Poiseuille's formula

Blood squeezes through capillaries thinner than a hair, an IV drip feeds fluid through a slender needle into a vein, and crude oil is pumped down kilometres of pipeline — in every case a viscous fluid is pushed through a narrow tube by a pressure difference. How much fluid gets through per second is fixed by just four things: the tube's radius, its length, the fluid's stickiness (viscosity), and the pressure difference driving it. Poiseuille's formula packages this into one equation, and its most striking feature is that flow rate depends on the tube's radius raised to the fourth power — a small change in radius produces a huge change in flow. That single fact explains why a slightly narrowed artery can choke blood supply, and why engineers reach for a wider pipe rather than a stronger pump. This article states the formula, unpacks what each term means, shows where the fourth power comes from using only dimensional reasoning, and applies it to typical numericals — exactly the level the ISC Class 11 syllabus asks for.

The core insight

The volume flow rate of a viscous fluid in streamline flow through a narrow tube is $Q = \pi P r^4/(8\eta l)$ , so it grows as the fourth power of the radius and directly with pressure difference, but shrinks as viscosity or tube length increase.

The core idea

Consider a viscous, incompressible fluid flowing steadily and slowly — in streamline (laminar) flow, at a Reynolds number below the critical value — through a narrow horizontal tube of uniform circular cross-section. A pressure difference between the two ends drives the flow. Under these conditions, the volume of fluid crossing any cross-section per second is given by Poiseuille's formula:

Q = \frac{dV}{dt} = \frac{\pi P r^4}{8\,\eta\, l}

where: $Q$ is the volume flow rate ( $\text{m}^3/\text{s}$ ), $P$ is the pressure difference across the ends of the tube ( $\text{Pa}$ ), $r$ is the tube's radius ( $\text{m}$ ), $\eta$ is the fluid's coefficient of viscosity ( $\text{Pa}\,\cdot\,\text{s}$ ), and $l$ is the tube's length ( $\text{m}$ ). Three proportionalities are worth holding onto separately. Flow rate rises directly with the pressure difference $P$ — a bigger push moves more fluid each second. It falls inversely with viscosity $\eta$ — a thicker, stickier fluid crawls through more slowly for the same pressure and geometry. It falls inversely with length $l$ — a longer tube offers proportionately more resistance, exactly as a longer wire offers more electrical resistance. The dominant term, though, is the radius. Because $Q \propto r^4$ , doubling the tube's radius (with everything else fixed) multiplies the flow rate by $2^4 = 16$ , and halving the radius cuts it to $1/16$ . No other everyday variable in physics swings a quantity so violently for so small a change in itself — which is precisely why the body's blood vessels, and industrial pipelines, are so sensitive to bore.

Figure

A horizontal tube of radius r and length l with higher pressure at the left end and lower pressure at the right end and a parabolic arrow pattern showing fluid moving fastest along the axis and slowest near the walls

Poiseuille's formula applies only to streamline flow of an incompressible viscous fluid through a narrow tube; once the flow turns turbulent it no longer holds.

Why it works

The ISC syllabus explicitly excludes deriving Poiseuille's formula from the underlying parabolic velocity profile — that requires integrating over the tube's cross-section, which lies beyond Class 11. But you can still see, without any of that machinery, why the radius must enter to the fourth power, using nothing more than dimensional analysis together with one physically reasonable assumption.

Derivation AFixing the radius exponent by dimensional analysis

The goal is to show, using only dimensional analysis and the physically reasonable assumption that a longer tube resists flow in direct proportion to its length, that Poiseuille's formula must take the form $Q \propto P\,r^4/(\eta\, l)$ — pinning down the fourth-power dependence on radius without integrating the fluid's velocity profile, which the syllabus excludes.

Step 1. Assume $Q$ depends on the four quantities as a power law, $Q = k\, P^{a}\, r^{b}\, \eta^{c}\, l^{d}$ , where $k$ is a dimensionless constant. Write the dimensions of each quantity: $[Q] = L^3T^{-1}$ , $[P] = ML^{-1}T^{-2}$ , $[r] = L$ , $[\eta] = ML^{-1}T^{-1}$ , $[l] = L$ .

Step 2. Match powers of mass ( $M$ ), length ( $L$ ) and time ( $T$ ) on both sides. Mass gives $a + c = 0$ . Time gives $-2a - c = -1$ . Length gives $-a + b - c + d = 3$ .

Step 3. Solve the mass and time equations together. From mass, $c = -a$ . Substituting into the time equation: $-2a - (-a) = -1$ , so $-a = -1$ , giving $a = 1$ and hence $c = -1$ . Dimensional analysis alone cannot separate $b$ and $d$ from the length equation, which only fixes $b + d = 3$ once $a$ and $c$ are known — a fourth relation is needed.

