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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 size 12{R} {} to laminar flow of an incompressible fluid having viscosity η size 12{η} {} through a horizontal tube of uniform radius r size 12{r} {} and length l size 12{l} {} , such as the one in [link] , is given by

R = 8 η l π r 4 . size 12{R= { {8η l} over {π r rSup { size 8{4} } } } "."} {}

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.

Part a of the diagram shows a fluid flow across a rectangular non viscous medium. The speed of the fluid is shown to be same across the tube represented as same length of vertical rising arrows. Part b of the diagram shows a fluid flow across a rectangular viscous medium. The speed of the fluid speed at the walls is zero, increasing steadily to its maximum at the center of the tube represented as wave like variation for length of vertical rising arrows. Part c of the figure shows a burning Bunsen burner.
(a) If fluid flow in a tube has negligible resistance, the speed is the same all across the tube. (b) When a viscous fluid flows through a tube, its speed at the walls is zero, increasing steadily to its maximum at the center of the tube. (c) The shape of the Bunsen burner flame is due to the velocity profile across the tube. (credit: Jason Woodhead)

Let us examine Poiseuille’s expression for R size 12{R} {} to see if it makes good intuitive sense. We see that resistance is directly proportional to both fluid viscosity η size 12{η} {} and the length l size 12{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 size 12{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 size 12{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 = 16 size 12{2 rSup { size 8{4} } ="16"} {} .

Taken together, Q = P 2 P 1 R size 12{Q= { {P rSub { size 8{2} } - P rSub { size 8{1} } } over {R} } } {} and R = 8 η l π r 4 size 12{R= { {8ηl} over {π`r rSup { size 8{4} } } } } {} give the following expression for flow rate:

Q = ( P 2 P 1 ) πr 4 8 η l . size 12{Q= { { \( P rSub { size 8{2} } - P rSub { size 8{1} } \) πr rSup { size 8{4} } } over {8ηl} } } {}

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 = ( P 2 P 1 ) πr 4 8 η l . size 12{Q= { { \( P rSub { size 8{2} } - P rSub { size 8{1} } \) πr rSup { size 8{4} } } over {8ηl} } } {}

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

Q 1 r 1 4 = Q 2 r 2 4 . size 12{ { {Q rSub { size 8{1} } } over {r rSub { size 8{1} } rSup { size 8{4} } } } = { {Q rSub { size 8{2} } } over {r rSub { size 8{2} } rSup { size 8{4} } } } } {}

So, given that Q 2 = 0 . 5 Q 1 size 12{Q rSub { size 8{2} } =0 "." 5Q rSub { size 8{1} } } {} , we find that r 2 4 = 0 . 5 r 1 4 size 12{r rSub { size 8{2} } rSup { size 8{4} } =0 "." 5r rSub { size 8{1} } rSup { size 8{4} } } {} .

Therefore, r 2 = 0 . 5 0 . 25 r 1 = 0 . 841 r 1 size 12{r rSub { size 8{2} } = left (0 "." 5 right ) rSup { size 8{0 "." "25"} } r rSub { size 8{1} } =0 "." "841"r rSub { size 8{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 P 2 P 1 size 12{ left (P rSub { size 8{2} } - P rSub { size 8{1} } right )} {} of a factor of two, with subsequent strain on the heart.

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Coefficients of viscosity of various fluids
Fluid Temperature (ºC) Viscosity η (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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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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