Fluids, Continuity, and Bernouli


 Carol Marshall
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1 Fluids, Continuity, and Bernouli Announcements: Exam Tomorrow at 7:30pm in same rooms as before. Web page:
2 Clicker question 1 A satellite, mass m, in uniform circular orbit around the earth, at a steady height H above the surface of the earth. At time t=0, it is above Manhattan, traveling northeast at 17,000 mi/hr. At this moment, what is the direction of the acceleration of the satellite? A) There is no direction, because the acceleration is zero. B) Straight down (towards the center of the earth.) C) Straight up (directly away from the center of the earth.) D) In the same direction as the velocity, northeast. E) Some other direction, significantly different from all the above answers.
3 Clicker question 1 A satellite, mass m, in uniform circular orbit around the earth, at a steady height H above the surface of the earth. At time t=0, it is above Manhattan, traveling northeast at 17,000 mi/hr. At this moment, what is the direction of the acceleration of the satellite? A) There is no direction, because the acceleration is zero. B) Straight down (towards the center of the earth.) C) Straight up (directly away from the center of the earth.) D) In the same direction as the velocity, northeast. E) Some other direction, significantly different from all the above answers. In uniform circular motion we always have centripetal acceleration towards the center of the circle.
4 Clicker question 2 A satellite, mass m, in uniform circular orbit around the earth, at a steady height H above the surface of the earth.
5 Clicker question 2 A satellite, mass m, in uniform circular orbit around the earth, at a steady height H above the surface of the earth.
6 Clicker question 3 The magnitude of the buoyant force on an object which is fully immersed in a liquid depends on A. The mass of the object. B. The depth of the object. C. The density of the object. D. The volume of the object. E. None of the above.
7 Clicker question 3 The magnitude of the buoyant force on an object which is fully immersed in a liquid depends on A. The mass of the object. The buoyant force is always B. The depth of the object. equal to the weight of the liquid C. The density of the object. displaced by the object. D. The volume of the object. E. None of the above. For a fully immersed object, the weight of the displaced water is the density of the water times the volume of the object times g. That is, the volume of the displace water is equal to the volume of the object: so.
8 Clicker question 4 A solid piece of plastic of volume V, and density ρ plastic is floating in a cup of water. (The density of water is ρ water.) What is the buoyant force on the plastic? A. 0 B. ρ plastic V C. ρ water V D. ρ plastic Vg E. ρ water Vg
9 Clicker question 4 What is the buoyant force on the plastic? A. 0 B. ρ plastic V C. ρ water V D. ρ plastic Vg E. ρ water Vg A solid piece of plastic of volume V, and density ρ plastic is floating in a cup of water. (The density of water is ρ water.) Since the plastic is floating, some of it is out of the water and so the volume of displaced water is less than the volume of the plastic object. Since the object is in equilibrium, the net force is zero. Therefore, the object weight must be exactly balanced by the buoyant force. The object weight is W = m plastic g = ρ plastic Vg. Can also use this to figure out the volume of displaced water.
10 Archimedes principle A body partially or fully immersed in a fluid feels an upward force equal to the weight of the displaced fluid. This force is called the buoyant force: As shown, it is due to the increase of pressure with depth in a fluid. If the object is fully immersed then the volume of the displaced fluid is equal to the volume of the object: Note that volume is related to mass and density: If an object is only partially submerged, the volume of the displaced fluid is less than the volume of the object:
11 Buoyancy example A 2 cm by 2 cm by 2 cm cube of iron (ρ=8 g/cm 3 ) is weighed with the iron outside, half in and fully in the water, as shown in the diagram. What is the measured weight in each case? Iron mass: Out of the water: In the water: so so so Archimedes used this principle to determine whether King Hiero s gold crown had any silver mixed in it
12 Mercury barometer Flip a tube filled with mercury upside down in a bucket of mercury. Some mercury will run out, creating a vacuum in the top. Let s examine the pressure at the dotted line. Outside the tube, pressure is atmospheric: Inside the tube, pressure is These two are at the same height so must be equal: Atmospheric pressure holds up the column of mercury The height of the column measures atmospheric pressure 1 atm = 10 5 Pa = 760 mmhg = 760 torr = 30 inhg = 14.7 psi Atmospheric pressure pushes! The vacuum does not pull!
