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The behavior of fluids

Up to now in the study of forces and motion we have examined the forces on solid objects. Now we consider the behavior of fluids. A fluid is matter in the form of a liquid or a gas. Fluid dynamics includes the study of phenomena such as ocean and air currents, interstellar gases, and blood flow.

Fluids differ from solids in that they flow. The fluidity of a fluid is how easily it flows, while the opposite term, viscosity, is how much the fluid resists flowing. Thick, tarry oil has a higher viscosity than water.

Daniel, Augsburg Cathedral, early 12th century. Some textbooks talk about glass as a fluid that flows very slowly, and may mention the fact that very old stained glass windows are thicker at the bottom, as though the glass has flowed down due to gravity. However, this is now disputed. The medieval glassmaking process produced glass sheets that were thicker at the edges. The sheets were cut into panes and the installers usually placed the thicker and heavier side at the bottom of the window. Learn more. Source: Wikipedia

Fluids also differ from solids in how they respond to a force applied to their surface. A force applied along the surface of a fluid — that is, a shear force — will cause the top layer of fluid to slide over the layer below. This does not happen with a solid. In fact, we can define a fluid as something that deforms when subjected to a shear force. A cupful of sugar is not a fluid, even though it can be poured, because the sugar can be piled up in a mound and in that case the grains of sugar on the sloping sides of the mound are resisting shear forces due to gravity. A cupful of coffee, on the other hand, is a fluid; you can't pile up the coffee!

Because they flow and cannot resist being deformed, fluids take the shape of the containers they are in.

Density

Solids, liquids, and gases have density. Density is the mass per unit volume:

$LaTeX: Density\:=\:\frac{mass}{volume}$

If students have trouble remembering the formula for density, one teacher suggests: Draw a heart with a horizontal line through it. Tell them, "Density is mass divided by volume. If you can't remember that, it will break my heart." The line through the heart makes the top half look like an M and the bottom like a V.

In the metric system, the mass is measured in grams or kilograms. The units of volume are cubic meters (m³) or cubic centimeters (cm³). Density therefore has units such as kg/m³.

The symbol for density is the Greek letter ρ (pronounced Ro).

The density of water is very close to 1 g/cm³. It is often useful to remember this number.

The density of air is about `1/1000 "g/cm"^3`.
The density of rock is generally 2 to 3 `"g/cm"^3`. (The exception is pumice, which has half the density of water!)

Pumice rock is very porous and will float in water. Source: Stan Celestian, Glendale Community College.

The density of metals varies from about 2.7 g/cm³ for aluminum to 10.5 g/cm³ for silver and 19.3 g/cm³ for gold. The densest natural element is osmium, with a density of 22.6 g/cm³.

Osmium pellets. Source: Photo by Justin Urgitis, Wikipedia

Measuring density

To measure the density of an object, we measure the mass and the volume, and use these values in the equation

$LaTeX: \rho=\frac{mass}{volume}$

Measuring the mass of an object is easy — we can put it on a balance. Measuring the volume can be trickier.

In the case of a regular solid, calculate the volume from the length, width, and height.

The volume of a cube is the length of one of its sides cubed.
The volume of a rectangular solid is length x height x width.
The volume of a sphere is $4 π r^{3}$ , where r is the radius of the sphere.
The volume of a cylinder is $π r^{2} l$ , where r is the radius of the circular end and l is the length.

To find the volume of an irregularly shaped object we can use the water displacement method.

Take a graduated cylinder and partially fill it with water. Note the level of the water — this is the length of the cylinder of water in the graduated cylinder. (The surface of the water, and thus the water line, is curved in a shape known as the meniscus. Measure the water level at the lowest point of the meniscus.) Multiply this height by the circular area of the base of the cylinder to get the volume of water.

Now insert the object whose volume is to be determined and make sure it is completely submerged in the water. The water in the cylinder will rise — this phenomenon is called the displacement of water. When the object is submerged, it takes up space previously occupied by the fluid. Note the new water level on the side of the graduated cylinder, and use this measurement to calculate the volume of the (water and object) combination.

Subtract the first measurement (volume of water) from the second measurement (volume of water plus object) to get the volume of the object.

