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058:0160 Professor Fred Stern Chapter 2 1 Fall 2009 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 (C.V.) bounded by a C.S. 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 C.S. 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 2009 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 Or (one value at a point, independent of direction, p is a scalar) px = py = pz = pn = p i.e. normal stress (pressure) is isotropic. This can be easily seen by considering the equilibrium of a wedge shaped fluid element z F x F : pn dA sin px dA sin 0 pn p x pndA α dl : pn dA cos pz dA cos W 0 pn p z 1 Where W gdAcos dl sin 2 = O dl pxdAsinα α dA=dldy z W=ρgV pzdAcosα 058:0160 Professor Fred Stern Chapter 2 3 Fall 2009 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 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). p dFx pdydz p dx dydz x p = dxdydz x net dy pdydz p p dx dydz x 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 4 Fall 2009 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 dF g or f g (per unit volume) grav grav Application of Newton’s law yields: ma F da f d a f f f per unit d body surface f body g and g gkˆ f body g kˆ f surface f pressure z g f viscous (due to viscous stresses, since in general ij p ij ij ) f pressure Viscous part p 2V 2V 2V f viscous 2 2 2 2 V y z x For ρ, μ=constant, the viscous force will have this form (chapter 4). a p g 2V inertial pressure gradient gravity with a viscous V V V t 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 V B( x, t ) 2 058:0160 Professor Fred Stern Chapter 2 5 Fall 2009 This is simply a first order p.d.e. for p and can be solved readily. For the general case (V and p unknown), one must solve the N-S and continuity equations, which is a formidable task since the N-S equations are a system of 2nd order nonlinear p.d.e.’s. We now consider the following special cases : 1) Hydrostatics ( a V 0 ) 2) Rigid body translation or rotation ( V 0 ) 2 3) Irrotational motion ( V 0 ) ( a) ( a) 2 a vector identity if cons tan t V 0 2 V 0 Euler equation Bernoulli eqn. also, 0 V 0 V & if const. 2 0 058:0160 Professor Fred Stern Chapter 2 6 Fall 2009 Case (1) Hydrostatic Pressure Distribution p g g k p p i.e. 0 x y z g p g z and dp gdz 2 or 2 2 1 1 p p gdz g ( z )dz 2 1 r g go o r const. near earth ' s surface ro liquids ρ = const. (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 2009 Chapter 2 7 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 8 Fall 2009 Hydrostatic forces on plane surfaces The force on a body due to a pressure distribution is: F p n 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 058:0160 Professor Fred Stern Chapter 2 9 Fall 2009 Where p is the pressure at the centroid. 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. ycp F ydF Note: dF y sin dA A sin y 2 dA A y dA I 2nd moment of Inertia about O O 2 o A y A I 2 = moment of inertia w.r.t 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 cp A I x x yA xy cp where I product of inertia xy I I x yA xy xy 058:0160 Professor Fred Stern Chapter 2 10 Fall 2009 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 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 perp. to x direction That is, the horizontal component of force acting on a curved surface is equal to the force acting on a vertical projection of that surface (which includes 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 058:0160 Professor Fred Stern Chapter 2 11 Fall 2009 Where h is the depth to any element area dA of the surface. That is, 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. Example Drum Gate p h R 1 cos h n sin i cos k dA lRd F R 1 cos sin i cos k lRd 0 p n F i Fx lR2 1 cos sin d 0 dA 058:0160 Professor Fred Stern Chapter 2 12 Fall 2009 1 lR 2 cos cos 2 2 lR 2 4 0 R 2 Rl p Same force as that on projection of gate onto vertical plane A Fz lR2 1 cos