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058:0160 Professor Fred Stern Chapter 2 1 Fall 2016 Chapter 2: Pressure Distribution in a Fluid Pressure and pressure gradient In fluid statics, as well as in fluid dynamics, the forces acting on a portion of fluid (CV) bounded by a CS are of two kinds: body forces and surface forces. Body Forces: act on the entire body of the fluid (force per unit volume). Surface Forces: act at the CS and are due to the surrounding medium (force/unit areastress). In general the surface forces can be resolved into two components: one normal and one tangential to the surface. Considering a cubical fluid element, we see that the stress in a moving fluid comprises a 2nd order tensor. y σxy σxz x z Face σxx Direction 058:0160 Professor Fred Stern Chapter 2 2 Fall 2016 ij xx yx zx xy yy zy xz yz zz Since by definition, a fluid cannot withstand a shear stress without moving (deformation), a stationary fluid must necessarily be completely free of shear stress (σij=0, i ≠ j). The only non-zero stress is the normal stress, which is referred to as pressure: σii=-p σn = -p, which is compressive, as it should be since fluid cannot withstand tension. (Sign convention based on the fact that p>0 and in the direction of –n) n (one value at a point, independent of direction; p is a scalar) i.e. normal stress (pressure) is isotropic. This can be easily seen by considering the equilibrium of a wedge shaped fluid element Or p x = p y = p z = pn = p 058:0160 Professor Fred Stern Chapter 2 3 Fall 2016 F : p dA sin p dA sin 0 x n x p p n x pndA α dl F : p dA cos p dA cos W 0 z n z z y x pxdAsinα α dA=dldy W=ρgV=γV Where: pzdAcosα W V V y 1 xz 2 x l cos W dA cos z l sin 1 yl dA y dA / dl dl sin 2 1 p dA cos p dA cos dA cos dl sin 0 n z 2 p p dl sin 0 n z 2 p p for dl 0 i.e. p p p p n z n x y z Note: For a fluid in motion, the normal stress is different on each face and not equal to p. σxx ≠ σyy ≠ σzz ≠ -p By convention p is defined as the average of the normal stresses p 1 1 xx yy zz ii 3 3 058:0160 Professor Fred Stern Chapter 2 4 Fall 2016 The fluid element experiences a force on it as a result of the fluid pressure distribution if it varies spatially. Consider the net force in the x direction due to p(x,t). dy dz dx = The result will be similar for dFy and dFz; consequently, we conclude: p p ˆ p ˆ dFpress iˆ j k x y z Or: f p force per unit volume due to p(x,t). Note: if p=constant, f 0 . 058:0160 Professor Fred Stern Chapter 2 5 Fall 2016 Equilibrium of a fluid element Consider now a fluid element which is acted upon by both surface forces and a body force due to gravity f g (per unit volume) dF gd or grav grav Application of Newton’s law yields: ma F da f d a f f body f surface f body g and g gkˆ f per unit d body g kˆ z g f f pressure viscous surface (includes f , since in general ij p ij ij ) viscous f f pressure Viscous part p 2V 2V 2V 2V f viscous x 2 y 2 z 2 For ρ, μ=constant, the viscous force will have this form (chapter 4). a p g 2V inertial pressure gradient gravity viscous with a V V V t 058:0160 Professor Fred Stern Chapter 2 6 Fall 2016 This is called the Navier-Stokes equation and will be discussed further in Chapter 4. Consider solving the N-S equation for p when a and V are known. p g a 2V B( x, t ) This is simply a first order PDE for p and can be solved readily. For the general case (V and p unknown), one must solve the NS and continuity equations, which is a formidable task since the NS equations are a system of 2nd order nonlinear PDEs. We now consider the following special cases: 1) Hydrostatics ( a V 0 ) 2) Rigid body translation or rotation ( 2V 0 ) 3) Irrotational motion ( V 0) ( a) ( a) 2 a vector identity if constant V 0 2V 0 Euler equation also, V 0 V & if const. 