57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 1 Chapter : 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. Normal stresses are referred to as pressure p. For the general case, the stress on a fluid element or at a point is a tensor τ ij stress tensor τ xx τ xy τ xz τ yx τ yy τ yz τ zx τ zy τ zz For a static fluid, τ ij 0 i j shear stresses 0 i force j direction τ ii p τ xx τ yy τ zz i j normal stresses -p Also shows that p is isotropic, one value at a point which is independent of direction, a scalar.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 Definition of Pressure: δf df p lim δa 0 δa da N/m Pa (Pascal) F normal force acting over A As already noted, p is a scalar, which can be easily demonstrated by considering the equilibrium of forces on a wedge-shaped fluid element Geometry ΔA Δ l Δy Δx Δ l cosα Δz Δ l sinα ΣF x 0 p n ΔA sin α - p x ΔA sin α 0 p n p x W mg ρvg γv V ½ ΔxΔzΔy ΣF z 0 -p n ΔA cos α + p z ΔA cos α - W 0 γ W ( Δl cosα)( Δlsin α) Δy Δx Δz p ΔlΔycosα+ p ΔlΔycosα n γ Δl cos αsin αδ y 0 ΔlΔycosα γ pn + pz Δlsinα 0 z
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 3 γ pn + pz Δlsinα 0 pn pz for Δl 0 i.e., p n p x p y p z p is single valued at a point and independent of direction. A body/surface in contact with a static fluid experiences a force due to p F p S B pnda Note: if p constant, F p 0 for a closed body. Scalar form of Green's Theorem: fnds fd f constant f 0 s
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 4 Pressure Transmission Pascal's law: in a closed system, a pressure change produced at one point in the system is transmitted throughout the entire system. Absolute Pressure, Gage Pressure, and Vacuum p A > p a p g > 0 p g < 0 p a atmospheric pressure 101.35 kpa p A < p a p A 0 absolute zero For p A >p a, For p A <p a, p g p A p a gage pressure p vac -p g p a p A vacuum pressure
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 5 Pressure Variation with Elevation Basic Differential Equation For a static fluid, pressure varies only with elevation within the fluid. This can be shown by consideration of equilibrium of forces on a fluid element 1 st order Taylor series estimate for pressure variation over dz Newton's law (momentum principle) applied to a static fluid ΣF ma 0 for a static fluid i.e., ΣF x ΣF y ΣF z 0 ΣF z 0 pdxdy ( p p + dz dxdy z ) ρ gdxdydz 0 p ρg γ z Basic equation for pressure variation with elevation
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 6 F y 0 F x 0 p pdxdz (p + dy)dxdz 0 y p 0 y p pdydz (p + dx)dydz 0 x p 0 x For a static fluid, the pressure only varies with elevation z and is constant in horizontal xy planes. The basic equation for pressure variation with elevation can be integrated depending on whether ρ constant or ρ ρ(z), i.e., whether the fluid is incompressible (liquid or low-speed gas) or compressible (high-speed gas) since g constant Pressure Variation for a Uniform-Density Fluid p ρg γ ρ constant for liquid Z z g p γ z Δ p γδ z ( ) p p γ z z 1 1 Alternate forms: i.e., p p1+ γ z1 p +γ z p + γ z constant p( z 0) 0 gage γ z p γ + z constant constant piezometric pressure increase linearly with depth decrease linearly with height piezometric head
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 7 7.7 7.06 p + γ z cons tan t p +γ z p +γz 1 1 ( ) p p +γ z z 1 1 p p 0 1 atm p γ Δ z.8 9810.9 7.06kPa oil ( ) p p +γ z z 3 water 3 7060 + 9810.1 7.7kPa
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 8 Pressure Variation for Compressible Fluids: Basic equation for pressure variation with elevation dp dz γ γ(p,z) ρg Pressure variation equation can be integrated for γ(p,z) known. For example, here we solve for the pressure in the atmosphere assuming ρ(p,t) given from ideal gas law, T(z) known, and g g(z). p ρrt R gas constant 87 J/kg K p,t in absolute scale dry air dp dz pg RT dp p g dz which can be integrated for T(z) known R T(z)
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 9 Pressure Variation in the Troposphere T T o α(z z o ) linear decrease T o T(z o ) where p p o (z o ) known α lapse rate 6.5 K/km dp p g R [T o dz α(z z o )] z' T o dz' αdz α(z z o ) ln p g R α ln[to α(z zo )] + constant use reference condition ln p g lnt αr o o + solve for constant constant z o earth surface 0 p o 101.3 kpa T 15 C α 6.5 K/km ln p p o g R α T ln o α(z z T o o ) p p o T o α(z z T o o ) g αr i.e., p decreases for increasing z
