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1 Fluids: Example: Application of Fluid Physics: The Sinking of the Titanic (15 April, 1912) Fact #1: The Titanic sailed at a mass of about 4.2x107 Kg. It was 268.85 m long, 28 m wide, 31.6 m high (bridge to keel) and it floated. Where was the waterline? Fact #2: The Titanic struck an iceberg which created damage to the side below the water surface. The bow became very heavy due to the ship taking on water. The Titanic sank after 2 hours, 40 minutes. How big was the hole in the side of the Titanic? Fact #3: The sinking was complex: First the boat tilted forward, and then it broke between the third and forth smoke stacks. The forward section sank bow first to the bottom and remains largely intact today. The rear of the vessel tilted back almost to level, then, as it filled in with water, it sank vertically with the stern high. It arrived at the bottom in many pieces strewn over a large area. On the way to the bottom of the ocean, why did the bow section remain intact while the stern exploded? What is a fluid? A Fluid is a collection of molecules that are randomly arranged and held together by weak cohesive forces or by forces exerted by the walls of its container. Fluids can be either liquids or gasses. A Fluid is a substance that flows Examples? Water, air, alcohol, oxygen. Fluid mechanics is the study of stationary (statics) and moving (dynamics) fluids. For the first 4 sections of chapter 15, we shall deal only with static fluids 2 Fluid Statics Fluid statics: Read Chapter 15, sections 15-1 to 15-4 Section 15-1: Fluids Density ρ for an object is defined as the mass per unit volume: ρ = m/v SI units (kg/m3) Density of fresh water = 1000 kg/m3 = 1000× (1000g/kg) × 1kg / [(1×106 cm3/m3) ×1 kg] = 1g/cm3 See Table 15-1 on page 446 of your text for densities of a variety of fluids. What is the mass of air in this lecture room? Assume the volume has a triangular shape with height = 6 m, length = 16m, width =25m Then we get M = ρV = ρ× ½ length×height×width = HW 1.28 kg/m3× 6m×16m×25m = 1229 Kg Section 15-2: Pressure How do we apply a force to a fluid? What happens when you push on a fluid? How does it push back on you? Fluids cannot sustain shearing (sliding) or tensile (pulling) stresses. The force exerted by a fluid is always perpendicular to the walls of the vessel. How do we deal with forces in fluids? Answer: we use the concept of pressure. Consider a cube immersed in a static fluid. What forces are exerted by the fluid on the cube? Note that pressure is a scalar, not a vector. A fluid exerts pressure in all directions. At any surface the pressure is exerted perpendicular to that surface- this is because a fluid cannot sustain shearing forces. For an area A on the top or bottom surface of the cube (preferably the bottom or the top since the force is constant across), the pressure is defined as P = F/A 3 Note that Pressure is a scalar quantity (i.e. it has a magnitude but no direction) whereas Force is a vector (with magnitude and direction). Since the force is often dependent on position, it is often better to use the definition: P=dF/dA for an infinitesimal area on the surface of the object. The standard units of pressure are the Pascal. 