Chapter 14 Fluids
Fluids Density Pressure in a Fluid Buoyancy and Archimedes Principle Fluids in Motion Fluid = Gas or Liquid MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01
Densities MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 3
Densities MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 4
Densities 1 Mass = Density * Volume m kg = ρv kg = m 3 m g g kg 10 cm 3 cm 3 cm 1000g m g g kg 10 cm 1 = 3 3 3 cm cm 1000g m g 1 cm = 3 10 kg = m 3 3 3 6 3 3 MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 5
Pressure Pressure arises from the collisions between the particles of a fluid with another object (container walls for example). There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 6
By Newton s 3 rd Law, there is a force on the wall due to the particle. F Pressure is defined as P =. A The units of pressure are N/m and are called Pascals (Pa). Note: 1 atmosphere (atm) = 101.3 kpa MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 7
Example (text problem 9.1): Someone steps on your toe, exerting a force of 500 N on an area of 1.0 cm. What is the average pressure on that area in atmospheres? 1.0 cm 1m 100 cm = 1.0 10 4 m A 500N person weighs about 113 lbs. P av F 500 N = = 4 A 1.0 10 m 6 1Pa = 5.0 10 N/m 1 N/m = 49 atm 1atm 1.013 10 5 Pa MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 8
Gravity s Effect on Fluid Pressure FBD for the fluid cylinder An imaginary cylinder of fluid y P 1 A Imaginary cylinder can be any size w P A x MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 9
Apply Newton s nd Law to the fluid cylinder. Since the fluids isn t moving the net force is zero. P A F P = P or P A P A 1 P P 1 P ( ρad ) ρgd = 1 P P A = w g = ρgd 1 1 0 + ρgd = = 0 0 If P 1 (the pressure at the top of the cylinder) is known, then the above expression can be used to find the variation of pressure with depth in a fluid. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 10
If the top of the fluid column is placed at the surface of the fluid, then P 1 = P atm if the container is open. P = P atm + ρgd You noticed on the previous slide that the areas canceled out. Only the height matters since that is the direction of gravity. Think of the pressure as a force density in N/m MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 11
Measuring Pressure A manometer is a U-shaped tube that is partially filled with liquid, usually Mercury (Hg). Both ends of the tube are open to the atmosphere. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 1
A container of gas is connected to one end of the U-tube Equal pressure Equal pressure If there is a pressure difference between the gas and the atmosphere, a force will be exerted on the fluid in the U-tube. This changes the equilibrium position of the fluid in the tube. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 13
From the figure: At point C Also P c = P atm P B = P B' The pressure at point B is the pressure of the gas. P P P B B = P P gauge B' C = = P P B = ρgd C + ρgd P atm = ρgd P gauge = Pmeas - Patm MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 14
Siphoning MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 15
Atmospheric Pressure Will Support a Column of Fluid The column is sealed at one end, filled with the fluid and then inverted into a container of the same fluid. The difference in pressure between the two ends of the column makes the process work MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 16
A Barometer The atmosphere pushes on the container of mercury which forces mercury up the closed, inverted tube. The distance d is called the barometric pressure. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 17
From the figure and P = P = A B P A = ρgd P atm Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 18
The Many Units of Pressure 1 ATM equals 1.013x10 5 N/m 14.7 lbs/in 1.013 bar 76 cm Hg 760 mm Hg 760 Torr 34 ft H O 9.9 in Hg MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 19
Pascal s Principle A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid. (This is useful in making a hydraulic lift.) MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 0
The applied force is transmitted to the piston of cross-sectional area A here. Apply a force F 1 here to a piston of crosssectional area A 1. In these problems neglect pressure due to columns of fluid. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 1
Mathematically, P at point 1= P at point F A 1 F 1 = F A A = A 1 F 1 MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01
Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A /A 1 ) are 1, 10, and 100. Using Pascal s Principle: A A 1 1 10 100 F 500 N 5000 N 50,000 N MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 3
Pressure Depends Only on the Vertical Height Pressure depends only on the depth of the fluid, not on the shape of the container. So the pressure is the same for all parts of the container that are at the same depth. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 4
Pressure Gauge P gauge = P - P atm When the tire is flat the pressure inside the tire is atmospheric pressure. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 5
Law of Atmosphere dp = -λp dy dp = - λdy P MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 6
Pressure Decrease with Height P(y) = P e 0 -λy ln() λ = 5.5(km) P(y) = P 0exp - y(km) ln() 5.5 MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 7
Archimedes s Principle A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 8
Archimedes s Principle MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 9
Archimedes s Principle m g = m g 1 m = m 1 ρ V =ρ V 1 1 The density and volume are tied together MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 30
