57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 1 Chapter 2: Pressure and Fluid Statics Pressure For a static fluid, the only stress is the normal stress since by definition a fluid subjected to a shear stress must deform and undergo motion. Normal stresses are referred to as pressure p. For the general case, the stress on a fluid element or at a point is a tensor For a static fluid, ij= 0 ij shear stresses = 0 ii= p = xx= yy= zz i = j normal stresses =-p Also shows that p is isotropic, one value at a point which is independent of direction, a scalar. *Tensor: A mathematical object analogus to but more general than a vector, represented by an array of components that are functions of the coordinates of a space (Oxford) ij = stress tensor* = xx xy xz yx yy yz zx zy zz i = face j = direction57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 2 x z Definition of Pressure: A0F dFpA dAlim N/m2 = Pa (Pascal) F = normal force acting over A As already noted, p is a scalar, which can be easily demonstrated by considering the equilibrium of forces on a wedge-shaped fluid element Geometry A =  y x =  cos z =  sin Fx = 0 pnA sin - pxA sin = 0 pn = px Fz = 0 -pnA cos + pzA cos - W = 0 y)sin)(cos(2W   0sincos2coscos2 yypypzn W = mg = Vg = V V = ½ xzy57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 3    p pn z20sin p p forn z  0 i.e., pn = px = py = pz p is single valued at a point and independent of direction. A body/surface in contact with a static fluid experiences a force due to p BSpdAnpF Note: if p = constant, Fp = 0 for a closed body. Scalar form of Green's Theorem: sfnds fd   f = constant f = 057:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 4 Pressure Transmission Pascal's law: in a closed system, a pressure change produced at one point in the system is transmitted throughout the entire system. Absolute Pressure, Gage Pressure, and Vacuum For pA>pa, pg = pA – pa = gage pressure For pA<pa, pvac = -pg = pa – pA = vacuum pressure pA < pa pg < 0 pg > 0 pA > pa pa = atmospheric pressure = 101.325 kPa pA = 0 = absolute zero57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 5 Pressure Variation with Elevation Basic Differential Equation For a static fluid, pressure varies only with elevation within the fluid. This can be shown by consideration of equilibrium of forces on a fluid element Newton's law (momentum principle) applied to a static fluid F = ma = 0 for a static fluid i.e., Fx = Fy = Fz = 0 Fz = 0 pdxdy ppzdz dxdy gdxdydz   ( ) 0  pzg    Basic equation for pressure variation with elevation 1st order Taylor series estimate for pressure variation over dz57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 6 0yp0dxdz)dyypp(pdxdz0Fy 0xp0dydz)dxxpp(pdydz0Fx For a static fluid, the pressure only varies with elevation z and is constant in horizontal xy planes. The basic equation for pressure variation with elevation can be integrated depending on whether  = constant or = (z), i.e., whether the fluid is incompressible (liquid or low-speed gas) or compressible (high-speed gas) since g  constant Pressure Variation for a Uniform-Density Fluid  pzg    pz    2 1 2 1p p z z    Alternate forms: 1 1 2 2p z p z      pz    p z 0 0 i.e., pz   = constant for liquid constant constant piezometric pressure gage constant piezometric head pz increase linearly with depth decrease linearly with height Z pz  g57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 7 7.06 27.7   1 1 2 22 1 1 21 atm2 oil3 2 water 2 3p z constan tp z p zp p z zp p 0p z .8 9810 .9 7.06kPap p z z7060 9810 2.127.7kPa                    57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 8 Pressure Variation for Compressible Fluids: Basic equation for pressure variation with elevation ( , )dpp z gdz        Pressure variation equation can be integrated for (p,z) known. For example, here we solve for the pressure in the atmosphere assuming (p,T) given from ideal gas law, T(z) known, and g  g(z). p = RT R = gas constant = 287 J/kg K p,T in absolute scale RTpgdzdp )z(TdzRgpdp  which can be integrated for T(z) known dry air57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 9 zo = earth surface = 0 po = 101.3 kPa T = 15C  = 6.5 K/km Pressure Variation in the Troposphere T = To  (z – zo) linear decrease To = T(zo) where p = po(zo) known  = lapse rate = 6.5 K/km )]zz(T[dzRgpdpoo dz'dz)zz(T'zoo constant)]zz(Tln[Rgplnoo use reference condition constantTlnRgplnoo solve for constant RgooooooooT)zz(TppT)zz(TlnRgppln i.e., p decreases for increasing z57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 10 Pressure Variation in the Stratosphere T = Ts = 55C dppgRdzTs  constantzRTgplns use reference condition to find constant ]RT/g)zz(exp[ppeppsooRT/g)zz(os0 i.e., p decreases exponentially for increasing z.57:020 Fluid Mechanics Chapter 2 Professor Fred Stern Fall 2013 11 Pressure Measurements Pressure is an important variable in fluid mechanics and many instruments have been devised for its measurement. Many devices are based on hydrostatics such as barometers and manometers, i.e., determine pressure through measurement of a column (or columns) of a liquid using the pressure