1 Pascal = 1 Pa = 1 N/m2 Variation of pressure with depth Consider a stationary incompressible fluid of density ρ in a container open to the atmosphere. (We want to measure the pressure as a function of depth in the fluid.) Note the pressure due to air in the atmosphere at sea level is Po = 1.013x105 Pa = 1 atm. This corresponds to the weight of air (in Newtons) above a 1m2 square. PoA HH h We marked a cylinder of water in the beaker with dark lines. These lines could be made from a very thin light material. What would happen to the system if we removed the box? Mg Why do we expect the pressure to vary with depth? Water in lower parts of the beaker has to bear the weight of water above it. PA + We consider an imaginary cylinder of fluid of depth h and cross sectional area A. The fluid enclosed by the cylinder has mass M = ρV = ρAh. Choosing upwards as the positive direction, we sum the forces acting on our imaginary cylinder Σ Fy = PA - Mg - PoA = 0. So PA - ρAhg - PoA = 0 Note that the pressure of the atmosphere is transmitted from the surface of the liquid all the way down to the bottom of the container. Then P = Po + ρgh Titanic Problem: The stern of the Titanic contained large water-tight compartments which contained air at atmospheric pressure. At the site of the sinking, the ocean floor was 3.7 km below the surface. At the ocean floor, what would be the force on a 100 m2 area (i.e. 10m x 10m) on the outside of one of the compartments? 4 Pressure in water at the ocean floor: P = Po + ρgh; Pressure in the compartment = Po (Note that the force on the inside walls of the compartment is PoA whereas that on the outside is PA. P is a lot bigger than Po.) Force F on 100 m2 = (P - Po) A= ρghA =(1030 kg/m3)(9.81 m/s2)(3700m)(100m2) = 3.7x109N Equivalent to 335 atm pressure. Po Weight of Titanic = (4.2x 107kg)(9.81 m/s2) = 4.1x108 N Then about 10 times the weight of the Titanic was exerted on a single 10mx10m section of the stern. It imploded, then exploded when the air escaped. One car = 10,000N, One Titanic = 40,000 cars Pascal's Principle: Pressure applied to a confined fluid is transmitted undiminished to every point of the fluid and to the walls of the containers. The classic example where Pascal's Principle is applied is the hydraulic lift. Fin, Ain, Pin Fout, Aout, Pout Remember the previous example where the force due to the atmospheric pressure was transmitted throughout the compartment. Here we replace the atmosphere with an applied force. Fin is typically applied by a pump and transmitted throughout the fluid. The pressures Pin and Pout are equal (provided the fluid level is the same on both). Then Fin/Ain = Fout/Aout and Fout = Fin(Aout/Ain). This acts like a force multiplier. Note also that the distance traveled by the fluid level on the right will be Ain/Aout that traveled by the fluid level on the left. (There is no free ride, the work done on both sides is equal.) Note also: This is how your car brakes work. Sealed and moveable Fixed Po P Definition: gauge pressure = P - Po is the absolute pressure less the pressure due to the atmosphere. Gauge pressure often comes up because when we measure pressure, it is often the difference between the absolute pressure and atmospheric pressure. (e.g in a bicycle or car tire gauge) 5 Section 15-4: Buoyant Forces and Archimedes Principle Consider a cube of unspecified solid material, side h, located inside a volume of a fluid of density ρ. Note: that here we are not concerned about the weight of the object, only in the net force exerted by the fluid. We can assume that the horizontal forces are balanced since the pressure changes only in the vertical direction. Also, if there was a right/left pressure difference, the box would move horizontally P1 P2 What is the net force exerted vertically on the object by the fluid? Net Force = P2A - P1A, where A = h2. Recall that P2 = P1 + ρgh, then net force = ρgh3 = ρgV = weight of water displaced =buoyant force. Generally, regardless of the object, Archimedes' Principle holds: The buoyant force exerted by the fluid on a body is equal to the weight of fluid displaced by the body. Titanic problem: Where was the waterline on the Titanic? The Titanic was L= 268.85 m long and W=28 m wide. We assume that it had a rectangular shape. (not a good assumption but easier for calculation of volume) L + h Vf ρVf g Mg 6 Sum forces Buoyant