Viscous flow — Poiseuille flow, velocity profile in pipe: Worked examples

Everything here rests on the parent: the topic note. Symbols used: $R$ = pipe radius, $L$ = pipe length, $Δ P$ = pressure drop across the length, $η$ = viscosity, $Q$ = volume flow rate (volume per second), $v (r)$ = speed at distance $r$ from the axis.

The scenario matrix

Before any numbers, here is the full map of case-classes this topic can throw at you. Every worked example below is tagged with the cell (C1–C12) it covers.

Cell	Case class	What makes it tricky
C1	Solve for $Q$ directly	Straight plug-in — the baseline
C2	Solve for a hidden variable ( $Δ P$ , $η$ , $R$ , $L$ )	Must rearrange the formula
C3	Velocity profile — speed at a point $r$	Use $v (r)$ , not $Q$
C4	Degenerate input: $r = 0$ (axis) and $r = R$ (wall)	Endpoints of the parabola
C5	Limiting / scaling: double $R$ , halve $L$ , etc.	Powers ( $R^{4}$ , $R^{2}$ ) dominate
C6	Zero / vanishing input: $Δ P = 0$ , or $R \to 0$	Flow stops — sanity of formula
C7	Pipes in series (same $Q$ , pressures add)	Continuity + additive $Δ P$
C8	Mean vs max speed	Factor of $\frac{1}{2}$
C9	Real-world word problem (blood / oil)	Translate story → symbols
C10	Exam twist: validity check (is it even laminar?)	Reynolds number gatekeeper
C11	Pipes in parallel (same $Δ P$ , flows add)	Conductances, not resistances, add
C12	Sign & singular limits: $Δ P < 0$ , $L \to 0$	Reversed flow; blow-up as $L \to 0$

We now clear the matrix, cell by cell.

C1 — Straight flow-rate calculation

Worked example Oil through a level pipe

Oil ( $η = 0.10 Pa\cdots$ ) flows through a horizontal pipe of radius $R = 1.0 cm$ and length $L = 5.0 m$ under a pressure drop $Δ P = 400 Pa$ . Find the volume flow rate $Q$ .

Forecast: Will $Q$ be a big number or a tiny one? Guess before reading — the $R^{4}$ with $R$ in metres is very small, so expect a very small $Q$ .

Convert units. $R = 1.0 cm = 1.0 \times 1 0^{- 2} m$ , so $R^{4} = (1 0^{- 2})^{4} = 1.0 \times 1 0^{- 8} m^{4}$ . Why this step? Every SI formula needs metres; a stray centimetre wrecks the $R^{4}$ by a factor of $1 0^{8}$ .
Plug into Hagen–Poiseuille. $Q = \frac{π R ^{4} Δ P}{8 η L} = \frac{π ( 1.0 \times 1 0 ^{- 8} ) ( 400 )}{8 ( 0.10 ) ( 5.0 )} .$ Why this step? We are asked for $Q$ and given all four inputs — direct substitution.
Arithmetic. Numerator $= π \cdot 4.0 \times 1 0^{- 6} = 1.257 \times 1 0^{- 5}$ . Denominator $= 4.0$ . $Q = \frac{1.257 \times 1 0 ^{- 5}}{4.0} \approx 3.14 \times 1 0^{- 6} m^{3} / s .$

Verify: Units — $\frac{m ^{4} \cdot Pa}{Pa\cdots \cdot m} = \frac{m ^{4}}{s \cdot m} = m^{3} / s$ . ✓ Correct dimensions for a flow rate. The tiny answer ( $\sim 3 mL/s$ ) matches the forecast: metres make $R^{4}$ minuscule.

C2 — Solve for a hidden variable

Worked example What viscosity does the pipe reveal?

A liquid is pushed through $R = 2.0 mm$ , $L = 1.0 m$ with $Δ P = 500 Pa$ , producing $Q = 1.0 \times 1 0^{- 6} m^{3} / s$ . Find its viscosity $η$ .

