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ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 CHAPTER 2 - FLUIDS 2.1 Mass Density Fluids are materials that can flow which include both gases and liquids. Example: i) gas: air ii) liquid: water Mass density of liquid or gas is important factor that determine its behaviour as a fluid. Example: Air (gas) has smallest densities compared to water (liquid) because gas molecules are relatively far apart and contains large fraction of empty space. While for liquids, the molecules are much more tightly packed that leads to larger densities. Equation of mass density, = mass = m SI unit of = kg / m3 volume v A convenient way to compare densities is to use the concept of specific gravity. specific gravity of substance = density of substance density of water at 4C (1000 kg/m3) EXERCISE 1 What is the approximate mass of air in a living room 4.8m x 3.8m x 2.8m? SOLUTION The mass is found from the density of air and the volume of air. m V 1.29 kg m 3 4.8 m 3.8 m 2.8 m 66 kg EXERCISE 2 Accomplished silver workers in India can pound silver into incredibly thin sheets as thin as 3.00 x 10-7 m. Find the area of such sheet that can be formed from 1.00 kg of silver. SOLUTION m 1.00 kg V 9.52 105 m3 10 500 kg/m 3 V 9.52 10–5 m3 The area of the silver, is, therefore, A 7 317 m2 d 3.00 10 m EXERCISE 3 A bottle has a mass of 35.00 g when empty and 98.44 g when filled with water. When filled with another fluid, the mass is 88.78 g. What is the specific gravity of this other fluid? SOLUTION To find the specific gravity of the fluid, take the ratio of the density of the fluid to that of water, noting that the same volume is used for both liquids. fluid m V fluid mfluid 88.78 g 35.00 g SJ fluid 0.8477 water m V water mwater 98.44 g 35.00 g DEPARTMENT OF ENGINEERING 23 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 2.2 Pressure The pressure P exerted by a fluid is the magnitude F of the force acting perpendicular to a surface in the fluid divided by the area A over which the force acts: P = F /A SI Unit : N/m2 or Pascals ( Pa ) where: 1 Pa = 1 Nm 2 Other units : atmosphere (atm) where 1 atm = 1.013 x 105 Pa EXERCISE 4 High heeled shoes can cause tremendous pressure to be applied to a floor. Suppose the radius of a heel is 6.00x10-3m. Find the pressure that is applied to the floor under the heel because of the weight of a 50 kg woman. SOLUTION F mg (50.0 kg)(9.80 m/s 2 ) P 4.33 106 Pa A r2 6.00 10 m 3 2 EXERCISE 5 A person who weighs 625 N is riding a 98 N mountain bike. Suppose the entire weight of the rider and bike is supported equally by the two tires. If the gauge pressure in each tire is 7.6 x 105 pa, what is the area of contact between each tire and the ground? SOLUTION Since the weight is distributed uniformly, each tire exerts one-half of the weight of the rider and bike on the ground. According to the definition of pressure, Equation 11.3, the force that each tire exerts on the ground is equal to the pressure P inside the tire times the area A of contact between the tire and the ground. From this relation, the area of contact can be found. The area of contact that each tire makes with the ground is F A 1 2 Wperson Wbike 12 625 N 98 N 4.76 104 m2 P P 7.60 105 Pa 2.3 Pressure and Depth in a Static Fluid How the pressure in a fluid varies with depth? Figure 1 shows a container of fluid (assumed to be liquid) in which one column of the liquid is outlined. The liquid is at rest. (If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium). This column has a cross sectional area, A and extends to a depth, h. Figure 1 DEPARTMENT OF ENGINEERING 24 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 3 external forces act on this volume of liquid: i) Force of gravity or weight, mg ii) Downward force, P1A iii) Upward force, P2A exerted by the liquid below it. Since the column is in equilibrium, the sum of vertical forces equal to zero. Fy = 0 P2A P1A – mg = 0 OR P2A = P1A + mg But mass, m = V where V = A x h therefore, P2A = P1A + Ahg OR where P1 = pressure at higher level P2 = P1 +hg P2 = pressure at deeper level h = vertical distance between 2 points according to the equation, pressure P2 at a depth h is greater than P1 Conclusion Pressure in a fluid increases with depth due to the weight of the fluid above the point of interest. EXERCISE 6 At a given