force: ρVf g = Mg Where Vf is the volume of the Titanic under the water line. (Recall that the buoyant force is equal to the weight of water displaced) Then ρLWhg = Mg And h = M/ρLW = (4.2 x 107 Kg)/[(1030 Kg/m3)(268.85m)(28m)] = 5.4m So the distance from the bottom of the Titanic to the waterline was 5.4m. Of course, for a more reasonable shape (which would be more triangular than rectangular), the width would not be constant with depth and the estimated water line would be higher. At the maximum designated displacement of the Titanic (66,000 tons), the height would have been 7.8m. This makes sense since another record indicates that the height from the ship bottom to the decks was 10.6m. The Ice Cube Question What happens to the water level when the ice cube melts? The Buoyant force is equal to the weight of water displaced. Let Vf be the volume of the ice cube below the water level. From Archimedes Law: ρwaterVf g = Mice g or ρwater Vf = ρice Vice so Vf = ρice/ρwater Vice or about 0.917 Vice Now, when the ice melts, it changes its density without changing its mass, so the new volume that it occupies is Vnew = Mice/ρwater = ρiceVice/ρwater = Vf so the melted ice just fills the space that it previously occupied under water. There will be no change in water level. Problem from Final Exam in 2001: An iron block containing a number of cavities weighs 6000N in air and 4000N in water. What is the volume of the cavities? The density of iron is 7870 kg/m3 and the density of water is 1000 kg/m3. We assume that water cannot enter the cavities of the iron block and we neglect the weight of the air in the cavities. Then the buoyant force is: FB = 6000N – 4000N = ρwaterVwater g and the volume of the block is Vwater = 2000N/1000kg/m3/9.81m/s2 = 0.2039 m3 7 The volume of iron is given by Miron = ρironViron; Viron = Miron/ρiron = 6000N/(9.81m/s2)/7870 kg/m3 = 0.0777 m3 and the volume of the cavities is 0.2039 m3 – 0.0777 m3 = 0.126m3. Rock in Boat Question: What happens to the water level when the rock is pitched out of the boat and falls to the bottom of the lake? (Note: the density of rock is about 3x that of H2O) The water level goes down The water level depends upon the water volume displaced by the boat and the rock. The volume displaced by the rock is as if the rock had the density of water (true for the boat also) The volume displaced by the rock is as if the rock had the density of rock. Buoyant Force: ρwVf1g = (Mboat+ Mrock)g Vf2 = Mboat/ρw = water volume displaced by boat alone. However, the Change in water displaced by boat = Vf1 - Vf2 = Mrock/ρw rock now displaces: Vrock = Mrock/ρrock which is ~ 1/3 Mrock/ρw. Vf1 = (Mboat + Mrock)/ρw = Mboat/ρw + Mrock/ρw = water volume displaced by boat + rock. Buoyant force: ρwVf2g = Mboatg Therefore, the boat plus rock displace less water and the water level falls. Fluid dynamics Read Section 15.5 Section 15-4: Flow Rate and the Equation of Continuity The Ideal Fluid: 1) Non viscous (no internal friction, doesn’t stick to itself or the container walls) 2) Incompressible (Constant density) (not always found in gasses but a good approximation in most liquids) 3) Steady flow (constant velocity at any point in the fluid) (as opposed to turbulence) 4) No rotational motion 8 Equation of continuity Consider a pipe of variable thickness containing a fluid undergoing steady flow. (The lines of flow (streamlines) are parallel to the sides of the tube.) Note that there doesn’t have to be rigid walls here: this could be inside a volume of liquid undergoing steady flow. Since mass is conserved, the amount of fluid going in, in time ∆t, is equal to the amount of fluid going out in the same time period. v2 A2 v1 A1 This tube of flow consists of a set of streamlines. A streamline is the path followed by a single particle undergoing