VIEWS: 600 PAGES: 13 CATEGORY: Other POSTED ON: 3/29/2009 Public Domain
Chapter 2/ Hydrostatic pressure Fluid mechanics Chapter 2: Hydrostatic Pressure 2.1 Pascal’s Triangle. Let’s look at the pressure forces acting on a small triangle of fluid at rest. The fluid element is of uniform thickness, with sides of length a, b and c. z Pc c b gravity Pb g a Pa x the pressure acts normal to the surfaces of the fluid volume assume gravity acts in the z-direction; later we’ll see that it doesn’t matter which direction gravity acts in. Force balance in the z-direction, per unit depth into the page: paa = pcc cos q + rg ab/2 but a = c cos q, so pa = pc + rg b/2 Force balance in the x-direction, per unit depth into the page: pbb = pcc sin q but b = c sin q, so pb = pc Now let a, b and c0 so that Pascal’s Triangle becomes a point in the fluid. The contribution due to gravity becomes negligible, so pa = pb= pc for all angles . 2.2 The hydrostatic equation Let’s look at a small volume of static fluid again: the only forces that can act are due to I) gravity and ii) pressure acting on the faces of the volume. There are no gravity components acting in the x and y directions. Therefore 1 Chapter 2/ Hydrostatic pressure Fluid mechanics The pressure acting on faces ABCD and EFGH must be equal. Similarly, the pressure acting on faces AEHD and BFHC also must be equal p+dp B F G= (0,0,g) A E C dz dy C G D H dx p Force balance in the z-direction (vertical) Pdxdy – (p+dp)dxdy = rg dx dy dz So dp/dz = -rg Hydrostatic equation Integrating p-p0 = Where the reference pressure at height z0 is equal to p0 Example 2.2.1 2 Chapter 2/ Hydrostatic pressure Fluid mechanics Open to atmosphere Po 17 ft = h1 P1 gasoline P2 water 3 ft = h2 0 water Find P1 and P2 in lb/ft2, lb/in2 and as a pressure head in feet of water given SG = 0.68 for gasoline. Solution P1 = gasoline h1 + Po = SG H2O h1 + Po = (0.68) (62.4 lb/ft3) (17 ft) + Po = 721 + Po (lb/ft2) absolute pressure rearrange P1 – P0 = 721 lb/ft2 (gauge pressure) = 721 /144 (lb/ft2/(in2/ft2) = 5.01 lb/in2 (gauge) = 5.01 psig or P1 Po 721 lb / ft 2 11.6 ft H O 2 62.4 lb / ft3 For P2 = H2O h2 + P1 = (62.4 lb/ft3) (3 ft) + 721 lb/ft2 + Po = 908 lb/ft2 + Po 3 Chapter 2/ Hydrostatic pressure Fluid mechanics P1-Po = 908 lb/ft 2 (gauge) = 6.31 lb/in2 P1 Po 908 lb / ft 2 14.6 ft H O2 62.4 lb / ft3 2.3 Gauge and Absolute Pressure Pressure 1 Gauge pressure at 1 local atm pressure ref. 2 Gauge pressure At 2 (suction or vacuum) Absolute absolute Pressure at 1 pressure at 2 Absolute zero pressure Gauge pressure is measured relatively to the local atmospheric pressure. Standard atmospheric pressure is 760 mm Hg Example 2.3.1- Measurement of atmospheric pressure -using mercury barometer P vapour A h Patm B mercury Patm = h + P vapour 4 Chapter 2/ Hydrostatic pressure Fluid mechanics For mercury P vapour = 0.000023 psia at 68 F Patm = h h mercury = Patm / mercury For Patm = 14.7 psia at 68 F (20 C) mercury = 847 lb/ft3 hmercury = 14.7 (lb/in2) (144 in2/ft2) / 847 (lb/ft3) = 2.5 ft water = 62.4 lb/ft3 h water = 14.7*144 /62.4 = 33.9 ft 2.4 Transmission of fluid pressure F1 = PA1 F2=PA2 F2 = (A2/A1) F1 A small force applied at the smaller piston can be used to develop a large force at the larger piston. 2.5 Compressible fluid dp g dz From ideal gas law 5 Chapter 2/ Hydrostatic pressure Fluid mechanics P RT dp g dz RT dp g P2 Z2 dz P R P T Z1 1 For isothermal condition, T=To ln P2 g (Z 2 Z1 ) P1 RT0 g ( Z 2 Z1 ) P2 P1 exp RTo 2.6 Manometry - manometers eg barometers for atmospheric pressure - other examples are piezometer tube, U-tube manometer, inclined tube manometer - are used as vertical or inclined liquid columns to measure pressure 2.6.1 Piezometer tube Open , Po 1, A H1 Pressure that we want to measure PA - 1h1 = Po PA = 1h1 + Po PA = 1h1 (in gauge) 6 Chapter 2/ Hydrostatic pressure Fluid mechanics 2.6.2 U-tube manometer open 1, Ax 1 H2 H1 2 3 2 ( gauge fluid) P2 = P 3 PA + 1h1 =2h2 + Po PA = Po + 2h2 -1h1 For A = gas 1h1 0 PA = 2h2 + Po PA = 2h2 (gauge) Example 2.6.1 Pressure gauge air open h1 oil h3 h3 1 2 Hg Sg oil = 0.9 7 Chapter 2/ Hydrostatic pressure Fluid mechanics