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 xy xz x z Face xx Direction 058 0160 Professor Fred Stern Chapter 2 2 Fall 2005 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 ij 0 i j The only stress is the normal stress which is referred to as the pressure ii p n 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 p x dA sin 0 F pn p x pndA dl pn dA cos pz dA cos W 0 pn p z 1 Where W gdA cos dl sin 2 O dl pxdAsin dA dldy z W gV pzdAcos 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 xx yy zz p By convention p is defined as the average of the normal stresses p 1 3 xx yy zz 1 3 ii 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 p dFpress i 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 g per unit volume dF g grav grav Application of Newton s law yields ma F d a f d a f f body f f body g g f surface f f pressure f viscous pressure surface k f viscous due to viscous stresses since in general ij p ij ij p 2V 2V 2V 2 2 2 2 V z y x For constant the viscous force will have this form chapter 4 a p g 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 g a 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 if cons tan t V 0 P 2V 0 Euler equation Bernoulli eqn also V 0 V if const 2 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 g g k p p i e 0 x y z p g z and g 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 const for one liquid p gz constant gases p t which is known from the equation of state p RT 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 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 sin 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 F pA sin yA I I y y yA F pA and cp sin 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 …
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