G12 Physics Unit 3 - Fluid Mechanics PDF

Unit 3 Fluid Mechanics Introduction Matter most commonly exists as solids, liquids or gases. Liquid and gas are both Brainstorming fluids: in contrast to solids they lack the ability to resist deformation. As a result question 3.1 fluid moves, or flows under the action of the force. Flu...

Unit 3 Fluid Mechanics Introduction Matter most commonly exists as solids, liquids or gases. Liquid and gas are both Brainstorming fluids: in contrast to solids they lack the ability to resist deformation. As a result question 3.1 fluid moves, or flows under the action of the force. Fluid mechanics refers to the 1. Why do over- study of fluid behavior at rest and in motion. Fluid mechanics has a wide range of inflated balloon applications in mechanical and aerodynamic engineering, in biological systems, bursts? and in many more fields. 2. How does a plane move upwards, At the end of this unit, you will be able to: against gravity? Develop knowledge and understanding of the concepts related to fluids and pressure. Gain knowledge and understanding of Pascal’s principle and Archimedes’ principle. Understand the behaviors of fluid flow. Develop skills of solving problems related to fluid mechanics. 3.1 Fluid Statics At the end of this section, you will be able to: Explain the properties of solids, liquids and gases. Define pressure. 69 70 Unit 3 Fluid Mechanics Convert values of pressure from one pressure unit to another. Explain causes of pressure in gas and liquids. Define absolute, atmospheric and gauge pressure. Define density and relative density. Solve problems related to pressure, density and specific gravity. Brainstorming question 3.2 Properties of solids, liquids and gases Why are gases easy Atoms in solids are very close to each other. The forces between them acts as a to compress, while spring that allow the atoms to vibrate without changing positions relative to their liquids and solids neighboring atoms (Figure 3.1 a). Thus, a solid resist all types of stress because are almost the atoms are not able to move about freely. Solids also resist compression, incompressible? because their atoms are relatively fixed distance apart. Under compression, the atoms would be forced into one another. Figure 3.1 (a) Atoms in a solid always have the same neighbors, held near equilibrium position by forces represented here by springs. (b) Atoms in a liquid are also in close contact but can slide over one another. Forces between them strongly resist attempts to push them closer together and also hold them in close contact. (c) Atoms in a gas are separated by distances that are considerably larger than the size of the atoms themselves, and they move about freely. The molecular spacing in the liquid phase is not much different from that of the 3.1 Fluid Statics 71 solid phase except the molecules are no longer at fixed positions relative to each other and they can rotate and translate freely. In a liquid, the intermolecular forces are weaker relative to solids, but still strong compared with gases. Liquids deform easily when stressed and do not spring back to their original shape once the force is removed because the atoms are free to slide about and change neighbors. That is, they flow (so they are a type of fluid), with the molecules held together by their mutual attraction (Figure 3.1b). Atoms in gases are separated by distances that are large compared with the size of the atoms (Figure 3.1c). The forces between gas atoms are therefore very weak, except when the atoms collide with one another. Gases thus not only flow but they are relatively easy to compress because there is much space and little force between atoms. A gas also expands until it encounters the walls of the container and fills the entire available space (Figure 3.2a). In contrast, if we move the liquid filling a small container to a much larger container, the liquid volume remains the same independent of the container’s shape (Figure 3.2b). Solids maintain not only their volume but also their shape. The properties of solids, liquids and gases can be observed from the following simulation link: https://phet.colorado.edu/ Figure 3.2 (a) The same gas fills sims/html/states-of-matter-basics/latest/states-of-matter-basics_en.html. completely different volume (b) The liquid volume remains the Activity 3.1: Compressibility of gases and liquids same regardless of volume and shape of container. Materials required: Two syringes of equal size and water. Procedures: Fill a first syringe with water by pushing the piston into the syringe casing until it stops. Put the open end of the syringe in to water and then pull water in to it by pulling the piston back to its maximum limit. Close the water filled syringe tightly by your finger and push the piston. Notice how far does the piston move. Fill a second syringe with air by pushing the piston into the syringe casing until it stops. 72 Unit 3 Fluid Mechanics Pull air in to it by pulling the piston back to its maximum limit. Close the air filled syringe tightly by your finger and push the piston. Notice how far does the piston move. Precaution: (i) Please do not use a syringe with needle to keep yourself safe. (ii) Closed the water and air filled syringe tightly so that no water and air comes out from the syringes while you are pushing the piston, Discussion question 3.1 Questions: (i) How far does a piston move in water and air filled syringe? (i) Explain the (ii) What can you say about the compressibility of water and air from the difference between solids, liquids distance moved by the piston? and gases in terms of distance Fluid statics deals with nature of fluids at rest. The fluid can be either gaseous or between particles, liquid. In fluid statics, there is no relative motion between adjacent fluid layers, intermolecular and no shear (tangential) stresses. The only stress in fluid statics is the normal force, their stress, which is the pressure. shape and their compressibility. Fluid statics is used to determine the forces acting on floating or submerged (ii) What is the bodies and the forces developed by devices like hydraulic presses and car jacks. difference between fluids and solids? The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on their surfaces using fluid statics. Pressure in Fluid Brainstorming question 3.3 (i) Why does a mattress sink more when a person is standing on it compared with when he is lying (Figure 3.4) ? (ii) Why Scissors, blade, saw and knives cut objects easily? (iii) Why nails, syringes and pins penetrate into objects easily? Figure 3.3 The normal stress The normal component of a force acting on a surface per unit area is called the and shear stress at the surface normal stress, and the tangential component of a force acting on a surface per of a fluid element. For fluids at rest, the shear stress is zero and unit area is called shear stress (Figure 3.3). For fluid at rest, the shear stress is zero pressure is the only normal stress. and the only existing stress is the normal stress and is called pressure. 