Poiseuille's Law & Pipe Flow

Set a radius, a length and a pressure to see how much a pipe carries, and where the law stops describing what is in it.

Worked example — Poiseuille's Law & Pipe Flow

Flow rate	120.4 L/min
Mean velocity	39.9 m/s
Reynolds number	318,404
Flow regime	Turbulent
Hydraulic resistance	9.96 × 10⁷ Pa·s/m³
Radius needed	3.3605 mm
Regime at that radius	Turbulent

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Poiseuille's Law & Pipe Flow

Poiseuille's law is what a pressure difference buys in a pipe: Q = πΔp r⁴ / (8ηL). The fourth power is the whole story and it comes from two places at once — a wider pipe has more cross-section to carry fluid, and the fluid in the middle of it is further from the wall that is slowing everything down. The first gives r², the second another r².

What each symbol means

Q: volume flow rate, what comes out per second
Δp: the pressure difference driving it
r: the radius of the pipe, the term that is raised to the fourth power
L: its length, which the flow is simply inversely proportional to
η: the viscosity of the fluid

Breaks when: On blood, which is the example everyone reaches for. Blood is not Newtonian — it thins as you shear it — an artery is elastic rather than rigid, and the flow is pulsed rather than steady. The 30% narrowing on this page is the right shape of answer and the wrong number for any particular patient.

Narrow an artery by 30% and 76% of the flow is gone 🖖

Flow through a round pipe goes as the fourth power of the radius. Halve the radius and the flow does not halve, and does not quarter: it divides by 16, and by 16.0000 rather than roughly 16, because r⁴ is the only radius-dependent term in Q = πΔP·r⁴ / 8ηL and nothing else in the formula has moved.

Press Thin tube — half a millimetre of radius, a metre of pipe, 5,000 Pa — and the flow-rate row reads 7.351 mL/min. Press Half the radius and it reads 0.459 mL/min, which is the first figure divided by 16 and then rounded.

Small changes land harder than they look. Shave 10% off a radius and 34.4% of the flow leaves with it. Shave 30% off and you keep 24.0%, which is what the tool's two blood presets do: 2 mm down to 1.4 mm at the same 200 Pa, a loss of 76.0%.

A water main 10% narrowed by scale still delivers about two thirds of its flow and nobody phones anybody. An artery 10% narrowed is running the same arithmetic with no two thirds to spare. Poiseuille was a physician. He measured the fourth power in the 1840s by pushing liquids through narrow tubes, looking for how blood moves.

18.92% more radius is worth twice the pressure 🖖

Q is linear in ΔP and quartic in r, so there are two ways to double a flow and they cost very different things.

Take the laminar water setting: 0.5 mm radius, 1 m of pipe, 5,000 Pa. The flow-rate row reads 7.351 mL/min. Raise the pressure to 10,000 Pa and it reads 14.703. Or leave the pressure where it is and widen the radius to 0.5946 mm, 18.92% more, and it reads 14.702. The missing thousandth is the rest of 2^1/4 = 1.1892071, which you rounded away by typing four decimals.

Which of the two you can reach for is not a physics question. Mains pressure belongs to the water company, and a heart that could double its output on demand would not be the one you were worried about.

Re 318,404: the flow this page opens on is fiction 🖖

This page opens on water through a 4 mm pipe, 10 m long, pushed by 2 bar. Poiseuille's law returns 120.4 L/min and the arithmetic behind it is beyond dispute. That flow implies a mean velocity of 39.9 m/s, roughly 144 km/h of water, and a Reynolds number of 318,404 — about 138 times the 2300 below which the law holds.

Above that ceiling the flow is no longer laminar. It loses energy into eddies as well as to viscous shear, so a real pipe under 2 bar carries considerably less than 120.4 L/min. The formula has no term that could notice.

So the flow-rate, Reynolds-number and flow-regime rows all go red, and the chart shades everything wider than 0.7732 mm out of the green. Narrow the pipe to 0.77 mm and the tool turns green again: Re 2,271, laminar, an answer you can keep. Every other preset on this page is already there.

Problem solved in full

What the heart would have to pay 7 steps
A 10 cm length of artery, 2 mm in radius, carries blood (η = 3.5 × 10⁻³ Pa·s, ρ = 1060 kg/m³) under a pressure drop of 200 Pa. Plaque narrows the bore by 30%, to 1.4 mm. Find the flow that is left, and then find the pressure drop that would put the original flow back through the narrowed vessel.

Open this problem in the tool
1. Start with the healthy vessel and keep everything in SI. Radius 2 × 10⁻³ m, length 0.1 m, 200 Pa across it, blood at 3.5 × 10⁻³ Pa·s. The fourth power of the radius is 1.6 × 10⁻¹¹ m⁴, and the whole expression comes to 3.590 × 10⁻⁶ m³/s, which is 0.2154 litres a minute.
  
