3-1 Pressure 3-2 Fluid Statics 3-3 Buoyancy and Stability 3-4 Rigid-Body Motion Chapter 3 Fluid Statics
3-1 Pressure (1) Pressure is defined as a normal force exerted by a fluid per unit area. Units of pressure are N/m 2, which is called a pascal (Pa). Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 kpa = 10 3 Pa) and megapascal (1 MPa = 10 6 Pa) are commonly used. Other units include bar, atm, kgf/cm 2, lbf/in 2 =psi. 3-1
3-1 Pressure (2) Absolute, gage, and vacuum pressures: Actual pressure at a give point is called the absolute pressure. Most pressure-measuring devices are calibrated to read zero in the atmosphere, and therefore indicate gage pressure, P gage =P abs -P atm. Pressure below atmospheric pressure are called vacuum pressure, P vac =P atm -P abs. 3-2
3-1 Pressure (3) Absolute, gage, and vacuum pressures: 3-3
3-1 Pressure (4) Pressure at a Point Pressure at any point in a fluid is the same in all directions. Pressure has a magnitude, but not a specific direction, and thus it is a scalar quantity. 3-4
3-1 Pressure (5) The Basic Equations of Fluid Statics: Body Force Surface Force Total Force 3-5
3-1 Pressure (6) 3-6
3-1 Pressure (7) Newton s Second Law 3-7
3-1 Pressure (8) Pressure-Height Relation: Pressure Variation in a Static Fluid: 3-8
3-1 Pressure (9) Variation of Pressure with Depth: In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers. To obtain a relation for the variation of pressure with depth, consider rectangular element Force balance in z-direction gives F = ma = 0 z z P2Δx P1Δx ρ gδxδ z = 0 Dividing by Dx and rearranging gives Δ P = P P = ρgδ z = γ Δz 2 1 s 3-9
3-1 Pressure (10) Variation of Pressure with Depth: Pressure in a fluid at rest is independent of the shape of the container. Pressure is the same at all points on a horizontal plane in a given fluid. 3-10
3-1 Pressure (11) Scuba Diving and Hydrostatic Pressure: 3-11
3-1 Pressure (12) Scuba Diving and Hydrostatic Pressure: 1 100 ft Pressure on diver at 100 ft? kg m 1m Pgage,2 = ρgz = 998 9.81 3 2 ( 100 ft) m s 3.28 ft 1atm = 298.5kPa = 2.95atm 101.325kPa P = P + P = 2.95atm+ 1atm= 3.95atm abs,2 gage,2 atm If you hold your breath on ascent, your lung volume would increase by a factor of 4, which would result in embolism and/or death. 2 Danger of emergency ascent? PV = 1 1 2 2 V1 P2 3.95atm = = 4 V P 1atm 2 1 PV Boyle s law 3-12
3-1 Pressure (13) Pascal s Law: P Pressure applied to a confined fluid increases the pressure throughout by the same amount. In picture, pistons are at same height: F F F A = P = = A A F A 1 2 1 2 2 2 1 2 1 1 Ratio A 2 /A 1 is called ideal mechanical advantage 3-13
3-1 Pressure (14) The Manometer: P1 = P2 P = P + ρgh 2 atm An elevation change of Dz in a fluid at rest corresponds to DP/rg. A device based on this is called a manometer. A manometer consists of a U-tube containing one or more fluids such as mercury, water, alcohol, or oil. Heavy fluids such as mercury are used if large pressure differences are anticipated. 3-14
3-1 Pressure (15) Mutlifluid Manometer: For multi-fluid systems Pressure change across a fluid column of height h is DP = rgh. Pressure increases downward, and decreases upward. Two points at the same elevation in a continuous fluid are at the same pressure. Pressure can be determined by adding and subtracting rgh terms. P + ρ gh + ρ gh + ρ gh = P 2 1 1 2 2 3 3 1 3-15
3-1 Pressure (16) Measuring Pressure Drops: Manometers are well--suited to measure pressure drops across valves, pipes, heat exchangers, etc. Relation for pressure drop P 1 -P 2 is obtained by starting at point 1 and adding or subtracting rgh terms until we reach point 2. If fluid in pipe is a gas, r 2 >>r 1 and P 1 -P 2 = rgh 3-16
