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ME3560 – Fluid Mechanics Summer 2016 ME 3560 Fluid Mechanics Chapter II. Fluid Statics 1 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics Examples of cases where forces on submerged surfaces must be calculated. • Fluid power cylinder. • Storage tank. • Aquarium observation windows. • Tank with curved surface. Chapter II. Fluid Statics • Retaining wall. • Fluid reservoir and 2 hatch. Summer 2016 ME3560 – Fluid Mechanics 2.1 Pressure at a Point • The term pressure is used to indicate the normal force per unit area at a given point acting on a given plane within the fluid mass of interest. ΣFy = p yδ xδ z − psδ xδ s sin θ = ρ δ xδ yδ z ay 2 δ xδ yδ z δ xδ yδ z ΣFz = p zδ xδ y − psδ xδ s cosθ − γ =ρ az 2 2 3 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics • Realizing that δ y = δ s cosθ ; δ z = δ s sin θ •Then the previous equations can be expressed as: p y − ps = δy 2 ρa y ; p z − ps = δz 2 ( ρa z + γ ) •Since we are really interested in what is happening at a point, we take the limit as δx, δy, and δz approach zero : p = p ; p =p y s z s p y = p z = ps • The pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present(Pascal's law). σyy p=− σ xx + σ yy + σ zz 3 Chapter II. Fluid Statics σzz σxx σxx σzz σyy 4 Summer 2016 ME3560 – Fluid Mechanics 2.2 Basic Equation for Pressure Field • Consider a small rectangular element of fluid. •There are two types of forces acting on this element: •surface forces due to the pressure •body force equal to the weight of the element. • Let p be the pressure at the center of the element be designated as p, then the surface forces in the y direction is: ∂p δy )δxδz − δF y = ( p − ∂y 2 ∂p δ y (p+ )δxδz ∂y 2 5 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics • Simplifying the previous expression we get: ∂p δF y = − δ xδyδ z ∂y •Following a similar procedure the surface forces for x and z are: ∂p δFx = − δxδyδz ∂x ∂p δFz = − δxδyδz ∂z •The surface forces exerted on the element of fluid can be expressed as: r δFs = δFx iˆ + δFy ˆj + δFz kˆ = r ∂p ˆ ∂p ˆ ∂p ˆ δFs = − ( i + j+ k )δxδyδz ∂x ∂y ∂z r ∂ () ˆ ∂ () ˆ ∂ () ˆ δFs = −∇ pδxδyδz ; ∇ () = i+ j+ k ∂x ∂y ∂z 6 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics • By considering the body forces, we can find that the weight of the element is: ˆ ˆ ˆ − δWk = − ρgδxδyδzk = −γδxδyδzk •Adding up body and surface forces, and applying Newton’s 2nd Law we r get: ˆ − (∇p + ρgk )δxδyδz = ρδxδyδz a •Which after simplification becomes: r ˆ − (∇p + ρgk ) = ρ a •This is the general equation of motion for a fluid in which there are no shearing stresses. 7 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics 2.3 Pressure Variation in a Fluid at Rest r ˆ • For a fluid at rest, the equation: − (∇p + ρgk ) = ρ a •Becomes: ∇p + ρgkˆ = 0 •Which can also be expressed as: ∂p =0 ∂x ∂p =0 ∂y ∂p = − ρg ∂z •These equations show that the pressure does not depend on x or y. •Therefore p depends only on z (vertical direction – depth) dp = − ρg = −γ dz •This is the fundamental equation for fluids at rest and can be used to determine how pressure changes with elevation. •This equation expresses that the pressure decreases as we move upward in a fluid at rest. 8 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics 2.3.1 Incompressible Fluid • A fluid with constant density is called an incompressible fluid. • For liquids the variation in density is usually negligible, even over large vertical distances (ρ and therefore γ are constant): ∫ p2 p1 z2 dp = − ρ g ∫ dz z1 p1 − p2 = ρ g ( z 2 − z1 ) p1 − p2 = ρ gh = γ h • The pressure head can be defined from the previous equation as: p1 − p2 p1 − p2 h= = ρg γ •If p0 is a reference pressure, the pressure at a depth h is p = γh + p 0 9 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 • As demonstrated, the pressure in a homogeneous, incompressible fluid at rest depends on the depth of the fluid relative to some reference plane, and it is not influenced by the size or shape of the tank or container in which the fluid is held. The actual value of the pressure along AB depends only on the depth, h, the surface pressure, p0, and the specific weight, γ, of the liquid in the container. 10 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics 2.3.2 Compressible Fluid • Gases such as air, oxygen, and nitrogen are compressible fluids. ρ gas = f (T , p ) •γgas<< γliq •Since γgas are comparatively small, then ∂p/∂z is small (even over distances of several hundred feet the pressure will remain essentially constant for a gas). •Thus the effect of elevation changes on the pressure in gases can be neglected in tanks, pipes. 