Notes for Lecture 2 Fluids Buoyancy Fluid Dynamics Bernoulli s Equation

Notes for Lecture 2 Fluids Buoyancy Fluid Dynamics Bernoulli s Equation Lana Sheridan De Anza College April 12, 2017

Last time introduction to static fluids pressure and depth Pascal s principle measurements of pressure

re at a given distance above level 1 in terms of the atmospheric pressure p 1 at level 1 suming that the atmospheric density is uniform over that distance). For example, to d the atmospheric pressure at a distance d above level 1 in Fig.14-3,we substitute Warm Up Question y 1 0, p 1 p 0 and y 2 d, p 2 p. The figure shows four containers of olive oil. Rank them according to the pressure at pdepth p h, greatest first. 1 0 r air gd. en with r r air,we obtain Fig. 14-3 The with depth h belo according to Eq. 1 CHECKPOINT 1 he figure shows four ontainers of olive oil. ank them according o the pressure at depth, greatest first. h (a) (b) (c) (d) A a, b, c, d B a, d, c, b C a, c, d, b D All the same 1 Halliday, Resnick, Walker, 9th ed, page 363.

Overview buoyancy and Archimedes principle fluid dynamics the continuity equation Bernoulli s equation Torricelli s law other implications of Bernoulli s equation

Barometers Barometers are devices for measuring local atmospheric pressure. Typically, simple barometers are filled with mercury, which is very dense. The weight of the mercury in the tube exerts the same pressure as the surrounding atmosphere. On low pressure days, the level of the mercury drops. On high pressure days it rises.

Mercury Barometer The pressure at points A and B is the same. The pressure at B is P 0. h P 0 Above the mercury in the tube is a vacuum, so pressure at A is ρ Hg gh. (ρ Hg = 13.6 kg/m 3 ) A P 0 B Therefore, P 0 = ρ Hg gh. h P 0 Pressure is sometimes quoted in inches of mercury. a 1 Diagrams from Serway & Jewett, 9th ed.

mon baromd Manometer at one end. 14.6a). The of the merpoint A, due to the atmold move mererefore, P 0 5 the mercury lumn varies, us determine 0 5 1 atm 5 be the pres- The pressure being measured, P, can be compared to atmospheric pressure P 0 by measuring the height of the incompressible fluid in the U-shaped tube. a P A h P 0 B b If h is positive, P > P 0, if negative, P < P 0. 5 0.760 m Figure 14.6 Two devices for P P 0 is calledmeasuring the gauge pressure: pressure. (a) a mercury barometer and (b) an open-tube

Buoyancy Astronauts training in their spacesuits: The total mass of NASA s EMU (extravehicular mobility unit) is 178 kg. Why does training underwater make maneuvering in the suits easier? 1 Picture from Hubblesite.org.

Buoyancy The apparent weight of submerged objects is less than its full weight. For an object that would float, but is held underwater, its apparent weight is negative! There is an upward force on an object in a fluid called the buoyant force.

Buoyancy Why does this force exist? Where does it come from?

Buoyancy Why does this force exist? Where does it come from? We know density depends on depth, so an object that s not completely flat will have different pressure on different parts of its surface. Consider a rectangular object of height h and base area A with its top edge at a depth d. The force on each of the four sides will be equal. The force on the bottom will be (P 0 + ρg(h + d))a. The force on the top will be (P 0 + ρgd)a.

Buoyancy How big will the upward force be? F buoy = F up F down

Buoyancy How big will the upward force be? F buoy = F up F down = (P 0 A + ρg(h + d)a) (P 0 A + ρgda) = ρgha = ρgv obj because the volume of the submerged block is V obj = ha.

Buoyancy How big will the upward force be? F buoy = F up F down = (P 0 A + ρg(h + d)a) (P 0 A + ρgda) = ρgha = ρgv obj because the volume of the submerged block is V obj = ha. Notice that ρv obj = m f, the mass of the displaced fluid.

Buoyancy and Archimedes Principle Archimedes Principle The buoyant force on an object is equal to the weight of the fluid that the object displaces. Logically, if a brick falls to the bottom of a pool it must push an amount water equal to its volume up and out of the way.

