Chapter 3: Fluid Statics
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 21Chapter 2: Pressure and Fluid StaticsPressureFor a static fluid, the only stress is the normal stress sinceby definition a fluid subjected to a shear stress must deformand undergo motion. Normal stresses are referred to aspressure p.For the general case, the stress on a fluid element or at apoint is a tensor ij stress tensor* xx xy xz yx yy yz zx zy zzi facej direction*Tensor: A mathematical objectanalogus to but more general than avector, represented by an array ofcomponents that are functions of thecoordinates of a space (Oxford)For a static fluid, ij 0i jshear stresses 0 ii p xx yy zz i jnormal stresses -pAlso shows that p is isotropic, one value at a point which isindependent of direction, a scalar.
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 22Definition of Pressure:p lim A 0 F dF A dAN/m2 Pa (Pascal)F normal force acting over AAs already noted, p is a scalar, which can be easilydemonstrated by considering the equilibrium of forces on awedge-shaped fluid elementGeometry A y x cos z sin Fx 0pn A sin - px A sin 0pn p xW mg Vg VV ½ x z y Fz 0-pn A cos pz A cos - W 0 W ( cos )( sin ) y2 x z pn y cos pz y cos 2 cos sin y 02
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 23 p n p z sin 02pn pzfor 0i.e., pn px py pzp is single valued at a point and independent of direction.A body/surface in contact with a static fluid experiences aforce due to pF p pndASBNote: if p constant, Fp 0 for a closed body.Scalar form of Green's Theorem: f nds fd f constant f 0s
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 24Pressure TransmissionPascal's law: in a closed system, a pressure changeproduced at one point in the system is transmittedthroughout the entire system.Absolute Pressure, Gage Pressure, and Vacuumpg 0pA papa atmosphericpressure pg 0101.325 kPapA papA 0 absolutezeroFor pA pa,pg pA – pa gage pressureFor pA pa,pvac -pg pa – pA vacuum pressure
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 25Pressure Variation with ElevationBasic Differential EquationFor a static fluid, pressure varies only with elevation withinthe fluid. This can be shown by consideration ofequilibrium of forces on a fluid element1st order Taylor seriesestimate for pressurevariation over dzNewton's law (momentum principle) applied to a staticfluid F ma 0 for a static fluidi.e., Fx Fy Fz 0 Fz 0pdxdy ( p pdz)dxdy gdxdydz 0 z p g zBasic equation for pressure variation with elevation
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 26 Fy 0 Fx 0 p pdy)dxdz 0 pdydz (p dx )dydz 0 x y p p 0 0 x ypdxdz (p For a static fluid, the pressure only varies with elevation zand is constant in horizontal xy planes.The basic equation for pressure variation with elevationcan be integrated depending on whether constant or (z), i.e., whether the fluid is incompressible (liquid orlow-speed gas) or compressible (high-speed gas) sinceg constantPressure Variation for a Uniform-Density Fluidg p g zZ constant for liquid p zp2 p1 z 2 z1 p zAlternate forms:p1 z1 p2 z2 constantpiezometric pressurep z constantp z 0 0 gagei.e.,p z increase linearly with depthdecrease linearly with heightp z constant piezometric head
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 27p z cons tan tp1 z1 p 2 z 2p 2 p1 z1 z 2 p1 p atm 07.06p 2 oil z .8 9810 .9 7.06kPap3 p 2 water z 2 z3 7060 9810 2.127.7 27.7kPa
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 28Pressure Variation for Compressible Fluids:Basic equation for pressure variation with elevationdp ( p, z ) gdzPressure variation equation can be integrated for (p,z)known. For example, here we solve for the pressure in theatmosphere assuming (p,T) given from ideal gas law, T(z)known, and g g(z).p RTR gas constant 287 J/kg K dry airp,T in absolute scaledppg dzRTdp g dz pR T(z)which can be integrated for T(z) known
