chapter 11

Chapter 11

Fluids

11.1 Mass Density

DEFINITION OF MASS DENSITY

The mass density of a substance is the mass of a substance divided by its volume:

V

m

SI Unit of Mass Density: kg/m3

11.1 Mass Density

11.1 Mass Density

Example 1 Blood as a Fraction of Body Weight

The body of a man whose weight is 690 N contains about5.2x10-3 m3 of blood.

(a) Find the blood’s weight and (b) express it as a percentage of the body weight.

kg 5.5mkg1060m102.5 333 Vm

N 54sm80.9kg 5.5 a) 2 mgW

%8.7%100N 690

N 54Percentage b)

11.2 Pressure

A

FP

SI Unit of Pressure: 1 N/m2 = 1Pa

Pascal

DEFINITION OF PRESSURE:

Pressure is a measure of the component of force acting perpendicular to the surface divided by the area of the surface.

11.2 Pressure

Example 2 The Force on a Swimmer

Suppose the pressure acting on the back of a swimmer’s hand is 1.2x105 Pa. The surface area of the back of the hand is 8.4x10-3m2.

(a)Determine the magnitude of the force that acts on it.

(b)Discuss the direction of the force.

a)

b)

N 100.1

m104.8mN102.13

2325

F

PAF

Since the water pushes perpendicularly against the back of the hand, the force is directed downward in the drawing.

11.2 Pressure

While you may not feel it, the force of the atmosphere constantly acts on you.

Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere

11.3 Pressure and Depth in a Static Fluid

012 mgAPAPFy

mgAPAP 12

Vm

Anyone who has gone swimming has experienced the increase in pressure that occurs as you descend to the bottom of a pool or lake.

We can use Newton’s Second Law of motion to summarize the forces acting on an object.


VgAPAP 12

AhV

AhgAPAP 12

hgPP 12

Since we know that the volume of a cube can be determined by multiplying the area of the base times the height (V = A h).

This relationship works for liquids since they are considered incompressible and their densities will not change with depth.


Conceptual Example 3 The Hoover Dam

Lake Mead is the largest wholly artificial reservoir in the United States. The water in the reservoir backs up behind the dam for a considerable distance (120 miles).

Suppose that all the water in Lake Mead were removed except a relatively narrow vertical column.

Would the Hoover Dam still be needed to contain the water, or could a much less massive structure do the job?

The answer is that the dam would need to be just as massive since the pressures at the greatest depths would be the same.


Conceptual Example 3 The Hoover Dam

Why would the dam need to be just as massive if it is holding back much less water?

The reason has to do with the relationship:

This formula only deals with pressure changes caused by changes in height in the vertical direction. Pressure increases as the depth increases, not due to horizontal changes in distance.

Pressure increases in depth are primarily dependent upon gh only.

hgPP 12


ghPP 12

Example 4 The Swimming Hole

Points A and B are located a distance of 5.50 m beneath the surface of the water. Find the pressure at each of these two locations.

Pa 1055.1

m 50.5sm80.9mkg1000.1Pa 1001.15

2

233

pressure catmospheri

52

P

P

Since points A and B are at the same depth, the pressures will the same at both points.

Note that P1 is often the barometric pressure.

The closed-tube manometer consists of a tube full of mercury inverted such that its open end is underneath a pool of mercury.

The pressure at the top of the tube is zero Pa as shown.

The pressure at points A and B are equal since they are at the same elevation, and are equal to the atmospheric pressure as shown.

What is h?

11.4 Pressure Gauges (The Closed-Tube Mercury Manometer)

ghPP 12

ghPatm

mm 760m 760.0

sm80.9mkg1013.6

Pa 1001.1233

5

h

g

Ph atm

ghP 2

The open-tube manometer consists of a tube, often full of mercury, that is open to atmospheric pressure as shown below.

The pressure, P2 is greater than the atmospheric pressure.

The pressure at points A and B, as well as P2 are all the same.

The pressure at P1 is equal to the atmospheric pressure.

11.4 Pressure Gauges (Open-Tube Manometer)

)( 2 AB PPP

ghPP atm pressure gauge

2

ghPP 12

Absolute vs. Gauge Pressure• The difference between P2 and the

atmospheric pressure is called the gauge pressure. This is the pressure often read from a gauge such as one on your water tank or boiler.

• The absolute pressure is the actual value of P2.

11.5 Pascal’s Principle

PASCAL’S PRINCIPLE

Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls.

In the diagram to the right, force F1 acting on A1 will cause an increase in pressure by the amount F1 /A1 throughout the whole enclosure.

If the heights of A1 and A2 are the same, then gh = 0.

The significance of this relationship is that a small force acting on a small area will result in a large force acting on a much larger area.

m 012 gPP

1

1

2

2

A

F

A

F

1

212 A

AFF

Example 5 A Car Lift

The input piston has a radius of 0.0120 m and the output plunger has a radius of 0.150 m.

The combined weight of the car and the plunger is 20500 N. Suppose that the input piston has a negligible weight and the bottom surfaces of the piston and plunger are at the same level. What is the required input force?

