43  Flow Rate: Poiseuille’s Law

Plaque has narrowed a long segment of artery to one quarter of its original diameter (Figure 42.1). If the pressure drop across the segment stays the same, how much does blood flow fall? Summing the parabolic velocity profile of steady laminar tube flow, \(v_z(r) = \Delta P\,(R^2 - r^2)/(4\mu L)\) (flow), over the cross-section gives the volumetric flow rate \(Q\).

Flow rate from the velocity profile

If all the fluid moved at one speed \(v\), the flow rate would be \(Q = vA\), the speed times the cross-sectional area. In a tube the velocity varies with \(r\), so the flow through each small piece of the cross-section is summed as an integral (integrals as accumulation), \(Q = \int v_z\,dA\). Because \(v_z\) is constant around a ring, the pieces are taken to be thin rings. A ring of radius \(r\) and width \(dr\) unrolls into a strip of length \(2\pi r\) and width \(dr\) (Figure 43.1), so its area is \(dA = 2\pi r\,dr\).

Figure 43.1: A thin ring of the cross-section (left), cut and unrolled into a strip of length \(2\pi r\) and width \(dr\) (right, to scale).

Summing the rings from the axis to the wall gives

\[Q = \int_0^R \frac{\Delta P}{4\mu L}\left(R^2 - r^2\right) 2\pi r\,dr = \frac{\pi\,\Delta P}{2\mu L}\left[\frac{R^2 r^2}{2} - \frac{r^4}{4}\right]_0^R = \frac{\pi\,\Delta P}{2\mu L}\cdot\frac{R^4}{4}.\]

NotePoiseuille’s law

\[Q = \frac{\pi R^4\,\Delta P}{8\mu L} \qquad\Longleftrightarrow\qquad \Delta P = \frac{8\mu L}{\pi R^4}\,Q\]

It holds for steady laminar flow of a Newtonian fluid through a tube of radius \(R\) and length \(L\) (Urone and Hinrichs 2022), and is also called the Hagen–Poiseuille equation.

Dividing \(Q\) by the area \(\pi R^2\) gives the mean velocity, half the velocity on the axis:

\[\bar v = \frac{Q}{\pi R^2} = \frac{\Delta P\,R^2}{8\mu L} = \frac{v_{\max}}{2}.\]

The fourth power and hydraulic resistance

The flow rate scales with the fourth power of the radius. Halving \(R\) at a fixed pressure drop divides the flow by \(2^4 = 16\), because the mean velocity falls fourfold (\(\bar v \propto R^2\)) and so does the area that carries it (Figure 43.2). Delivering the same flow through a tube of half the radius takes 16 times the pressure drop.

Figure 43.2: Narrowing a tube from diameter \(D_1\) to \(D_2\) at a fixed pressure drop. The area and the mean velocity each fall as \((D_2/D_1)^2\), so the flow rate, their product, falls as \((D_2/D_1)^4\). The point marks the plaque example, \(D_2 = D_1/4\), where \(Q_2 = Q_1/256\).

In an electrical circuit, Ohm’s law states that the voltage \(V\) across a resistor equals the current \(i\) through it times its resistance \(R_{\text{elec}}\): \(V = iR_{\text{elec}}\) (Ling et al. 2016). Rearranged, Poiseuille’s law has the same form:

\[\Delta P = R_{\text{hyd}}\,Q, \qquad R_{\text{hyd}} = \frac{8\mu L}{\pi R^4}.\]

Pressure drop plays the role of voltage, flow rate the role of current, and the hydraulic resistance \(R_{\text{hyd}}\) (Pa·s/m³) the role of electrical resistance (Urone and Hinrichs 2022). Doubling the pressure drop doubles the flow, while widening the tube by 20% multiplies it by \(1.2^4 = 2.07\). Tubes joined end to end carry the same \(Q\), and their pressure drops add, so their hydraulic resistances add, like electrical resistors in a series circuit.

Wall shear stress

At the wall, \(r = R\), the shear stress is

\[\tau_w = \frac{\Delta P\,R}{2L} = \frac{4\mu Q}{\pi R^3},\]

where the second form uses Poiseuille’s law (Problem 2). Endothelial cells, the cells that line the inside of a blood vessel in a single layer (Betts et al. 2022), sit at the wall, where the shear stress is largest, and respond to it. The wall shear stress typical of healthy arteries, called arterial-level shear stress, is above about 1.5 Pa (Malek et al. 1999). Atherosclerosis, the build-up of plaque, develops preferentially at sites of low wall shear stress, typically below about 0.4 Pa, such as the outer edges of vessel branch points (Malek et al. 1999).