11 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics • When the variations in heights are large, on the order of thousands of feet, the variation in γ must be taken into account. p • For an ideal (or perfect) gas the eqn. of state is: ρ = which when RT combined with dp = − ρg = −γ dz dp pg Results in the following relation: =− dz RT •This equation can be integrated to obtain: ∫ p2 p1 dp p2 g z2 dz = ln =− ∫ p p1 R z1 T •Finally, it is necessary to specify the nature of the variation of temperature with elevation. If the temperature has a constant value T0 over the range z2 to z1 (isothermal conditions), then g ( z2 − z1 ) p2 = p1 exp− RT0 12 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 2.5 Measurement of Pressure •The pressure at a point within a fluid mass is designated as either an absolute pressure (pabs) or a gage pressure (pgage). •pabs is measured relative to a perfect vacuum (absolute zero pressure). •pgage is measured relative to the local atmospheric pressure. •Absolute pressures are always positive. •Gage pressures can be positive or negative. •A negative gage pressure is also referred to as a suction or vacuum pressure. 13 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics • Since p=F/A, its units are lb/ft2 (psf), lb/in2 (psi) and N/m2 (Pa) •Pressure can also be expressed as the height of a column of liquid. •The units will refer to the height of the column (in., ft, mm, m), also the liquid in the column must be specified (H2O, Hg). •The measurement patm is usually done with a mercury barometer. •The column of mercury will come to an equilibrium position where its weight plus the force due to the vapor pressure (which develops in the space above the column) balances the force due to the atmospheric pressure. patm = ρgh + pvapor = γh + pvapor Chapter II. Fluid Statics 14 https://www.youtube.com/watch?v=EkDhlzA-lwI Summer 2016 ME3560 – Fluid Mechanics 2.6 Manometry •A standard technique for measuring pressure involves the use of liquid columns in vertical or inclined tubes. •Pressure measuring devices based on this technique are called manometers. •Three common types of manometers are: the piezometer tube, the Utube manometer, and the inclined-tube manometer. 2.6.1 Piezometer Tube •The simplest type of manometer consists of a vertical tube, open at the top, and attached to the container in which the pressure is desired. •Manometers involve columns of fluids at rest, thus p = γh + p0 gives the pressure at any elevation within a homogeneous fluid in terms p0 and h •Applying this equation to the piezometer tube Chapter II. Fluid Statics p A = γ 1h1 15 Summer 2016 ME3560 – Fluid Mechanics 2.6.2 U–Tube Manometer •The U–Tube manometer consists of a tube formed into the shape of a U. •The fluid in the manometer is called the gage fluid. •To find the pressure pA in terms of the various column heights, an analysis is started at one end of the system towards the other end. p A + γ 1h1 − γ 2 h2 = 0 16 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics •The U-tube manometer is also used to measure the difference in pressure between two containers or two points in a given system. •Consider a manometer connected between containers A and B. The difference in pressure between A and B can be found by again starting at one end of the system and working around to the other end. Thus: p A + γ 1h1 − γ 2 h2 − γ 3h3 = pB •Therefore, the pressure difference is: p A − pB = γ 2 h2 + γ 3h3 − γ 1h1 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 2.6.3 Inclined–Tube Manometer •To measure small pressure changes, an inclined–tube manometer is frequently used. p A + γ 1h1 − γ 2l2 sin θ − γ 3h3 = pB p A − pB = γ 2l2 sin θ + γ 3h3 − γ 1h1 18 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics Examples of cases where forces on submerged surfaces must be calculated. • Fluid power cylinder. • Storage tank. • Aquarium observation windows. • Tank with curved surface. Chapter II. Fluid Statics • Retaining wall. • Fluid reservoir and hatch. 19 ME3560 – Fluid Mechanics Summer 2016 2.8 Hydrostatic Force on a Plane Surface •When a surface is submerged in a fluid, forces develop on the surface due to the fluid. •The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures. •For fluids at rest we know that the force must be perpendicular to the surface since there are no shearing stresses present. •For incompressible fluids, the pressure varies linearly with depth. •For horizontal surfaces, magnitude of the resultant force is FR=pA. •If patm acts on both sides of the bottom, the FR on the bottom is only due to the liquid in the tank. •Since the pressure is constant and uniformly distributed over the bottom, FR acts through the centroid of the 20 area. Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •When a submerged plane surface is inclined and assuming that the fluid surface is open to the atmosphere. •Let the plane in which the surface lies intersect the free surface at 0 and make an angle θ with this surface. •The x–y coordinate system is defined so that 0 is the origin and y = 0 (i.e., the x-axis) is directed along the surface as shown. •The objective is to determine the direction, location, and magnitude of the resultant force acting on one side of the submerged area. 21 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics •At any given depth, h, the force acting on dA is dF = h dA and is perpendicular to the surface. •Thus, the magnitude of the resultant force can be found by summing these differential forces over the entire surface. In equation form: FR = ∫ γ hdA A FR = ∫ γ y sin θdA A •Since γ and θ are constant: FR = γ sin θ ∫ ydA A 22 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics •The integral appearing in the previous equation is the first moment of the area with respect to the x axis ∫ A ydA = yc A •yc is the y coordinate of the centroid of area A measured from the x axis. The equation for FR can be written as FR = γ Ayc sin θ = γ hc A •hc is the vertical distance from the fluid surface to the centroid of the area. The magnitude of the force depends only on the γ, A, and hc (it is independent of θ). •The magnitude of the resultant fluid force is equal to the pressure acting at the centroid of the area multiplied by the total area. •The y coordinate, yR, of the resultant force can be determined by summation of moments around the x axis. That is, the moment of the resultant force must equal the moment of the distributed pressure force: FR yR = ∫ ydF = ∫ γ sin θ y dA 2 Chapter II. Fluid Statics A A 23 ME3560 – Fluid Mechanics •Therefore, since FR= γAyc sinθ: yR ∫ = A 2 y dA Summer 2016 yc A •The integral in the numerator is the moment of inertia, Ix, with respect to an axis formed by the intersection of the plane containing the surface and the free surface (x axis). Thus Ix yR = yc A •By using the parallel axis theorem to express Ix: I x = I xc + Ayc •Ixc is the moment of inertia of the area with respect to an axis passing through its centroid and parallel to the x axis: I xc 2 yR = ∫ = •In a similar analysis xR Chapter II. Fluid Statics A xydA yc A = yc A + yc I xy yc A c 24 ME3560 – Fluid Mechanics •In a similar analysis: xR ∫ = A xydA yc A Summer 2016 = I xy yc A c •Ixy is the product of inertia with respect to the x and y axes. Again, using the parallel axis theorem I xyc xR = yc Ac + xc •Ixyc is the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area and formed by a translation of the x–y coordinate system. •If the submerged area is symmetrical with respect to an axis passing through the centroid and parallel to either the x or y axes, the resultant force must lie along the line x =xc. •The point through which FR acts is called the center of pressure. •As yc increases the center of pressure gets closer to the centroid of the area. •Read section 2.9. Pressure Prism 25 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 2.10 Hydrostatic Force on a Curved Surface •To determine the resultant force exerted by a fluid on a curved surface, it is necessary to determine independently the vertical and horizontal components of FR 26 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •The horizontal force, FH, is determined by projecting the curved surface onto the vertical plane (y–z) and calculating the force that the fluid would exert on this imaginary vertical surface. •The point of application of FH corresponds to the location where the force would act on the imaginary (projection) vertical surface. 27 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •The vertical force, FV, is the weight of the fluid above the curved surface up to the free surface. •The line of action of FV passes through the centroid of the volume of fluid above the curved surface. 28 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 2.11 Buoyancy, Flotation and Stability 2.11.1 Archimedes’ Principle •If a stationary body is completely or partially (floating) submerged in a fluid, the force exerted by the fluid on the body called buoyant force. •A net upward vertical force results because pressure increases with depth and the pressure forces acting from below are larger than the pressure forces acting from above. 