Lecture Outlines Chapter 15 Physics, 3rd Edition James S. Walker

Lecture Outlines Chapter 15 Physics, 3rd Edition James S. Walker
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Chapter 15 Fluids

Units of Chapter 15 Density Pressure
Static Equilibrium in Fluids: Pressure and Depth Archimedes’ Principle and Buoyancy [Fb= ρfluidVg] (flytförmåga) Applications of Archimedes’ Principle

Units of Chapter 15 Fluid Flow and Continuity Bernoulli’s Equation
Applications of Bernoulli’s Equation Viscosity and Surface Tension

15-1 Density The density of a material is its mass per unit volume:

Conceptual Checkpoint 15-1
I ett kylskåp med måtten 1,0m•0,60m•0,75m, finns förutom luft, bara ett dussin ägg (44 g/styck). Väger luften << ≈ >> än äggen? Vluft = 1,0m•0,60m•0,75m = 0,45 m3 mluft = ρ • V = 1,29 kg/m3 • 0,45 m3 = 0,58 kg mägg = 12 • 0,044 kg = 0,53 kg Luften väger 10% mer än äggen dvs ≈

15-2 Pressure Pressure is force per unit area:

15-2 Pressure The same force applied over a smaller area results in greater pressure – think of poking a balloon with your finger and then with a needle.

Example 15-1 Popping a Balloon
Vad blir trycket på ytan av en balIong om man trycker med kraften 2,1 N och använder fingertoppen (A = 1,0•10-4 m2)? P = F/A = 2,1 N/(1,0•10-4 m2) = 21 kN/m2 en nål (A = 2,5•10-7m2)? P = F/A = 2,1 N/(2,5•10-7m2) = 8,4 MN/m2 Bestäm den minsta kraften som smäller ballongen (Pmin = 3,0•105 N/m2) om man använder nålen. Fmin = P•A = (3,0•105 N/m2)•(2,5•10-7m2) = 0,075 N

15-2 Pressure Atmospheric pressure is due to the weight of the atmosphere above us. The pascal (Pa) is 1 N/m2. Pressure is often measured in pascals.

15-2 Pressure There are a number of different ways to describe atmospheric pressure. In pascals: In bars: Exercise 15-1 (p.479) The force on the palm of your hand (8 cmx10 cm) F = PatA = 101 kPa (0,08 m • 0,10 m) = 810 N

Photo 15-2 Atmospheric pressure
The air around you exerts a force of about 14.7 pounds on every square inch of your body. Because this force is the same in all directions and is opposed by an equal pressure inside your body, you are generally unaware of it. An air splint utilizes the same principle of unequal pressure. A plastic sleeve is placed around an injured limb and inflated to a pressure greater than that of the atmosphere—and thus of the body’s internal pressure. The increased external pressure retards bleeding from the injured area, and also tends to immobilize the limb in case it might be fractured. (Air splints are carried by hikers and others who might be in need of emergency treatment far from professional medical facilities. Before it is inflated, the air splint is about the size and weight of a credit card.)

15-2 Pressure Since atmospheric pressure acts uniformly in all directions, we don’t usually notice it. Therefore, if you want to, say, add air to your tires to the manufacturer’s specification, you are not interested in the total pressure. What you are interested in is the gauge pressure – how much more pressure is there in the tire than in the atmosphere?

Example 15-2 Pressuring the Ball: Estimate the Gauge Pressure
Estimate the gauge pressure in a basketball by pushing down on it and noting the area of contact it makes with the surface on which it rests.

Example 15-2 Pressuring the Ball: Estimate the Gauge Pressure
”Rimliga”uppskattningar 22 N kraft och d = 2 cm? A = π•10-4 m2 Pg = F/A = 22 N/(π •10-4 m2) = 7,0•104 N/m2 Detta är trycket, som är ”utöver” lufttrycket, så det totala trycket är Ptotal = Pg + Pat = 7,0•104 N/m2 + 1,0(1)•105 N/m2

15-3 Static Equilibrium in Fluids: Pressure and Depth
The increased pressure as an object descends through a fluid is due to the increasing mass of the fluid above it.

Example 15-3 Pressure and Depth
A cubical box cm on a side is completely immersed in a fluid. At the top of the box the pressure is kPa; at the bottom the pressure is kPa. What is the density of the fluid?