3.1 Fluid Statics 73 Activity 3.2: Hold a sharp pencil between your two fingers and start pressing it from both sides. Notice how you feel on Figure 3.4 Effect of lying and standing on a mattress. both the fingers. Do Pressure is defined as a normal force exerted by a fluid (or a solid) per unit area. you feel pain in any If F is the magnitude of the force exerted on the fluid (or solid) at a particular one of your fingers? point and A is the surface area at which this force is applied, the pressure P at Notice the finger in this particular point is defined as the ratio of the force to the contact area A over which pencil leaves a deeper mark. which that force is exerted: F P= (3.1) A Pressure is a scalar quantity because it is proportional to the magnitude of the force. If a large force acts on a small area, the pressure is large. The unit of pressure is newtons per square meter (N/m2 ) in the SI system. Another name for the SI unit of pressure is the pascal (Pa): 1P a = 1N /m 2 (3.2) In addition to pascal there are other units of pressure such as millimeter mercury (mmHg), torr, atmosphere (atm) and pounds per square meter (psi) with their Figure 3.5 Effect of force varies relation shown as follow: with area 1 at m = 760 mmH g = 760 t or r = 101.3 K P a = 14.7 psi (3.3) Example 3.1 As a woman walks, her entire weight is momentarily placed on one of her shoes. Calculate the pressure exerted on the floor by the shoe if it has an average width 10 cm and average length of 30 cm and the woman’s mass is 55.0 kg. 74 Unit 3 Fluid Mechanics Solution: (a) Assuming the shoe is rectangular, its are is obtained by the relation A = w × l = 10 cm × 30 cm = 300 cm 2 = 0.03 m 2 The force exerted on the floor is the wait of the woman, which is obtained from the relation F = mg = 55.0 kg × 9.8 m/s 2 = 539.00 N Therefore, the pressure F 539.00 N P = = = 1.8 × 101 K P a A 0.03 m 2 Exercise 3.1 Repeat example 3.1, Example 3.2 if the woman stands Express the pressure obtained in example 3.1 in atm, torr, mmHG and psi. on her two shoes. Compare the result Solution: with example 3.1. To express the pressure in other pressure units we have to use relations in Eq 3.3. What can you say about the difference To convert pressure in pascal to other units: between the results? P cons × P pasc al Assume the shoes P ot her = 101.3 K P a have equal area where P cons is the constant value of pressure in other system of units excluding and her weight is pascal, P pasc al is the given value of pressure in pascal, P ot her is the value of balanced on her two pressure in the new system of unit. Here P pasc al is 1.8 × 101 K P a. Therefore, shoes. 1 at m × P pasc al 1 at m × 1.8 × 101 K P a P at m = = = 0.18 at m; 101.3 K P a 101.3 K P a 760 mmH g × P pasc al 760 mmH g × 1.8 × 101 K P a P mm H g = = = 135.04 mmH g ; 101.3 K P a 101.3 K P a 760 t or r × P pasc al 760 t or r × 1.8 × 101 K P a P t or r = = = 135.04 t or r ; 101.3 K P a 101.3 K P a 14.7 psi × P pasc al 14.7 psi × 1.8 × 101 K P a P psi = = = 2.61 psi ; 101.3 K P a 101.3 K P a 3.1 Fluid Statics 75 Example 3.3 Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of 3.00 × 109 N /m 2. Solution: From the definition of pressure P = F /A, the force F is F = P A. Here, the pressure P is given, and the area of the end of the nail A is given by ³ ´2 d A = π 2. Since d = 1.00 mm = 1x10−3 m. Thus, A = (3.14)(5x10−4 m)2 = 7.85x10−7 m 2 F = P A = (3.00 × 109 N /m 2 )(7.85x10−7 m 2 ) = 2.36 × 103 N The result shows that the magnitude of pressure is far greater than the force applied. Brainstorming The reason is that the area of the tip of the nail is very small. This is why nails are question 3.4 easily hammered into a wood and concrete. Try to crush air- filled balloon a little bit. Can you feel the Pressure in Gases balloon resisting the crushing, as if As air particles move randomly in space, they eventually collide with the solid something inside surfaces of any objects in that space. In each of these collisions, the particle it pushes back on exerts an impulsive force on the object (see Figure 3.6). your fingers. How can we explain this “resistance” of However, when a huge number of particles bombard a solid surface at a constant the air inside the rate, these collisions collectively exert an approximately constant force on the balloon,and how object (Figure 3.6 b). This impulsive force must be what we feel when we are pressure of the gas is trying to squeeze the balloon. formed? 76 Unit 3 Fluid Mechanics Figure 3.6 a) Each particle exerts an impulsive force on the wall (b) collective collision of gas particles on a wall. As we blow a balloon, we add air particles to the interior of the balloon; thus there are more particles inside colliding with the walls. This greater collision rate results in a larger outward average force on each part of the balloon’s surface, causing it to expand outward. The force exerted by the gas on the walls of the container per unit contact area gives the pressure of the gas. Figure 3.7 Absolute, gauge, and vacuum pressures. The actual pressure at a given position is called the absolute pressure, and it is 3.1 Fluid Statics 77 measured relative to absolute vacuum (i.e., absolute zero pressure). Most pressure-measuring devices, however, are calibrated to read zero in the atmosphere, and so they indicate the difference between the absolute pressure and the local atmospheric pressure. This difference is called the gauge pressure ( P g ag e ). Absolute and gauge pressures are related to each other by P g ag e = P abs − P at m (3.4) This is illustrated in Figure 3.7. Like other pressure gauges, the gauge used to measure the air pressure in an automobile tire reads the gauge pressure. Therefore, the common reading of 32.0 psi indicates a pressure of 32.0 psi above the atmospheric pressure. Figure 3.8 Measuring a gauge pressure of a car. Example 3.4 What is the absolute pressure at a location where the atmospheric pressure is 14.3 psi and the gauge pressure of an automobile tire is 32.0 psi? Solution: The absolute pressure is determined by rearranging Eq. 3.4 as: Exercise 3.2 P abs = P at m + P g ag e = 14.3 psi + 32.0 psi = 46.3 psi Determine the absolute pressure where gauge pressure is 61.152 Example 3.5 KPa and atmospheric pressure is 14.0 psi. The absolute pressure in water at a depth of 8 m is read to be 175 kPa. Determine the local atmospheric pressure if the guage pressure at this depth is 78.4 KPa. 78 Unit 3 Fluid Mechanics Solution: The absolute pressure in a water at a depth of 8 m is given as 175 K P a The local atmospheric pressure is to be determined. It can be obtained from Eq. 3.4 P l oc.at m = P abs − P g aug e = 175 K pa − 78.4 K P a = 96.6 K P a We can observe that the local atmospheric pressure is less than the atmospherics pressure at sea level. Brainstorming question 3.5 Density Why some objects Density is an important characteristic of substances. It is crucial, for example, in float and others determining whether an object sinks or floats in a fluid. It directly affects sink? pressure of fluids (gases and liquids). Density is a much more useful physical quantity for gases. Density, ρ, is the mass, m, per unit volume, V , of any object. It is calculated by dividing the mass of an object by its volume. m ρ= (3.5) V The unit of density is kg/m3. Discussion Exercise 3.3 question 3.2 Density measures the mass of one cubic meter of a substance. For example, at t An iron ball with sea level and 0◦C the mass of 1.0 m3 of air is 1.3 kg. We say that the density of air is radiusthree When 5.0 cm has liquids a mass of 2.0 kg. 1.3 kg/m3. If we had 2.0 m3 of air at sea level, its mass would be 2.6 kg. Its density with densities Determine ρ1 < ρ2 < theρball’s is still 1.3 kg/m3. 3 are density.together, mixed which liquid will be The density of most gases is proportional to pressure and inversely proportional to at the top, middle temperature. Liquids and solids, on the other hand, are essentially incompressible and bottom layer? substances, and the variation of their density with pressure is usually negligible. Example 3.6 Assume the following about a person: mass is 80 kg; dimensions are 1.8 m tall, 0.3 m wide, and 0.1 m thick; and volume is V = 1.8 m × 0.3 m × 0.1 m = 0.054 m 3 , 3.1 Fluid Statics 79 What is the density of this person. Solution: The person’s density is given by Eq. 3.5: m ρ= V Substitute the person’s mass and volume into the above to get 80 kg ρ= = 1500 kg /m 3 0.054 m 3 Table 3.1 Densities of some common substance at standard temperature (◦C) and pressure (1 atm) Substance ρ (kg /m 3 ) substance ρ(kg /m 3 ) Air 1.29 Iron 7.86 x103 Aluminum 2.70x103 carbon Steel 7.84 x103 Benzene 0.879x103 mercury 13.6 x103 Brass 8.4 x103 oxygen 1.43 copper 8.92 x103 nitrogen 1.25 Fresh water 1.00 x103 Sea water 1.03 x103 Gold 19.3 x103 silver 10.5 x103 Ice 0.917 x103 Tin 7.30 x103 Figure 3.9 Less dense materials As you can see from Table 3.1, the density of an object may help identify its floats on denser matter. composition. The density of gold, for example, is about 2.5 times the density of iron. The density of iron is about 2.5 times the density of aluminum. Density also indicates about the phase of the matter and its particles arrangement. The densities of liquids and solids are roughly comparable, consistent with the fact that their atoms are in close contact. Understanding density allows us to answer questions like: “why does oil form a film on water?”