2 0 Bernoulli equation 058:0160 Professor Fred Stern Chapter 2 7 Fall 2016 Case (1) Hydrostatic Pressure Distribution p g g k i.e. or p p 0 x y and z p g z 2 2 p p gdz g ( z )dz 2 1 1 1 g dp gdz 2 r g g 0 0 r constant near earth' s surface r 0 liquids ρ = constant (for one liquid) p = -ρgz + constant gases ρ = ρ(p,t) which is known from the equation of state: p = ρRT ρ = p/RT dp g dz p R T (z ) which can be integrated if T =T(z) is known as it is for the atmosphere. 058:0160 Professor Fred Stern Fall 2016 Chapter 2 8 Manometry Manometers are devices that use liquid columns for measuring differences in pressure. A general procedure may be followed in working all manometer problems: 1.) Start at one end (or a meniscus if the circuit is continuous) and write the pressure there in an appropriate unit or symbol if it is unknown. 2.) Add to this the change in pressure (in the same unit) from one meniscus to the next (plus if the next meniscus is lower, minus if higher). 3.) Continue until the other end of the gage (or starting meniscus) is reached and equate the expression to the pressure at that point, known or unknown. 058:0160 Professor Fred Stern Chapter 2 9 Fall 2016 Hydrostatic forces on plane surfaces The force on a body due to a pressure distribution is: F pn dA A where for a plane surface n = constant and we need only consider |F| noting that its direction is always towards the surface: | F | p dA . A Consider a plane surface AB entirely submerged in a liquid such that the plane of the surface intersects the freesurface with an angle α. The centroid of the surface is denoted ( x, y ). F sin yA pA Where p is the pressure at the centroid. 058:0160 Professor Fred Stern Chapter 2 10 Fall 2016 To find the line of action of the force which we call the center of pressure (xcp, ycp) we equate the moment of the resultant force to that of the distributed force about any arbitrary axis. y F ydF Note: dF y sin dA sin y dA cp A 2 A 2 y dA I moment of Inertia about O O o A 2 y A I = moment of inertia WRT horizontal centroidal axis F pA sin yA I 2 ycp sin yA sin y A I y y cp I yA and similarly for xcp x F xdF I product of inertia where cp xy A I I x yA I x x yA xy xy xy cp Note that the coordinate system in the text has its origin at the centroid and is related to the one just used by: x xx text and y y y text 058:0160 Professor Fred Stern Chapter 2 11 Fall 2016 Hydrostatic Forces on Curved Surfaces z x y In general, Horizontal Components: Fx F i p n i dA F p n dA dAx A Fy p dAy Ay dAx = projection of n dA onto a plane perpendicular to x direction dAy = projection of n dA onto a plane perpendicular to y direction 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 and can be determined by the methods developed for plane surfaces. Fz pn k dA p dAz h dAz Az Az Where h is the depth to any element area dA of the surface. The vertical component of force acting on a curved surface is equal to the net weight of the total column of fluid directly above the curved surface and has a line of action through the centroid of the fluid volume. 