058:0160 Chapter 2Professor Fred Stern Fall 2006 1Chapter 2: Pressure Distribution in a FluidPressure and pressure gradient 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: bodyforces 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 a cubical fluid element, we see that the stressin a moving fluid comprises a 2nd order tensor.σxzσxyyxzσxxFaceDirection058:0160 Chapter 2Professor Fred Stern Fall 2006 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 non-zero stress is the normal stress, which is referred to as 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 nz(one value at a point, independent of direction, p is a scalar)Where = pndAαdlpxdAsinαW=ρgVpzdAcosαdA=dldyασn = -p, which is compressive, as it should be since fluid cannot withstand tension. (Sign convention based on the fact that p>0 and in thedirection of –n)058:0160 Chapter 2Professor Fred Stern Fall 2006 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 ( )1 13 3xx yy zz iip s s s s=- + + =-The 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.dxdzdypdydzdydzdxxpp= viscouspressuresurfacebodysurfacebodyfffggffffadfadk058:0160 Chapter 2Professor Fred Stern Fall 2006 4Equilibrium of a fluid elementConsider 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 ij ij ijps d t=- +)VzVyVxVfpfviscouspressure2222222 For ρ, μ=constant, the viscous force will have this form (chapter 4).2a p g Vr r m=- � + + � 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 Viscous partinertial pressure gradientgravity viscous058:0160 Chapter 2Professor Fred Stern Fall 2006 5This 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 (0Va)2) Rigid body translation or rotation (02 V)3) Irrotational motion (0 V).00tan2eqnBernoulliequationEulerVVtconsifalso,20 & . 0V V if constj r jѴ= �=�=��= identityvectoraaa2)()( Case (1) Hydrostatic Pressure Distribution058:0160 Chapter 2Professor Fred Stern Fall 2006 6$p g g kr r� = =- z gi.e.0ypxpandpgzr�=-�gdzdpor212112)( 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.ManometryManometers are devices that use liquid columns for measuring differences in pressure. A general procedure may be followed in working all manometer problems:)(zTdzRgpdp058:0160 Chapter 2Professor Fred Stern Fall 2006 71.) 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.Hydrostatic forces on plane surfacesThe force on a body due to a pressure distribution is:AdAnpF058:0160 Chapter 2Professor Fred Stern Fall 2006 8where for a plane surface n = constant and we need only consider |F| noting that its direction is always towards the surface: | |AF p 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,). sinF yA pAg a= =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 inertia of the resultant force to that of the distributed forceabout any arbitrary axis.058:0160 Chapter 2Professor Fred Stern Fall 2006 92sincpAAy F ydFy dAg a==��IAyOOaboutInertiaofmomentndIdAyoA222I = moment of inertia w.r.t horizontal centroidal axissinF pA yAg a= = ()2sin sincpy yA y A Ig a g a= +AyIyycp and similarly for xcpwhereNote that the coordinate system in the text has its origin at the centroid and is related to the one just used by: yyyandxxxtexttextHydrostatic Forces on Curved SurfacesxAyIxxdFFxxycpAcpAyxIIinertiaofproductIxyxyxy058:0160 Chapter 2Professor Fred Stern Fall 2006 10In general, Horizontal
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