All that begins... peace be upon you

All that begins... peace be upon you

School of Mechanical Engineering Mechanics of Fluids & Engineering Computing SKMM 2313 Mechanics of Fluids 1 Pressure Distribution in a Fluid «An excerpt (mostly) from White (2011)» Abu Hasan ABDULLAH September 2018 ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 0 / 40

Outline 1 Pressure and Pressure Gradient 2 Pressure Force on a Fluid Element 3 Equilibrium of a Fluid Element 4 Hydrostatic Pressure Distributions 5 Application to Manometry 6 Hydrostatic Forces on Plane Surfaces 7 Hydrostatic Forces in Layered Fluids 8 Buoyancy and Stability 9 Pressure Distribution in Rigid-Body Motion 10 Bibliography ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 1 / 40

Pressure and Pressure Gradient Normal stress on any plane through a fluid element at rest is equal to a unique value called the fluid pressure p, taken positive for compression by common convention. Consider a small wedge of fluid at rest of size x by z by s and depth b into the paper. There is no shear by definition, but we postulate that the pressures p x, p z, and p n may be different on each face. Figure 1: Equilibrium of a small wedge of fluid at rest. Summation of forces must equal zero (no acceleration) in both the x and z directions. Fx = 0 = p xb z p nb s sin θ Fz = 0 = p zb x p nb s cos θ 1 2 γb x z (1) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 2 / 40

Pressure and Pressure Gradient The geometry of the wedge is such that s sin θ = z s cos θ = x (2) and substitution into Eq.(1) and rearrangement give p x = p n p z = p n + 1 γ z (3) 2 These relations illustrate two important principles of the hydrostatic, or shear-free, condition: 1 there is no pressure change in the horizontal direction, and 2 there is a vertical change in pressure proportional to the density, gravity, and depth change. In the limit as the fluid wedge shrinks to a point z 0 and Eqs. (3) become p x = p z = p n = p (4) Since θ is arbitrary, we conclude that the pressure p at a point in a static fluid is independent of orientation. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 3 / 40

Pressure and Pressure Gradient In the case of a moving fluid, the pressure is defined as the average of the three normal stresses σ ii on the element p = 1 (σxx + σyy + σzz) 3 (5) Minus sign on LHS because a compression stress is taken as ve whereas p is +ve. The great majority of viscous flows have negligible viscous normal stresses, so Eq.(5) is rarely needed. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 4 / 40

Pressure Force on a Fluid Element Pressure causes no net force on a fluid element unless it varies with space. Figure 2 shows pressure acting on the two x faces of the element: p = p(x, y, z, t) (6) Net force in the x-direction is given by ( df x = pdydz p + p ) x dx dy dz = p x dx dy dz (7) Figure 2: Net x force on an element due to pressure variation. In like manner the net force df y involves p/ y, and the net force df z concerns p/ z. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 5 / 40

Pressure Force on a Fluid Element Total net-force vector on the element due to pressure is ( df press = i p x j p y k p ) dx dy dz (8) z Denoting f as the net force per unit element volume, Eq.(8) can be rewritten as ( f press = dfpress dx dy dz = i x + j y + k ) p = p (9) z Thus it is not the pressure p but the pressure gradient p (which is a surface force) that is causing a net force which must be balanced by body forces (e.g. gravity, e.m.f.), surface forces (e.g. viscous stresses gradients), or some other effect in the fluid. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 6 / 40

Equilibrium of a Fluid Element Body forces due to electromagnet or gravity could be acting on the entire mass of the element. The gravity force (i.e. weight) of the element is df grav = ρg dx dy dz or f grav = ρg (10) where g is acceleration of gravity; on earth g 9.807 m/s 2. Surface force due to the gradient of the viscous stresses is assumed to be present, given below without derivation, ( ) 2 V f visc = µ x + 2 V 2 y + 2 V = µ 2 V (11) 2 z 2 The total vector resultant of these three forces pressure, gravity, and viscous stress must either keep the element in equilibrium or cause it to move with acceleration a. From Newton s law, we have f = fpress + f grav + f visc = p + ρg + µ 2 V = ρa (12) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 7 / 40