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 29Pressure Variation in the TroposphereT To (z – zo)linear decreaseTo T(zo)where p po(zo) known lapse rate 6.5 K/kmdpgdz pR [To (z z o )]ln p dz' dzgln[ To (z z o )] constant Ruse reference conditionln p o gln To constant Rsolve for constantlnz' To (z z o )po 101.3 kPaT 15 C 6.5 K/kmT (z z o )pg ln op o RTop To (z z o ) p o To zo earth surface 0g Ri.e., p decreases for increasing z
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Pressure Variation in the StratosphereT Ts 55 Cdpg dz pR Tsgln p z constantRTsuse reference condition to find constantp e ( z z0 ) g / RTspop p o exp[ (z z o )g / RTs ]i.e., p decreases exponentially for increasing z.Chapter 210
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 211Pressure MeasurementsPressure is an important variable in fluid mechanics andmany instruments have been devised for its measurement.Many devices are based on hydrostatics such as barometersand manometers, i.e., determine pressure throughmeasurement of a column (or columns) of a liquid usingthe pressure variation with elevation equation for anincompressible fluid.DifferentialmanometerMore modern devices include Bourdon-Tube Gage(mechanical device based on deflection of a spring) andpressure transducers (based on deflection of a flexiblediaphragm/membrane). The deflection can be monitoredby a strain gage such that voltage output is p acrossdiaphragm, which enables electronic data acquisition withcomputers.Bourdon-TubeGageIn this course we will use both manometers and pressuretransducers in EFD labs 2 and 3.
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 212Manometry1. Barometerpv Hgh patm Hg 13.6 kN/m3patm Hghpv 0 i.e., vapor pressure Hgnearly zero at normal Th 76 cmpatm 101 kPa (or 14.6 psia) Note:patm is relative to absolute zero, i.e., absolutepressure. patm patm(location, weather)Consider why water barometer is impractical Hg h Hg H2O h H2O H2O 9.80 kN/m3hH O 2 HghHg S Hg hHg 13.6 77 1047.2 cm 34 ft. H O2
57:020 Fluid MechanicsProfessor Fred Stern Fall 20132. PiezometerChapter 213patmpatm h ppipe pabsolutep hgageSimple but impractical for large p and vacuum pressures(i.e., pabs patm). Also for small p and small d, due to largesurface tension effects, which could be corrected using h 4 d , but accuracy may be problem if p/ h.3. U-tube or differential manometer patmp1 m h l p4p1 patmp4 m h lgage w[Sm h S l]for gases S Sm and can be neglected, i.e., can neglect pin gas compared to p in liquid in determining p4 ppipe.
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 214Example:Air at 20 C is in pipe with a water manometer. For givenconditions compute gage pressure in pipe. airl 140 cm h 70 cmp4 ? hgage (i.e., p1 0)Pressure same at 2&3 sincesame elevation & Pascal’sp1 h p3step-by-step method law: in closed systemp3 - airl p4pressure change produce atone part transmittedthroughout entire systemp1 h - airl p4complete circuit method h - airl p4gage water(20 C) 9790 N/m3 p3 h 6853 Pa [N/m2] air gpabs p p atm 6853 101300p 3 1.286 kg / m 3RT R C 273 287(20 273) K air 1.286 9.81m/s 12.62 N/m32note air waterp4 p3 - airl 6853 – 12.62 1.4 6835 Pa17.668if neglect effect of air columnp4 6853 Pa
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 215A differential manometer determines the difference inpressures at two points ①and ② when the actual pressureat any point in the system cannot be determined.p m h ( h) p1 f 12f 2p p ( ) ( m ) h1 2 f 2 1f p1 p 1 2 2 m 1 h f f f difference in piezometric head if fluid is a gas f m : p1 – p2 m h if fluid is liquid & pipe horizontalp1 – p2 ( m - f) h1 2:
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 216Hydrostatic Forces on Plane SurfacesFor a static fluid, the shear stress is zero and the only stressis the normal stress, i.e., pressure p. Recall that p is ascalar, which when in contact with a solid surface exerts anormal force towards the surface.Fp pndAAFor a plane surface n constant such that we canseparately consider the magnitude and line of action of Fp.F p F pdAALine of action is towards and normal to A through thecenter of pressure (xcp, ycp).
