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Lecture PowerPoints

Physics for Scientists & Engineers, with Modern

Physics, 4th edition

Giancoli

Chapter 13


Chapter 13Fluids

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• Phases of Matter

• Density and Specific Gravity

• Pressure in Fluids

• Atmospheric Pressure and Gauge Pressure

• Pascal’s Principle

• Measurement of Pressure; Gauges and the Barometer

• Buoyancy and Archimedes’ Principle

Units of Chapter 13


• Fluids in Motion; Flow Rate and the Equation of Continuity

• Bernoulli’s Equation

• Applications of Bernoulli’s Principle: Torricelli, Airplanes, Baseballs, TIA

• Viscosity

• Flow in Tubes: Poiseuille’s Equation, Blood Flow

Units of Chapter 13

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The three common phases of matter are solid, liquid, and gas.

A solid has a definite shape and size.

A liquid has a fixed volume but can be any shape.

A gas can be any shape and also can be easily compressed.

Liquids and gases both flow, and are called fluids.

13-1 Phases of Matter


The density ρ of a substance is its mass per unit volume:

The SI unit for density is kg/m3. Density is also sometimes given in g/cm3; to convert g/cm3 to kg/m3, multiply by 1000.

Water at 4°C has a density of 1 g/cm3 = 1000 kg/m3.

The specific gravity of a substance is the ratio of its density to that of water.

13-2 Density and Specific Gravity

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13-2 Density and Specific Gravity

Example 13-1: Mass, given volume and density.

What is the mass of a solid iron wrecking ball of radius 18 cm?

kg 1900.024107.8

m 0.0243

)18.0(4

m 0.18cm 18

kg/m 107.8

3

33

334

33iron

VmV

m

rV

r

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Pressure is defined as the force per unit area.

Pressure is a scalar; the units of pressure in the SI system is the pascal:

1 Pa = 1 N/m2.

13-3 Pressure in Fluids



Example 13-2: Calculating pressure.

The two feet of a 60 kg person cover an area of 500 cm2.

(a) Determine the pressure exerted by the two feet on the ground.

(b) If the person stands on one foot, what will the pressure be under that foot?

(a) A = 500 cm2 = 500 × 10-4 = 5 × 10-2 m2

F = mg = 60 × 9.8 = 588 N

P = F/A = 588/(5 × 10-2) = 1.2 × 104 Pa

(b) Area halved Pressure doubledP = 2.4 × 104 Pa

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Pressure is a scalar (it has a magnitude but no direction) and is exerted equally in all directions.

Example:

A living room has floor dimensions of 3.5 m x 4.2 m and height of 2.4 m.

(a) What does the air in the room weigh?

(b) What force does the atmosphere exert on the floor of theroom?

(a) What does the air in the room weigh?

Volume of room: V = 3.5 x 4.2 x 2.4 = 35.28 m3

Density of air: ρair = 1.29 kg/m3

Mass of air: ρ = m/V → m = ρV = 1.29 x 35.28 = 45.51 kgWeight of air:W = mg = 45.51 x 9.8 = 446 N

(b) What force does the atmosphere exert on the floor of the room?

Atmospheric pressure: Patm = 1.013 x 105 PaFloor area: A = 3.5 x 4.2 = 14.7 m2

Force on floor: A

FP

→ F = PA = 1.013 x 105 x 14.7 = 1.5 x 106 N

Much greater than the weight of air in the room. → It is theweight of a column of air (same area as floor) extending fromground level to the top of the atmosphere.

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Pressure is the same in every direction in a static fluid at a given depth; if it were not, the fluid would flow.


For a fluid at rest, there is also no component of force parallel to any solid surface—once again, if there were, the fluid would flow.


F|| = 0

So force is perpendicular to the surface

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The pressure at a depth h below the surface of the liquid is due to the weight of the liquid above it. We can quickly calculate:

This relation is valid for any liquid whose density does not change with depth.




If there is external pressure in addition to the weight of the fluid itself, or if the density of the fluid is not constant, we calculate the pressure at a height y in the fluid; the negative sign indicates that the pressure decreases with height (increases with depth):

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13-3 Pressure in FluidsWe then integrate to find the pressure:

Pressure at a depth h = (y2 – y1) in a liquid of density ρ is

P = P0 + ρghwhere P0 is the external pressure at the liquid’s top surface.


13-3 Pressure in FluidsExample 13-3: Pressure at a faucet.

The surface of the water in a storage tank is 30 m above a water faucet in the kitchen of a house. Calculate the difference in water pressure between the faucet and the surface of the water in the tank.

P = P0 + ghP = P – P0 = gh= 1000×9.8×30 = 2.9×105 Pa

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13-3 Pressure in FluidsExample 13-4: Force on aquarium window.

Calculate the force due to water pressure exerted on a 1.0 m x 3.0 m aquarium viewing window whose top edge is 1.0 m below the water surface.



Example 13-4: Force on aquarium window.

N 440002

8.910009

2

9

2

1

2

43

233

3gy :Force

water) todue(:depth At

2

1

22

1

g

g

ygdyygF

dy PdAdF

gyPy

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13-3 Pressure in FluidsExample 13-5: Elevation effect on atmospheric

pressure.

(a) Determine the variation in pressure in the Earth’s atmosphere as a function of height y above sea level, assuming g is constant and that the density of the air is proportional to the pressure. (This last assumption is not terribly accurate, in part because temperature and other weather effects are important.)

(b) At what elevation is the air pressure equal to half the pressure at sea level?


13-3 Pressure in FluidsExample 13-5: Elevation effect on atmospheric pressure.

(a)

then, )ln(ln ln since

ln

height) with pressure as (-ve

level seaat Density kg/m 1.29

level seaat pressure cAtmospheri Pa 101.013

As

00

000

0

0

0

30

50

00

0

P/PPP

gyP

Pdyg

PP

dP

gP

Pgdy

dPdygdP

P

P

PP

yP

P

yPgePP 00 /0

Pressure decreases approximately exponentially with height.

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Example 13-5: Elevation effect on atmospheric pressure.

(b) At what elevation is the air pressure equal to half the pressure at sea level?

m 555010248.1

693.0

693.02ln10248.1

2

2/For

level seaat Density kg/m 1.29

level seaat pressure cAtmospheri Pa 101.013

4

4

10248.1

10248.110013.1

8.929.1

21

0

/0

30

50

4

45

00

y

y

e

ee

PP

ePP

P

y

yy

yPg


At sea level the atmospheric pressure is about 1.013 x 105 Pa ; this is called 1 atmosphere (atm).

Another unit of pressure is the bar:

1 bar = 1.00 x 105 Pa.

Standard atmospheric pressure is just over 1 bar.

This pressure does not crush us, as our cells maintain an internal pressure that balances it.

13-4 Atmospheric Pressure and Gauge Pressure

Meteorologists use the milli bar = 100 Pa on weather maps.(Standard atmospheric pressure = 1013 milli bar)

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13-4 Atmospheric Pressure and Gauge PressureConceptual Example 13-6: Finger holds water in a straw.

You insert a straw of length l into a tall glass of water. You place your finger over the top of the straw, capturing some air above the water but preventing any additional air from getting in or out, and then you lift the straw from the water. You find that the straw retains most of the water. Does the air in the space between your finger and the top of the water have a pressure P that is greater than, equal to, or less than the atmospheric pressure P0 outside the straw?There must be a net upward force on the water in the straw to keep it from falling out; therefore the pressure in the space above the water must be less than atmospheric pressure


Most pressure gauges measure the pressure above the atmospheric pressure—this is called the gauge pressure.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

13-4 Atmospheric Pressure and Gauge Pressure

Determine the gauge pressure at a house which is situated100xm below the level of the storage dam? (9.8 x 105 Pa)

( Use P = gh )

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More Sample Problems

1. In water the pressure increases with depth according toP = gh. At what depth below the surface would thepressure on a scuba diver be 1 atmosphere above thepressure at the surface?