Sg mercury =13.6 H1 = 36 in H2 = 6 in H3 = 9 in Find reading in the pressure gauge? Pair = Po P1 = P2 Pair + oilh1 + oilh2 = Po + hgh3 Pair = Hglh3 - oil(h1 +h2 ) = (SG hg) (h20 )h3 – (SG oil) (h20 ) (h1 + h2) = (13.6) (62.4 lb/ft3) (9/12 ft) – (0.9) (62.4 lb/ft3) ((36+6)/12) ft = 440 lb/ft2 = 440/144 = 3.06 psig 2.6.3 Differential U-tube manometer (see MYO p 56 Fig 2.11) Two common gauge fluids are mercury and water. Both give a well-defined meniscus and have well known properties. To avoid capillary effects, use relatively large bore tubes (diameter> 0.5 “). Also the gauge fluid must be immiscible w.r.t. the measured fluids. 2.6.4 Inclined tube manometer - see MYO, p58 - for measuring small change in pressure Example 2.6.2 1 4 5 h2 1 2 3 h1 flow A B 8 Chapter 2/ Hydrostatic pressure Fluid mechanics Fluid in the columns of the manometer are at rest. Find and explain expression to calculate (PA – PB ) Solution PA PB PA – 1 h1 - 2 h2 + 1 (h 1+ h2 ) = PB PA – PB = h2 (2 - 1) 2.7 Hydrostatic Force on Surfaces (Noel de Neves, p39) Static fluid can exet only pressure forces on surfaces adjacent to them. Some pressure is the normal (perpendicular) force per unti area, the pressure forces must act normal to the surfaces.For an infinitesimal surface area, dA, the force exerted, dF; dF = pdA 2.7.1 This dF is a vector quantity. It has both direction (perpendicular to the surface) and magnitude. For a plane surface, all the differential dF vector point in the same direction F PdA 2.7.2 If the pressure over an entire surface is constant , Equation 2.7.2 becomes F = pA 2.7.3 Equation above applies to hotrizontal plane surfaces exposed to static fluids. Patm F, A Patm For vertical plane surfaces, the pressure is not constant over the whole surface. (Equation 2.7.3) must be used to find the force. 9 Chapter 2/ Hydrostatic pressure Fluid mechanics Example 2.7 Patm Lock gate h Pressure of Patm Water on front of gate Fwater F air Lock gate is 20 m wide and 10 m high What is the net force on the lock gate? Solution Fnet= Net Force= Force exerted by the water on the front of the gate – force exerted by the atmosphere on the back of the gate Fnet= Fwater- Fair Fwater= pdA ( P atm h)dA = P atm A + hWdh 10 h2 = Patm A + W 2 0 Where W – width of the gate And h – depth below the free surfaces Fnet = Fwater-Fair 10 h2 = P atmA + W – Patm A 2 0 and = g = 1000 kg/m3 x 9.8 m/s2 Fnet = 1000 (kg/m3) x 9.8 (m/s2) x 20 x (100-0)/2 = 9.81 MN 10 Chapter 2/ Hydrostatic pressure Fluid mechanics For more complicated shapes, the integration in equation 2.7.2 is more difficult. We could simplify the problem using the idea of a centroid. Write equation 2.7.2 for a constant density fluid. F= PdA = g hdA h = g dA A = gA hc 2.7.4 where hdA hc = A 2.7.5 hc is the centroid of the depth measured from the free surface. The centroids of many geometries are tabulated in texts on mechanics and strength of materials. 2.8 Buoyancy Force (or upthrust) - upward vertical force due to the fluid - archimedes principles Recall that pressure force always acts perpendicular to a body and it is a function of depth. For any solid body completely submerged in a fluid or floating, the pressure acting around the surface of a body give a net upward vertical force known as buoyancy force. Increasing P buoyancy force Consider an arbitrary shape solid body having a volume immersed in a fluid. 11 Chapter 2/ Hydrostatic pressure Fluid mechanics F1=gz Volume, Surface, h dA C, F2= g(z+h) Centrod of a sphere FB volume For any sphere, we can divide the body up into a number of vertical prisms of area dA. The net force acting on the body is FB = g z h gz dA = gh dA = gV The buoyancy force is equal to the weight of fluid displaced. The line of action of the buoyancy force may be found by taking moments about a point . _ FB x = gh xdA _ g x = g xdV _ dV x = x V Centroid of the displaced volume of fluid or the centre of buoyancy 12 Chapter 2/ Hydrostatic pressure Fluid mechanics Therefore the buoyancy force acts through the centre of buoyancy. Note: Floating body more complicated 2.9 Stability (p 76-78, MYO) FB W C Cg Cg C W FB stable unstable 13