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 217Unless otherwise stated, throughout the chapter assume patmacts at liquid surface. Also, we will use gage pressure sothat p 0 at the liquid surface.Horizontal Surfaceshorizontal surface with area Ap constantFF pdA pALine of action is through centroid of A,i.e., (xcp, ycp) x, y
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 218Inclined SurfacesgzFx(x,y) centroid of A(xcp,ycp) center of pressureydp dz p zp – p0 - (z – z0) where p0 0 & z0 0p - z and y sin -zp y sin dF pdA y sin dA and sin are constantspF pdA sin ydAAAyAF sin y Ap pressure at centroid of Ay 1 ydAA1st moment of area
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 219F pAMagnitude of resultant hydrostatic force on plane surface isproduct of pressure at centroid of area and area of surface.Center of PressureCenter of pressure is in general below centroid sincepressure increases with depth. Center of pressure isdetermined by equating the moments of the resultant anddistributed forces about any arbitrary axis.Determine ycp by taking moments about horizontal axis 0-0ycpF y dFA y pdAA y( y sin )dAA sin y 2 dAAIo 2nd moment of area about 0-0 moment of inertiatransfer equation:I 2Io y A Imoment of inertia with respect to horizontalcentroidal axis
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 2202y cp F sin ( y A I)2y cp (pA ) sin ( y A I)2y cp sin yA sin ( y A I)2y cp yA y A Iycp y IyAycp is below centroid by I / yAycp y for large yFor po 0, y must be measured from an equivalent freesurface located po/ above y .
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 221Determine xcp by taking moment about y axisxcpF xdFA xpdAAx cp ( y sin A) x ( y sin )dAAx cp yA xydAAIxy product of inertia I xy x yAtransfer equationx cp yA I xy x yAx cp I xy xyAFor plane surfaces with symmetry about an axis normal to0-0, I xy 0 and xcp x .
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 222
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 223Hydrostatic Forces on Curved SurfacesFree surfacep hF pndAAHorizontal ComponentsFx F î pn îdAh distance belowfree surface(x and y components)A pdA xdAx projection of ndA ontovertical plane to x-directionAxFy F ĵ pdA ydA y n ĵdAAy projection ndAonto vertical plane toy-directionTherefore, the horizontal components can be determined bysome methods developed for submerged plane surfaces.
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 224The horizontal component of force acting on a curvedsurface is equal to the force acting on a vertical projectionof that surface including both magnitude and line of action.Vertical ComponentsFz F k̂ pn k̂dAA pdA zp hAzh distancebelow freesurface hdA z VAz weight offluid abovesurface AThe vertical component of force acting on a curved surfaceis equal to the net weight of the column of fluid above thecurved surface with line of action through the centroid ofthat fluid volume.
