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058:0160 Professor Fred Stern Chapter 2 1 Fall 2005 Chapter 2: Pressure Distribution in a Fluid 2.1) 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 then 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 2005 ⎡σ σ = ⎢⎢σ ⎢⎣σ ij xx yx zx σ σ σ σ σ σ 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 stress is the normal stress, which is referred to as the pressure. σii = -p σn = -p, which is compressive, as it should be since fluid cannot withstand tension. (Sign convention for p>0 is 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 : pn dA sin α − p x dA sin α = 0 ∑F pn = p x pndA dl : − pn dA cosα + pz dA cosα − W = 0 pn = p z 1 Where W = ρgdA cos α dl sin α 2 = O ( dl ) pxdAsinα α dA=dldy z W=ρgV pzdAcosα 058:0160 Professor Fred Stern Chapter 2 3 Fall 2005 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 = -1/3(σxx + σyy + σzz ) = -1/3 σii 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 2005 2.2) 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 or f = ρ g (per unit volume) dF = ρ g∀ grav grav Application of Newton’s law yields: ma = ∑ F ρd∀ a = (∑ f )d∀ ρ a = ∑ f = f body + f f body = ρ g = − ρg f surface = f f pressure f viscous pressure surface k + f viscous (due to viscous stresses, since in general σ ij = − pσ ij + τ ij ) = −∇p ⎡ ∂ 2V ∂ 2V ∂ 2V ⎤ = µ ⎢ 2 + 2 + 2 ⎥ = µ∇ 2 V ∂z ⎦ ∂y ⎣ ∂x For ρ=constant, the viscous force will have this form (chapter 4). ρ a = −∇p + ρ g + µ∇ 2 V with motn. pres. grav. visc. a= ∂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 = ρ g − a + µ∇V = B( x, t ) 2 058:0160 Professor Fred Stern Chapter 2 5 Fall 2005 This is simply a first order p.d.e. for p and can be solved readily. For the general case (v and p unkown), 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 know consider the following special cases : 1) Hydrostatics ( a = V = 0 ) 2) Rigid body translation or rotation ( ∇ V = 0 ) 2 3) Irrotational motion ( ∇ × V = 0 ) ifρ =cons tan t ∇ ×V = 0 ⇒ P ∇ 2V = 0 ⇒ Euler equation ⇒ ∫ ⇒ Bernoulli eqn. also, ∇ × V = 0 ⇒ V = ∇θ & if ρ = const. ⇒ ∇ 2θ = 0 ∇ × (∇ × a ) = ∇(∇ ⋅ a ) − ∇ a vector identity 2 2.3) Case (1) Hydrostatic Pressure Distribution 058:0160 Professor Fred Stern Chapter 2 6 Fall 2005 ^ ∇p = ρ g = − ρg k ∂p ∂p i.e. = =0 ∂x ∂y z ∂p = −ρ g ∂z and g dp = − ρgdz 2 or 2 2 1 1 p − p = − ∫ ρgdz = − g ∫ ρ ( z )dz 2 1 ⎛r ⎞ g = go ⎜ o ⎟ ⎝r⎠ ≅ const. near earth ' s surface ro liquids Æ ρ = const. (for one liquid) p = -ρgz + constant gases Æ ρ = ρ(p,t) which is known from the equation of state: p = ρRT Æ ρ = p/RT g dz dp =− p R T (z ) which can be integrated if T =T(z) is known as it is for the atmosphere. 2.4) Manometry Manometers are devices that use liquid columns for measuring differences in pressure. A general procedure may be followed in working all manometer problems: 058:0160 Professor Fred Stern Chapter 2 7 Fall 2005 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. 2.5) Hydrostatic forces on plane surfaces The force on a body due to a pressure distribution is: F = − ∫ p n dA A 058:0160 Professor Fred Stern Fall 2005 Chapter 2 8 where for a plane surface n = constant and we need only consider |F| noting that its direction is always towards the surface: F = ∫ pn 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 ). To find the line of action of the force which we call the center of pressure (xcp,ycp) we equate the moment of inertia of the resultant force to that of the distributed force about any arbitrary