Subject Name Fluid Mechanics Subject Code 10 ME

Subject Name: Fluid Mechanics Subject Code: 10 ME 36 B Prepared By: R. Punith Department: AE Date: 22 -08 -2014 08/22/2014

Contents • • • Fluid Statics Manometers Forces on plane areas Horizontal surfaces Inclined surfaces Curved surfaces 08/22/2014

FLUID STATICS • The science of fluid statics will be treat in two parts: – the study of pressure and its variation throughout a fluid; – the study of pressure forces on finite surfaces. • Special cases of fluids moving as solids are included in the treatment of statics because of the similarity of forces involved. • Since there is no motion of a fluid layer relative to an adjacent layer, there are no shear stresses in the fluid. • Hence, all free bodies in fluid statics have only normal pressure forces acting on their surfaces. 08/22/2014

PRESSURE AT A POINT • The average pressure is calculated by dividing the normal force pushing against a plane area by the area. • The pressure at a point is the limit of the ratio of normal force to area as the area approaches zero size at the point. • At a point a fluid at rest has the same pressure in all directions. This means that an element δA of very small area, free to rotate about its center when submerged in a fluid at rest, will have a force of constant magnitude acting on either side of it, regardless of its orientation. • To demonstrate this, a small wedge-shaped free body of unit width is taken at the point (x, y) in a fluid at rest (Fig. 1) 08/22/2014

Figure 1. Free-body diagram of wedge-shaped particle 08/22/2014

• There can be no shear forces, the only forces are the normal surface forces and gravity. The equations of motion in the x and y directions px, py, ps are the average pressures on the three faces, γ is the unit gravity force of the fluid, ρ is its density, and ax, ay are the accelerations. • When the limit is taken as the free body is reduced to zero size by allowing the inclined face to approach (x, y) while maintaining the same angle θ, and using the geometric relations • The equations simplify to Last term of the second equation is an infinitestimal of higher of smallness, may be neglected. 08/22/2014

• When divided by δy and δx, respectively, the equations can be combined: • Since θ is any arbitrary angle, this equation proves that the pressure is the same in all directions at a point in a static fluid. 08/22/2014

BASIC EQUATION OF FLUID STATICS Pressure Variation in a Static Fluid • Force balance: • The forces acting on an element of fluid at rest (Fig. 2) consist of surface forces and body forces. • With gravity the only body force acting, and by taking the y axis vertically upward, it is - γ δx δy δz in the y direction. • With pressure p at its center (x, y, z) the approximate force exerted on the side normal to the y axis closest to the origin and on the opposite side are approximately where δy/2 is the distance from center to a face normal to y. 08/22/2014

Figure 2. Rectangular parallelepiped element of fluid at rest 08/22/2014

• Summing the forces acting on the element in the y direction gives • For the x and z directions, since no body forces act, • If the element is reduced to zero size, alter dividing through by δx δy δz = δV, the expression becomes exact. This is the resultant force per unit volume at a point, which must be equated to zero for a fluid at rest. • • The quantity in parentheses is the gradient, called ∇ (del) 08/22/2014

• The negative gradient of , -∇p, is the vector field f of the surface pressure force per unit volume • The fluid static law of variation of pressure is then • For an inviscid fluid in motion, or a fluid so moving that the shear stress is everywhere zero, Newton's second law takes the form where a is the acceleration of the fluid element, f - jγ is the resultant fluid force when gravity is the only body force acting. 08/22/2014

• In component form, it becomes The partials, for variation in horizontal directions, are one form of Pascal's law; they state that two points at the same elevation in the same continuous mass of fluid at rest have the same pressure. • Since p is a function of y only, relates the change of pressure to unit gravity force and change of elevation and holds for both compressible and incompressible fluids. • For fluids that may be considered homogeneous and incompressible, γ is constant, and the above equation, when integrated, becomes in which c is the constant of integration. The hydrostatic law of variation of pressure is frequently written in the form h = -y, p is the increase in pressure from that at the free surface. 08/22/2014

UNITS AND SCALES OF PRESSURE MEASUREMENT • Pressure may be expressed with reference to any arbitrary datum. The usual data are – absolute zero; – local atmospheric pressure. • Absolute pressure - difference between its value and a complete vacuum. • Gauge pressure - difference between its value and the local atmospheric pressure. 08/22/2014

• Figure 3. illustrates the data and the relations of the common units of pressure measurement. • Standard atmospheric pressure is the mean pressure at sea level, 760 mm Hg. • A pressure expressed in terms of the length of a column of liquid is equivalent to the force per unit area at the base of the column. The relation for variation of pressure with altitude in a liquid p = γh shows the relation between head h, in length of a fluid column of unit gravity force γ, and the pressure p (p is in pascals, γ in newtons per cubic metre, and h in metres). • With the unit gravity force of any liquid expressed as its relative density S times the unit gravity force of water: • For water γw may be taken as 9806 N/m 3. 08/22/2014