instant the blood pressure in the heart is 1.6x10 4 Pa. If an artery in the brain is 0.45m above the heart, what is the pressure in the artery? SOLUTION Pressure P1 in the brain (the higher point), P P2 gh 1.6 104 Pa 1060 kg/m3 9.80 m/s 2 0.45 m 1.110 4 Pa 1 EXERCISE 7 Measured along the surface of the water, a rectangular swimming pool has a length of 15m. Along this length, the flat bottom of the pool slopes downward at an angle of 11 below the horizontal, from one end to the other. By how much does the pressure at the bottom of the deep end exceed the pressure at the bottom of the shallow end? SOLUTION As the depth h increases, the pressure increases according to Equation 11.4 (P2 = P1 + gh). In this equation, P1 is the pressure at the shallow end, P2 is the 3 3 pressure at the deep end, and is the density of water (1.00 10 kg/m , see Table 11.1). We seek a value for the pressure at the deep end minus the pressure at the shallow end. PDeep PShallow gh or PDeep PShallow gh The drawing at the right shows that a value for h can be obtained from the 15-m length of the pool by using the tangent of the 11 angle: 15 m h 11 DEPARTMENT OF ENGINEERING 25 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 h tan11 or h 15 m tan11 15 m PDeep PShallow g 15 m tan11 1.00 103 kg/m3 9.80 m/s 2 15 m tan11 2.9 10 4 Pa 2.4 Pressure Gauges It is possible to measure pressure by relating pressure differences to the variation in the height of a column of liquid. Two basic types of pressure gauges (used to measure pressure) are the mercury barometer and the open tube manometer. 1. Mercury barometer-used for measuring atmospheric pressure, Patm. How is the Patm can be measured using this device? Pressure P2 at point A (at the bottom of mercury tube) is same as at point B, which is called Patm. So, P2 = Patm and P1=0 Pa (because the closed end of the barometer is vacuum) Using the equation P2 = P1 +gh Patm = 0 + gh Patm = gh So, Usually weather forecasters report the pressure in terms of h, expresses it in millimetres (mm) Example: Using Patm = 1.013 x 105 and = 13.6 x 103 kg /m3 (density of mercury), we find that h = 760 mm 2. Open tube manometer - to determine the pressure of container Open tube refers to the fact that one end of a U shaped tube is open to atmosphere and the other end is connected to a system of unknown P. How is the P in the container can be measured using this device? P1=Patm while P2 (pressure at point B) is equal with pressure at point A. PB = PA P2 = Patm + gh P2 – Patm = Pg = gh P2absolute pressure: the actual value for P. Pg = P2 – Patmgauge pressure: is the amount by which a pressure P differs from atmospheric pressure. DEPARTMENT OF ENGINEERING 26 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 Gauge pressure, Pg, is defined to be how much the given pressure is above atmospheric pressure. That is, if the pressure at some point is P, then the gauge pressure at that point is given by Pg = P2 – Patm. It is important, however, to keep in mind that, to find the force exerted on an object due to a given pressure, the absolute pressure (that is, the pressure P) must be used, and not just the gauge pressure, Pg. That is, F = PA, not F = PgA EXERCISE 8 A mercury barometer reads 747.0 mm on the roof of a building and 760.0 mm on the ground. Determine the height of the building .Density of air is 1.29kg/m 3. SOLUTION P = gh 133 Pa 13.0 mm Hg h 1 mm Hg 137 m *Note that : 133 Pa = 1 mm Hg 1.29 kg/m3 9.80 m/s2 EXERCISE 9 An open tube mercury manometer is used to measure the pressure in an oxygen tank, as shown in figure 2. What is the pressure in the tank if the height of the mercury in the right-hand column is 28 cm lower than the mercury in the left- hand column? Patn = 1.013 105 Pa, mercury = 13.6 103 kg/m3 SOLUTION P2 = P oxygen in the tank P1 = Patm P2 = Patm –h = (1.013 105)-[(28x10-2)( 13.6 103 kg/m3)(9.8)] = 6.39 104 Pa EXERCISE 10 A water tower is a familiar sight in many towns. The purpose of such tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. Figure below shows a spherical reservoir that contains 5.25 x 105 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes. DEPARTMENT OF ENGINEERING 27 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 SOLUTION The pressure at the level of house A is given by equation P P gh . atm Since the tank is spherical, its full mass is given