Pressure and Fluid Statics

Chapter 3 PRESSURE AND FLUID STATICS 3-1_Pressure Pressure: A normal force exerted by a fluid per unit area 1 Pa  1 N/m 2 Pascal 1 bar  105 Pa  0.1 MPa  100 kPa 1 atm  101,325 Pa  101.325 kPa  1.01325 bars 1 kgf/cm 2  9.807 N/cm 2  9.807  104 N/m 2  9.807  104 Pa  0.9807 bar  0.9679 atm © Ashcroft Inc. Used by permission. Some basic pressure gages. The normal stress (or “pressure”) on the feet of a chubby person is much greater than on the feet of a slim person. Absolute pressure: The actual pressure at a given position. It is measured relative to absolute vacuum (i.e., absolute zero pressure). Gage pressure: The difference between the absolute pressure and the local atmospheric pressure. Most pressure-measuring devices are calibrated to read zero in the atmosphere, and so they indicate gage pressure. Vacuum pressures: Pressures below atmospheric pressure. Pgage  Pabs  Patm Pvac  Patm  Pabs Pressure at a Point Pressure is the compressive force per unit area but it is not a vector. Pressure at any point in a fluid is the same in all directions. Pressure has magnitude but not a specific direction, and thus it is a scalar quantity. Pressure is a scalar quantity, not a vector; the pressure at a point in a fluid is the same in all directions.  Fx  max  0 : P1 yz  P3 yl sin   0 1 P2 yx  P3 yl cos    g x y z  0 2 W  mg   g x y z /2  Fz  maz  0 : z  l sin  x  l cos  P1  P3  0 1 P2  P3   g z  0 2 P1  P2  P3  P Forces acting on a wedge-shaped fluid element in equilibrium. Variation of Pressure with Depth P  P2  P1    g z   s z Pbelow  Pabove   g z  Pabove   s z P  Patm   gh or Pgage   gh The pressure of a fluid at rest increases with depth (as a result of added weight). P  P2  P1     g dz 2 1 When the variation of density with elevation is known Free-body diagram of a rectangular fluid element in equilibrium. In a room filled with a gas, the variation of pressure with height is negligible. Pressure in a liquid at rest increases linearly with distance from the free surface. The pressure is the same at all points on a horizontal plane in a given fluid regardless of geometry, provided that the points are interconnected by the same fluid. Lifting of a large weight by a small force by the application of Pascal’s law.  F1 F2  A1 A2  F2 A2  F1 A1 (Top) © Stockbyte/Getty Images RF Pascal’s law: The pressure P1  P2 applied to a confined fluid increases the pressure throughout by the same amount. The area ratio A2/A1 is called the ideal mechanical advantage of the hydraulic lift. 3-2_Pressure Measurement Devices The Barometer Patm   gh Atmospheric pressure is measured by a device called a barometer. The atmospheric pressure is often referred to as the barometric pressure. A frequently used pressure unit is the standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0°C (Hg = 13,595 kg/m3) under standard gravitational acceleration (g = 9.807 m/s2). The basic barometer The length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer, provided that the tube diameter is large enough to avoid surface tension (capillary) effects. At high altitudes, a car engine generates less power and a person gets less oxygen because of the lower density of air. The Manometer It is commonly used to measure small and moderate pressure differences. A manometer contains one or more fluids such as mercury, water, alcohol, or oil. P2  Patm   gh The basic manometer Photo by John M. Cimbala. Some manometers use a slanted or inclined tube in order to increase the resolution (precision) when reading the fluid height. Such devices are called inclined manometers. A simple U-tube manometer, with high pressure applied to the right side. Dwyer Inclined Manometer Source: https://www.coleparmer.com In stacked-up fluid layers, the pressure change across a fluid layer of density  and height h is gh. Patm  1 gh1  2 gh2  3 gh3  P1 P1  1 g (a  h)  2 gh  1 ga  P2 P1  P2  (2  1 ) gh Measuring the pressure drop across a flow section or a flow device by a differential manometer. Example - Manometer 3–16E A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.40, and the differential height between the two arms of the manometer is 20 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the Page 116manometer arm with the (a) higher and (b) lower fluid level being attached to the tank. Example - Manometer 3–17 The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Fig. P3–17. Determine the gage pressure of air in the tank if h1 = 0.4 m, h2 = 0.6 m, and h3 = 0.8 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. Example - Manometer 3–50 The pressure difference between an oil pipe and water pipe is measured by a doublefluid manometer, as shown in Fig. P3–50. For the