. If you pour oil into water or water into oil, they form layers (see Figure 3.9 a). Independently of which fluid is poured first, the layer of oil is always on top of the water. The density of oil is less than the density of water. Similar phenomena occur with gases. Helium-filled balloons accelerate upward in air while air-filled balloons accelerate (slowly) downward. The mass of helium 80 Unit 3 Fluid Mechanics atoms is much smaller than the mass of any other molecules in the air. At the same pressure and temperature, atoms and molecules of gas have the same concentration; because helium atoms have much lower mass, their density is lower. The air-filled balloon must be denser than air. The rubber with which the skin of any balloon is made is denser than air. We can ignore the slight compression of the gas by the balloon, because even though it increases the density of the gas, the effect is the same for both the air and helium in the balloons. You can study more about density from the simulation: https://phet.colorado.edu/sims/html/density/latest/density_en.html. Relative density Sometimes the density of a substance is given relative to the density of an another substance. Then it is called specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4◦ C, for which ρ H2 O = 1000kg /m 3 ). That is; ρ specific gravity = SG = ρ H2 O Note that the specific gravity of a substance does have any unit. In SI units, the numerical value of the specific gravity of a substance is exactly equal to its density in g/cm3. For example the density of mercury at 20◦ C is 13.6 x103 kg/m3 ). Its relative density is 13.6 x103 kg/m3 divided by density of water, 1 x103 kg/m3 , which is 13.6. The density of mercury in g/cm3 is also 13.6. The specific gravities of some substances at 20◦ C are given in Table 3.2. Note that substances with specific gravities less than 1 are lighter than water, and thus they would float on water. Ideal gas equation The simplest and best-known equation in the gas phase that relate density and pressure of gases is the ideal-gas equation of state, expressed as m PV = nRT = M RT = mR speci f i c T 3.1 Fluid Statics 81 Table 3.2 The specific gravity of some substances at 20◦C and 1 atm unless stated otherwise Substance SG Substance SG Air 0.00129 Seawater 1.025 Blood (at 37◦C) 1.06 Gasoline 0.68 Ethyl alcohol 0.790 mercury 13.6 Gold 19.3 bones 1.7-2.0 Ice (0◦C) 0.916 This gives us m P= R speci f i c T = ρR speci f i c T (3.6) V where P is the absolute pressure, V is the gas volume, n is number of mole, T is m the thermodynamic (absolute) temperature, ρ = V is the density, and R speci f i c = R/M is the specific gas constant. The specific gas constant is different for different gases and R is the universal gas constant whose value is R = 8.314 J/mol. K and M is the molar mass of the gases. The relation of pressure, volume, temperature and number of moles of gases can be observed from the simulation: https://phet.colorado.edu/sims/html/ gas-properties/latest/gas-properties_en.html. Example 3.7 Determine the density, specific gravity, and mass of the air in a room whose Figure 3.10 Dimension of air at ◦ dimensions are 4 m × 5 m × 6 m at 100 kPa and 25 C (Figure 3.10). 25◦C and 100 kPa Solution: Air can be treated as an ideal gas. The gas constant of air is R/M = 0.287 kPa·m3 /kg·K. The density of the air is determined from the ideal-gas relation P = ??(R/M)T to be P 100 K pa ρ= = = 1.17 kg /m 3 (R/M ) T (0.287 K P a.m 3 /kg.K ) (25 + 273.15) K Note that the temperature is converted to (absolute) unit K from (relative) unit ◦ C before using it in the ideal-gas relation. 82 Unit 3 Fluid Mechanics Then the specific gravity of the air becomes ρ 1.17kg /m 3 SG = = = 0.00117 ρ H2 O 1000 kg /m 3 The volume of the air is equal to the volume of the room V = (4 m) (5 m) (6 m) = 120 m 3 Finally, the mass of the air in the room is m = ρV = (1.17 kg /m 3 ) 120 m 3 = 140 kg ¡ ¢ 3.2 Pressure in fluids at rest At the end of this section, you will be able to: State Pascal’s principle Explain how Pascal’s principle is used in hydraulic press machine. Discuss how pressure depends on the depth of fluid. Discuss the working principles of pressure measuring devices like barometers and manometers. Perform simple experiments related to pressure. Solve problems related to Pascal’s principle, fluid depth and atmospheric pressure. As gas particles collide with the walls of the container in which they reside, they exert pressure. In fact, if you place any object inside a gas, the gas particles exert the same pressure on the object as the gas exerts on the walls of the container. Do liquids behave in a similar way? The particles in a liquid are in continual random motion, somewhat similar to particles in gases. 3.2 Pressure in fluids at rest 83 Activity 3.3: Testing direction of pressure Materials required: A plastic bottle, five nails and water Procedures: Pock five holes at the middle of the bottle along its perimeter using the nail as shown in (Figure 3.11). Close all the holes by nails. Fill the bottle with water while the it is opened. Remove the nails and observe what happens. Precaution: Take care to avoid the damage while pocking the bottle using nail. Figure 3.11 Arcs of water leaving Question holes at the same level in a bottle. Try to explain your observation. Pascal’s principle Pushing the piston in to a confined fluid in one direction causes a greater pressure in the fluid close to the piston. Almost immediately the pressure throughout the fluid increases uniformly, fluid is pushed out of all of the holes in the container (Figure 3.12 a and b)). This phenomenon was first discovered by French scientist Blaise Pascal in 1653 and is called P asc al 0 s principle. Pascal’s principle: States that a change in the pressure applied to a static fluid is transmitted undiminished to every point of the fluid and to the walls of the container. Figure 3.12 Pascal’s principle: Increasing the pressure of a fluid How can we explain this observation? The pressure applied at one point is sooner at one location causes a uniform transmitted to the whole part of the fluid by a continuous collisions of neighboring pressure increase throughout the molecules of the fluid. fluid. Activity 3.4: Pressure in a pair of connected syringes Materials required: Two syringes of different size, a piece of flexible plastic tube, two identical 250 ml transparent beakers and bucket partly filled 84 Unit 3 Fluid Mechanics with water Procedures: Submerge the plastic tube and let it fill completely with water. Push the plunger of the small syringe all the way into it so there is no air in it. Fill it partially with water. Then insert its tip into one end of the submerged, water-filled plastic tube. Push the plunger of the larger syringe all the way into it so there is no air in it. Fill it partially with water. Then insert its tip into an other end of the submerged, water-filled plastic tube. Then take the entire assembly out of the water and arrange the syringes vertically as shown in the Figure below with their initial water level at the same height. Put identical beakers on the top of the syringes’ plungers. Draw some water into the beaker on the plunger of small syringe. What do you observe about the water level in the two syringes? Draw water into the beaker on the plunger of the large syringe until the water level on the two syringes is the same. Precaution: Syringes should not have needle to avoid damage. Question: How much water is require on both beakers to balance the level of water in the syringes. Try to explain your observation. 