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 10 Pressure Variation in the Stratosphere T T s 55 C dp g dz p R T s g ln p z RT s + constant use reference condition to find constant p p o e (z z )g / RT 0 s p p o exp[ (z z o )g / RT s ] i.e., p decreases exponentially for increasing z. Pressure Measurements Pressure is an important variable in fluid mechanics and many instruments have been devised for its measurement. Many devices are based on hydrostatics such as barometers and manometers, i.e., determine pressure through measurement of a column (or columns) of a liquid using the
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 11 pressure variation with elevation equation for an incompressible fluid. Differential manometer More modern devices include Bourdon-Tube Gage (mechanical device based on deflection of a spring) and pressure transducers (based on deflection of a flexible diaphragm/membrane). The deflection can be monitored by a strain gage such that voltage output is Δp across diaphragm, which enables electronic data acquisition with computers. Bourdon-Tube Gage In this course we will use both manometers and pressure transducers in EFD labs and 3.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 1 Manometry 1. Barometer p v + γ Hg h p atm p atm γ Hg h p v 0 i.e., vapor pressure Hg nearly zero at normal T h 76 cm p atm 101 kpa (or 14.6 psia) Note: p atm is relative to absolute zero, i.e., absolute pressure. p atm p atm (location, weather) Consider why water barometer is impractical γ h γ h Hg Hg HO HO h H O γ γ Hg H O h Hg S Hg h Hg 13.6 76 1033.6cm 34ft.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 13. Piezometer p atm p atm + γh p pipe p p γh absolute gage Simple but impractical for large p and vacuum pressures (i.e., p abs < p atm ). Also for small p and small d, due to large surface tension effects, could be corrected using Δ h 4σ γ d, but accuracy may be problem if p γ Δ 3. U-tube or differential manometer γ h σ p atm p 1 + γ m Δh γl p 4 p 1 p atm p 4 γ m Δh γl gage γ w [S m Δh S l] for gases S << S m and can be neglected, i.e., can neglect Δp in gas compared to Δp in liquid in determining p 4 p pipe. Example:
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 14 Air at 0 C is in pipe with a water manometer. For given conditions compute gage pressure in pipe. γ air l 140 cm Δh 70 cm γδh p 4? gage (i.e., p 1 0) Pressure same at &3 since same elevation & Pascal s p 1 + γδh p 3 step-by-step method law: in closed system p 3 - γ air l p 4 pressure change produce at one part transmitted throughout entire system p 1 + γδh - γ air l p 4 complete circuit method γδh - γ air l p 4 gage γ water (0 C) 9790 N/m 3 p 3 γδh 6853 Pa [N/m ] γ air ρg ( p + patm ) ( C + 73) p abs p 6853 + 101300 ρ RT R 87(0 + 73) K γ air 1.86 9.81m/s 1.6 N/m 3 3 m 3 1.86 kg / note γ air << γ water p 4 p 3 - γ air l 6853 1.6 1.4 6835 Pa 17.668 if neglect effect of air column p 4 6853 Pa or could use Table A.3
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 15 A differential manometer determines the difference in pressures at two points 1and when the actual pressure at any point in the system cannot be determined. p +γ l γmδh γ ( l Δ h) p 1 f 1 f p p γ ( l l ) + ( γm γ ) Δh 1 f 1 f p1 p γ m + l 1 + l 1 Δh γ f γ f γ f difference in piezometric head if fluid is a gas γ f << γ m : p 1 p γ m Δh if fluid is liquid & pipe horizontal l l : 1 p 1 p (γ m - γ f ) Δh
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 16 Hydrostatic Forces on Plane Surfaces For a static fluid, the shear stress is zero and the only stress is the normal stress, i.e., pressure p. Recall that p is a scalar, which when in contact with a solid surface exerts a normal force towards the surface. F p pnda A For a plane surface n constant such that we can separately consider the magnitude and line of action of F p. F p F A pda Line of action is towards and normal to A through the center of pressure (x cp, y cp ).