058:0160 Professor Fred Stern Chapter 2 12 Fall 2016 Example Drum Gate h=R-Rcosθ=R(1-cosθ) p h R 1 cos h n sin i cos k dA lRd F R1 cos sin iˆ coskˆ lRd dA 0 n p F .iˆ Fx lR 2 1 cos sin d 0 1 lR 2 cos 0 cos 2 0 2lR 2 4 Same force as that on projection of gate R 2 Rl p A onto vertical plane perpendicular direction 058:0160 Professor Fred Stern Fz lR 2 Chapter 2 13 Fall 2016 1 cos cos d 0 1 lR 2 sin sin 2 2lR 2 2 4 0 R 2 lR l 2 2 2 Net weight of water above curved surface Another approach: 1 F1 2 R 2l R 2l / 2 2 1 lR 2 2 / 2 2 1 F2 R 2l F1 2 1 F F2 F1 R 2l 2 058:0160 Professor Fred Stern Fall 2016 Chapter 2 14 Hydrostatic Forces in Layered Fluids See textbook 2.7 Buoyancy and Stability Archimedes Principle F F F B V (2) V (1) = fluid weight above 2ABC – fluid weight above 1ADC = weight of fluid equivalent to the body volume In general, FB = ρg ( = submerged volume). The line of action is through the centroid of the displaced volume, which is called the center of buoyancy. 058:0160 Professor Fred Stern Chapter 2 15 Fall 2016 Example: Floating body in heave motion L y y G ρb h d=draft B b hg mg 0 where is Weight of the block W b Lb 0 A wp displaced water volume by the block and is the specific weight of the liquid. W B b Lbhg w Lbdg d b h Sb h w W B b w : d h b w : d h sink b w : d h floating Instantaneous displaced water volume: 0 yAwp F V .. m y B W 0 Awp y 058:0160 Professor Fred Stern Chapter 2 16 Fall 2016 .. m y Awp y 0 .. y Awp m y0 y A cos nt B sin nt . Use initial condition ( t 0, and B: y y0 cos nt . y y0 y y 0 ) to determine A . y0 n sin nt Where n period T 2 Awp m 2 m Awp Spar Buoy T is tuned to decrease response to ambient waves: we can increase T by increasing block mass m and/or decreasing waterline area Awp . 058:0160 Professor Fred Stern Chapter 2 17 Fall 2016 Stability: Immersed Bodies Stable Neutral Unstable Condition for static equilibrium: (1) ∑Fv=0 and (2) ∑M=0 Condition (2) is met only when C and G coincide, otherwise we can have either a righting moment (stable) or a heeling moment (unstable) when the body is heeled. Stability: Floating Bodies For a floating body the situation is 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. 058:0160 Professor Fred Stern Fall 2016 Chapter 2 18 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, then the ship is unstable. 058:0160 Professor Fred Stern Fall 2016 Chapter 2 19 Consider a ship which has taken a small angle of heel α 1. α=small heel angle 2. evaluate the lateral displacement of the center of buoyancy, x = CC 3. then from trigonometry, we can solve for GM and evaluate the stability of the ship: x =centroid; CM=GM+CG Recall that the center of buoyancy is at the centroid of the displaced volume of fluid (moment of volume about yaxis – ship centerplane) x x d x i i This can be evaluated conveniently as follows: moment of before heel (goes to zero due to x = symmetry of original buoyant volume AKKD about centerplane) - moment of AOB + moment of EOD 058:0160 Professor Fred Stern Chapter 2 20 Fall 2016 x x d ( x) d AOB tan y DOE d y dA x tan dA x x x tan dA x tan dA 2 2 OA OD tan ship waterplane area x 2 dA (moment of inertia of ship waterplane = I00, i.e, Izz) x I 00 tan CC ' x I 00 tan CC ' CM tan (from Section View) CM I 00 058:0160 Professor Fred Stern Chapter 2 21 Fall 2016 GM CM CG , i.e. CM=GM+CG GM I 00 CG This equation is used to determine the stability of floating bodies: If GM is positive, the body is stable If GM is negative, the body is unstable 058:0160 Professor Fred Stern Roll: Chapter 2 22 Fall 2016 The rotation of a ship about the longitudinal axis through the center of gravity. Consider symmetrical ship heeled to a very small angle θ. Solve for the subsequent motion due only to hydrostatic and gravitational forces. F b cosˆj sin iˆ g M g r Fb ( g = Δ) M g GCˆj CC iˆ cosˆj sin iˆ GC sin