Floating and Sinking

Most people will guess that an object made of wood will float on water, and one made of metal will sink. How, then, does one account for ships made of steel? Clearly, it is not just a matter of heavy objects sinking while light objects float. Another demonstration of the complexity of the problem is that a ball of clay will sink, while the same clay made into the shape of a canoe will float.

Students can be led to discover that neither mass nor volume alone can be used to predict whether an object will sink or float. What matters is the average density of the object — the total mass divided by the total volume, including any volume of air — compared to the density of the fluid:

If the density of the object is greater than the density of the fluid it is in, the object will sink.
If the density of the object is less than the density of the fluid, the object will float. It will be partially submerged.
If the density of the object is the same as the density of the fluid, the object will float just under the surface of the fluid.

The fluid in the comparison of the two densities need not be water. A simple experiment will show that a wax candle will float in water but sink in rubbing alcohol, which has a density of 0.786 g/cm³ — less than water. A raw egg (density 1.06 to 1.09 g/cm³) will sink in tap water but float in salt water, which has a greater density than fresh water.

To float an egg, dissolve at least 4 T Kosher or pickling salt in 2 C water. (Table salt will work, but the water will be cloudy.) The density of a saturated salt solution is about 1.2 g/cm³.

The Buoyant Force

The buoyant force is what you feel when you push down on a floating object. Take a floating block of wood. Try pushing straight down on it. Or try to submerge an inflated pool toy. It's hard!

Notice, too, that the harder you push down on a partially submerged object — the more fluid you cause it to displace — the more the buoyant force resists. Once an object is completely submerged, the buoyant force remains the same no matter what depth the object is at.

The buoyant force acting on an object that is partially or fully submerged is equal to the weight of the fluid that is displaced.

Another way to say this is,

An object that is floating, at rest, has a buoyant force that exactly balances its weight.

Another statement that follows from setting the buoyant force equal to the weight of the object is that the mass of the displaced water is the mass of the entire object. This applies to objects that are floating partially submerged.

Let's look at the force in more detail. Consider a block of wood that is partially submerged in water. Say the block has an area of 10 cm by 4 cm, and a height of 2 cm. The block is partially submerged so that 0.5 cm of its height is underwater.

Step 1: What is the volume of the block that is underwater? It is just the area times a height of 0.5 cm, i.e.

$LaTeX: submerged\:volume=\left(10cm\:\times4cm\times0.5cm\right)=20cm^3$

Step 2: The volume of water displaced is equal to the volume of the block that is underwater. How much mass of water is there in this displaced volume?

We know the volume of the water, and we know the density of water (1 g/cm³) so we can calculate the mass of water by turning around the equation for density:

$LaTeX: \rho\:=\:\frac{mass}{volume}$

$LaTeX: mass\:=\rho\times volume$

In other words, the mass of water displaced is the density of water (1 g/cm³) times the volume that has been displaced.

$LaTeX: Mass\:of\:water\:displaced\:=\:\left(1\frac{g}{cm^3}\right)\times20g$

Step 3: What is the weight of the water that has been displaced?

The weight of the water displaced is the mass times the acceleration due to gravity. (To make sure the units are consistent with the units of force (kg m s^-2) we convert the mass from grams to kilograms: 20 g = 0.02 kg)

$Weight of water displaced = (0.02 k g) (9.8 m s^{- 2}) = 0.196 N$

Therefore the buoyant force is 0.196 N, directed upward.

Now imagine that we push on the block so that half of it is submerged (10 cm x 4 cm, up to a height of 1 cm). More of the block is now submerged, compared to how it was for the calculations we just did.

Repeating the steps, we find that the volume of block submerged is 40 cm³. The mass of water displaced is 40 g or 0.04 kg. The buoyant force upward has now increased to (0.04 kg) (9.8 ms^-2) = 0.392 N. As we noted above, the more you push down on a partially submerged object, the more buoyant force there is pushing back.

Now if we completely submerge the block, the volume of block that is underwater is (10 cm x 4 cm x 2 cm) = 80 cm³. That volume of water has a mass of (1 g/cm³) (80 cm³) = 80 g = 0.08 kg. The weight of water displaced, and hence the buoyant force,
is (0.08 kg) (9.8 ms^-2) = 0.784 N.

This is the limit of the buoyant force, because further submerging the block does not increase the volume of block that is underwater or the volume of displaced water.