Example 15-3 En kub med 20,00 cm sida är helt nedsänkt i en vätska. Trycket på ovansidan är 105,0 kPa och på undersidan 106,8 kPa. Vad är vätskans täthet? p2 = p1 + ρgh ρgh = N/m2 ρ = 1800 N/m2 / (9,81 m/s2•0,20 m) = 920 kg/m3 (skillnaden mellan trycket ger bara två siffrors noggranhet)

Conceptual Checkpoint 15–2 The Size of Bubbles Beror deras radie/volym på djupet (> < = ) ?
One day, while swimming below the surface of the ocean, you let out a small bubble of air from your mouth. As the bubble rises toward the surface, does its diameter (a) increase, (b) decrease, or (c) stay the same?

A barometer compares the pressure due to the atmosphere to the pressure due to a column of fluid, typically mercury. The mercury column has a vacuum above it, so the only pressure is due to the mercury itself.

This leads to the definition of atmospheric pressure in terms of millimeters of mercury: In the barometer, the level of mercury is such that the pressure due to the column of mercury is equal to the atmospheric pressure.

This is true in any container where the fluid can flow freely – the pressure will be the same throughout.

Example 15-4 Oil and Water Don’t Mix
A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has been added to one side, as shown in the sketch. The density of the water is 1.00 x 103 kg/m3, and the density of the vegetable oil 9.20 x 102 kg/m3. is If the depth of the oil is 5.00 cm, what is the difference in level h between the top of the oil on one side of the U and the top of the water on the other side?

Example 15-4 Ett U-rör har vatten (med tätheten 1000 kg/m3) och olja (med tätheten 920 kg/m3) i sina skänklar, enligt figur. Om h2 = 5,00 cm, vad är då h och h1? Öppna ändar innebär att trycket är detsamma pA + ρvgh1 = pB + ρogh2 h1 = h2ρo/ρv = 4,60 cm och då blir h = 0,40 cm

Pascal’s principle: An external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid.

Exercise 15-3 För att undersöka en bil med vikten N används en hydralisk lift. Om r1 = 4,0 cm och r2 = 17 cm, vad blir då F1? Trycket är detsamma så F1 /A1 = F2 /A2 F1= F2A1/A2 = F2(r1/r2)2 = 800 N

15-4 Archimedes’ Principle and Buoyancy
A fluid exerts a net upward force on any object it surrounds, called the buoyant force. This force is due to the increased pressure at the bottom of the object compared to the top. F1 = p1•A = p1•L2 p2 = p1 + ρgL F2 = p2•A = F1 + ρgL3

15-4 Archimedes’ Principle and Buoyancy
Archimedes’ Principle: An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object.

Figure 15-9 Buoyant force equals the weight of displaced fluid (den beror sålunda på föremålets VOLYM och inte dess massa) The buoyant force acting on the object in (a) is equal to the weight of the “fluid object” (with the same size and shape) in (b). This is because the fluid in (b) is at rest; hence the buoyant force acting on the fluid object must cancel its weight. The same forces act on the original object in (a), however, and therefore the buoyant force it experiences is also equal to the weight of the fluid object.

Conceptual Checkpoint 15–3 How Is the Scale Reading Affected
Conceptual Checkpoint 15–3 How Is the Scale Reading Affected? Om man stoppar sitt finger i vätskan enligt figur vad händer på skalan? Ökar, minskar eller oförändrad? A flask of water rests on a scale. If you dip your finger into the water, without touching the flask, does the reading on the scale (a) increase, (b) decrease, or (c) stay the same?

15-5 Applications of Archimedes’ Principle
An object floats when it displaces an amount of fluid equal to its weight.

Example 15-5 Measuring the Body’s Density
A person who weighs N in air is lowered into a tank of water to about chin level. He sits in a harness of negligible mass suspended from a scale that reads his apparent weight. He now exhales as much air as possible and dunks his head underwater, submerging his entire body. If his apparent weight while submerged is 34.3 N, find (a) his volume and (b) his density.

Example 15-5 Measuring the Body’s density
En person vars vikt (W) är 720,0 N i luft sänks ner helt och hållet i vatten. Hans vikt (Wa) mäts då till 34,3 N. Vad är a) personens volym och b) täthet? Standardkoordinatsystem ger Wa = W - Fb Fb = (personens volym)•(vattnets täthet)•g = 686 N Låt personens volym vara Vp och hans täthet ρp Vp= 686 N/(1000 kg m3 • 9,81 m/s2) = 0,0699 m3 ρp = (720,0 N/9,81 m/s2) / Vp = 1050 kg/m3 Men hur flyter man då? (Märk också 900/1100)

Active Example 15–1 Find the Tension in the String
A piece of wood with a density of 706 kg/m3 is tied with a string to the bottom of a water-filled flask. The wood is completely immersed, and has a volume of 8.00 x 10-6 m3. What is the tension in the string?