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 17 Unless otherwise stated, throughout the chapter assume p atm acts at liquid surface. Also, we will use gage pressure so that p 0 at the liquid surface. Horizontal Surfaces horizontal surface with area A p constant F pda pa Line of action is through centroid of A, x, y i.e., (x cp, y cp ) ( ) F
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 18 Inclined Surfaces g z dp γ dz Δ p γδz F (x,y) centroid of A (x cp,y cp ) center of pressure x y df pda γy sin α da F pda γsin α A p A yda y A γ and sin α are constants 1 y yda A 1 st moment of area
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 19 F γsin α ya p pressure at centroid of A F pa Magnitude of resultant hydrostatic force on plane surface is product of pressure at centroid of area and area of surface. Center of Pressure Center of pressure is in general below centroid since pressure increases with depth. Center of pressure is determined by equating the moments of the resultant and distributed forces about any arbitrary axis. Determine y cp by taking moments about horizontal axis 0-0 y cp F ydf A y pda A A y ( γysin α)da γsin α y da A I o nd moment of area about 0-0 moment of inertia
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 0 transfer equation: Io y A + I y cp I moment of inertia with respect to horizontal centroidal axis F γsin α(y A + I) y cp (pa) γsin α(y A + I) y cp γsin α ya γsin α(y A + I) ycp ya y A + I y cp y+ I ya y cp is below centroid by I / ya y cp y for large y For p o 0, y must be measured from an equivalent free surface located p o /γ above y.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 1 Determine x cp by taking moment about y axis x cp F xdf A A xpda x cp ( γysin αa) x( γysin α) da A x cp ya xyda A I xy product of inertia Ixy + xya transfer equation x cp ya Ixy + xya x Ixy ya cp + x For plane surfaces with symmetry about an axis normal to 0-0, 0 Ixy and x cp x.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 3 Hydrostatic Forces on Curved Surfaces Free surface p γh F pnda A h distance below free surface Horizontal Components F F î pn îda x A A x pda x (x and y components) da x projection of nda onto plane to x-direction F y F ĵ pda y da y n ĵda A y projection nda onto plane to y-direction Therefore, the horizontal components can be determined by some methods developed for submerged plane surfaces. The horizontal component of force acting on a curved surface is equal to the force acting on a vertical projection of that surface including both magnitude and line of action.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 4 Vertical Components F z F kˆ pn kˆ da A pda z p γh A z hdistance below free surface γ hda γv A z z weight of fluid above surface A The vertical component of force acting on a curved surface is equal to the net weight of the column of fluid above the curved surface with line of action through the centroid of that fluid volume.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 5 Example: Drum Gate Pressure Diagram p γh γr(1-cosθ) n sin θî + cosθkˆ da l Rdθ F γr(1 cosθ)( sin θî + cosθkˆ) l Rdθ π 0 F î F +γ l R x p n da π (1 cosθ)sin θdθ 0 1 γ lr cosθ + cosθ γlr 4 0 (γr)(r l ) same force as that on projection of p A area onto vertical plane F γ l R z π (1 cosθ)cosθdθ 0 π θ sin θ γ l R sin θ 4 0 π πr γ l R γ l γv net weight of water above surface π
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 6 Buoyancy Archimedes Principle F B F v F v1 fluid weight above Surface (ABC) fluid weight above Surface 1 (ADC) fluid weight equivalent to body volume V F B ρgv V submerged volume Line of action is through centroid of V center of buoyancy Net Horizontal forces are zero since F BAD F BCD