Solution

If P = Patm = 1.013 x 105 Pa, sea water = 1025 kg/m3,and g = 9.8 m/s2,

then the depth h is given by:

h = P/g = 1.013 x 105/(1025 x 9.8) = 10.08 m.

When you are diving (in the sea) the pressure on your bodyincreases by about 1 atm for every 10 m under water.

When scuba diving, the demand valve supplies air at waterpressure so that the pressure in the chest cavity is almost thesame as the water pressure so that your chest is not crushed.

Note that the solubility of nitrogen in blood plasma increases asthe pressure of the air breathed by a scuba diver increases withdepth.

When diving at greater depths it is essential to ascend slowly,pausing at regular intervals to avoid the bends (gas bubblesforming in blood vessels and tissue).

If the diver surfaces too rapidly, nitrogen dissolved in the bloodwill come out of solution and form bubbles in the blood vessels.

This causes occlusion of the capillaries and causes the bends.

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This is life threatening situation and is normally treated in ahyperbaric chamber.

The air pressure in the chamber is increased to the equivalentof the pressure at the maximum depth that the diver attainedand slowly decreased to atmospheric pressure so that thenitrogen can be eliminated through the lungs.

For very deep dives this problem can largely be avoided bybreathing a helium – oxygen mixture as helium is almostinsoluble in blood plasma.

2. The Russian submarine “Kursk” sank in a terrible tragedy in 2001. After taking on water, the submarine sank to the continental shelf only 100 m below the surface.

(a) What was the absolute pressure at this depth?(b) How many times greater is this than atmospheric

pressure?

Solution:

Pressure at surface: P0 = Patm = 1.013 x 105 PaDensity of sea water: ρ = 1025 kg/m3

Depth of the water: d = 100 m

P = P0 + ρgh = 1.013 x 105 + 1025 x 9.8 x 100 = 1.106 x 106 Pa= 1106 kPa

Since atmospheric pressure Patm = 1.013 x 105 Pa = 101.3 kPa

This is 1106/101.3 = 10.9 x atmospheric pressure

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3. The water in a marine aquarium is 5 m deep.(a) What is the pressure at the surface?(b) What is the water pressure at the bottom?

Solution:

(a) Pressure at the surface: Patm = 1.013 x 105 Pa = 101.3 kPa

(b) Pressure at bottom:

P = P0 + ρgh = 1.013 x 105 + 1025 x 9.8 x 5 = 1.515 x 105 Pa= 151.5 kPa

4. Vacuum cleaner salespeople use the “bowling ball”demonstration to “prove” how strong “suction” is. In factthere is no “suction” force rather atmospheric pressure ispushing on the ball from all sides.

In this problem let’s imagine we have a flat square metal plate100 mm x 100 mm in size. On one side is vacuum (P = 0), onthe other is atmospheric pressure of 1.0 atm = 1.013 x 105 Pa.

What is the mass of the plate that can be supported by“suction” if the vacuum seal is perfect?

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Solution:

For the maximum supportable mass the weight equals thepressure force:

FP = PA = FW = mg

mg = PA

& Area of metal plate: A = 10-1 x 10-1 = 10-2 m2

m = PA/g = (1.013 x 105 x 10-2)/9.8 = 103.4 kg

So you can see that air pressure can hold 103 kg if the vacuumon the other side is perfect.

5. Water has a density of 998 kg/m3 (at 20C). What is thedensity of water in g/cm3?(0.998 g/cm3)

6. The density of water at 4C is 1000 kg/m3. What is themass of one litre of pure water? What are thedimensions in mm of a cube which has a volume of 1 L?(1 kg, 100 mm).

7. Air pressure at sea-level is 101.3 kPa. The surface areaof an adult human is ~ 2.5 m2. What force is the airexerting on us at all times? Why don’t we feel it? Whathappens to us in a vacuum?(253 kN)

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mmHg 2941101.013

760103.92P

(mmHg) P

(mmHg) 760

(Pa) 103.92

(Pa) 101.0135

5

5

5

(Pa) P107.503

(Pa) P101.013

760

(Pa) P

(Pa) P(mmHg)P(mmHg) P

(mmHg) P

(mmHg) P

(Pa) P

(Pa) P

3

5atm

atmatmatm

8. The gravity fed water reservoir for a city is located 40 mabove the city. Calculate the water pressure at homes inthe city in units of N/m2, atm and mmHg.(Use ρwater = 1000 kg/m3)(3.92 x 105 N/m2, 3.9 atm, 2941 mmHg)

Note: 1. N/m2 = Pa

2. 1 atm = 1.013 x 105 Pa = 760 mmHg

3. Using ratios is the easiest way to convertbetween Pa and mmHg.


If an external pressure is applied to a confinedfluid, the pressure at every point within the fluid increases by that amount.

This principle is used, for example, in hydraulic lifts and hydraulic brakes.

13-5 Pascal’s Principle

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Hydraulic Lift

Two cylinders connected bya tube are filled with anincompressible fluid (eg. oil).The fluid is confined in thesystem by two pistons, asshown at right.

When force F1 is exerted on the small piston, fluid flows fromthe small cylinder to the large one, moving the large piston upand exerting an upward force F2. Pascal’s principle states thatthe pressure created by exerting the force F1 on the smallpiston, P1 = F1/A1, is transmitted undiminished to all parts of thefluid.

2

2

1

1

A

F

A

F 1

1

22 F

A

AF Therefore, P1 = P2 and

The force is magnified by the ratio of the areas. Often a leveris used to magnify the force F1 even further so that very smallforces can lift heavy objects, eg. hydraulic car jack.

Hydraulic Lift (Cont)

Problem:

In a hydraulic press the smaller piston has a diameter of 40 mmand the larger piston has a diameter of 500 mm. Whatforce must be applied to the smaller cylinder to lift a massof 2000 kg? If it is required to lift the 2000 kg mass 100mm, how far must the small piston move?

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Solution:

N 1250.5

196000.042

2

22

221

22

221

2

211

2

2

1

1

d

Fd

kd

Fkd

A

FAF

A

F

A

F

The volume must be the same on both sides.

m 15.6250.04

0.10.52

2

21

222

12221

21

2221

21

21

d

hdhhdhd

4

hπd

4

hπdVV

d1 = 40 mm = 0.04 m d2 = 500 mm = 0.5 m Since A = d2/4 → A = kd2 A1 = kd1

2 A2 = kd2

2 F2 = m2g = 2000 x 9.8 = 19600 N h2 100 mm = 0.1 m

Note that this is impractical. In practice a lever would be usedto apply force F1 and valves used so that the smaller cylinderwould refill after each pump of the handle.

If each time you pumped the handle, piston 1 moved 50 mm,how many times would you have to pump the handle to lift the2000 kg mass by 100 mm?


Solution:

Each pump of the handle pushes volume

The volume moved by piston 2

Number of handle pumps = ( x 6.25 x 10-3)/( x 2 x 10-5)= 313

2piston into m 1024/05.0)04.0(4/ 35221

hdV

332222 m 1025.64/1.0)5.0(4/ hdV

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There are a number of different types of pressure gauges. This one is an open-tube manometer. The pressure in the open end is atmospheric pressure; the pressure being measured will cause the fluid to riseuntil the pressures on both sides at the same height are equal.

