Physics 106 Lecture 13 Fluid Mechanics SJ 7th Ed.: Chap 14.1 to 14.5
• What is a fluid? • Pressure • Pressure varies with depth • Pascal’s principle • Methods for measuring pressure • Buoyant forces • Archimedes principle • Fluid dyypnamics assumptions • An ideal fluid • Continuity Equation • Bernoulli’s Equation
What is a fluid? Solids: strong intermolecular forces definite volume and shape rigid crystal lattices, as if atoms on stiff springs deforms elastically (strain) due to moderate stress (pressure) in any direction
Fluids: substances that can “flow” no definite shape molecules are randomly arranged, held by weak cohesive intermolecular forces and by the walls of a container liquids and gases are both fluids Liquids: definite volume but no definite shape often almost incompressible under pressure (from all sides) can not resist tension or shearing (crosswise) stress no long range ordering but near neighbor molecules can be held weakly together Gases: neither volume nor shape are fixed molecules move independently of each other comparatively easy to compress: density depends on temperature and pressure
Fluid Statics - fluids at rest (mechanical equilibrium) Fluid Dynamics – fluid flow (continuity, energy conservation)
1 Mass and Density • Density is mass per unit volume at a point: Δm m ρ ≡ or ρ ≡ • scalar V V • units are kg/m3, gm/cm3.. Δ 3 3 • ρwater= 1000 kg/m = 1.0 gm/cm
• Volume and density vary with temperature - slightly in liquids • The average molecular spacing in gases is much greater than in liquids.
Force & Pressure • The pressure P on a “small” area ΔA is the ratio of the magnitude of the net force to the area ΔF P = or P = F / A ΔA G G PΔA nˆ ΔF = PΔA = PΔAnˆ • Pressure is a scalar while force is a vector • The direction of the force producing a pressure is perpendicular to some area of interest
• At a point in a fluid (in mechanical equilibrium) the pressure is the same in h any direction
Pressure units: 1 Pascal (Pa) = 1 Newton/m2 (SI) 1 PSI (Pound/sq. in) = 6894 Pa. 1 milli-bar = 100 Pa.
2 Forces/Stresses in Fluids • Fluids do not allow shearing stresses or tensile stresses (pressures)
Tension Shear Compression • The only stress that can be exerted on an object submerged in a static fluid is one that tends to compress the object from all sides • The force exerted by a static fluid on an object is always perpendicular to the surfaces of the object
Question: Why can you push a pin easily into a potato, say, using very little force, but your finger alone can not push into the skin even if you push very hard?
A sensor to measure absolute pressure
• The spring is calibrated by a known force • Vacuum is inside • The force due to fluid pressure presses on the top of the piston and compresses the spring until the spring force and F are equal. • The force the fluid exerts on the piston is then measured
3 Pressure in a fluid varies with depth
Fluid is in static equilibrium y=0 The net force on the shaded volume = 0
• Incompressible liquid - constant density ρ y1 F P • Horizontal surface areas = A 1 1 h • Forces on the shaded region: F 2 P2 – Weight of shaded fluid: Mg y 2 Mg – Downward force on top: F1 =P1A
– Upward force on bottom: F2 = P2A
∑Fy = 0 = P2A − P1A − Mg
• In terms of density, the mass of the The extra pressure at shaded fluid is: extra depth h is: M = ρΔV = ρAh ΔP = P2 − P1 = ρgh
∴ P2A ≡ P1A + ρghA h ≡ y1 − y2
Pressure relative to the surface of a liquid
air Example: The pressure at depth h is: P0 Ph = P0 + ρgh h P0 is the local atmospheric (or ambient) pressure
Ph is the absolute pressure at depth h Ph The difference is called the gauge pressure All points at the same depth are at the liquid same pressure; otherwise, the fluid could not be in equilibrium The pressure at depth h does not depend PdiPreceding equations also on the shape of the container holding the hold approximately for fluid gases such as air if the P density does not vary 0 much across h
Ph
4 Atmospheric pressure and units conversions
• P0 is the atmospheric pressure if the liquid is open to the atmosphere. • Atmospheric pressure varies locally due to altitude, temperature, motion of air masses, other factors.