Archimedes’ principle

From the discussion in the last section, we can see that if two objects of equal mass are submerged, the one with greater buoyancy is the one with greater volume. A 1 kg block of gold (very dense) has a smaller volume than a 1 kg block of aluminum. Therefore, if both are fully submerged, the aluminum block will displace more water.

Another way to look at it is: a dense object displaces less water than a more rarified object of the same mass.

This is the basis of Archimedes’ discovery. According to the story, Archimedes had some gold that was known to be pure. He noted how much water it displaced. He also had some silver, and noted how much more water it displaced. (Silver is less dense than gold, so the same mass occupies and displaces more volume of water.) By measuring the water displaced by the king’s crown, he was able to show that some fraction of it was silver, reducing the density of the crown.

Learn more about Archimedes here, here or here .

The depth to which an object is submerged

Consider an object in a container of water.

The density of the object is some fraction of the density of water. Say the fraction is Z. In other words, if Z is $\frac{1}{4}$ , then the density of the object is $\frac{1}{4}$ that of water or

$LaTeX: density\:of\:object\:=\:Z\:\times\left(density\:of\:water\right)$

Remember, the density of water is 1 g/cm³!

The problem is: what fraction of the volume of the object is underwater? Call the underwater fraction X.

The floating object is at rest so the buoyant force is balanced by the weight of the object. We can use the equality of these two forces to solve for X, the fraction of the object that is underwater. Let’s calculate the weight of the object first:

$LaTeX: mass\:of\:the\:object\:=\:\left(density\:of\:the\:object\right)\:\times\left(volume\:of\:the\:object\right)=\:\left(Z\:\times density\:of\:water\right)\times\left(volume\:of\:the\:object\right)$

The weight of the object, and hence the force required to support it, is

$LaTeX: weight\:of\:the\:object\:=\:\left(mass\:of\:the\:object\right)\times g$

where g is the acceleration due to gravity.

So the weight of the object is

$LaTeX: weight\:=\:\left(Z\times1\frac{g}{cm^3}\right)\times\left(volume\:of\:object\right)\times g$

Now let’s calculate the buoyant force:

Step 1: The volume of the part of the object that is underwater is X times the volume of the object.

Step 2: What is the mass of water in this volume?

$LaTeX: mass\:of\:water\:=\:\left(density\:of\:water\right)\:\times volume$

$LaTeX: weight\:=\:mass\:\times\left(acceleration\:due\:to\:gravity\right)=\:\left(density\:of\:water\right)\:\times\left(volume\:of\:the\:object\right)\times accerlation$

Therefore the buoyant force:

$LaTeX: F_{buoyant}\:=\:\left(density\:of\:water\right)\times g\times\left(volume\:of\:object\right)$

Set the weight equal to the buoyant force: $W = Z â € ¢ (density of water) â € ¢ (volume of object) â € ¢ g = F_{buoyant} = (density of water) â € ¢ g â € ¢ X â € ¢ (volume of the object)$ $LaTeX: W=Z\times\left(density\:of\:water\right)\times\left(volume\:of\:object\right)\times g=F_{buoyant}=\left(density\:of\:water\right)\times g\times\left(volume\:of\:object\right)$

Notice we have the density of water, the volume of the object, and g on both sides of the equation! They cancel out.

That leaves $Z = X!$

In other words, the fraction of the object that is underwater is the same as the ratio of the object's density to the density of water:

If the object has a density $\frac{1}{2}$

that of water, then $\frac{1}{2}$

the volume of the object will be underwater.
If the object has a density of $\frac{1}{27}$

that of water, then

\frac{1}{27}

of the volume of the object will be underwater.
If the object is floating in another liquid, like corn syrup, then the same calculation applies. If you are told the object has a density $\frac{1}{3}$

that of corn syrup, then

\frac{1}{3}

of the object will be under the corn syrup level.

The case of the clay boat

As noted earlier, a ball of clay will sink, but the same ball of clay made into a boat shape will float. Let’s examine that situation in more detail.

The calculation done here may go beyond what would be required for the CSET Science I test. However, it is interesting to illustrate how one can turn a “sinker” into a “floater.” Some lesson plans have students turning sinkers into floaters by attaching the sinking object to a non-sinking one — for example, attaching a marble to a ping-pong ball. Here we more simply decrease the density of a sphere by hollowing it out.