Buoyancy and Archimedes Principle For a fully submerged object the buoyant force is: F buoy = ρ f V obj g where ρ f is the mass density of the fluid and V obj is the volume of the object. ρ f V obj is the mass of the water moved aside by the object.

Buoyancy and Archimedes Principle An object that floats will displace less fluid than its entire volume. For a floating object: F buoy = ρ f V sub g where V sub is the volume of the part of the object underneath the fluid level only.

Sinking and Floating Will a particular object sink or float in a particular fluid? If the object is less dense than the fluid it will float. If the object is more dense than the fluid it will sink. If the object and the fluid have the same density if will neither float or sink, but drift at equilibrium.

Sinking and Floating Since the relative density of the object to the fluid determines whether it will sink or float, we sometimes use the notion of specific gravity. The specific gravity of an object relates its density to the density of water (or occasionally other liquids): Specific gravity, SG of a sample is the ratio of its density to that of water. SG = ρ sample ρ water Often referenced in brewing!

Sinking and Floating A floating object displaces a mass of water equal to its own mass! (Equivalently, a weight of water equal to its own weight.) This also means that ρ f V sub = m obj.

Questions Military ships are often compared by their displacements, the weight (or mass, depending on context) of water they displace. The USS Enterprise was an aircraft carrier (now decommissioned). Displacement: 94,781 tonnes (metric tons), fully loaded. 1 tonne = 1000 kg What is the mass of the fully loaded USS Enterprise in kgs? 2 Hewitt, page 246.

Question Quick Quiz 14.4 3 You are shipwrecked and floating in the middle of the ocean on a raft. Your cargo on the raft includes a treasure chest full of gold that you found before your ship sank, and the raft is just barely afloat. To keep you floating as high as possible in the water, should you (i) leave the treasure chest on top of the raft, (ii) secure the treasure chest to the underside of the raft, or (iii) hang the treasure chest in the water with a rope attached to the raft? (Assume throwing the treasure chest overboard is not an option you wish to consider.) A option (ii) is the best B option (iii) is the best C options (ii) and (iii) would be the same, better than (i) D All would be the same 3 Serway & Jewett, page 425.

Buoyancy in Air Buoyancy in air works the same way as in liquids: F buoy = ρ f V obj g If an object is less dense than air, it will float upwards. However, in the atmosphere, the density of air varies with height.

Buoyancy in Air 1 Photo by Derek Jensen, Wikipedia.

Buoyancy in Air By roughly how much is your weight reduced by the effects of the air you are submerged in?

Buoyancy in Air By roughly how much is your weight reduced by the effects of the air you are submerged in? Suppose you have a mass of 100 kg and volume of 0.1 m 3. ρ air = 1.20 kg/m 3 (at room temperature and atmospheric pressure)

Fluid Dynamics When fluids are in motion, their behavior can be very complex. We will only consider smooth, laminar flow.

Fluid Dynamics When fluids are in motion, their behavior can be very complex. We will only consider smooth, laminar flow. Laminar flow is composed of streamlines that do not cross or curl into vortices. Streamline The lines traced out by the velocities of individual particles over time. Streamlines are always tangent to the velocity vectors in the flow.

Fluid Dynamics 1 Image by Dario Isola, using MatLab.

Fluid Dynamics A diagram of streamlines can be compared to Faraday s representation of the electric field with field lines. In fluids, the vector field is instead a field of velocity vectors in the fluid at every point in space and time, and streamlines are the field lines. 1 Image by Dario Isola, using MatLab.

Fluid Dynamics We will make some simplifying assumptions: 1 the fluid is nonviscous, ie. not sticky, it has no internal friction between layers 2 the fluid is incompressible, its density is constant 3 the flow is laminar, ie. the streamlines are constant in time 4 the flow is irrotational, there is no curl

Question (14-25) (kg/s). The Equation figure shows 14-25 a pipe and gives the volume flow rate (in cm 3 /s) 4-15 each and the second direction must of flow for all but one section. What are the ond. volume flow rate and the direction of flow for that section? (Assume that the fluid in the pipe is an ideal fluid.) s) and the dirate and the 2 5 4 8 6 4 A 11 cm 3 /s, outward le Problem B 13 cm 3 /s, outward C 3 cm m narrows as 3 /s, inward it falls D cannot be determined m KEY IDEA 1 Halliday, Resnick, Walker, 9th ed, page 373.