• Sea level atmospheric pressure P0 = 1.00 atm = 1.01325 x 105 Pa = 101.325 kPa = 1013.25 mb (millibars) = 29.9213” Hg = 760.00 mmHg ~ 760.00 Torr = 14.696 psi (pounds per square inch)
Pascal bar (bar) atmosphere torr pound-force per (Pa) (atm) (Torr) square inch (psi) 2 −5 −6 −3 −6 1Pa1 Pa ≡ 11N/m N/m 10 9. 8692×10 7. 5006×10 145. 04×10 1 bar 100,000 ≡ 0.98692 750.06 14.5037744 106 dyn/cm2 1 atm 101,325 1.01325 ≡ 1 atm 760 14.696 1 torr 133.322 1.3332×10−3 1.3158×10−3 ≡ 1 Torr; 19.337×10−3 ≈ 1 mmHg 1 psi 6.894×103 68.948×10−3 68.046×10−3 51.715 ≡ 1 lbf/in2
Pressure Measurement: Barometer
• Invented by Torricelli (1608-47) near-vacuum • Measures atmospheric pressure P0 as it varies with the weather • The closed end is nearly a vacuum (P = 0) • One standard atm = 1.013 x 105 Pa. Mercury (Hg) P0 = ρHggh How high is the Mercury column? P 1.013 ×105 Pa h 0 0.760 m = = 3 3 2 = ρHgg (13.6 ×10 kg / m )(9.80 m/s ) One 1 atm = 760 mm of Hg = 29.92 inches of Hg
How high would a water column be? P 1.013 ×105 Pa h = 0 = = 10.34 m 3 3 2 ρwaterg (1.0 ×10 kg /m )(9.80 m/s ) Height limit for a suction pump
5 Pressure Measurements: Manometer
• Measures the pressure of the gas in the closed vessel • One end of the U-shaped tube is open to
the atmosphere at pressure P0 • The other end is connected to the
pressure PA to be measured • Points A and B are at the same elevation.
• The pressure PA supports a liquid column of height h • The Pressure of the gas is:
PA = PB = P0 + ρgh What effect would
increasing P0 have on Absolute vs. Gauge Pressure the column height?
• The gauge pressure is PA –P0= ρgh – What you measure in your tires • Gauge pressure can be positive or negative.
• PA is the absolute pressure – you have to add the ambient pressure P0 to the gauge pressure to find it.
Example: Blood Pressure
• Normal Blood pressure is: 120 (systolic = peak) over 70 (diastolic = resting) – in mmHg. • What do these numbers mean? • How much pressure do they correspond to in Pa. and psi? Recall: 760.00 mmHg = 1.01325 x 105 Pa =14.696 psi
(101,325/760.00) = 133.322 Pa./mmHg
(101,325/14.696) = 6894. Pa./psi
Systolic pressure of 120 = 16,000 Pa. =232psi= 2.32 psi
Diastolic pressure of 70 = 9333 Pa. = 1.354 psi
Mercury Aneroid Gauge or absolute? Sphygmomanometer Sphygmomanometer
6 Pascal’s Principle A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and to the walls of the container. ΔF ΔP
Example: open container • The pressure in a fluid depends on depth h and on the p0 value of P0 at the surface • All points at the same depth have the same pressure.
ph Ph = P0 + ρgh • AdditAdd piston of area A with ithl lead dbll balls on it it& & wei ihtWght W. Pressure at surface increases by ΔP = W/A
Pext = P0 + ΔP
• Pressure at every other point in the fluid (Pascal’s law), increases by the same amount, including all locations at P depth h. h Ph = Pext + ρgh
Pascal’s Law Device - Hydraulic press A small input force generates a large output force • Assume the working fluid is incompressible • Neglect the (small here) effect of height on pressure
• The volume of liquid pushed down on the l eft equal s th e vol ume push ed up on the right, so:
A1Δx1 = A2Δx2 Δx A ∴ 2 = 1 Δx1 A2 • Assume no loss of energy in the fluid, no friction, etc .
Other hydraulic lever devices Work = F Δx = Work = F Δx using Pascal’s Law: 1 1 1 2 2 2
• Squeezing a toothpaste tube mechanical advantage • Hydraulic brakes • Hydraulic jacks F2 Δx1 A2 • Forklifts, backhoes ∴ = = F1 Δx2 A1
7 Archimedes Principle C. 287 – 212 BC • Greek mathematician, physicist and engineer • Computed π and volumes of solids • Inventor of catapults, levers, screws, etc. • Discovered nature of buoyant force – Eureka! Why do ships float and sometimes sink? Why do objects weigh less when submerged in a fluid? identical pressures at every point
hollow ball ball of liquid same in equilibrium upward force
• An object immersed in a fluid feels an upward buoyant force that equals the weight of the fluid displaced by the object. Archimedes’s Principle – The fluid pressure increases with depth and exerts forces that are the same whether the submerged object is there or not. – Buoyant forces do not depend on the composition of submerged objects. – Buoyant forces depend on the density of the liquid and g.
Archimedes’s principle - submerged cube A cube that may be hollow or made of some material is submerged in a fluid Does it float up or sink down in the liquid?