Equilibrium of a Fluid Element In this chapter, V and a are known, we need to solve for p. See next slide for nomenclature of the pressure terminology. Note: In later chapters, V, a and p are all unknown. We rewrite Eq.(12) as p = ρ(g a) + µ 2 V = B(x, y, z, t) (13) where B is a short notation for the vector sum on the RHS. If V and a = dv/dt, time, density and viscosity are known, we can solve Eq.(13) for p(x, y, z, t) by direct integration. We could use Eq.(13) in at least four special cases: 1 Flow at rest or at constant velocity Section 2.3 [White (2002)]. 2 Rigid-body translation and rotation Section 2.9 [White (2002)]. 3 Irrotational motion ( V 0) Section 4.9 [White (2002)]. 4 Arbitrary viscous motion Section 6.4 [White (2002)]. Cases 1 and 2 will be treated in this chapter. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 8 / 40

Equilibrium of a Fluid Element Absolute, Atmospheric, Gauge and Vacuum Pressures Nomenclature Absolute Pressure ( p abs or p) The actual pressure at a given position is called the absolute pressure, and it is measured relative to absolute vacuum (i.e., absolute zero pressure). Gauge Pressure ( p gage) Gage pressure is the pressure relative to the atmospheric pressure. In other words, how much above or below is the pressure with respect to the atmospheric pressure. Vacuum Pressure ( p vac) Pressures below atmospheric pressure are called vacuum pressures and are measured by vacuum gauges that indicate the difference between the atmospheric pressure and the absolute pressure. Atmospheric Pressure ( p atm) The atmospheric pressure is the pressure that an area experiences due to the force exerted by the atmosphere, easily measured using mercury barometer. For engineering calculations p atm used is the pressure at sea level, i.e. p atm = 101350 Pa. These four pressures are related through p gage = p abs p atm p vac = p atm p abs p abs = p atm + p gage Figure 3: Absolute, gauge, and vacuum pressure readings. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 9 / 40

Hydrostatic Pressure Distributions If the fluid is at rest or at constant velocity, a = 0 and f visc = 0. Eq.(12) for the pressure distribution reduces to p = ρg (14) This is hydrostatic distribution and is correct for all fluids at rest, regardless of their viscosity, because the viscous term vanishes. A fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere normal to the local-gravity vector: g = gk (15) where g = 9.807 m/s 2. The maximum pressure increase will be in the direction of gravity i.e. down. For these coordinates Eq.(14) has the components p x = 0 p y = 0 p = ρg = γ (16) z ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 10 / 40

Hydrostatic Pressure Distributions Eq.(16) shows that p is independent of x and y. Hence the partial derivative p/ z can be replaced by the total derivative dp/dz, and the hydrostatic condition reduces to dp = γ = p2 p1 = dz which is the solution to the hydrostatic problem. 2 1 γ dz (17) In Figure 4 points a, b, c, and d are at equal depths in water and therefore have identical pressures. Points A, B, and C are also at equal depths in water and have identical pressures higher than a, b, c, and d. Point D has a different pressure from A, B, and C because it is not connected to them by a water path. Figure 4: Hydrostatic-pressure distribution. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 11 / 40

Hydrostatic Pressure Distributions Hydrostatic Pressure in Liquids Since density is constant in liquid hydrostatic calculations, Eq.(17) integrates to p 2 p 1 = γ(z 2 z 1) or z 2 z 1 = p2 γ p1 γ (18) with the first form used in most problems; γ is the specific weight of the fluid and p/γ is the pressure head of the fluid whose SI unit is m. Exercise: Work through Example 2.1 (White, 2011, p. 71). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 12 / 40

Hydrostatic Pressure Distributions Mercury Barometer Mercury (Hg) barometer, Figure 5, measures atmospheric pressure. It uses Eq.(18). It s a tube filled with Hg and inverted while submerged in a reservoir. Hg has extremely small vapour pressure at room temperatures (0.16 Pa at 20 C); this causes a near vacuum in the closed upper end. Atmospheric pressure p a forces a mercury column to rise a distance h into the tube, the upper mercury surface is at p 1 0. Figure 5: Mercury barometer. With p 1 = 0 at z 1 = h and p 2 = p a at z 2 = 0, Eq.(18) reduces to p a 0 = γ Hg(0 h) or h = pa γ Hg (19) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 13 / 40

