Chapter 11 Fluids
11.1 Mass Density DEFINITION OF MASS DENSITY The mass density of a substance is the mass of a substance divided by its volume: ρ m V SI Unit of Mass Density: kg/m 3
11.1 Mass Density
11.1 Mass Density Example 1 Blood as a Fraction of Body Weight The body of a man whose weight is 690 N contains about 5.x10-3 m 3 of blood. (ρ oo bbbbb ii 1060 kk/m 3 ) (a) Find the blood s weight and m Vρ W mg ( 3 3 )( 3 5. 10 m 1060kg m ) 5.5 kg ( )( 5.5 kg 9.80m s ) 54 N (b) express it as a percentage of the body weight. 54 N Percentage 100% 690 N 7.8%
11. Pressure P F A SI Unit of Pressure? 1 N/m 1Pa Pascal
11. Pressure Example The Force on a Swimmer Suppose the pressure acting on the back of a swimmer s hand is 1.x10 5 Pa. The surface area of the back of the hand is 8.4x10-3 m. (a) Determine the magnitude of the force that acts on it. F P A F PA 5 3 ( 1. 10 N m )( 8.4 10 m ) (b) Discuss the direction of the force. 1000 N Since the water pushes perpendicularly against the back of the hand, the force is directed downward in the drawing. Equilibrium
11. Pressure Atmospheric Pressure at Sea Level: 1.013x10 5 Pa 1 atmosphere
11.3 Pressure and Depth in a Static Fluid F y P A P A mg 1 0 P A P A + 1 mg m Vρ P A 1 P A + ρ Vg
11.3 Pressure and Depth in a Static Fluid V Ah P A P A + ρ Vg 1 P A 1 P A + ρ Ahg P P 1 + ρ hg
11.3 Pressure and Depth in a Static Fluid Example 4 The Swimming Hole Points A and B are located a distance of 5.50 m beneath the surface of the water. Find the pressure at each of these two locations.
11.3 Pressure and Depth in a Static Fluid P P 1 + ρ gh P atmospheric pressure ( 5 ) ( 3 3 )( 1.01 10 Pa + 1.00 10 kg m 9.80 m s )( 5.50 m) 1.55 10 5 Pa
11.4 Pressure Gauges P P 1 + ρ gh P atm ρ gh h Patm ρ g 5 ( 1.01 10 Pa) 3 3 ( 13.6 10 kg m )( 9.80m s ) 0.760 m 760 mm
11.4 Pressure Gauges P P B P A P A P 1 + ρ gh absolute pressure P Patm ρ gh gauge pressure
11.4 Pressure Gauges Systolic Pressure When heart is at peak of its beating cycle Diastolic pressure When heart is at low point of its beating cycle Both are reported as mm of mercury and are relative to atmospheric pressure
11.5 Pascal s Principle PASCAL S PRINCIPLE Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls.
11.5 Pascal s Principle P P1 + ρ g ( 0 m) F A F A 1 1 F F1 A A 1
11.5 Pascal s Principle Example 7 A Car Lift The input piston has a radius of 0.010 m and the output plunger has a radius of 0.150 m. The combined weight of the car and the plunger is 0500 N. Suppose that the input piston has a negligible weight and the bottom surfaces of the piston and plunger are at the same level. What is the required input force?
11.5 Pascal s Principle F F 1 A A 1 F ( 0.010 m) ( 0.150 m) ( N) π 0500 π 131 N
11.6 Archimedes Principle P P 1 ρ gh F ( B P A P 1 A P P 1 )A F B ρ gha V ha F B ρ V g mass of displaced fluid
11.6 Archimedes Principle ARCHIMEDES PRINCIPLE Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces: F B Magnitude of buoyant force W fluid displaced Weight of fluid
11.6 Archimedes Principle If the object is floating then the magnitude of the buoyant force is equal to the magnitude of its weight.
11.6 Archimedes Principle Example 9 A Swimming Raft The raft is made of solid square pinewood. Determine whether the raft floats in water and if so, how much of the raft is beneath the surface.
