A. The Equation of Continuity – Conservation
of Mass

B. Bernoulli’s Equation – Conservation of Energy

C. The Magnus Effect

D. Fluid Resistance

Problems

Back to content

In this final chapter we will discuss the properties of a moving fluid and what happens to objects that travel in a fluid. Two types of fluid flow will be considered: laminar flow and turbulent flow. In laminar flow the particles in the fluid follow streamlines, and the motion of particles in the fluid is predictable. If the flow rate is very large, or if objects obstruct the flow, the fluid starts to swirl in an erratic motion. No longer can one predict the exact path a particle on the fluid will follow. This region of constantly changing flow lines is said to consist of turbulent flow.

We begin our discussion of fluid dynamics by examining the simplest case of fluid flow: laminar flow with constant velocity in a uniform pipe. In the diagram below a fluid of density d flows through a uniform pipe of Area A with velocity v.

One property of interest is the amount of mass passing through area A over unit time. Since this is simply the rate at which mass passes through the pipe, we can immediately write

_{ }

To find what determines this flow rate examine the following diagram. Let us assume that in time t all mass contained in the shaded volume V will pass through area A. Let the length of this cylindrical section of fluid be L.

Now the flow rate is simply m/t. We can use the expression for density to write m = dV. This gives the relation

_{ }

Now we know that the volume of water in this cylinder is simply the product of the length of the cylinder and its base area, V = AL. So we can write

_{ }

But the quantity L/t is simply the rate at which distance is covered by the fluid, that is, the fluid’s velocity.

_{ }

So we have an expression for the rate at which mass flows in terms of the velocity of fluid flow, density of the fluid and area of the pipe in which the fluid is flowing. This result is very reasonable.

_{ }
(9.1)

We now complicate our analysis of fluid flow by examining what
happens to the fluid if the size of the tubing through which it flows changes.
We will allow the change to be gradual and continuous so that laminas flow is
maintained. Consider the following diagram which shows the pipe slowly constricting
from area A_{1} to area A_{2}. From practical experience we
know that the velocity of fluid through the small area is larger than the velocity
of the fluid through the large area.

Many of us have heard the expression “still water runs deep.”
This phenomenon can be explained and quantified by examining the flow rate of
mass through the tubing. Because no fluid can leave through the walls and there
are no “sources” or “sinks” wherein the fluid can be created or destroyed, the
mass crossing each section of the tube per unit time must be the same. This
is simply the principle of **conservation of mass**. This principle is embodied
in **the equation of continuity**.

_{ }

or

_{ }
(9.2)

This equation expresses the law of conservation of mass in fluid dynamics.

If fluid is incompressible, as will be the case with all examples
considered here, then the density is constant (d_{1} = d_{2}),
and Eq. 9.2 takes on simpler form

_{ }

or

_{ }

__Example__:
Water flows through a 1 inch diameter hose with a speed of 2 ft/sec. Find the
speed of water through the nozzle of the diameter is reduced to 1/8 inch.

We use the principle of conservation of mass to solve this problem.

Av = constant

Reducing the diameter of the hose will reduce the area. Consequently the velocity must be increases by the same factor that the area is decreased. We must find by what amount the area is decreased. For a circle

Area
= π r^{2}

where r is the radius. In this problem the diameter is reduced by a factor of eight. Subsequently, the radius is also reduced by a factor of eight. But the area is reduced by a factor of 64. This results in an increase in velocity by a factor of 64.

_{
}

__Note__: Streamlines
are very convenient for representing fluid flow. Notice in particular, that
closely spaced streamlines indicate regions where the velocity of the fluid
is great.

Back to the top

Let us continue to observe what happens to a fluid as it flows through a pipe of varying area. We have already determined that if the flow is laminar and the fluid is incompressible then the product Av is constant. Now use Newton’s second law of motion and consider the pressure acting on a flowing fluid. Let us begin by considering the following question.