Forecast: Since $Q \propto 1/ η$ , a small $Q$ hints at a large $η$ . Guess whether it's more or less viscous than water ( $1 0^{- 3} Pa\cdots$ ).

Rearrange for $η$ . From $Q = \frac{π R ^{4} Δ P}{8 η L}$ , isolate: $η = \frac{π R ^{4} Δ P}{8 L Q} .$ Why this step? $η$ sits in the denominator; multiply both sides by $η$ and divide by $Q$ .
Compute $R^{4}$ . $R = 2.0 \times 1 0^{- 3} m$ , so $R^{4} = (2 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . Why this step? The fourth power must be evaluated carefully — $2^{4} = 16$ and $(1 0^{- 3})^{4} = 1 0^{- 12}$ .
Plug in. $η = \frac{π ( 1.6 \times 1 0 ^{- 11} ) ( 500 )}{8 ( 1.0 ) ( 1.0 \times 1 0 ^{- 6} )} = \frac{2.513 \times 1 0 ^{- 8}}{8.0 \times 1 0 ^{- 6}} \approx 3.14 \times 1 0^{- 3} Pa\cdots .$

Verify: $η \approx 3.1 \times 1 0^{- 3} Pa\cdots$ — about three times water's viscosity, so more viscous, matching the "small $Q$ → thick fluid" forecast. Units: $\frac{m ^{4} \cdot Pa}{m \cdot m ^{3} / s} = Pa\cdots$ . ✓

C3 & C4 — The velocity profile and its endpoints

Worked example Speed across the pipe (a picture problem)

A pipe has $R = 6.0 mm$ , and the centre speed is $v_{m a x} = 0.90 m/s$ . Find the speed at (a) the axis $r = 0$ , (b) $r = 3.0 mm$ , (c) the wall $r = R$ . This covers C3 (a point) and C4 (both degenerate endpoints).

Forecast: The profile is a parabola. Before computing, sketch in your head: highest in the middle, zero at the wall, and at half-radius it should still be quite fast (recall the parent's "75% at halfway" fact).

Where $v_{m a x}$ comes from. Start from the parent's derived profile $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2})$ . At the axis $r = 0$ this is largest: $v_{m a x} = v (0) = \frac{Δ P R ^{2}}{4 η L} .$ Divide the profile by this peak and the messy $Δ P / (4 η L)$ cancels: $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2}) = v_{m a x} \frac{Δ P R ^{2}}{4 η L} (1 - \frac{r ^{2}}{R ^{2}}) = v_{m a x} (1 - \frac{r ^{2}}{R ^{2}}) .$ Why this step? The ratio form is not a new law — it is the same parabola, rewritten so that all of $Δ P$ , $η$ , $L$ are packaged inside the single measured number $v_{m a x}$ . Since the problem hands us $v_{m a x}$ directly, we never need those three separately.
(a) Axis, $r = 0$ : $v = v_{m a x} (1 - 0) = 0.90 m/s$ . Why this step? At the centre nothing is subtracted — this is the definition of the peak. In the figure this is the tip of the red arrow.
(b) $r = 3.0 mm$ : $\frac{r ^{2}}{R ^{2}} = \frac{3 ^{2}}{6 ^{2}} = \frac{9}{36} = 0.25$ , so $v = 0.90 (1 - 0.25) = 0.675 \approx 0.68 m/s .$ Why this step? Halfway out ( $r = R /2$ ), only a quarter of the speed is lost — the parabola is flat near the top.
(c) Wall, $r = R$ : $\frac{R ^{2}}{R ^{2}} = 1$ , so $v = 0.90 (1 - 1) = 0$ . Why this step? This is the no-slip condition — fluid touching the wall is glued in place. In the figure the arrow shrinks to nothing at the edge.

Verify: Endpoints match the physics: max at centre, zero at wall. The half-radius value $0.68 m/s$ is $75%$ of $0.90$ , exactly the parent's fact. ✓

C5 — Scaling / limiting behaviour

Worked example Change several things at once

A pipe's radius is doubled, its length is tripled, and the pressure drop is halved, with the same fluid. By what factor does $Q$ change?