steady flow. The particle velocity is tangent to the streamline at all points. ∆x2 ∆x1 Fluid that goes in in time ∆t. The rest of the tube is in steady flow. Since nothing is changing, we need not consider it further. Fluid that leaves in time ∆t. In the time interval ∆t, the fluid moves a distance ∆x1 at the entrance and ∆x2 at the exit. v1 = ∆x1/∆t and v2 = ∆x2/∆t. How much fluid enters the tube in time ∆t? What is the flow rate? A1 ∆x1 = A1v1∆t (a volume) A1v1 (volume per unit time) How much fluid leaves the tube in time ∆t? What is the flow rate? A2 ∆x2 = A2v2∆t A2v2 Note that v1 does not equal v2 Since the fluid is incompressible, the amount of fluid entering the tube must be equal to the amount of fluid leaving the tube. Then, A1v1 = A2 v2 the Equation of Continuity Problem: Blood flows from arterioles to the capillaries. A typical arteriole has diameter 0.030 mm and carries blood at a rate of 5.5x10-6 cm3/s (a) What is the speed of blood in an arteriole? (b) Suppose that an arteriole branches into 340 capillaries, each of diameter 4.0x10-6m. What is the blood speed in the capillaries? Arteriole Equation of continuity Capillaries 9 (a) Aarteriolevarteriole = flow rate = 5.5x10-6 cm3/s = 5.5x10-12m3/s then varteriole = (5.5x1012 3 m /s)/(π(3.0x10-5m/2)2) = 7.8x10-3m/s = 7.8 mm/s (b) Aarteriolevarteriole = Acapillariesvcapillaries where Acapillaries = NAcapillary = 340 (π (4.0x10-6m/2)2) = 4.3x10-9m2 Then vcapillaries = Aarteriolevarteriole/ NAcapillary= (5.5x10-12m3/s)/4.3x10-9m2 = 1.3x10-3m/s = 1.3 mm/s Issues with blood: 1) blood flow is not constant- it follows the heart cycle; 2) Blood is a viscous fluid; it is also non-Newtonian- its viscosity depends upon its speed. 3) Blood flow is not necessarily laminar- can have turbulence at certain locations at certain times. More on this later…. Bernoulli's Equation v2 Once again, in time ∆t, at the entrance the fluid moves to the right a distance ∆l1 with a velocity V1 and under a pressure P1. At the exit, the fluid moves a distance ∆l2 and against a pressure P2. P2A2 ∆x2 v1 y2 P1A1 y1 ∆x1 + We aren’t worried about what happens inside; only external forces do work on the system Work done = change in kinetic energy + change in potential energy At 2, the force is opposite the distance travelled. Work done = Force x distance = P1A1∆x1 - P2A2∆x2 = (P1 - P2)V Net change in kinetic energy = 1/2 mv22 - 1/2mv12 Net change in potential energy = mgy2 - mgy1 The tube itself is in steady flow and hence constant kinetic energy and potential energy. In the time ∆t, the rest of the world lost a mass m of fluid at the left end and gained an equal mass m at the right end. We consider only these two masses and the change in energy of the external world. Then (P1 - P2)V = 1/2 mv22 - 1/2 mv12 + mgy2 - mgy1 (Let m = ρV, divide everything by V and collect the 1's on the left and the 2's on the right to get) P1 + 1/2 ρv12 + ρgy1 = P2 + 1/2ρv22 + ρgy2 Or more generally, P + 1/2ρv2 + ρgy = constant throughout the fluid. 10 Titanic Problem (last one). Given that it took 2 Hr 40 min for the Titanic to sink, how large was the hole in the side of the boat? Previously, we learned that at its load of 4.2x107Kg, the waterline was 5.4 m from the bottom. The decks were about 10.6 m from the bottom, so if we add another 4.2x107 Kg, the water will be over the decks and the boat will sink. We wish to know how large a hole in the side of the Titanic will let in 4.2x107kg of water in 2 Hr 40 min. Surface Po We shall assume that h is 2m. Note that 2 hr 40 min is 2x60+40 =160 min and 60x160 =9600 s. 1 h v Po A 2 Apply Bernoulli's equation at 1 and 2 Po + ρgh + 0 = Po + 0 + 1/2ρv2 v = √2gh Torcelli's law Flow rate = Av We choose the centre of the hole as the reference point for height. Note that the difference in atmospheric pressure