3.2 Pressure in fluids at rest 85 Hydraulic press One of the technical applications of Pascal’s Principle is a hydraulic press which is a form of simple machine that converts small forces into larger forces, or vice versa. Automobile mechanics use hydraulic presses to lift cars, and dentists and barbers use them to raise and lower their clients’ chairs. The hydraulic brakes of an automobile are also a form of hydraulic press. Most of these devices work on the simple principle illustrated in Figure 3.13, although the actual devices are usually more complicated in construction. Figure 3.13 Schematic of a hydraulic press. → − In Figure 3.13, a downward force F 1 on L is exerted by piston 1 (with small area A 1 ) on the liquid. This piston compresses a liquid (usually oil) in the lift. The pressure in the fluid just under piston 1 is F 1 on L P1 = (3.7) A1 Because the pressure changes uniformly throughout the liquid, the pressure under piston 2 is also P = F 1 on L /A 1 , assuming the pistons are at the same height. Since piston 2 has a greater area A 2 than piston 1, the liquid exerts a greater upward force on piston 2 than the downward force on piston 1: F 1 on L A2 µ ¶ µ ¶ F L on 2 = P A 2 = A2 = F 1 on L (3.8) A1 A1 Since A 2 is greater than A 1 , the lift provides a significantly greater upward force 86 Unit 3 Fluid Mechanics F L on 2 on piston 2 than the downward push of the smaller piston 1 on the liquid F 1 on L. Figure 3.14 Hyraulic press Example 3.8 A hydraulic lift has a small piston with surface area 0.0020 m2 and a larger piston with surface area 0.20 m2. Piston 2 and the car placed on piston 2 have a combined mass of 1800 kg. What is the minimal force that piston 1 needs to exert on the fluid to slowly lift the car? Solution: Assume that the levels of the two pistons are the same and that the car is being lifted at constant velocity. Use the force diagram for the car and piston 2 (see Figure 3.14) and Newton’s second law to determine F L on 2. Note that the force that the liquid exerts on the piston 2, F L on 2 , is equal in magnitude to the force that piston 2 and the car exert on the liquid F 2 on L , which equals the downward gravitational force that Earth exerts on the car and piston: Exercise 3.4 FE on C ar +P i st on2 = F 2 on L = mC ar +P i st on g. Dear students, We rewrite the hydraulic press Eq. (3.8) to determine the unknown force: in example 3.8, if ³ ´ ³ ´ F 1 on L = AA 21 F 2 on L = AA 21 mC ar +pi st on g you needed to lift substituting the givens and solving the car about 0.10 m above the ground, µ 0.0020 m 2 £ ¶ ¤ what distance would F 1 on L = 2 (1800 kg )(9.8 N /kg ) = 180 N 0.20 m you have to push down on the small That is the force equal to lifting an object of mass 18 kg. This example shows piston? that a six year old child of average mass 20 kg is more than capable of raising the 3.2 Pressure in fluids at rest 87 car simply by sitting on the small piston. A hydraulic press is a force multiplying machine. It reduces the amount of force needed to lift a load. A small force is applied at small piston to raise a heavy load at the larger piston. However, the work done by the two pistons is the same. That is Wb y P i st on1 = Won P i st on2 or F b y P i st on1 (d 1 ) = F on P i st on2 (d 2 ), where W , F and d refer work done, force and distance traveled respectively. Variation of pressure with depth Activity 3.5: Variation of pressure with depth Materials required: Three identical plastic bottles, three identical nails, a table and water. Procedures: Make a hole near to the bottom of each plastic bottle by identical nails. The holes should have identical height. Fill the three bottles with water to different height while the holes are closed by the nails. Put the bottles on the edge of a table facing the holes in the same direction, away the table. Remove the nails gently simultaneously and observe the distance the water shots hits the ground. Precaution: Take care to avoid damage while making a hole using the nails. Keep the lids of the bottles open throughout your experiment. Question Which water shot has the largest distance (wider arc) from the bottom of the table? 88 Unit 3 Fluid Mechanics What do you think the reason for that? How the distance of the water shots from each bottles vary with time and explain the result. Consider a liquid of density ρ at rest as shown in Figure 3.15. We assume ρ is uniform throughout the liquid, which means the liquid is incompressible. Let us select a parcel of the liquid contained within an imaginary block of cross- sectional area A extending from depth d to depth d + h. The liquid external to our parcel exerts forces at all points on the surface of the parcel, perpendicular to the surface. The pressure exerted by the liquid on the bottom face of the parcel is P , and the pressure on the top face is P o. Therefore, the upward force exerted Figure 3.15 A parcel of fluid in a by the outside fluid on the bottom of the parcel has a magnitude P A, and the larger volume of fluid. downward force exerted on the top has a magnitude is P o A. The mass of liquid in the parcel is M = ρV = ρ Ah; therefore, the weight of the liquid in the parcel is W = M g = ρ Ahg. Because the parcel is at rest and remains at rest, it can be modeled as a particle in equilibrium, so that the net force acting on it must be zero. The forces are the upward force at the bottom (PA), the downward force at the top (P o A) and weight of the parcel (Mg). Choosing upward to be the positive y direction, we see that X→ − F = P A bj −P o A jb−M g bj =0 Or P A−P o A−ρ Ahg =0 P = P o +ρhg (3.9) That is, as the depth increases by h, the pressure increases by ρg h. It depends on the depth as well as the density of the liquid. If the liquid is open to the atmosphere and P o is the pressure at the surface of the liquid, then P o is atmospheric pressure. P o = 1.00 at m = 1.013 3 × 105 P a Equation 3.9 implies that the pressure is the same at all points having the same depth, whatever shape the container posses. Because the pressure in a fluid depends on depth and on the value of P o , any increase in pressure at the surface 3.2 Pressure in fluids at rest 89 must be transmitted to every other point in the fluid by the Pascal’s principle. Thus the equation P = hρg represents the pressure due to the weight of any fluid of average density ρ at any depth h below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. The pressure due depth of liquid is independent of the shape and size of the container, it only depends on the depth. Activity 3.6: Independence of liquid pressure with size and shape of container Materials required: Three plastic bottles of different diameter, three identical nails and a table Procedures: Make a hole near to the bottom of each plastic bottle by identical nails. The holes should have identical height. Fill the three bottles with water to the same level. Put the bottles on the edge of a table facing the holes in the same direction, away the table. Remove the nails gently simultaneously and observe the distance the water shots hits the ground. Precaution: Take care to avoid damage while making a hole using the nails. Keep the lids of the bottles open throughout your experiment. Question Which water shot has the largest distance (wider arc) from the bottom of the table? Explain the reason for your observation? 