Forecast: $R^{4}$ is the giant here — doubling $R$ multiplies $Q$ by $16$ . The length and pressure changes are gentler. Guess: net increase, but how much?

Write $Q$ as a proportionality. $Q \propto \frac{R ^{4} Δ P}{L} .$ Why this step? $π$ , $8$ , $η$ are unchanged, so they cancel in a ratio — only the varied quantities matter.
Apply each factor. New over old: $\frac{Q _{new}}{Q _{old}} = \frac{( 2 ) ^{4} \cdot \frac{1}{2}}{3} = \frac{16 \cdot 0.5}{3} = \frac{8}{3} \approx 2.67.$ Why this step? Each variable enters with its own power — $R$ to the 4th, $Δ P$ to the 1st, $L$ to the $- 1$ .

Verify: $\frac{8}{3} \approx 2.67$ . Flow rises by ~2.7×. The dominant factor really is the radius: alone it would have given ×16, and the halved pressure (÷2) and tripled length (÷3) only tame it. ✓

C6 — Zero and vanishing inputs (degenerate limits)

Worked example What if the push disappears — or the pipe pinches shut?

(a) If $Δ P = 0$ , what is $Q$ and the profile $v (r)$ ? (b) As $R \to 0$ , what happens to $Q$ ?

Forecast: No push should mean no flow. A pipe closing to nothing should choke flow to zero — but how fast?

(a) Set $Δ P = 0$ in both formulas. $Q = \frac{π R ^{4} ( 0 )}{8 η L} = 0, v (r) = \frac{0}{4 η L} (R^{2} - r^{2}) = 0 everywhere .$ Why this step? Poiseuille flow is driven by the pressure difference. Remove the driver and the sticky fluid simply stops — no residual motion. This is a crucial sanity check on the whole derivation.
(b) Take $R \to 0$ . Since $Q \propto R^{4}$ , $lim_{R \to 0} Q = lim_{R \to 0} \frac{π R ^{4} Δ P}{8 η L} = 0.$ Why this step? $R^{4} \to 0$ faster than any lower power. A pipe that pinches to a point passes nothing — and it chokes quartically, dramatically faster than the cross-sectional area ( $\propto R^{2}$ ) alone shrinks.

Verify: Both limits give $0$ , exactly as physical intuition demands. The formula is well-behaved at its degenerate corners — no division-by-zero, no negative flow. ✓

C7 — Two pipes in series

Worked example Pressure drop across a joined pipe

Two horizontal pipes carry the same $Q = 2.0 \times 1 0^{- 6} m^{3} / s$ of a fluid with $η = 1.0 \times 1 0^{- 3} Pa\cdots$ , joined end to end. Pipe A: $R_{A} = 2.0 mm$ , $L_{A} = 1.0 m$ . Pipe B: $R_{B} = 1.0 mm$ , $L_{B} = 1.0 m$ . Find the total pressure drop.

Forecast: Same $Q$ flows through both (continuity). The narrow pipe B, with $R^{4}$ sixteen times smaller, must demand a much bigger $Δ P$ . Guess which pipe dominates.

Read the figure first. The wide teal box is pipe A, the narrow orange box is pipe B, joined end to end. The single plum arrow running straight through both is the flow $Q$ — the same arrow, because whatever volume enters A per second must leave B per second (continuity). Below each pipe the figure labels its own pressure drop; notice pipe B's is far larger even though both pipes are the same length. The bottom line shows how the two add.