13-6 Measurement of Pressure; Gauges and the Barometer


Here are two more devices for measuring pressure: the aneroid gauge and the tire pressure gauge.


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Pressure is measured in a variety of different units. This table gives the conversion factors.


This is a mercury barometer, developed by Torricelli to measure atmospheric pressure. The height of the column of mercury is such that the pressure in the tube at the surface level is 1 atm.

Therefore, pressure is often quoted in millimeters (or inches) of mercury.


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Any liquid can serve in a Torricelli-style barometer, but the most dense ones are the most convenient. This barometer uses water.


A water barometer: a full tube of water is inserted into a tub of water, keeping the tube’s spigot at the top closed. When the bottom end of the tube is unplugged, some water flows out of the tube into the tub, leaving a vacuum between the water’s upper surface and the spigot. Why? Because air pressure can not support a column of water more than 10 m high



Conceptual Example 13-7: Suction.

A student suggests suction-cup shoes for Space Shuttle astronauts working on the exterior of a spacecraft. Having just studied this Chapter, you gently remind him of the fallacy of this plan. What is it?

Suction cups work because of air pressure, and there isn’t any air where the shuttle orbits

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The pressure difference between the container and theatmospheric pressure is shown by difference in the height ofthe working fluid between the two sides of the manometer.

Problems:

1. In the previous diagram an oil ( = 1900 kg/m3) is used asworking fluid. The difference in height h of the workingfluid is 100 mm. What is the pressure in the container?What is the gauge pressure?

Solution:

The pressure in the container is given by: ghρPP wf0container

As atmospheric pressure is 101.3 kPa and the density of oil is 1.9 x103 kg/m3

P = 1.013 x 105 + 1.9 x 103 x 9.8 x 0.1 = 1.032 x 105 Pa= 103.2 kPa

Pgauge = P – P0 = 103.2 – 101.3 = 1.9 kPa

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2. Compare the heights of a water column and mercurycolumn when used in a barometer.

Solution:

ρwater = 1000 kg/m3

ρHg = 13600 kg/m3

Water column will be 13600/1000 = 13.6 times the height ofmercury column.


This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different.

The buoyant force is found to be the upwardforce on the same volume of water:

13-7 Buoyancy and Archimedes’ Principle

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Archimedes’ principle:

The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object.



Conceptual Example 13-8: Two pails of water.

Consider two identical pails of water filled to the brim. One pail contains only water, the other has a piece of wood floating in it. Which pail has the greater weight?

Both weigh the same; if both pails were full to the brim before the wood was put in, some water will have spilled out.

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Example 13-9: Recovering a submerged statue.

A 70 kg ancient statue lies at the bottom of the sea. Its volume is 3.0 x 104 cm3. How much force is needed to lift it?

N 3902948.970

N 294891031000

m 103 cm 103.0

waterdisplacedtheofweight thetoequalisforcebuoyancy The

B

2

waterwaterB

32-34

FmgF

.

VggmF

V



Example 13-10: Archimedes: Is the crown gold?

When a crown of mass 14.7 kg is submerged in water, an accurate scale reads only 13.4 kg. Is the crown made of gold?

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Example 13-10: Archimedes: Is the crown gold?

33

333crowncrown

33OH

B

kg/m 1019.3 is gold ofdensity The

lead. of made iscrown that thesuggestswhich

kg/m 103.11103.1/7.14/

m 101.31000/3.1//

kg 1.3 water displaced of Mass

N 1.3N )4.13(14.7'

waterdisplaced ofweight ForceBuoyancy

2

Vm

mVVm

ggww

F

A widely known anecdote about Archimedes tells how he invented a method for determining the volume of an object with an irregular shape. According to Vitruvius, a votive crown for a temple had been made for King Hiero II of Syracuse(308 - 215 BC), who had supplied the pure gold to be used, and Archimedes was asked to determine whether some silver had been substituted by the dishonest goldsmith.

Archimedes had to solve the problem without damaging the crown, so he could not melt it down into a regularly shaped body in order to calculate its density. He deduced that if an object was immersed in water, it would displace a volume of water equal to the volume if the object.

From mass and volume the density of the crown could be determined. The density would be lower than that of gold if cheaper and less dense metals had been added. The test was conducted successfully, proving that silver had indeed been mixed in (Read goldsmith was executed).

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This story does not appear in the known works of Archimedesand the practicality of the method has been called into question,due to the extreme accuracy with which one would have tomeasure the water displacement.

Archimedes may have used a method based on his principle(described in his treatise ‘On Floating Bodies’. Using thisprinciple, it would have been possible to compare the density ofthe golden crown to that of solid gold by balancing the crown ona scale with a gold reference sample, then immersing theapparatus in water. The difference in density between the twosamples would cause the scale to tip accordingly. Galileoconsidered it "probable that this method is the same thatArchimedes followed, since, besides being very accurate, it isbased on demonstrations found by Archimedes himself."

As the crown is not pure gold(has a larger volume than thesame mass of gold) it willdisplace a larger volume ofwater than the gold andexperience a greater buoyancyforce. ie. It will appear lighterthan an equal mass of pure goldwhen immersed in water.

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If an object’s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water.



For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid.


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Example 13-11: Hydrometer calibration.

A hydrometer is a simple instrument used to measure the specific gravity of a liquid by indicating how deeply the instrument sinks in the liquid. This hydrometer consists of a glass tube, weighted at the bottom, which is 25.0 cm long and 2.00 cm2 in cross-sectional area, and has a mass of 45.0 g. How far from the end should the 1.000 mark be placed?


Example 13-11: Hydrometer calibration.

bottom from cm 22.525.00.90

surfaceunder 0.90900/1000 float with willHydrometer

kg/m 900102.00.25

0.045

:hydrometer ofdensity Average

34H

x

V

mρ

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Example 13-12: Helium balloon.

What volume V of helium is needed if a balloon is to lift a load of 180 kg (including the weight of the empty balloon)?


Example 13-12: Helium balloon. mload = 180 kg

3

Heair

load

loadHeair

loadHeloadHeB

airairB

3air

3He

m 160179.029.1

180

air displaced ofweight

kg/m 1.29kg/m 0.179

mV

mVV

gmVggmgmF

VggmF

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More Problems:

14. What is the buoyancy force due to the earth’satmosphere on an average human (height 1.83 m,weight = 70 kg, density = 1030 kg/m3).

Solution:

ρ = m/V → V = m/ρ = 70/1030 = 0.068 m3

Weight of air displaced (ρair = 1.29 kg/m3)

→ W = mg = ρVg = 1.29 x 0.068 x 9.8 = 0.860 N

The buoyancy force due to the air is only 0.860 N.Since W = mg → m = W/g = 0.860/9.8 = 0.088 kg which is onlyan apparent reduction in mass of

(0.088/70) x 100 = 0.1% of our total mass.

If we were immersed in a more dense fluid e.g. water:ρ = 1000 kg/m3) the buoyancy force would be greater and wewould feel lighter.

Weight of displaced water = ρwaterVg = 1000 x 0.068 x 9.8= 666 N

Total weight of the person is: W = mg = 70 x 9.8 = 686 N, this isa reduction in apparent weight of (668 – 666)/668 = 0.003. i.e.When immersed in water the apparent weight of the personwould have only 0.3% of their weight in air.

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2. There is a common saying “just the tip of the iceberg”,because icebergs are ≈90% underwater. Given thedensities of ice and sea-water we can calculate thevolume of an iceberg that lies below the water line.

Density of ice: ρice = 917 kg/m3

Density sea water: ρsw = 1025 kg/m3

iceberg

Solution:

The solution to this problem is (slightly) difficult.