  $Q_{0} = \frac{π Δ P r _{0}^{4}}{8 η L} = \frac{π \cdot 200 \cdot ( 2 \times 1 0 ^{- 3} ) ^{4}}{8 \cdot 3.5 \times 1 0 ^{- 3} \cdot 0.1} = 3.590 \times 1 0^{- 6} m^{3} /s$
2. Now narrow it. Length, pressure and blood are untouched, so the ratio of the two flows is the ratio of r⁴ and of nothing else in the formula. 1.4 divided by 2 is 0.7, and 0.7⁴ is 0.2401. A bore that still looks nearly right is carrying 24.01% of what it did. The Flow rate row reads 51.723 mL/min.
  
  $\frac{Q _{1}}{Q _{0}} = (\frac{r _{1}}{r _{0}})^{4} = 0. 7^{4} = 0.2401 \Rightarrow Q_{1} = 8.621 \times 1 0^{- 7} m^{3} /s = 51.723 mL/min$
3. Mean velocity is that flow spread over the bore. Dividing Q by πr² cancels two of the four powers and leaves v̄ = ΔP r²/(8ηL), so velocity falls with the square while the flow falls with the fourth power. Here it comes out at 0.140 m/s exactly, and the Mean velocity row agrees.
  
  $\overset{v}{ˉ} = \frac{Q _{1}}{π r _{1}^{2}} = \frac{Δ P r _{1}^{2}}{8 η L} = \frac{200 \cdot ( 1.4 \times 1 0 ^{- 3} ) ^{2}}{8 \cdot 3.5 \times 1 0 ^{- 3} \cdot 0.1} = 0.140 m/s$
4. Before trusting any of it, check that the law is allowed to speak. Poiseuille's derivation assumes laminar flow, and the test is the Reynolds number built on the diameter: 119, against a ceiling of 2300. The Flow regime row reads Laminar. Notice which way the narrowing pushed it. Put 2 back in the radius field and Re is 346, so choking the flow made the arithmetic safer rather than shakier.
  
  $Re = \frac{ρ v ˉ D}{η} = \frac{1060 \cdot 0.140 \cdot 2.8 \times 1 0 ^{- 3}}{3.5 \times 1 0 ^{- 3}} = 118.7 < 2300$
5. The same fourth power drives the resistance. R = 8ηL/πr⁴ is the pressure a vessel demands per unit of flow, and the Hydraulic resistance row reads 2.32 × 10⁸ Pa·s/m³ where the healthy artery read 5.57 × 10⁷. The factor between them is 4.1649, which is 0.2401 turned over.
  
  $R = \frac{8 η L}{π r ^{4}} : R_{0} = 5.57 \times 1 0^{7}, R_{1} = 2.32 \times 1 0^{8} Pa s/ m^{3}, \frac{R _{1}}{R _{0}} = \frac{1}{0.2401} = 4.1649$
6. Q = ΔP/R is Poiseuille's law rearranged, and it is the line the circulation has to live on. Holding Q at the healthy 3.590 × 10⁻⁶ m³/s through the new resistance costs ΔP = Q₀R₁ = 833 Pa. Flow is linear in pressure, so the drop has to rise by exactly the factor the resistance rose by: 200 × (10/7)⁴, and (10/7)⁴ is 10 000/2401.
  
  $Δ P_{1} = Q_{0} R_{1} = 3.590 \times 1 0^{- 6} \cdot 2.320 \times 1 0^{8} = 833 Pa$
7. Check the answer against its own assumption. At 833 Pa the mean velocity in the narrowed vessel is 0.583 m/s and the Reynolds number is 494, still well under 2300, so the law that produced the figure still holds at the state it predicts. Type 833 into the pressure field with the radius at 1.4 mm and the Flow rate row returns to 0.22 L/min, where the healthy artery began.
  
  $\overset{v}{ˉ}_{1} = 0.583 m/s, Re_{1} = \frac{1060 \cdot 0.583 \cdot 2.8 \times 1 0 ^{- 3}}{3.5 \times 1 0 ^{- 3}} = 494 < 2300$
Answer

833 Pa, against the 200 Pa the healthy vessel needed. Four and a sixth times the driving pressure, to move the same litres per minute through a bore 0.6 mm narrower. The tool will answer the other inversion for you: type 0.2154 into the target flow box and the Radius needed row says 1.9999 mm, which is the healthy artery back, and which is a surgeon's answer rather than a heart's. Nothing on the page solves for the pressure, and the pressure is the half a body has to find.

$Δ P_{1} = Δ P_{0} (\frac{r _{0}}{r _{1}})^{4} = 200 \cdot \frac{1 0 ^{4}}{2401} = 833 Pa$

References (1)

The measurements the fourth power came out of, made by a physician looking for how blood moves: J. L. M. Poiseuille, "Recherches expérimentales sur le mouvement des liquides dans les tubes de très petits diamètres." Mémoires présentés par divers savants à l'Académie Royale des Sciences de l'Institut de France 9, 1846, pp. 433–544.

By the 42Math team · Reviewed against published sources. Corrections welcome.