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 225Example: Drum GatePressure Diagramp h R(1-cos )n sin î cos k̂dA Rd : Area p acts over (Note: Rd arc length)F R (1 cos )( sin î cos k̂ ) Rd 0p ndAF î Fx R (1 cos ) sin d 202 1 R cos cos 2 2 R 24 0 ( R)(2R ) same force as that on projection ofAarea onto vertical planep Fz R (1 cos ) cos d 20 sin2 R 2 sin 24 0 R 2 2 V R 2 2 net weight of water above surface
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Another approach:1 F1 R 2 R 2 4 R 2 1 4 F2 R 22F F2 F1 R 22 F1Chapter 226
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 227BuoyancyArchimedes PrincipleFB Fv2 – Fv1 fluid weight above Surface 2 (ABC)– fluid weight above Surface 1 (ADC) fluid weight equivalent to body volume VFB gVV submerged volumeLine of action is through centroid of V center ofbuoyancyNet Horizontal forces are zero sinceFBAD FBCD
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 228HydrometryA hydrometer uses the buoyancy principle to determinespecific weights of liquids.StemBulbFB wV oW mg fV S wV S w(Vo V) S w(Vo a h) fVFB W at equilibrium: wV o S w(Vo a h)a cross section area stem h stem height above waterlineVo/S Vo a ha h Vo – Vo/S 1 1 h(S); Calibrate scale using fluids of S known SVoS S( h); Convert scale to directly read SV0 a h h Voa
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 229Example (apparent weight)King Hiero ordered a new crown to be made from puregold. When he received the crown he suspected that othermetals had been used in its construction. Archimedesdiscovered that the crown required a force of 4.7# tosuspend it when immersed in water, and that it displaced18.9 in3 of water. He concluded that the crown was notpure gold. Do you agree? Fvert 0 Wa Fb – W 0 Wa W – Fb ( c - w)VW cV, Fb wVWW wVor c a w aVV c 4.7 62.4 18.9 / 1728 492.1 c g18.9 / 1728 c 15.3 slugs/ft3 steel and since gold is heavier than steel the crowncan not be pure gold
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 230Stability of Immersed and Floating BodiesHere we’ll consider transverse stability. In actualapplications both transverse and longitudinal stability areimportant.Immersed BodiesStableNeutralUnstableStatic equilibrium requires: Fv 0 and M 0 M 0 requires that the centers of gravity and buoyancycoincide, i.e., C G and body is neutrally stableIf C is above G, then the body is stable (righting momentwhen heeled)If G is above C, then the body is unstable (heeling momentwhen heeled)
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 231Floating BodiesFor a floating body the situation is slightly morecomplicated since the center of buoyancy will generallyshift when the body is rotated depending upon the shape ofthe body and the position in which it is floating.Positive GMNegative GMThe center of buoyancy (centroid of the displaced volume)shifts laterally to the right for the case shown because partof the original buoyant volume AOB is transferred to a newbuoyant volume EOD.The point of intersection of the lines of action of thebuoyant force before and after heel is called the metacenterM and the distance GM is called the metacentric height. IfGM is positive, that is, if M is above G, then the ship isstable; however, if GM is negative, the ship is unstable.
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 232Floating Bodies small heel anglex CC lateral displacementof CC center of buoyancyi.e., centroid of displacedvolume VSolve for GM: find x using(1) basic definition for centroid of V; and(2) trigonometryFig. 3.17(1) Basic definition of centroid of volume VxV xd V x i Vimoment about centerplanexV moment V before heel – moment of VAOB moment of VEOD 0 due to symmetry oforiginal V about y axisi.e., ship centerplanexV ( x)dV xdVAOBEODtan y/xdV ydA x tan dAxV x2 tan dA x 2 tan dAAOBEOD
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013xV tan x 2 dAship waterplane areamoment of inertia of ship waterplaneabout z axis O-O; i.e., IOOIOO moment of inertia of waterplanearea about centerplane axis(2) TrigonometryxV tan I OOCC x tan I OO CM tan VCM IOO / VGM CM – CGGM I OO CGVGM 0StableGM 0UnstableChapter 233
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 234Fluids in Rigid-Body MotionFor fluids in motion, the pressure variation is no longerhydrostatic and is determined from application of Newton’s2nd Law to a fluid element. ij viscous stressesdirectionp pressureMa inertia forceW weight (body force)net surface force in X yx zx p VX net xx y z x xNewton’s 2nd LawpressureMa F FB FSper unit volume ( V) a fb fsThe acceleration of fluid particleDV V V Va Dt tfb body force gk̂fs surface force fp fvviscous