by M V [(4 / 3) r 3 ] . Therefore, 1/ 3 1/3 3M 3M 3(5.25 105 kg) r 3 or r 3 5.00 m 4 4 4 (1.000 10 kg/m ) 3 Therefore, the diameter of the tank is 10.0 m, and the height h is given by h 10.0 m + 15.0 m = 25.0 m (a) the gauge pressure in house A is, therefore, Pg= P Patm gh (1.000 103 kg/m3 )(9.80 m/s 2 )(25.0 m) = 2.45 105 Pa (b) The pressure at house B is P P gh , where atm h 15.0 m 10.0 m 7.30 m 17.7 m According to Equation 11.4, the gauge pressure in house B is Pg = P Patm gh (1.000 103 kg/m3 )(9.80 m/s2 )(17.7 m) = 1.73 105 Pa EXERCISE 11 Mercury is poured into a tall glass. Ethyl alcohol is then poured on top of the mercury until the height of the ethyl alcohol itself is 110 cm. The two fluids do not mix and the air pressure at the top of the ethyl alcohol is one atm. What is the absolute pressure at a point that is 7.1 cm below the ethyl alcohol-mercury interface? SOLUTION DEPARTMENT OF ENGINEERING 28 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 2.5 Pascal’s Principle PASCAL’S PRINCIPLE States that any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls. Other words: an external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid. Application : Hydraulic lift Refer to figure (a): (a) A downward force, F1 is applied to a small piston of area A 1. The pressure is transmitted through a fluid to a larger piston of area A2. The pressure is the same on both sides (because same depth). Because pressure is same on both sides, the equation for pressure will be, P P2 1 F1 F2 P OR A1 A2 F1 A2 F2 A1 (applies only when the points 1 and 2 lie at the same depth in the fluid) Since A2 > A1, then F2 > F1 (The input force is greatly multiplied) Note that the greater the area of piston 2 the greater will be the upward force. As a result, it is a good machine for lifting objects. Refer to Figure (b) The bottom surface of the plunger at (b) point B is at the same level as point A, which is at a depth h, beneath the input piston. PA PB P gh P2 1 F1 F gh 2 A1 A2 DEPARTMENT OF ENGINEERING 29 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 Conclusion: Compare figure (a) and (b): F1 in part (b) less than in part (a) because the weight of the h column of hydraulic oil provides some of the input forces to support the car. EXERCISE 12 In a hydraulic car lift the input piston has an area of 1.2x10-2m2 and output plunger has an area of 0.15m2.The combined weight of car and output plunger is 20500N.The hydraulic oil of density 800 kg/m 3 is used in this lift. Calculate the input force needed to support the car and the output plunger if the faces of the input piston and the output plunger area (a) At the same level (b) At different level h=1.10 m as shown in figure. SOLUTION F1 F2 (a) = A1 A 2 A1 (0.012m 2 ) F1 = F2 =(20500N) = 1640 N A2 (0.15m 2 ) (b) P2 = P1 + ρ gh F2 F1 = + ρ gh A 2 A1 F1 F2 = - ρ gh A1 A 2 F2 F1 = A1 ( - ρ gh) = A2 (20500) (0.012) [ - (800)(9.8)(1.1)] (0.15) = 1537 N DEPARTMENT OF ENGINEERING 30 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 EXERCISE 13 In the Hydraulic press used in the trash comparator, the radii of the input and output piston are 6.4x10-3m and 5.1x10-2m respectively. The height difference between input and output piston can be neglected. What force is applied to the trash if input force is 330N? SOLUTION F F Using 1 2 F2 = (A2/A1)F1 A1 A2 2 2 R2 5.1 102 m 6.4 103 m 330 N 2.1 10 N F2 2 F1 4 R1 EXERCISE 14 The hydraulic oil in a lift has a density of 8.3 x 10 2 kg/m3. The weight of the input piston is negligible. The radii of the input piston and the output plunger are 7.7 x 10-3m and 0.125m respectively. What input force F is needed to support the 24500 N combined weight of a car and the output plunger when (a) the bottom surfaces of the piston and plunger are at the same level? (b) the bottom surfaces of the output plunger is 1.3 m above the input piston? SOLUTION 2 (a) Using A = r for the circular areas of the piston and plunger, the input force required to support the 24 500-N weight is 3 2 A2 7.70 10 m F2 F1 24 500 N 93.0 N A 0.125 m 2 1 (b) The pressure P2 at the input piston is related to the pressure P1 at the bottom of the output plunger by equation, P2 = P1 + gh, where h is the difference in heights. Setting P2 F2 / A2 F2 / r22 , P F1 / r12 , and 1 solving for F2, we have r2 F2 F1 22 gh r22 r 1 2 3 7.70 10 m 24 500 N 0.125 m 2 2 8.30 102 