Say the clay has a density of 1.1 times the density of water (i.e., 110%). The fraction of the clay ball that will be underwater is more than the entire ball, which means the clay ball will sink.

To get the clay object to float so that ½ of it is below water, we must reduce the density of the clay object. We can do this by giving it a shape that includes air, so that the overall density is reduced. We will make the clay into a shape that has a large volume, but some of this volume will be occupied by air (which has so little density we will ignore it here).

We want the density of the new clay object to be 0.5 times the density of water, or 0.5 g/cm³. What should the new volume be?

$Density = \frac{total mass}{total volume}$

$= \frac{mass of clay + mass of air}{volume of object}$

$= \frac{mass of clay}{volume of object}$

Set this density equal to $LaTeX: 0.5\times\left(1\frac{g}{cm^3}\right)$ :

$LaTeX: 0.5\frac{g}{cm^3}=\frac{mass\:of\:clay}{volume\:of\:object}$

Solve for the volume of the object:

$LaTeX: Volume\:of\:object\:=\frac{\:mass\:of\:clay}{0.5\frac{g}{cm^3}}$

or $Volume of object = 2 \frac{mass of clay}{density of water}$

Say the mass of the clay is 37 g. As a sphere, such a mass would have a radius of about 2 cm and a volume of about 33 cm³. (A sphere has volume $(\frac{4}{3}) π {radius}^{3}$ , and the mass is the density times this volume.)

Our calculation shows that the new volume of clay must be

$2 \frac{mass of clay}{density of water}$

$= 2 \frac{37 g}{1 {g/cm}^{3}} = 74 c m^{3}$

That is more than twice the volume of the original ball of clay. More precisely, the radius of the new volume of clay is given by

$LaTeX: radius=\sqrt[3]{\left(new\:volume\right)\times\frac{3}{4\pi}}=2.6cm$

To recapitulate, if we have about 37 g of clay in the form of a solid sphere of radius about 2 cm, it will sink in water. If we form the clay into a hollow sphere of radius about 2.6 cm, the new sphere will float in water, about halfway submerged.

Pressure in a fluid at rest

What is pressure?

The pressure of a fluid in a container is the force per unit area that the fluid exerts perpendicular or normal to the surface. (The surface could be the free surface of the fluid, or the surface of the container.)

$LaTeX: Pressure\:=\frac{\:Force}{Area}$

The force can only be exerted normal to the surface; a fluid cannot exert a force parallel to the surface it is in contact with.

Although the force exerted by the fluid on the container has a direction (the force is a vector), the force per unit area is defined as a scalar quantity. The pressure felt by a "test particle" that we might place in a fluid comes from all directions and is the same in all directions.

Example:

A cylinder, open at the top, contains oil. A piston fitted into the top of the cylinder has a mass of 10 kg. The area of the cylinder or piston is 30 cm². By how much does the pressure of the oil inside the cylinder increase with the piston resting on it?

The force applied is the weight of the piston:

$LaTeX: F=\left(mass\:of\:piston\right)\times g=\left(10kg\right)\times\left(9.8ms^{-2}\right)=980\:N$
The area of the piston is $30 c m^{2} = 3 x 10^{- 3} m^{2}$ (converting to SI units)
The increase of pressure, or force per unit area, is
$p = 980 \frac{N}{3 x 10^{- 3} m^{2}} = 3.26 x 10^{5} \frac{N}{m^{2}}$

The unit of $\frac{N}{m^{2}}$ is called the Pascal, abbreviation Pa. For convenience, some multiples of the Pascal have their own name. One hundred thousand Pascals ( $10^{5} Pa$ ) equals one bar. The pressure of 1 atmosphere (1 atm) is numerically close to the bar: $1 atm = 1.01 x 10^{5} P a$ .

Pressure can also be measured in pounds per square inch, or PSI.

(Pressure can also be measured in terms of the height of a column of liquid that pressure can support).

Atmospheric pressure

Atmospheric pressure is the pressure exerted by the Earth’s atmosphere on the surface of the Earth. The atmosphere is relatively dense down near the surface of the Earth, and becomes less dense with altitude. At about 1,000 km above the Earth’s surface the atmosphere tapers off into empty space.