• The extra pressure at the lower surface compared to the top is:
ΔP = Pbtbot − Ptop = ρflgh • The pressure at the top of the cube causes
a downward force of Ptop A • The larger pressure at the bottom of the
cube causes an upward force of Pbot A • The upward buoyant force B is the weight of the fluid displaced by the cube:
B = ΔPA = ρflghA = ρflgV = Mflg • The weight of the actual cube is:
Fg = Mcubeg = ρcubegV
The cube rises if B > F (ρ > ρ ) g fl cube Similarly for irregularly shaped The cube sinks if Fg>B (ρcube > ρfl) objects
ρcube is the average density
8 Archimedes's Principle: totally submerged object
Object – any shape - is totally submerged in a fluid of density ρfluid The upward buoyant force is the weight of displaced fluid: B = ρfluidgVfluid The downward gravitational force on the object is: Fg = Mobjectg = ρobjectgVobject The volume of fluid displaced and the object’s volume are equal for a totally submerged object. The net force is:
Fnet = B − Fg = [ρfluid − ρobject ] g Vobject
The direction of the motion of an object in a fluid is determined only by the densities of the fluid and the object The apparent weight is the – If the density of the object is less than external force needed to the density of the fluid, the unsupported object accelerates upward restore equilibrium, i.e. – If the density of the object is more than the density of the fluid, the unsupported Wapparent = Fg − B = − Fnet object sinks
Archimedes’s Principle: floating object An object sinks or rises in the fluid until it reaches equilibrium. The fluid displaced is a fraction of the object’s volume.
At equilibrium the upward buoyant force is balanced by the downward weight of the object:
Fnet = B − Fg = 0 ⇒ B = Fg
The volume of fluid displaced Vfluid corresponds to the portion of the object’s volume below the fluid level and is always less than the object’s volume. Equate:
B = ρfluidgVfluid = Fg = ρobjectgVobject
Solving: Objects float when their average ρfluidVfluid = ρobjectVobject density is less than the density of the fluid they are in. The ratio of densities equals the Vfluid ρobject = fraction of the object’s volume that Vobject ρfluid is below the surface
9 Archimedes’s Principle: King’s Crown
King to Archimedes: “Is the crown made of pure gold?” 3 • The crown’s density ρcrown = mcrown/ Vcrown should = ρgold = 19,300 kg/m • Weigh the crown in air to find mcrown • But: how to find Vcrown?? Solution:
• Weigh crown in air to find mcrown • Weigh it again fully submerged in a fluid
• The apparent weight T2 is:
T2 = Fg − B • The Buoyant force equals the difference in scale readings...
B = Fg − T2 = ρflgVcrown • solve: Fg − T2 F V = g crown ρ g ∴ ρcrown = ρfl fl Fg − T2
Example: What fraction of an iceberg is underwater?
Apply Archimedes’ Principle displaced seawater
Vunderwater Vfluid ρobject = = Vtotal Vobject ρfluid
glacial fresh water ice From table: 3 3 ρobject = ρice = 0.917 ×10 kg/m 3 3 ρfluid = ρseawater = 1.03×10 kg/m Water expands when it freezes. If not...... ponds, lakes, seas freeze to the bottom in winter
Vunderwater ∴ = 89.03% Vttltotal What if iceberg is in a freshwater lake? 3 3 ρfluid = ρfreshwater = 1.00×10 kg/m Vunderwater = 91.7% Vtotal Floating objects are more buoyant in saltwater Freshwater tends to float on top of seawater...
10 Water has special properties
Maximum density at 4o C. 3 3 • ρwater= 1000 kg/m = 1.0 gm/cm • turnover of ponds and lakes • ice floats, so bodies of water do not freeze to the bottom
Fluids’ Flow is affected by their viscosity
• Viscosity measures the internal friction in a fluid.
Low viscosity Medium viscosity High viscosity • gases • water • honey • other fluids that pour • oil and grease and flow easily • glass • Viscous forces depend on the resistance that two adjacent layers of fluid have to relative motion. • Part of the kinetic energy of a fluid is converted to internal energy, analogous to friction for sliding surfaces
Ideal Fluids – four approximations to simplify the analysis of fluid flow: • The flu id is nonviscous – ilinternal fiifriction is negldlected • The flow is laminar (steady, streamline flow) – all particles passing through a point have the same velocity at any time. • The fluid is incompressible – the density remains constant • The flow is irrotational – the fluid has no angular momentum about any point. A small paddle wheel placed anywhere does not feel a torque and rotate
11 Laminar Flow - Streamlines Steady state flow – at any point the velocity of particles is constant Each particle of the fluid follows a smooth path called a streamline. Particles’ velocities are tangent to streamlines (field lines). The paths of different particles never cross each other Every given fluid particle arriving at a given point has the same velocity
velocity field lines
Turbulent flow
An irregular flow characterized by small whirlpool-like regions Turbulent flow occurs when the particles go above some critical (Mach) speed Energy is dissipated, more work must be done to maintain flow
12 Flow of an ideal fluid through a short section of pipe Constant density and velocity within volume element dV Incompressible fluid means dρ/dt = 0 Mass flow rate = amount of mass crossing area A per unit time = a “current”sometimes called a “mass flux”
cross-section area A
velocity v volume of fluid in cylinder = dV = Adx dM d I = mass flow rate = = (ρV) mass dt dt dx = ρA = ρAv dt length dx Imass ≡ mass flow rate = ρAv
Ivol ≡ volume flow rate = Av
Jmass ≡ mass flow/unit area = ρv
Equation of Continuity: conservation of mass • An ideal fluid is moving through a pipe of nonuniform diameter • The particles move along streamlines in steady-state flow • The mass entering at point 1 cannot disappear or collect in the pipe
• The mass that crosses A1 in some time interval is the same as the mass that crosses A2 in the same time interval.