CC cos kˆ GC GM sin kˆ GM sin kˆ Note: recall that M F d , o where d is the perpendicular distance from O to the line of d action of F . M G GZ GM sin O F 058:0160 Professor Fred Stern Chapter 2 23 Fall 2016 CC ' sin CM cos .. M I tan G I = mass moment of inertia about long axis through G = angular acceleration .. .. I GM sin 0 .. GM for small : 0 I GM g GM mgGM I I I k I k I m definition of radius of gyration GM gGM I k mk I 2 2 m 2 The solution to this equation is, . (t ) cos t o n sin t o n n = the initial heel angle where o n = natural frequency gGM k2 gGM k 0 for no initial velocity 058:0160 Professor Fred Stern Chapter 2 24 Fall 2016 Simple (undamped) harmonic oscillation: The period of the motion is T 2 n T 2k gGM Note that large GM decreases the period of roll, which would make for an uncomfortable boat ride (high frequency oscillation). Earlier we found that GM should be positive if a ship is to have transverse stability and, generally speaking, the stability is increased for larger positive GM. However, the present example shows that one encounters a “design tradeoff” since large GM decreases the period of roll, which makes for an uncomfortable ride. 058:0160 Professor Fred Stern Chapter 2 25 Fall 2016 Parametric Roll: The periodicity of the encounter wave causes variations of the metacentric height i.e. GM=GM (t). Therefore: .. I GM (t ) 0 Assuming GM (t ) GM 0 GM1 cos(t ) : .. I GM 0 GM1 cos(t ) 0 .. n2 Cn2 cos(et ) 0 where n gGM 0 k ; C GM1 ; mg ; I mk 2 ; and e encounter wave freq. GM 0 By changing of variables ( et ): .. n2 ( ) 1 C cos ( ) 0 and 2 e This ordinary 2nd order differential equation where the restoring moment varies sinusoidally, is known as the Mathieu equation. This equation gives unbounded solution (i.e. it is unstable) when n2 2n 1 2 n 0,1, 2,3,.. e 2 For the principle parametric roll resonance, n=0 i.e. 2 e 2n 2 2 2 Tn 2Te Te Tn 058:0160 Professor Fred Stern Chapter 2 26 Fall 2016 Case (2) Rigid Body Translation or Rotation In rigid body motion, all particles are in combined translation and/or rotation and there is no relative motion between particles; consequently, there are no strains or strain rates and the viscous term drops out of the N-S equation V 0 . 2 p g a from which we see that p acts in the direction of g a , and lines of constant pressure must be perpendicular to this direction (by definition, f is perpendicular to f = constant). Rigid body of fluid translating or rotating 058:0160 Professor Fred Stern Chapter 2 27 Fall 2016 The general case of rigid body translation/rotation is as shown. If the center of rotation is at O where V V 0 , the velocity of any arbitrary point P is: V V 0 r0 where = the angular velocity vector and the acceleration is: dV0 dV d a r 0 r0 dt dt dt 2 1 3 1 = acceleration of O 2 = centripetal acceleration of P relative to O 3 = linear acceleration of P due to Ω Usually, all these terms are not present. In fact, fluids can rarely move in rigid body motion unless restrained by confining walls. 058:0160 Professor Fred Stern Chapter 2 28 Fall 2016 1.) Uniform Linear Acceleration p=constant p g a Constant g a k a i ^ z ^ x p ax x 1. ax 0 2. ax 0 p increase in +x p decrease in +x p g az z p decrease in +z 1. az 0 2. az 0 and az g p decrease in +z but slower than g 3. az 0 and | az | g p increase in +z 058:0160 Professor Fred Stern Chapter 2 29 Fall 2016 unit vector in the direction of p : s p | p | g az k ax i g a z a x 2 2 1 2 lines of constant pressure are perpendicular to p . n s j ax k g az i a g az 2 2 x 1 2 unit vector in direction