When the surface of the Earth is at sea level, the atmosphere lying above this level exerts a standard pressure of 1.013 x 10⁵ Pa or 101.3 kPa. The unit called 1 atmosphere, or 1 atm, is equal to this value, 1.013 x 10⁵ Pa.

In the mountains one is above the densest part of the atmosphere, and there is less of the atmosphere above us than there would be at sea level.

The summit of Mount San Jacinto lies at 10,834 feet above sea level. At this altitude, we expect the atmospheric pressure to be about 0.69 x 10⁵ Pa, or 0.68 atm.

Weather causes deviations from the standard levels of atmospheric pressure. At a given temperature, humid air is less dense than dry air because water vapor occupies the space that would have been occupied by nitrogen and oxygen, and water vapor molecules are less massive than these. Low pressure areas are associated with higher relative humidity.

For an informative display of information about the atmosphere, see Earthguide. You can click to show the pressure in units of millibars (mbar).

View from Mount San Jacinto. Source: California State Parks

Blaise Pascal

The Pascal unit of pressure is named after the French mathematician Blaise Pascal (1623-1662). Pascal lost his mother at a young age and was home-schooled by his father, a local judge who was interested in mathematics. Blaise himself was a mathematical prodigy. At the age of 12, he rediscovered for himself some of Euclid’s propositions in geometry. At 17, he published a paper on projective geometry which the great mathematician Leibnitz called “brilliant.” At 19 he invented a sort of mechanical calculator to help his father figure the taxes he was then charged with collecting — although the invention was not commercially successful. In his 20s he was in full swing as a scientist, challenging Aristotle’s rejection of the existence of a vacuum, demonstrating that air has weight and inventing the hydraulic press.

As well as these and other contributions to mathematics and physical sciences, Pascal devoted a large part of his life to religious exercises. He was in poor health and pain much of his life. He died at 39.

Blaise Pascal (Dibner Library of the History of Science and Technology) Source: Smithsonian Institutions.

Pascal's law

Pascal's law tells us what happens when you apply a force per unit area to a fluid that is confined in some kind of container: the force per unit area will be transmitted everywhere in the container. The force per unit area — or pressure — must be isotropic (having equal magnitude in all directions), for if it were not, the fluid would move until the pressure was equalized.

Pascal's law: A change in the pressure of an enclosed incompressible fluid is conveyed undiminished to every part of the fluid and to the surfaces of its container.

This means that a force can be transmitted through fluid-filled pipes or tubes in some kind of connected system.

Note that Pascal's law applied to an incompressible fluid. An incompressible fluid is one whose density does not change. (If you could compress the same amount of fluid into a smaller volume, it would have a higher density.) Liquids such as oil and water can usually be taken to be incompressible, while gases can not.

The hydraulic press

An important application of Pascal’s law is the hydraulic press. (“Hydraulics” is the use of fluids to do work.)
The hydraulic press is the equivalent of a lever, for a small force on one piston can produce a large force on another.

A car lift is an application of the hydraulic press.

The hydraulic press consists of two connected cylinders or tubes, one narrow (with a small cross-sectional area) and one wide (with a large cross-sectional area). An incompressible fluid such as oil fills the vessel. Pistons cap the tops of the cylinders, as shown in the diagram.

The force multiplication of a hydraulic press can seem impossible: A given force, say 100 N, exerted on the small piston of area A1, can translate to a force of several hundred N on piston of area A2. Let's work out the details.

A force of 100 N is exerted on the piston of area A1. The pressure, or force per unit area, is

$p = \frac{100 N}{A 1}$

According to Pascal's law, this pressure is felt throughout the system. At the piston A2, we have the same force per unit area pushing up on the piston.

What is the force on the second piston?

$LaTeX: Force\:on\:A2=\left(force\:per\:unit\:area\:prevailing\:in\:the\:system\right)\times\left(area\:A2\right)$

$LaTeX: Force\:on\:A2=\left(\frac{F1}{A1}\right)\times A2=\left(100N\right)\left(\frac{A2}{A1}\right)$

If A2 is, say, 4 times bigger than A1, then the force on A2 is 4 times the force on A1. The force is multiplied according to the ratio of the areas. In this case, since the force on the first piston was 100 N, the force on the second piston must be 400 N.