First assume the iceberg has volume VB. Using the density ofice calculate the mass of ice:

ρB = mB/VB → mB = ρBVB = 917VB

As the ice floats, we know the weight of the water displacedmust be the same as the weight of the entire iceberg.

Weight of water displaced: WW = WB → mWg = mBg → mW = mB

Since m = ρV → mW = ρWVW = mB = ρBVB → ρWVW = ρBVB

→ VW/VB = ρB/ρW = 917/1025 = 0.895

This shows us that ≈90% of the volume of the iceberg isunderwater.

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If the flow of a fluid is smooth, it is called streamline or laminar flow (a).

Above a certain speed, the flow becomes turbulent (b). Turbulent flow has eddies; the viscosity of the fluid is much greater when eddies are present.

13-8 Fluids in Motion; Flow Rate and the Equation of Continuity

(a)

(b)

Airflow around an airplane wing

Streamlines

A streamline is a path traced out by a tiny fluid element whenfluid flows

The local fluid velocity is always at a tangent to the streamline.

v

Streamlines cannot cross. If they did the fluid would be goingin two directions at once.

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We will deal first with laminar flow.

The mass flow rate is the mass that passes a given point per unit time. The flow rates at any two points must be equal, as long as no fluid is being added or taken away.

This gives us the equation of continuity:


Since

then

Avt

A

t

V

t

m

l

Note:


If the density doesn’t change—typical for

liquids—this simplifies to A1v1 = A2v2 . Where

the pipe is wider, the flow is slower.


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Example 13-13: Blood flow.

In humans, blood flows from the heart into the aorta, from which it passes into the major arteries. These branch into the small arteries (arterioles), which in turn branch into myriads of tiny capillaries. The blood returns to the heart via the veins. The radius of the aorta is about 1.2 cm, and the blood passing through it has a speed of about 40 cm/s. A typical capillary has a radius of about 4 x 10-4 cm, and blood flows through it at a speed of about 5 x 10-4 m/s. Estimate the number of capillaries that are in the body.


Example 13-13: Blood flow.

9

2

6

2

4

2

C

A

C

A

C2

A2

CC

AA

CCAA

4C

C

2CC

64C

A

2AA

2A

107104

102.1

105

4.0

N

N

m/s 105

Nscapillarie of area Total

m 104cm 104

m/s 0.4cm/s 40

m 101.2cm 1.2

r

r

v

v

vr

vr

vA

vA

vAvA

v

A

rA

r

v

rA

r

c

A

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Example 13-14: Heating duct to a room.

What area must a heating duct have if air moving 3.0 m/s along it can replenish the air every 15 minutes in a room of volume 300 m3? Assume the air’s density remains constant.

2

21

21

2

2

2

222211

m 0.1160150.3

300

duct wider a as room Treat the

tv

VA

t

V

tAvAvA

l

Problems:

1. Convert a flow rate of 150 L/min into m3/s volume flow rate.

Solution:

1 L is the volume of a cube with sides of 100 mm

1 mm = 10-3 m

1 L = (100 mm)3 = (100 x 10-3 m)3 = (10-1 m)3 = 10-3 m3

/sm 102.5s

m

60

10150

min

L150 33

33

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39

2. A pipe changes from a diameter of 500 mm to 250 mm. Ifthe flow rate in the 500 mm section is 1 m3/s:

(a) What is the flow rate in the narrower section?(b) What is the flow velocity in both sections of the pipe?

Solution:

(a) Equation of Continuity → Av = constant so the flow rate isstill 1 m3/s

(b) Larger section: r = 0.25 m→ A = r2 = G0.252 = 0.1963 m2

Flow Rate: Q = Av → v = Q/A = 1/0.1963 = 5.09 m/s

Smaller section: r = 0.125 m→ A = r2 = G0.1252 = 0.0491 m2

Flow Rate: Q = Av → v = Q/A = 1/0.0491 = 20.37 m/s

Additional note on Question 2

2

2

1

1

22

221

212211 r

r

v

vvπrvπrvAvA

If the diameter (radius) of the tube is halved

→ r1/r2 = 2 and v2/v1 = 4

i.e. the flow velocity in the smaller section will be four times the velocity in the larger section if the diameter is halved

As

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Bernoulli’s principle:

Where the velocity of a fluid is high, the pressure is low, and where the velocity is low, the pressure is high.

13-9 Bernoulli’s Equation

This makes sense, as a force is required to accelerate the fluid to a higher velocity.



Consider the work it takes to move a small volume of fluid from one point to another while its flow is laminar. Work must be done to accelerate the fluid, and also to increase its height. Conservation of energy gives Bernoulli’s equation:

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41

Bernoulli’s Equationdensity ρ

v1

v2

F1 = P1A1

F2 = P2A2

x1

x2

y1

y2 P1

P2

1 2

Consider a volume (V) of fluid moving from position 1 toposition 2 in time t

The mass of the fluid is m = ρV = ρA1x1 = ρA2x2

Using Conservation of Energy– The work done (W) by theforces F1 and F2 on the mass mwill equal the increase in totalenergy EK + EP

Note that the work done by F1 is positive and the work done byF2 is negative as it opposes the motion

density ρ

v1

v2

F1 = P1A1

F2 = P2A2

x1

x2

y1

y2 P1

P2

1 2

Work done on m:

W = W1 – W2 = F1x1 – F2x2 (Work = Force x Distance)

W = F1v1t – F2v2t (distance = velocity x time)

= P1A1v1t – P2A2v2t But Av = volume flow rate = V/t

ΔVPPΔVPΔVPΔtΔt

ΔVPΔt

Δt

ΔVPΔW 212121

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42

21

222

121

222

1212

1222

1K vvρΔvvvΔmΔmvΔmvΔE

121212P yyρΔVgyyΔmgΔmgyΔmgyΔE

Change in Kinetic Energy:

Change in Potential Energy:

1221

222

121

PK

yyρΔVgvvρΔVΔVPP

ΔEΔEΔW

Cancelling V and rearranging gives:

So

constantρgyρvPor

ρgyρvPρgyρvP2

21

2222

121

212

11

Bernoulli’s Equation

If there is no change in height (y = 0), then

P + ½ρv2 = constant

This shows that a fluid has its maximum pressure when it isstationary (v = 0) and as the flow velocity increases thepressure decreases.

Using Bernoulli’s Principle we can explain some commonphenomena.

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The ball is in a moving air stream where pressure is lower thanatmospheric pressure. If the ball moves slightly out of the moving airstream it will be moving to a region of higher pressure and the pressureforces will push the ball back into the stream – this is a stableequilibrium. Note that the vertical component of the upwards force dueto air pressure supports the weight of the ball.

Demonstration of Bernoulli’s Principle(Click image to show video)



Example 13-15: Flow and pressure in a hot-water heating system.

Water circulates throughout a house in a hot-water heating system. If the water is pumped at a speed of 0.5 m/s through a 4.0-cm-diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6-cm-diameter pipe on the second floor 5.0 m above? Assume the pipes do not divide into branches.

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m/s 1.183.1

0.25.0

Continuity ofEquation

m 0.5

?

cm 3.1

?2:Floor 2nd

0

Pa 103.0atm 3.0

cm 2.0

m/s 0.5:Basement

Pa 10Pa 101.013kg/m 1000

2

2

2

111

2

122211

2

2

22

22

1

51

12

11

1

55atm

3OH2

r

rvv

A

AvvAvA

y

P

rrA

v

y

P

rrA

v

P



atm 2.510

102.5

Pa 25000049000571300000

)0.50(8.910002

18.15.01000100.3

:sEquationBernoulli

5

5

225

2122

212

112

2222

121

212

11

yygvvPP

gyvPgyvP

'

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45


Using Bernoulli’s principle, we find that the speed of fluid coming from a spigot on an open tank is:

(Note: P1 = P2)

This is called Torricelli’s theorem.