Hydrostatic Pressure Distributions Hydrostatic Pressure in Gases Gases are compressible, with ρ nearly proportional to p. Thus ρ must be considered as a variable in Eq.(17) if the integration carries over large p changes. We introduce the perfect-gas law p = ρrt in Eq.(17) dp p = γ = ρg = dz RT g then separate the variables and integrate between points 1 and 2: 2 1 dp p = ln p2 p 1 = g R 2 1 dz T The integral over z requires an assumption about the temperature variation T(z). Case 1: We assume isothermal atmosphere (20) T = T 0 (21) and integration of Eq.(20) lead to [ ] g(z2 z1 p 2 = p 1 exp RT 0 (22) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 14 / 40

Hydrostatic Pressure Distributions Hydrostatic Pressure in Gases Case 2: We assume earth s mean atmospheric temperature drops off linearly with z within the troposphere (0 < z 11000 m) T T 0 Bz (23) Introducing Eq.(23) into Eq.(20) ln p2 p 1 = g R 2 and integrating yields 1 dz T 0 Bz (24) ( p = p a 1 Bz ) g/(rb) ( and ρ = ρ 0 1 Bz ) g/(rb) 1 (25) T 0 T 0 where g/rb = 5.26 (air), ρ 0 = 1.2255 kg/m 3, p 0 = 101350 Pa. By international agreement these standard values are assumed to apply: T 0 = 288.15 K= 15 C and B = 0.00650 K/m. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 15 / 40

Hydrostatic Pressure Distributions Hydrostatic Pressure in Gases The linear approximation from Eq.(18) δp = ρgδz = p 2 p 1 = γ(z 2 z 1) is satisfactory for liquids, which are nearly incompressible. For gases, it is inaccurate unless z is very small. Note: Calculation of p through binomial expansion of Eq.(18) incurs an error of less than 1% if z < 200 m. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 16 / 40

Application to Manometry Column of Multiple Fluids A change in elevation z 2 z 1 of a liquid is equivalent to a change in pressure (p 2 p 1)/γ. Thus a static column of one or more liquids or gases can be used to measure pressure differences between two points. Such a device is called a manometer. If multiple fluids are used, the density in the formula is changed as we move from one fluid to another. Figure 6 illustrates the use of the formula with column of multiple fluids. Figure 6: Pressure changes through a column of multiple fluids. Total change p 5 p 1 is addition of successive changes p 2 p 1, p 3 p 2, p 4 p 3 and p 5 p 4. p 5 p 1 = γ o(z 2 z 1) γ w(z 3 z 2) γ G(z 4 z 3) γ M(z 5 p z) (26) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 17 / 40

Application to Manometry A Simple Manometer Figure 7 shows a simple U-tube open manometer that measures the gauge pressure p A relative to the atmosphere, p a. The chamber fluid ρ 1 is separated from the atmosphere by a second, heavier fluid ρ 2. We first apply the hydrostatic formula, Eq.(18) from A down to z 1. Figure 7: Simple open manometer. Note: Note that we can then go down to the bottom of the U-tube and back up on the right side to z 1, and the pressure will be the same, p = p 1. Instead, we jump across and then up to level z 2: p A + γ 1 z A z 1 γ 2 z 1 z 2 = p 2 = p a (27) Exercise: Work through Example 2.3 (White, 2011, p. 76). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 18 / 40

Application to Manometry Pascal s Law Another physical reason that we can jump across at section 1 is due to the Pascal s law or the principle of transmission of fluid-pressure. Pascal s law Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure. This law is applied in the hydraulic lift, Figure 8, to multiply the force, which for the two pistons implies p 1 = p 2 This allows the lifting of a heavy load with a small force, as in an auto hydraulic lift, but of course there can be no multiplication of work, W, so in an ideal case with no frictional loss: W 1 = W 2 See http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html Figure 8: Hydraulic lift. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 19 / 40

Application to Manometry Multiple-fluid Manometer Figure 9 shows a multiple-fluid manometer for finding the difference in pressure between two chambers A and B. Figure 9: Multiple-fluid manometer. We compute four pressure differences while making three jumps: p A p B = (p A p 1) + (p 1 p 2) + (p 2 p 3) + (p 3 p B) = γ 1(z A z 1) γ 2(z 1 z 2) γ 3(z 2 z 3) γ 4(z 3 z B) (28) Exercise: Work through Example 2.4 (White, 2011, p. 78). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 20 / 40