13 Clicker question 5 What happens when the air is evacuated from a 55 gallon drum? A. Nothing B. Squashed like stepping on a soda can C. Implodes from all sides D. Explodes Atmospheric pressure is 14.7 psi = 14.7 lb/in 2.
14 Clicker question 5 What happens when the air is evacuated from a 55 gallon drum? A. Nothing B. Squashed like stepping on a soda can C. Implodes from all sides D. Explodes I will use English units for this exercise. Atmospheric pressure is 14.7 psi = 14.7 lb/in 2. What is the force on the top due to the atmosphere? This is the weight of an H2 Hummer! Usually have the same air pressure underneath the top so the net force is 0. Applies to the sides and bottom as well. What happens when that air pressure is removed?
15 Fluid flow So far, all of our fluid studies have dealt with static situations (the fluid was in static equilibrium). Now we will investigate flowing fluids. We only consider ideal fluids. Ideal fluids are incompressible good assumption for liquids, not for gases. are nonviscous no kinetic friction between liquid parts have laminar flow steady flow with no turbulence
16 Streamlines Consider water flowing through a tube. One can imagine paths taken by a particle in the fluid. These paths are streamlines. In an ideal fluid, streamlines have two important properties: 1. Streamlines never cross. 2. Fluid speed is higher where streamlines are closer together.
17 Continuity equation For a steady flow, the amount of water entering point 1 in time Δt must be equal to the amount of 1 2 water exiting point 2 in time Δt. How much fluid goes through points 1 and 2 in time Δt? An area A traveling at a speed of v will travel d = vδt in time Δt. A 1 A 2 For steady flow, the two volumes A 1 d 1 & A 2 d 2 must be equal: Therefore: d 1 = v 1 Δt d 2 = v 2 Δt This is a continuity equation which says whatever goes in must come out; water is not created or lost inside.
18 Clicker question 6 The end of a hose has a diameter of 4 cm. If one wants the velocity of the water coming out to be 4 times higher, what should be the diameter of the nozzle on the end? A. ¼ cm B. ½ cm C. 1 cm D. 2 cm E. 4 cm
19 Clicker question 6 The end of a hose has a diameter of 4 cm. If one wants the velocity of the water coming out to be 4 times higher, what should be the diameter of the nozzle on the end? A. ¼ cm B. ½ cm C. 1 cm D. 2 cm E. 4 cm Solving the continuity equation for the unknown final area gives us: If we want v 2 /v 1 = 4 then we need A 2 = A 1 /4. Since area depends on the diameter squared, the diameter needs to be reduced by a factor of 2 to get a factor of 4 reduction in area. In equations:
20 About the continuity equation The continuity equation says that the same amount (volume) of liquid goes in as comes out. Note that area times velocity (Av) has units of volume/second. Thus, knowing this could tell us how quickly a vessel of a given volume may be filled or emptied. Continuity equations show up where there is a conserved current. Here it is liquid but next semester it will be electrical current.
21 Bernoulli s equation Our general conservation of energy equation looks like or We consider each term with regards to fluid flow starting with W other. Consider two equal volumes V of water entering and leaving a tube. p 1 p 2 A 1 A 2 The water entering feels a force from the pressure of p 1 A 1 over a distance of d 2 d 1 d 1 so the work done is W=p 1 A 1 d 1 =p 1 V The water exiting feels a force of p 2 A 2 over a distance of d 2 but force and displacement are in opposite directions so the work done is negative: W = p 2 A 2 d 2 = p 2 V The net work done by pressure is W = p 1 V p 2 V
22 Bernoulli s equation We are dealing with the same volumes of water: V = A 1 d 1 = A 2 d 2. Gravitational potential energy is mgy. For a volume V of liquid m = ρv A 2 y 2 So at points 1 and 2 we have and A 1 d 2 y 1 Kinetic energy is ½mv 2 and m = ρv so and d 1 Now that we have all three components, we can combine them.
23 Bernoulli s equation Conservation of energy equation: The net work done by pressure is Gravitational potential energy: Kinetic energy: and and Conservation of energy: Dividing through by the volume V and rearranging gives Bernoulli s equation: Bernoulli s equation is a restatement of conservation of energy! Problems often require both Bernoulli & continuity equations.
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