kg/m3 9.80 m/s 2 1.30 m 7.70 10 3 m 94.9 N DEPARTMENT OF ENGINEERING 31 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 2.6 Archimedes’ Principle Why things float??? An object that is either partially or completely submerged in a fluid will experience an upward buoyant force exerted by the fluid. The buoyant force is the upward force that a fluid applies to an object that is partially or completely immersed in it. ARCHIMEDES’S PRINCIPLE States that an object wholly or partially immersed in a fluid is buoyed up by a force equal in magnitude to the weight of the fluid displaced by the object. FB Wf m f g f gV f The subscript refer to, B = buoyant force and f = fluid displaced How to relate buoyant force, FB and Weight of object, WO ? 1. If FB > Wo Object will float 2. If FB < Wo Object will sink 3. If FB = Wo Object will be in equilibrium at any submerged depth in a fluid. Usually for the object that is totally submerged Vf = Vo The buoyant force is = FB = Wf = f V f g f Vo g Vo Usually for the object that is partially submerged Vf = Vsub The buoyant force is = FB = Wf = f V f g f Vsubg Vsub (Vsub is the volume of the object submerged beneath the surface of the water) An object will float in a liquid which is more dense than that object. The greater the density of the liquid the more volume of the object will be seen above the surface of the liquid. (For example, very little volume of an iceberg is seen above the surface of the water since the density of the water is not much bigger than the density of the ice.) EXERCISE 15 What percent volume of an iceberg floats beneath the surface of the water? SOLUTION Using Archimedes’s Principle : The buoyant force is = FB = Wf = f Vf g Since the object partially submerged Vf = Vsub FB = Wf = f V f g f Vsubg …(1) DEPARTMENT OF ENGINEERING 32 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 Applying Newton’s 2nd law to the y-direction, we get that Fy may 0 FB mg 0 FB = mo g …(2) Combine (1) and (2) f Vsub g f Vobjectg Replace fluid seawater and object iceberg seawaterVsubmerged iceVice sea water = 1024 kg/m3 and ice = 917 kg/m3 . The percent volume submerged beneath the surface of the water. This answers the famous question of how much of the iceberg is unseen beneath the water’s surface! (Poor Titanic didn’t have a chance!) EXERCISE 16 A scuba diver and her gear displace a volume of 65.0 L and have a total mass of 68.0 kg. (a) What is the buoyant force on the diver in sea water? (b) Will the diver sink or float? SOLUTION (a) The buoyant force is the weight of the water displaced, using the density of sea water. Fbuoyant mwater g waterVdisplaced g displaced 1 10 3 m 3 1.025 10 3 kg m 3 65.0 L 1L 9.80 m s 653 N 2 (b) The weight of the diver is mdiver g 68.0 kg 9.80 m s 2 666 N. Since the buoyant force is not as large as her weight, she will sink , although it will be very gradual since the two forces are almost the same. EXERCISE 17 A flat-bottomed rectangular boat is 4.0 m long and 1.5 m wide. If the load is 2000kg including the mass of the boat, how much height of the boat will be submerged when it floats in a lake? (density of lake water = 1000 kg/m 3) SOLUTION FB = mg = ρ fluid gVfluid = WFluids displaced by the boat mg = ρ fluid g( LxWxH) m 2000 H= = = 0.33m ρxLxW (1000)(4)(1.5) DEPARTMENT OF ENGINEERING 33 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 EXERCISE 18 A spherical balloon has a radius of 7.35 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself. SOLUTION The buoyant force of the balloon must equal the weight of the balloon plus the weight of the helium in the balloon plus the weight of the load. For calculating the weight of the helium, we assume it is at 0 oC and 1 atm pressure. The buoyant force is the weight of the air displaced by the volume of the balloon. Fbuoyant airVballoon g mHe g mballoon g mcargo g mcargo airVballoon mHe mballoon airVballoon HeVballoon mballoon air He Vballloon mballoon 1.29 kg m 3 0.179 kg m 3 7.35 m 930 kg 4 3 3 920 kg 9.0 10 3N EXERCISE 19 A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock? SOLUTION The difference in the actual mass and the apparent mass is the mass of the water displaced by the rock. The mass of the water displaced is the volume of the rock times the density of water, and the volume of the rock is the mass of the rock divided by its density. Combining these relationships yields an expression for the density of the rock. mrock mactual mapparent m waterVrock water rock mrock 9.28 kg rock water m 1.00 10 3 kg m 3 9.28 kg 6.18 kg 2.99 10 3 kg m 3 2.7 Fluids in Motion Basic types of fluid flow: 1. Fluid flow can be steady or unsteady: A fluid is said to display steady flow (or laminar flow) if, at any point in the liquid, the velocity of the flow is always the same. Unsteady flow (or turbulent flow) exists whenever the velocity at a point in the fluid changes as time passes. 