2 Fluids 11.1 Mass Density 11.2 Pressure 11.3 Pressure and Depth in a Static Fluid11.4 Pressure Guages11.5 Pascal’e Principle11.6 Archimedes’ Principle11.7 Fluid in Motion11.8 The Equation of Continuity11.9 Bernoulli’s EquationApplications of Bernoulli’s EquationViscous Flow
3 Definition of Mass Density The mass density of a substance is the mass of asubstance per unit volume:SI Unit of Mass Density: kg / m3
5 Example-1: Blood as a Fraction of Body Weight 11.1 Mass DensityExample-1: Blood as a Fraction of Body WeightThe body of a man whose weight is 690 N contains about5.2x10-3 m3 of blood.(a) Find the blood’s weight and (b) express it as apercentage of the body weight.
7 Definition of Weight Density 11.1 Mass DensityDefinition of Weight DensityThe Weight density of a substance is the Weight of asubstance per unit volume:SI Unit of Weight Density: N / m3
8 Specific Gravity Is relative density. It is a ratio of the densities of two materials.Dimensionless.Using water as a reference can be convenient, since density of water is (approximately) 1000 kg/m³ or 1 g/cm³ at 4C.Specific gravity of Mercury is 13.6, means it is 13.6 times massive or heavier than water.
10 Pressure Pressure is force per unit area. P = F/A Pressure is not a vector quantity.To calculate the pressure we only consider the magnitude of the force.The force generated by the pressure of the static fluid is always perpendicular to the surface of the contact.Pa = N/m2 = x 10-4 lb/in2 (very small quantity)bar = 105 Pa = 100 kPaAtmospheric pressure is kPa = bar = lb/in2 = 760 torrlb/in2 or psi = x 103 Pa
11 Pressure causes Perpendicular Force The forces of a liquid pressing against a surface add up to a net force that is perpendicular to the surface.Components parallel tothe surface cancels outComponents vertical tothe surface adds up
12 Example-2: The Force on a Swimmer 11.2 PressureExample-2: The Force on a SwimmerSuppose the pressure acting on the backof a swimmer’s hand is 1.2x105 Pa. Thesurface area of the back of the hand is8.4x10-3 m2.Determine the magnitude of the forcethat acts on it.(b) Discuss the direction of the force.
13 Since the water pushes perpendicularly 11.2 PressureSince the water pushes perpendicularlyagainst the back of the hand, the forceis directed downward in the drawing.
14 Shape of the Dam Dam-B Dam-A Would you prefer Dam-A or Dam-B? Justify your answer.Dam-BDam-A
15 Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere Weight of the column of the air of 1m2 = x105 NMass of the column of the air of 1m2 = x105 N/g = kg
16 11.3 Pressure and Depth in a Static Fluid Relation between Pressure and DepthSince the column is in equilibrium,The sum of the vertical forcesequal to zero
17 11.3 Pressure and Depth in a Static Fluid The pressure increment = gh = weight density x heightPressure at a depth, h = P1 + gh = Atmospheric Pressure + Liquid PressureSo, Liquid Pressure at a depth, h = gh = weight density x depth
18 PressureWater pressure acts perpendicular to the sides of a container, and increases with increasing depth.
19 11.3 Pressure and Depth in a Static Fluid Conceptual Example-3: The Hoover DamLake Mead is the largest wholly artificialreservoir in the United States. The waterin the reservoir backs up behind the damfor a considerable distance (120 miles).Suppose that all the water in Lake Meadwere removed except a relatively narrowvertical column.Would the Hoover Dam still be neededto contain the water, or could a much lessmassive structure do the job?
20 11.3 Pressure and Depth in a Static Fluid Example-4: The Swimming HolePoints A and B are located a distance of 5.50 m beneath the surfaceof the water. Find the pressure at each of these two locations.
