Chapter 12 Fluid Mechanics - University Of Minnesota

Chapter 12Fluid Mechanics12.1DensityWe have already seen (9.58) that the local destiny of a material can bedefined asdm.(12.1)ρ dVWhen the object has uniform (i.e. position independent) density, then thelocal density is the same as average density defined asρ m.V(12.2)For the same substance this number does not change even if the mass andvolume might be different. For example both a steel wrench and a steel nailhave the same density which the density of steel.165

CHAPTER 12. FLUID MECHANICS166In SI the units of density are given by kilogram per cubic meter1 kg1 m3but gram per unit centimeter are also widely used1 kg/m3 1 g/cm3 1000 kg/m3 .In the following table we summarize densities of common substances:(12.3)(12.4)

CHAPTER 12. FLUID MECHANICS167Another useful (but dimensionless) measure of density is specific density also known as relative density. It is defined as a ratio of density of agiven substance to density of water (at temperature 4.0 C),ρsubstance.(12.5)specific density ρwaterExample 12.2. Find the mass and weight of the air (at 1 atm and 20 C)in a living room with 4.0m 5.0m floor and a ceiling 3.0m high, and the massand weight of an equal volume of water.Volume of the living room isV 3.0m 4.0m 5.0m 60 m3(12.6)From definition of density!"!"mair ρair V 1.20 kg/m3 60 m3 72 kg!"!"mwater ρwater V 1000 kg/m3 60 m3 6.0 104 kg(12.7)and thus the corresponding weights are!"wair mair g (72 kg) 9.8 m/s2 700 N!"!"wwater mwater g 6.0 104 kg 9.8 m/s2 5.9 105 N. (12.8)Example. Rank the following objects in order from highest to lowestaverage density:(i) mass 4.00 kg, volume 1.60 10 3 m3 ;(ii) mass 8.00 kg, volume 1.60 10 3 m3 ;(iii) mass 8.00 kg, volume 3.20 10 3 m3 ;(iv) mass 2560 kg, volume 0.640 m3 ;(v) mass 2560 kg, volume 1.28 m3 .The densities of these objects areρi ρii ρiii ρiv ρv 4.00 kg 2500 kg/m31.60 10 3 m38.00 kg 5000 kg/m3 331.60 10 m8.00 kg 2500 kg/m33.20 10 3 m32560 kg 4000 kg/m30.64 m32560 kg 2000 kg/m31.28 m3(12.9)And so the order is(ii) (iv) (i, iii) (v).(12.10)

CHAPTER 12. FLUID MECHANICS12.2168Pressure in a FluidPressure. In fluids pressure might change from one place to another andthus it is convenient to define a local pressure asp dF dA(12.11)which reduces to (11.36) of average pressurep F A(12.12)for uniform (i.e. position independent) pressures. Units of pressure werealready introduced in the previous chapter,1 Pa 1 N/m21 atm 1.013 105 Pa1 psi 6900 Pa.(12.13)Example 12.2. In the living room with 4.0m 5.0m floor what is thetotal downward force on the floor due to air pressure of 1.00 atm?From definition of pressureF pA (1.00 atm)1.013 105 N(4.0m 5.0m) 2.0 106 N. (12.14)1.00 atmPressure with depth. Consider an infinitesimal volume element offluid dV dxdydz, where y-axis points upwards. In the equilibrium allforces acting on the object must add up to zero and thus along y-axis wehave (ρdV ) g (p dp) dA pdA ρ (dxdydz) g (p dp) (dxdz) p (dxdz) ρgdy (p dp) pdpdy 0 0 0 ρg(12.15)and thus the pressure must change linearly with y. The above equation canbe solved by direct integration, i.e.# p(y2 )# y2dp ρgdyp(y1 )y1p(y2 ) p(y1 ) ρg (y1 y2 ) .(12.16)

CHAPTER 12. FLUID MECHANICS169where we used an assumption that ρ is constant or in other words thatthe fluid is non-compressible. (This is a good assumption for liquids (e.g.water), but is not a very good assumption for gases (e.g. air) whose densitycan change considerably. )In terms of depthd y2 y1(12.17)and reference pressurep0 p(y2 )(12.18)pressure at arbitrary depth is given byp p0 ρgd.(12.19)As we see if the density is constant, the pressure p depends only on thepressure at the surface p0 and depth d. Thus if p0 (atmospheric pressure)and p(pressure at the bottom of liquid) is the same than d must be the same:If we change pressure p0 at the surface of fluid, the pressure will change bythe same amount everywhere in the fluid. For example one can use this resultto construct a hydraulic lift to measure large weights:

CHAPTER 12. FLUID MECHANICS170Because the pressure is the same at all point on the same heightp0 F1F2 A1A2(12.20)orA2F1 .(12.21)A1More generally one formulate what is known as Pascal’s law: Pressureapplied to an enclosed fluid is transmitted undiminished to every portion ofthe fluid and the walls of the containing vessel.Gauge pressure. It often useful to measure relative pressure comparedto atmospheric pressure,F2 p0 1 atm 14.7 psi 1.01 105 Pa.(12.22)For example, if the absolute pressure of a car tire isp 47 psi(12.23)then it is often said that the gauge pressure ispgauge p p0 32 psi.(12.24)Example 12.3. Water stands 12.0 m deep in a storage tank whose topis open to the atmosphere. What are the absolute and gauge pressure at thebottom of the tank?

CHAPTER 12. FLUID MECHANICS171The absolute pressure at the bottom of the tank isp p0 ρgh!" !"!" 1.01 105 Pa 1000 kg/m3 9.8 m/s2 (12.0 m) 2.19 105 Pa(12.25)and so the gauge pressure isp p0 ρgh 1.18 105 Pa.(12.26)To measure the gauge pressure directly, one can use an open-tube manometer, where the difference in heights tells you what the gauge pressure ispgauge p p0 ρg(y2 y1 )(12.27)To measure air-pressure one can use a barometer where the difference inheights tells you what the atmospheric ispatm p0 ρg(y2 y1 ) ρg(y2 y1 ).(12.28)

CHAPTER 12. FLUID MECHANICS172The latter example suggest another unit of measuring pressure in “millimeters of mercury” which is also called torr after the inventor of mercurybarometer Evangelista Torricelli.Example 12.4. A manometer tube is partially filled with water. Oil(which does not mix with water) is poured into the left arm of the tube untilthe oil-water interface is at the midpoint of the tube as shown. Both arms ofthe tube are open to the air. Find a relationship between the heights hoil andhwater .The pressure in both fluids at the surface and at the bottom are the samep p0 ρwater ghwaterp p0 ρoil ghoil(12.29)

CHAPTER 12. FLUID MECHANICSand thus,orhoil 12.3173ρwater ghwater 1ρoil ghoil(12.30)ρwater1000 kg/m3hwater hwater 1.2 hwater .ρoil850 kg/m3(12.31)BuoyancyAny object placed in a fluid experiences a force (buoyant force) arisingdue to changes of the pressure inside fluid. This phenomena is known asArchimedes’s principle: When a body is completely or partially immersedin a fluid, the fluid exerts an upward force on the body equal to the weight ofthe fluid displaced by the body.To prove Archimedes’s principle we consider an element of fluid of arbitrary shape. If the fluid is in equilibrium then the sum of all forces (due towater pressure) have to be the same as the force of gravityB Fgravity(12.32)B V ρfluid g(12.33)orNow if we fill the shape with some other material, then the equilibriumcondition might not be satisfied, but the buoyant force due to water pressurewould not change.

CHAPTER 12. FLUID MECHANICS174Example 12.5. A 15.0 kg solid gold statue is raised from the seabottom. What is the tension in the hosting cable (assumed massless) whenthe statues isa) at rest and completely underwater.b) at rest and completely out of water.We can first find volume of the statueV m15.0 kg 7.77 10 4 m3 .ρgold19.3 103 kg/m3(12.34)Then the equilibrium condition impliesT Bfluid mg 0(12.35)or in waterT mg V ρwater g!" !"!"!" (15.0 kg) 9.8 m/s2 7.77 10 4 m3 1000 kg/m3 9.8 m/s2 139 N(12.36)and in airT mg V ρair g" !"!"!"! (15.0 kg) 9.8 m/s2 7.77 10 4 m3 1.2 kg/m3 9.8 m/s2 147 N.(12.37)

CHAPTER 12. FLUID MECHANICS175Example. You place a container of seawater on a scale and note readingon the scale. You now suspend the statue of Example 12.5 in the water. Howdoes the scale reading change?(i) it increases by 7.84 N;(ii) it decreases by 7.84 N;(iii) it remains the same;(iv) none of these.In addition to buoyant force there is a force of surface tension whichacts on the object at the surface of fluids, but this force is subdominant forsufficiently large objects.12.4Fluid FlowConsider a simple model of fluid which is incompressible (density is constant) and inviscid (vanishing internal friction). Then one can follow trajectories of small elements of water (we call particles) which will flow alongalong these trajectories. We call these trajectories flow lines and say thatthe flow is steady if the flow line do no change with time. This does notmean that the velocities on any given particle does not change with time.

