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Chapter 14. Fluid Mechanics Dr. Yousef Abou-Ali [email protected] Syllabus 14.1 Density 14.2 Pressure in a Fluid 14.3 Buoyancy 14.4 Fluid Flow 14.5 Bernoulli's Equation 14.6 Viscosity and Turbulence Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Goals for Chapter 14 • To study density and pressure in a fluid. • To apply Archimedes Principle of buoyancy. • To describe surface tension and capillary action. • To study and solve Bernoulli's Equation for fluid flow. • To see how real fluids differ from ideal fluids (turbulence and viscosity). Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Introduction • Fluids play a vital role in many aspects of everyday life. • We drink them, breathe them, and swim in them. • They circulate through our bodies and control our weather. • Airplanes fly through them; ships float in them. • A fluid is any substance that can flow; we use the term for both liquids and gases. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Introduction • We usually think of a gas as easily compressed and a liquid as nearly incompressible. • We begin our study with fluid statics, the study of fluids at rest in equilibrium situations. • Than will move to talk about fluid dynamics. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.1 Density • An important property of any material is its density, defined as its mass per unit volume. • A homogeneous material such as ice or iron has the same density throughout. • We use the Greek letter ρ (“rho”) for density. • If a mass m of homogeneous material has volume V , the density ρ is: m ρ = V Dr. Y. Abou-Ali, IUST (d efin ition of d en sity ) (14.1) University Physics, Chapter 14 14.1 Density • At right, liquids of different densities separate with denser liquids lower in the glass. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.1 Density • The density of some materials varies from point to point within the material; some examples are the earth’s atmosphere, which is less dense at high altitude. and oceans (which are denser at greater depths). • For these materials, Eq. (14.1) describes the average density. • In general, the density of a material depends on environmental factors such as temperature and pressure. • The SI unit of density is the kilogram per cubic meter (1 kg/m3). 3 3 1 g/cm = 1000 kg/m Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.1 Density • The specific gravity of a material is the ratio of its density to the density of water at 4.0o C, 1000 kg/m3. • It is a pure number without units. • Since it has nothing to do with gravity; “relative density” would have been better. Example 14.1 (The weight of a roomful of air): Find the mass and weight of the air in a living room with a 4.0 m × 5.0 m floor and a ceiling 3.0 m high. What is the mass and weight of an equal volume of water? Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.1 Density Solution: • Identify : We assume that air is homogeneous, so that the density is the same throughout the room. • Set Up : We use Eq. (14.1) to relate the mass (the target variable) to the volume (which we calculate from the dimensions of the room) and the density (from Table 14.1, in your book). • Execute: The volume of the room is: 3 V = (3.0m)(4.0m)(5.0m) = 60 m The mass m air of air is given by Eq. (14.1): Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.1 Density 3 3 mair = ρair V = (1.2 kg / m )(60m ) = 72 kg • The weight of the air is: 2 wair = mair g = (72 kg) (9.8 m/s ) = 700 N • The mass of an equal volume of water is: mwater = ρ water V = (1000 kg / m 3 ) (60 m 3 ) = 6.0 × 10 4 kg • The weight is: wwater = mwater g = (6.0 × 10 4 kg) (9.8 m/s 2 ) = 5.9 × 105 N Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid • We define the pressure p at the point as the normal force per unit area, that is, the ratio dF┴ to dA : d F⊥ p = dA (d e f in itio n o f p r e s s u r e ) (1 4 .2 ) • If the pressure is the same at all points of a finite plane surface with area A , then: F⊥ p= A (14.3) • The SI unit of pressure in the pascal, where: Dr. Y. Abou-Ali, IUST 1 pascal = 1 Pa = N/m2 University Physics, Chapter 14 14.2 Pressure in a Fluid • Two related units, used principally in meteorology, are the bar , equal to 105 Pa, and millibar , equal to 100 Pa. • Atmospheric pressure p a is the pressure of the earth’s atmospheric, the pressure at the bottom of this sea of air which we live. ( pa )av = 1atm = 1.013 ×105 Pa = 1.013 bar = 1013 millibar Example 14.2 ( The force of air): In the room described in Example 14.1, what is the total downward force on the surface of the floor due to air pressure of 1.00 atm? Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Solution: • Identify and Set Up : The pressure is uniform, so we use Eq. (14.3) to determine the force from the pressure and area. • Execute: The floor area is A = (4.0 m) (5.0 m) = 20 m2, so from Eq. (14.3) the total downward force is 5 2 2 F⊥ = p A= (1.013 × 10 N/m )(20 m ) = 2.0 × 106 N Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law • If the weight of the fluid can be neglected, the pressure in a fluid is the same throughout its volume. But often the fluid’s weight is not negligible. • Atmospheric pressure is less at high altitude than at sea level, which is why an airplane cabin has to pressurized when flying at 35.000 feet. • When you dive in deep water, your ears tell you that the pressure increases rapidly with increasing depth below the surface. • We can derive a general relation between the pressure p at any point in a fluid at rest and the elevation y of the point. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law • We will assume that the density ρ and the acceleration due to gravity g are the same throughout the fluid. • If the fluid in equilibrium, every volume element is in equilibrium. • Consider a thin element of fluid with height dy (figure below). The bottom and top surface have area A , they are at elevations y & y + dy , above some reference level where y = 0. • The volume of the fluid element is dV = A dy , its mass is dm = ρ dV = ρ A dy , and its weight is dw = dm g = ρgA dy . Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law (p + dp) A dy A dw pA y Element of fluid thickness dy (b) 0 (a) Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law • What are the other forces on this fluid element? The pressure at the bottom surface p ; the total y -component of upward force on this surface is pA . The pressure at the top surface is p + dp , and the total y -component of (downward) force on the top surface is – (p + dp )A . • The fluid element is in equilibrium, so ∑ Fy = 0 so pA − ( p + dp ) A − ρ gA dy = 0 dp = −ρ g dy Dr. Y. Abou-Ali, IUST (14.4) University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law • If p 1 and p 2 are the pressure at elevations y 1 and y 2, respectively, and if ρ and g are constant, then p2 − p1 = −ρ g ( y2 − y1) ( pressure in a fluid of uniform density) (14.5) • It is often convenient to express Eq. (14.5) in terms of the depth below the surface of a fluid (figure below). Fluid, density ρ 2 P2= y2 – y1 = P1 = Dr. Y. Abou-Ali, IUST y2 – y1 y1 1 y2 University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law po − p = − ρ g ( y2 − y1) = − ρ gh p = po + ρ g h ( pressure in a fluid of uniform density ) (14.6) • The pressure p at a depth h is greater than the pressure p o at the surface by an amount ρgh. • Note that the pressure is the same at any two points at the same level in the fluid. The shape of the container does not matter. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law • The pressure in any fluid at the same elevation will be the same regardless of the shape or size of the container. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law Pascal’s law: Pressure applied to an enclosed fluid is transmitted to every portion of the fluid and the walls of the containing vessel. • The hydraulic lift shown schematically in figure below illustrates Pascal’s law. • A piston with small cross-sectional area A 1 exerts a force F1 on the surface of a liquid such as oil. The applied pressure p = F1/A 1 transmitted through the connecting pipe to a larger piston of area A 2. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Pressure, Depth, and Pascal’s Law • The applied pressure is the same in both cylinders, so: F1 F2 p= = A1 A2 & A2 F2 = F1 A1 (14.7) 14.7) • In a room with a ceiling height 3.0 m filled with air of uniform density 1.2 kg/m3, the difference in pressure between floor and ceiling given by Eq. (14.6): ρgh = (1.2 kg/m3) (9.8 m/s2) (3.0 m) = 35 Pa Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • If the pressure inside a car tire is equal to atmospheric pressure, the tire is flat. • The pressure has to be greater than atmospheric to support the car, so significant quantity is the difference between the inside and outside pressure. • When we say that the pressure in a car tire is (220 kPa or 2.2 × 10 Pa), 5 we mean it is greater than atmospheric pressure (1.01 × 105 Pa) by this amount. • The total pressure in the tire is then 320 kPa. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • The excess pressure above atmospheric pressure is usually called gauge pressure. • The total pressure is called absolute pressure. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges Example 14.3 ( Finding absolute and gauge pressures): A solar water-heating system uses solar panels on the roof, 12.0 m above the storage tank. The water pressure at the level of the panels is one atmosphere. What is the absolute pressure in the tank? The gauge pressure? Solution: • Identify : Water is nearly incompressible. Hence we can treat it as a fluid of uniform density. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • Set Up : The level of the panels corresponds to point 2 in the figure above, and the level of the tank corresponds to point 1. Hence our target variable is p in Eq. (14.6); we are given p o= 1 atm = 1.01 × 10-5 Pa and h = 12.0 m. • Execute: From Eq. (14.6), the absolute pressure is: p = po + ρ gh = (1.01× 105 Pa )+(1000 kg/m3 )(9.80 m/s 2 )(12.0 m) = 2.19 × 105 Pa = 2.16 atm Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges The gauge pressure is: 5 p − po = (2.19 − 1.01) ×10 Pa 5 = 1.18 ×10 Pa = 1.16 atm Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • The simplest pressure gauge is the open-tube manometer (figure below). The U-shaped tube contains a liquid of density ρ. Pa= y2 - y1= y2 Pressure p y1 P + ρgy1 Dr. Y. Abou-Ali, IUST Pa + ρgy2 University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • The pressure at the bottom of the tube due to the fluid in the left column is p + ρgy 1, and the pressure at the bottom due to the fluid in the right column is p a + ρgy 2. • These pressure are measured at the same point, so they must be equal: p + ρ gy1 = p a + ρ gy 2 p − pa = ρ g ( y2 − y1 ) = ρ gh (14.8 ) • In Eq. (14.8), p is the absolute pressure, and the difference p – p a between absolute and atmospheric pressure is the gauge pressure. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • Another common pressure gauge is the mercury barometer. It consists of a long glass tube, closed at one end, that has to be filled with mercury and then inverted in a dish of mercury. p0 = 0 p = pa y2 – y1= y2 y1 Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • The pressure p o at the top of the mercury column is practically zero. From Eq. (14.6), pa = p = 0 + ρ g ( y2 − y1 ) = ρ gh (14.9 ) • Thus the mercury barometer reads the atmospheric pressure p a directly from the height of the mercury column. • In many applications, pressures are still commonly described in terms of the height of the corresponding mercury column (mm Hg). • A pressure of 1 mm Hg is called 1 torr. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • One common type of blood-pressure gauge, sphygmomanometer, uses a mercury-filled manometer. called a • Blood-pressure readings, such as 130/80, give the maximum and minimum gauge pressures in the arteries, measured in mm Hg or torr. • Blood pressure varies with height; the standard reference point is the upper arm, level with the heart. • We distinguish high-and low pressure sections of the cardiovascular system because the blood pressure varies significantly as illustrated in figure below. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • The high-pressure part includes the aorta, the arteries and arterioles and the capillaries of the systemic circulation. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • The blood pressure in the arteries varies and 16.0 KPa (systolic pressure equal to 120 mm Hg). • The low-pressure part includes the veins and the pulmonary circulation in this circulation the pressure varies only between 1.3 KPa and 3.3 KPa (10 - 25 mm Hg). • The pressure in our cardiovascular system exceeds the ambient air pressure everywhere, this is correct only while lying down negative gauge pressures can occur while standing. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges Example 14.4 ( A tale of two fluids): A manometer tube is partially filled with water. Oil (which does not mix with water, and has a lower density than water) is then poured into the left arm of the tube until the oil-water interface is at the midpoint of the tube (figure below). Both arms of the tube are open to the air. Find a relationship between the heights hoil and hwater. Solution: • Identify : the relationship between pressure and depth in a fluid applies only to fluids of uniform density. Hence we cant write a single equation for the oil and the water together. We can write a pressure-depth relation for each fluid separately. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges hoil hwater Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges • Set Up : Let p o be atmospheric pressure and let p be the pressure at the bottom of the tube. The densities of the two fluids are ρwater and ρoil (which is less than water). We use Eq. (14.6) for each fluid. • Execute: For the two fluids, Eq. (14.6) becomes, p = po + ρ water g hwater p = p o + ρ oil g hoil Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Absolute Pressure, Gauge Pressure, and Pressure Gauges Since the pressure p at the bottom of the tube is the same for both fluids, ρ water hoil = hwater ρoil Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Example: Relative to the blood pressure in supine position calculate the additional blood pressure difference between the brain and the feet in a standing standard man. Use ρ = 1.06 g/cm3. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid Solution: • To quantify the additional difference for the standing standard man we use pascal’s law: • For the pressure difference: h = ybrain − yfeet • For the height of the person. this choice for ∆p eliminates the extra minus sign on the right hand: Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid ∆p = ρblood gh • This difference is about 20% of the atmospheric pressure. • For physiological applications it is more useful to refer to pressure that are measured relative to the pressure at the height of the heart. • For a standing person of 1.73 m height the heart is at a height of 1.22 m and the arterial venous pressure in the feet are increased relative to the pressure at the height of the heart by: ∆p = ρblood gh = (1.06 × 103 )(9.80 )(1.22 ) = 12.7 KPa = 95 mmHg Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.2 Pressure in a Fluid • The pressure is increased to an average arterial value of 190 mm Hg and an average venous value of 100 mm Hg. • In the scalp the pressures decrease for a standing person by: ∆p = ρblood gh = (1.06 × 103 )(9.80)( −0.51) = −5.3 KPa = -40 mmHg • The average arterial pressure drops to a value of 55 mm Hg and the average venous pressure becomes -35 mm Hg. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Blood Pressure Measurement Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Blood Pressure Measurement • The figure above, illustrated what the health practitioner hears with a stethoscope as a function of pressure. • Each sketch shows the pressure relative to systolic (ps) and diastolic (pd) pressure at the right the normal variation of the blood pressure (red curve) and the audible sound (orange curve). • Initially (Fig. a) the pressure in the cuff is increased until the flow of the blood through the brachial artery is below the cuff. A valve on the bulb is then opened to lower the pressure in the cuff. • When the pressure in the brachial artery falls just below the maximum pressure generated by the heart (which is the systolic pressure, where systole refers to the contraction of the heart muscle , the artery opens momentarily on each beat of the heart . Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Blood Pressure Measurement • The velocity of the blood in the artery is high and the blood flow is turbulent during these events this leads to a noisy blood flow easily recognized by the health practitioner. The manometer now reads 120 mm Hg for a standard man with a healthy heart. • When the pressure in the cuff is lowered further (Fig.c) intermittent sounds are heard until the pressure in the cuff falls below the minimum heart pressure (which is the diastolic pressure where diastolic refers to the expansion of the heart muscle) the a continuous background sound is heard as illustrated in (Fig.d) the transition to the continuous sound occurs at about 80 mm Hg for the standard man with a healthy heart. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy • Buoyancy is a familiar phenomenon: a body immersed in water seems to weight less than when it is in air. • When the body is less dense than the fluid, it floats. • The human body usually floats in water. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Archimedes’s principle states: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. • The upward force exerted on the body by the fluid is called buoyant force. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Example 14.5 ( Buoyancy): A 15.0-kg solid gold statue is being raised from a sunken ship (figure below). What is the tension in the hoisting cable when the statue is at rest and a) completely immersed; b) out of the water? Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Solution: • Identify : When the statue is immersed, it experiences an upward buoyant force = in magnitude to the weight of the fluid displaced. Three forces are acting on the statue: weight, the buoyant force, and the tension in the cable. • Set Up : Our target variable is the tension T . We are given mg, we can calculate the buoyant force B by using Archimedes’s principle. • Execute: a) To find B, we first find the volume of the statue, using the density of gold from table 14.1: V= m ρgold Dr. Y. Abou-Ali, IUST = 15.0 kg 19.3 ×103 kg/m3 − 4 3 = 7.77 ×10 m University Physics, Chapter 14 14.3 Buoyancy • Using table 14.1 again, we find the weight of this volume of seawater: wsw = msw g = ρ sw V g = (1.03×103 kg/m3)(7.77 ×10−4 m3)( 9.80 m / s2 ) = 7.84 N This equals the buoyant force B. • The statue is at rest, so the net external force acting on it is zero. ∑ Fy = B + T + (−mg ) = 0 T = mg − B = (15.0kg)( 9.80 m / s2 ) − 7.84 N = 147 N - 7.84 N = 139 N Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy • If a spring scale is attached to the upper end of the cable, it will indicate 7.84 N less than if the statue were not immersed in seawater. • Hence the submerged statue seems to weight 139 N, about 5% less than its actual weight of 147 N. b) The density of air is about 1.2 kg/m3, so the buoyant force of air on the statue is B = ρair V g = (1.2 kg/m3 )( 7.77 ×10−4 m3 )( 9.80m / s2 ) = 9.1×10−3 N • Thus the tension in the cable with the statue in air is equal to the statue’s weight, 147 N. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Example: The human brain is immersed in cerebrospinal fluid of density 1.007 g/cm3, density is slightly less than the average density of the brain 1.04 g/cm3. Thus most of the weight of the brain is supported by the buoyant force of the surrounding fluid. What fraction of the weight of the brain is not supported by this force? Solution: • This problem is an application of the Archimedes principle the weight of the brain is: w = ρbrainVbrain g • The buoyant force in turn for the brain fully impressed in the cerebrospinal fluid: Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy B = ρcerebrospinalVbrain g • We know that these two force do not balance each medulla oblongata to the spinal cord which exerts an additional force on the brain to determine the fraction of the weight of the brain that is not balanced by buoyant force we calculate: w − B ρbrain − ρcerebrospinal 1.04 − 1.007 = = = 0.032 w ρbrain 1.04 • Just 3.2% of the brain’s weight is not balanced by the cerebrospinal fluid, requiring only a small force to be exerted by the spinal cord on the brain. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Surface Tension • The surface of the liquid behaves like a membrane under tension. • Surface tension arises because the molecules of the liquid exert attractive forces on each other. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Surface Tension • There is zero net force on a molecule inside the volume of the liquid. • But a surface molecule is drawn into the volume (figure below). Thus the liquid tends to minimize its surface area, just as a stretched membrane does. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Surface Tension • Surface tension explains why freely falling raindrops are spherical (not teardrop-shaped): a sphere has a smaller surface area for its volume than any other shape. • It also explains why hot, soapy water is used for washing. To wash clothing thoroughly, water must be forced through the tiny spaces between the fibers. • To do so requires increasing the surface area of the water, which is difficult to achieve because of surface tension. • The job is made easier by increasing the temperature of the water and adding soap, both of which decrease the surface tension. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.3 Buoyancy Surface Tension • Surface tension is important for a millimeter-sized water drop, which has a relatively large surface area for its volume. • For large quantities of liquid, the ratio of surface area to volume is relatively small, and surface tension is negligible compared to pressure forces. • For the remainder of this chapter, we will consider only fluids in bulk and hence will ignore the effects of surface tension. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow • We will now consider motion of a fluid. Fluid flow can be extremely complex, as shown by the currents in river rapids. • An ideal fluid is a fluid that is incompressible (that is, its density cant change) and has no internal frication (called viscosity). • The path of an individual particle in a moving fluid is called a flow line. • If the overall flow pattern does not change with time, the flow is called steady flow. •A streamline is a curve whose tangent at any point is in the direction of the fluid velocity at that point. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow • We will consider only steady-flow situations, for which flow lines and streamlines are identical. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow • The flow lines passing through the edge of an imaginary element of area A , form a tube called a flow tube (see figure 14.19/466). • From the definition of a flow line, in steady flow no fluid can cross the side walls of a flow tube; the fluids in different flow tubes cant mix. • Laminar flow, in which adjacent layers of fluid slide smoothly past each other and the flow is steady. • At sufficiently high flow rates, or when boundary surface cause abrupt changes in velocity, the flow can become irregular and chaotic. This is called turbulent flow. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow Laminar flow Dr. Y. Abou-Ali, IUST Turbulent flow University Physics, Chapter 14 14.4 Fluid Flow The Continuity Equation • The mass of a moving fluid does not change as it flows. This leads to an important quantitative relationship called the continuity equation. v2 dV2 = A2v2dt dV1 = A1v1dt Dr. Y. Abou-Ali, IUST v1 University Physics, Chapter 14 14.4 Fluid Flow The Continuity Equation • Let us consider the case of an incompressible fluid so that the density ρ has the same value at all point. • The mass dm 1 flowing into the tube across A 1 in time dt is: dm1 = ρ A1v1 dt • The mass dm 2 flowing into the tube across A 2 in time dt is: dm 2 = ρ A2 v 2 dt • In steady flow the total mass in the tube is constant, so dm 1 = dm 2, Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow The Continuity Equation ρ A1 v1 d t = ρ A 2 v 2 d t A1 v 1 = A 2 v 2 (1 4 .1 0 ) ( c o n tin u ity e q u a tio n , in c o m p r e s s ib le f lu id ) • The product Av is the volume flow rate dV /dt , the rate at which volume crosses a section of the tube: dV = Av ( v o lu m e flo w ra te ) (1 4 .1 1 ) dt • The mass flow rate is the mass flow per unit time through a cross section. This is equal to density ρ times the volume flow rate dV /dt . Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow The Continuity Equation • We can generalise Eq. (14.10): ρ1 A1v1 = ρ 2 A2 v2 (continuity equation, compressible fluid) (14.12) Example 14.6 ( Incompressible fluid flow): As part of a lubricating system for heavy machinery, oil of density 850 kg/m3 is pumped through a cylindrical pipe of diameter 8.0 cm at a rate of 9.5 liters per second. a) What is the speed of the oil? What is the mass flow rate? b) If the pipe diameter is reduced to 4.0 cm, what are the new values of the speed and volume flow rate? Assume that the oil is incompressible. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow Solution: • Identify and Set Up : we use Eq. (14.11), to determine the speed v 1 in the 8.0 cm-diameter section. Using Eq. (14.10), to find the speed v 2 in the 4.0 cm-diameter section. • Execute: a) The volume flow rate dV /dt equals the product A 1v 1, where A 1 is the cross-sectional area of the pipe of diameter 8.0 cm and radius 4.0 cm. Hence: dV / dt v1 = = A1 − 3 3 ( 9.5 L/s)( 10 m /L) π (4.0 ×10−2 m)2 = 1.9 m/s The mass flow rate is: ρ dV /dt = (850 kg/m3)(9.5 × 10-3 m3/s) = 8.1 kg/s. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.4 Fluid Flow b) Since the oil is incompressible, the volume flow rate has the same value (9.5 L/s) in both sections of pipe. From Eq. (14.10), A1 π (4.0 ×10−2 m)2 v2 = v1 = (1.9 m/s) = 7.6 m/s − 2 2 A2 π (2.0 ×10 m) Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Speed of Blood in Capillaries Example: The heart of the standard man pumps 5 liters of blood per minute into the aorta: A) What is the volume flow rate in the cardiovascular system B) What is the speed of the blood in the aorta C) If we assume that the blood passes through all systemic capillaries in our body in series, how fast would it have to flow through each capillary D) What is the speed of the blood in parallel through the system capillaries in the human body. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Speed of Blood in Capillaries Solution: A) The amount of the blood flowing: m3/s B) A typical inner diameter of the aorta is d aorta= 2.2 cm this leads to cross-section area: m2 The speed of the blood in aorta is obtained from it’s inner crosssectional area and the volume passing per second using: Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Speed of Blood in Capillaries We obtain: This is the frequently used results: blood flows through the aorta at an average speed of about 20 cm/sec. C) Let’s assume that blood passes through each single systemic capillary at the rate found in part (a). For the outer diameter a capillary we use 9 µm this value leads to an inner diameter of d capillary = 7 µm the cross section area is: Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Speed of Blood in Capillaries We use the equation of the continuity to derive the speed for blood: Then it would no longer be possible to exchange oxygen and nutrients with the surrounding tissue. The physiological purpose of the cardiovascular system would be lost. D) A slow flow of blood in the systemic capillaries is achieved by arranging them parallel to each other with a combined cross-section that is large than the cross- section of the aorta. The equation of continuity allows us to determine the actual speed of blood in the capillaries once we have know the compined cross-section and the is 2100\cm2 so: Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 Speed of Blood in Capillaries vcapillary ∆V ∆t aorta = = 8.3 ×10 −5 / 0.21m 2 = 4 ×10 − 4 m / s Acapillary Again this is a frequently used value blood flows very slowly through the capillaries at less than 1 mm/s. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.5 Bernoulli’s Equation • We can derive an important relationship called Bernoulli's equation that relates the pressure, flow speed, and height for flow of an ideal, incompressible fluid. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.5 Bernoulli’s Equation • To derive Bernoulli’s equation, we apply the work-energy theorem to the fluid in a section of a flow tube. 2 2 1 1 p1 + ρ gy1 + ρ v1 = p 2 + ρ gy 2 + ρ v 2 2 2 (Bernoulli’s equation) (14.17) • The subscripts 1 & 2 refer to any two points along the flow tube, so we can also write: 2 1 p + ρ gy + ρ v = constant 2 Dr. Y. Abou-Ali, IUST (14.18) University Physics, Chapter 14 14.5 Bernoulli’s Equation • Blood flow characteristics are changed dramatically by plaque. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.5 Bernoulli’s Equation Example 14.7 ( Water pressure in the home): Water enters a house through a pipe with an inside diameter of 2.0 cm at an absolute pressure of 4.0 × 105 Pa (about 4 atm). A 1.0 cm diameter pipe leads to the second-floor bathroom 5.0 m above (figure 14.24). When the flow speed at the inlet pipe is 1.5 m/s, find the flow speed, pressure, and volume flow rate in the bathroom. Solution: • Identify and Set Up : Let points 1 & 2 be at the inlet pipe and at the bathroom, respectively. We take y 1 = 0 (at the inlet) and y 2 = 5.0 m (at the bathroom). Our first two targets are the speed v 1 and pressure p 1. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.5 Bernoulli’s Equation • Execute: From continuity equation, Eq. (14.10): A1 π (1.0 cm) 2 (1.5 m / s) = 6.0 m / s v2 = v1 = A2 π (0.50 cm) 2 We are given p 1 and v 1, and can find p 2 from Bernoulli’s equation: p2 = p1 − 1 ρ (v22 − v12 ) − ρ g ( y2 − y1) = 4.0 ×105 Pa 2 − 1 (1.0 ×103 kg / m3 )( 36 m2 / s2 − 2.25 m2 / s2 ) 2 − (1.0 ×103 kg / m3 )( 9.8 m / s2 )( 5.0 m) Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.5 Bernoulli’s Equation 5 5 5 = 4.0 ×10 Pa - 0.17 ×10 Pa - 0.49 ×10 Pa = 3.3 ×105 Pa = 3.3 atm The volume flow rate is: dV = A2v2 = π (0.50 ×10−2 m2 )2 (6.0 m / s) dt = 4.7 ×10−4 m3 / s = 0.47 L / s • Read Example: 14.8/470 & 14.9/471. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.5 Viscosity and Turbulence Viscosity • Viscosity is internal friction in a fluid. Viscous forces oppose the motion of one portion of a fluid relative to another. • Viscosity is the reason why it takes effort to paddle a canoe through calm water. • Viscous effects are important in the flow of fluids in pipes, the flow of blood, and the lubrication of engine parts. • Fluids that flow readily, such as water or gasoline, have smaller viscosities than do “thick” liquids such as honey or motor oil. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.6 Viscosity and Turbulence Turbulence • When the speed of a flowing fluid exceeds a certain critical value, the flow is no longer laminar. • Instead, the flow pattern becomes extremely irregular and complex, and it changes continuously with time; there is no steadystate pattern.. • This irregular, chaotic flow is called turbulence. Dr. Y. Abou-Ali, IUST University Physics, Chapter 14 14.6 Viscosity and Turbulence Dr. Y. Abou-Ali, IUST University Physics, Chapter 14