Invert Hagen–Poiseuille to get $Δ P$ . Start from $Q = \frac{π R ^{4} Δ P}{8 η L}$ . We want $Δ P$ , so multiply both sides by $8 η L$ and divide by $π R^{4}$ : $Q = \frac{π R ^{4} Δ P}{8 η L} ⟹ Q (8 η L) = π R^{4} Δ P ⟹ Δ P = \frac{8 η L Q}{π R ^{4}} .$ Why this step? Each segment carries a known $Q$ but an unknown $Δ P$ , so we solve the master formula for the pressure drop rather than the flow. This inverted form is the tool for every "given the flow, find the push" problem.
Series rule — the drops add. Same $Q$ through each; the pressure needed to push through A plus the pressure needed to push through B is the total: $Δ P_{total} = Δ P_{A} + Δ P_{B} .$ Why this step? Continuity (the intuition box) forces one common $Q$ ; going down the joined pipe you lose pressure in A, then lose more in B, and losses in a line simply sum.
Pipe A. $R_{A}^{4} = (2 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . $Δ P_{A} = \frac{8 ( 1 0 ^{- 3} ) ( 1.0 ) ( 2.0 \times 1 0 ^{- 6} )}{π ( 1.6 \times 1 0 ^{- 11} )} = \frac{1.6 \times 1 0 ^{- 8}}{5.027 \times 1 0 ^{- 11}} \approx 318 Pa .$ Why this step? Direct application of the inverted formula to segment A.
Pipe B. $R_{B}^{4} = (1 \times 1 0^{- 3})^{4} = 1.0 \times 1 0^{- 12} m^{4}$ . $Δ P_{B} = \frac{1.6 \times 1 0 ^{- 8}}{π ( 1.0 \times 1 0 ^{- 12} )} = \frac{1.6 \times 1 0 ^{- 8}}{3.1416 \times 1 0 ^{- 12}} \approx 5093 Pa .$ Why this step? Same numerator (same $η L Q$ ), but $R_{B}^{4}$ is $16 \times$ smaller, so $Δ P_{B}$ is $16 \times$ larger than $Δ P_{A}$ .
Add. $Δ P_{total} = 318 + 5093 \approx 5411 Pa$ .

Verify: $Δ P_{B} /Δ P_{A} = 5093/318 \approx 16.0 = 2^{4}$ ✓ — exactly the ratio of $R^{4}$ s. The narrow pipe carries ~94% of the total resistance, matching the forecast. ✓

C8 — Mean speed vs maximum speed

C9 — Real-world word problem (blood)

Worked example A narrowed artery

Blood ( $η = 3.5 \times 1 0^{- 3} Pa\cdots$ ) flows through an artery of radius $R = 4.0 mm$ , length $L = 0.10 m$ , under $Δ P = 300 Pa$ . (a) Find the healthy flow rate $Q$ . (b) If plaque narrows the radius by $20%$ (to $0.80 R$ ), what fraction of the original flow remains?

Forecast: A "mere" 20% narrowing feels minor — but $R^{4}$ punishes it. Guess the surviving fraction before computing.

(a) Healthy $Q$ . First compute $R^{4} = (4.0 \times 1 0^{- 3})^{4} = 2.56 \times 1 0^{- 10} m^{4}$ , then substitute all four given quantities into Hagen–Poiseuille: $Q = \frac{π R ^{4} Δ P}{8 η L} = \frac{π ( 2.56 \times 1 0 ^{- 10} ) ( 300 )}{8 ( 3.5 \times 1 0 ^{- 3} ) ( 0.10 )} = \frac{2.413 \times 1 0 ^{- 7}}{2.8 \times 1 0 ^{- 3}} \approx 8.62 \times 1 0^{- 5} m^{3} / s .$ Why this step? The problem gives $η$ , $R$ , $L$ and $Δ P$ directly and asks for the flow rate, so this is a straight application of the master formula — the same move as C1, now dressed as physiology (see Blood flow and circulatory system). The only care needed is converting the radius to metres before raising it to the fourth power.
(b) Fraction after narrowing. Only $R$ changes, so $\frac{Q _{new}}{Q _{old}} = (\frac{0.80 R}{R})^{4} = (0.80)^{4} = 0.4096 \approx 0.41.$ Why this step? Everything but $R$ cancels; the ratio is purely the fourth power of the radius ratio.