What's the catch?

Just like in the lever problems, we find that the work done by each force is the same. A small force acting over a larger distance balances a large force acting over a smaller distance.

Consider the situation if the small piston is pushed down a distance of 25 cm (or 0.25 m). The work done is the force x displacement.

$LaTeX: Work=100N\times0.25m$

How far does the piston of area A2 move up?

The volume of oil pushed out of the way by the first piston (with area A1) must be equal to the volume of oil pushed up under the second piston (with area A2). (The oil can't go anywhere else, and remember, the fluid is incompressible, so the oil doesn't just get squished up into a smaller volume.)

The volume of oil displaced by the first piston is just the area times the distance of displacement: $V = (A 1) â € ¢ 0.25 m$

At the piston with area A2, the same volume fits into a space of area A2 and undetermined height. $A 1 â € ¢ 0.25 m = A 2 â € ¢ (distance moved up by second piston)$

$LaTeX: A1\times0.25m=A2\times\left(distance\:moved\:up\:by\:second\:piston\right)$

Therefore the distance moved by the second piston = $LaTeX: \left(\frac{A1}{A2}\right)\times0.25m$

We stated earlier that area A2 is four times bigger than area A1. Then the distance moved by the second piston is $LaTeX: \frac{1}{4}\times0.25m$ . In other words, when we push down on the first piston with a force of 100 N, and the piston goes down 0.25 m, the second piston experiences a force of 400 N, but it only moves $LaTeX: \frac{1}{4}$ the distance of the first piston.

This NASA site has more details and some practice questions with answers.

Static fluid pressure

Consider a column of fluid of area A and height h. In such a static fluid, the pressure at any level arises from the weight of the fluid above that level, i.e., from the influence of gravity.

The pressure at the base of this column is given by the force per unit area, as we saw in an earlier section, and we are considering the force due to the weight of the fluid.

The weight of the fluid is the mass times g, the acceleration due to gravity:

$LaTeX: W=m\times g$

The mass of the fluid is the density of the fluid times its volume: $LaTeX: m=\rho\times V$

So the weight $LaTeX: W=\rho\times V\times g$ $W = ρ â € ¢ V â € ¢ g$

Putting these together, we have:

$pressure = \frac{force}{area}$

$= \frac{weight of fluid}{area}$

= $LaTeX: \rho\times V\times\left(\frac{g}{A}\right)$

But note that the volume V is the area A times the height h. Substituting, we get $LaTeX: pressure\:=\:\rho\times g\times h$

where the height h is the height of fluid above the level at which we are calculating the pressure.

The surprising thing about static fluid pressure

Static fluid pressure at a given level in the container of fluid doesn't depend on the total mass or total volume of the fluid. It does not depend on the shape or surface area of the fluid. This is counterintuitive, but true.

It means that the pressure at a given level in each of the three differently shaped containers shown is the same.

The pressure at a given level in a container of fluid depends only on the height of the fluid above it. One consequence of this fact is that one can express a pressure value in terms of the corresponding pressure due to a column of liquid of a particular height.

The pressure at a given level in a fluid depends on the height of the fluid above it.

The mercury barometer

The mercury barometer is a device for measuring air pressure. It works by measuring the length of a column of mercury which is supported by the weight of the atmosphere.

The balance between the force per unit are of mercury in the glass tube and the atmospheric pressure is indicated by the level of the mercury.

At sea level on a day with “standard” pressure, the weight of the atmosphere exerts enough force on the surface of the mercury in the reservoir to push it up into the tube to a height of 29.92 inches or 76 cm. If atmospheric pressure decreases, the atmosphere is not pushing down as much on the mercury in the reservoir so the height in the tube falls. If the atmospheric pressure increases, the mercury can rise further in the glass tube because there is a vacuum at the top. (If there were air, the mercury, in rising, would be compressing the air, which would exert a pressure against the mercury.)

Schematic of a mercury barometer. Source: About.com

Blood pressure

The heart pumps blood around the body. The heart’s main purpose is to generate enough pressure to pump blood through the arteries and to the lungs and body.