of p=constant angle between n and x axes: tan 1 a (g a ) x z In general the pressure variation with depth is greater than in ordinary hydrostatics; that is: 1 dp 2 2 2 p s ax ( g az ) ds G which is > ρg p Gs constant Gs gage pressure 058:0160 Professor Fred Stern Chapter 2 30 Fall 2016 2). Rigid Body Rotation Consider a cylindrical tank of liquid rotating at a constant rate Ω = Ω k : p g a a r 0 ^ r e 2 r p g a gkˆ r2eˆr i.e. p r2 r p g z 058:0160 Professor Fred Stern and p p 2 2 Chapter 2 31 Fall 2016 p f ' g r f ( z) c 2 2 z f ( z ) gz C r gz Constant 2 2 The constant is determined by specifying the pressure at one point; say, p = p0 at (r,z) = (0,0). p p0 gz r 2 2 2 (Note: Pressure is linear in z and parabolic in r) Curves of constant pressure are given by: p p r z a br g 2g 2 0 2 2 which are paraboloids of revolution, concave upward, with their minimum points on the axis of rotation. The position of the free surface is found, as it is for linear acceleration, by conserving the volume of fluid. 058:0160 Professor Fred Stern Chapter 2 32 Fall 2016 The unit vector in the direction of p is: gkˆ r2eˆr sˆ 1/ 2 ( g )2 ( r2 )2 dz tan g 2 r dr z r slope of s θ 2 dr 2 z dz ln r g r g z i.e. r C exp g 2 1 equation of p surfaces s 058:0160 Professor Fred Stern Chapter 2 33 Fall 2016 Case (3) Pressure Distribution in Irrotational Flow; Bernoulli Equation a ( p ) gkˆ 2 V ( p z ) 2 V V V V ( p z ) ( V ) ( V ) t Viscous term=0 for =constant and =0, i.e., Potential flow solutions also solutions NS under such conditions! 1. Assuming inviscid flow: =0 V 1 / 2V V V ( V ) ( p z ) t Euler Equation 2. Assuming incompressible flow: =constant V 2 p V gz V t 2 V 2 V V 058:0160 Professor Fred Stern Fall 2016 Chapter 2 34 3. Assuming steady flow: t 0 B V 2 V p B gz 2 Consider: B perpendicular B= constant V perpendicular V and Therefore, B=constant contains streamlines and vortex lines: 058:0160 Professor Fred Stern Fall 2016 4. Assuming irrotational flow: =0 B 0 B= constant (everywhere same constant) 5. Unsteady irrotational flow V V 2 p gz 0 2 t p gz B(t ) t 2 B(t)= time dependent constant Chapter 2 35 058:0160 Professor Fred Stern Fall 2016 Chapter 2 36 058:0160 Professor Fred Stern Fall 2016 Chapter 2 37 Larger speed/density or smaller R require larger pressure gradient or elevation gradient normal to streamline. 058:0160 Professor Fred Stern Fall 2016 Chapter 2 38 Flow Patterns: Streamlines, Streaklines, Pathlines 1) A streamline is a line everywhere tangent to the velocity vector at a given instant. 2) A pathline is the actual path traveled by a given fluid particle. 3) A streakline is the locus of particles which have earlier passed through a particular point. 058:0160 Professor Fred Stern Chapter 2 39 Fall 2016 Note: 1. For steady flow, all 3 coincide. 2. For unsteady flow, ψ(t) pattern changes with time, whereas pathlines and streaklines are generated as the passage of time. Streamline By definition we must have V dr 0 which upon expansion yields the equation of the streamlines for a given time t t1 dx dy dz ds u v w s= integration parameter So if (u,v,w) known, integrate with respect to s for t=t1 with IC (x0,y0,z0,t1) at s=0 and then eliminate s. Pathline The pathline is defined by integration of the relationship between velocity and displacement. dx u dt dy v dt dz w dt Integrate u,v,w with respect to t using IC ( x0 , y0 , z0 , t0 ) then eliminate t. 058:0160 Professor Fred Stern Chapter 2 40 Fall 2016 Streakline To find the streakline, use the integrated result for the pathline retaining time as a parameter. Now, find the integration constant which causes the pathline to pass through ( x0 , y0 , z0 ) for a sequence of times t . Then eliminate . Example: by: u x 1 t an idealized velocity distribution is given v y 1 2t w0 calculate and plot: 1) the streamlines 2) the pathlines 3) the streaklines which pass through ( x0 , y0 , z0 ) at t=0. 