CHAPTER 12. FLUID MECHANICS176More generally the flow becomes irregular where small scale mode and largescale modes interact with each other which gives rise to turbulence. It isinteresting to note that for 3D fluids the energy is transferred from largescales to small scale, when in 2D fluids the energy is transferred from smallscales to large scales. Richard Feynman called turbulence “the last unsolvedproblem of classical physics”. It is also related to one of seven Millenniumproblems formulated by Clay Mathematics Institute in 2000, six of which(including the turbulence problem) remain unsolved.Continuity. Consider a flow of fluid through a pipe with changing crosssectional area:

CHAPTER 12. FLUID MECHANICS177If the fluid is incompressible (i.e. constant density), then the amount offluid passing through each cross-sectional area per unit time must be thesameρA1 ds1 ρA2 ds2(12.38)orA1 v1 dt A2 v2 dt(12.39)which give us (1D) the continuity equation for non-compressible fluidA1 v1 A2 v2 .(12.40)The continuity equation equates the volume flow rate across different crosssectional areasdV Av.(12.41)dtNote that(12.40) can be easily generalized to the case when densities dochangeρ1 A1 v1 ρ2 A2 v2 .(12.42)Example 12.6. Incompressible oil of density 850 kg/m3 is pumped througha cylindrical pipe at a rate of 9.5 liters per second.(a) The first section of the pipe has a diameter of 8.0 cm. What is theflow speed of the oil? What is the mass flow rate?(b) The second section of the pipe has a diameter of 4.0 cm. What is theflow speed and the mass flow rate in that section?From continuity equation!"dV A1 v1 A2 v2 (9.5 L/s) 10 3 m3 /L 9.5 10 3 m3 /s.dt(12.43)and thus the flow speeds are9.5 10 3 m3 /s 1.9 m/s3.14 (4.0 10 2 m)29.5 10 3 m3 /s 7.6 m/s 3.14 (2.0 10 2 m)2v1 v2(12.44)From the definition of densitydmdV ρdtdtand thus the mass flow rates are the same for both sections"!"dm ! 850 kg/m3 9.5 10 3 m3 /s 8.1 kg/s.dt(12.45)(12.46)

CHAPTER 12. FLUID MECHANICS178Example. A maintenance crew is aworking on a section of a three-lanehighway, leaving only one lane open to traffic. The result is much slowertraffic flow (a traffic jam.) Do cars on a highway behave like:(i) the molecules of an incompressible fluid or(ii) the molecules of compressible fluid?12.5Bernoulli’s EquationAs fluid moves through pipe external forces such as gravitational force cando work on the fluid.This can be described by computing the total work done on (an incompressible) fluid element between sections a and c as they move to sections band ddW p1 A1 ds1 p2 A2 ds2 (p1 p2 ) dV(12.47)This must be equal to the change in mechanical energy for fluid. The changein kinetic and potential energies is due to the difference of kinetic and potential energies of the fluid between sections a and b to fluid between sections c

CHAPTER 12. FLUID MECHANICS179and d!"1ρdV v22 v122dU ρdV g (y2 y1 ) .dK (12.48)By equating the work and change in mechanical energy we arrive at theBernoulli’s equation:dW dK dU!"1(p1 p2 ) dV ρdV v22 v12 ρdV g (y2 y1 )211 2p1 ρgy1 ρv1 p2 ρgy2 ρv2222(12.49)which only applies to incompressible and inviscid fluids. More general fluidsare described by the Navier-Stokes equation.Example 12.7. Water enters a house through a pipe with an insidediameter of 2.0 cm at an absolute pressure of 4.0 105 Pa. A 1.0 cm diameterpipe leads to the second-floor bathroom 5.0 m above. When the flow speed atthe inlet pipe is 1.5 m/s, find the flow speed, pressure and volume flow in thebathroom.