Blood from the entire body (except the lungs) enters the heart in the right upper chamber. From the right upper chamber (the right auricle) the blood goes to the right lower chamber (the ventricle), to the lungs, to the left upper chamber and left lower chamber, and finally to the aorta, the largest blood vessel leading away from the heart. The aorta carries the blood to the arteries, so that it can reach the entire body (except the lungs).

The upper chambers receive blood (from the body, in the case of the right upper chamber, and from the lungs, in the case of the left upper chamber). The job of the ventricles — the lower chambers — is to force blood out of the heart.

Diagram of the human heart. Source: NIAAA.

The role of valves

Valves in the heart prevent backflow.

On the right side, the tricuspid valve prevents blood from flowing back into the atrium or upper chamber once it has started filling the ventricle.
On the left side, in a similar position, is the mitral or bicuspid valve.
The pulmonary semilunar valve prevents backflow from the lungs to the right ventricle.
The aortic valve prevents blood from seeping back from the aorta to the left ventricle when the left ventricle relaxes.

The cardiac cycle

The cardiac cycle is the sequence of events from the start of one set of ventricular contractions to the start of the next.

The upper chambers fill with blood and experience an increased pressure. Some of the blood starts to fill the ventricles, and then a contraction of the upper chambers only squeezes the remaining blood from the atria down into the ventricles.

As the ventricles fill, the pressure increases there. The tricuspid and mitral valves close, the atria relax, and the pressure is high in the ventricles. When the pressure in the left ventricle is greater than that in the aorta, the aortic valve opens and blood can go into circulation.

Then the ventricles contract — in ventricular systole — and blood goes to the lungs and to the whole circulatory system.

Then the cycle starts again. As the ventricles relax after systole, the blood starts flowing down into the ventricles from the atria. The aortic valve is closed. The cycle repeats some 70 to 80 times per minute in average resting adults.

The meaning of blood pressure

The cardiac cycle can be divided into two parts: systole and diastole:

During systole the ventricles are contracting. The systolic pressure is the pressure on the walls of the arteries when the ventricles contract. It is the peak pressure on the arteries. When the blood pressure is measured by a sphygmomanometer or blood pressure meter, the systolic pressure is the number “on top” (e.g., if the result is 120/68, the systolic pressure is 120).
During diastole, the ventricles are relaxed. Initially the pressure in the ventricles is very low, and they start filling with blood. The diastolic pressure is the blood pressure during the period when the ventricles are relaxed. It is the lower of the two numbers quoted in a blood pressure reading.

The danger associated with narrowed arteries

The blood pressure is fairly constant throughout the circulatory system. Because the pressure is constant, the rate of blood flow adjusts according to the cross-sectional area of the arteries. A narrowing or constriction in an artery causes the blood to flow faster than in wider sections.

This phenomenon of faster flow in constricted vessels arises from the equation of continuity, which in turn follows from the principle of the conservation of mass.

Consider a fluid flowing through a narrowing tube, as shown. The cross-sectional area is A1 for the area on the left, and A2 for the area on the right. The velocity of the fluid is v1 as it crosses A1 and v2 as it crosses A2.

The equation of continuity states that

$A_{1} v_{1} = A_{2} v_{2}$

Now find $v_{2}$ as a function of $A_{1}$ , $A_{2}$ and $v_{1}$ :

$v_{2} = (\frac{A_{1}}{A_{2}}) v_{1}$

If $A_{2}$ is half the size of $A_{1}$ , the velocity $v_{2}$ will be $2 v_{2}$ .

Figure 5 at this site has an animation of blood flow through a constricting vessel (as well as the following example).

The danger of this increased rate of blood flow is that the higher velocity may break off bits of plaque tissue lining the blood vessels. A broken-off bit of plaque can completely block up a narrowed blood vessel.

Say a normal artery has radius $r_{1} = 500 micrometers$ , and blood flows at a rate of 20 cm/s. If an artery with plaque build-up has a radius of $r_{2} = 250 micrometers$ , the velocity of blood through the constricted area is

$v_{2} = (\frac{π {(r_{1})}^{2}}{π {(r_{2})}^{2}}) v_{1}$

Plugging in the numbers, we get a flow through the constricted area of 80 cm/s. This is so much larger than the original flow rate of 20 cm/s because the velocity is proportional to the square of the radius ratio.