13-10 Applications of Bernoulli’s Principle: Torricelli, Airplanes, Baseballs, TIA

or


Lift on an airplane wing is due to the different air speeds and pressures on the two surfaces of the wing.

13-10 Applications of Bernoulli’s Principle: Torricelli, Airplanes,

Baseballs, TIA

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46


A sailboat can move againstthe wind, using the pressure differences on each side of the sail, and using the keel to keep from going sideways.


Baseballs, TIA


A ball’s path will curve due to its spin, which results in the air speeds on the two sides of the ball not being equal; therefore there is a pressure difference.


Baseballs, TIA

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E.g. 1 – Curving of spinning ball in flight.

Diagrams show the air flow relative to the ball

(a) shows streamlines moving around the non-spinning ball

As the streamlines are compressed the air velocity increasesand the pressure decreases, however the reduction in pressureis the same on both sides of the ball and there is no net forcedue to pressure

(b) shows the situation when the ball is spinning

The spinning ball drags air in contact with its surface,increasing the air velocity at the top of the diagram and thusreducing the pressure.

At the bottom the air velocity will be reduced with acorresponding increase in pressure.

This difference in pressure causes a net deflection force asshown and the ball curves in flight.

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A person with constrictedarteries may experience a temporary lack of blood to the brain (TIA) as blood speeds up to get past the constriction, thereby reducing the pressure.


Baseballs, TIA


A venturi meter can be used to measure fluid flow by measuring pressure differences.


Baseballs, TIA

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Bernoulli’s principle also applies to any case of a spinning ball

e.g. cricket ball, golf ball tennis ball

Explain what happens if you hit a tennis ball with either top spinor back spin.

Can you explain the following?

Low Pressure System

Image courtesy the SeaWiFS Project, NASA/Goddard Space Flight Center, and ORBIMAGE URL:http://earthobservatory.nasa.gov/NaturalHazards/natural_hazards_v2.php3?img_id=2108

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Problems:

1. A severe low pressure system is known as a cyclone. Ifthe fastest winds in a cyclone are travelling at 200 km/h,what must be the pressure difference between that air andstationary air at 101.3 kPa?(The density of air is 1.29 kg/m3)For simplicity assume that the density of air is constantand that there is no variation in height.

m/s 55.6s3600

m 1000200

hr

km200

2222

121

212

11 ρgyρvPρgyρvP

Solution:

Convert 200 km/hr into m/s 1 km = 1000 m1 hr = 60 min = 60 x 60 = 3600 s

Using Bernoulli’s Equation:

The difference in pressure is (P1 - P2) if we consider v1 = 0 to bestationary air and v2 = 55.6 m/s to be the fast moving air.

Also assuming y1 = y2 gives:

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51

kPa 1.99Pa 1994055.61.29 221 2

1222

121 vvρPP

Note normal atmospheric pressure is = 101.3 kPa

1.99 kPa represents a drop in pressure of about 1.8%

Pressure where the air velocity is 200 km/h is about995 millibar (standard pressure = 1013 mb)

Image courtesy the SeaWiFS Project,NASA/Goddard Space Flight Center, and ORBIMAGEURL:http://earthobservatory.nasa.gov/NaturalHazards/natural_hazards_v2.php3?img_id=2108

Image Credit: ©BOM Commonwealth of Australia 2002, Bureau of Meteorology.URL:http://www.bom.gov.au/weather/national/charts/synoptic.shtml

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2. In order to maintain water pressure, some houses storetap water in a tank in the ceiling. If a water tank is in theceiling of a two storey house (9.0 m from ground level),what is the pressure of the water at ground level if thevolume flow rate is 1.5 litres per minute? The pipes in thehouse can be assumed to have a constant cross-section.

Solution:

Using Bernoulli’s equation:

2222

121

212

11 ρgyρvPρgyρvP

Since the flow rate is constant we can remove the velocity termfrom the equation. This gives:

P2 – P1 = ρg(y1 – y2) = 1000 x 9.8 (9 – 0) = 88200 Pa = 88.2 kPa

3.

2.00m

2.00m

2.00m

2.00m

A circular water tank(diameter 2.00 m) hasa small circular pipenear the base with adiameter of 15 mm.

(a) If the water tank contains fresh water with a density of998 kg/m3 what is the initial flow rate of water out of thepipe?

(b) When the water level has dropped to 1.00 m, what is theflow rate then?

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Solution:

(a) Equation of Continuity→ Flow Rate at Top of Tank = Flow Rate at Outlet

A1v1 = A2v2 (v2 is velocity at outlet)

112

112 vv

A

vAv 17778

0150

22

2

.

Bernoulli’s Equation: 2222

121

212

11 ρgyρvPρgyρvP

Pressure difference is very small so we can assume P1 = P2

Also v1 << v2 → v12 << v2

2 → can ignore v1 term

Relative to the outlet: y1 = 2 m and y2 = 0

m/s 6.329.822 : toreducesEquation sBernoulli' 12222

11 gyvρvρgy

(b) When y1 = 1 m:

m/s 4.419.822 : toreducesEquation sBernoulli' 12 gyv

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Further Examples in Fluid Flow

Aerodynamic Lift – Airflow over wing

A small plane has a mass of 300 kg and a wing surface area of30 m2. What is the pressure difference required to produceenough lift to enable the aeroplane to fly? Note that thepressure is lower on top of the wing as the air velocity is higher(streamlines are closer together).

You do not need to use Bernoulli’s Equation to calculate thepressures, rather you should work out the weight of the aircraftand the pressure difference required to lift that weight.

Weight of aircraft: W = mg = 300 x 9.8 = 2940 N

Pressure difference: P = F/A = 2940/30 = 98 Pa

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In this example assume the plane is cruising at 180 km/h= (180 x 1000)/(60 x 60) = 50 m/s.

The air velocity below the wing will be 180 km/h and the densityof the air is 1.29 kg/m3. What is the air velocity above the wingsurface to provide the necessary pressure difference of 98 Pa?

Using Bernoulli’s Equation:

2222

121

212

11 ρgyρvPρgyρvP

and noting that the height difference is negligible (y1 = y2), then Bernoulli’s Equation gives:

29.1

5029.1982 2

ρ

ρvPP2v

2121

2

km/hr 18560601051.5m/s 51.5 3

Stagnation Pressure

Stagnation streamline

Stagnation point

If we follow a streamline that ends on the nose of an aircraft wediscover the stagnation pressure – the maximum pressureobtainable along a streamline – given by P + ½ρv2

This is the point where all the kinetic energy (per unit volume) isconverted to pressure increase

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Dynamic Pressure

In Bernoulli’s equation ½ρv2 is often referred to as thedynamic pressure

As demonstrated in the previous questions, the dynamicpressure dominates lift in most circumstances (elevation effectsare small enough to be neglected)

The lift force is ~ the dynamic pressure x the planform area(area of wing viewed from above)

Some examples of lift force per unit area:

72 N/m2 for the Wright Brother’s plane

7200 N/m2 for a Boeing 747

50 N/m2 for a bee

Venturi Flow Meter

The principle of the venturi flow meter can be understood from the above diagram. Using Bernoulli’s Equation (with y1 = y2) and the Equation of Continuity gives:

222

12

212

11 ρvPρvP and since 1

2

122211 v

A

AvvAvA

1A

Aρ

ΔPv1

A

AρvPPΔP

2

1

122

212

121

21 2

2

P can be calculated from: P = ρgh

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A more practical version is shown at right

Problem:

A venturi flow-meter like the one shown above is used tomeasure the fluid velocity of air (density 1.29 kg/m3) as itpasses around an aircraft. What is the velocity of theaircraft given that:

The U-Tube is filled with mercury (ρHg = 13.6 x 103 kg/m3)h = 350 mm.The cross-sectional areas of the flow-meter are 10.0 cm2 and3.00 cm2 respectively.