Verify: A 20% radius loss cuts flow to $41%$ — nearly halved by a small pinch. This is exactly why arterial narrowing is so dangerous, and the $R^{4}$ law is the reason. $(0.8)^{4} = 0.4096$ ✓.

C10 — Exam twist: is Poiseuille even valid?

Worked example Check before you compute

Water ( $η = 1.0 \times 1 0^{- 3} Pa\cdots$ , density $ρ = 1000 kg/m^{3}$ ) flows with mean speed $\overset{v}{ˉ} = 2.0 m/s$ through a pipe of radius $R = 5.0 mm$ . Is Poiseuille's law applicable? The Reynolds number is $Re = \frac{ρ v ˉ ( 2 R )}{η}$ , and flow is laminar only if $Re ≲ 2000$ .

Forecast: High speed in a not-tiny pipe smells turbulent. Guess whether $Re$ clears 2000.

Compute the diameter. $2 R = 2 (5.0 \times 1 0^{- 3}) = 1.0 \times 1 0^{- 2} m$ . Why this step? The Reynolds number uses the diameter as the characteristic length, not the radius — a classic exam trap.
Evaluate $Re$ . $Re = \frac{ρ v ˉ ( 2 R )}{η} = \frac{( 1000 ) ( 2.0 ) ( 1.0 \times 1 0 ^{- 2} )}{1.0 \times 1 0 ^{- 3}} = \frac{20}{1.0 \times 1 0 ^{- 3}} = 20000.$ Why this step? Plug the four quantities into the definition; see Reynolds number and turbulence.
Decide. $Re = 20000 ≫ 2000$ , so the flow is turbulent — Poiseuille's law (which assumes laminar flow) does not apply. Any $Q$ you compute from it would be physically wrong. Why this step? Poiseuille's whole derivation rests on smooth, layered (laminar) flow with no mixing. Once turbulence sets in, the neat parabolic profile is destroyed and flow obeys a different, non-linear pressure law.

Verify: $Re = 20000$ , ten times the laminar ceiling of $2000$ . The correct exam move is to state that Poiseuille is invalid here rather than blindly plug numbers into $Q = π R^{4} Δ P / (8 η L)$ . ✓

C11 — Two pipes in parallel

Worked example Splitting the flow between two branches

A pressure drop $Δ P = 200 Pa$ is applied across two pipes connected side by side (both ends shared), each of length $L = 1.0 m$ , fluid $η = 1.0 \times 1 0^{- 3} Pa\cdots$ . Pipe A: $R_{A} = 2.0 mm$ . Pipe B: $R_{B} = 1.0 mm$ . Find the total flow rate.

Forecast: In parallel, both pipes feel the same push $Δ P$ (their ends are joined), and their flows add. The fat pipe should dominate — guess by how much.

Parallel rule. Same $Δ P$ across each branch; total flow is the sum: $Q_{total} = Q_{A} + Q_{B}, Q = \frac{π R ^{4} Δ P}{8 η L} .$ Why this step? Unlike series (where $Q$ was shared and $Δ P$ added), here the pressure is shared and the flows add — the two branches offer independent paths, so their carrying capacities (conductances) add, not their resistances.
Pipe A. $R_{A}^{4} = (2 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . $Q_{A} = \frac{π ( 1.6 \times 1 0 ^{- 11} ) ( 200 )}{8 ( 1 0 ^{- 3} ) ( 1.0 )} = \frac{1.005 \times 1 0 ^{- 8}}{8.0 \times 1 0 ^{- 3}} \approx 1.257 \times 1 0^{- 6} m^{3} / s .$ Why this step? Direct plug-in for branch A.
Pipe B. $R_{B}^{4} = 1.0 \times 1 0^{- 12} m^{4}$ , which is $16 \times$ smaller, so $Q_{B} = \frac{Q _{A}}{16} \approx 7.85 \times 1 0^{- 8} m^{3} / s .$ Why this step? Same $Δ P$ , $L$ , $η$ ; only $R^{4}$ differs by the factor $16$ .
Add. $Q_{total} = 1.257 \times 1 0^{- 6} + 7.85 \times 1 0^{- 8} \approx 1.335 \times 1 0^{- 6} m^{3} / s$ .