given fluid heights and specific gravities, calculate the pressure difference ΔP = PB − PA. Example – Hydraulic Lift 3–56 A hydraulic lift is to be used to lift a 2500 kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determine the diameter of the piston on which the weight is to be placed. (Bottom) Photo by John M. Cimbala. Other Pressure Measurement Devices Bourdon tube: Consists of a hollow metal tube bent like a hook whose end is closed and connected to a dial indicator needle. Pressure transducers: Use various techniques to convert the pressure effect to an electrical effect such as a change in voltage, resistance, or capacitance. Pressure transducers are smaller and faster, and they can be more sensitive, reliable, and precise than their mechanical counterparts. Strain-gage pressure transducers: Work by having a diaphragm deflect between two chambers open to the pressure inputs. Piezoelectric transducers: Also called solid-state pressure transducers, work on the principle that an electric potential is generated in a crystalline substance when it is subjected to mechanical pressure. Various types of Bourdon tubes used to measure pressure. Deadweight tester: Another type of mechanical pressure gage. It is used primarily for calibration and can measure extremely high pressures. A deadweight tester measures pressure directly through application of a weight that provides a force per unit area. It is constructed with an internal chamber filled with a fluid (usually oil), along with a tight-fitting piston, cylinder, and plunger. Weights are applied to the top of the piston, which exerts a force on the oil in the chamber. The total force F acting on the oil at the piston–oil interface is the sum of the weight of the piston plus the applied weights. A deadweight tester is able to measure extremely high pressures (up to 10,000 psi in some applications). 3-3_Introduction to Fluid Statics Fluid statics: Deals with problems associated with fluids at rest. The fluid can be either gaseous or liquid. Hydrostatics: When the fluid is a liquid. Aerostatics: When the fluid is a gas. In fluid statics, there is no relative motion between adjacent fluid layers, and thus there are no shear (tangential) stresses in the fluid trying to deform it. The only stress we deal with in fluid statics is the normal stress, which is the pressure, and the variation of pressure is due only to the weight of the fluid. The topic of fluid statics has significance only in gravity fields. The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on the surfaces using fluid statics. 3-4_Hydrostatic Forces on Submerged Plane Surfaces On a plane surface, the hydrostatic forces form a system of parallel forces, and we often need to determine the magnitude of the force and its point of application, which is called the center of pressure. © Comstock Images/Jupiterimages RF A plate, such as a gate valve in a dam, the wall of a liquid storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface when exposed to a liquid. Hoover Dam. Cross Section of a Dam Source:http://www.virtualmuseum.ca/sgc-cms/expositions-exhibitions/hydro/en/resources/crosssection.php and https://gifer.com/en/BbvH When analyzing hydrostatic forces on submerged surfaces, the atmospheric pressure can be subtracted for simplicity when it acts on both sides of the structure. 3-4_Hydrostatic Forces on Submerged Plane Surfaces The pressure at the centroid of a surface is equivalent to the average pressure on the surface. Hydrostatic force on an inclined plane surface completely submerged in a liquid. P  P0   gh  P0   gy sin  FR   P dA   ( P0   gy sin  ) dA  P0 A   g sin  A A  A ydA FR  ( P0   gyC sin   A  ( P0   ghC  A  PC A  Pavg A 1 yC   y dA A A PC  P0   ghC The resultant force acting on a plane surface is equal to the product of the pressure at the centroid of the surface and the surface area, and its line of action passes through the center of pressure. FR  PC A yP  yC  yP  yC  I xx ,C [ yC  P0 /( g sin  )]A I xx ,C yC A I xx ,O   y 2 dA A hP  yP sin  I xx ,O  I xx ,C  yC2 A second moment of area (area moment of inertia) about the x-axis. The centroid and the centroidal moments of inertia for some common geometries. Pressure acts normal to the surface, and the hydrostatic forces acting on a flat plate of any shape form a volume whose base is the plate area and whose length is the linearly varying pressure. This virtual pressure prism has an interesting physical interpretation: its volume is equal to the magnitude of the resultant hydrostatic force acting on the plate since FR =  PdA, and the line of action of this force passes through the centroid of this homogeneous prism. The projection of the centroid