between 1 and 2 is negligible since ρair = 1.28 kg/m3 whereas ρwater = 1000kg/m3. (m/s) For h=2m, v = 6.3 m/s (m3/s) Accumulated mass of water inside the Titanic = Av ∆t ρ = 4.2x107 kg at the time of sinking. Area of damage A = 4.2x107kg/[v ρ(∆t)] =(4.2x107kg)/[(6.3m/s)(1030kg/m3)(9600s)] = 0.7 m2 Note that the above is an approximation since, 1) we don't really know h, 2) once the water level inside reaches the hole, the flow will slow down, 3) there were likely several holes in the boat, perhaps at different depths. However, experts agree that the size of the holes added up to around 1 m2. It turns out that the evaluated area is not very dependent on h. h=1 A=0.95m2, h=5, A =0.42m2 11 Additional problem using Bernoulli's Eqn. 1 Note: The fluid in the h2 tube is stationary. We also assume that the fluid in the main reservoir is stationary. Also note that the fluid is water. H= 5.0m h2=? 2 3 A2=4.0 cm2 v2 = ? Note that this drawing is not to scale! A3=2.0cm2 v3 =? Apply Bernoulli's equation at 1 and 3 to get v3 Po + ρgH + 0 = Po + 0 + 1/2ρv32; v3 = √2gH = √2(9.81 m/s2)(5m) = 9.9 m/s Apply the Equation of Continuity to get v2: A2v2 = A3v3; Then v2 = v3A3/A2 = (9.9m/s)(2.0 cm2)/(4.0cm2) = 4.9 m/s Apply Bernoulli's at 2 and 3 to get P2 P2 + 0 + 1/2ρv22 = Po + 0 + 1/2 ρv32 ; P2 = Po + 1/2ρ(v32- v22) = Po + 1/2(1000 kg/m3){(9.9m/s)2- (4.9m/s)2} = Po + 3.7x104 Pa Height h2. Apply Bernoulli's at bottom and top of pipe: Po + 3.7x104 Pa +0+0 = Po + ρgh2 Then h2 = (3.7x104 Pa)/{(1000kg/m3)(9.81m/s2)} = 3.7m Note: In order to be able to apply Bernoulli’s equation, the entire region of fluid must be participating in the flow. The fluid in h2 is not participating in the flow, therefore, we cannot simply apply Bernoulli’s equation at the top of h2 and the end of the pipe (3). This is a subtle but important point. The h2 pipe is acting as a sort of pressure probe. 12 Problem 15.65: Air flows through the tube shown in the figure. Assume air is an ideal fluid. a. What are the air speeds v1 and v2 at points 1 and 2? b. What is the volume flow rate? 2 mm 1 2 1.0 cm The pressure here is ρHggh less than the pressure on the other side, neglecting the pressure gradient in air 10 cm The pressures are equal across at the tip of the arrow. Hg We don’t know the velocities at 1 or 2 but we do know that they follow the equation of continuity: A1v1 = A2v2. We know the difference in pressure between 1 and 2 since it is able to support a 10 cm column of mercury. We neglect changes in air pressure with height. Why? ρair(height) = ρairgh = 1.28 kg/m3 gh ρHg(height) = ρairgh = 13,600kg/m3 gh We could of course calculate the pressure gradient in air and a problem might arise where this is important! There is a ratio of 10,625 (greater than 10,000 times!) Then P2 – P1 = ρHggh – ρairgh = ρHg gh to the 4th decimal place. Apply Bernoulli’s at 1 and 2 to get: P1 + ½ ρairv12 + 0 = P2 + ½ ρairv22 Then: v12 – v22 = v12(1 – A12/A22) = 2/ρair (P2 – P1) Or: v12 = (2ρHg/ρair)gh/(1- A12/A22) = 2×13,600 kg/m3/1.28 kg/m3× 9.80× .1m/(1- ((1mm)2/((5mm)2))2) = 20858m2/s2 and v1 = 144 m/s Then v2 = A1/A2v1 = 1/25v1 = 5.78m/s Flow rate = A1v1 = π(1×10-3m)2×144 m/s = 4.54×10-4 m3/s 13 Giancoli Sections 13-10,11: Real Fluids: Viscosity Viscosity is caused by internal friction (units N.s/m2) What causes internal friction? intermolecular electrical attractive forces, entanglements between neighbours, .. Consider two flat plates of area A separated by a fluid of thickness L. The upper plate moves to the right with velocity v Note that the viscous force is determined F by the velocity gradient. L Viscous Force: F = ηAv/L Coefficient of viscosity, η; for water, η = 0.001 N.s/m2 at 20oC Velocity Gradient (=v/L) For an ideal fluid flowing through a cylindrical tube, the flow speed is constant throughout. P η=o P Flow