Hydrostatic Forces on Plane Surfaces Many design of containment structures requires computation of hydrostatic forces on solid surfaces adjacent to the fluid; these forces relate to the weight of fluid bearing on the surface. Figure 10 shows a plane panel of arbitrary shape completely submerged in a liquid making an arbitrary angle θ with the horizontal free surface so that depth varies over the panel surface. If h is depth to elemental area da of the plate, Eq.(18) yields pressure p = p a + γh. To derive formulae involving the plate shape we need to establish an xy coordinate system in the plane of the plate with the origin at its centroid, a dummy coordinate ξ down from the surface in the plane of the plate. Figure 10: Hydrostatic force and centre of pressure on an arbitrary plane. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 21 / 40

Hydrostatic Forces on Plane Surfaces Notice that in Figure 10, depth h = ξ sin θ and slant distance from the surface to the centroid of elemental area da is ξ CG = 1 ξ da A The total hydrostatic force on one side of the plate is given by F = p da = (p a + γh)da = p aa + γ h da (29) Since θ is constant along the plate, Eq.(29) becomes F = p aa + γ sin θ ξ da = p aa + γ sin θ ξ CG A (30) Straight down from the surface to the plate centroid is depth h CG = ξ CG sin θ, thus F = p aa + γh CG A = (p a + γh CG)A = p CGA (31) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 22 / 40

Hydrostatic Forces on Plane Surfaces Eq.(31) can be visualized physically in Figure 11 as the resultant of a linear stress distribution over the plate area. This simulates combined compression and bending of a beam of the same cross section. To balance the bending-moment portion of the stress, the resultant force F acts NOT through the centroid but through the centre of pressure CP toward the high-pressure side. Figure 11: Hydrostatic pressure on a plane surface. To find the coordinates (x CP, y CP), we sum moments of the elemental force pda about the centroid and equate to the moment of the resultant F. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 23 / 40

Hydrostatic Forces on Plane Surfaces To compute y CP, we equate F y CP = yp da = y(p a + γξ sin θ)da = γ sin θ yξda The term p ayda vanishes by definition of centroidal axes and introducing ξ = ξ CG y, we obtain ( ) F y CP = γ sin θ ξ yda y 2 da = γ sin θi xx where again yda = 0 and I xx is the area moment of inertia of the plate area about its centroidal x axis, computed in the plane of the plate. Substituting for F from Eq.(31) results in y CP = γ sin θ Ixx p CGA (32) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 24 / 40

Hydrostatic Forces on Plane Surfaces The determination of x CP is exactly similar: F x CP = xp da = x[p a + γ(ξ CG y) sin θ] da = γ sin θ xy da = γ sin θi xy Substituting for F from Eq.(31) results in x CP = γ sin θ Ixy p CGA (33) For positive I xy, x CP is negative because the dominant pressure force acts in the third, or lower left, quadrant of the panel. If I xy = 0, usually implying symmetry, x CP = 0 and the center of pressure lies directly below the centroid on the y axis. Exercise: Work through Example 2.7 (White, 2011, p. 85). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 25 / 40

Hydrostatic Forces on Plane Surfaces Area Moment of Inertia & Product Moment of Area The area moment of inertia is a property of a two-dimensional plane shape which characterizes its deflection under loading, a.k.a. second moment of area or second moment of inertia. It has dimensions of L 4. Not to be confused with the usual (mass) moment of inertia (which characterizes the angular acceleration undergone by a solids when subjected to a torque and has dimensions of ML 2 ) The area moment of inertia about the x-axis is defined by I x = I xx = y 2 dx dy = y 2 da... while more generally, the product moment of area a.k.a. product of inertia is defined by I xy = x y dx dy = x y da I xy of an area symmetrical about an axis will cancel out each other and becomes zero, and the integral will reduces to zero also. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 26 / 40

Hydrostatic Forces on Plane Surfaces Area Moment of Inertia & Product Moment of Area Figure 12: I xx and I xy for: (a) rectangle, (b) circle, (c) triangle, and (d) semicircle. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 27 / 40