1.) First, note that since w=0 there is no motion in the z direction and the flow is 2-D dx x ds 1 t dy y ds 1 2t s s x C exp( ) y C exp( ) 1 2 1 t 1 2t s 0 at ( x , y ) : C x C y 0 0 1 0 2 0 and eliminating s 058:0160 Professor Fred Stern Chapter 2 41 Fall 2016 x y (1 2t ) ln x y 0 0 x 1 t y y ( )n where n 0 x 1 2t 0 s (1 t ) ln This is the equation of the streamlines which pass through ( x0 , y0 ) for all times t. 2.) To find the pathlines we integrate dx x dt 1 t dy y dt 1 2t 1 x C (1 t ) y C (1 2t ) 2 1 2 t 0 ( x, y ) ( x , y ) : C x 0 0 1 0 C y 2 0 now eliminate t between the equations for (x, y) 1 x y y [1 2( 1)] 2 0 x 0 This is the pathline through ( x , y ) at t=0 and does not coincide with the streamline at t=0. 0 0 058:0160 Professor Fred Stern Chapter 2 42 Fall 2016 3.) To find the streakline, we use the pathline equations to find the family of particles that have passed through the point ( x0 , y0 ) for all times t . x C1 (1 t ) C1 x0 1 y C2 (1 2t ) C2 1 2 y0 (1 2 ) 1 2 x0 1 y 1 (1 2t )( 0 ) 2 1 x 2 y y 1 2t ( )2 x y0 1 2[(1 t )( 0 ) 1] x (1 t ) t 0: y x 1 2 0 1 y0 x 1 2 058:0160 Professor Fred Stern Fall 2016 Chapter 2 43 The Stream Function Powerful tool for 2-D flown in which V is obtained by differentiation of a scalar which automatically satisfies the continuity equation. Continuity boundary conditions (4 required): at infinity : on body : u U v 0 y x u v 0 y x 058:0160 Professor Fred Stern Fall 2016 Irrotational Flow 2 0 2nd order linear Laplace equation on S : U y const. on S : const. B u y x v x y Ψ and φ are orthogonal. d dx dy udx vdy x y d dx dy vdx udy x y dy u 1 i.e. dx const v dy dx const Chapter 2 44 058:0160 Professor Fred Stern Chapter 2 45 Fall 2016 Geometric Interpretation of Besides its importance mathematically important geometric significance. also has = constant = streamline Recall definition of a streamline: V dr 0 dr dxiˆ dyˆj dx dy u v udy vdx 0 compare with d dx dy vdx udy x y i.e. d 0 along a streamline Or =constant along a streamline and curves of constant are the flow streamlines. If we know (x, y) then we can plot = constant curves to show streamlines. 058:0160 Professor Fred Stern Fall 2016 Chapter 2 46 Physical Interpretation dQ V .ndA ˆ dy ˆ dx ˆ (iˆ j ).( i j ) ds 1 y x ds ds y dy x dx d (note that ψ and Q have same dimensions: m3/s) i.e. change in d is volume flux and across streamline 2 2 1 1 Q12 V .ndA d 2 1 Consider flow between two streamlines: dQ 0 . 058:0160 Professor Fred Stern Fall 2016 Chapter 2 47 Incompressible Plane Flow in Polar Coordinates 1 rv r 1 v 0 r r r rv r v 0 or : r continuity : 1 v r r 1 then (r ) ( )0 r r r as before d 0 along a streamline and dQ d volume flux change in stream function say: vr Incompressible axisymmetric flow 1 rv v 0 r r r z z 1 1 say : v v r z r r r z 1 1 1 then : r 0 r r r z z r r as before d 0 along a streamline and dQ d continuity : 058:0160 Professor Fred Stern Chapter 2 48 Fall 2016 Generalization Steady plane compressible flow: ( u ) ( v) 0 x y define : u v compressib le flow stream function y x SL udy vdx 0 1 1 compare with y dy x dx 0 d x dx y dy 1 (d ) 0 i.e. d 0 and constant is a streamline Now: dm (V .n)dA d 2 m 12 (V .n)dA 2 1 1 Change in is equivalent to the mass flux.