rate = Q = Av Viscous flow through a cylindrical tube η P1, Radius, r P2 Parabolic flow profile. We shall not derive the exact expression for the viscous velocity profile through a tube but rather we shall present an expression for the flow rate of a viscous fluid In an ideal fluid, P1 = P2 and the flow profile is flat. L The actual flow is maximum in the tube centre and stopped at the walls. An important equation for understanding viscous flow is Poiseuilles Equation. (We shall not derive it here.) Flow rate (m3/s) = Q = (P1 - P2)πr4 8ηL Note that for a real fluid with viscosity, when the tube is narrowed but the pressure gradient is maintained constant the flow rate decreases. In fact it decreases as the fourth power of the radius! This is most important for small radii. 14 Bernoulli's Equation Equation of Continuity Poiseuille’s Equation Ideal fluids, low viscosity Ideal Fluids, low viscosity Viscous fluids Blood Flow Head and Arms Lungs Heart Abdomen, legs Average Pressure at the Left Ventricle is 100 torr. (pressure units used in medicine.) (~120 torr at systole and 80 torr at diastole) One torr = 1mm of mercury; the pressure required to support a 1mm column of mercury. One atmosphere (1.013x105Pa) will support P=0 5 3 2 Po = ρHggh where h = 1.013x10 Pa/(13,600kg/m )(9.81m/s ) = .760m Then 100 torr corresponds to a pressure of (100/760)(1.013x105Pa) = 1.33x104 Pa Note that this is a 'gauge pressure' since atmospheric pressure acts inward on the arteries P=Po There is a gravitational gradient in this pressure. The head is about 50 cm above the heart. Phead =Pheart - ρgh = 1.33x104Pa - (1000kg/m3)(9.81m/s2)(0.5m) = 8.42x103Pa = 63 torr The feet are about 1.30 m below the heart. Pfeet = Pheart + ρgh = 1.33x104Pa + (1000kg/m3)(9.81 m/s2)(1.3m) = 2.60x104Pa = 195 torr Blood has a viscosity of about 4x10-3 N.s/m2 (water has η= 1x10-3 N.s/m2) (The actual situation is more complicated since the viscosity of blood depends upon the hematocrit (i.e. the number of red blood cells) and also on the blood speed) 15 For non-viscous fluid, the pressure is constant across a cylinder of flow tube of constant diameter. For a real fluid, like blood, pressure drops along a flow tube. This is because work must be done to overcome viscous forces in the fluid. Hence the 100 torr pressure provided by the heart has the job of pushing blood all the way around the body. The flow velocity for blood in the aorta (leaving the heart) can approach 1 m/s. It depends upon the phase in the heart cycle and the heart output. v r P1 P2 L Consider an artery of length 30 cm and radius 5mm which carries 8 l/m of blood. What is the pressure drop along this artery? Use Poiseille's Equation Q = (P1 - P2) πr4 8ηL Then P1 - P2 = 8QηL/{πr4) = 8(8x10-3/60m3/s)(0.004N.s/m2)(0.3m)/{π(0.005m)4} P1 - P2 = 652 Pa. Note that this is about 1/20 the total pressure available for blood flow around the circulation (100 torr or 1.33×104 Pa Vessel diameter range Approximate blood pressure drop(%) Arteries < 1cm 10 Arterioles < 20 µm 50 Capillaries veins < 1cm 4 to 9 µm 30 10 Atherosclerotic Plaques: In atherosclerotic plaques, buildup of calcium causes severe restriction in artery diameter. For the same pressure gradient, a halving of the vessel radius results in the flow being restricted by a factor 1/16! Many older people these days undergo calcium scoring tests whereby the heart is imaged by x-rays in synchronization with the cardiac cycle. If there is a lot of calcium in the coronary arteries, the person is advised to change lifestyle and perhaps undergo a cardiac catheterisation procedure. Section 15.6 (Knight) Read this chapter. we shall not cover it in detail in the lectures but we shall make use of the definition of bulk modulus later when we cover the propagation velocities of longitudinal waves.