Hydrostatic Forces on Curved Surfaces The resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components. Figure 13 shows a curved surface and a free-body diagram of the column of fluid contained in the vertical projection above it. Figure 13: Hydrostatic force on a curved surface: (a) submerged curved surface; (b) free-body diagram of fluid above the curved surface. Forces F H and F V are exerted by the surface on the fluid column. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 28 / 40

Hydrostatic Forces on Curved Surfaces On the upper part of the column bcde, the horizontal components F 1 exactly balance and are not relevant to the discussion. F 1,vertical side cd = F 1,vertical side be On the lower, irregular portion of fluid abc, summation of horizontal forces shows that force F H due to the curved surface is exactly equal to force F H on the vertical left side of the fluid column, which can be computed using Eq.(9), based on a vertical projection of the area of the curved surface. F H,vertical side ac = F H,curved side ab Summation of vertical forces on the fluid free body is F V = W 1 + W 2 + W air (34) Thus the calculation of F V involves little more than finding centres of mass of a column of fluid. Exercise: Work through Examples 2.8 & 2.9 (White, 2011, pp.87 89). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 29 / 40

Hydrostatic Forces in Layered Fluids If the fluid is layered with different densities, as in Figure 14, a single formula cannot solve the problem because the slope of the linear pressure distribution changes between layers, i.e. (total force on the plate) (pressure at the centroid) (plate area) Different formula applies separately to each layer, and the remedy is to compute and sum the separate layer forces and moments. F = F i = p CG,i A i (35) Figure 14: Hydrostatic forces on a surface immersed in a layered fluid. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 30 / 40

Hydrostatic Forces in Layered Fluids The centroid of the plate portion in each layer can be used to locate the centre of pressure on that portion: y CP,i = ρ ig sin θ i I xx,i p CG A i and x CP,i = ρ ig sin θ i I xy,i p CG A i (36) Exercise: Work through Example 2.10 (White, 2011, p.90). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 31 / 40

Buoyancy and Stability Archimedes Principle Archimedes Laws of Buoyancy: 1 a body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces, 2 a floating body displaces its own weight in the fluid in which it floats, Figure 15: Two different approaches to the buoyant force on an arbitrary immersed body. are shown in Figure 15 the body lies between upper curved surface 1 & lower curved surface 2. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 32 / 40

Buoyancy and Stability Buoyant Force There two ways to calculate buoyant force F B acting on the body. Method 1: Vertically, the body experiences a net upward force F B = F V(2) F V(1) = (fluid weight above 2) (fluid weight above 1) = weight of fluid equivalent to body volume (37) Method 2: We sum the vertical forces on elemental vertical slices through the immersed body, assuming fluid has uniform specific weight: F B = (p 2 p 1) da H = γ (z 2 z 1) da H = (γ)(body volume) (38) body Buoyant force F B passes through the centre of volume of the displaced body, i.e. its centre of mass computed as if it had uniform density. This point through which F B acts is called the centre of buoyancy. Eqs.(38) and (39) are identical results and equivalent to Archimedes law 1. Eq.(38) can be generalized to a layered fluid (LF) by summing the weights of each layer of density ρ i displaced by the immersed body: (F B) LF = ρ i g (displaced volume) i (39) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 33 / 40

Buoyancy and Stability Floating Bodies Floating bodies are a special case a portion of the body is submerged, the remainder poking up out of the free surface see Figure 16. Eq.(38) is modified to apply to the submerged volume: F B = (γ)(displaced volume) = floating-body weight = W (40) F B and W are collinear since there can be no net moments for static equilibrium. Figure 16: Static equilibrium of a floating body. Eq.(40) is the mathematical equivalent of Archimedes law 2. Exercise: Work through Example 2.11 (White, 2011, pp.93 94). ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 34 / 40

Buoyancy and Stability Stability Figure 17 outlines the basic principle of the static stability calculation. (a) The basic floating position is calculated from Eq.(40). The body s centre of mass G and centre of buoyancy B are computed. (b) The body is tilted a small angle θ, and a new waterline is established for the body to float at this angle. The new position B of the center of buoyancy is calculated. A vertical line drawn upward from B intersects the line of symmetry at a point M, called the metacenter, which is independent of θ for small angles & MG is metacentric height. (c) If M is above G (positive MG): a restoring moment is present & the original position is stable. If M is below G (negative MG): body is unstable and will overturn if disturbed. Stability increases with increasing MG. Figure 17: Static equilibrium of a floating body. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 35 / 40

