone?

Balancing pressure force (

\propto r

acting over area) against viscous drag gives

d v / d r \propto r

; integrating a term linear in

r

produces an

r^{2}

term, and

v \propto (R^{2} - r^{2})

is a parabola.

Why must there be a pressure drop at all for steady flow?

Viscosity continuously drains momentum into wall friction; to keep the fluid moving at constant speed (no acceleration in steady flow), that loss must be paid for continuously by a sustained push

Δ P

Why is the mean speed exactly half the maximum and not, say, two-thirds?

Because averaging the parabola

v_{m a x} (1 - r^{2} / R^{2})

over the disc (weighting outer rings more, since ring area grows with

r

) works out to exactly

\frac{1}{2} v_{m a x}

— a geometric property of that specific profile.

Why does the wall fluid feel the greatest viscous stress even though it moves slowest?

Stress depends on the gradient

d v / d r

, not on speed. Near the wall the speed changes most steeply over a small distance, so the layers there rub hardest — the wall is the maximum-gradient region.

Why can't we simply use Bernoulli's principle to describe flow in a real pipe?

Bernoulli's principle assumes no viscosity, so it conserves mechanical energy along a streamline. Real pipe flow loses energy to friction, producing the very pressure drop Poiseuille's law accounts for.

Why does a slightly narrowed artery cause such a large drop in blood flow?

The

R^{4}

dependence: even a small reduction in radius shrinks

Q

dramatically (a 20% narrowing cuts flow to about

0. 8^{4} \approx 41%

). This is central to Blood flow and circulatory system.

Why is

τ = η d v / d r

proportional to the gradient and not to the speed

v

itself?

Friction arises from relative sliding between adjacent layers; if two neighbouring layers move at the same speed there is no rubbing even if that speed is huge. Only the difference — the gradient — creates shear. See Viscosity and Newton's law of viscosity.

Edge cases

At the exact centre (

r = 0

), what is the shear stress, and is that physically sensible?

Zero.

τ = η d v / d r \propto r

, so at

r = 0

there is no shear — the centre layer moves with its neighbours essentially in lockstep at the profile's flat top, so no rubbing occurs.

What does the velocity profile become if

Δ P = 0

v (r) = 0

everywhere — no push means no flow. The parabola collapses to a flat zero line; nothing drives the fluid against friction.

η \to 0

(an ideal, frictionless fluid), what does the formula predict for a fixed

Δ P

Both

v_{m a x}

and

Q

blow up to infinity, signalling the model breaks: with zero viscosity there is no drag to balance the push, so no steady finite-speed solution exists — you must switch to the non-viscous picture entirely.

Does the flow rate change if you increase

R

but decrease

Δ P

so their product stays fixed?

Yes, flow still rises.

Q \propto R^{4} Δ P

, so if

R

grows and

Δ P

shrinks to keep

R \cdot Δ P

constant, the extra three powers of

R

dominate and

Q

increases sharply — the radius always wins.

What happens to the parabolic profile just as the flow starts to become turbulent?

It stops being a clean parabola: turbulence flattens the core (a "plug-like" middle) and steepens the wall gradient. Once past the critical Reynolds number, Poiseuille's parabola no longer describes the flow at all.

v (r)

ever negative anywhere in the pipe under normal Poiseuille flow?

No. Since

R^{2} - r^{2} \geq 0

for all

0 \leq r \leq R

, and

Δ P / (4 η L) > 0

, the speed is non-negative everywhere — zero only at the wall, positive everywhere inside.

Active Recall

Recall One-line trap summary (cover and recall)

Where is shear stress maximum? → At the wall (max gradient), zero at centre.
$Q$ vs $R$ ? → $Q \propto R^{4}$ — radius helps twice.
$Q$ vs $η$ and $L$ ? → $Q \propto 1/ η$ and $Q \propto 1/ L$ .
$\overset{v}{ˉ}$ vs $v_{m a x}$ ? → Exactly half.
Boundary condition fixing the profile? → No-slip, $v (R) = 0$ .
When does the law fail? → Turbulent, non-steady, non-Newtonian, or short-pipe flow.

Connections

Parent: Poiseuille flow — full derivations these traps target.
Viscosity and Newton's law of viscosity — why $τ$ depends on the gradient.
Reynolds number and turbulence — when the parabola breaks down.
Bernoulli's principle — the frictionless contrast case.
Blood flow and circulatory system — the $R^{4}$ trap in real arteries.