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 7 Hydrometry A hydrometer uses the buoyancy principle to determine specific weights of liquids. Stem Bulb W mg γ f V Sγ w V W γ w V o Sγ w (V o ΔV) Sγ w (V o aδh) γ f V a cross section area stem V o /S V o aδh aδh V o V o /S V Δh o 1 1 Δh(S) a S Δh V o a S 1 calibrate scale using fluids of known S S S V 0 Vo aδh
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 8 Example (apparent weight) King Hero ordered a new crown to be made from pure gold. When he received the crown he suspected that other metals had been used in its construction. Archimedes discovered that the crown required a force of 4.7# to suspend it when immersed in water, and that it displaced 18.9 in 3 of water. He concluded that the crown was not pure gold. Do you agree? F vert 0 W a + F b W 0 W a W F b (γ c - γ w )V Wγ c V, F b γ w V Wa Wa + γ wv or γ c + γ w V V γ 4.7 + 6.4 18.9/178 18.9/178 49.1 c ρ c ρ c 15.3 slugs/ft 3 ρ steel and since gold is heavier than steel the crown can not be pure gold g
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 9 Stability of Immersed and Floating Bodies Here we ll consider transverse stability. In actual applications both transverse and longitudinal stability are important. Immersed Bodies Static equilibrium requires: 0 and M 0 M 0 requires that the centers of gravity and buoyancy coincide, i.e., C G and body is neutrally stable If C is above G, then the body is stable (righting moment when heeled) If G is above C, then the body is unstable (heeling moment when heeled) F v
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 30 Floating Bodies For a floating body the situation is slightly more complicated since the center of buoyancy will generally shift when the body is rotated depending upon the shape of the body and the position in which it is floating. Positive GM Negative GM The center of buoyancy (centroid of the displaced volume) shifts laterally to the right for the case shown because part of the original buoyant volume AOB is transferred to a new buoyant volume EOD. The point of intersection of the lines of action of the buoyant force before and after heel is called the metacenter M and the distance GM is called the metacentric height. If GM is positive, that is, if M is above G, then the ship is stable; however, if GM is negative, the ship is unstable.
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 31 Floating Bodies α small heel angle x CC lateral displacement of C C center of buoyancy i.e., centroid of displaced volume V Solve for GM: find x using (1) basic definition for centroid of V; and () trigonometry Fig. 3.17 (1) Basic definition of centroid of volume V x V xdv x i ΔVi moment about centerplane x V moment V before heel moment of V AOB + moment of V EOD 0 due to symmetry of original V about y axis i.e., ship centerplane xv ( x)dv + xdv tan α y/x AOB EOD dv yda x tan α da xv xtan da x α + tanαda AOB EOD
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 3 xv tan α x da ship waterplane area moment of inertia of ship waterplane about z axis O-O; i.e., I OO I OO moment of inertia of waterplane area about centerplane axis () Trigonometry xv tan αi CC x OO tan αi V OO CM tan α CM I OO / V GM CM CG GM I OO V CG GM > 0 GM < 0 Stable Unstable