on the plate is the pressure center. Therefore, with the concept of pressure prism, the problem of describing the resultant hydrostatic force on a plane surface reduces to finding the volume and the two coordinates of the centroid of this pressure prism. The hydrostatic forces acting on a plane surface form a pressure prism whose base (left face) is the surface and whose length is the pressure. Special Case: Submerged Rectangular Plate b ab3 /12 yP  s   2 [ s  b /2  P0 /(  g sin  )]ab b b2  s  2 12[ s  b /2  P0 /(  g sin  )] Tilted rectangular plate: FR  PC A  [ P0   g ( s  b /2) sin  ]ab Tilted rectangular plate ( s  0): FR  [ P0   g (b sin  )/2]ab Hydrostatic force acting on the top surface of a submerged tilted rectangular plate. Vertical rectangular plate: Vertical rectangular plate ( s  0): FR  [ P0   g ( s  b /2)]ab FR  ( P0   gb /2)ab Hydrostatic force acting on the top surface of a submerged vertical rectangular plate. Horizontal rectangular plate: FR  ( P0   gh)ab Hydrostatic force acting on the top surface of a submerged horizontal rectangular plate. Example – Submerged Surface 3–68 Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hydrostatic force on each wall and the distance of the line of action of this force from the ground. (b) If the height of the walls of the pool is doubled and the pool is filled, will the hydrostatic force on each wall double or quadruple? Why? Example – Submerged Surface 3–76E The flow of water from a reservoir is controlled by a 5-ftwide L-shaped gate hinged at point A, as shown in Fig. P3–76E. If it is desired that the gate open when the water height is 12 ft, determine the mass of the required weight W. © Corbis RF 3-4_Hydrostatic Forces on Submerged Curved Surfaces In many structures of practical application, the submerged surfaces are not flat, but curved as here at Glen Canyon Dam in Utah and Arizona. 3-4_Hydrostatic Forces on Submerged Curved Surfaces FR  FH2  FV2   FV /FH Determination of the hydrostatic force acting on a submerged curved surface. Horizontal force component on curved surface: Vertical force component on curved surface: FH  Fx FV  Fy  W When a curved surface is above the liquid, the weight of the liquid and the vertical component of the hydrostatic force act in the opposite directions. The hydrostatic force acting on a circular surface always passes through the center of the circle since the pressure forces are normal to the surface and they all pass through the center. In a multilayered fluid of different densities can be determined by considering different parts of surfaces in different fluids as different surfaces, finding the force on each part, and then adding them using vector addition. For a plane surface, it can be expressed as Plane surface in a multilayered fluid : PC , i  P0  i ghC , i The hydrostatic force on a surface submerged in a multilayered fluid can be determined by considering parts of the surface in different fluids as different surfaces. FR   FR , i   PC , i Ai Example – Curved Surface 3–9 A Gravity-Controlled Cylindrical Gate A long solid cylinder of radius 0.8 m hinged at point A is used as an automatic gate, as shown in Fig. 3–41. When the water level reaches 5 m, the gate opens by turning about the hinge at point A. Determine (a) the hydrostatic force acting on the cylinder and its line of action when the gate opens and (b) the weight of the cylinder per m length of the cylinder. Example – Curved Surface 3–71 The water side of the wall of a 70-m-long dam is a quarter circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. 3-6_Buoyancy and Stability Buoyant force: The upward force a fluid exerts on a body immersed in it. The buoyant force is caused by the increase of pressure with depth in a fluid. The buoyant force acting on the plate is equal to the weight of the liquid displaced by the plate. For a fluid with constant density, the buoyant force is independent of the distance of the body from the free surface. It is also independent of the A flat plate of uniform thickness h submerged density of the solid body. in a liquid parallel to the free surface. FB  Fbottom  Ftop  f g ( s  h) A  f gsA  f ghA  f gV Archimedes’ principle: The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume. The buoyant forces acting on a solid body submerged in a fluid and on a fluid body of the same shape at the same depth are identical. The buoyant force FB acts upward through the centroid C of the displaced volume and is equal in magnitude to the weight W of the displaced fluid, but is opposite in direction. For a solid of uniform density, its weight Ws also acts through the centroid, but its magnitude is not necessarily equal to that of the fluid it