Archimedes s Principle BF > BF 1 ρ V g >ρv g L 1 L V > V 1 Since ρ V =ρ V ρ <ρ 1 1 1 The crown isn t 100% gold. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 31
Archimedes s Principle MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 3
Archimedes Principle y F 1 An FBD for an object floating submerged in a fluid. x w F The total force on the block due to the fluid is called the buoyant force. F B = where F F F > 1 F 1 MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 33
Buoyant force = the weight of the fluid displaced The magnitude of the buoyant force is: From before: F B P = = = F F P A P A ( P P )A 1 1 1 P 1 = ρgd The result is F B = ρ gda = ρgv MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 34
Example: A flat-bottomed barge loaded with coal has a mass of 3.0 10 5 kg. The barge is 0.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline? FBD for the barge y F B w x Apply Newton s nd Law to the barge: ρ m w F w ρ V F = B g w = = = F w B w( Ad ) = mb ( ρ V ) m b w w = 0 w g = m b g d = mb ρ A w = 3.0 10 kg ( 3 1000 kg/m )( 0.0 m*10.0 m) 5 = 1.5 m MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 35
Example: A piece of metal is released under water. The volume of the metal is 50.0 cm 3 and its specific gravity is 5.0. What is its initial acceleration? (Note: when v = 0, there is no drag force.) FBD for the metal y Solve for a: F B w x Apply Newton s nd Law to the piece of metal: F = FB w = ma The magnitude of the buoyant force equals the weight of the fluid displaced by the metal. F B = ρ water Vg F B ρwatervg = = = ρwaterv a g g g 1 m ρ objectvobject ρobjectvobject MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 36
Example continued: Since the object is completely submerged V=V object. specific gravity = ρ ρ water where ρ water = 1000 kg/m 3 is the density of water at 4 C. ρobject Given specific gravity = = 5. 0 ρ water a ρ ρ water = g 1 = g 1 = g object V V object 1 S. G. 1 5.0 1 = 7.8 m/s The sign is minus because gravity acts down. BF causes a < g. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 37
Fluid Flow A moving fluid will exert forces parallel to the surface over which it moves, unlike a static fluid. This gives rise to a viscous force that impedes the forward motion of the fluid. A steady flow is one where the velocity at a given point in a fluid is constant. V 1 = constant v 1 v V = constant MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 38
Steady flow is laminar; the fluid flows in layers. The path that the fluid in these layers takes is called a streamline. Streamlines do not cross. Crossing streamlines would indicate a volume of fluid with two different velocities at the same time. An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 39
The Continuity Equation Conservation of Mass Faster Slower The amount of mass that flows though the cross-sectional area A 1 is the same as the mass that flows through cross-sectional area A. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 40
m t = ρ Av is the mass flow rate (units kg/s) V = Av = I V t = Q is the volumetric flow rate (m 3 /s) In general the continuity equation is dm 1 I M1 - I M = dt dm 1 = 0 If then dt 1A1 v1 ρ Av ρ = If the fluid is incompressible, then ρ 1 = ρ. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 41
Example: A garden hose of inner radius 1.0 cm carries water at.0 m/s. The nozzle at the end has radius 0.0 cm. How fast does the water move through the constriction? A v = A v 1 1 A π r v = v = v 1 1 1 1 A πr Simple ratios 1.0 cm = (.0 m / s ) = 50 m / s 0.0 cm MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 4
Bernoulli s Equation Bernoulli s equation is a statement of energy conservation. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 43
This is the most general equation P 1 + 1 ρ gy1 + ρv1 = P + ρgy + 1 ρv Work per unit volume done by the fluid Potential energy per unit volume Kinetic energy per unit volume Points 1 and must be on the same streamline MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 44
Torricelli s Law Application of Bernoulli s Law Torricelli s Law states that the water exiting the hole in the side of the beaker has a speed equal to that it would have had after falling a verticle distance h. ρgh =ρgh + v = b ρv 1 a b b g h MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 45
Venturi Meter Poor transition zones Closed design; Two fluids MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 46
Venturi Meters - Good Transitions Closed design; One fluid Open design; One fluid MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 47
Venturi Meter What about the rest of the fluid in the column? P + ρ v = P + ρ v 1 1 1 F 1 F v = A v = rv 1 1 1 A P - P = ρ g h - ρ g h 1 L F v = (ρ -ρ )g h L F 1 F ρ (r - 1) MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 48
End of Chapter Problems MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 49
Cartesian Diver In reality it is essential that the large bottle be filled with water. http://www.lon-capa.org/~mmp/applist/f/f.htm MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 50
Extra Slides MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 51
Low pressures such as natural gas lines are sometimes specified in inches of water, typically written as w.c. (water column) or W.G. (inches water gauge). A typical gas using residential appliance is rated for a maximum of 14 w.c. which is approximately 0.034 atmosphere. In the United States the accepted unit of pressure measurement for the HVAC industry is inches of water column. MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 5
Application of Pascal s Principle Point 1 Point MFMcGraw-PHY45 Chap_14Ha-Fluids-Revised 10/13/01 53