22 Where to put the pump? Pump-X Pump-Y Pump-X pushed the water up. Pump-Y removes the air from the pipe and air pressure outside pushes the water up.Pump-Y has a limit, what is that?Pump-XPump-Y
23 Pressure GaugesOne of the simplest pressure gauges is the mercury barometer used for measuring atmospheric pressure.760 mm of mercury
24 Point A and B are at the same depth, so, PB = PA Patm = P1 + gh 11.4 Pressure GaugesPoint A and B are at the same depth, so,PB = PAPatm = P1 + gh= 0 + gh
25 Open-tube manometerThe phrase “open-tube” refers to the fact that one side of the U-tube is open to atmospheric pressure.The tube contains mercuryIts other side is connected to the container whose pressure P2 is to be measured.PB = PAP2 = Patm + ghP2 - Patm = ghP2 – Patm is called the guage pressureP2 is called the absolute pressure
27 sphygmomanometerblood-pressure instrument: an instrument used to measure blood pressure in an artery that consists of a pressure gauge, an inflatable cuff placed around the upper arm, and an inflator bulb or pressure pump.Systolic pressure is 120 mm of mercuryDiastolic pressure is 80 mm of mercury
29 Any change in the pressure applied 11.5 Pascal’s PrinciplePascal’s PrincipleAny change in the pressure appliedto a completely enclosed fluid is transmittedundiminished to all parts of the fluid andenclosing walls.
31 The input piston has a radius of 0.0120 m 11.5 Pascal’s PrincipleExample-7: A Car LiftThe input piston has a radius of mand the output plunger has a radius of0.150 m.The combined weight of the car and theplunger is N. Suppose that the inputpiston has a negligible weight and the bottomsurfaces of the piston and plunger are atthe same level. What is the required inputforce?
34 11.6 Archimedes’ Principle Any fluid applies a buoyant force to an object that is partiallyor completely immersed in it; the magnitude of the buoyantforce equals the weight of the fluid that the object displaces:
35 11.6 Archimedes’ Principle If the object is floating then themagnitude of the buoyant forceis equal to the magnitude of itsweight.
36 11.6 Archimedes’ Principle Example-9: A Swimming RaftThe raft is made of solid squarepinewood. Determine whetherthe raft floats in water and ifso, how much of the raft is beneaththe surface.
39 11.6 Archimedes’ Principle If the raft is floating:
40 11.6 Archimedes’ Principle Conceptual Example-10: How Much Water is Neededto Float a Ship?A ship floating in the ocean is a familiar sight. But is allthat water really necessary? Can an ocean vessel floatin the amount of water than a swimming pool contains?
42 Unsteady flow exists whenever the velocity of the fluid particles at a 11.7 Fluids in MotionIn steady flow the velocity of the fluid particles at any point is constantas time passes.Unsteady flow exists whenever the velocity of the fluid particles at apoint changes as time passes.Turbulent flow is an extreme kind of unsteady flow in which the velocityof the fluid particles at a point change erratically in both magnitude anddirection.
43 Fluid flow can be compressible or incompressible. Most liquids are 11.7 Fluids in MotionFluid flow can be compressible or incompressible. Most liquids arenearly incompressible.Fluid flow can be viscous or nonviscous.An incompressible, nonviscous fluid is called an ideal fluid.
44 When the flow is steady, streamlines are often used to represent 11.7 Fluids in MotionWhen the flow is steady, streamlines are often used to representthe trajectories of the fluid particles.
45 Making streamlines with dye and smoke. 11.7 Fluids in MotionMaking streamlines with dyeand smoke.
46 11.8 The Equation of Continuity The mass of fluid per second that flows through a tube is calledthe mass flow rate.
48 11.8 The Equation of Continuity The mass flow rate has the same value at every position along atube that has a single entry and a single exit for fluid flow.SI Unit of Mass Flow Rate: kg/s
49 11.8 The Equation of Continuity Incompressible fluid:Volume flow rate Q:
50 11.8 The Equation of Continuity Example-12: A Garden HoseA garden hose has an unobstructed openingwith a cross sectional area of 2.85x10-4m2.It fills a bucket with a volume of 8.00x10-3m3in 30 seconds.Find the speed of the water that leaves the hosethrough (a) the unobstructed opening and (b) an obstructedopening with half as much area.