Solution:

The pressure difference is: P = ρHg g h

= 13.6 x 103 x 9.8 x 350 x 10-3

= 46.6 x 103 Pa = 14.6 kPa

13

1029.1

106.46222

3

2

1AA

ρ

ΔPv

2

1

1

km/h 30460601084.5m/s 84.5 3

Note that in this case you can get away without convertingareas to m2 as they are only used as a ratio

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The Pitot tube is used as an air speed indicator on aircraft

ρ

ΔP2v

Sample Multichoice Questions – Pressure, Ideal Fluid Flow

1. Usually, atmospheric pressure(a) is equal to zero absolute pressure.(b) Is equal to negative gauge pressure.(c) Is equal to zero gauge pressure.(d) Is equal to positive gauge pressure.

2. Which of the following pressures is the highest?(a) Atmospheric pressure on the earth’s surface at

sea level.(b) Blood pressure in a major artery.(c) Atmospheric pressure on the moon’s surface.(d) Atmospheric pressure on top of Mount Everest.(e) Pressure in the pleural cavity.

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3. A nurse exerts a force of 30 N onto the top surface of asyringe plunger. If the contact area is 2.0 cm2 , then thepressure on the plunger as exerted by the nurse is:Note: 2 cm2 is equal to 0.0002 m2

(a) 30 / (0.00020) Pa. (b) 30 / (0.020) Pa.(c) 30 / (0.0002) N. (d) 30 / 2.0 Pa.(e) 30 / 2.0 newton per cm.

4. Aortal blood pressure has a typical value of 120 mmHg to130 mmHg at systole. This value(a) is a pressure above atmospheric pressure.(b) is a pressure less than atmospheric pressure.(c) is a negative gauge pressure.(d) is a static atmospheric gauge pressure.(e) can only be measured by use of a mercury based

device.

Note (c) has the wrong units

5. A force of 120 N is applied to a surface of area 3 m2.What is the pressure on the surface.(a) 40 Pa (b) 60 Pa(c) 3 / 120 Pa (d) 360 Pa(e) 3 x 120 x 10 Pa

6. The static pressure increases with depth in a fluidbecause:(a) the density of fluid increases with depth(b) of the weight of the fluid above(c) of the difference in the buoyancy force(d) the viscosity of fluid becomes larger(e) the increased diffusion produces a greater force.

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7. Blood flow speed in the capillaries is less than in the aortabecause:(a) the narrow capillaries have a higher flow resistance.(b) the narrow capillaries exert a higher friction force on

the blood.(c) the aorta is closer to the pulsing heart.(d) the total cross section area of ALL capillaries is greater

than the cross section area of the aorta.(e) the capillary has a smaller cross-sectional area.

8. Water moves along a tube which increases in cross-sectional area. As the water moves from narrow to widersections of the tube, then(a) increased friction between the water and tube wall

will decrease the speed in the narrow section.(b) water (volume) flow rate will be less in the narrow

section of the tube.(c) water speed will be greater in the narrow section.(d) water (volume) flow rate will be greater in the narrow

section.(e) water speed will not change from narrow to wider

section.

Equation of Continuity: A1v1 = A2v2

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9. Imagine holding two bricks under water. Brick A is justbelow the surface and Brick B is at a greater depth. Theforce required to hold brick B in place is:(a) Larger(b) The same as(c) Smallerthan the force required to hold brick A in place.

FTOT = weight – FB (FB does not depend on depth)

10. When a hole is made in the side of a container holdingwater, water flows out and follows a parabolic trajectory.If the container is dropped in free fall, the water flow:(a) Diminishes(b) Stops altogether(c) Goes out in a straight line(d) Curves upwards

There is no pressure difference in free fall

11. A container is filled with oil and fitted on both ends withpistons. The area of the left piston is 10 mm2 while theright is 10 000 mm2. What force must be exerted on theleft piston to keep a 10 000 N car on the right at the sameheight.

(a) 10 N(b) 100 N(c) 10,000 N(d) 106 N

10,000 N?F1

A1

F2

A2

N 1010000

1010000

2

121

2

2

1

1

A

AFF

A

F

A

F

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12. A 200 tonne ship enters the lock of a canal. The fitbetween the sides of the ship and the lock is so tight, thatless than 200 tonnes of water is left in the lock. Can theship still float if the quantity of water left in the lock ismuch less than the ships weight?(a) Yes, as long as the level of the water reaches the

waterline of the ship(b) No, the ship touches the bottom because it

weighs more than the water in the lock

13. Two cups are filled to the same level with water. One ofthe two cups has two ice cubes in it. Which weighs more?(a) The cup without ice cubes(b) The glass with two ice cubes(c) The two weigh the same

14. A lead weight is fastened on top of a large solid piece ofstyrofoam that floats in a container of water. Because ofthe weight of the lead, the water line is flush with the topsurface of the styrofoam. If the piece of styrofoam isturned up-sidedown so the weight is suspended beneath it:(a) The arrangement sinks(b) The water line is below the top surface of the

styrofoam(c) The water line is still flush with the top surface

15. A siphon has been started, fluid is now flowing through it.What is the velocity at the top of the tube in relation to thevelocity of the exiting fluid:

(a) The velocity is greater at the top(b) The velocity is less (but not zero) at the top(c) The velocity is zero at the top(d) The velocity is the same at the bottom

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Notes on question 14:

With the lead on top the buoyancy force is due to the weight of the water displaced by the styrofoam and this is equal to the total weight of the styrofoam + lead.

When the styrofoam is turned over (and initially held so that its surface is level with the water surface) the total weight is unchanged however the buoyancy force will increase due to the increased displacement due to the volume of the lead. There will therefore be a net upwards force as buoyancy > weight and the top of the styrofoam will rise above the water surface until equilibrium is attained.

In practice these equations are only an approximation as theyare based on an ‘ideal’ fluid

Real fluids have internal frictional forces called viscosity andare not incompressible

For low viscosity fluids this is a good approximation for mostliquids however gases are compressible

Bernoulli’s Equation gives qualitative results for gas flow;however the approximations are only good for low pressuredifferences, i.e. low flow velocities

In a real fluid the flow may be turbulent which furthercomplicates the issue

Real flow meters often use a pressure gauge to measurepressure difference and need to be calibrated in order to obtainaccurate results

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Real Fluid Flow

So far we have only studied the properties of ‘ideal’ fluids.Additional forces come into play when ‘real’ fluids are in motion.

The most important of these is viscosity.

Viscosity is the internal friction in a fluid which tends to preventit from flowing when subjected to an applied force

High-viscosity fluids resist flow; low-viscosity fluids flow easily.

The tendency of a fluid to drag adjacent layers of fluid alongwith it means that within a tube the velocity is greatest at thecentre and least (zero) at the tube wall (fluid molecules adhereto the wall).

Velocity profile in a tube

The flow can be considered in terms of concentric cylinders (a). A short time later the central cylinders have moved further (b).

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Real fluids have some internal friction, called viscosity.

The viscosity can be measured; it is found from the relation

13-11 Viscosity

gradient velocity theis lv

lv

lv

ηAF

The force F is proportional to the area A and the velocity gradient

The constant (Greek letter “eta”) is called the viscosity of the liquid and is measured in pascal second (Pa.s).