Bernoulli s Principle A law discovered by the 18th-century Swiss scientist, Daniel Bernoulli. Bernoulli s Principle As the speed of a fluid s flow increases, the pressure in the fluid decreases. This leads to a surprising effect: for liquids flowing in pipes, the pressure drops as the pipes get narrower.

Bernoulli s Principle Why should this principle hold? Where does it come from? 1 Something similar can be argued for compressible fluids also.

Bernoulli s Principle Why should this principle hold? Where does it come from? Actually, it just comes from the conservation of energy, and an assumption that the fluid is incompressible. 4 Consider a fixed volume of fluid, V. In a narrower pipe, this volume flows by a particular point 1 in time t. However, it must push the same volume of fluid past a point 2 in the same time. If the pipe is wider at point 2, it flows more slowly. 1 Something similar can be argued for compressible fluids also.

Bernoulli s Principle V = A 1 v 1 t also, V = A 2 v 2 t

Bernoulli s Principle V = A 1 v 1 t also, V = A 2 v 2 t This means The Continuity equation. A 1 v 1 = A 2 v 2

Bernoulli s Equation Bernoulli s equation is just the conservation of energy for this fluid. The system here is all of the fluid in the pipe shown. Both light blue cylinders of fluid have the same volume, V, and same mass m. We imagine that in a time t, volume V of fluid enters the left end of the pipe, and another V exits the right.

Bernoulli s Equation At t 0, fluid in the blue portion is moving past It makes sense that the energy of the S fluid might change: the fluid point 1 at velocity v 1. is moved along, and some is lifted up. A 1 a x 1 S v 1 Point 2 Point 1 x 2 A 2 S v 2 How does it change? Depends on the work done: After a time interval t, the fluid in the blue portion is moving past W = K + U S The path taken by velocity of the particl A set of streamlines particles cannot flow lines would cross one Consider ideal flui ure 14.16. Let s focus shows the segment at

a and 14.18b are equal because the fluid is incompressible.) This work ecause Bernoulli s the force on the Equation segment of fluid is to the left and the displacepoint of application of the force is to the right. Therefore, the net work segment by these forces in the time interval Dt is The pressure at point 1 is P 1. W 5 (P 1 2 P 2 )V The work done is the sum of the work done on each end of the fluid by more fluid that is on either side of it: P 1 A 1 ˆi W = F 1 x 1 F 2 x 2 y 1 x 1 S v 1 Point 2 = P 1 A 1 x 1 P 2 A 2 x 2 a b Point 1 The pressure at point 2 is P 2. x 2 S v 2 y 2 P 2 A 2 ˆi (The environment fluid just to the right of the system fluid does negative work on the system as it must be pushed aside by the system fluid.) 1 Diagram from Serway & Jewett.

Bernoulli s Equation Notice that V = A 1 x 1 = A 2 x 2 W = P 1 A 1 x 1 P 2 A 2 x 2 = (P 1 P 2 )V Conservation of energy: W = K + U

Bernoulli s Equation Notice that V = A 1 x 1 = A 2 x 2 W = P 1 A 1 x 1 P 2 A 2 x 2 = (P 1 P 2 )V Conservation of energy: W = K + U (P 1 P 2 )V = 1 2 m(v 2 2 v 2 1 ) + mg(h 2 h 1 )

Bernoulli s Equation Notice that V = A 1 x 1 = A 2 x 2 W = P 1 A 1 x 1 P 2 A 2 x 2 = (P 1 P 2 )V Conservation of energy: W = K + U (P 1 P 2 )V = 1 2 m(v 2 2 v 2 1 ) + mg(h 2 h 1 ) Dividing by V : P 1 P 2 = 1 2 ρv 2 2 + ρg(h 2 h 1 ) P 1 1 2 ρv 2 1 + ρgh 1 = P 2 + 1 2 ρv 2 2 + ρgh 2

Bernoulli s Equation P 1 1 2 ρv 2 1 + ρgh 1 = P 2 + 1 2 ρv 2 2 + ρgh 2 This expression is true for any two points along a streamline. Therefore, P + 1 2 ρv 2 + ρgh = const is constant along a streamline in the fluid. This is Bernoulli s equation.