Active Example 15-1 Find the Tension in the String
Ett stycke trä med tätheten 706 kg/m3 och volymen 8,00 •10-6 m3 är nedsänkt i en vattentank enligt figur. Vad är spänning i tråden? Standardkoordinatsystem ger Fb = T + mg Fb = (trävolymen)•(vattnets täthet)•g = 78,5 •10-3 N m = ρ V = 5,65•10-3 kg T = Fb – mg = 78,5•10-3 N - 5,65 •10-3 kg • 9,81 m/s2) = = 23,1 mN

An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water.

Active Example 15-2 Floating a Block of Wood
Ett kubiskt stycke trä har tätheten 655 kg/m3 och sidan 15,0 cm. Hur mycket vatten undantrycks för att stycket skall flyta? Trävolymen V är 3,375 •10-3 m3 m = ρ V = 2,21 kg (så dess vikt är 21,7 N) Den (undanträngda) vattenvolym som krävs är den som precis kompenserar föremålets vikt dvs ρf Vsub g = 21,7 N Vsub = 2,21 kg/(1000 kg/m3) = 2,21•10-3 m3

Conceptual Checkpoint 15-4 The Plimsoll Mark Vilken av markeringarna gäller “Maximum load for saltwater”? On the side of a cargo ship you may see a horizontal line indicating “maximum load.” (It is sometimes known as the “Plimsoll mark,” after the nineteenth-century British legislator who caused it to be adopted.) When a ship is loaded to capacity, the maximum load line is at water level. The ship shown here has two maximum load lines, one for fresh water and one for salt water. Which line should be marked “maximum load for salt water”: (a) the top line or (b) the bottom line?

The fraction of an object that is submerged when it is floating depends on the densities of the object and of the fluid.

Example 15-6 The Tip of the Iceberg
What percentage of a floating chunk of ice projects above the level of the water? Assume a density of 917 kg/m3 for the ice and 1.00 x 103 kg/m3 for the water.

Tip of Iceberg (p.492) Hur mycket av ett flytande föremål är nedsänkt i en vätska? Anta att föremålet har tätheten ρs och flyter i en vätska med tätheten ρf. Om föremålet har volymen Vs är dess totala vikt Ws = ρs Vs g På samma sätt är den undantryckta vätskans vikt Wf = ρf Vf g om de sätts lika så att den (undanträngda) vattenvolym Vsub som krävs precis kompenserar föremålets vikt gäller 15-10 (volymen)•(tätheten) = (vätskans täthet)•(vätskevolymen) ρs Vs = ρf Vsub

Tip of Iceberg (p.492) Hur stor del av ett flytande isberg är ovan vattenytan? Isens tätheten (ρs) är 917 kg/m3 och den flyter i vatten med tätheten (ρf) 1000kg/m3. Ekvation ger direkt (volymen)•(tätheten) = (vätskans täthet)•(vätskevolymen) ρs Vs = ρf Vsub Vsub/Vs = 91,7 % (så 8,3 % ligger ovanför vattenytan)

Conceptual Checkpoint 15–5 The New Water Level I (> = <)?
A cup is filled to the brim with water and a floating ice cube. When the ice melts, which of the following occurs? (a) Water overflows the cup, (b) the water level decreases, or (c) the water level remains the same.

Conceptual Checkpoint 15–6 The New Water Level II (>=<)?
A cup is filled to the brim with water and a floating ice cube. Resting on top of the ice cube is a small pebble. When the ice melts, which of the following occurs? (a) Water overflows the cup, (b) the water level decreases, or (c) the water level remains the same.

15-6 Fluid Flow and Continuity
Continuity tells us that whatever the volume of fluid in a pipe passing a particular point per second, the same volume must pass every other point in a second. The fluid is not accumulating or vanishing along the way. This means that where the pipe is narrower, the flow is faster, as everyone who has played with the spray from a drinking fountain well knows.

15-6 Fluid Flow and Continuity (Δm=ρΔV=ρAvΔt)

15-6 Fluid Flow and Continuity
Most gases are easily compressible; most liquids are not. Therefore, the density of a liquid may be treated as constant, but not that of a gas.