90 Unit 3 Fluid Mechanics Example 3.9 Calculate the force on a circular area of diameter 0.40 m on the bottom of the ocean which is 25.0 m below the surface. Take atmospheric pressure and density of sea water at the bottom to be 1 atm and 1.03 K kg /m 3 respectively. Solution: The absolute pressure is obtained from the relation: P abs = P amt + P g aug e = P amt + ρg h where P amt = 1 at m = 101.3 K P a, ρ = 1.03 × 103 k/m 3 , and h = 25 m. Therefore, P abs = 101.3 K P a + (1.03 × 103 k/m 3 )(9.8 m/s 2 ) (25 m) = 101.3 K P a + 252.35 K P a = 353.65 K P a ³ ´2 ¡ 0.4 m ¢2 d Similarly, the area is obtained as A = π 2 = (3.14) 2 = 0.1256 m 2 F net We can use the relation P abs = A. Then, the force exerted is F net = P abs A = 353.65 K P a × 0.1256 m 2 = 4.44 × 104 N Atmospheric pressure Discussion question 3.3 A sealed empty plastic water bottle collapses as a plane descends from a higher elevation to a lower elevation as shown in Figure 3.16. Why this happens? Figure 3.16 The plastic bottle (a) sealed at 4,300 m altitude, (b) the same sealed plastic bottle at an altitude of 2,700 m and (c) at the 300 m. 3.2 Pressure in fluids at rest 91 Activity 3.7: Effect of temperature on Pressure Materials required: Two identical plastic bottles, and hot water Procedures: Fill one of the plastic bottle by hot water and cover tightly; then keep it to cool down to room temperature. Fill the other plastic bottle by hot water and keep it to cool down to room temperature while the lid is open. Precaution: Take care to avoid the damage of hot water by using heat resistance glove and using moderately hot water. Question What happens to the closed plastic bottles at room temperature? Explain the reason. What happens to the opened plastic bottles at room temperature? Explain the reason. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. There are two reasons why air pressure decreases as altitude increases: density and depth of the atmosphere. Most gas molecules in the atmosphere are pulled close to Earth’s surface by gravity, so gas particles are denser near the surface. With more gas particles in a given volume, there are more collisions of particles and therefore greater pressure. The depth (distance from top to bottom) of the atmosphere is greatest at sea level and decreases at higher altitudes. With greater depth of the atmosphere, more air is pressing down from above. Therefore, air pressure is greatest at sea level and falls with increasing altitude. On top of Mount Everest, which is the tallest mountain on Earth, air pressure is only about one-third of the pressure at sea level. 92 Unit 3 Fluid Mechanics Atmospheric pressure P at m changes from 101.325 KPa at sea level to 89.88, 79.50, 54.05, 26.5, and 5.53 KPa at altitudes of 1000, 2000, 5000, 10,000, and 20,000 meters, respectively. Remember that the atmospheric pressure at a location is simply the weight of the air above that location per unit surface area. Activity 3.8: Effect of atmospheric pressure on the flow of confined water Materials required: A Plastic bottle and nail or pin Procedures: Fill a plastic bottle with water and cover the lid tightly. Make a small hole at the bottom of this bottle by a nail or a pin and take the nail/pin out. Precaution: Take care to avoid the damage while using the nail/ pin. Question (i) Does all the water leave the plastic bottle? Why? (ii) What will happen if the bottle is open? The decline of atmospheric pressure with elevation has far-reaching consequences in daily life. For example, cooking takes longer at high altitudes since water boils at a lower temperature at lower atmospheric pressures. For a given temperature, the density of air is lower at high altitudes, and thus a given volume contains less air and less oxygen. So it is no surprise that we tire more easily and experience breathing problems at high altitudes. To compensate for this effect, people living at higher altitudes develop more efficient lungs. Similarly, a 2.0-L car engine will act like a 1.7-L car engine at 1500 m altitude (unless it is turbocharged) because of the 15 percent drop in pressure and thus 15 percent drop in the density of air. Measuring pressure Under the water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight 3.2 Pressure in fluids at rest 93 of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many floors, but you need only dive one meter or below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense. The Barometer Atmospheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is often referred to as the barometric pressure. The Italian Evangelista Torricelli (1608–1647) was the first to conclusively prove that the atmospheric pressure can be measured by inverting a mercury-filled tube into a mercury container that is open to the atmosphere, as shown in Figure 3.17(a). The pressure at point B is equal to the atmospheric pressure, and the pressure at point C can be taken to be zero since there is only mercury vapor above point C and the pressure is very low relative to P at m and can be neglected to an excellent approximation. Writing a force balance in the vertical direction gives P at m = ρg h (3.10) where ρ is the density of mercury, g is the local gravitational acceleration, Discussion and h is the height of the mercury column above the free surface. Note that the question 3.4 length and the cross-sectional area of the tube have no effect on the height of the fluid column of a barometer (Figure 3.17b). (i) Why nosebleed is a common experience at high altitudes? (ii) Federation International de Football Association (FIFA) had banned international matches played above an altitude of 2,800m in 2007. Why do you think the reason for this? Figure 3.17 (a) The basic barometer. (b) The length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer, provided that the tube diameter is large enough to avoid surface tension (capillary) effects. 94 Unit 3 Fluid Mechanics A frequently used pressure unit is the standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0◦ C (ρ H g = 13, 595 kg /m 3 ) under standard gravitational acceleration 2 (g = 9.807 m/s ). If water instead of mercury were used to measure the standard atmospheric pressure, a water column of about 10.3 m or 10300 mm would be needed. Pressure is sometimes expressed (especially by weather forecasters) in terms of the height of the mercury column. The standard atmospheric pressure, Exercise 3.5 for example, is 760 mmHg at 0◦ C. The unit mmHg is also called the torr in honor of Torricelli. Therefore, 1 atm = 760 torr and 1 torr = 133.3 Pa. (i) The reading of a barometer in your room is 700 mm Example 3.10 Hg. What does this Determine the atmospheric pressure at a location where the barometric reading mean? What is the is 740 mmH g and the gravitational acceleration is g = 9.805 m/s 2. Assume the pressure in pascals? temperature of mercury to be 10◦C , at which its density is 13, 595 kg /m 3 ). (ii) If oil of density 950 kg /m 3 is used Solution: in the barometer instead of mercury, The temperature of mercury is assumed to be 10◦C. what would be the The density of mercury is given to be 13, 595 kg /m 3. height of the oil in From Eq. 3.11, the atmospheric pressure is determined to be the tube at 1 atm? P at m = ρg h = (13, 570 kg /m 3 )(9.805m/s 2 )(0.740 m) = 98.5 kP a The manometer A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm and the other arm connected to the gas whose pressure is to be measured, and a nonvolatile liquid (usually mercury) in between. The distance between the liquid levels in the two arms of the tube (h in Figure 3.18) is proportional to the pressure of the gas in the container. An open-end manometer (Figure 3.18) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and 3.2 Pressure in fluids at rest 95 the atmosphere. Figure 3.18 Measurement of gas pressure with manometer (a) closed end manometer (b) open end manometer with atmospheric pressure greater than gas pressure (c) open end manometer