058:0160 Chapter 2 Professor Fred Stern Fall 2005 1 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 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. σxzσxyy Face σxxDirection x z058:0160 Chapter 2 Professor Fred Stern Fall 2005 2 ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=zzzyzxyzyyyxxzxyxxijσσσσσσσσσσ 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 i.e. normal stress (pressure) is isotropic. This can be easily seen by considering the equilibrium of a wedge shaped fluid element Or px = py = pz = pn = p nσn = -p, which is compressive, as it should be since fluid cannot withstand tension. (Sign convention for p>0 is direction of n) (one value at a point, independent of direction, p is a scalar) z∑=− 0sinsin:ααdApdApFxnx xnpp = ∑=−+− 0coscos: WdApdApFznzαα znpp =Where ααρsin21cos dlgdAW =l = Od ()pndA α dl pxdAsinα W=ρgV dA=dldy pzdAcosα058:0160 Chapter 2 Professor Fred Stern Fall 2005 3 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). dy The result will be similar for dFy and dFz; consequently, we conclude: ˆˆˆpresspp pdF i j kxy z⎡⎤∂∂∂=− − − ∆∀⎢⎥∂∂ ∂⎣⎦ Or: pf −∇= force per unit volume due to p(x,t). Note: if p=constant, 0=f . dx dz pdydzdydzdxxpp⎟⎠⎞⎜⎝⎛∂∂+−= pdydzdFnetxdydzdxxpp⎟⎠⎞⎜⎝⎛∂∂+ = dxdydzxp∂∂−058:0160 Chapter 2 Professor Fred Stern Fall 2005 4 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 ∀= gdFgravρ or gfgravρ= (per unit volume) Application of Newton’s law yields: ∑=Fam ()viscouspressuresurfacebodysurfacebodyfffggffffadfadk+=−==+==∀=∀∑∑ρρρρ (due to viscous stresses, since in general ijijijpτσσ+−= ) VzVyVxVfpfviscouspressure2222222∇=⎥⎦⎤⎢⎣⎡∂∂+∂∂+∂∂=−∇=µµ For ρ=constant, the viscous force will have this form (chapter 4). ....2viscgravpresmotnVgpa ∇++−∇=µρρ 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 =∇+−=∇µρ058:0160 Chapter 2 Professor Fred Stern Fall 2005 5 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 (0==Va ) 2) Rigid body translation or rotation (02=∇ V ) 3) Irrotational motion (0=×∇V ) P.00tan2eqnBernoulliequationEulerVVtconsif⇒⇒⇒=∇⇒=×∇∫=ρ also, 0.&02=∇⇒=∇=⇒=×∇θρθconstifVV identityvectoraaa2)()(∇−⋅∇∇=×∇×∇ 2.3) Case (1) Hydrostatic Pressure Distribution058:0160 Chapter 2 Professor Fred Stern Fall 2005 6 kggp^ρρ−==∇ z g i.e. 0=∂∂=∂∂ypxp and pgzρ∂=−∂ gdzdpρ−= or ∫−=∫−=−212112)( dzzggdzppρρ2.'ooorggrconst nearearth s surface r⎛⎞=⎜⎟⎝⎠≅ 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 which can be integrated if T =T(z) is )(zTdzRgpdp−=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 Chapter 2 Professor Fred Stern Fall 2005 7 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: ∫−=AdAnpF058:0160 Chapter 2 Professor Fred Stern Fall 2005 8 where for a plane surface n = constant and we need only consider |F| noting that its direction is always towards the surface: AFpndA=∫. 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, ). 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 Chapter 2 Professor Fred Stern Fall 2005 9 2sincpAAyF ydFydAγα==∫∫ IAyOOaboutInertiaofmomentndIdAyoA+=−→=∫222 I = moment of inertia w.r.t horizontal centroidal axis Æ sinFpA yAγα== Æ AyIyycp+= and AyApFαδsin== and similarly for xcp where xAyIxxdFFxxycpAcp+=∫=AyxIIinertiaofproductIxyxyxy+== Note that the coordinate system in the text has its origin at the centroid and is
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