# Chapter 11. Fluids. continued

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1 Chapter 11 Fluids continued

2 11.2 Pressure Pressure is the amount of force acting on an area: Example 2 The Force on a Swimmer P = F A SI unit: N/m 2 (1 Pa = 1 N/m 2 ) Suppose the pressure acting on the back of a swimmer s hand is 1.2x10 5 Pa. The surface area of the back of the hand is 8.4x10-3 m 2. (a) Determine the magnitude of the force that acts on it. (b) Discuss the direction of the force. F = PA = ( N m )( m ) 2 = N Since the water pushes perpendicularly against the back of the hand, the force is directed downward in the drawing. Pressure on the underside of the hand is somewhat greater (greater depth). So force upward is somewhat greater - bouyancy

3 11.2 Pressure Atmospheric Pressure at Sea Level: 1.013x10 5 Pa = 1 atmosphere

4 11.3 Pressure and Depth in a Static Fluid Fluid density is ρ Equilibrium of a volume of fluid F 2 = F 1 + mg with F = PA, m = ρv P 2 A = P 1 A + ρvg with V = Ah P 2 = P 1 + ρ gh Pressure grows linearly with depth (h)

5 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. Atmospheric pressure P 1 = N/m 2 P 2 = P 1 + ρ gh P 2 = P 1 + ρ gh ( ) = ( Pa) + ( kg m 3 )( 9.80m s 2 ) 5.50 m = Pa

6 11.4 Pressure Gauges ρ Hg = kg m 3 P 2 = P 1 + ρ gh P 2 = ρ gh P 1 = 0 (vacuum) P atm = ρ gh h = P atm ρ g ( Pa) = ( kg m 3 ) 9.80m s 2 ( ) = m = 760 mm of Mercury

7 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. P 2 = P 1 + ρ gh, h = 0 P 2 = P 1 P 2 = F 2 A 2 ; P 1 = F 1 A 1 F 2 A 2 = F 1 A 1 A F 1 = F 1 2 A 2

8 11.5 Pascal s Principle Example 7 A Car Lift The input piston has a radius of m and the output plunger has a radius of m. The combined weight of the car and the plunger is 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? A F 1 = F 1 2 A 2 ( ) π ( m ) 2 = 131 N π ( m) 2 = N

9 11.6 Archimedes Principle Buoyant Force F B = P 2 A P 1 A = ( P 2 P ) 1 A = ρ gha = ρv g mass of displaced fluid P 2 = P 1 + ρ gh V = ha Buoyant force = Weight of displaced fluid

10 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: CORROLARY If an object is floating then the magnitude of the buoyant force is equal to the magnitude of its weight.

11 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.

12 11.6 Archimedes Principle W raft = m raft g = ρ pine V raft g = ( 550kg m 3 )( 4.8m 3 )( 9.80m s 2 ) = N Raft properties V raft = ( 4.0) ( 4.0) ( 0.30)m 3 = 4.8 m 3 ρ pine = 550 kg/m 3 If W raft < F B max, raft floats F B max = W fluid (full volume) F B max = ρvg = ρ water V water g = ( 1000kg m 3 )( 4.8m 3 )( 9.80m s 2 ) = N W raft < F B max Raft floats Part of the raft is above water

13 11.6 Archimedes Principle How much of raft below water? Floating object F B = W raft F B = ρ water gv water = ρ water g( A water h) h = = W raft ρ water ga water 26000N ( 1000kg m 3 ) 9.80m s 2 = 0.17 m W raft = N ( )( 16.0 m 2 )

14 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. 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.

15 11.7 Fluids in Motion When the flow is steady, streamlines are often used to represent the trajectories of the fluid particles. The mass of fluid per second that flows through a tube is called the mass flow rate.

16 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. SI Unit of Mass Flow Rate: kg/s Δm 2 Δt = ρ 2 A 2 v 2 Incompressible fluid: ρ 1 = ρ 2 Volume flow rate Q:

17 11.8 The Equation of Continuity Example 12 A Garden Hose A garden hose has an unobstructed opening with a cross sectional area of 2.85x10-4 m 2. 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. a) Q = Av ( ) / 30.0 s v = Q A = m 3 ( ) m 2 = 0.936m s b) A 1 v 1 = A 2 v 2 v 2 = A 1 v A 1 = ( 2) ( 0.936m s) = 1.87m s 2

18 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. Apply Work-Energy theorem to determine relationship between pressure, height, velocity, of the fluid.

19 11.9 Bernoulli s Equation Work done by tiny pressure piston W ΔP = ( F )s = ( ΔF )s = ( ΔPA)s; V = As Work (NC) done by pressure difference from 2 to 1 W NC = ( P 2 P 1 )V E 1 = 1 mv 2 + mgy E 2 = 1 mv 2 + mgy W NC = E 1 E 2 = 1 mv 2 + mgy ( ) ( 1 mv 2 + mgy ) 2 2 2

20 11.9 Bernoulli s Equation W NC = ( P 2 P 1 )V W NC = E 1 E 2 = 1 mv 2 + mgy ( ) ( 1 mv 2 + mgy ) NC Work yields a total Energy change. Equating the two expressions for the work done, ( ) ( 1 mv 2 + mgy ) ( P 2 P 1 )V = 1 mv 2 + mgy m = ρv ( P 2 P ) 1 = 1 ρv 2 + ρgy BERNOULLI S EQUATION ( ) ( 1 ρv 2 + ρgy ) Rearrange to obtain 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 ρv 2 + ρgy = P ρv 2 + ρgy 2 2 2

21 11.10 Applications of Bernoulli s Equation Conceptual Example 14 Tarpaulins and Bernoulli s Equation When the truck is stationary, the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway. Account for this behavior. Bernoulli s Equation P ρv 2 + ρgy = P ρv 2 + ρgy Relative to moving truck v 1 = 0 under the tarp v 2 air flow over top P 1 = P ρv P 1 > P 2

22 11.10 Applications of Bernoulli s Equation

23 11.10 Applications of Bernoulli s Equation Example 16 Efflux Speed The tank is open to the atmosphere at the top. Find and expression for the speed of the liquid leaving the pipe at the bottom. P 1 = P 2 = P atmosphere ( N/m 2 ) v 2 = 0, y 2 = h, y 1 = 0 P ρv 2 + ρgy = P ρv 2 + ρgy ρv 2 = ρgh 2 1 v 1 = 2gh

24 11.11 Viscous Flow Flow of an ideal fluid. Flow of a viscous fluid. FORCE NEEDED TO MOVE A LAYER OF VISCOUS FLUID WITH CONSTANT VELOCITY The magnitude of the tangential force required to move a fluid layer at a constant speed is given by: F = ηav y η, is the coefficient of viscosity SI Unit: Pa s; 1 poise (P) = 0.1 Pa s POISEUILLE S LAW (flow of viscous fluid) The volume flow rate is given by: Pressure drop in a straight uniform diamater pipe.

25 11.11 Viscous Flow Example 17 Giving and Injection A syringe is filled with a solution whose viscosity is 1.5x10-3 Pa s. The internal radius of the needle is 4.0x10-4 m. The gauge pressure in the vein is 1900 Pa. What force must be applied to the plunger, so that 1.0x10-6 m 3 of fluid can be injected in 3.0 s? P 2 P 1 = 8ηLQ π R 4 ( )( m) ( m s) = 1200 Pa π ( m) 4 = Pa s P 2 = ( P 1 )Pa = ( )Pa = 3100 Pa F = P 2 A = ( 3100 Pa) ( m 2 ) = 0.25N

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