3-1 Pressure (17) The Barometer: P + ρgh = P P C atm = ρgh atm Atmospheric pressure is measured by a device called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure. P C can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to P atm. Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance. 3-17
3-2 Fluid Statics (1) Fluid Statics deals with problems associated with fluids at rest. In fluid statics, there is no relative motion between adjacent fluid layers. Therefore, there is no shear stress in the fluid trying to deform it. The only stress in fluid statics is normal stress Normal stress is due to pressure Variation of pressure is due only to the weight of the fluid fluid statics is only relevant in presence of gravity fields. Applications: Floating or submerged bodies, water dams and gates, liquid storage tanks, etc. 3-18
3-2 Fluid Statics (2) Hoover Dam: 3-19
3-2 Fluid Statics (3) Hoover Dam: Example of elevation head z converted to velocity head V 2 /2g. We'll discuss this in more detail later (Bernoulli equation). 3-20
3-2 Fluid Statics (4) Hydrostatic Forces on Plane Surfaces: On a plane surface, the hydrostatic forces form a system of parallel forces For many applications, magnitude and location of application, which is called center of pressure, must be determined. Atmospheric pressure P atm can be neglected when it acts on both sides of the surface. 3-21
3-3 Buoyancy and Stability (1) Buoyancy is due to the fluid displaced by a body. F B =r f gv. Archimedes principal : 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. 3-22
3-3 Buoyancy and Stability (2) Buoyancy force F B is equal only to the displaced volume ρ f gv displaced. Three scenarios possible 1. ρ body <ρ fluid : Floating body 2. ρ body =ρ fluid : Neutrally buoyant 3. ρ body >ρ fluid : Sinking body 3-23
3-3 Buoyancy and Stability (3) Buoyancy: 3-24
3-3 Buoyancy and Stability (4) Example: Floating Drydock: Auxiliary Floating Dry Dock Resolute (AFDM-10) partially submerged Submarine undergoing repair work on board the AFDM-10 Using buoyancy, a submarine with a displacement of 6,000 tons can be lifted! 3-25
3-3 Buoyancy and Stability (5) Example: Submarine Buoyancy and Ballast: Normal surface trim SSN 711 nose down after accident which damaged fore ballast tanks 3-26
3-4 Rigid-Body Motion (1) There are special cases where a body of fluid can undergo rigid-body motion: linear acceleration, and rotation of a cylindrical container. In these cases, no shear is developed. Newton's 2nd law of motion can be used to derive an equation of motion for a fluid that acts as a rigid body r r P + ρgk = ρa In Cartesian coordinates: P a, P = ρ = ρa, P = ρ g+ a x y z ( ) x y x 3-27
3-4 Rigid-Body Motion (2) Linear Acceleration: Container is moving on a straight path a 0, a = a = 0 x y z P P P = ρax, = 0, = ρg x y z Total differential of P dp = ρa dx ρgdz x Pressure difference between 2 points ( ) ρ ( ) P P = ρa x x g z z 2 1 x 2 1 2 1 Find the rise by selecting 2 points on free surface P 2 = P 1 ax Δ zs = zs2 zs 1 = ( x2 x1) g 3-28
3-4 Rigid-Body Motion (3) Rotation in a Cylindrical Container: Container is rotating about the z-axis 2 ar rω aθ az =, = = 0 P P P r, 0, r θ z 2 = ρ ω = = Total differential of P 2 dp = ρω r dr ρgdz ρg On an isobar, dp = 0 dzisobar rω ω = = + dr g 2g 2 2 2 zisobar r C1 Equation of the free surface 2 ω 2 2 zs = h0 ( R 2r ) 4g 3-29