__Question__: In which region, A B, or C, in the figure
below would you expect the pressure on the walls of the pipe to be the greatest?
(Region A > region C > region B)

The law of conservation of mass tells us that the velocity is greatest in region B and smallest in region A. In going from a wide area to a narrow area the velocity increases. This represents an acceleration of the fluid. Acceleration requires an unbalanced force, which in this case is supplied by the pressure on the fluid by the walls. Thus the pressure in region A must be greater than the pressure in region B to accelerate the fluid. Conversely, the pressure must be larger in region C to accomplish this deceleration.

In the previous discussion we found a qualitative relationship between pressure and velocity in a fluid flow. We now seek to quantify this relation. In addition, we will generalize the problem by allowing the elevation of the fluid to change. Consider the flow of fluid on the pipe shown below.

Let us examine the energy of the fluid as it flows. Because
the fluid is moving it has kinetic energy (KE = ½ mv^{2}). In addition,
because of its position in the earth’s gravitational filed it possesses potential
energy (PE = mgh). Finally there is a force on the fluid due to the pressure
(P = F/A), and this force moves the fluid a certain distance. Thus work is
being done on the fluid. If there are no frictional losses in the system we
can apply the law of conservation of energy.

Work in + Potential energy at the bottom + kinetic energy and the bottom =

Work out + Potential energy at the top + Kinetic energy at the top.

We would like to write this conservation law in terms of the
quantities which are related to fluid flow (pressure, density, velocity, etc.).
This law, written in terms of fluid variables, is called __Bernoulli’s equation__.
Bernoulli’s equation states that at any point in the channel of a flowing fluid
the following relationship holds.

P
+ dgh + ½ dv^{2} = constant
(9.4)

Here P is the pressure, h is the height, v is the velocity, and d is the density at any point in the flow channel.

If the elevation of the fluid remains constant, or if the change in elevation is small enough to not change the gravitational potential energy of the fluid appreciably, then the potential energy term can be ignored. We then have

P
+ ½ dv^{2} = constant

This relationship will apply to all the problems and physical phenomena we will examine in this chapter.

__Examples__:

- Water (density = 1000 kg/m
^{3}) flows through a hose with a velocity of 1 m/sec. As it leaves the nozzle the constricted area increases the velocity to 20 m/sec. The pressure on the water as it leaves is atmospheric pressure (1 Atm = 100,000 N/m^{2}). What is the pressure on the water in the hose? Express the answer in N/m^{2}and Atm.

First let us list the information we are given.

Inside: v_{1}= 1 m/sec

P_{1}= ?

Outside: v_{2}= 20 m/sec

P_{2}= 100,000 N/m^{2}

Density of water = d = 1000 kg/m^{3}

Notice that care has been taken to put all values in mks units. We can now use Eq. 9.5 to find the unknown pressure P_{1}.

_{ }

Express the result in atmospheres.

_{ } - Bernoulli’s equation can also be used to show how the
design of an airplane wing results in an upward lift. The flow of air around
an airplane wing is illustrated below. In this case you will notice that
the air is traveling faster on the upper side of the wing than on the lower.

As a result the pressure will be greater on the bottom of the wing, and the wing will be forced upward. Let us consider an airplane wing where the flow of air (density = 1.3 kg/m^{3}) is 250 m/sec over the top of the wing and 220 m/sec over the bottom.

- Calculate the pressure difference (P
_{1}– P_{2}) between the bottom and the top wing.

Begin by listing the information given and checking that the units are consistent.

Air velocity on bottom = v_{1}= 220 m/sec

Air velocity on top = v_{2}= 250 m/sec

Air density = 1.3 kg/m^{3}

Pressure difference P_{1}– P_{2}= ?

Now use Bernoulli’s equation

_{ } - If the area of the two wings of the airplane is 10 m
^{2}what is the upward force?

The result of the first part of this example told us that the net upward pressure is 9165 N/m^{2}. Using the definition of pressure, P = F/A, we can write

F = PA = 9165 N/m^{2}x 10 m^{2}= 91,650 N

If we convert this force to pounds we obtain an upward force of approximately 20,000 lbs. This means that this air foil in question is capable of supporting a 10 ton airplane in level flight.

