Laminar Flow Confined to Tubes—Poiseuille’s Law
Laminar Flow Confined to Tubes—Poiseuille’s Law
What causes flow? The answer, not surprisingly, is pressure difference. In fact, there is a very simple relationship between horizontal flow and pressure. Flow rate \(Q\) is in the direction from high to low pressure. The greater the pressure differential between two points, the greater the flow rate. This relationship can be stated as
\(Q=\cfrac{{P}_{2}-{P}_{1}}{R}\text{,}\)
where \({P}_{1}\) and \({P}_{2}\) are the pressures at two points, such as at either end of a tube, and \(R\) is the resistance to flow. The resistance \(R\) includes everything, except pressure, that affects flow rate. For example, \(R\) is greater for a long tube than for a short one. The greater the viscosity of a fluid, the greater the value of \(R\). Turbulence greatly increases \(R\), whereas increasing the diameter of a tube decreases \(R\).
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 the figure below, 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 the figure below, is given by
\(R=\cfrac{8\eta l}{\pi {r}^{4}}\text{.}\)
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.
(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 \(R\) 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, \(Q=\cfrac{{P}_{2}-{P}_{1}}{R}\) and \(R=\cfrac{8\eta l}{\pi {r}^{4}}\) give the following expression for flow rate:
\(Q=\cfrac{({P}_{2}-{P}_{1}){\mathrm{\pi r}}^{4}}{8\eta l}\text{.}\)
This equation describes laminar flow through a tube. It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.
Example: 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
\(Q=\cfrac{({P}_{2}-{P}_{1}){\mathrm{\pi r}}^{4}}{8\eta l}\text{.}\)
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
\(\cfrac{{Q}_{1}}{{r}_{1}^{4}}=\cfrac{{Q}_{2}}{{r}_{2}^{4}}\text{.}\)
So, given that \({Q}_{2}=0\text{.}\text{5}{Q}_{1}\), we find that \({r}_{2}^{4}=0\text{.}{5r}_{1}^{4}\).
Therefore, \({r}_{2}={(0\text{.}5)}^{0\text{.}\text{25}}{r}_{1}=0\text{.}\text{841}{r}_{1}\), 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 \(({P}_{2}-{P}_{1})\) 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 | 20 | 3.015 |
| 37 | 2.084 | |
| Blood plasma | 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 |
The circulatory system provides many examples of Poiseuille’s law in action—with blood flow regulated by changes in vessel size and blood pressure. Blood vessels are not rigid but elastic. Adjustments to blood flow are primarily made by varying the size of the vessels, since the resistance is so sensitive to the radius. During vigorous exercise, blood vessels are selectively dilated to important muscles and organs and blood pressure increases. This creates both greater overall blood flow and increased flow to specific areas. Conversely, decreases in vessel radii, perhaps from plaques in the arteries, can greatly reduce blood flow.
If a vessel’s radius is reduced by only 5% (to 0.95 of its original value), the flow rate is reduced to about \((0\text{.}\text{95}{)}^{4}=0\text{.}\text{81}\) of its original value. A 19% decrease in flow is caused by a 5% decrease in radius. The body may compensate by increasing blood pressure by 19%, but this presents hazards to the heart and any vessel that has weakened walls. Another example comes from automobile engine oil. If you have a car with an oil pressure gauge, you may notice that oil pressure is high when the engine is cold. Motor oil has greater viscosity when cold than when warm, and so pressure must be greater to pump the same amount of cold oil.
Poiseuille’s law applies to laminar flow of an incompressible fluid of viscosity \(\eta \) through a tube of length \(l\) and radius \(r\). The direction of flow is from greater to lower pressure. Flow rate \(Q\) is directly proportional to the pressure difference \({P}_{2}-{P}_{1}\), and inversely proportional to the length \(l\) of the tube and viscosity \(\eta \) of the fluid. Flow rate increases with \({r}^{4}\), the fourth power of the radius.
Example: What Pressure Produces This Flow Rate?
An intravenous (IV) system is supplying saline solution to a patient at the rate of \(0\text{.}\text{120}\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}\text{/s}\) through a needle of radius 0.150 mm and length 2.50 cm. What pressure is needed at the entrance of the needle to cause this flow, assuming the viscosity of the saline solution to be the same as that of water? The gauge pressure of the blood in the patient’s vein is 8.00 mm Hg. (Assume that the temperature is\(\text{20ºC}\).)
Strategy
Assuming laminar flow, Poiseuille’s law applies. This is given by
\(Q=\cfrac{({P}_{2}-{P}_{1})\pi {r}^{4}}{8\eta l}\text{,}\)
where \({P}_{2}\) is the pressure at the entrance of the needle and \({P}_{1}\) is the pressure in the vein. The only unknown is \({P}_{2}\).
Solution
Solving for \({P}_{2}\) yields
\({P}_{2}=\cfrac{8\eta l}{{\mathrm{\pi r}}^{4}}Q+{P}_{1\text{.}}\)
\({P}_{1}\) is given as 8.00 mm Hg, which converts to \(1\text{.}\text{066}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\). Substituting this and the other known values yields
\(\begin{array}{lll}{P}_{2}& =& [\cfrac{8(1\text{.}\text{00}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{s/m}}^{2})(2\text{.}\text{50}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m})}{\pi (0\text{.}\text{150}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{4}}](1\text{.}\text{20}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s})+1\text{.}\text{066}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\\ & =& 1\text{.}\text{62}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\text{.}\end{array}\)
Discussion
This pressure could be supplied by an IV bottle with the surface of the saline solution 1.61 m above the entrance to the needle (this is left for you to solve in this tutorial’s Problems and Exercises), assuming that there is negligible pressure drop in the tubing leading to the needle.
This lesson is part of:
Fluid Dynamics and Applications