Example: shear stress in a microfluidic channel

Water is pumped at \(Q = 1~\mu\)L/min through a round channel of radius \(R = 50~\mu\)m and length \(L = 2\) cm. What pressure drop must the pump supply, and what shear stress acts on the wall? With \(1~\mu\text{L} = 10^{-9}~\text{m}^3 = 10^{-6}\) L, the flow rate is \(Q = (10^{-9}~\text{m}^3)/(60~\text{s}) = 1.67\times10^{-11}\) m³/s, and \(1~\text{m}^3/\text{s} = 6\times10^{10}~\mu\)L/min.

import numpy as np

mu = 1.0e-3                            # viscosity of water, Pa*s
Q = 1e-9 / 60                          # 1 uL/min in m^3/s
R, L = 50e-6, 0.02                     # radius and length, m

dP = 8 * mu * L * Q / (np.pi * R**4)   # Poiseuille's law, Pa
tau_w = dP * R / (2 * L)               # wall shear stress, Pa
print(f"dP = {dP:.0f} Pa, tau_w = {tau_w:.2f} Pa")

which prints

dP = 136 Pa, tau_w = 0.17 Pa

The pump must supply about 136 Pa. The wall shear stress, 0.17 Pa, is below the 0.4 Pa low-shear level. Endothelial cells grown in this channel would therefore experience the shear stress of plaque-prone regions rather than that of a healthy artery (Malek et al. 1999). Because \(\tau_w \propto Q\), raising the flow rate raises the shear stress in proportion (Problem 4).

The sliders in Figure 43.3 change \(\Delta P\), \(\mu\) and \(L\). They start at the example’s values (water, \(\mu = 1\) mPa·s, at \(\Delta P = 136\) Pa in a channel 2 cm long), where the readout below the figure gives \(Q = 1.00\) µL/min and \(\tau_w = 0.170\) Pa. Raising \(\mu\) slows the flow but leaves \(\tau_w\) unchanged.

viewof dP = Inputs.range([10, 500], {value: 136, step: 1, label: "pressure drop ΔP (Pa)"})
viewof muMPa = Inputs.range([0.5, 5], {value: 1, step: 0.1, label: "viscosity μ (mPa·s)"})
viewof Lcm = Inputs.range([1, 5], {value: 2, step: 0.1, label: "tube length L (cm)"})

tube = {
  const R = 50e-6;                                        // channel radius, m
  const flow = (dp, mu, L) => ({                          // SI: Pa, Pa s, m
    vmax: dp * R * R / (4 * mu * L),                      // m/s
    Q: Math.PI * R ** 4 * dp / (8 * mu * L),              // m^3/s
    tauw: dp * R / (2 * L),                               // Pa
    pts: d3.range(-50, 50.5, 1).map(r => ({r,             // r in um
      v: 1e3 * dp * (R * R - (r * 1e-6) ** 2) / (4 * mu * L),   // mm/s
      tau: dp * Math.abs(r) * 1e-6 / (2 * L)}))           // Pa
  });
  return {now: flow(dP, muMPa * 1e-3, Lcm * 1e-2), ref: flow(136, 1e-3, 0.02)};
}