with atmospheric pressure less than gas pressure This device is mainly used to measure low pressure differences accurately. In the Figure 3.18, let us use the principle of hydrostatic equilibrium. The pressure in the gas supply (P g as ) for the situations shown in Figure 3.18 are given as follow: (a) For closed end manometer: P g as = ρg h (b) For open end manometer with atmospheric pressure greater than gas pressure: P g as = P at m − ρg h (c) open end manometer with atmospheric pressure less than gas pressure: P g as = P at m + ρg h where ρ denotes density of the liquid, g denotes gravitational constant, h is the height of the liquid column. Note that the cross-sectional area of the tube has no effect on the differential height h, and thus the pressure exerted by the fluid. However, the diameter of the tube should be large enough (more than several 96 Unit 3 Fluid Mechanics Exercise 3.6 millimeters) to ensure that the surface tension effect and thus the capillary rise is The pressure of negligible. a sample of gas is measured with a closed-end Example 3.11 manometer, as An open manometer is used to measure the pressure of a gas in a tank. The fluid shown in Figure used has a specific gravity of 0.85, and the fluid column in the open arm is 55 cm 3.19. The liquid in above the gas connected arm, as shown in Figure 3.18. If the local atmospheric the manometer is pressure is 96 kPa, determine the absolute pressure within the tank. mercury. Determine the pressure of the Solution: gas. The reading of a manometer attached to a tank and the atmospheric pressure are given. The absolute pressure in the tank is to be determined. The density of the gas in the tank is much lower than the density of the manometer fluid. The specific gravity of the manometer fluid is given to be 0.85. We take the standard density of water to be 1000 kg/m3. The density of fluid is obtained by multiplying its specific gravity by the density of water, ρ = SG(ρ H2 O ) = (0.85)(1000 kg /m 3 ) = 850 kg /m 3 Then from Eq. 3.12, Figure 3.19 Fluid level in closed end manometer P = P at m + ρg h = 96 kP a + (850 kg /m 3 )(9.81 m/s 2 )(0.55 m) = 100.6 kPa Brainstorming question 3.6 Note that the gauge pressure in the tank is 4.6 kPa. Dear students, can you guess why objects like empty plastic bottle, plastic 3.3 Archimedes’ principle balls, some wooden and metallic objects At the end of this section, you will be able to: like boats is difficult Define the buoyant force. to submerge in to a water or other Classify floating and submerging objects comparing the objects and the liquids? fluid density. Solve problems related to buoyant force. 3.3 Archimedes’ principle 97 Buoyant force It is extremely difficult to push a ball down under water because of the large upward force exerted by the water on the ball. The upward force exerted by a fluid on any immersed object is called a buoyant force. Where does this buoyant force come from? Buoyant force is based on the fact that pressure increases with depth in a fluid. This means that the upward force on the bottom of an object in a fluid is greater than the downward force on the top of the object. There is a net upward, or buoyant force on any object in any fluid (See Figure 3.20). Buoyant force is also called up thrust force. How a buoyance force is exerted by gas particle can be observed from the link: https://phet.colorado.edu/en/simulations/balloons-and-buoyancy. How large is this buoyant force? To answer this question, think about what happens when a submerged object is removed from a fluid. The space it occupied is filled by fluid having a weight W f l. Since this weight is supported by surrounding fluid, the magnitude of buoyant force on an object must equal the weight of the fluid displaced by the object. It is a tribute to the Greek mathematician and inventor Archimedes (287–212 B.C.) that he stated this principle long before concepts of force were well established. Archimedes’ principle is Stated as follows: The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is given as: F B = W f l ui d (3.11) Figure 3.20 Pressure variation with depth resulting a buoyant force Activities 3.9: Buoyancy Materials required: Two glass cups, table salt, spoon, two eggs and water. Procedures: Fill 3/4 of both glass cups with ordinary water. Dissolve 4 to 5 spoons of salt in one of water filled glass cups. Put one of the egg into fresh water and the other in to a salt solution. Question 98 Unit 3 Fluid Mechanics Which of the eggs floats and which one is sinking? Reason out your observation in both experiments. To further understand the origin of the buoyant force, consider a cylinder of solid material immersed in a liquid as in Figure 3.20. The pressure P 2 at the bottom of the cylinder is greater than the pressure P 1 at the top by an amount ρ f l ui d g h, where h is the height of the cylinder and ρ f l ui d is the density of the fluid. The pressure at the bottom of the cylinder causes an upward force equal to P 2 A , where A is the area of the bottom face. The pressure at the top of the cylinder causes a downward force equal to P 1 A. The resultant of these two forces is the buoyant force F B with magnitude F B = P 2 A − P 1 A = ρ f l ui d g h A F B = ρ f l ui d g Vd i sp (3.12) where Vd i sp = Ah is the volume of the fluid displaced by the cylinder. Because the product ρ f l ui d Vd i sp is equal to the mass of fluid displaced by the object, F B = M d i sp g (3.13) where M d i sp g is the weight of the fluid displaced by the cylinder. Totally submerged object When an object is totally submerged in a fluid of density ρ f l ui d , the volume Vd i sp of the displaced fluid is equal to the volume of the object Vob j ; so, the magnitude of the upward buoyant force is F B = ρ f l ui d g Vob j. If the object has a mass M and density ρ ob j , its weight is equal to F g = M g = ρ ob j g Vob j , and the net force on the object is F B − F g = (ρ f l ui d − ρ ob j )g Vob j. Hence, if the density of the object is less than the density of the fluid, the downward gravitational force is less than the buoyant force and the unsupported object accelerates upward. If the density of the object is greater than the density of the fluid, the upward buoyant force is less than the downward gravitational force and the unsupported object sinks. If the density of the submerged object equals the density of the fluid, the net force on the object is zero and the object remains in equilibrium. It can be anywhere inside the fluid. Therefore, the direction of motion of an object submerged in a 3.3 Archimedes’ principle 99 fluid is determined only by the densities of the object and the fluid. Floating object Now consider an object of volume Vob j and density ρ ob j < ρ f l ui d in static equilibrium floating on the surface of a fluid, that is, an object that is only partially submerged. In this case, the upward buoyant force is balanced by the downward gravitational force acting on the object. If Vd i sp is the volume of the fluid displaced by the object (this volume is the same as the volume of that part of the object beneath the surface of the fluid), the buoyant force has a magnitude F B = ρ f l ui d g Vd i sp. Because the weight of the object is F g = M g = ρ ob j g V ob j and F g = F B , we see that ρ f l ui d g Vd i sp = ρ ob j g Vob j , or Vd i sp ρ ob j = (3.14) Vob j ρ f l ui d This equation shows that the fraction of the volume of a floating object that is below the fluid surface is equal to the ratio of the density of the object to that of the fluid. Example 3.12 An iceberg floats in seawater as shown in Figure ??. What fraction of the iceberg lies below the water level? Givens: Density of sea water and icebergs are 1030 kg /m 3 and 917 kg /m 3 respectively. Solution: From equation (3.14) Vdi sp ρ i ce 917 kg /m 3 = = = 0.89 = 89% Vi ce ρ sea w ast er 1030 kg /m 3 Therefore, the visible fraction of ice above the water’s surface is about 11%. It is the unseen 89% below the water that represents the danger to a passing ship. This hidden ice can damage a ship that is still at a considerable distance from the visible ice. 100 Unit 3 Fluid Mechanics Figure 3.21 (a) Much of the volume of this iceberg is beneath the water. (b) A ship can be damaged even when it is not near the visible ice. Activities 3.10. Measuring buoyant force and density of liquid Materials required: two 100 ml beakers, a pendulum bob, spring balance and water. Procedures: (i) Measure the weight of the pendulum using spring balance. (ii) Fill a 100 ml beaker by pure water. (iii) Measure the weight of a pendulum bob inside the water by suspending it on a spring balance while collecting the overflowing water. (iv) Measure the weight of overflow water you have collected. (v) Subtract the weight of the bob measured in pure water from its weight measured in air. This is equal to buoyant force. Questions (a)Compare the weight of overflow water you measured in step (iv) and the buoyant force in step (v). Does your result verify Archimedes’ principle? 