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 33 Fluids in Rigid-Body Motion For fluids in motion, the pressure variation is no longer hydrostatic and is determined from application of Newton s nd Law to a fluid element. τ ij viscous stresses p pressure Ma inertia force W weight (body force) net surface force in X direction X net p τ + x x xx τ + y yx τ + z zx V Newton s nd Law pressure viscous Ma F F B + F S per unit volume ( V) ρa f b + f s The acceleration of fluid particle DV V a + V V Dt t f b body force ρgkˆ f s surface force f p + f v
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 34 f p surface force due to p p f v surface force due to viscous stresses τ ij ρ a f + f + f b p v ρa ρgkˆ p Neglected in this chapter and included later in Chapter 6 when deriving complete Navier-Stokes equations inertia force body force due + surface force due to to gravity pressure gradients Where for general fluid motion, i.e. relative motion between fluid particles: DV V a + V V Dt t 13 substantial derivative x: { convective local acceleration acceleration Du p ρ Dt x u u u u ρ + u + v + w t x y z p x y: Dv p ρ Dt y v v ρ + u t x v + v y v + w z p y
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 35 Dw p Dt z z w w w w ρ + u + v + w t x y z z p + γz z: ρ ρg ( p + γz) ( ) But in this chapter rigid body motion, i.e., no relative motion between fluid particles Note: for V 0 p ρgkˆ p x p z p y ρg 0 γ ρa (p + γz) Euler s equation for inviscid flow V 0 Continuity equation for incompressible flow (See Chapter 6) 4 equations in four unknowns V and p For rigid body translation: a a ˆ ˆ xi + azk For rigid body rotating: a rω eˆr If a 0, the motion equation reduces to hydrostatic equation: p p 0 x y p γ z
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 36 Examples of Pressure Variation From Acceleration Uniform Linear Acceleration: ρa ρgkˆ p p ρ p ρ p ρa x ( a + gkˆ ) ρ( g a) g gkˆ [ a î + ( g + a ) kˆ ] a a î a kˆ x z x + p ρ z ( g + ) x a z z p ρ a x x 1. a x < 0 p increase in +x. a x > 0 p decrease in +x p ρ z ( g + ) a z 1. a z > 0 p decrease in +z. a z < 0 and az 3. a z < 0 and az < g p decrease in +z but slower than g > g p increase in +z
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 37 ŝ unit vector in direction of p p / p [ a xî + ( g + a z ) kˆ ] + ( g + a ) 1/ [ a ] x z nˆ unit vector in direction of p constant ŝ ĵ ijkijk a [ a + (g + a ) ] 1/ x x kˆ + (g + a z z )î to p by definition lines of constant p are normal to p θ tan -1 a x / (g + a z ) angle between nˆ and x dp ds [ a + ( g + a ) ] 1/ p ŝ ρ x z > ρg G p ρgs + constant p gage ρgs
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 38 Rigid Body Rotation: Consider a cylindrical tank of liquid rotating at a constant rate Ω Ωkˆ ( Ω ) a Ω r o centripetal acceleration rω êr V êr r p ρ(g a) 1 êr + êθ + ê r r θ z ρgkˆ + ρrω ê grad in cylindrical coordinates z r p i.e., r and p ρrω C (r) ρ r Ω + f (z) + c p ρg z p 0 θ pressure distribution is hydrostatic in z direction p z -ρg p -ρgz + C(r) + c p ρ r Ω ρgz p V + constant + z constant γ g V rω
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 39 The constant is determined by specifying the pressure at one point; say, p p o at (r, z) (0, 0) p p o ρgz + 1 r Ω ρ Note: pressure is linear in z and parabolic in r Curves of constant pressure are given by p p ρg 0 z + a + br r Ω g which are paraboloids of revolution, concave upward, with their minimum point on the axis of rotation Free surface is found by requiring volume of liquid to be constant (before and after rotation) The unit vector in the direction of p is ρgkˆ + ρrω êr ŝ 1/ ( ρg) + ( ρrω ) dz g tan θ slope of ŝ dr rω Ω i.e., r C 1 exp z equation of p surfaces g
57:00 Fluid Mechanics Chapter Professor Fred Stern Fall 008 40