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 235fp surface force due to p pfv surface force due to viscous stresses ij a f b f p f v a gkˆ pNeglected in this chapter andincluded later in Chapter 6when deriving completeNavier-Stokes equationsinertia force body force due surface force due toto gravitypressure gradientsWhere for general fluid motion, i.e. relative motionbetween fluid particles:a DV Dt V tlocalaccelerationx:y: V Vconvectiveaccelerationsubstantial derivativeDu p Dt x u u u u p u v w x y z x t Dv p Dt y v v v v p u v w x y z y t
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013z:Chapter 236Dw p g p z Dt z z w w w w u v w p z x y z z t But in this chapter rigid body motion, i.e., norelative motion between fluid particles a (p z) V 0Note: for V 0 p gk̂ p p 0 x y p g zEuler’s equation for inviscid flowContinuity equation forincompressible flow (See Chapter 6)4 equations in four unknowns V and pFor rigid body translation: a axiˆ az kˆ2For rigid body rotating: a r eˆrIf a 0 , the motion equation reduces to hydrostaticequation: p p 0 x y p z
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 237Examples of Pressure Variation From AccelerationUniform Linear Acceleration: a gk̂ p p a p a gk̂ g a x î g a z k̂ p a x xg gk̂a a x î a z k̂ p g a z z p ax x1. ax 0p increase in x2. ax 0p decrease in x p g az z1. az 0p decrease in z2. az 0 and az gp decrease in z but slower than g3. az 0 and az gp increase in z
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 238ŝ unit vector in direction of p p / p a x î g a z k̂ a2x g a z 2 1/ 2n̂ unit vector in direction of p constant ŝ ĵ ijkijk to pby definition linesof constant p arenormal to p a x k̂ (g a z )î a2x (g a z ) 2 1/ 2 tan-1 ax / (g az) angle between n̂ and x 1/ 2dp p ŝ a 2x g a z 2 gdsGp Gs constant pgage Gs
57:020 Fluid MechanicsProfessor Fred Stern Fall 2013Chapter 239Rigid Body Rotation:Consider a cylindrical tank of liquid rotating at a constantrate k̂a ro centripetal acceleration p (g a ) gk̂ r 2 ê rcoordinates r 2 ê rV2ê r r 1 ê r ê ê z rr zgrad in cylindrical p p p r 2 g 0 r z C (r)pressure distribution is hydrostatic in z direction and p r 2 2 f (z) cpz - g2i.e.,p - gz C(r) c p r 2
57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 1 Chapter 2: Pressure and Fluid Statics Pressure For a static fluid, the only stress is the normal stress since by definition a fluid subjected to a shear stress must deform and undergo motion.
ARET 3400 Chapter 3 – Fluid Statics Page 17 . Chapter 3 – Fluid Statics . 3.1 Pressure . Consider a small cylinder of fluid at rest as shown in Figure 3.1. The cylinder has a length L and a cross-sectional area A. Since the cylinder is at rest, the sum of
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
Fluid statics, static pressure/1 Two types of forces act on a fluid volume element: surface (pressure) forcesand body (gravitational) forces: see Figure Pressure (a scalar!) is defined as surface force / area, for example pb Fb / (d·w) p @ z z1 Picture: KJ05 Fluid volume h·d·w with
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
L M A B CVT Revision: December 2006 2007 Sentra CVT FLUID PFP:KLE50 Checking CVT Fluid UCS005XN FLUID LEVEL CHECK Fluid level should be checked with the fluid warmed up to 50 to 80 C (122 to 176 F). 1. Check for fluid leakage. 2. With the engine warmed up, drive the vehicle to warm up the CVT fluid. When ambient temperature is 20 C (68 F .
Engineering Mechanics Statics 8th Edition meriam Solutions Manual Author: meriam Subject: Engineering Mechanics Statics 8th Edition meriam Solutions ManualInstant Download Keywords: Engineering Mechanics Statics
The statics and strength of materials course is both a foundation and a framewor k for most of the following advanced MET courses. Many of the advanced courses have a prerequisite of statics and strength of materials. Thus, this statics and strength of materials cours e is critical to the MET curriculum.
READING COMPREHENSION PRACTICE EXAM. GENERAL INSTRUCTIONS: You will have 90 minutes for this test. Work rapidly but carefully. Do no spend too much time on any one question. If you have time after you have finished the test, go back to the questions you have left unanswered. The three parts of this test are English Usage, Sentence Correction, and Reading Comprehension. When you have finished .