11.6 Archimedes Principle V raft 3 ( 4.0 m)( 4.0 m)( 0.30 m) 4.8 m F ρvg V g max B ρ water water ( 3 )( 3 )( 1000 kg m 4.8m 9.80m s ) 47000 N
11.6 Archimedes Principle W raft m g ρ V g raft pine raft ( 3 )( 3 )( 550kg m 4.8m 9.80m s ) 6000 N < 47000 N The raft floats!
11.6 Archimedes Principle If the raft is floating: W raft F B 6000 N ρ water V water g 3 ( 1000kg m )( 4.0 m)( 4.0 m) ( 9.80m s ) 6000 N h h 6000 N ( 1000 kg m )( 4.0 m)( 4.0 m)( 9.80m s ) 3 0.17 m
11.7 Fluids in Motion In steady flow the velocity of the fluid particles at any point is constant as time passes. Unsteady flow exists whenever the velocity of the fluid particles at a point changes as time passes. Turbulent flow is an extreme kind of unsteady flow in which the velocity of the fluid particles at a point change erratically in both magnitude and direction.
11.7 Fluids in Motion Fluid flow can be compressible or incompressible. Most liquids are nearly incompressible. Fluid flow can be viscous or nonviscous. An incompressible, nonviscous fluid is called an ideal fluid.
11.7 Fluids in Motion When the flow is steady, streamlines are often used to represent the trajectories of the fluid particles.
11.7 Fluids in Motion Making streamlines with dye and smoke.
11.8 The Equation of Continuity The mass of fluid per second that flows through a tube is called the mass flow rate.
11.8 The Equation of Continuity m ρv ρ A v t distance m t ρ Av m t 1 ρ 1A1 v1
11.8 The Equation of Continuity EQUATION OF CONTINUITY The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow. ρ 1A1 v1 ρ Av SI Unit of Mass Flow Rate: kg/s
11.8 The Equation of Continuity A v A Incompressible fluid: 1 1 v Volume flow rate Q: Q Av
11.8 The Equation of Continuity Example 1 A Garden Hose A garden hose has an unobstructed opening with a cross sectional area of.85x10-4 m. It fills a bucket with a volume of 8.00x10-3 m 3 in 30 seconds. Find the speed of the water that leaves the hose through (a) the unobstructed opening and (b) an obstructed opening with half as much area.
11.8 The Equation of Continuity (a) Q Av v Q A 3 3 ( 8.00 10 m ) ( 30.0 s).85 10-4 m 0.936 m s (b) A 1v1 Av v A v ( )( 0.936m s) 1.87 m s 1 1 A
11.9 Bernoulli s Equation The fluid accelerates toward the lower pressure regions. According to the pressure-depth relationship, the pressure is lower at higher levels, provided the area of the pipe does not change.
11.9 Bernoulli s Equation ( F ) s ( F ) s ( P) As ( P P )V W 1 nc ( ) ( ) 1 1 mv + mgy mv mgy W + 1 1
11.9 Bernoulli s Equation ( ) ( ) ( ) 1 1 P P V mv + mgy mv + 1 1 1 mgy ( ) ( ) ( ) 1 1 P P ρ v + ρgy ρv + ρ 1 1 1 gy BERNOULLI S EQUATION In steady flow of a nonviscous, incompressible fluid, the pressure, the fluid speed, and the elevation at two points are related by: P 1 1 1 + ρ v1 + ρgy1 P + ρv + ρgy
11.10 Applications of Bernoulli s Equation Example 16 Efflux Speed The tank is open to the atmosphere at the top. Find an expression for the speed of the liquid leaving the pipe at the bottom.
11.10 Applications of Bernoulli s Equation P v 0 1 P P atm P 1 1 1 + ρ v1 + ρgy1 P + ρv + ρgy 1 y y h 1 ρ v ρgh 1 v1 gh
For Practice FOC Questions: 1, 4, 9 and 0. Problems: 7, 6, 43, 44 and 65.