3.3 Archimedes’ principle 101 (b)Calculate the density of water using the relation F B = ρ sol g Vsol. where F B is the buoyant force you got in step (v), Vsol is the volume of the overflow water. ρ sol is the density of water you are going to calculate and g is gravitational acceleration. Compare your result with the density of water in Table 3.1. Example 3.13 (a) Calculate the buoyant force on 10,000 metric tons (1.00×107 kg) of solid carbon steel completely submerged in water, and compare this with the steel’s weight. (b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace 1.00×105 m3 of water? Solution: for (a) To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table 3.1. We note that, since the steel is completely submerged, its volume and the water’s volume are the same. m From the definition of density ρ = V m st 1.00x107 kg Vst = = = 1.28x103 m 3 ρ st 7.84x103 kg /m 3 Because the steel is completely submerged, this is also the volume of water displaced, Vw. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives m w = ρ w Vw = 1.000x103 kg /m 3 1.28x103 m 3 ¡ ¢¡ ¢ = 1.28x106 kg By Archimedes’ principle, the weight of water displaced is m w g , so the buoyant force is F B = Ww = m w g = (1.28x106 kg )(9.80 m/s 2 ) = 1.3x107 N The steel’s weight is m st g = 9.80 × 107 N , which is much greater than the buoyant 102 Unit 3 Fluid Mechanics force, so the steel will remain submerged. Solution for (b) The mass of water displaced is found from its relationship to density and volume, Exercise 3.7: both of which are known. That is, Suppose your mass is 70.0 kg and m w = ρ w Vw = (1.000x103 kg /m 3 )(1.00x105 m 3 ) your density is 970 kg /m 3. If you could = 1.00x108 kg. stand on a scale in The maximum buoyant force is the weight of this much water, or a vacuum chamber on Earth’s surface, F B = m w g = (1.00x108 )(9.80m/s 2 ) the reading of the scale would be mg = = 9.80x108 N (70.0 kg )(9.80 N/kg) The maximum buoyant force is ten times the weight of the steel, meaning the ship = 686 N. What will can carry a load nine times its own weight without sinking. the scale read when you are completely submerged in air of density 1.29 Example 3.14 kg /m 3 ? Compare your answer with The mass of an ancient Greek coin is determined in air to be 8.630 g. When the coin your weight in the is submerged in water, its apparent mass is 7.800 g. Calculate its density, given that vacuum chamber water has a density of 1.0 g /cm 3 and that effects caused by the wire suspending and decide whether the coin are negligible. the buoyant force of air is to be Solution: considered or to The volume of water displaced Vw can be found by solving the equation for density be ignored. ρ = m/V. The volume of water is Vw = m w / ρ w where m w is the mass of water displaced. As noted, the mass of the water displaced equals the apparent mass loss, which is m w 0.830 g = 8.630 g − 7.800 g = 0.830 g. Thus the volume of water is Vw = 1.000 g /cm 3 = 0.830 cm 3. This is also the volume of the coin, since it is completely submerged. We can now find the density of the coin using the definition of density: mc 8.630 g ρc = = = 10.4 g /cm 3 Vc 0.830 cm 3 You can see from Table 3.1 that this density is very close to that of pure silver. 3.4 Fluid flow 103 Exercise 3.8: Exercise 3.9: A crown weighs 25.0 N when it is measured in air and 22.6 N when it is A rectangular submerged in water. Check whether the crown is made from pure gold or wooden block some less valuable metal assuming the density of gold is 19, 300kg /m 3. floats with 75 % of its volume inside a water. What is 3.4 Fluid flow the density of this block? Fluid flow is caused by differences in pressure. When the pressure in one region of the fluid is lower than in another region, the fluid tends to flow from the higher pressure region toward the lower pressure region. For example, large masses of air in Earth’s atmosphere move from regions of high pressure into regions of low Brainstorming pressure creating what we call wind. question 3.7 At the end of this section, you will be able to: i) What mechanism maintains the Blood Differentiate steady and turbulent fluid flow. flows through our circulatory system? Define flow rate. ii) How can a plane Derive equation of continuity from flow rate. fly against gravity? Explain the cause and working principle of some common activities or experiences. Solve problems on equation of continuity. Steady and turbulant fluid flow When fluid is in motion, its flow can be characterized as being steady fluid flow if each particle of the fluid follows a smooth path such that the paths of different particles never cross each other as shown in Figure 3.22. In steady flow, every fluid particle arriving at a given point in space has the same velocity. The laminar flow always occurs when the fluid flow with low velocity and in small diameter pipes. Above a certain critical speed, turbulent fluid flow occurs. Turbulent flow is irregular flow characterized by small whirlpool-like regions as shown in Figure 3.22. The adjacent layers of the fluid cross each other and move randomly in a 104 Unit 3 Fluid Mechanics zigzag manner. The turbulent flow occurs when the velocity of the fluid is high and it flows through larger diameter pipes. As a solid surface is acted by a frictional force when it slides over another solid, there is also an internal frictional force in liquid, called viscosity, when two adjacent layers of fluid try to move relative to each other. Viscosity causes part of the fluid’s kinetic energy to be transformed to internal energy. This mechanism is similar to the one by which the kinetic energy of an object sliding over a rough, horizontal surface decreases. Figure 3.22 Laminar and turbulent flow. The path taken by a fluid particle under steady flow is called a streamline. The velocity of the particle is always tangent to the streamline as shown in Figure 3.23. A set of streamlines like the ones shown in Figure 3.23 form a tube of flow. In laminar flow, fluid particles cannot flow into or out of the sides of this tube; if they could, the streamlines would cross one another and results in turbulent flow. Flow rate Flow rate Q is defined to be the volume of fluid passing by some location through an area during a period of time, as seen in Figure 3.24. In symbols, this can be Figure 3.23 A particle in laminar written as flow follows a streamline. V Q = (3.15) t where V is the volume and t is the elapsed time. The SI unit for flow rate is m 3 /s , but a number of other units for Q are in common use. Example 3.15 How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min? Figure 3.24 Flow rate of fluid. Solution: Time and flow rate Q are given, and so the volume V can be calculated from: V Solving Q = t for