Bernoulli s Equation P + 1 2 ρv 2 + ρgh = const Even though we derived this expression for the case of an incompressible fluid, this is also true (to first order) for compressible fluids, like air and other gases. The constraint is that the densities should not vary too much from the ambient density ρ.

Bernoulli s Principle from Bernoulli s Equation For two different points in the fluid, we have: 1 2 ρv 2 1 + ρgh 1 + P 1 = 1 2 ρv 2 2 + ρgh 2 + P 2

Bernoulli s Principle from Bernoulli s Equation For two different points in the fluid, we have: 1 2 ρv 2 1 + ρgh 1 + P 1 = 1 2 ρv 2 2 + ρgh 2 + P 2 Suppose the height of the fluid does not change, so h 1 = h 2 : 1 2 ρv 2 1 + P 1 = 1 2 ρv 2 2 + P 2

Bernoulli s Principle However, rom the continuity equation A 1 v 1 = A 2 v 2 we can see that if A 2 is smaller than A 1, v 2 is bigger than v 1. So the pressure really does fall as the pipe contracts!

s. 14-32, 14-33, and 14-34 yields Question V(y 2 y 1 ) V(p 2 p 1 ) 1 2 V(v 2 2 v 2 1). arrangement, Water flows matches smoothly Eq. 14-28, through which the pipe we set shown out to in prove. the figure, descending in the process. Rank the four numbered sections of pipe according to the volume flow rate through them, greatest first. 4 ly through the ure, descending the four nume according to ate R V through eed v through ater pressure p first. A 4, 3, 2, 1 1 B 1, (2 and 3), 4 C 4, (2 and 3), 1 D All the same Flow 1 Halliday, Resnick, Walker, 9th ed, page 375. 2 3 4

s. 14-32, 14-33, and 14-34 yields Question V(y 2 y 1 ) V(p 2 p 1 ) 1 2 V(v 2 2 v 2 1). arrangement, Water flows matches smoothly Eq. 14-28, through which the pipe we set shown out to in prove. the figure, descending in the process. Rank the four numbered sections of pipe according to the flow speed v through them, greatest first. 4 ly through the ure, descending the four nume according to ate R V through eed v through ater pressure p first. A 4, 3, 2, 1 1 B 1, (2 and 3), 4 C 4, (2 and 3), 1 D All the same Flow 1 Halliday, Resnick, Walker, 9th ed, page 375. 2 3 4

s. 14-32, 14-33, and 14-34 yields Question V(y 2 y 1 ) V(p 2 p 1 ) 1 2 V(v 2 2 v 2 1). arrangement, Water flows matches smoothly Eq. 14-28, through which the pipe we set shown out to in prove. the figure, descending in the process. Rank the four numbered sections of pipe according to the water pressure P within them, greatest first. 4 ly through the ure, descending the four nume according to ate R V through eed v through ater pressure p first. A 4, 3, 2, 1 1 B 1, (2 and 3), 4 C 4, (2 and 3), 1 D All the same Flow 1 Halliday, Resnick, Walker, 9th ed, page 375. 2 3 4

Torricelli s Law from Bernoulli s Equation Bernoulli s equation can also be used to predict the velocity of streams of water from holes in a container at different depths.