axis. 058:0160 Professor Fred Stern Chapter 2 9 Fall 2005 ycp F = ∫ ydF A = γ sin α ∫ y 2 dA A ∫ y dA = I → 2nd moment of Inertia about O − O 2 o A 2 = y A+ I = moment of inertia w.r.t horizontal centroidal axis Æ F = pA = γ sin α yA I Æ I y = y+ yA F = pA and cp = δ sin α y A and similarly for xcp x F = ∫ xdF cp where I = product of inertia 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 = x−x text and y = −( y − y ) text 2.6) Hydrostatic Forces on Curved Surfaces z x y 058:0160 Professor Fred Stern Chapter 2 10 Fall 2005 In general, Horizontal Components: F = − ∫ p dA F = − ∫ p n dA y A y Ay ^ ^ F = F ⋅ i = − ∫ p n ⋅ i dA dA x x dAx = projection of n A 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. ^ γ∀ Fz = − ∫ pn ⋅ k dA = − ∫ p dAz = γ ∫ h dAz = δ∀ Az Az 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. 058:0160 Professor Fred Stern Fall 2005 Chapter 2 11 2.7) Hydrostatic Forces in Layered Fluids 2.8) Buoyancy and Stability Archimedes Principle FB = FV ( 2 ) − FV (1) = 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. 058:0160 Professor Fred Stern Fall 2005 Chapter 2 12 Stability: Immersed Bodies 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 2005 Chapter 2 13 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 Chapter 2 14 Fall 2005 Consider a ship which has taken a small angle of heel α 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: x∀ = moment of ∀ before heel (goes to zero due to symmetry of original buoyant volume AKKD about centerplane) - moment of ∀ AOB + moment of ∀ EOD 058:0160 Professor Fred Stern Chapter 2 15 Fall 2005 x∀ = − ∫ (− x ) d∀ + ∫ ( x) d∀ AOB tan α = y DOE d∀ = y dA = x tan α dA x x∀ = ∫ x tan α dA + ∫ x tan α dA 2 2 OA OD = tan α ∫ x dA 2 ship waterplane area (moment of inertia of ship waterplane = I00, i.e, Izz) x∀ = I 00 tan α CC ' = x = I 00 tan α ∀ 058:0160 Professor Fred Stern Chapter 2 16 Fall 2005 CC ' = CM tan α (from Section View) CM = I 00 ∀ GM = CM − 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 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 hyrdostatic and gravitational forces. ^ ^ ⎛ ⎞ Fb = ⎜ cos θ j − sin θ i ⎟ ρg∀ ⎝ ⎠ r Fb ( ρg∀ = ∆) M = r×F g b 058:0160 Professor Fred Stern Chapter 2 17 Fall 2005 M g = ⎛⎜ − GC j + CC ' i ⎞⎟ × ∆⎛⎜ cos θ j − sin θ i ⎞⎟ ⎠ ⎠ ⎝ ⎝ ^ ^ ^ ^ = (− GC sin θ + CC ' cos θ )∆ k ^ = (− GC + CM ) sin θ ∆ Note: recall that M = F ⋅ d , = GM sin θ ∆ tan θ = o where d is the perpendicular distance from O to the line of action of F . CC ' sin θ = CM cos θ M = Gz ∆ .. ∑ M = −I θ G = GM sin θ ∆ G I = mass moment of inertia about long axis through G θ = angular acceleration .. .. I θ + ∆GM sin θ = 0 .. ∆GM θ =0 for small θ : θ + I ∆GM ρg∀GM mgGM = + I I I k= k =I 2 I m m definition of radius of gyration mk = I 2 ∆GM gGM = I k 2 O d F 058:0160 Professor Fred Stern Chapter 2 18 Fall 2005 The solution to this equation is, 0 for no initial velocity . θ (t ) = θ cos ω t + o n θ sin ω t ω o n n θ = the initial heel angle where ω o n = natural frequency = gGm k 2 = gGM k Simple (undamped) harmonic oscillation: The period of the motion is T= T= 2π ω n 2πk 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 058:0160 Professor Fred Stern Chapter 2 19 Fall 2005 tradeoff” since large GM decreases the period of roll, which makes for an uncomfortable ride. 2.9) 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 2 equation (µ∇ V = 0) . ∇p = ρ (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 Fall 2005 Chapter 2 20 