displaces. (Here Ws > W and thus Ws > FB; this solid body would sink.) For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body: Vsub avg, body FB  W  f gVsub  avg, body gVtotal   Vtotal f A solid body dropped into a fluid will sink, float, or remain at rest at any point in the fluid, depending on its average density relative to the density of the fluid. • For a body floating on the surface of a liquid, the total weight of the body must be less than that of the liquid it displaces. • A portion of the body volume is submerged, while the remaining portion is positioned above the surface of the liquid. Since the system is stationary, the two vertical forces W and FB must still balance. • For a body of known weight W, we see that as the liquid density f increases, a smaller percentage of the body volume is submerged. Photo by Andy Cimbala. Used with permission. W  FB  f gVsubmerged  Vsubmerged  W /f g The density of the water in the Dead Sea is about 24% higher than that of pure water. Therefore, people float much more easily (with more of their bodies above the water) in the Dead Sea than in fresh water or in normal seawater. The altitude of a hot air balloon is controlled by the temperature difference between the air inside and outside the balloon, since warm air is less dense than cold air. When the balloon is neither rising nor falling, the upward buoyant force exactly balances the downward weight. © PhotoLink/Getty Images RF © Corbis RF Stability of Immersed and Floating Bodies For floating bodies such as ships, stability is an important consideration for safety. Stability is easily understood by analyzing a ball on the floor. A floating body possesses vertical stability, while an immersed neutrally buoyant body is neutrally stable since it does not return to its original position after a disturbance. An immersed neutrally buoyant body is (a) stable if the center of gravity G is directly below the center of buoyancy B of the body, (b) neutrally stable if G and B are coincident, and (c) unstable if G is directly above B. When the center of gravity G of an immersed neutrally buoyant body is not vertically aligned with the center of buoyancy B of the body, it is not in an equilibrium state and would rotate to its stable state, even without any disturbance. A ball in a trough between two hills is stable for small disturbances, but unstable for large disturbances. 3–6 ■ BUOYANCY AND STABILITY (8) A floating body is stable if the body is bottom-heavy and thus the center of gravity G is below the centroid B of the body, or if the metacenter M is above point G. However, the body is unstable if point M is below point G. Metacentric height GM: The distance between the center of gravity G and the metacenter M—the intersection point of the lines of action of the buoyant force through the body before and after rotation. The length of the metacentric height GM above G is a measure of the stability: the larger it is, the more stable is the floating body. Example – Buoyancy 3–97 A 200-kg granite rock (ρ = 2700 kg/m3) is dropped into a lake. A man dives in and tries to lift the rock. Determine how much force the man needs to apply to lift it from the bottom of the lake. Do you think he can do it? Example – Buoyancy 3–98 The hull of a boat has a volume of 180 m3, and the total mass of the boat when empty is 8560 kg. Determine how much load this boat can carry without sinking (a) in a lake and (b) in seawater with a specific gravity of 1.03. 3-7_Fluids in Rigid Body Motion Pressure at a given point has the same magnitude in all directions, and thus it is a scalar function. In this section we obtain relations for the variation of pressure in fluids moving like a solid body with or without acceleration in the absence of any shear stresses (i.e., no motion between fluid layers relative to each other).    F  m  a  m   dV   dx dy dz  FS , z  FS , x P dz  P dz  P   P dx dy   P  dx dy   dx dy dz     z 2  z 2  z P P  dx dy dz and  FS , y   dx dy dz x y      FS   FS , x i   FS , y j   FS , z k   P  P  P     i j k  dx dy dz   P dx dy dz  x y z  P  P  P  P  i j k x y z      FB , z   g m k    g dx dy dzk        F  m  a  F   FS   FB   (P   g k ) dx dy dz  Rigid -body motion of fluids:    m   dV   dx dy dz  P   g k    a     P  P  P  i j k   g k    (ax i  a y j  az k ) x y z Accelerating fluids: P P P   a x ,   a y , and    ( g  az ) x y z For fluids at rest or moving on a straight path at constant velocity, all components of acceleration are zero, and the relations reduce to Fluids at rest: P P dP  0,  0, and  g x y dz The pressure remains constant in any horizontal direction (P is independent of x and y) and varies only in the vertical direction as a result of