29 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics •Consider a flat plate of thickness h submerged in a liquid of density ρf parallel to the free surface. •The area of the top and bottom surface of the plate is A and its distance to the free surface is s. •The pressure at the top and bottom surfaces are ρf gs and ρf g(s+h). •The hydrostatic force acting on the top surface is Ftop=ρf gsA, the larger force Fbottom=ρfg(s+h)A acts upward on the bottom surface of the plate. •The difference between these two forces is a net upward force (buoyancy force) FB = Fbottom − Ftop FB = ρ f g ( s + h) A − ρ f gsA FB = ρ f ghA = ρ f gV 30 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •The relation ρf gV is the weight of the liquid whose volume is equal to the volume of the plate. •The previous relation is developed for a simple geometry, but it is valid for any body regardless of its shape. •This can be shown mathematically by a force balance or simply by this argument: •Consider an arbitrary shaped solid body submerged in a fluid at rest and compare it to a body of fluid of the same shape indicated by dotted lines at the same distance from the free surface. •The buoyant forces acting on these two bodies, which depend only on depth, are the same at the boundaries of both. •The imaginary fluid body is in static equilibrium, and thus, the net force and net moment acting on it are zero. 31 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •Thus, the upward buoyant force must be equal to the weight of the imaginary fluid body, whose volume is equal to the volume of the solid body. •Further, the weight and the buoyant force must have the same line of action to have a zero moment. •This is known as 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. 2.11.2 Stability •A body is in a stable equilibrium position if, when displaced, it returns to its equilibrium position. •It is in an unstable equilibrium if, when displaced (even slightly), it goes to a new equilibrium position. 32 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •Stability considerations are particularly important for submerged or floating bodies since the centers of buoyancy and gravity do not necessarily coincide. •A small rotation can result in either a restoring or overturning couple. •For a completely submerged body with CG below the center of buoyancy. •A rotation from its equilibrium position will create a restoring couple formed by the weight, W, and the buoyant force, FB, causing the body to rotate back to its original position. Thus, for this configuration the body is stable. •The body will be stable as long as the center of gravity falls below the center of buoyancy. 33 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •For the completely submerged body shown, which has a center of gravity below the center of buoyancy, a rotation from its equilibrium position will create a restoring couple formed by the weight, W, and the buoyant force, FB, which causes the body to rotate back to its original position. Thus, for this configuration the body is stable. It is to be noted that the body will be stable as long as the center of gravity falls below the center of buoyancy, thus the body is in a stable equilibrium position with respect to small rotations. However, as is illustrated in Fig. 2.26, if the center of gravity of the completely submerged body is above the center of buoyancy, the resulting couple formed by the weight and the buoyant force will cause the body to overturn and move to a new equilibrium position. Thus, a completely submerged body with its center of gravity above its center of buoyancy is in an unstable equilibrium position. 34 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •If the center of gravity of the completely submerged body is above the center of buoyancy, the resulting couple formed by the weight and the buoyant force will cause the body to overturn and move to a new equilibrium position. •A completely submerged body with its center of gravity above its center of buoyancy is in an unstable equilibrium position.. 