Torricelli s Law from Bernoulli s Equation The liquid at point 2 is at rest, at a height y 2 and pressure P. At point 1 is leaves with a velocity v 1, at a height y 1 and pressure P 0. blow more strongly and watch the increased pressure difin its side at a distance he atmosphere, and its air above the liquid is id as it leaves the hole er. When the hole is pressure P at the top speed. If the pressure extinguisher must be 1 Point 2 is the surface of the liquid. P y 2 h A 2 A 1 Point 1 is the exit point of the hole. S v 1 P y 0 1 Figure 14.20 (Example 14.9) A liquid leaves a hole in a tank at speed v 1. ρv2 2 + ρgy 2 + P 2 ρv 2 1 + ρgy 1 + P 0 = 1 2 0

Torricelli s Law from Bernoulli s Equation in its side at a distance he atmosphere, and its air above the liquid is id as it leaves the hole Point 2 is the surface of the liquid. P h A 2 A 1 Point 1 is the exit point of the hole. er. When the hole is pressure P at the top speed. If the pressure extinguisher must be y 2 S v 1 P y 0 1 Figure 14.20 (Example 14.9) 1A liquid leaves a hole in a tank at 2speed ρv 1 2 + ρgy v 1. 1 + P 0 = ρgy 2 + P Rearranging, and using y = h 2 h 1, 2(P P 0 ) v 1 = + 2gh ρ

Torricelli s Law from Bernoulli s Equation Notice that if the container is open to the air (P = P 0 ), then the speed of each jet is v = 2gh where h is the depth of the hole below the surface.

Implications of Bernoulli s Principle Bernoulli s principle also explains why in a tornado, hurricane, or other extreme weather with high speed winds, windows blow outward on closed buildings.

Implications of Bernoulli s Principle Bernoulli s principle also explains why in a tornado, hurricane, or other extreme weather with high speed winds, windows blow outward on closed buildings. The high windspeed outside the building corresponds to low pressure. The pressure inside remains higher, and the pressure difference can break the windows. It can also blow off the roof! It makes sense to allow air a bit of air to flow in or out of a building in extreme weather, so that the pressure equalizes.

Implications of Bernoulli s Principle Bernoulli s principle can help explain why airplanes can fly. Air travels faster over the top of the wing, reducing pressure there. That means the air beneath the wing pushes upward on the wing more strongly than the air on the top of the wing pushes down. This is called lift. 1 Diagram from HyperPhysics.

Air Flow over a Wing In fact, the air flows over the wing much faster than under it: not just because it travels a longer distance than over the top. This is the result of circulation of air around the wing. 1 Diagram by John S. Denker, av8n.com.

Air Flow over the Top of the Wing: Bound Vortex A starting vortex trails the wing. The bound vortex appears over the wing. Those two vortices counter rotate because angular momentum is conserved. The bound vortex is important to establish the high velocity of the air over the top of the wing. 1 Image by Ludwig Prandtl, 1934, using water channel & aluminum particles.

Wingtip Vortecies Other vortices also form at the ends of the wingtips. 1 Photo by NASA Langley Research Center.

Vortices around an Airplane 1 Diagram by John S. Denker, av8n.com.

Airflow at different Angles of Attack 1 Diagram by John S. Denker, av8n.com.

Implications of Bernoulli s Principle A stall occurs when turbulence behind the wing leads to a sudden loss of lift. The streamlines over the wing detach from the wing surface. This happens when the plane climbs too rapidly and can be dangerous. 1 Photo by user Jaganath, Wikipedia.

Implications of Bernoulli s Principle Spoilers on cars reduce lift and promote laminar flow. 1 Photo from http://oppositelock.kinja.com.

Implications of Bernoulli s Principle Wings on racing cars are inverted airfoils that produce downforce at the expense of increased drag. This downforce increases the maximum possible static friction force turns can be taken at higher speed. 1 Photo from http://oppositelock.kinja.com.

Implications of Bernoulli s Principle A curveball pitch in baseball also makes use of Bernoulli s principle. The ball rotates as it moves through the air. Its rotation pulls the air around the ball, so the air moving over one side of the ball moves faster. This causes the ball to deviate from a parabolic trajectory. 1 Diagram by user Gang65, Wikipedia.

Summary buoyancy and Archimedes principle fluid dynamics the continuity equation Bernoulli s equation Torricelli s law other implications of Bernoulli s equation Test Wednesday, April 19, in class. Collected Homework due Wednesday, April 19. (Uncollected) Homework Serway & Jewett: Ch 14, onward from page 435, OQs: 3, 5, 7, 9, 13; CQs: 5, 9, 14; Probs: 25, 27, 29, 35, 36, 43, 49, 53, 60, 65, 71, 77, 85