058 0160 Professor Fred Stern Chapter 2 1 Fall 2009 Chapter 2 Pressure Distribution in a Fluid 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 a cubical fluid element we see that the stress in a moving fluid comprises a 2nd order tensor y xy x z xz Face xx Direction 058 0160 Professor Fred Stern Chapter 2 2 Fall 2009 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 non zero stress is the normal stress which is referred to as pressure ii p n p which is compressive as it should be since fluid cannot withstand tension Sign convention based on the fact that p 0 and in the 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 pndA dl pxdAsin dA dldy Where W gV pzdAcos 058 0160 Professor Fred Stern Chapter 2 3 Fall 2009 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 1 s xx s yy s zz s ii 3 3 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 pdydz dz p p dx dydz x dx The result will be similar for dFy and dFz consequently we conclude p p p dFpress ijk D x y z Or f p Note if p constant force per unit volume due to p x t f 0 058 0160 Professor Fred Stern Chapter 2 4 Fall 2009 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 dF g or f g per unit volume grav grav Application of Newton s law yields d a f d a f f body f surface f body g and g gk f surface f pressure f ma F per unit d f body g k z g viscous due to viscous stresses since in general s ij pdij t ij f f pressure viscous Viscous part p 2V 2V 2V 2 2 V 2 2 x y z For constant the viscous force will have this form chapter 4 V r a p r g m 2V inertial pressure gradient gravity with a t V V viscous 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 2 B x t This is simply a first order p d e for p and can be solved readily For the general case V and p unknown one 058 0160 Professor Fred Stern Chapter 2 5 Fall 2009 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 a V 0 2 Rigid body translation or rotation V 0 2 3 Irrotational motion V 0 a a 2 a vector identity if cons tan t V 0 2 V 0 Euler equation Bernoulli eqn also 0 V 0 V j if r const Case 1 Hydrostatic Pressure Distribution 2 j 0 058 0160 Professor Fred Stern Chapter 2 6 Fall 2009 p r g r g k p r g z z g i e p p 0 x y or r g g o o 2 2 r p2 p1 gdz g z dz 1 1 const near earth s surface ro and dp gdz 2 liquids const for one liquid p gz constant gases p t which is known from the equation of state p RT p RT dp g dz p R T z Manometry which can be integrated if T T z is known as it is for the atmosphere 058 0160 Professor Fred Stern Fall 2009 Chapter 2 7 Manometers 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 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 surfaces 058 0160 Professor Fred Stern Chapter 2 8 Fall 2009 The force on a body due to a pressure distribution is F p n dA A where for a plane surface n constant and we need only consider F noting that its direction is always towards the p dA surface F 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 F g sin a yA pA Where p is the pressure at the centroid 058 0160 Professor Fred Stern Chapter 2 9 Fall 2009 To find the line of action of the force which we call the center of pressure xcp ycp we equate the moment of the resultant force to that of the distributed force about any arbitrary axis ycp F ydF A Note dF g sin a y dA 2 y sin 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 g sin a yA I 2 ycpg sin a yA g sin a y A I y y cp I yA and similarly for xcp x F xdF cp where A I xy product of inertia I I I x x yA xy xy x yA xy cp Note that the coordinate system in the text has its origin at the centroid and is related to the …
View Full Document
Unlocking...