35 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •For floating bodies the stability problem is more complicated. •As the body rotates the location of the center of buoyancy (which passes through the centroid of the displaced volume) may change. •A floating body such as a barge that rides low in the water can be stable even though the center of gravity lies above the center of buoyancy. •This is true since as the body rotates FB shifts to pass through the centroid of the newly formed displaced volume and combines with W to form a couple which will cause the body to return to its original equilibrium position. 36 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 •However, for a relatively tall, slender body, a small rotational displacement can cause the buoyant force and the weight to form an overturning couple as illustrated. 37 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics 2.12 Pressure Variation in a Fluid with Rigid– Body Motion r ˆ •The general equation of motion: − ∇p − γ k = ρ a •was developed for both fluids at rest and fluids in motion, for situations where there were no shearing stresses present. This equation can be expressed as ∂p ∂p ∂p − ∂x = ρa x − ∂y = ρa y − ∂z = ρa z + γ •Problems involving fluid motion with no shearing stresses appear when a mass of fluid undergoes rigid-body motion. •For example, if a container of fluid accelerates along a straight path, the fluid will move as a rigid mass (after the initial sloshing motion has died out) with each particle having the same acceleration. Since there is no deformation, there will be no shearing stresses. Thus the previous equations apply. 38 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics 2.12.1 Linear Motion •Consider an open container of a liquid that is translating along a straight path with a constant acceleration a (ax = 0), therefore ∂p/∂x = 0. •In the y and z directions: ∂p − = ρa y ∂y ∂p − = ρ (a z + g ) ∂z •The change in pressure between two closely spaced points located at y, z, and y + dy, z + dz can be expressed as ∂p ∂p dp = dy + ∂y dp = − ρa y dy − ρ ( g + a z ) dz •Or in terms of the problem studied: Chapter II. Fluid Statics ∂z dz 39 Summer 2016 ME3560 – Fluid Mechanics •Along a line of constant pressure (for example the free surface), dp = 0, then the slope of this line ay dz is: dy =− g + az •Thus, the free surface of the accelerating mass will be inclined if ay ≠ 0. •In addition, all lines of constant pressure will be parallel to the free surface. •For the case when ay = 0, az ≠ 0 (the mass of fluid accelerating in the vertical direction), the fluid surface will be horizontal with pressure distribution different than hydrostatic, and given by dp dz = − ρ ( g + az ) 40 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics 2.12.2 Rigid–Body Rotation •After an initial “start-up” transient, a fluid contained in a tank that rotates with a constant angular velocity ω about an axis will rotate with the tank as a rigid body. •The acceleration of a fluid particle located at a distance r from the axis of rotation is equal in magnitude to rω2, and the direction of the acceleration is toward the axis of rotation. •The pressure gradient ∇p can be expressed in cylindrical polar coordinates r, θ, and z as: ∂p ∂p 1 ∂p ∇p = eˆr + eˆθ + eˆz ∂r r ∂θ ∂z 2 a = − r ω eˆr ; aθ = 0; a z = 0 •In terms of this coordinate system: r •It follows from r ˆ − ∇p − γ k = ρ a ∂p •That the eqns. are: ∂r Chapter II. Fluid Statics = ρ rω ; 2 ∂p ∂θ = 0; ∂p ∂z = −γ 41 Summer 2016 ME3560 – Fluid Mechanics •Thus, for this type of rigid-body rotation, p = p(r, z), therefore the differential pressure is ∂p ∂p 2 dp = ∂r dr + ∂z dz = ρ rω dr − γdz •On a horizontal plane (dz = 0), then dp/dr = ρ ω2r, (greater than zero). That is: due to the centrifugal acc., p increases in the radial direction. 42 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics •Along a surface of constant pressure, such as the free surface, dp = 0, dp = ρ rω dr − γdz 2 dz rω = dr g 2 •Integration of this result gives the equation for surfaces of constant 2 2 pressure as rω z= 2g + constant •This expression reveals that these surfaces of constant pressure are parabolic. 43 Chapter II. Fluid Statics Summer 2016 ME3560 – Fluid Mechanics •Integration of •Yields dp = ρ rω 2 dr − γdz ∫ dp = ρ ω ∫ rdr − γ ∫ dz p= 2 ρω r 2 2 2 − γ z + constant •The constant of integration can be expressed in terms of a specified pressure at some arbitrary point r0, z0. •This result shows that the pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies hydrostatically in the vertical direction. Chapter II. Fluid Statics 44 Summer 2016 ME3560 – Fluid Mechanics Read Sections: 2.3.2 Compressible Fluid 2.6.3 Inclined–Tube Manometer 2.9. Pressure Prism 45 Chapter II. Fluid Statics ME3560 – Fluid Mechanics Summer 2016 46 Chapter II. Fluid Statics