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 + Ω × r0 where Ω = the angular velocity vector and the acceleration is: dV0 dΩ dV =a= + Ω × (Ω × r 0 ) + × r0 dt dt dt N 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 21 Fall 2005 1.) Uniform Linear Acceleration For uniform rigid body acceleration, a has same magnitude and direction for all particles. Vector sum of g and –a gives direction of pressure gradient ∇p , which is the direction of greatest rate of increase of p. ∇p = ρ (g − a ) = constant [ = − ρ (g + a ) k + a i ^ z ^ x ] Consider for example specific one dimensional cases: dp = − ρ ax , if a is positive, dp is negative; this means x dx dx acceleration occurs with decreasing pressure in x direction, and vice-versa, deceleration occurs with increasing pressure in x direction. 058:0160 Professor Fred Stern Chapter 2 22 Fall 2005 Similarly in z: dp = − ρ ( g − az ) dz Which means pressure increases with depth unless az>g For 2-D cases, unit vector in the direction of ∇p (i.e. direction of greatest rate of increase of p) is given by ^ ∇p s= ^ | ∇p | == ^ (g + a z ) k + a x i [(g + a ) 2 z + ax ] 1 2 2 Lines of constant pressure are perpendicular to ∇p . Unit vector in direction of p = constant is given by ^ ^ ^ a k − (g + a z ) i n = sˆ × ˆj = x 1 (Where ĵ - unit normal to paper) 2 2 2 a x + (g + a z ) ^ angle between n and x axes: [ ] θ = tan −1 a (g + a ) x z In general the pressure variation with depth is greater than in ordinary hydrostatics; that is: 058:0160 Professor Fred Stern Chapter 2 23 Fall 2005 ^ dp = ∇p ⋅ s = ρ [a + ( g + a ) ] ds G 2 2 x 1 2 which is > ρg z p = ρGs + cons. = ρGs gage pressure 2.) Rigid Body Rotation Consider a cylindrical tank of liquid rotating at a constant ^ rate Ω = Ω k : a = Ω × (Ω × r ) 0 ^ = − rΩ e 2 r 058:0160 Professor Fred Stern Chapter 2 24 Fall 2005 ∇p = ρ ( g − a ) ^ ^ = − ρg k + ρrΩ e 2 r ∂p = ρr Ω 2 ∂r i.e. Integrating, we get ρ 2 2 p = 2 and r Ω + f ( z ) + c and ∂p = − ρg ∂z p = − ρgz + f (r ) + c ρ 2 2 Hence f (r ) = 2 r Ω and f ( z ) = − ρgz Thus, p= ρ 2 r Ω − ρgz + 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 − ρ gz + r 2Ω2 2 (Note: Pressure is linear in z and parabolic in r) Curves of constant pressure are given by: 058:0160 Professor Fred Stern Chapter 2 25 Fall 2005 p −p rΩ z= + = a + br ρg 2g 2 2 0 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. The unit vector in the direction of ∇p is: ^ ^ s= See Fig 2.23 on page 85 tan θ = ^ − ρg k + ρrΩ e [( ρg ) 2 r 2 + ( ρrΩ ) 2 dz = −g rΩ dr 2 z ] 1 2 ^ 2 slope of s r θ ^ ⎛ Ω z⎞ i.e. r = C exp⎜ − ⎟ g ⎝ ⎠ 2 1 s equation of ∇p surfaces 2.10) Pressure Distribution in Irrotational Flow If the flow is irrotational, ∇ ×V = 0 ( i.e. ) ∇p = ρ g − a + µ∇ 2 V ω=0 (vorticity) incompressible flow 058:0160 Professor Fred Stern Chapter 2 26 Fall 2005 now use the vector identity to show that the viscous terms dropout automatically for incompressible irrotational flow ∇ 2 Vv = ∇ (N ∇ ⋅ Vv) − ∇ × (∇ × Vv) =0 =0 ρ = const. ω =0 ^ ∇p = − ρg k − ρ a a = −∇ ⎛⎜ p + gz ⎞⎟ ρ ⎝ ⎠ piezometric pressure Or Euler’s equation DV ∂V = + V ⋅ ∇ V = −∇⎛⎜ p + gz ⎞⎟ ⎝ ρ ⎠ dt ∂t Or 1 V ⋅ ∇V = ∇ VN ⋅ V − V × (∇ × V ) 2 V2 =0 vector identity Lastly, for steady flow: ⎛1 ⎞ ∇⎜ V 2 + p + gz ⎟ = 0 ρ ⎝2 ⎠ Or p+ ρ 2 V 2 + ρgz = constant (Bernoulli eq.) 058:0160 Professor Fred Stern Chapter 2 27 Fall 2005 = B For steady rotational flow, it is also possible to derive a Bernoulli equation; however, in this case B = B(ψ) that is, the Bernoulli constant varies from streamline to streamline.