Since the pistons are cylindrical, we can simply calculate their areas using A = r2.


1

212 A

AFF

N 131m 150.0

m 0120.0N 20500 2

2

1

F

Example 6 A Car Elevated on a Lift

Using the same lift as in the previous example, the vehicle is now raised such that the difference in height between the two pistons is 1.5 m. The density of the hydraulic oil is 830 kg/m3. Everything else is the same as before.


ghA

F

A

F 2

2

1

1ghA

F

A

F 1

1

2

2

NF

mmsmmkgm

NF

126

)012.0()50.1)(/81.9)(/830()150.0(

500,20

1

2232

11

Start with: P2 = P1 + gh and substitute F/A for P.

12

21 Agh

A

FF

When an object is immersed in a fluid, it will experience a net force in the upward direction (toward the surface) called the buoyant force (FB).

Consider the cylinder immersed in the diagram to the left. The buoyant force results from the difference between the force acting on the bottom of the cylinder and the one on top.

11.6 Archimedes’ Principle

12 FFFB

APPAPAPFB 1212

ghPP 12

ghAFB

hAV

gVFB

fluiddisplaced

of mass

PAF


ARCHIMEDES’ PRINCIPLE

Any fluid applies a buoyant force to an object that is partiallyor completely immersed in it; the magnitude of the buoyantforce equals the weight of the fluid that the object displaces:

• Buoyancy• Eureka!

fluid displaced

ofWeight

fluid

forcebuoyant of Magnitude

WFB

Since V = mass, and mg = Fg, or the weight of the fluid, we get:


• If the weight of the object is less than the weight of the fluid displaced, the object will float, and the magnitude of the buoyant force is equal to the magnitude of its weight (object < fluid).

• If the weight of the object is the same as the weight of the fluid displaced, then the object will neither sink or float (object = fluid).

• If the weight of the object is greater than the weight of the fluid displaced, the object will sink (object > fluid).

1N 2N 8N


Example 7 A Swimming Raft

The raft is made of solid square pinewood. Determine whether the raft floats in water and if so, how much of the raft is beneath the surface.

The density of water is 1000 kg/m3 and the density of pine is 550 kg/m3.

Start by finding the buoyant force that the raft would experience if it were to be completely submerged.

In order to do this, we will need to start by finding the volume of the raft.


N 47000

sm80.9m8.4mkg1000

max

233max

B

B

F

F

3m 8.4m 30.0m 0.4m 0.4 raftV

gVVgF waterwaterB max


N 47000N 26000

sm80.9m8.4mkg550 233

gVgmW raftpineraftraft

Hence, the raft floats!

We will now find the weight of the raft, given a known density for pine of 550 kg/m3.

We will then compare this weight with that of the maximum buoyant force of the water.


gVwaterwaterN 26000

Braft FW

Now that we know the raft is floating, how much of it is submerged? i.e. what is h?

23 sm80.9m 0.4m 0.4mkg1000N 26000 h

m 17.0sm80.9m 0.4m 0.4mkg1000

N 2600023

h


Conceptual Example 8 How Much Water is Neededto Float a Ship?

A ship floating in the ocean is a familiar sight. But is allthat water really necessary? Can an ocean vessel floatin the amount of water than a swimming pool contains?

Yes, as long as the water maintains a gap between the ship and the walls of the canal. Remember, the pressure increases with depth.

11.7 Fluids in Motion

In steady flow the velocity of the fluid particles at any point is constant as time passes.

Unsteady flow exists whenever the velocity of the fluid particles at a point changes as time passes.

Turbulent flow is an extreme kind of unsteady flow in which the velocity of the fluid particles at a point change erratically in both magnitude and direction.

Types of Fluid Flow:


Fluid flow can be compressible or incompressible, though most liquids are nearly incompressible. (Incompressible means that you cannot squeeze the fluid into a smaller volume)

Viscous fluids are those types of fluids where drag forces exist. Examples of such fluids include honey, syrup and motor oil products.

An incompressible, nonviscous fluid is called an ideal fluid. Water, alcohol and gasoline are a few examples of ideal fluids.

Types of Fluid Flow (cont.):


When the flow is steady, streamlines are often used to representthe trajectories of the fluid particles. Note that the velocity vector is always tangent to the streamline.

11.8 The Equation of Continuity

The mass of fluid per second that flows through a tube is called the mass flow rate.

Whether the thumb is over the end to create a sharp stream, or the fluid is allowed to flow freely out of the end of the hose, the mass flow rates are equal.

The figure below shows fluid flowing through a tube with two different cross-sectional areas. How much mass passes a given point during a period of time, t?


2222 vA

t

m

111

1 vAt

m

distance

tvAVm

If 1 kg of liquid passes by point 2 in time t, then 1 kg of fluid will pass by point 1 in the same amount of time.


222111 vAvA

EQUATION OF CONTINUITY

The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow.