An apparatus for measuring viscosity is represented below

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Table of ViscositiesFluid Temperature (C) /10-3 (Pa·s) Air 0 0.0171 20 0.0182 40 0.0193 Carbon Dioxide 20 0.0147 Helium 20 0.0196 Whole Blood 37 4.0 Glycerine 20 1490 Methanol 20 0.584 Castor Oil 20 986 Mercury 20 1550 Water 0 1.78 20 1.00 40 0.656

The listing of values as /10-3 Pa.s indicates that the viscosity of mercury is 1550 x 10–3 Pa.s

Problem:

A metal plate of area 200 cm2 = 0.02 m2 moves at a constant speed of 100 mm/s = 0.1 m/s parallel to another much larger plate. The 0.1 mm = 10-4 m space between the plates is filled with an oil of viscosity 0.15 Pa.s. What is the force driving the moving plate?

v

oil

Solution: N 310

1.002.015.0

4 Δy

ΔvηAF

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The rate of flow in a fluid in a round tube depends on the viscosity of the fluid, the pressure difference, and the dimensions of the tube.

The volume flow rate is proportional to the pressure difference, inversely proportional to the length of the tube and to the pressuredifference, and proportional to the fourth power of the radius of the tube.

13-12 Flow in Tubes; Poiseuille’s Equation, Blood Flow


This has consequences for blood flow—if the radius of the artery is halfwhat it should be, the pressure has to increase by a factor of 16 to keep the same flow.

Usually the heart cannot work that hard, but blood pressure goes up as it tries.

13-12 Flow in Tubes; Poiseuille’s Equation, Blood Flow

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More on Flow in Tubes

Laminar Flow

The simplest type of flow is laminar flow

Laminar means “in layers”, and we imagine layers of fluidsliding smoothly over each other in laminar flow. Laminar flowis also called streamline flow

We know from experience that the flow of a fast flowing fluidoften breaks up and becomes turbulent

For example, water flowing in a river can develop eddies(whirlpools) when flowing past a submerged tree trunk, insteadof flowing in smooth curves

The diagram above and (a) at left showstreamlines. The arrows give the direction ofthe motion of the fluid particles.

Diagram (b) represents turbulent flow

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Flow in Tubes

Bernoulli’s equation shows that the pressure in a tube dependson the flow velocity and is greatest where the velocity is lowest.When viscosity is taken into account there is also a pressuredrop along the tube due to energy losses. These situations areillustrated below:

(a) represents an ideal (non-viscous) fluid and pressure (asindicated by the rise of liquid in the manometer tubes)depends on fluid velocity

(b) represents the case of a viscous fluid showing pressuredrop due to viscosity.

Poiseuille’s Equation

When a pressure difference causes fluid flow through ahorizontal cylindrical tube, the flow rate is limited by theviscosity of the fluid

If there was no viscosity, fluid would flow through a level tubewithout an applied force. Due to viscosity, a pressuredifference is required for flow

The pressure in a fluid at rest depends only on the depth.When the fluid is flowing the situation is more complex.Viscosity () causes the pressure to drop along the length ofthe tube. E.g. blood flow along an artery the pressure dropsalong the length of the artery. For a tube of radius R, length L,with pressure difference P = P1 – P2 across the ends, the flowrate (Q) is give by Poiseuille’s Equation:

L8η

PPπRQ

421

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Poiseuille’s Law could be applied to the case ofatherosclerosis (build up of fatty acid deposits in arteries)

If the deposits reduce the effective radius (or diameter) of (forexample) a coronary artery to ½ its normal size, then the flow ofblood is reduced to

rate flow normal theofth 161

21

21

21

21

In practice this will be partially compensated for by an increasein blood pressure to maintain sufficient flow through the artery.

Problem:

24. Engine oil (assume SAE 10) at 25°C with a viscosity of0.1 Pa.s passes through a fine 1.80 mm diameter tube ina prototype engine. The tube is 5.5 cm long. Whatpressure difference is required to maintain a flow rate of5.6 mL/min?

Solution:

L8η

PPπRQ

421

Pa 1992109π

109.330.0550.1844

8

421 πR

LQη8PPΔP

= 0.1 Pa·sD = 1.8 mm → R = 0.9 mm = 9 G 10-4 mL = 55 mm = 0.055 mQ = 5.6 mL/min →

s60

m 105.6

min

mL 5.6 36

Q = 9.33 G 10-8 m3/s

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SAE Viscosity

SAE represents the Society of Automotive Engineers viscosityclassification. It is the most widely used system to classify theviscosity of engine lubricants

There are two series of SAE viscosity, those with the suffix “W”for winter conditions and those without the “W”.Multigrade oils can meet two different temperaturespecifications (eg. 20W50) meets spec for ambienttemperatures of 20 and 50°C.

Concept Tests:

1. Water flows from a large tank into a pipe and then into a bucket.The pipe is then replaced by another one with the same lengthbut twice the diameter. The pressure difference between the twoends of the pipe is then:(a) the same as previously (b) doubled(c) halved (d) 1/16 of the originalvalue(e) 16 times the original value

P depends only on P1 at bottom of tank and P2 = Patm at outlet

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2. A piece of stainless steel tubing that delivers fluid at aconstant flow rate is replaced with a tube that has threetimes the diameter. What length must the new tube befor the flow rate to remain the same if the pressure dropalong the tube is the same?

(a) 3 times the length of the original piece(b) 9 times the length of the original piece(c) 81 times the length of the original piece(d) less than the length of the original piece(e) none of the other answers

114

1

4

1

22

2

42

1

41

2

42

21

41

1

813

88

LLLR

RL

L

R

L

R

L

PRQ

L

PRQ

Transition from Laminar to Turbulent Flow- Reynolds Number

The Reynolds Number is a single number that depends on allthe relevant flow parameters and can tell us approximatelywhen flow instabilities such as eddies and turbulence will occur

At low velocities the flow in a tube will be laminar. If thepressure difference (P) is increased the flow rate will increase,however at a critical point eddies are formed, and the flowbecomes turbulent

Once turbulence starts energy is lost to the eddies - theflow becomes noisy and much greater pressuredifferences are required to produce even a small increasein flow rate

Reynolds number is often expressed in different formsdepending on the situation being studied

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Flow Through Tubes

For flow through tubes Reynolds number is often expressed as

η

DvρN R

where: is the density of the fluid,

v is the average velocity of fluid flow,

D the diameter of the tube the viscosity of the fluid

In tubes it is found experimentally that:

NR < 2000 flow is laminarNR > 3000 flow is turbulent

2000 < NR < 3000 flow is unstable(may change from laminar to turbulent or vice versa)

Note that the Reynolds number is dimensionless (has no units)so the same value would be obtained in any consistent set ofunits

Problem:

The average speed of blood in the aorta (D = 20 mm) duringthe resting part of the heart’s cycle is about 300 mm/s.

Is the flow laminar or turbulent?

Solution:

1590104

02.03.010603

η

DvρNR

Since NR < 2000 flow would be laminar

ρblood = 1060 kg/m3

vD = 0.02 m

blood = 4 x 10-3 Pa·s

= 0.3 m/s

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The graph at left showshow the flow rate Qvaries with an increasein pressure across theends of a tube.The horizontal axis is thepressure gradient P/L.

The critical velocity vc at which the flow may become turbulentis given by:

ρD

ηv

ρD

η

ρD

ηNv

η

DvρN R

2000

2000c

c

cR

Once turbulence sets in the flow resistance increasesdramatically.

