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Physics 111 Thursday, November 11, 2004 • Ch 15: Fluids Pascal’s Principle Archimede’s Principle Fluid Flows Continuity Equation Bernoulli’s Equation Toricelli’s Theorem Thurs Phys Nov Announcements 111 .11. • Wednesday, 8 - 9 pm in NSC 118/119 • Sunday, 6:30 - 8 pm in CCLIR 468 Thurs Phys Nov Announcements 111 .11. This week’s lab will be another physics workshop - on fluids this time. No quiz this week. Bring PHYSLETS book & CD. Thurs Phys Nov Announcements 111 .11. Friday 3:20 pm in Neils 224 Free cookies in Neils 231 at 3 pm 5 class participation points Tom Morris on forensic accident reconstruction. See how the physics you’ve learned this semester contributes to determining how accidents occur and who’s responsible. Thurs Phys Nov Announcements 111 .11. Thursday, Nov. 18, 2004 11:50 am – 1:05 pm Ch. 7, 8, 9.5 - 9.6, 10 - 13 • Hint: Be able to do the homework (graded AND recommended) and you’ll do fine on the exam! • You may bring one 3”X5” index card (hand- written on both sides), a pencil or pen, and a scientific calculator with you. • I will put any constants and mathematical formulas that you might need on a single page attached to the back of the exam. Thurs Phys Nov Announcements 111 .11. Thursday, Nov. 18, 2004 11:50 am – 1:05 pm Ch. 7, 8, 9.5 - 9.6, 10 - 13 Format: • Three sections (like homework). Do 2 of 3. Topics: Energy, Collisions, SHM, Rotation • One essay question. Required. Topics: Work, Energy • One section of multiple choice. Required. Topics: Gravity/Orbits, Torque/Rotation, SHM, Collisions, Energy/Work Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. We found last time that in hydrostatic equilibrium… P2 = P + ρ g(Δh) 1 ΔP = ρg ( Δh) Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. NOTE: our derivation here assumes a uniform density of molecules at a given layer in the atmosphere. In the real atmosphere, density Decreases with altitude. Nevertheless, our pressure and force balance diagram applies so long as our layer is sufficiently thin so that within it, the density is approximately constant. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. If the atmosphere is in equilibrium (which would imply a uniform temperature and no winds blowing), the pressure at a given height above the surface would be the same around the Earth. The same arguments can be made for pressure under water. All other things being equal, the pressure at a given depth below the surface is the same. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. A scuba diver explores a reef 10 m below the surface. The density of water is 1 g/cc. What is the external water pressure on the diver? Worksheet #1 We know the pressure at the surface of the water is 1 atm = 101.3 kPa. The change in pressure as the diver drops under a 10 m column of water is given by ΔP = ρ g(Δh) = (10 kg/m )(9.8 m/s )(10 m) 3 3 2 ΔP = 9.8 × 10 kg/m/s = 9.8 × 10 N/m 4 2 4 2 P = P0 + ΔP = (101.3 + 98) kPa = 199.3 kPa Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. In solving the last problem, we applied a principle that we haven’t even defined yet, but that probably made good sense to us. We said that the surface pressure at the bottom of the atmosphere equaled the pressure in the surface layer of water. If this weren’t true, the water would fly out of the oceans or sink rapidly toward the ocean floor! Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. In fact, Pascal’s Principle guarantees this will be true. It states: The pressure applied to an enclosed liquid is transmitted undiminished to every point in the fluid and to the walls of the container. Which means, that the pressure below the surface of the water is equal to the surface pressure + the pressure due to the column of water above a given level. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Worksheet #2 ≥ Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Worksheet #2 A container is filled with oil and fitted on both ends with pistons. The area of the left piston is 10 mm2; that of the right piston 10,000 mm2.What force must be exerted on the left piston to keep the 10,000-N car on the right at the same height? 1. 10 N 2. 100 N 3. 10,000 N 4. 106 N 5. 108 N 6. insufficient information PI, Mazur (1997) Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. What happens to a cork when we try to submerge it in water? It shoots right back up to the surface. What’s responsible for the motion of the cork? There must be a force acting upward on the cork greater in magnitude than gravity. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. But what happens to the cork when it gets to the It floats! surface? So what must be the net force on the cork as it’s floating on the surface? ZERO! What’s changed? Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. What have we noticed about our strange underwater force on the cork? • It’s greater than gravity when the cork is completely submerged. • It’s equal to gravity when the cork floats on the surface, only partially submerged. • Our new force relates to the volume of the cork that’s underwater! Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Archimedes had this whole process figured out some 2000 years ago! He said, A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid. So, the cork naturally float with just the right portion of its volume under the water’s surface so that the buoyant force upward from the water equals the gravitational force. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. If we have a cork with density of 0.8 g/cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? Let’s look at the free Worksheet #3 body diagram for B our cork. The gravitational force: W = mg = ( ρcorkVtot )g W W = (7840 kg )Vtot B =W m2s2 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. If we have a cork with density of 0.8 g/cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? The buoyant force is given by B the weight of the displaced water. B = mH 2 O g = ( ρ H 2OVsub )g kg W B = (9800 )Vsub m2s2 Now, set this equal to B =W the gravitational force... Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. If we have a cork with density of 0.8 g/cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? kg W = (7840 )Vtot B m2s2 kg = (9800 )Vsub = B m2s2 kg W le! Vsub (7840 ) V kab = m2s2 sub = 0.8 ar em B =W kg Vtot (9800 ) V R 2 2 m s tot Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Worksheet #4 ≥ Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Worksheet #4 A lead weight is fastened on top of a large solid piece of Styrofoam that floats in a container of water. Because of the weight of the lead, the water line is flush with the top surface of the Styrofoam. If the piece of Styrofoam is turned upside down so that the weight is now suspended underneath it, 1. the arrangement sinks. 2. the water line is below the top surface of the Styrofoam. 3. the water line is still flush with the top surface of the Styrofoam. PI, Mazur (1997) Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. We’re now going to examine the behavior of liquids as they flow or move through pipes, the atmosphere, the ocean,... Let’s trace out the motion of a given piece or parcel of water as if flows through a channel. These lines, which tell us where a parcel has been and in which direction it is going, are called trajectories. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. If the flow is in a condition known as steady state (not varying) then the trajectories are the same as the streamlines. The streamlines tell us the instantaneous direction of motion of a parcel in a flow, whereas the trajectories trace out exactly where the parcel has actually been. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Real flows often result in turbulence -- a condition in which the flow becomes irregular. turbulent region Real flows are also often viscous. Viscosity describes the internal “friction” of a fluid, or how well one layer of fluid slips past another. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. To simplify our problems, we’re going to study the behavior of a class of fluids known as “ideal” fluids. 1) The fluid is nonviscous (no internal friction) 2) The fluid is incompressible (constant density) 3) The fluid motion is steady (velocity, density and pressure at each point remain constant) 4) The fluid moves without turbulence. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. This is really just a conservation of mass argument. It says that if I put in 5 g of water each second at the left end of my hose, then under steady-state flow conditions, I must get out 5 g of water each second at the right end of the hose. Δmin Δmout 5 g/s 5 g/s = Δt Δt Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. For ideal fluids in steady-state (unchanging) flows, this must be true regardless of the shape of the hose. For instance, I could have a hose that’s narrower at the left end where the fluid enters the hose than it is at the right end where fluid leaves. 5 g/s 5 g/s Nevertheless, the mass entering at the left each second must equal the mass exiting at the right. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. v1 A1 A2 v2 The mass / time entering at the left side is Δmin ΔV1 A1Δx1 =ρ =ρ = ρ A1v1 Δt Δt Δt And similarly, the mass / time leaving at right is Δmout ΔV2 A2 Δx2 =ρ =ρ = ρ A2 v2 Δt Δt Δt Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. v1 A1 A2 v2 These two quantities must be equal, leaving us with the relationship ρ A1v1 = ρ A2v2 A1v1 = A2 v2 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. An ideal fluid flows through a pipe of cross-sectional area A. Suddenly, the pipe narrows to half it’s original width. What is the ratio of the final to the initial speed of the fluid flow? 1) 4:1 Worksheet #5 2) 2:1 2 Ar∝ 3) 1:1 4) 1:2 5) 1:4 A1v1 = A2v2 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. In examining flows through pipes in the Earth’s gravitational field, Bernoulli found a relationship between the pressure in the fluid, the speed of the fluid, and the height off the ground of the fluid. The sum of the pressure (P), the kinetic energy per unit volume (0.5ρv2), and the potential energy per unit volume (ρgy) has the same value at all points along a streamline. Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. 1m 2 m P+ v + gy = constant 2V V 1 2 P + ρv + ρgy = constant 2 We can derive Bernoulli’s Equation using a conservation of energy argument. Skip Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Δx2 A2 P2 v2 Δx1 The change in kinetic P1 A1 h2 g energy of the fluid v1 h1 between the two ends must equal the net The hatched regions have work done on the fluid. the same mass of fluid. 1 1 ΔK = mv − mv 2 2 2 2 2 1 Wg = −ΔU And the work done by = −(mgh2 − mgh1 ) gravity on the fluid is given by = mg(h1 − h2 ) Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Δx2 A2 P2 v2 Δx1 Next, we consider the P1 A1 h2 g work done by the v1 h1 pressure forces at each end of the pipe: The hatched regions have W1 = P A1Δx1 = P ΔV the same mass of fluid. 1 1 W2 = −P2 A2 Δx2 = − P2 ΔV Giving a net Wnet = W1 + W2 + Wg work on the fluid of = P ΔV − P2 ΔV + mg(h1 − h2 ) 1 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Δx2 A2 P2 v2 Now, putting it all Δx1 P1 A1 h2 g together, we have v1 h1 Wnet = ΔK 1 P ΔV − P2 ΔV + mg(h1 − h2 ) = m(v − v ) 1 2 2 2 2 1 Rearranging we get… 1 1 P ΔV + mgh1 + mv = P2 ΔV + mgh2 + mv 1 2 2 1 2 2 2 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Δx2 A2 P2 v2 Δx1 P1 A1 h2 g Dividing by ΔV… v1 h1 2 2 mgh1 mv mgh2 mv P+ + = P2 + 1 + 2 1 ΔV 2ΔV ΔV 2ΔV Finally, identifying density… P1 + ρgh1 + ρv = P2 + ρgh2 + ρv 1 2 2 1 1 2 2 2 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Airplane wings Figures copyrighted by ALLSTAR: www.allstar.fiu.edu/aero Bernoulli’s Principle: P + ρv = P2 + ρv 1 2 1 2 1 2 1 2 2 If v1 > v2, then P1 < P2. That creates a pressure gradient force known as “lift.” Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Curveballs Figure copyrighted by Cislunar Aerospace: muttley.ucdavis.edu/Book/Sports/instructor/curveball-01.html P2 v2 Bernoulli’s Principle: P + ρv = P2 + ρv 1 1 2 2 1 1 2 2 2 v1 P1 If v1 > v2, then P1 < P2. That creates a pressure gradient force that causes the “curveball.” Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. A large water tower is drained by a pipe of cross section A through a valve a distance 15 m below the surface of the water in the tower. If the velocity of the fluid in the bottom pipe is 16 m/s and the pressure at the surface of the water is 1 atm, what is the pressure of the fluid in the pipe at the bottom? Assume that the cross-sectional area of the tank is much bigger than that of the drain pipe. Worksheet #6 1 2 P + ρv + ρ gy = constant 2 Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. Pictorial Representation: Knowns: vtop = 0 vtop, Ptop Ptop = 101.5 kPa g vpipe = 16 m/s H = 15.0 m A y H ρ = 1000 kg/m3 x g = 9.81 sm 2 vpipe, Ppipe Unknowns: Ppipe Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. v = 0 m/s Mathematical Representation: Let’s look at the conditions 15 m A at the top of the tower: 1 2 v = 16 m/s Ptop + ρvtop + ρ gytop = 2 1 3 kg 101.3 × 10 Pa + (10 3 )(0 m/s) 2 + (103 kg )(9.8 m/s2 )(15 m) = 3 2 m m3 = 248 kPa This must match the conditions for the pipe at the bottom of the tower... Thurs Phys Nov Ch 15: Fluid Mechanics 111 .11. v = 0 m/s Mathematical Representation: Let’s look at the conditions in the 15 m A pipe at the bottom of the tower: 1 2 v = 16 m/s Ppipe + ρv pipe + ρ gy pipe = 2 1 3 kg 3 kg Ppipe + (10 3 )(16 m/s) + (10 3 )(9.8 m/s2 )(0 m) = 2 2 m m Ppipe + 128 kPa = 248 kPa Ppipe = 120 kPa

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the fluid, bernoulli's equation, pascal's principle, equation of continuity, buoyant force, fluid flow, gauge pressure, surface tension, atmospheric pressure, archimedes’ principle, fluid mechanics, pressure difference, ideal fluid, fluid dynamics, mass density

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