52 Bernoulli’s Equation For steady flow, The speed, Pressure, and Elevationof an incompressible and nonviscous fluid are related by Bernoulli’s equation.
53 The fluid accelerates toward the lower pressure regions. 11.9 Bernoulli’s EquationThe fluid accelerates toward thelower pressure regions.According to the pressure-depthrelationship, the pressure is lowerat higher levels, provided the areaof the pipe does not change.
54 Bernoulli’s Equation Bernoulli’s Equation from Work-energy theorem: Work done on an object by external non-conservative forces is equal to the change in total mechanical energy.The pressure within a fluid is caused by collisional forces, which are non-conservative.Therefore, when a fluid is accelerated because of a difference in pressures, work is being done by non-conservative forces.
56 Bernoulli’s EquationWhenever a fluid is flowing in a horizontal pipe and encounters a region of reduced cross-sectional area, the pressure of the fluid drops.Can be explained by the Newton’s 2nd law.When moving from the wider region 2 to the narrower region 1, the fluid speeds up according to the equation of continuity.According to the 2nd law, th eaccelerating fluid must be subjected to an unbalanced force.There can be an unbalanced force only if the pressure in region 2 is higher than the pressure in region 1.
57 Bernoulli’s Equation Explanation: In figure-b on the top surface of the blue fluid element, the surrounding fluid exerts a pressure P.This pressure give rise to a force, F = P/AOn the botom of the blue fluid element, the pressure is slightly higher, P+P.As a result, the force on the bottom surface has a magnitude of F+ F = (P+P)/A.The net force pushing the fluid element up the pipe is F=P/A
58 Bernoulli’s EquationWhen the fluid element moves through its own length s, the work done is theW = F s = P A s = P VThe total work done on the fluid element in moving it from region 2 to region 1is the sum of the small increments of work P V done as the element moves along the pipe.This sum amounts to Wnc = (P2 – P1)VWnc = (P2 – P1)V = E1 – E2
59 fluid speed, and the elevation at two points are related by: 11.9 Bernoulli’s EquationBernoulli’s EquationIn steady flow of a nonviscous, incompressible fluid, the pressure, thefluid speed, and the elevation at two points are related by:
60 Bernoulli’s Equation The term has a constant value at all positions in the flow.
61 Bernoulli’s Equation Bernoulli’s Equation reduces to the result for static fluids (v1=v2), as it is when the cross-sectional area remains constant.
62 11.10 Applications of Bernoulli’s Equation Conceptual Example-14: Tarpaulins and Bernoulli’s EquationWhen the truck is stationary, thetarpaulin lies flat, but it bulges outwardwhen the truck is speeding downthe highway.Account for this behavior.
63 Bernoulli’s Equation Bernoulli’s Equation reduces to When all parts have the same elevation (y1 = y2).The term P+ ½ remains constant throughout a horizontal pipe.If increases, P decreases and vice versa.
69 11.11 Viscous FlowFlow of an ideal fluid.Flow of a viscous fluid.
70 Force Needed to Move a Layer of Viscous Fluid with Constant Velocity 11.11 Viscous FlowForce Needed to Move a Layer of Viscous Fluid withConstant VelocityThe magnitude of the tangential force required to move a fluidlayer at a constant speed is given by:coefficientof viscositySI Unit of Viscosity: Pa·sCommon Unit of Viscosity: poise (P)1 poise (P) = 0.1 Pa·s
71 The volume flow rate is given by: 11.11 Viscous FlowPoiseuille’s LawThe volume flow rate is given by:
72 Example-17: Giving and Injection 11.11 Viscous FlowExample-17: Giving and InjectionA syringe is filled with a solution whoseviscosity is 1.5x10-3 Pa·s. The internalradius of the needle is 4.0x10-4m.The gauge pressure in the vein is 1900 Pa.What force must be applied to the plunger,so that 1.0x10-6m3 of fluid can be injectedin 3.0 s?