In a hospital ward, a bottle of saline hangs above a patient's bed. A thin tube carries it down to a needle in the arm. On the tube is a small plastic wheel, the roller clamp. The nurse rolls it by a millimetre or two, and the drops in the chamber change from a quick patter to a slow count. A small squeeze of the tube makes a large change in the flow.
The flow through a narrow tube
Lesson 2 found that a liquid in a pipe moves in cylindrical layers, with a parabolic profile. Now find that profile, and from it the volume flow rate \(Q\), the volume of liquid passing through the tube each second. Take a horizontal tube of radius \(r\) and length \(l\), with a pressure difference \(P\) between its ends. In steady flow nothing accelerates, so the forces on every part of the liquid balance.
\[ P\,\pi s^2 = -\,\eta\,(2\pi s l)\,\dfrac{dv}{ds} \]
Why. take the cylinder of liquid of radius \(s\) along the axis. The pressure difference pushes on its ends with \(P\pi s^2\). The slower liquid outside drags on its curved surface, of area \(2\pi s l\), with Newton's law of viscosity. The speed falls outward, so \(dv/ds\) is negative.
\[ \dfrac{dv}{ds} = -\dfrac{P s}{2 \eta l} \]
Why. divide both sides by \(-2\pi s l\,\eta\).
\[ v(s) = \dfrac{P}{4 \eta l}\left(r^2 - s^2\right) \]
Why. integrate from the wall, where \(v = 0\) at \(s = r\), inward to \(s\). This is the parabola of Lesson 2, with \(v_{max} = Pr^2/4\eta l\) on the axis.
\[ Q = \int_0^r v(s)\,2\pi s\,ds = \dfrac{\pi P}{2 \eta l}\int_0^r \left(r^2 s - s^3\right) ds \]
Why. the ring between \(s\) and \(s + ds\) has area \(2\pi s\,ds\), and each second a volume \(v\,2\pi s\,ds\) passes through it.
\[ Q = \dfrac{\pi P}{2\eta l}\left(\dfrac{r^4}{2} - \dfrac{r^4}{4}\right) = \dfrac{\pi P r^4}{8 \eta l} \]
Read it. carry out the integral.
In steady streamline flow of a liquid through a long, narrow horizontal tube, the volume flowing each second is \(Q = \pi P r^4/8\eta l\). It is proportional to the pressure difference and to the fourth power of the radius, and inversely proportional to the viscosity and the length.
A formula this unusual deserves a check by dimensions. The left side, \(Q\), is in \(\text{m}^3/\text{s}\). On the right, \(\pi\) and \(8\) are pure numbers. \[ \frac{[P][r^4]}{[\eta][l]} = \frac{\text{Pa} \times \text{m}^4}{\text{Pa s} \times \text{m}} = \frac{\text{m}^3}{\text{s}} \] The two sides agree. Dimensions cannot confirm the \(\pi/8\). But they do show that no power of \(r\) other than four could fit with \(P/l\) and \(\eta\).
Water, with \(\eta = 10^{-3}\,\text{Pa s}\), flows through a tube of radius \(1\,\text{mm}\) and length \(1\,\text{m}\) under a pressure difference of \(800\,\text{Pa}\). Find the flow rate. Then find the speed on the axis and the mean speed.
(a) \(Q = \dfrac{\pi \times 800 \times (10^{-3})^4}{8 \times 10^{-3} \times 1} = \pi \times 10^{-7} \approx 3.1 \times 10^{-7}\,\text{m}^3\text{/s}\), about \(0.3\,\text{mL}\) each second.
(b) \(v_{max} = Pr^2/4\eta l = 800 \times 10^{-6}/(4 \times 10^{-3}) = 0.2\,\text{m/s}\). The mean speed is \(Q/\pi r^2 = 0.1\,\text{m/s}\), half of \(v_{max}\), as Lesson 2 said.
Water (\(\eta = 10^{-3}\,\text{Pa s}\)) flows through a tube of radius \(0.5\,\text{mm}\) and length \(10\,\text{cm}\) under a pressure difference of \(2000\,\text{Pa}\). Find the flow rate, in mL/s.
The fourth power of the radius
Why \(r^4\), and not \(r^2\), the area? Widen a tube, and two things grow together. There is more area for the liquid to pass through, which grows as \(r^2\). And the liquid on the axis is farther from the walls that hold it back, so it moves faster, as \(v_{max} = Pr^2/4\eta l\) shows. The flow rate is the area times the mean speed, so it grows as \(r^2 \times r^2 = r^4\).
Move the slider in the sim above. At \(1\,\text{mm}\) the flow is \(Q_1\). At \(2\,\text{mm}\) the speed on the axis is four times as large and the area four times as large, so the flow is \(16\,Q_1\). At \(0.5\,\text{mm}\) it is \(Q_1/16\). A change of radius that the eye can barely see changes the flow enormously.
A liquid flows through a tube in streamline flow. Which changes, each made alone, would double the flow rate?
A narrowed artery
Blood flows through arteries much as Poiseuille's law describes, though blood is not a simple liquid. Suppose fatty deposits narrow an artery so that its radius falls to half. At the same pressure, the flow falls to \((1/2)^4 = 1/16\) of its old value. To restore the flow, the pressure difference must rise sixteen times. Even a narrowing of \(20\,\%\) cuts the flow to \(0.8^4 = 0.41\) of its value. This is why narrowed arteries make the heart work harder and raise the blood pressure.
An artery narrows so that its radius is \(80\,\%\) of its former value. By what factor must the pressure difference across it rise to keep the same flow?