2. Fluid flow can be Compressible or Incompressible: Most liquids are nearly incompressible that is the density of a liquid remains almost constant as the pressure changes. To a good approximation, liquids flow in an incompressible manner In contrast, gases are highly compressible. 3. Fluid flow can be viscous on nonviscous Viscosity – degree of internal friction in the fluid, associated with the resistance between 2 adjacent layers. Honey does not flow readily due to higher viscosity. Water is less viscous and flows more readily (no internal friction force between adjacent fluid layers) DEPARTMENT OF ENGINEERING 34 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 Conclusion: An incompressible, non viscous and steady flow of fluid is called as an ideal fluid. Fluids which have the ideal properties obey two important equations. 1. The Equation in Continuity 2. Bernoulli’s Equation 2.8 The Equation in Continuity The mass flow rate of a fluid with a density , flowing with a speed in a pipe of cross-sectional area A, is the mass per second (kg/s) flowing past a point and is given by Mass flow rate = A where : A is the cross-sectional area of the pipe at that point SI unit : m3/s Idea: If the fluid is incompressible, the flow rate must be the same everywhere along the pipe (Conservation of fluid particles: what goes in must come out). This leads directly to the Continuity Equation: Equation of continuity express the fact that mass is conserved: what flows into one end of a pipe flows out the other end, assuming there are no additional entry or exit points in between. Expressed in terms of the mass flow rate, the equation of continuity is 1 A11 2 A22 where the subscripts 1 and 2 denote two points along the pipe. If a fluid is incompressible, the density at any points is the same, 1 2 A11 A2 2 The product A is known as the volume flow rate Q: Q = volume flow rate = A SI unit : kg/s The equation shows that where the tube’s cross-sectional area is large, the fluid speed is small, and conversely, where the tube’s cross-sectional area is small, the speed is large. Refer to the figure V smaller V higher Conclusion This equation tells us that the product of the area and the speed or (Av) is constant. The only way this can be true is if, when the area decreases, the speed increases. Example: DEPARTMENT OF ENGINEERING 35 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 if you are holding a garden hose without a nozzle on the end, and the water is flowing out of the end of the hose, you can see that the flow is a rather slow but steady flow (mass of water per unit time). However, if you place your thumb over the end of the hose where the water is flowing out and press your thumb down on the opening, the water suddenly sprays out very quickly from the small opening that is left. – That is, a smaller area through which the water can flow means the faster the speed of the fluid flow. Refer to figure Another example is water flow in a river. If the river is very wide, then the flow will be relatively slow. However, if the river narrows appreciably, then the water speed will greatly increase, resulting in what is known as “rapids”. Again – less area for the fluid to flow through means a faster resulting speed of fluid flow EXERCISE 20 Water flows from left to right through the five section A,B,C,D and E of a pipe. In which sections does the water speed (a) Increase? (b) Decrease? (c) Remain constant? SOLUTION (a) Increase? [Ans: B] (b) Decrease? [Ans: D] (c) Remain constant? [Ans: A,C,E] DEPARTMENT OF ENGINEERING 36 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 EXERCISE 21 A water hose 1.00 cm in radius is used to fill a 20 liter bucket. If it takes 1.0 minute to fill the bucket, what is the speed at which the water leaves the hose? [1 liter = 103cm3] SOLUTION 20 x10 3 cm 3 Volume flow