Example 15-7 Spray I Water travels through a 9.6-cm diameter fire hose with a speed of 1.3 m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

Example 15-7 Spray I Vatten går genom en brandspruta, 9,6 cm i diameter med hastigheten 1,3 m/s. I änden på slangen passerar vattnet genom ett munstycke med diametern 2,5 cm. Vad är vattnets hastighet genom munstycket? Ekvation ger direkt v2 A2 = v1A1 v2 = v1A1/A2 eftersom A = πd2/4 v2 = v1(d1/d2)2 v2 = 1,3 m/s (9,6/2,5)2 = 19 m/s

15-7 Bernoulli’s Equation
When a fluid moves from a wider area of a pipe to a narrower one, its speed increases; therefore, work has been done on it.

Figure Work done on a fluid element ΔW1 = F1Δx1 = P1A1Δx1 på samma sätt fås ΔW2= - P2A2Δx2 Volymen ändras inte på en inkompressibel vätska så att V1 = V2 = V Detta innebär att det totala arbetet på ett volymselement V är ΔWtotal = ΔW1 + ΔW2 = P1ΔV – P2ΔV = (P1- P2)ΔV som ju är lika stort som ändringen i kinetisk energi dvs K2 –K1 As an incompressible fluid element of volume ΔV moves from pipe 1 to pipe 2, the pressure P1 does a positive work P1 ΔV and the pressure P2 does a negative work P2 ΔV. Since P1 is greater than P2 the net result is that positive work is done, and the fluid element speeds up.

The kinetic energy of a fluid element is: Equating the work done to the increase in kinetic energy gives: (15-14)

Example 15-8 Spray II Referring to Example 15–7, suppose the pressure in the fire hose is 350 kPa. What is the pressure in the nozzle?

Example 15-8 Spray II Vatten går genom en brandspruta, 9,6 cm i diameter med hastigheten 1,3 m/s. I änden på slangen passerar vattnet genom ett munstycke med diametern 2,5 cm. Trycket i slangen är 350 kPa? Hur stort är trycket i munstycket? Exemple Spray I gav hastigheten 19 m/s i munstycket. Ekvation ger direkt p1 + ρv12/2 = p2 + ρv22/2 p2 = p1+ ρ(v12 - v22)/2 p2 = 350 kPa – 180 kPa = 170 kPa

If a fluid flows in a pipe of constant diameter, but changes its height, there is also work done on it against the force of gravity. Equating the work done with the change in potential energy gives: (15-15)

Exercise 15-4 Vatten går genom en trädgårdsslang som går över ett 20,0 cm högt trappsteg. Om trycket i slangen är 143 kPa före steget, vad är det då efter? Ekvation ger direkt p1 + ρgy1 = p2 + ρgy2 p2 = p1 - ρg(y2 – y1) p2 = 143 kPa kg/m3 • 9,81 m/s2 • 0,200 m = 141 kPa

The general case, where both height and speed may change, is described by Bernoulli’s equation: This equation is essentially a statement of conservation of energy in a fluid.

Active Example 15–3 Find the Pressure Information som Exercise 15-4 dvs Vatten går genom en trädgårdsslang som går över ett 20,0 cm högt trappsteg. Om trycket i slangen är 143 kPa före steget, vad är det då efter? Dessutom att arean på den övre slangen är hälften av den undre, samt att hastigheten vid det nedre steget är 1,20 m/s. Repeat Exercise 15–4 with the following additional information: (a) the cross-sectional area of the hose on top of the step is half that at the bottom of the step, and (b) the speed of the water at the bottom of the step is 1.20 m/s.

Active Example 15-3 Kontinuitetsekvationen (15-12) ger direkt att
v2 = v1(A1/A2) = 2v1 = 2,4 m/s Bernoullis ekvation (15-16) ger sedan att p1 + ρgy1 + ρv12/2 = p2 + ρgy2+ ρv22/2 p2 = p1+ ρg(y1 - y2) + ρ(v12 - v22)/2 p2 = 143 kPa – 2,0 kPa – 2,2 kPa = 139 kPa

15-8 Applications of Bernoulli’s Equation
The Bernoulli effect is simple to demonstrate – all you need is a sheet of paper. Hold it by its end, so that it would be horizontal if it were stiff, and blow across the top. The paper will rise, due to the higher speed, and therefore lower pressure, above the sheet.

Figure 15-18 Airflow and lift in an airplane wing
Cross section of an airplane wing with air flowing past it. The wing is shaped so that air flows more rapidly over the top of the wing than along the bottom. As a result, the pressure on top of the wing is reduced, and a net upward force (lift) is generated.

Figure 15-19 Force on a roof due to wind speed
Wind blows across the roof of a house, but the air inside is at rest. The pressure over the roof is therefore less than the pressure inside, resulting in a net upward force on the roof.

