# 12.4 Viscosity and laminar flow; poiseuille’s law  (Page 3/12)

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If viscosity is zero, the fluid is frictionless and the resistance to flow is also zero. Comparing frictionless flow in a tube to viscous flow, as in [link] , we see that for a viscous fluid, speed is greatest at midstream because of drag at the boundaries. We can see the effect of viscosity in a Bunsen burner flame, even though the viscosity of natural gas is small.

The resistance $R$ to laminar flow of an incompressible fluid having viscosity $\eta$ through a horizontal tube of uniform radius $r$ and length $l$ , such as the one in [link] , is given by

This equation is called Poiseuille’s law for resistance    after the French scientist J. L. Poiseuille (1799–1869), who derived it in an attempt to understand the flow of blood, an often turbulent fluid.

Let us examine Poiseuille’s expression for $R$ to see if it makes good intuitive sense. We see that resistance is directly proportional to both fluid viscosity $\eta$ and the length $l$ of a tube. After all, both of these directly affect the amount of friction encountered—the greater either is, the greater the resistance and the smaller the flow. The radius $r$ of a tube affects the resistance, which again makes sense, because the greater the radius, the greater the flow (all other factors remaining the same). But it is surprising that $r$ is raised to the fourth power in Poiseuille’s law. This exponent means that any change in the radius of a tube has a very large effect on resistance. For example, doubling the radius of a tube decreases resistance by a factor of ${2}^{4}=\text{16}$ .

Taken together, $Q=\frac{{P}_{2}-{P}_{1}}{R}$ and $R=\frac{8\eta l}{\pi {r}^{4}}$ give the following expression for flow rate:

$Q=\frac{\left({P}_{2}-{P}_{1}\right){\mathrm{\pi r}}^{4}}{8\eta l}\text{.}$

This equation describes laminar flow through a tube. It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law    .

## Using flow rate: plaque deposits reduce blood flow

Suppose the flow rate of blood in a coronary artery has been reduced to half its normal value by plaque deposits. By what factor has the radius of the artery been reduced, assuming no turbulence occurs?

Strategy

Assuming laminar flow, Poiseuille’s law states that

$Q=\frac{\left({P}_{2}-{P}_{1}\right){\mathrm{\pi r}}^{4}}{8\eta l}\text{.}$

We need to compare the artery radius before and after the flow rate reduction.

Solution

With a constant pressure difference assumed and the same length and viscosity, along the artery we have

$\frac{{Q}_{1}}{{r}_{1}^{4}}=\frac{{Q}_{2}}{{r}_{2}^{4}}\text{.}$

So, given that ${Q}_{2}=0\text{.}\text{5}{Q}_{1}$ , we find that ${r}_{2}^{4}=0\text{.}{5r}_{1}^{4}$ .

Therefore, ${r}_{2}={\left(0\text{.}5\right)}^{0\text{.}\text{25}}{r}_{1}=0\text{.}\text{841}{r}_{1}$ , a decrease in the artery radius of 16%.

Discussion

This decrease in radius is surprisingly small for this situation. To restore the blood flow in spite of this buildup would require an increase in the pressure difference $\left({P}_{2}-{P}_{1}\right)$ of a factor of two, with subsequent strain on the heart.

Coefficients of viscosity of various fluids
Fluid Temperature (ºC) Viscosity $\eta \phantom{\rule{0.25em}{0ex}}\text{(mPa·s)}$
Gases
Air 0 0.0171
20 0.0181
40 0.0190
100 0.0218
Ammonia 20 0.00974
Carbon dioxide 20 0.0147
Helium 20 0.0196
Hydrogen 0 0.0090
Mercury 20 0.0450
Oxygen 20 0.0203
Steam 100 0.0130
Liquids
Water 0 1.792
20 1.002
37 0.6947
40 0.653
100 0.282
Whole blood The ratios of the viscosities of blood to water are nearly constant between 0°C and 37°C. 20 3.015
37 2.084
Blood plasma See note on Whole Blood. 20 1.810
37 1.257
Ethyl alcohol 20 1.20
Methanol 20 0.584
Oil (heavy machine) 20 660
Oil (motor, SAE 10) 30 200
Oil (olive) 20 138
Glycerin 20 1500
Honey 20 2000–10000
Maple Syrup 20 2000–3000
Milk 20 3.0
Oil (Corn) 20 65

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