rate = Q = Av = 20 liter/minute = 1x60 s 20 x10 3 cm 3 1 v= x 1x60 s r 2 20 x10 3 cm 3 1 = x 1x60 s (1cm) 2 = 106 cm/s EXERCISE 22 A 15-cm-radius air duct is used to replenish the air of a room 9.2 m 5.0 m 4.5 m every 16 min. How fast does air flow in the duct? SOLUTION We apply the equation of continuity at constant density, Eq. 10-4b. Flow rate out of duct Flow rate into room V Vroom Aduct vduct r 2 vduct room vduct 2 9.2 m 5.0 m 4.5 m 3.1 m s t to fill r t to fill 60 s 0.15 m 16 min 2 room room 1 min EXERCISE 23 A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.90cm. The flow rate through hose and nozzle is 0.5 liters per second. Calculate the velocity of water : a) in the hose and b) in the nozzle SOLUTION L 10 3 m 3 Q 0.5 x = 5 x 10-4 m3/s s 1L 5 x104 Q A1v1 v1 = 1.96 m/s velocity in the hose r 2 From equation of continuity A1 A1v1=A2v2 => v2 xv1 = 25.46 m/s velocity in nozzle A2 DEPARTMENT OF ENGINEERING 37 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 2.9 Bernoulli’s Equation and it’s applications Bernoulli’s equation is the direct result of the application of conservation energy to fluid. It is written as : p +½ ρv2 + ρgh = constant Or in other words: The sum of the pressure P, KE per unit volume ( 2 / 2 ), and the PE per unit volume ( gh ) has a constant value at all points along a streamline. For any two points: 2 P1 +½ ρv1 + ρgh1= P2 +½ ρv22 + ρgh2 You should recognize the second term as a kinetic energy term, and the third term as a potential energy term!) Note: If the flow is at a constant height, h = 0 (horizontal pipe, as shown in figure)) Then, P + ½ ρv2 = constant If there is no height difference, Bernoulli's equation relates the pressure at two points along the flow to the fluid velocities: P1 - P2 = ½ ρ(v22 - v12) EXERCISE 24 a) An ideal fluid flows at 4.0 m/s in a horizontal circular pipe. If the pipe narrows to half of its original radius, what is the flow speed in narrow section? b) If the fluid is water and the pressure at the narrow section is 1.8 x 10 5Pa, what is the pressure at the wide section? SOLUTION a) From the equation of continuity, A1v1=A2v2 A v r v 2 v2 1 1 1 21 16 m s A2 r2 DEPARTMENT OF ENGINEERING 38 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 b) Since the fluid is flowing horizontally, h is constant, so Bernoulli’s Equation can be written as : p1 +½ ρv12 = p2 +½ ρv2 2 So, p1 = p2 + ½ (v22-v12) = 1.8 x105 + ½(1000)(162-42) = 3.0 x 105 Pa. EXERCISE 25 A 6.0-cm-diameter pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow? SOLUTION Use the equation of continuity (Eq. 10-4) to relate the volume flow of water at the two locations, and use Bernoulli’s equation (Eq. 10-5) to relate the pressure conditions at the two locations. We assume that the two locations are at the same height. Express the pressures as atmospheric pressure plus gauge pressure. We use subscript “1” for the larger diameter, and subscript “2” for the smaller diameter. A1 r2 r2 A1v1 A2 v2 v2 v1 v1 12 v1 12 A2 r2 r2 P0 P1 1 v12 gy1 P0 P2 1 v2 gy2 2 2 2 r14 2 P1 P2 P1 1 v12 P2 1 v2 P2 1 v12 2 v1 2 2 2 r24 r14 r 4 1 2 2 P1 P2 2 32.0 10 3 Pa 24.0 10 3 Pa 2 A1v1 r12 3.0 10 2 m r 4 3.0 10 m 2 4 1 1.0 10 1 r 4 3 kg m 3 1 2.0 10 2 m 4 2 5.6 10 3 m 3 s EXERCISE 26 A small crack occurs at the base of a 15m high dam. The effective crack area through which water leaves is 1.3 x 10-3 m2. (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam? SOLUTION P1 = 1 atmosphere (a) The drawing shows two points, 1 v1 = 0 m/s labeled 1 and 2, in the fluid. Point 1 is at the top of the water, and point 2 is where it flows out of the dam at the bottom. Bernoulli’s equation can be used to determine the speed v2 of the y1 water exiting the dam. 2 P2 = 1 atmosphere According to Bernoulli’s equation, we have v2 y2 P 1 1 2 v1 2 gy1 P2 1 v2 2 2 gy2 DEPARTMENT OF ENGINEERING 39 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 2 Setting P1 = P2, v1 = 0 m/s, and solving for v2, we obtain v2 2 g y1 y2 2 9.80 m/s 2 15.0 m 17.1 m/s (b) The number of cubic meters per second of water that leaves the dam is the volume flow rate Q. According to Equation 11.10, the volume flow rate is the product of the cross-sectional area A2 of the crack and the speed v2 of the water; Q = A2v2. The volume flow rate of the water leaving the dam is Q A2v2 1.30 103 m2 17.1 m/s 2.22 102 m3/s DEPARTMENT OF ENGINEERING 40