{
  const sub = svg => {                                    // label "τ_w" -> τ with subscript w
    svg.querySelectorAll("text").forEach(t => {
      const m = t.textContent.match(/^(v|τ)_(max|w)$/);
      if (!m) return;
      t.innerHTML = `${m[1]}<tspan dy="0.35em" font-size="0.75em"${m[2] === "max" ? ' font-style="normal"' : ""}>${m[2]}</tspan>`;
      t.parentNode.setAttribute("paint-order", "stroke");   // white halo behind the letters
    });
    return svg;
  };
  const fit = (x, tops) => tops.find(top => x <= 0.8 * top) || tops[tops.length - 1];   // smallest axis that holds x
  // A curve near zero hugs the axis, so its label moves further right as the value shrinks.
  // Below 3% of the axis the curve lies on the axis line: the label is dropped and the readout gives the value.
  const label = (x, top, r, dy, text) => x < 0.03 * top ? [] : [Plot.text([[x, r]], {text: [text], dy,
    dx: 8 + 10 * Math.max(0, 1 - x / (0.15 * top)),
    fontStyle: "italic", textAnchor: "start", fill: "#333333", stroke: "white", strokeWidth: 4})];
  const panel = (key, xlabel, max, color, ariaLabel, extra) => sub(Plot.plot({
    ariaLabel, width: 300, height: 320, marginLeft: 52, marginRight: 36, marginTop: 32, marginBottom: 48,
    style: {fontSize: "16px"},
    x: {domain: [0, max], label: xlabel, grid: true, ticks: 4},
    y: {domain: [-56, 56], ticks: [-50, -25, 0, 25, 50], label: "r (µm)", grid: true},
    marks: [
      Plot.ruleY([-50, 50], {stroke: "#333333", strokeWidth: 2}),
      Plot.line(tube.ref.pts, {x: key, y: "r", stroke: color, strokeWidth: 2, strokeOpacity: 0.3}),
      Plot.line(tube.now.pts, {x: key, y: "r", stroke: color, strokeWidth: 3}),
      ...extra]
  }));
  const v = tube.now.vmax * 1e3, t = tube.now.tauw;
  const vTop = fit(v, [18, 36, 72]), tTop = fit(t, [0.7, 1.4]);
  const vPlot = panel("v", "velocity (mm/s)", vTop, "#3B7EA1",
    "Velocity across the channel: a parabola, largest on the axis and zero at the walls", [
    Plot.dot([[v, 0]], {fill: "#3B7EA1", r: 5}),
    ...label(v, vTop, 0, -14, "v_max")]);                                  // right of the dot, clear of the curve
  const tPlot = panel("tau", "shear stress |τ| (Pa)", tTop, "#FDB515",
    "Shear-stress magnitude across the channel: a V shape, zero on the axis and largest at the walls", [
    Plot.ruleX([0.4], {stroke: "#9A9A9A", strokeDasharray: "5,4"}),
    Plot.dot([[t, 50], [t, -50]], {fill: "#FDB515", stroke: "#333333", r: 5}),
    ...label(t, tTop, 50, 18, "τ_w")]);                                    // inside the channel, below the wall
  const cell = p => html`<div style="flex: 1 1 0; min-width: 0;">${p}</div>`;   // panels shrink, never wrap
  // padding-bottom keeps the axis titles clear of the figure caption
  return html`<div style="display: flex; gap: 16px; max-width: 616px; margin: 0 auto; padding-bottom: 16px;">${cell(vPlot)}${cell(tPlot)}</div>`;
}

Figure 43.3: Velocity and shear stress across a tube, wall to wall. Left: the velocity \(v_z(r)\). Right: the magnitude of the shear stress, with \(\tau_w\) marked at the walls. The channel radius is \(R = 50\) µm, and negative \(r\) is the far side of the axis. Faint curves show the starting settings; the dashed line marks the 0.4 Pa low-shear level. A horizontal axis doubles its range when a curve outgrows it.

{
  const f = x => x.toPrecision(3);
  return html`<p>With these settings the fluid on the axis moves at <b><i>v</i><sub>max</sub> = ${f(tube.now.vmax * 1e3)} mm/s</b>, the channel carries <b><i>Q</i> = ${f(tube.now.Q * 6e10)} µL/min</b>, and the shear stress at the wall is <b><i>τ</i><sub>w</sub> = Δ<i>P R</i>/(2<i>L</i>) = ${f(tube.now.tauw)} Pa</b>.</p>`;
}

The narrowed artery

The narrowed segment is modeled as a uniform tube of diameter \(D_2\), with the same length, blood, and pressure drop as the original segment of diameter \(D_1\) (Figure 42.1). With \(R = D/2\) in Poiseuille’s law, every factor except the diameter cancels:

\[\frac{Q_1}{Q_2} = \frac{\pi\,(D_1/2)^4\,\Delta P/(8\mu L)}{\pi\,(D_2/2)^4\,\Delta P/(8\mu L)} = \left(\frac{D_1}{D_2}\right)^4 = 4^4 = 256 .\]

The narrowed segment carries \(1/256\) of its original flow, about 0.4% (Figure 43.2, at \(D_2/D_1 = 0.25\)); restoring that flow takes a pressure drop 256 times larger. The uniform-tube model gives the scaling with diameter, not a clinical prediction. Flow through the pores of a test strip has the same fourth-power dependence on radius, but there the pore radius also sets the driving pressure (capillary flow).

Exercises

Unless stated otherwise, the fluid is water, with \(\mu = 1.0\times10^{-3}\) Pa·s and \(\rho = 1000\) kg/m³.

TipProblem 1

Plaque narrows a segment of artery to half its original diameter.

1. At the same pressure drop, by what factor does the flow fall?
2. A milder plaque reduces the diameter by 20%. What fraction of the original flow remains at the same pressure drop?

NoteAnswer 1

1. \(Q_1/Q_2 = (D_1/D_2)^4 = 2^4 = 16\): the flow falls to \(1/16\).
2. \(Q_2/Q_1 = 0.8^4 = 0.41\). A 20% narrowing leaves 41% of the flow.