gravity [and thus P = P(z)]. These relations are applicable for both compressible and incompressible fluids. A glass of water at rest is a special case of a fluid in rigid-body motion. If the glass of water were moving at constant velocity in any direction, the hydrostatic equations would still apply. © Imagestate Media (John Foxx)/Imagestate RF Special Case 1: Fluids at Rest Special Case 2: Free Fall of a Fluid Body A freely falling body accelerates under the influence of gravity. When the air resistance is negligible, the acceleration of the body equals the gravitational acceleration, and acceleration in any horizontal direction is zero. Therefore, ax = ay = 0 and az = −g Free - falling fluids: P P P   0 x y z  P  constant In a frame of reference moving with the fluid, it behaves like it is in an environment with zero gravity. Also, the gage pressure in a drop of liquid in free fall is zero throughout. The effect of acceleration on the pressure of a liquid during free fall and upward acceleration. Acceleration on a Straight Path P P   a x ,  0, and x y P  P ( x, z ) P   ( g  az ) z dP  (P /x) dx  (P /z ) dz dP   ax dx  ( g  az ) dz P2  P1   ax ( x2  x1 )  ( g  az )( z2  z1 ) Pressure at a point: P  P0  ax x  ( g  az ) z The system in the figure behaves like a fluid at rest except that g − a replaces g in the hydrostatic equations. Rigid-body motion of a liquid in a linearly accelerating tank. Vertical rise of surface: ax zs  zs 2  zs1   ( x2  x1 ) g  az Surfaces of constant pressure: Slope of isobars: dzisobar ax   constant dx g  az dzisobar ax Slope     tan  dx g  az Lines of constant pressure (which are the projections of the surfaces of constant pressure on the xz-plane) in a linearly accelerating liquid. Also shown is the vertical rise. Example – Acceleration on straight path 3–115 A fish tank that contains 60-cm-high water is moved in the cabin of an elevator. Determine the pressure at the bottom of the tank when the elevator is (a) stationary, (b) moving up with an upward acceleration of 3 m/s2, and (c) moving down with a downward acceleration of 3 m/s2. Example – Acceleration on straight path 3–116E A 15-ft-long, 6-ft-high rectangular tank open to the atmosphere is towed by a truck on a level road. The tank is filled with water to a depth of 5 ft. Determine the maximum acceleration or deceleration allowed if no water is to spill during towing. Rotation in a Cylindrical Container Consider a vertical cylindrical container partially filled with a liquid. The container is now rotated about its axis at a constant angular velocity of . After initial transients, the liquid will move as a rigid body together with the container. There is no deformation, and thus there can be no shear stress, and every fluid particle in the container moves with the same angular velocity. P  r 2 , r P  0,  and P   g z P  P (r , z ) dP  (P /r )dr  (P /z )dz dP  r 2 dr   g dz dzisobar r 2  dr g Surfaces of constant pressure: zisobar  2 g r 2  C1 Rigid-body motion of a liquid in a rotating vertical cylindrical container. zs  2 2g V R r 0 r 2  hc 2 zs r dr  2  R r 0 2 2  2 2     R 2  2 g r  hc  r dr   R  4 g  hc  V   R 2 h0 hc  h0   2 R2 4g Free surface: zs  h0  2 4g ( R 2  2r 2 ) Maximum height difference: zs ,max  zs ( R)  zs (0)  2 2g R2 dP  r 2 dr   g dz P2  P1   2 2 (r22  r12 )   g ( z2  z1 ) Pressure at that point: P  P0   2 2 r 2   gz At a fixed radius, the pressure varies hydrostatically in the vertical direction, as in a fluid at rest. For a fixed vertical distance z, the pressure varies with the square of the radial distance r, increasing from the centerline toward the outer edge. In any horizontal plane, the pressure difference between Surfaces of constant the center and edge of the container of radius R is pressure in a rotating liquid. P   2 R 2 /2 Photo courtesy of Paul Hickson, The University of British Columbia. The 6-meter spinning liquid-mercury mirror of the Large Zenith Telescope located near Vancouver, British Columbia. Example – Rotation in cylinder 3–118E A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the centerline, and the water level drops at the center while it rises at the edges. Determine the angular velocity at which the bottom of the tank will first be exposed. Also determine the maximum water height at this moment. Example – Rotation in cylinder 3–122 A 1.2-m-diameter, 3-m-high sealed vertical cylinder is completely filled with gasoline whose density is 740 kg/m3. The tank is now rotated about its vertical axis at a rate of 70 rpm. Determine (a) the difference between the pressures at the centers of the bottom and top surfaces and (b) the difference between the pressures at the center and the edge of the bottom surface.