Exercise 15-5 Under en storm blåser vinden med hastigheten 35,5 m/s över ett hus med platt tak. Vad är tryckskillnaden mellan insidan av taket och ovanpå detsamma? Bernoullis ekvation (15-16) ger att p1 + ρgy1 + ρv12/2 = p2 + ρgy2+ ρv22/2 lägeskoordinaten är (nästan) densamma och v1 ≈ 0 så p1 – p2 = ρ(v22)/2 = 630 kPa

This lower pressure at high speeds is what rips roofs off houses in hurricanes and tornadoes, and causes the roof of a convertible to expand upward. It even helps prairie dogs with air circulation!

Figure 15-21 An atomizer (gust = vindstöt, orifice = mynning, öppning)
The operation of an “atomizer” can be understood in terms of Bernoulli’s equation. The high-speed jet of air created by squeezing the bulb creates a low pressure at the top of the vertical tube. This causes fluid to be drawn up the tube and expelled with the air jet as a fine spray.

Conceptual Checkpoint 15–7 A Ragtop Roof (buktar upp, =, ned?)
A small ranger vehicle has a soft, ragtop roof. When the car is at rest, the roof is flat. When the car is cruising at highway speeds with its windows rolled up, does the roof (a) bow upward, (b) remain flat, or (c) bow downward?

If a hole is punched in the side of an open container, the outside of the hole and the top of the fluid are both at atmospheric pressure. Since the fluid inside the container at the level of the hole is at higher pressure, the fluid has a horizontal velocity as it exits.

Torricellis lag (p.500) Vi vill bestämma hastigheten på en vätska som strömmar ut ur ett hål i en behållare enligt figur Bernoullis ekvation (15-16) tillämpad vid punkterna 1 och 2 ger att p1 + ρgy1 + ρv12/2 = p2 + ρgy2+ ρv22/2 p1 är atmosfärstryck men det är också p2 och v1 ≈ 0 så ρgh = ρ(v22)/2 som ger som kallas Torricellis lag v2 = (2gh)1/2

If the fluid is directed upwards instead, it will reach the height of the surface level of the fluid in the container.

Example 15-9 A Water Fountain
In designing a backyard water fountain, a gardener wants a stream of water to exit from the bottom of one tub and land in a second one, as shown in the sketch. The top of the second tub is m below the hole in the first tub, which has water in it to a depth of m. How far to the right of the first tub must the second one be placed to catch the stream of water?

Example 15-0 A Water Fountain
En trädgårdsmästare vill designa en vattenfontän enligt figur. H = 0,500 m, h = 0,150 m så hur lång är sträckan D? Standard y/x Toricellis lag ger hastigheten på den utströmmande vätskan. vx = (2gh)1/2 = 1,72 m/s y = y0 + vyt + ayt2/2 ger att 0 = H – gt2/2 t = (2H/g)1/2 = 0,319 s x = x0 + vxt + axt2/2 ger för x = D = 0,549 m

15-9* Viscosity and Surface Tension
Viscosity is a form of friction felt by fluids as they flow along surfaces. We have been dealing with nonviscous fluids, but real fluids have some viscosity. A viscous fluid will have zero velocity next to the walls and maximum velocity in the center.

It takes a force to maintain a viscous flow, just as it takes a force to maintain motion in the presence of friction. A fluid is characterized by its coefficient of viscosity, η. It is defined so that the pressure difference in the fluid is given by:

Using this to calculate the volume flow rate yields: Note the dependence on the fourth power of the radius of the tube!

A molecule in the center of a liquid drop experiences forces in all directions from other molecules. A molecule on the surface, however, experiences a net force toward the drop. This pulls the surface inward so that its area is a minimum.

Since there are forces tending to keep the surface area at a minimum, it tends to act somewhat like a spring – the surface acts as though it were elastic.

This means that small, dense objects such as insects and needles can stay on top of water even though they are too dense to float.

Summary of Chapter 15 Density: Pressure: Atmospheric pressure:
Gauge pressure: Pressure with depth:

Summary of Chapter 15 Archimedes’ principle:
An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object. Volume of submerged part of object:

Summary of Chapter 15 Equation of continuity: Bernoulli’s equation:
Speed of fluid exiting a hole in a container a depth h below the surface:

Summary of Chapter 15 A pressure difference is required to keep a viscous fluid moving:

Lecture Outlines Chapter 15 Physics, 3rd Edition James S. Walker

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Lecture Outlines Chapter 15 Physics, 3rd Edition James S. Walker

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