SI Unit of Mass Flow Rate: kg/s


Incompressible fluid: 2211 vAvA

Volume flow rate Q: AvQ

21

Example 9 A Garden Hose

A garden hose has an unobstructed opening with a cross sectional area of 2.85x10-4m2. It fills a bucket with a volume of 8.00x10-3m3 in 30 seconds.

Find the speed of the water that leaves the hose through:

a. the unobstructed opening and b. an obstructed opening with half as much

area.


sm936.0

m102.85

s 30.0m1000.824-

33

A

Qv

AvQ a)

2211 b) vAvA

sm87.1sm936.0212

12 v

A

Av

11.9 Bernoulli’s Equation

Using the Equation of Continuity, we know that where the velocity is higher, the pressure is lower. The fluid experiences a net unbalanced force that causes it to accelerate from region 2 to region 1.

According to the pressure-depth relationship, the pressure is also lower at higher levels, provided the cross-sectional area of the pipe does not change.

If you think of the blue fluid element of mass m as an object moving in the pipe from region 2 to region 1, we can analyze it from the perspective of Conservation of Energy and the Work-Energy Theorem where we will consider only changes in mechanical energy (kinetic and potential energy).

The work done on the fluid as it moves from region 2 to region 1 can be written as:

Wnc = E1 – E2

11.9 Bernoulli’s Equation (Conservation of Energy & the Work-Energy Theorem)

Non-Conservative Work

Since the total mechanical energy of the fluid element in each region can be defined by its potential and kinetic energies:

Wnc = (KE1 + PE1) - (KE2 + PE2)

Wnc = (½ mv12 + mgy1) (½ mv2

2 + mgy2)

11.9 Bernoulli’s Equation (Work done on the fluid element)

Total mechanicalEnergy in region 1

Total mechanicalEnergy in region 2

We will now focus on the left side of the expression, Wnc. The pressure in region 2 is greater than it is in region 1, which gives rise to a net force, F, that causes the fluid to flow through a displacement, s, as seen in (b) below.

On the top surface: the pressure is: PT

the force is: FT = PTAOn the bottom surface: the pressure is: PB = PT + P

the force is: FB = FT + F = (PT + P)AThe net force that pushes the fluid element up the pipe is: F = (P)A

Knowing that Work = Fd, it follows that:

Wnc = Fs = (P)As = (P2 – P1)V

11.9 Bernoulli’s Equation (Work done on the fluid element)

11.9 Bernoulli’s Equation

Bernoulli’s Equation is an alternative representation for the Law of Conservation of Energy. It is true in cases where there is a steady flow of a nonviscous, incompressible fluid.

By combining the two relationships, we get:

Wnc = (P2 – P1)V = (½ mv12 + mgy1) (½ mv2

2 + mgy2)

Knowing that = m/V, we will divide both side by V to get:

Wnc = (P2 – P1) = (½ v12 + gy1) (½ v2

2 + gy2)

A rearrangement results in BERNOULLI’S EQUATION

P1 + ½ v12 + gy1 = P2 + ½ v2

2 + gy2

11.10 Applications of Bernoulli’s Equation

Conceptual Example 10 Tarpaulins and Bernoulli’s Equation

When the truck is stationary, the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway.

Account for this behavior.

The air is stationary both inside and outside the cargo hold while the vehicle is not moving. However, as it drives down the highway, the air flowing over the top of the tarp will result in lower pressure outside the cargo hold than inside, causing the tarp to bulge outward.


Lift in aircraft is a result of pressure differences caused by differences in the speed of the air flowing over and under the wings. As per Bernoulli’s Equation, as velocity increases, pressure decreases. An aircraft wing is designed such that the air flow will be faster over the top of the wing rather than the bottom.


A curveball in baseball is a pitch where the pitcher causes the ball to spin as it travels towards the batter and catcher. As a result of this rotation, the ball will follow a curved path. The deflection occurs in the direction where the airflow over the surface of the ball is the greatest.


Example 11 Efflux Speed

A large storage tank in a commercial winery is cylindrical in shape. The height, h, of the fluid is two meters above the spigot. The opening of the spigot has a radius of 0.5 cm.

a) Determine the speed of the fluid leaving the tank.

b) Determine how long it will take to fill a 1 liter bottle of wine.


2222

121

212

11 gyvPgyvP

We will start by making a few assumptions:1. The fluid at the top of the liquid and at

that leaving the spigot will experience the same pressure (Patm).

2. The velocity at the top of the liquid can be considered to be zero.

3. We will also consider wine to be an ideal incompressible fluid with negligible viscosity.

11.10 Applications of Bernoulli’s Equation (a)

2222

121

212

11 gyvPgyvP

atmPPP 2102 v

hyy 12ghv 212

1

ghv 21

smmsmv /4.6)0.2)(/81.9(2 21

11.10 Applications of Bernoulli’s Equation (b)

sAV

2rA tvs 1

tvrV 12

12 vr

Vt

sscmcm

cmt 0.2

)/640()5.0)((

10002

3

chapter 11

Documents

pressure changes

atmospheric pressure

barometric pressure

papascaldefinition of

pressuresi unit of pressure

depth increases

mass density11

mass densityexample