Figure 3. Units and scales for pressure measurement 08/22/2014

• Local atmospheric pressure is measured by – mercury barometer (Fig. 4) or – aneroid barometer, which measures the difference in pressure between the atmosphere and an evacuated box or tube in a manner analogous to the bourdon gauge except that the tube is evacuated and sealed. • Mercury barometer: glass tube closed at one end, filled with mercury, and inverted so that the open end is submerged in mercury. • It has a scale so arranged that the height of column R can be determined. • The space above the mercury contains mercury vapor. If the pressure of the mercury vapor hv is given in millimetres of mercury and R is measured in the same units, the pressure at A may be expressed as mm Hg 08/22/2014 Figure 4. Mercury barometer

• In Figure 3. a pressure may be located vertically on the chart, which indicates its relation to absolute zero and to local atmospheric pressure. • If the point is below the local-atmospheric-pressure line and is referred to gauge datum, it is called negative, suction, or vacuum. • For example, the pressure 460 mm Hg abs, as at 1, with barometer reading 720 mm, may be expressed as -260 mm Hg, 260 mm Hg suction, or 260 mm Hg vacuum. • It should be noted that 08/22/2014 Pabs = pbar + pgauge

MANOMETERS • Manometers are devices that employ liquid columns for determining differences in pressure. • The most elementary manometer – piezometer (Figure 5 a). – It measures the pressure in a liquid when it is above zero gauge. – Glass tube is mounted vertically so that it is connected to the space within the container. – Liquid rises in the tube until equilibrium is reached. – The pressure is then given by the vertical distance h from the meniscus (liquid surface) to the point where the pressure is to be measured, expressed in units of length of the liquid in the container. – Piezometer would not work for negative gauge pressures, because air would flow into the container through the tube. 08/22/2014

Figure 5. Simple manometers 08/22/2014

• Figure 5 b shows that for small negative or positive gauge pressures in a liquid the tube may take the form. • With this arrangement the meniscus may come to rest below A, as shown. Since the pressure at the meniscus is zero gauge and since pressure decreases with elevation, units of length H 2 O • Figure 5 c - for greater negative or positive gauge pressures a second liquid of greater relative density is employed. • It must be immiscible in the first fluid, which may now be a gas. • If the relative density of the fluid at A is S 1 (based on water) and the relative density of the manometer liquid is S 2, the equation for pressure at A may be written thus, starting at either A or the upper meniscus and proceeding through the manometer, h. A - the unknown pressure, expressed in length units of water, h 1, h 2 - in length units. 08/22/2014

• 1. 2. 3. • A general procedure should be followed in working all manometer problems: Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown. Add to this the change in pressure, in the same unit, from one meniscus to the next (plus if the next meniscus is lower, minus if higher). For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre. Continue until the other end of the gauge (or the starting meniscus) is reached and equate the expression to the pressure at that point, known or unknown. The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer. In equation form, 08/22/2014

• A differential manometer (Fig. 6) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined. • Application of the procedure outlined above to Fig. 6 a produces • Similarly, for Fig. 6 b: • If the pressures at A and B are expressed in length of the water column, the above results can be written, for Fig. 6 a, • Similarly, for Fig 6 b: in which S 1, S 2, and S 3 are the applicable relative densities of the liquids in the system. 08/22/2014

Figure 6. Differential manometers 08/22/2014

Figure 7. Inclined manometer • The inclined manometer (Fig. 7) is frequently used for measuring small differences in gas pressures. • It is adjusted to read zero, by moving the inclined scale, when A and B are open. Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube, it affords greater accuracy in reading the scale. • Surface tension causes a capillary rise in small tubes. If a U tube is used with a meniscus in each leg, the surface-tension effects cancel. 08/22/2014

FORCES ON PLANE AREAS • In the preceding sections variations of pressure throughout a fluid have been considered. • The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force, insofar as external reactions to the force system are concerned. • In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration, by formula, and by use of the concept of the pressure prism. 08/22/2014

Horizontal Surfaces • • • A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure. The magnitude of the force acting on one side of the surface is The elemental forces p d. A acting on A are all parallel and in the same sense; therefore a scalar summation of all such elements yields the magnitude of the resultant force. Its direction is normal to the surface and toward the surface if p is positive. To find the line of action of the resultant, i. e. , the point in the area where the moment of the distributed force about any axis through the point is zero, arbitrary xy axes may be selected, as in Fig. 8. Then, since the moment of the resultant must equal the moment of the distributed force system about any axis, say the y axis, x’ is the distance from the y axis to the resultant. 08/22/2014