- Calculate the pressure difference (P

Through our experience in sports, we all know that the path of a ball if flight is altered when the ball spins. We are perhaps most familiar with this phenomena in baseball. When a baseball pitcher throws the ball he releases it with spin. Depending upon the orientation of this spin the ball will “break” in some direction. The term “curve” and “screwball” refer to the direction in which a pitched ball “breaks”. The curving of balls in flight is also frequently observed in lesser sports such as golf, tennis, and soccer.

The phenomenon of spinning balls curving in flight is called
the **Magnus Effect**, after the German scientist who first studied it.
The Magnus Effect is described with Bernoulli’s equation. Examine the flow
of air about a baseball under three different conditions.

Ball moving with no spin

The air flow is the same on both sides. Now consider a stationary ball that is spinning counterclockwise.

Spinning Ball Without Froward Motion

As the ball spins air is dragged around with the ball. We notice here that the amount of air dragged around the spinning ball depends on the smoothness of the ball. The seams on a baseball will enhance this air flow. The next picture illustrates what happens if the spinning ball is also moving to the right.

Ball Moving to Right While Spinning Counterclockwise

The motion of air past the ball is most rapid above the ball and lowest below the ball. According to Bernoulli’s equation,

P
+ ½ dv^{2} = constant

When the fluid velocity is large, the pressure is small; and
when the fluid velocity is small, the pressure is large. So the air pressure
is least above the ball and greatest below. The ball will experience a net
force toward the top of the page and therefore follow a curved path. A baseball
thrown in this manner is called a rising fastball.

Back to the top

In Chapter 3 we examined a problem where a skier raced straight down a one kilometer hill inclined at an angle of 30 degrees. His velocity at the bottom of the hill was found to be an incredible 222 mph. We argued that this result was not believable, and we used this example to introduce friction, a force that opposes motion.

If the skier example is examined more closely it appears that
frictional force resulting from contact between the skis and the snow is not
large enough to reduce the skier’s velocity to a level we would expect. This
suspicion becomes more obvious when we see that the coefficient if friction
between wood and snow is a relatively small value (μ_{k} = 0.06).
We are forced to conclude that other resistive forces must be acting on the
skier.

With the exception of objects that move in a vacuum, we know
that motion always involves a resistive medium such as air or water. Let us
now investigate the phenomenon of fluid resistance. We begin by looking at
the most obvious source of fluid resistance, called __surface drag__.

Examine the flow of air around a streamlined airfoil that bears a striking resemblance to a flying saucer. The air follows streamlines of simple laminar flow as shown below. The air is disturbed very little by this motion.

Now examine what happens when the airfoil is rotated by 90 degrees and moves to the left.

In this orientation the wing is no longer a streamlined airfoil. Instead of laminar flow around the wing, the result is turbulent flow. The motion of the wing has caused violent agitation of the air through which it passes. In the process, the wing is doing work by moving great amounts of air around its surface. This work is done at the expense of the wing’s kinetic energy. The result is a force that slows the motion of the wing. This type of resistive force is called surface drag.

Surface drag is influenced by the following factors:

Cross sectional area of the object.

Smoothness of the object

Velocity of the object.

Flow characteristics of the fluid.

The cross sectional area determines the amount of fluid that is disturbed as the object flows through it. Clearly, in order to reduce surface drag it is important to minimize cross sectional area. This principle is demonstrated by the human body positions used in sports such as ski jumping, speed skating, and downhill racing.

During the America’s Cup races the boats were removed from the water every night and their hulls were scrubbed and polished. This was done to remove any barnacles, roughness or protrusion from the hull that could increase turbulence, hence increasing surface drag.

The speed at which an object moves through a fluid also affects the amount of turbulence. You can demonstrate this by simply moving your hand in water. If you move your hand slowly, little resistance is experienced. However, when your hand moves rapidly through the water, you feel a very strong force resisting its motion.