Buoyancy and Stability Stability Metacentric height is a property of the cross section for the given weight, and its value gives an indication of the stability of the body. Its computation can be very involved if a body has varying cross section and draft, such as a ship. A less involved computation requires area moment of inertia of the waterline area about the axis of tilt. Figure 18: A floating body tilted through a small angle θ. In Figure 18, y axis of the body is assumed to be a line of symmetry. Tilting the body a small angle θ then submerges the small wedge Obd and uncovers an equal wedge coa. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 36 / 40

Buoyancy and Stability Stability Related to Waterline Area The new position B of the centre of buoyancy is calculated as the centroid of the submerged portion aobde of the body: xv aobde = x dv + x dv x dv codea Obd coa = 0 + x (L da) x (L da) Obd coa = 0 + x L(x tan θ dx) x L( x tan θ dx) (41) Obd coa = tan θ x 2 da waterline = I O tan θ waterline where I O is the area moment of inertia of the waterline footprint of the body about its tilt axis O. Note: The first integral vanishes because of the symmetry of the original submerged portion codea. The remaining two wedge integrals combine into I O when we notice that Ldx equals an element of waterline area. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 37 / 40

Buoyancy and Stability Stability Related to Waterline Area Thus we determine the desired distance from M to B: or x tan θ = MB = I O = MG + GB V submerged MG = I O V submerged GB (42a) (42b) Exercise: Work through Example 2.12 (White, 2011, p.96). Note: Determining the floating stability of a buoyant body of irregular shape is difficult, even to an expert. Such bodies may have two or more stable positions. A floating iceberg (average density 1025 kg/m 3 ) in seawater (density is 1025 kg/m 3 ) has approximately 900/1025 7 of its volume lies below the waterline. It may 8 overturn after becoming unstable when its shape changes due to melting. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 38 / 40

Pressure Distribution in Rigid-Body Motion In rigid-body motion, all particles are in combined translation and rotation, and there is no relative motion between particles. With no relative motion, there are no strains or strain rates, so that the viscous term in Eq. (12) vanishes, leaving a balance between pressure, gravity, and particle acceleration: p = ρ(g a) (43) Fluids can rarely move in rigid-body motion unless restrained by confining walls for a long time. Water inside a tank in a car that starts a constant acceleration would begin to slosh about, and that sloshing would damp out very slowly until finally the particles of water would be in approximately rigid-body acceleration. ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 39 / 40

Bibliography 1 FRANK M. WHITE (2011): Fluid Mechanics, 7ed, McGraw-Hill (ISBN 0073529346) 2 MERLE C. POTTER, DAVID C. WIGGERT, BASSEM RAMADAN & TOM I-P. SHIH (2012): Mechanics of Fluids, 4ed, Cengage Learning (ISBN-13: 978-0-495-66773-5) 3 BRUCE R. MUNSON, THEODORE H. OKIISHI, WADE W. HUEBSCH & ALRIC P. ROTHMAYER (2013): Fundamentals of Fluid Mechanics, 7ed, John Wiley & Sons (ISBN 9781118116135) 4 YUNUS A. CENGEL & JOHN M. CIMBALA (2014): Fundamentals of Fluid Mechanics, 6ed, McGraw-Hill (ISBN 9780073380322) 5 JOHN F. DOUGLAS, ET AL. (2013): Fluid Mechanics, 5ed, Pearson (ISBN 9780131292932) 6 DONALD F. ELGER, ET AL. (2013): Engineering Fluid Mechanics, 5ed, John Wiley & Sons (ISBN 9781118164297) 7 PHILIP J. PRITCHARD & JOHN C. LEYLEGIAN (2011): Fox and MacDonald s Introduction to Fluid Mechanics, 8ed, John Wiley & Sons (ISBN 9780470547557) ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 40 / 40

... must end... and I end my presentation with two supplications my Lord! increase me in knowledge (TAA-HAA (20):114) O Allah! We ask You for knowledge that is of benefit (IBN MAJAH)