ics Announcements day, ember 11, 011 C5: Fluids Pascal s Principle Archimede s Principle Fluid Flows Continuity Equation Bernoulli s Equation Toricelli s Theorem Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468 Announcements If our system is in equilibrium, the net force must be 0. So... P A = A + m This week s lab will be another physics workshop - on fluids this time. No quiz this week. P A = A + ρ( AΔh P = + ρ(δh ΔP = ρ ( Δh NOTE: our derivation here assumes a uniform density of molecules at a iven layer in the atmosphere. In the real atmosphere, density reases with altitude. Nevertheless, our pressure and force balance diaram applies so lon as our layer is sufficiently thin so that within it, the density is approximately constant. If the atmosphere is in equilibrium (which would imply a uniform temperature and no winds blowin, the pressure at a iven heiht above the surface would be the same around the Earth. The same aruments can be made for pressure under water. All other thins bein equal, the pressure at a iven depth below the surface is the same. 1
A scuba diver explores a reef 10 m below the surface. The density of water is 1 /cc. What is the external water pressure on the diver? Worksheet #1 In solvin the last problem, we applied a principle that we haven t even defined yet, but that probably made ood 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! Worksheet # In fact, Pascal s Principle uarantees 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 iven level. CQ Pascal s Principle What happens to a cork when we try to submere it in water? It shoots riht back up to the surface. But what happens to the cork when it ets to the surface? It floats! What s responsible for the motion of the cork? So what must be the net force on the cork as it s floatin on the surface? There must be a force actin upward on the cork reater in manitude than ravity. Archimede s Principle ZERO! What s chaned? con t
What have we noticed about our strane underwater force on the cork? Archimedes had this whole process fiured out some 000 years ao! e said, It s reater than ravity when the cork is completely submered. It s equal to ravity when the cork floats on the surface, only partially submered. Our new force relates to the volume of the cork that s underwater! A body wholly or partially submered in a fluid is buoyed up by a force equal to the weiht of the displaced fluid. So, the cork naturally float with just the riht portion of its volume under the water s surface so that the buoyant force upward from the water equals the ravitational force. If we have a cork with density of 0.8 /cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? Worksheet #4 Worksheet #3 CQ4: Buoyant Force We re now oin to examine the behavior of liquids as they flow or move throuh pipes, the atmosphere, the ocean,... Let s trace out the motion of a iven piece or parcel of water as if flows throuh a channel. These lines, which tell us where a parcel has been and in which direction it is oin, are called trajectories. Fluid Flows: trajectories If the flow is in a condition known as steady state (not varyin 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. streamlines 3
Real flows often result in turbulence -- a condition in which the flow becomes irreular. Real flows are also often viscous. Viscosity describes the internal friction of a fluid, or how well one layer of fluid slips past another. turbulent reion To simplify our problems, we re oin to study the behavior of a class of fluids known as ideal fluids. 1 The fluid is nonviscous (no internal friction 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. This is really just a conservation of mass arument. It says that if I put in 5 of water each second at the left end of my hose, then under steady-state flow conditions, I must et out 5 of water each second at the riht end of the hose. 5 /s 5 /s Δm in = Δm out For ideal fluids in steady-state (unchanin flows, this must be true reardless 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 riht end where fluid leaves. 5 /s 5 /s Nevertheless, the mass enterin at the left each second must equal the mass exitin at the riht. A 1 A The mass / time enterin at the left side is Δm in = ρ ΔV 1 = ρ A 1 = ρa 1 And similarly, the mass / time leavin at riht is Δm out = ρ ΔV = ρ A Δx = ρa A 1 A These two quantities must be equal, leavin us with the relationship ρa 1 = ρa A1 = Av 4
An ideal fluid flows throuh a pipe of cross-sectional area A. Suddenly, the pipe narrows to half it s oriinal width. What is the ratio of the final to the initial speed of the fluid flow? 1 4:1 :1 3 1:1 4 1: 5 1:4 Worksheet #5 Ar In examinin flows throuh pipes in the Earth s ravitational field, Bernoulli found a relationship between the pressure in the fluid, the speed of the fluid, and the heiht off the round of the fluid. The sum of the pressure (P, the kinetic enery per unit volume (0.5ρ, and the potential enery per unit volume (ρy has the same value at all points alon a streamline. Δx A P P + 1 m V v + m y = constant V 1 P + ρv + ρy = constant The chane in kinetic enery of the fluid between the two ends must equal the net work done on the fluid. ΔK = 1 m 1 m 1 A 1 h The hatched reions have the same mass of fluid. W = ΔU We can derive Bernoulli s Equation usin a conservation of enery arument. Skip And the work done by ravity on the fluid is iven by = (mh m = m( h Δx A P Δx A P Next, we consider the work done by the pressure forces at each end of the pipe: W 1 = A 1 = ΔV W = P A Δx = P ΔV Givin a net work on the fluid of A 1 W net = W 1 + W + W derivation con t h The hatched reions have the same mass of fluid. = ΔV P ΔV + m( h Now, puttin it all toether, we have W net = ΔK A 1 derivation con t h ΔV P ΔV + m( h = 1 m( 1 Rearranin we et ΔV + m + 1 m = P 1 ΔV + mh + 1 m 5
Δx A P Airplane wins Dividin by ΔV A 1 h Fiures copyrihted by ALLSTAR: www.allstar.fiu.edu/aero + m ΔV + m 1 ΔV = P + mh ΔV + m ΔV Finally, identifyin density 1 1 P + ρh + ρv = P + ρh + ρv 1 1 1 Bernoulli s Principle: + 1 ρ = P 1 + 1 ρ If >, then < P. That creates a pressure radient force known as lift. Curveballs Fiure copyrihted by Cislunar Aerospace: muttley.ucdavis.edu/book/sports/instructor/curveball-01.html P Bernoulli s Principle: + 1 ρ = P 1 + 1 ρ A lare water tower is drained by a pipe of cross section A throuh 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 bier than that of the drain pipe. Worksheet #6 If >, then < P. That creates a pressure radient force that causes the curveball. v = 0 m/s v exit =? P + 1 ρv + ρy = C Bernoulli s Equation now reduces to a rather simple form The pressure in both these places is approximately the atmospheric pressure And since the crosssectional area of the container is much reater than that of the hole, we can approximate the speed of the fluid descent in the container as v = 0. v = 0 m/s v exit =? Bernoulli s Equation now reduces to a rather simple form + 1 ρv + ρ = + 1 ρ exit vexit = The pressure in both these places is approximately the atmospheric pressure 6
ow a mercury barometer works! Evacuated chamber (i.e., P = 0 Bernoulli s Principle now says ow tall will the mercury column be when the air pressure is at the standard value of 101.3 kpa? The density of mercury is 13.6 /cc. + 1 ρ + ρy 0 0 = P + 1 ρ + ρ( + y 0 apρ= The heiht of the mercury column is usually reported in the nihtly weather! = ρ Worksheet #7 7