The last factor influencing surface drag depends on the fluid
through which a body moves. As the body moves it displaces fluid around it.
The measure of how much force is required to slide the fluid around the body
is called the __viscosity__ of the fluid. Fluids that do nor flow easily,
such as syrup, have a large viscosity. Substances (like water and air) that
flow easily have small viscosity.

Secondary to surface drag, there is another type of fluid resistance
called __form drag__. To examine this phenomenon let us look at a spinless
ball in flight with laminar flow. The laminar flow of air around the ball is
shown below. As the ball moves through the air a partial vacuum is created
behind the ball. As a result the air pressure is greater in front of the ball.
The unbalanced pressures result in unbalanced forces that slow the ball down.

From drag can be reduced by streamlining a body. This streamlining eliminates the partial vacuum behind the body. Airplane wings are streamlined in such a manner so as to keep form drag at a minimum.

It is obvious that objects such as baseballs and golf balls
cannot be streamlined. Form drag will have a large effect on their motion if
it is not reduced. This is accomplished by the rotation of the balls. As a
ball spins it drags air around its surface. This air will help to fill the
partial vacuum behind the ball. Subsequently, the unbalanced pressures on the
ball will be reduced. The dimples on a golf ball and threads on a baseball
will increase the amount of air dragged around them, thus reducing the pressure
differences even more.

Back to the top

Laminar Flow | Magnus Effect |

Turbulent Flow | Surface Drag |

Flow Rate | Viscosity |

The Equation of Continuity | Form Drag |

Bernoulli’s Equation |

- Water is flowing at a rate
of 3 m/sec in a horizontal pipe under a pressure of 200,000 N/m
^{2}. If the pipe narrows to half its original diameter - What is the new speed of flow?
- What is the new pressure?
- How does the flow rate through the narrow section compare with the flow rate through the wider section?

- Water flows through a 2 cm diameter hose with a speed of 0.5 m/sec. Find the speed of water through the nozzle if the diameter is reduced to 0.4 cm.

- Blood flows through the
aorta at a rate of 5 liter/min. Determine the flow rate in gm/sec. (The
density of blood is 1.06 gm/cm
^{3}).

- Compute the speed of blood
flow in the aorta of radius 1 cm (Area = 3.14 cm
^{2}), if the flow rare is 5 liter/min (88 gm/sec).

- Hardening of the arteries
results in a constriction of the arteries. Blood flows with a speed of 26.5
cm/sec through an aorta of radius 1 cm. (Area = 3.14 cm
^{2}). If the flow of blood is to remain constant, calculate the velocity of blood in a region where the aorta has constricted to a radius of 0.8 cm (Area = 2.01 cm^{2}) dues to hardening of the arteries.

- If the flow of blood is
to remain constant through the aorta the velocity of blood must increase from
26.5 cm/sec to 40 cm/sec as the radius decreases from 1 cm to 0.8 cm. Calculate
the pressure difference between these two areas of the aorta in N/m
^{2}. (Density of blood = 1060 kg/m^{3}).

- The following questions are simply applications of Bernoulli’s principle.
- Why is the draft in a fireplace better on a windy day?
- What keeps a Frisbee in flight?
- Why does you car lurch toward an oncoming truck as it passes by?
- Why does the wind increase the size of water waves?

- The air above the wing of
a plane travels at a speed of 320 m/sec; below the wing the speed of air is
300 m/sec. The density of air is 1.3 kg/m
^{3}. - Calculate the pressure difference between the bottom and the top of the wing.
- If the wing area of the
plane is 8 m
^{2}calculate the upward force (lift) on the plane.

- The luge is a small sled in which the racer lies on his back and rides down the course feet first. This is a very dangerous sport and participants always wear helmets. In 1983 the East German luge team introduced helmets which were cone shaped. Honestly, it looked like they had just stepped off the set of “Saturday Night Live.” What competitive advantage would these helmet give s luge racer?

- These helmets were quickly outlawed because they were considered as dangerous as no helmet at all. Why were these helmets considered to be so dangerous?