Return to the drip in the ward. The roller clamp squeezes the soft tube and narrows the channel through it. Because the flow goes as the fourth power of the width, a squeeze the eye can hardly see changes the flow a great deal. Narrow the channel to \(80\,\%\), and the drip falls to \(41\,\%\) of its rate. That is why a nurse turns the clamp by a millimetre at a time and counts the drops again.
Tubes as resistors
Poiseuille's law can be written \(P = R_f\,Q\), with \(R_f = 8\eta l/\pi r^4\). This is Ohm's law for a liquid. The pressure difference plays the part of the voltage, the flow rate that of the current, and \(R_f\) is the fluid resistance of the tube. A long, narrow tube full of a sticky liquid has a large resistance.
Join two tubes end to end. The same volume flows through both each second, so the same \(Q\) passes through each. The pressure differences add, and so the resistances add, just as for resistors in series. Now connect two tubes side by side between the same two pressures. Each carries its own flow, and the flows add. The combined resistance follows the rule for resistors in parallel.
Two tubes of equal length, of radii \(1\,\text{mm}\) and \(2\,\text{mm}\), are joined end to end. What fraction of the total pressure difference appears across the narrow tube? Then two identical tubes are connected side by side between the same two pressures. How does their total flow compare with the flow through one tube?
(a) \(R_f \propto 1/r^4\), so the narrow tube has \(2^4 = 16\) times the resistance of the wide one. The same \(Q\) flows through both, so the pressure divides as \(16 : 1\). The narrow tube takes \(16/17\) of it.
(b) Each tube has the full pressure difference across it, so each carries the same \(Q\) as before. The total is \(2Q\), and the combined resistance is \(R_f/2\).
Fig 5.1 shows the example with real numbers. Of the \(170\,\text{Pa}\), only \(10\,\text{Pa}\) is used up in the wide tube and \(160\,\text{Pa}\) in the narrow one. The pressure falls gently along the wide tube and steeply along the narrow one. The flow is \(6.3 \times 10^{-7}\,\text{m}^3\text{/s}\) in both, since the water has nowhere else to go.
A tube of radius \(r\) and length \(l\) is joined end to end to a tube of radius \(r/2\) and length \(l/2\). What fraction of the total pressure difference appears across the second tube?
Three identical tubes, each of fluid resistance \(R_f\), are connected side by side between the same two pressures. Their combined fluid resistance is
When the law holds
The derivation assumed that the liquid moves in smooth cylindrical layers. That needs streamline flow, a small Reynolds number. A narrow tube and a small pressure difference keep the speed, and so \(R_e\), small. The tube must also be long. Near its entrance the liquid has not yet settled into its parabolic profile, and the law holds only beyond that.
Raise the pressure too far, and the flow becomes turbulent. The eddies carry momentum across the tube and waste energy far faster than smooth layers do. The flow rate is then less than Poiseuille's law predicts, and no longer grows in proportion to the pressure.
A tank keeps water \(10\,\text{cm}\) above a horizontal capillary of radius \(0.5\,\text{mm}\) and length \(20\,\text{cm}\). Take \(\eta = 10^{-3}\,\text{Pa s}\) and \(g = 10\,\text{m/s}^2\). Find the flow rate, the time to collect \(12\,\text{mL}\), and check that the flow is streamline.
The water level stays fixed, so the pressure difference across the capillary is the head: \(P = h\rho g = 0.10 \times 1000 \times 10 = 1000\,\text{Pa}\).
\(Q = \dfrac{\pi P r^4}{8\eta l}\), with \(r^4 = (5 \times 10^{-4})^4 = 6.25 \times 10^{-14}\,\text{m}^4\).
\(Q = \dfrac{\pi \times 1000 \times 6.25 \times 10^{-14}}{8 \times 10^{-3} \times 0.2} = 1.23 \times 10^{-7}\,\text{m}^3\text{/s} = 0.123\,\text{mL/s}\). Then \(12\,\text{mL}\) takes \(12/0.123 \approx 98\,\text{s}\).
The mean speed is \(Q/\pi r^2 = 0.16\,\text{m/s}\), so \(R_e = 1000 \times 0.16 \times 10^{-3}/10^{-3} \approx 160\). The flow is well within the streamline range, and the law is fair to use.
In an experiment to verify Poiseuille's law, why is the tube long and narrow and the pressure difference small? What happens to the flow if the pressure difference is made very large?
The law assumes steady streamline flow in parabolic layers. A narrow tube and a small pressure difference keep the speed, and so the Reynolds number, small, which keeps the flow streamline. A long tube lets the flow settle into its parabolic profile, so the short entrance region matters little. At a very large pressure difference the flow turns turbulent; eddies waste energy, and the flow rate is less than the law predicts.
Jean Poiseuille was a physician in Paris who wanted to understand how blood moves through the finest vessels of the body. Blood was too hard to study directly. So he forced water through glass capillaries, some of them finer than a hair, and measured the flow with great care. He found the flow grew as the pressure and as the fourth power of the diameter. The German engineer Gotthilf Hagen had found the same law independently, in 1839.
- Taking the flow through a tube to grow as the area, \(r^2\).The flow grows as \(r^4\): the area grows as \(r^2\) and the speed on the axis as \(r^2\) as well.
- Adding fluid resistances for tubes side by side.Side by side the pressure is shared and the flows add, like resistors in parallel. Only tubes end to end add their resistances.
- Using Poiseuille's law for fast flow in a wide pipe.It needs streamline flow. Check the Reynolds number first.