Step 4. Supply the one physical fact that dimensional analysis cannot: a longer tube resists flow in direct proportion to its length (double the length, half the flow, exactly as doubling a wire's length doubles its electrical resistance). This fixes $d = -1$ . The length equation then gives $b = 3 - d = 3 - (-1) = 4$ . Collecting the exponents, $Q = k\,P^{1} r^{4} \eta^{-1} l^{-1} = k\,\dfrac{P r^4}{\eta l}$ — the fourth-power dependence on radius falls out directly. The full theory (integrating the parabolic velocity profile, beyond this syllabus) fixes the dimensionless constant at $k = \pi/8$ , recovering exactly $Q = \pi P r^4/(8\eta l)$ .

Worked example

Water ( $\eta = 1.0\times10^{-3}\ \text{Pa}\,\cdot\,\text{s}$ ) flows through a horizontal capillary tube of radius $r = 0.50\ \text{mm}$ and length $l = 0.20\ \text{m}$ , driven by a pressure difference $P = 4000\ \text{Pa}$ . Find the volume flow rate.

Step 1. Convert the radius to metres and raise it to the fourth power: $r = 0.50\times10^{-3}\ \text{m}$ , so $r^4 = (0.50\times10^{-3})^4 = 6.25\times10^{-14}\ \text{m}^4$ .

Step 2. Substitute into Poiseuille's formula: $Q = \dfrac{\pi P r^4}{8\eta l} = \dfrac{\pi \times 4000 \times 6.25\times10^{-14}}{8 \times 1.0\times10^{-3} \times 0.20}$ .

Step 3. Evaluate numerator and denominator separately. Numerator: $\pi \times 4000 \times 6.25\times10^{-14} \approx 7.854\times10^{-10}$ . Denominator: $8 \times 1.0\times10^{-3} \times 0.20 = 1.6\times10^{-3}$ .

Step 4. Divide: $Q \approx \dfrac{7.854\times10^{-10}}{1.6\times10^{-3}} \approx 4.9\times10^{-7}\ \text{m}^3/\text{s}$ , which is about $0.49\ \text{mL/s}$ . Answer: $Q \approx 4.9\times10^{-7}\ \text{m}^3/\text{s} \approx 0.49\ \text{mL/s}$ — a modest trickle, consistent with water in a narrow capillary.

Competitive stretch

A tube carries flow rate $Q_0$ under a fixed pressure difference. If its radius is increased by $50\%$ (to $1.5$ times the original) with everything else unchanged, find the new flow rate as a multiple of $Q_0$ .

Step 1. Since $Q \propto r^4$ , the ratio of new to old flow rate is $\dfrac{Q_{\text{new}}}{Q_0} = \left(\dfrac{r_{\text{new}}}{r_0}\right)^4 = (1.5)^4$ .

Step 2. Compute $(1.5)^4 = 1.5 \times 1.5 \times 1.5 \times 1.5 = 5.0625$ . Answer: $Q_{\text{new}} \approx 5.06\,Q_0$ — a mere $50\%$ increase in radius produces roughly a five-fold increase in flow, the fourth-power sensitivity in action.

Common traps

Treating flow rate as proportional to $r^2$ (the pipe's cross-sectional area) instead of $r^4$ — this drastically underestimates how sensitive flow is to radius.
Attempting to derive Poiseuille's formula from the parabolic velocity profile in an exam — the syllabus explicitly excludes this derivation; state and apply the formula, do not derive it from scratch.
Applying the formula to turbulent flow — it holds only for streamline (laminar) flow through a narrow tube; check the flow conditions first.
Mixing up numerator and denominator quantities — bigger $P$ or $r$ means more flow; bigger $\eta$ or $l$ means less flow (more resistance).

Ask the tutor: Want to see how Poiseuille's formula connects to Stokes' law and terminal velocity, or get extra practice scaling flow rate with different variables? Ask the AI tutor for more worked problems.

Why does my textbook say we do not need to derive Poiseuille's formula?

Because the full derivation requires integrating the fluid's velocity across the tube's parabolic profile, which uses calculus and physics beyond the Class 11 syllabus. You are expected to know the formula, understand what each symbol means physically, and use it correctly in numericals — not reconstruct it from first principles. The dimensional-analysis argument in this article is a legitimate shortcut that shows why $r^4$ appears, without needing that excluded integral.

Why does flow depend on r to the fourth power instead of r squared like the cross-sectional area?

Area alone would suggest $Q \propto r^2$ , but that ignores how the flow speed itself changes with radius. In a wider tube, fluid near the axis also moves faster because it is farther from the slowing effect of the walls, so both the area available for flow and the typical speed of the fluid increase together as the tube widens. That combined effect is what pushes the exponent from $2$ up to $4$ , and it is exactly what the dimensional-analysis derivation captures without needing the detailed velocity profile.

Does Poiseuille's formula literally apply to blood flowing through arteries and capillaries?

It applies well as a first approximation, especially in narrow capillaries where flow is genuinely slow and streamline, which is why doctors reason about vessel narrowing using the $r^4$ dependence. In larger arteries near the heart, blood flow can become turbulent and blood itself is not a perfectly simple viscous fluid, so the formula becomes only approximate there. For the syllabus, though, you only need the idealised streamline case described here.