cos d 0 1 lR sin sin 2 2 4 2 0 R2 lR l 2 2 2 Net weight of water above surface Another approach: 1 F1 2 R 2 L R 2 L / 2 2 1 LR 2 2 / 2 2 1 1 F2 R 2 L F1 F F2 F1 R 2 L 2 2 Hydrostatic Forces in Layered Fluids See textbook 2.7 in page 86. Buoyancy and Stability Archimedes Principle FB FV ( 2 ) FV (1) 058:0160 Professor Fred Stern Chapter 2 13 Fall 2009 = fluid wt. above 2ABC – fluid wt. above 1ADC = wt. of fluid equivalent to 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. Example L y y G ρb h d B b hg mg 0 where Weight of the block W b Lb 0 A wp is displaced water volume by the block and is the specific weight of the liquid. W B b Lbhg w Lbdg d b h w W B Instantaneous displaced water volume: 058:0160 Professor Fred Stern Chapter 2 14 Fall 2009 0 yAwp F V .. m y B W 0 Awp y .. 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 We can increase period T by increasing block mass m and/or decreasing waterline area Awp . 058:0160 Professor Fred Stern Chapter 2 15 Fall 2009 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. 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. 058:0160 Professor Fred Stern Fall 2009 Chapter 2 16 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. Consider a ship which has taken a small angle of heel α 058:0160 Professor Fred Stern Chapter 2 17 Fall 2009 1. evaluate the lateral displacement of the center of buoyancy, CC 2. then from trigonometry, we can solve for GM and evaluate the stability of the ship 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 x x d ( x) d AOB tan y x DOE d y dA x tan dA 058:0160 Professor Fred Stern Chapter 2 18 Fall 2009 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 GM CM CG GM I 00 CG 058:0160 Professor Fred Stern Fall 2009 Chapter 2 19 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 Roll: 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. Fb cos j sin i g ( g = Δ) M rF g b 058:0160 Professor Fred Stern Chapter 2 20 Fall 2009 M g GC j CC ' i cos j sin i GC sin CC 'cos k GC CM sin k GM sin k tan CC ' sin CM cos Note: recall that M F d , o where d is the perpendicular distance from O to the line of action of F . M Gz G .. GM sin M I 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 mk I 2 2 m GM gGM I k 2 d O F 058:0160 Professor Fred Stern Chapter 2 21 Fall 2009 The solution to this equation is, . (t ) cos t o n sin t 0 for no initial velocity o n n = the initial heel angle where o n = natural frequency gGM k2 gGM k 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 22 Fall 2009 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 23 Fall 2009 The general case of rigid body translation/rotation is as shown. If the center of rotation is at O where V V , the velocity of any arbitrary point P is: 0 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 for a long time. 058:0160 Professor Fred Stern Chapter 2 24 Fall 2009 1.) Uniform Linear Acceleration p=constant p g a constant g a k a i ^ z p ax x 1. ax 0 2. ax 0 p increase in +x p decrease in +x ^ x 058:0160 Professor Fred Stern Chapter 2 25 Fall 2009 p g az z 1. az 0 p decrease in +z 2. az 0 and | az | g p decrease in +z but slower than g 3. az 0 and | az | g p increase in +z 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 058:0160 Professor Fred Stern Chapter 2 26 Fall 2009 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 cons. Gs gage pressure 2). Rigid Body Rotation Consider a cylindrical tank of liquid rotating at a constant rate Ω = Ω k : p g a 058:0160 Professor Fred Stern Chapter 2 27 Fall 2009 a r 0 ^ r e 2 r p ( g a ) g k r 2 er p r2 r i.e. and p p 2 2 p g z p f ' g r f ( z) c 2 2 z p gz f (r ) c f ( z ) gz r gz const. 2 2 The constant is determined by specifying the pressure at one point; say, p = p0 at (r,z) = (0,0). 