058:0160 Chapter 2Professor Fred Stern Fall 2005 1Chapter 2: Pressure Distribution in a Fluid2.1) Pressure and pressuregradient In fluid statics, as well as in fluiddynamics, the forces acting on aportion of fluid (C.V.) bounded bya 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 area- stress).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.σxzσxyyxzσxxFaceDirection058:0160 Chapter 2Professor Fred Stern Fall 2005 2zzzyzxyzyyyxxzxyxxijSince, 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 = -pi.e. normal stress (pressure) is isotropic. This can be easily seen by considering the equilibrium of a wedge shaped fluid elementOr px = py = pz = pn = p n(one value at a point, independent of direction, p is a scalar)zWhere = pndAαdlpxdAsinαW=ρgVpzdAcosαdA=dldyσn = -p, which is compressive, as it should be since fluid cannot withstand tension. (Sign convention for p>0 is direction of n)058:0160 Chapter 2Professor Fred Stern Fall 2005 3Note: For a fluid in motion, the normal stress is different on each face and not equal to p.σxx ≠ σyy ≠ σzz ≠ -pBy convention, p is defined as the average of the normal stresses:p = -1/3(σxx + σyy + σzz ) = -1/3 σiiThe fluid element experiences a force on it as a result ofthe fluid pressure distribution if it varies spatially. Consider the net force in the x direction due to p(x,t).The result will be similar for dFy and dFz; consequently, we conclude:ˆˆ ˆpressp p pdF i j kx y z� �� � �= - - - D"� �� � �� �Or:pf force per unit volume due to p(x,t).Note: if p=constant, 0f.2.2) Equilibrium of a fluid elementdxdzdypdydzdydzdxxpp= viscouspressuresurfacebodysurfacebodyfffggffffadfadk058:0160 Chapter 2Professor Fred Stern Fall 2005 4Consider now a fluid element which is acted upon by bothsurface forces and a body force due to gravity gdFgravorgfgrav (per unit volume)Application of Newton’s law yields: Fam (due to viscous stresses, since in general ijijijp)VzVyVxVfpfviscouspressure2222222 For ρ=constant, the viscous force will have this form (chapter 4).....2viscgravpresmotnVgpa with VVtVa 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. ),(2txBVagp 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 Viscous part058:0160 Chapter 2Professor Fred Stern Fall 2005 5formidable 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 (0Va)2) Rigid body translation or rotation (02 V)3) Irrotational motion (0 V).00tan2eqnBernoulliequationEulerVVtconsifalso,0.&02constifVV identityvectoraaa2)()( 2.3) Case (1) Hydrostatic Pressure Distributionkggp^ z gi.e.0ypxpandpgzr�=-�gdzdp058:0160 Chapter 2Professor Fred Stern Fall 2005 6or212112)( dzzggdzpp2.'ooorg grconst nearearth s surface r� �=� �� �@liquids ρ = const. (for one liquid)p = -ρgz + constantgases ρ = ρ(p,t) which is known from the equation of state: p = ρRT ρ = p/RTwhich can be integrated if T =T(z) is known as it is for the atmosphere.2.4) ManometryManometers 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).)(zTdzRgpdp058:0160 Chapter 2Professor Fred Stern Fall 2005 73.) 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 surfacesThe force on a body due to a pressure distribution is:AdAnpFwhere for a plane surface n = constant and we need only consider |F| noting that its direction is always towards the surface: AF pn dA=�.Consider a plane surface AB entirely submerged in a liquid such that the plane of the surface intersects the free-surface with an angle α. The centroid of the surface is denoted (yx,).058:0160 Chapter 2Professor Fred Stern Fall 2005 8To 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 forceabout any arbitrary axis.2sincpAAy F ydFy dAg a==��IAyOOaboutInertiaofmomentndIdAyoA222I = moment of inertia w.r.t horizontal centroidal axissinF pA yAg a= =AyIyycp andAyApFsinand similarly for xcp058:0160 Chapter 2Professor Fred Stern Fall 2005 9whereNote that the coordinate system in the text has its origin at the centroid and is related to the one just used by: yyyandxxxtexttext2.6) Hydrostatic Forces on Curved SurfacesIn general, Horizontal Components:xxAyydAdAinpiFFdApFy^^ dAx = projection of n A onto a plane perp. to x directionThat is, the horizontal component of force acting on a
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