Verify: $Q_{A} / Q_{B} = 16 = 2^{4}$ ✓. The fat pipe carries about $94%$ of the total flow — the mirror image of the series case, where the thin pipe hogged $94%$ of the resistance. ✓

C12 — Sign of $Δ P$ and the singular short-pipe limit

Worked example Reversed push and a vanishing pipe

(a) A pipe normally has $P_{1} > P_{2}$ so $Δ P = P_{1} - P_{2} > 0$ . What does the formula predict if we instead make $P_{1} < P_{2}$ , i.e. $Δ P < 0$ ? (b) What happens to $Q$ as $L \to 0$ (an infinitesimally short pipe)?

Forecast: Flipping which end is at higher pressure should just reverse the direction of flow — the physics is symmetric. And a pipe of zero length offers zero resistance, so guess what $Q$ does.

(a) Sign of the flow. With $Δ P < 0$ , the formula gives $Q = \frac{π R ^{4} Δ P}{8 η L} < 0.$ A negative $Q$ simply means the flow runs the other way — from $P_{2}$ toward $P_{1}$ . Likewise every velocity $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2})$ flips sign, so the whole parabola points backward. Why this step? The formula is linear and odd in $Δ P$ : replace $Δ P \to - Δ P$ and everything just negates. Fluid always flows down the pressure gradient, from high to low — the sign of $Δ P$ is only a bookkeeping choice of which end you called "1". This is the sign-convention check that guarantees the law is physically consistent.
(b) The limit $L \to 0$ . Since $L$ sits in the denominator, $lim_{L \to 0^{+}} Q = lim_{L \to 0^{+}} \frac{π R ^{4} Δ P}{8 η L} = + \infty.$ Mathematically the flow rate blows up — a zero-length pipe has zero viscous resistance, so a finite push would drive infinite flow. Why this step? This is a genuine singularity of the formula, and it flags where the model stops being physical: for a very short pipe the assumption of "fully developed" parabolic flow (fluid needs a length of pipe to settle into the parabola) fails, and inertial/entry effects — not captured by Poiseuille — take over. The blow-up is the formula honestly telling you it is out of its domain.

Verify: Odd symmetry: $Q (- Δ P) = - Q (Δ P)$ ✓ (flow reverses, magnitude unchanged). Divergence: $Q \to \infty$ as $L \to 0^{+}$ ✓ — the resistance $8 η L / (π R^{4}) \to 0$ . Both behaviours are exactly what the algebra of the formula demands.

Active recall

Recall Which formula for which question? (cover answers)

Given $v_{m a x}$ and asked for speed at a point, which form avoids $η$ and $Δ P$ ? ::: The ratio form $v (r) = v_{m a x} (1 - r^{2} / R^{2})$ , where $v_{m a x} = Δ P R^{2} / (4 η L)$ . Two pipes in series carry the same what, and add what? ::: Same $Q$ (continuity); pressure drops add. Two pipes in parallel share the same what, and add what? ::: Same $Δ P$ ; the flow rates $Q$ add. A 20% radius reduction leaves what fraction of flow? ::: $(0.8)^{4} \approx 0.41$ — about 41%. What does a negative $Δ P$ mean physically? ::: Flow simply reverses direction; magnitude unchanged. Before applying Poiseuille, which number must you check? ::: The Reynolds number ( $≲ 2000$ for laminar). Mean speed relates to max speed how? ::: $\overset{v}{ˉ} = \frac{1}{2} v_{m a x}$ .

Connections

Parent topic — the derivations these examples exercise.
Equation of continuity — the "same $Q$ " rule used in the series problem (C7).
Reynolds number and turbulence — the validity gate in C10.
Blood flow and circulatory system — the physiology behind C9.
Viscosity and Newton's law of viscosity — where $η$ and shear stress come from.