1 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 2 058:0160 Professor Fred Stern Fall 2009 Chapter 2 28 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 29 Fall 2009 The unit vector in the direction of p is: g k r2 er s ( g ) ( r ) 2 See Fig 2.23 on page 99 2 2 dz tan g 2 r dr z i.e. r C exp g 2 1 1 2 z r θ slope of s equation of p surfaces s Case (3) Pressure Distribution in Irrotational Flow (Bernoulli Equation) a ( P) gkˆ ( 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 058:0160 Professor Fred Stern Fall 2009 Chapter 2 30 2. Assuming incompressible flow: =constant V 2 p V gz V t 2 3. Assuming steady flow: ddt=constant 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 2009 4. Assuming irrotational flow: =0 B 0 B= constant (everywhere same constant) 5. Assuming irrotational flow: =0 V V . 2 V 2 p gz 0 t 2 V p gz B(t ) t 2 2 B(t)= time dependent constant Chapter 2 31 058:0160 Professor Fred Stern Fall 2009 Chapter 2 32 058:0160 Professor Fred Stern Fall 2009 Chapter 2 33 058:0160 Professor Fred Stern Fall 2009 Chapter 2 34 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 35 Fall 2009 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 I.C. (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 I.C. ( x , y , z , t )then eliminate t. 0 0 0 0 058:0160 Professor Fred Stern Chapter 2 36 Fall 2009 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 ( x , y , z ) at t=0. 0 0 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 C1 exp( ) y C2 exp( ) 1 t 1 2t s 0 at ( x0 , y0 ) : C1 x0 C 2 y0 and eliminating s 058:0160 Professor Fred Stern s (1 t ) ln y y0 ( x n ) x0 Chapter 2 37 Fall 2009 x y (1 2t ) ln x0 y0 where n 1 t 1 2t 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 x C1 (1 t ) t0 y C2 (1 2t ) 1 2 ( x, y ) ( x0 , y0 ) : C1 x0 C 2 y0 now eliminate t between the equations for (x, y) y y0 [1 2( 1 x 1)] 2 x0 058:0160 Professor Fred Stern Chapter 2 38 Fall 2009 This is the pathline through ( x , y ) at t=0 and does not coincide with the streamline at t=0. 0 0 3.) To find the streakline, we use the pathline equations to find the family of particles that have passed through the point ( x , y ) for all times t . 0 x C1 (1 t ) C1 x0 1 y C2 (1 2t ) C2 0 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 Chapter 2 39 Fall 2009 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. cont.: ux vy 0 say: u y and v x ( y ) ( x ) yx xy 0 x y V y iˆ x ˆj then: by definition! curl V kˆ z kˆ 2 DV ( p z ) 2 V) Dt curl (V. V) 2curl V steady Steady flow (u v )(kˆ 2 ) 2 ( kˆ 2 ) x y curl ( ( 2 ) ( 2 ) 4 x y y x ary conditions (4 required): at infinity : u y U v x 0 on body : u v 0 y x single scalar equation, but 4th order! bound 058:0160 Professor Fred Stern Chapter 2 40 Fall 2009 Irrotational Flow 2 0 2nd order linear Laplace equation on S : U y const. on S B : const. Geometric Interpretation of Besides its importance mathematically important geometric significance. also has = constant = streamline Recall definition of a streamline: V dr 0 dx dy u v udy vdx 0 compare with i.e. d 0 dr dxiˆ dyˆj d x dx y dy vdx udy 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 2009 Chapter 2 41 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 i.e. change in d is volume flux and across streamline dQ 0 . 2 2 1 1 Q12 V .ndA d 2 1 Consider flow between two streamlines: 058:0160 Professor Fred Stern Fall 2009 Chapter 2 42 Incompressible Plane Flow in Polar Coordinates cont.: or 1 1 (rvr ) (v ) 0 r r z ( rvr ) (v ) 0 r 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 058:0160 Professor Fred Stern Chapter 2 43 Fall 2009 Incompressible axisymmetric flow 1 (rvr ) (vz ) 0 r r t 1 1 say: vr vz r z r r 1 1 1 then (r ) ( )0 r r r z z r r as before d 0 along a streamline and dQ d cont.: Generalization Steady plane compressible flow: ( u ) ( v) 0 x y define : u v y x S.L. udy vdx 0 compare with 1 1 dy x dx 0 y d x dx y dy 1 (d ) 0 i.e. d 0 and constant is a streamline 058:0160 Professor Fred Stern Fall 2009 Now: dm (V .n)dA d 2 m 12 (V .n)dA 2 1 1 Change in is equivalent to the mass flux. Chapter 2 44