volume gives V = Qt 3.4 Fluid flow 105 Substituting the known values gives: 5.00 L ¡ ¢ 1 m3 5 mi n µ ¶ µ ¶µ ¶ V= 75 y 5.26 × 10 1 mi n 103 L y = 2.0 × 105 m 3 Figure 3.25 Flow of incompressible fluid through a pipe of decreasing radius Equation of continuity Figure 3.25 shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow Discussion past any point in the tube in a given time to ensure continuity of flow. In this question 3.5: case, because the cross-sectional area of the pipe decreases, the velocity must What happens when necessarily increase. This logic can be extended to say that the flow rate must be water flows from a the same at all points along the pipe. In particular, for points 1 and 2, hose into a narrow ) spray nozzle? Q1 = Q2 (3.16) A1 v1 = A2 v2 This is called the equation of continuity and is valid for any incompressible fluid where v 1 is the average speed of the fluid passing cross section A 1 and v 2 is the average speed of the fluid passing cross section A 2. The equation of continuity is used to relate the cross-sectional area and average speed of fluid flow in different parts of a rigid vessel carrying an incompressible fluid. You can visualize effect of cross-sectional area on velocity of fluid flow from the link: https://phet.colorado.edu/en/simulations/fluid-pressure-and-flow. Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, and so the equation must be applied with caution to gases if they are subjected to compression or expansion. Example 3.16 A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle. Solution: for (a) 106 Unit 3 Fluid Mechanics First, we solve Q = Av for v 1 and note that the cross-sectional area is A = πr 2 , yielding Q Q v1 = = A 1 πr 12 Substituting known values and making appropriate unit conversions yields (0.500 L/s(m −3 /L) v1 = = 1.96m/s π(9.00 × 10−3 m)2 Solution for (b) Discussion We could repeat this calculation to find the speed in the nozzle v 2 , but we will use question 3.6 the equation of continuity to give a somewhat different insight. Using the equation What happens which states when you blow air A1 v1 = A2 v2 hard across the top solving for v 2 and substituting πr 2 for the cross-sectional area yields surface of a piece A1 πr 12 r 12 v2 = A2 v 1 = πr 22 v1 = r 22 v 1 Substituting known values, of paper held at the corners as shown in (0.900)2 Figure 3.26. What is v2 = 1.96m/s = 25.5 m/s (0.250 cm )2 the reason for that? Can you relate this to how airplane flies As a fluid’s speed increases, the pressure that the moving fluid exerts on the against gravity? surface decreases. This is called Bernoulli’s principle and stated as: The pressure that a fluid exerts on a surface decreases as the speed with which the fluid moves across the surface increases. Fluid flow has important implications in biological systems for example, in the flow of blood through blood vessels. The blood pressure against the wall of a vessel depends on how fast the blood is moving. Pressure is lower when the blood is moving faster. Similarly, a snoring sound occurs when air moving through the narrow opening above the soft palate at the back of the roof of the mouth has Figure 3.26 A piece of paper held lower pressure than nonmoving air below the palate (Figure 3.27). The normal at the corner before air is blown air pressure below the soft palate, where the air is not moving, pushes the palate across the top. closed. When airflow stops, the pressures equalize and the passage reopens. The rhythmic opening and closing of the soft palate against the throat leads to the snoring sound. Pressure is also very important in wind musical instruments. 3.5 Safety and high pressure 107 Figure 3.27 Snoring occurs when the soft palate opens and closes due to the starting and stopping of air flow across it. Brainstorming question 3.8 3.5 Safety and high pressure Dear students, have you ever seen At the end of this section, you will be able to: while metal and List high- pressure systems. wood technicians spraying ink List application of high-pressure system or varnish on Identify the common causes of risk in the high-pressure systems wooden or metallic furniture? How Familiarized the safety measures related to high pressure systems do these spraying machines work? Have you observed High pressure systems other equipment having similar Pressure far greater than 1 atmosphere (most of the time greater than 50 atm) is purpose? considered as high pressure. High pressure is used for many applications. High pressure cookers are used in a kitchen to cook food. Gas cylinders containing 108 Unit 3 Fluid Mechanics liquid petroleum gas at high pressure are used as fuel. Gas cylinder are also used to seal different types of gases at high pressure for laboratory or medical use. The bicycle and car tires are inflated by high pressure tire inflator. High pressure is also used in high pressure washers. In physical science (physics and chemistry) high pressure is important to study physical properties of various materials (mainly solids) and to transform their nature. Many materials undergo fascinating changes in their physical and chemical characteristics when subjected to high pressure. The application of high pressure to biological samples is also of technological relevance because it is known that the microorganism activity is diminished or canceled by application of high pressures, a process called pascalization. Pascalization can be used to increase the shelf lives of perishable foodstuffs: juice, fish, meat, dairy products, etc. High pressure affects many scientific and technological fields, like biology, chemistry, environmental engineering, food technology, material science, pharmacy, and physics. High pressure equipment may consist of high-pressure compressors (or pumps), high pressure piping (fittings, seals, tubing and valves), high pressure vessels, Steam Generator, Safety Accessories and high-pressure instrumentation (Figure 3.29). Figure 3.28 A compressor with a storage tank. High pressure compressors (pumps): A compressor is a mechanical device that increases the pressure of a gas by reducing its volume (Figure 3.28). Compressors are similar to pumps: both increase the pressure on a fluid and both can transport the fluid through a pipe. Heated pressure equipment are intended for generation of steam or super-heated water at temperatures higher than 110◦ C and having a volume greater than two liters. This includes all pressure cookers. Because of severe stresses, it is essential to design and fabricate high-pressure machines very carefully. 3.5 Safety and high pressure 109 Figure 3.29 Components of high pressure equipments. High pressure vessels: Pressure vessel means a housing designed and built to contain fluids under pressure including its direct attachments up to the coupling point connecting it to other equipment. A vessel may be composed of more than one chamber. Safety Accessories: These accessories include safety valves and bursting discs, as well as limiting devices. Limiting devices can either activate the means for correction or shutdown and lock-out, such as pressure switches and temperature switches. High-pressure instrumentation: To operate high-pressure plants, adequate control- and measuring devices are required. High-pressure instrumentation is available for purposes like pressure, temperature, flow and level measurements. Safety for high pressure equipment If pressure systems or equipment fails and bursts violently apart, it can seriously injure or kill people and cause serious damage to property. The main hazards from pressure are: impact from the blast of an explosion, impact from parts of equipment that fail or any flying debris, impact with the released liquid or gas (such as steam), fire resulting from the escape of flammable liquids or gases. 110 Unit 3 Fluid Mechanics Common causes of pressure system and equipment risks includes