TipProblem 2

1. Substitute Poiseuille’s law into \(\tau_w = \Delta P\,R/(2L)\) to show that \(\tau_w = 4\mu Q/(\pi R^3)\).
2. A channel’s radius is halved. By what factor does \(\tau_w\) change if a syringe pump holds \(Q\) fixed? If a pressure controller holds \(\Delta P\) fixed?

NoteAnswer 2

1. \(\tau_w = \dfrac{R}{2L}\cdot\dfrac{8\mu L Q}{\pi R^4} = \dfrac{4\mu Q}{\pi R^3}\).
2. At fixed \(Q\), \(\tau_w \propto 1/R^3\), so it rises \(2^3 = 8\)-fold. At fixed \(\Delta P\), \(\tau_w \propto R\), so it halves.

TipProblem 3

The glass capillary tube of Problem 2 in flow (inner radius 0.25 mm, length 10 cm) carries water under a pressure drop of 100 Pa.

1. Find the volumetric flow rate in µL/min.
2. Find the mean velocity, and confirm that it is half the velocity on the axis.
3. Find the Reynolds number, \(\mathrm{Re} = \rho\bar v D/\mu\), and check that the flow is laminar (Re below about 2000).

NoteAnswer 3

1. \(Q = \dfrac{\pi R^4\,\Delta P}{8\mu L} = \dfrac{\pi(2.5\times10^{-4})^4(100)}{8(10^{-3})(0.10)} = 1.53\times10^{-9}\) m³/s. With \(1~\text{m}^3 = 10^9~\mu\)L, this is 1.53 µL/s, or 92 µL/min.
2. \(\bar v = Q/(\pi R^2) = 1.53\times10^{-9}/[\pi(2.5\times10^{-4})^2] = 7.8\) mm/s. The velocity on the axis is \(v_{\max} = \Delta P\,R^2/(4\mu L) = 15.6\) mm/s, twice \(\bar v\).
3. \(\mathrm{Re} = (1000)(7.8\times10^{-3})(5.0\times10^{-4})/10^{-3} = 3.9\), far below 2000: laminar.

TipProblem 4

Endothelial cells are grown on the walls of the channel in the example (\(R = 50~\mu\)m, \(L = 2\) cm).

1. What flow rate gives an arterial-level wall shear stress of 1.5 Pa?
2. What pressure drop must the pump then supply?

NoteAnswer 4

1. \(Q = \pi R^3\tau_w/(4\mu) = \pi(5.0\times10^{-5})^3(1.5)/(4\times10^{-3}) = 1.47\times10^{-10}\) m³/s \(= 8.8~\mu\)L/min, which is \(1.5/0.17\) times the example’s flow, since \(\tau_w \propto Q\).
2. \(\Delta P = 2L\tau_w/R = 2(0.02)(1.5)/(5.0\times10^{-5}) = 1200\) Pa.

TipProblem 5

A syringe pump drives water at \(Q = 1~\mu\)L/min through a supply tube (radius 0.5 mm, length 20 cm) joined end to end with a channel (radius 50 µm, length 2 cm).

1. Find the hydraulic resistance of each.
2. Find the total pressure the pump must supply.
3. What fraction of that pressure drop occurs across the channel?

NoteAnswer 5

1. \(R_{\text{hyd}} = 8\mu L/(\pi R^4)\) gives \(8.15\times10^{9}\) Pa·s/m³ for the tube and \(8.15\times10^{12}\) Pa·s/m³ for the channel. A 10-fold smaller radius multiplies resistance by \(10^4\), and a 10-fold shorter length divides it by 10, a net factor of \(10^3\).
2. The tube and channel are joined end to end, so their resistances add: \(\Delta P = (8.16\times10^{12})(1.67\times10^{-11}) = 136\) Pa.
3. \(1000/1001 = 99.9\%\). The narrow channel controls the flow; the tube accounts for 0.14 Pa.

References

Betts, J. Gordon, Kelly A. Young, James A. Wise, et al. 2022. Anatomy and Physiology 2e. OpenStax, Rice University. https://med.libretexts.org/Bookshelves/Anatomy_and_Physiology/Anatomy_and_Physiology_2e_(OpenStax).

Ling, Samuel J., Jeff Sanny, and William Moebs. 2016. University Physics Volume 2. OpenStax, Rice University. https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/10:_Direct-Current_Circuits/10.06:_RC_Circuits.

Malek, Adel M., Seth L. Alper, and Seigo Izumo. 1999. “Hemodynamic Shear Stress and Its Role in Atherosclerosis.” JAMA 282 (21): 2035–42. https://doi.org/10.1001/jama.282.21.2035.

Urone, Paul Peter, and Roger Hinrichs. 2022. College Physics 2e. OpenStax, Rice University. https://openstax.org/books/college-physics-2e/pages/12-4-viscosity-and-laminar-flow-poiseuilles-law.