mass flow in = ρ1A1v1 = mass flow out = ρ2A2v2 • The fluid is incompressible so: ρ2 ρ1 = ρ2 = a cons tan t
∴ A1v1 = A2v2
• This is called the equation of continuity for an incompressible fluid ρ1 • The product of the area and the fluid speed (volume flux) at all points along a pipe is constant.
The rate of fluid volume entering one end equals the volume leaving at the other end Where the pipe narrows (constriction), the fluid moves faster, and vice versa
13 Bernoulli’s Equation – energy conservation OPTIONAL TOPIC
Daniel Bernoulli, Swiss physicist, 1700 – 1782 • An ideal fluid flows through a region where its speed, cross- sectional area, and/or altitude changes. • The pressure in the fluid varies with these changes. • Application of energy conservation yields the Bernoulli Equation
ΔW = work done by pressure = ΔEmech = E2 − E1
Pressure P1 Pressure P2 speed v Equal volumes flow in equal times speed v 1 due to continuity equation 2 height y1 height y2
Volume Volume ΔV = A1Δx1 ΔV = A2Δx2
time t time t + Δt
Bernoulli’s Equation – energy conservation OPTIONAL TOPIC The fluid is incompressible. The continuity equation (no leaks) requires:
ΔV = A1Δx1 = A2Δx2 m1 = m2 ≡ m = ρΔV The net work done on fluid volume by external pressures:
ΔW = F1dx1 − F2dx2 = P1A1dx1 − P2A2dx2 = [P1 − P2 ] ΔV The change in the mechanical energy: ΔE = E − E = 1 mv2 − 1 mv2 + mgy − mgy 2 1 2 2 2 1 2 1 Equate to the net work and divide through by the volume ΔV: 1 2 1 2 P − P = ρv − ρv + ρg(y − y ) 1 2 2 2 2 1 2 1 Rearrange: 1 2 1 2 P + ρv + ρgy = P + ρv + ρgy 1 2 1 1 2 2 2 2
Pressure P 1 Equal volumes flow in equal times Pressure P2 speed v 1 diiidue to continuity equation speed v2 height y 1 height y2 Volume Volume ΔV = A1Δx1 ΔV = A2Δx2
time t time t + Δt
14 OPTIONAL TOPIC Bernoulli’s Equation – energy conservation
1 2 1 2 P + ρv + ρgy = P + ρv + ρgy 1 2 1 1 2 2 2 2 Bernoulli’s Equation is often expressed as: 1 2 P + ρv + ρgy = Constant 2 For a fluid at rest Bernoulli’s Equation reduces to the earlier result for variation of pressure with depth:
P1 = P2 + ρg[y2 - y1]
For flow at constant altitude (y1 = y2) the pressure is lower where ther fluid flows faster: 1 2 1 2 P + ρv = P + ρv 1 2 1 2 2 2 Gases are compressible, but the above pressure / speed relation holds true.
OPTIONAL TOPIC Bernoulli effect causes airplane wings to work
• Streamlines flow around a moving airplane wing • Lift is the upward force on the wing from the air • Drag is the resistance • The lift depends on the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal
Low Pressure where flow velocity is high
Higher Pressure where flow velocity is low
15 OPTIONAL TOPIC Lift – General
• In general, an object moving through a fluid experiences lift as a result of any effect that causes the fluid to change its direction as it flows past the object • Some factors that influence lift are: – The shape of the object – The object’s orientation with respect to the fluid flow – Any spinning of the object – The texture of the object’s surface
Golf Ball
• The b all i s gi ven a rapid b ack spi n • The dimples increase friction – Increases lift • It travels farther than if it was not spinning
OPTIONAL TOPIC Atomizer
• A stream of air passes over one end of an open tube • The other end is immersed in a liquid • The moving air reduces the pressure above the tube • The fluid rises into the air stream • The liquid is dispersed into a fine spray of droplets
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