An example fromphysiology:

Curve I represents a normal artery while Curve II is the casefor an artery partly constricted by atherosclerosis (build up offatty acid deposits in arteries). Because of the presence of thedisk-shaped cells in whole blood, the real flow behaviour ismore complicated; but, to a first approximation the diagram isvalid and will serve as the basis for a discussion on themechanics of blood flow.

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Flow rates: Qa → Resting flow rateQb → Flow rate during exercise

For the normal healthy person the blood flow is laminar at bothQa and Qb

For the person suffering atherosclerosis the blood flow islaminar at Qa (resting) however the blood pressure required tomaintain the flow rate is greater than in the normal person.

If the flow rate is increased to Qb then the flow becomesturbulent and the heart must work even harder to maintain therequired blood pressure.

Flow Around Objects:

In considering flow around objects the Reynolds number isusually expressed in the form:

η

ρLvN R

where: ρ = densityL = characteristic lengthv = characteristic velocity = viscosity

For example, for flow past a post in a very wide channel, Lwould be given by the post diameter and v would be the flowvelocity far from the post.

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For low values of NR flow isessentially laminar

Illustration at right: NR = 0.16

Slight instability

Illustration at right: NR = 1.54

Once NR exceeds about 20, small vortices appear behind thecylinder. When NR is about 100, these vortices (eddies)periodically peel off and move with the flowing fluid.

Note that for every “clockwise” spiral (eddy) there is an“anticlockwise” spiral (eddy). These are more correctly called avortex and are remarkably stable entities.

For example, the shedding of vortices from the wings of a 747can affect aircraft travelling several hundreds of metres behind.

NR = 26 →

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When NR is larger still, the flow becomes chaotic (turbulent)behind the cylinder and eventually, a boundary layer betweenthe turbulent region and the laminar region develops

At still higher values of NR, more periodicities and morecomplicated behaviour emerges. The true nature of turbulentflow remains poorly understood, even by the best physicists!

In everyday life we see turbulent flow (e.g. smoke rising, waterfalling) which are both effects of increasing v rather than length L

NR = 2000 →

Drag Forces:

For low velocity flow (NR < 1) the drag force is proportional tovelocity → Fdrag v

For high velocity flow (NR > 1000) the drag force isapproximately proportional to v2

Stokes’ Law

When a spherical particle moves through a fluid, a frictionalforce due to the viscosity of the fluid acts on the particle.Stokes showed that under certain conditions, the drag force ona spherical particle is given by:

avFDrag 6In this equation F is a drag force on the particle, is the fluidviscosity, a is the radius of the sphere and v is the velocity ofthe particle (relative to the fluid).

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Limitations of Stokes’ Law

This form of Stokes’ Law assumes that the fluid velocity is thesame as the particle velocity at the surface of the particle – the‘stick’ boundary condition

Stokes law is only valid for spherical particles in a containermuch larger than the size of the sphere and for smallvelocities

Stokes law can be used to measure the viscosity of a liquid bydropping a small sphere into the liquid. The sphere must reachterminal velocity

v

Fw

FB Fdrag

2R

v

Fw

FB Fdrag

2R

Using Stokes’ Law to Measure Viscosity

Consider forces actingon a falling sphere asshown in diagram

FW = weight of sphere

FB = buoyancy force(Archimedes’ Principle)

FDrag = drag force(Stokes’ Law)

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v

Fw

FB Fdrag

2R

v

Fw

FB Fdrag

2R

Using Stokes’ Law to Measure Viscosity - 2

FW and FB are constant and FDrag increases as the velocityincreases until equilibrium is obtained when FW - FB - FDrag = 0

At equilibrium there is no acceleration and the sphere reachesterminal velocity vT

VgρmgF sW

VgρFB l

vaηπ6FDrag

Weight:

Buoyancy:

Stokes' Law:


ρs = density of sphere

ρl = density of liquid

Apply Newton’s Second Law: maFFF DragBW

at terminal velocity: v = vT and a = 0

Vgρρavηπ60avηπ6VgρVgρ sTTs ll

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80

334 πaV

gaρρ2vη93

gaπ4ρρvaηπ6 2

sT

3s

T ll

T

2s

v9

gaρρ2η l

η9

gaρρ2v

2s

Tl

Note that

Viscosity:

Terminal Velocity:


Aerodynamic drag

When an object moves through air so quickly that theassumption of laminar flow is violated, the drag force is nolonger directly proportional to the velocity

At high enough velocities, the fluid flow becomes turbulent andthe drag force is more nearly proportional to the square of thevelocity

In this case, the drag force is often written as

2AvρCD 21

where: A is the cross-sectional area of the body(perpendicular to the velocity vector)ρ is the density of airC is the drag coefficient

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Drag Coefficient

Typical values of drag coefficient (C)

Thin disk 1.10small truck 0.701988 BMW 735i 0.32

Problems:

1. A steel ball-bearing, 8.0 mm in diameter, is dropped into acylinder of glycerine, viscosity 1.49 Pa.s.The densities of steel and glycerine are 7.8 x 103 kg/m3

and 1.26 x 103 kg/m3 respectively. What is the terminalvelocity of the ball-bearing?

Solution:

m/s 0.15349.19

8.9)104)(12607800(2

9

2

23

η

gaρρv l

2s

T

ρs = 7800 kg/m3

ρl = 1260 kg/m3

a = 4 mm = 4 G 10-3 m = 1.49 Pa·s

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2. A bottle of light corn-syrup is taken from a refrigerator(T = 5°C) and a glass marble of density 2.5 x 103 kg/m3

is dropped into it. The marble takes 45 s to sink to thebottom.The diameter of the marble is 1.57 cm, the depth of theliquid is 12.1 cm, and its density is 1.2 x 103 kg/m3.Assuming that the average velocity of the marble isequal to its terminal velocity, what is the viscosity of thesyrup at that temperature?

Solution:

sPa 64.910689.29

8.9)1085.7)(12002500(2

9

2

3

23

T

2s

v

gaρρη l

At 5Cρs = 2500 kg/m3

ρl = 1200 kg/m3

a = 0.0157/2 = 7.85 G 10-3 ms = 0.121 m and t = 45 s vT = s/t = 0.121/45

= 2.689 G 10-3 m/s

3. Using the same data from the previous problem: If thebottle is kept out of the fridge for several hours and thenthe experiment is repeated, the marble takes 5 s to fallthrough the liquid. What is the viscosity of the syrup atroom temperature (T = 20°C)?

Solution:

You could do this the hard way by calculating vT andsubstituting data into equation as in previous question.

The easy way is to realise that the terminal velocity hasincreased by a factor of 9 and hence the viscosity will be1/9 of the previous answer. → = 69.4/9 = 7.2 Pa·s

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• Phases of matter: solid, liquid, gas

• Liquids and gases are called fluids.

• Density is mass per unit volume.

•Specific gravity is the ratio of the density of the material to that of water.

• Pressure is force per unit area.

• Pressure at a depth h is ρgh.

•External pressure applied to a confined fluid is transmitted throughout the fluid.

Summary of Chapter 13


• Atmospheric pressure is measured with a barometer.

• Gauge pressure is the total pressure minus the atmospheric pressure.

• An object submerged partly or wholly in a fluid is buoyed up by a force equal to the weight of the fluid it displaces.

• Fluid flow can be laminar or turbulent.

• The product of the cross-sectional area and the speed is constant for horizontal flow.


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• Where the velocity of a fluid is high, the pressure is low, and vice versa.

• Viscosity is an internal frictional force within fluids.

• Additional Topics:

• Transition from laminar to turbulent flow:

• Reynolds Number

• Stokes’ Law


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