ics Thursday, ember 11, 2004 Ch 15: Fluids Pascal s Principle Archimede s Principle Fluid Flows Continuity Equation Bernoulli s Equation Toricelli s Theorem
Announcements Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468
Announcements This week s lab will be another physics workshop - on fluids this time. No quiz this week. Bring PHYSLETS book & CD.
Announcements 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.
Announcements Thursday,. 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 (handwritten 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.
Announcements Thursday,. 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
We found last time that in hydrostatic equilibrium P 2 = P 1 + ρg(δh) ΔP = ρ g( Δh)
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.
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.
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 = ρg(δh) = (10 3 kg/m 3 )(9.8 m/s 2 )(10 m) ΔP = 9.8 10 4 kg/m/s 2 = 9.8 10 4 N/m 2 P = P 0 + ΔP = (101.3 + 98) kpa = 199.3 kpa
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!
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.
Worksheet #2
Thurs Worksheet #2 A container is filled with oil and fitted on both ends with pistons. The area of the left piston is 10 mm 2 ; that of the right piston 10,000 mm 2.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)
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.
But what happens to the cork when it gets to the surface? It floats! So what must be the net force on the cork as it s floating on the surface? ZERO! What s changed?
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!
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.
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 body diagram for our cork. Worksheet #3 B The gravitational force: W = mg = (ρ cork V tot )g W W = (7840 kg m 2 s 2 )V tot B = W
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 the weight of the displaced water. B B = m H 2O g = (ρ H 2O V sub )g B = (9800 kg m 2 s 2 )V sub Now, set this equal to the gravitational force... B = W W
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? W = (7840 V sub V tot = = (9800 (7840 (9800 kg m 2 s 2 )V tot kg m 2 s 2 )V sub = B kg m 2 s 2 ) V sub kg ) m 2 s 2 V tot = 0.8 B = W W B Remarkable!
Worksheet #4
Thurs 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)
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.
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.
Real flows often result in turbulence -- a condition in which the flow becomes irregular. 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 region
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.
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. 5 g/s 5 g/s Δm in Δt = Δm out Δt
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.
v 1 v 2 A 1 A 2 The mass / time entering at the left side is Δm in Δt = ρ ΔV 1 Δt = ρ A 1 Δx 1 Δt = ρa 1 v 1 And similarly, the mass / time leaving at right is Δm out Δt = ρ ΔV 2 Δt = ρ A 2 Δx 2 Δt = ρa 2 v 2
v 1 v 2 A 1 A 2 These two quantities must be equal, leaving us with the relationship ρa v = ρa v A v = A v 1 1 2 2 1 1 2 2
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 2) 2:1 3) 1:1 4) 1:2 5) 1:4 Worksheet #5 2Ar A 1 v 1 = A 2 v 2
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 (0.5ρv 2 ), and the potential energy per unit volume (ρgy) has the same value at all points along a streamline.
P + 1 2 m V v2 + m V gy = constant 1 P + ρv 2 + ρgy = constant 2 We can derive Bernoulli s Equation using a conservation of energy argument. Skip
Δx 2 A 2 P 2 The change in kinetic energy of the fluid between the two ends must equal the net work done on the fluid. ΔK = 1 2 mv 2 1 2 2 mv 2 1 And the work done by gravity on the fluid is given by P 1 A 1 v 1 Δx 1 h 1 h 2 v 2 The hatched regions have the same mass of fluid. W g = ΔU g = (mgh 2 mgh 1 ) = mg(h 1 h 2 )
Δx 2 A 2 P 2 Next, we consider the work done by the pressure forces at each end of the pipe: W 1 = P 1 A 1 Δx 1 = P 1 ΔV W 2 = P 2 A 2 Δx 2 = P 2 ΔV P 1 A 1 v 1 Δx 1 h 1 h 2 v 2 The hatched regions have the same mass of fluid. g Giving a net work on the fluid of W net = W 1 + W 2 + W g = P 1 ΔV P 2 ΔV + mg(h 1 h 2 )
Δx 2 A 2 P 2 Now, putting it all together, we have P 1 A 1 Δx 1 h 2 v 2 g W net = ΔK v 1 h 1 PΔV P ΔV + mg(h h ) = 1 1 2 1 2 2 m(v 2 v 2 ) 2 1 Rearranging we get PΔV + mgh + 1 1 1 2 mv 2 = P ΔV + mgh + 1 1 2 2 2 mv 2 2
Δx 2 A 2 P 2 v 2 Dividing by ΔV P 1 A 1 v 1 Δx 1 h 1 h 2 g P + mgh 1 1 ΔV + mv 2 1 2ΔV = P 2 + mgh 2 ΔV + mv 2 2 2ΔV Finally, identifying density P + ρgh + ρv = P + ρgh + ρv 1 1 1 2 2 1 2 2 1 2 2 2
Airplane wings Figures copyrighted by ALLSTAR: www.allstar.fiu.edu/aero Bernoulli s Principle: P 1 + 1 ρv 2 = P 2 1 2 + 1 ρv 2 2 2 If v 1 > v 2, then P 1 < P 2. That creates a pressure gradient force known as lift.
Curveballs Figure copyrighted by Cislunar Aerospace: muttley.ucdavis.edu/book/sports/instructor/curveball-01.html v 2 P 2 Bernoulli s Principle: P 1 + 1 ρv 2 = P 2 1 2 + 1 ρv 2 2 2 v 1 P 1 If v 1 > v 2, then P 1 < P 2. That creates a pressure gradient force that causes the curveball.
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 P + 1 2 ρv2 + ρgy = constant
Pictorial Representation: H v top, P top A g v pipe, P pipe y x Knowns: v top = 0 P top = 101.5 kpa v pipe = 16 m/s H = 15.0 m ρ = 1000 kg/m 3 g = 9.81 m s 2 Unknowns: P pipe
v = 0 m/s Mathematical Representation: 15 m A Let s look at the conditions at the top of the tower: v = 16 m/s P + 1 top 2 ρv 2 top + ρgy top = 101.3 10 3 Pa + 1 2 (103 kg m 3 )(0 m/s)2 + (10 3 kg = 248 kpa m 3 )(9.8 m/s2 )(15 m) = This must match the conditions for the pipe at the bottom of the tower...
v = 0 m/s Mathematical Representation: 15 m A Let s look at the conditions in the pipe at the bottom of the tower: v = 16 m/s P + 1 pipe 2 ρv 2 pipe + ρgy pipe = P + 1 pipe 2 (103 kg )(16 m 3 m/s)2 + (10 3 kg )(9.8 m 3 m/s2 )(0 m) = P pipe + 128 kpa = 248 kpa P pipe = 120 kpa