Figure 8. Notation for determining the line of action of a force 08/22/2014

Inclined Surfaces • • In Fig. 9 a plane surface is indicated by its trace A'B'; it is inclined θo from the horizontal. The intersection of the plane of the area and the free surface is taken as x axis. The y axis is taken in the plane of the area, with origin O in the free surface. The xy plane portrays the arbitrary inclined area. The magnitude, direction, and line of action of the resultant force due to the liquid, acting on one side of the area, are sought. • For an element with area δA: • Since all such elemental forces are parallel, the integral over the area yields the magnitude of force F, acting on one side of the area, • The magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid. In this form the presence of a free surface is unnecessary. 08/22/2014

Figure 9. Notation force of liquid on one side of a plane inclined area 08/22/2014

Center of Pressure • The line of action of the resultant force has its piercing point in the surface at a point called the pressure center, with coordinates (xp, yp) (Fig. 9). The center of pressure of an inclined surface is not at the centroid. To find the pressure center, the moments of the resultant xp. F, yp. F are equated to the moment of the distributed forces about the y axis and x axis, respectively; thus: Equations may be evaluated conveniently through graphical integration; for simple areas they may be transformed into general formulas: 08/22/2014

• In the parallel-axis theorem for moments of inertia in which IG is the second moment of the area about its horizontal centroidal axis. 08/22/2014

If Ix is eliminated or IG is always positive; hence, yp – yˉ is always positive and the pressure center is always below the centroid of the surface. 08/22/2014

FORCE COMPONENTS ON CURVED SURFACES • When the elemental forces p δA vary in direction, as in the case of a curved surface, they must be added as vector quantities; – their components in three mutually perpendicular directions are added as scalars, and then the three components are added vectorially. • With two horizontal components at right angles and with the vertical component - all easily computed for a curved surface - the resultant can be determined. • The lines of action of the components also are readily determined. 08/22/2014

Horizontal Component of Force on a Curved Surface • The horizontal component pressure force on a curved surface is equal to the pressure force exerted on a projection of the curved surface. The vertical plane of projection is normal to the direction of the component. • The surface of Fig. 10 represents any three-dimensional surface, and δA an element of its area, its normal making the angle θ with the negative x direction. Then • Projecting each element on a plane perpendicular to x is equivalent to projecting the curved surface as a whole onto the vertical plane. • Hence, force acting on this projection of the curved surface is the horizontal component of force exerted on the curved surface in the direction normal to the plane of projection. • To find the horizontal component at right angles to the x direction, the curved surface is projected onto a vertical plane parallel to x and the force on the projection is determined. 08/22/2014

Figure 10. Horizontal component of force on a curved surface Figure 11. Projections of area elements on opposite sides of a body 08/22/2014

• When the horizontal component of pressure force on a closed body is to be found, the projection of the curved surface on a vertical plane is always zero, since on opposite sides of the body the area-element projections have opposite signs (see Fig. 11). • Let a small cylinder of cross section δA with axis parallel to x intersect the closed body at B and C. If the element of area of the body cut by the prism at B is δAB and at C is δAC, then and similarly for all other area elements. • To find the line of action of a horizontal component of force on a curved surface, the resultant of the parallel force system composed of the force components from each area element is required. • This is exactly the resultant of the force on the projected area, since the two force systems have an identical distribution of elemental horizontal force components. 08/22/2014

Vertical Component of Force on a Curved Surface • • • The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface. Can be determined by summing up the vertical components of pressure force on elemental areas δA of the surface. In Fig. 12. an area element is shown with the force p δA acting normal to it. Let θ be the angle the normal to the area element makes with the vertical. Then the vertical component of force acting on the area element is p cos θ δA, and the vertical component of force on the curved surface is given by When p replaced by its equivalent γh, and it is noted that cos θ δA is the projection of δA on a horizontal plane in which δV is the volume of the prism of height h and base cos θ δA, or the volume of liquid vertically above the area element. 08/22/2014

Figure 12. Vertical component of force on a curved surface Figure 13. Liquid with equivalent free surface 08/22/2014

• When the liquid is below the curved surface (Fig. 13. ) and the pressure magnitude is known at some point (e. g. , O), an imaginary or equivalent free surface s-s can be constructed p/γ above O, so that the product of unit gravity force and vertical distance to any point in the tank is the pressure at the point. • The weight of the imaginary volume of liquid vertically above the curved surface is then the vertical component of pressure force on the curved surface. In constructing an imaginary free surface, the imaginary liquid must be of the same unit gravity force as the liquid in contact with the curved surface; otherwise, the pressure distribution over the surface will not be correctly represented. • • • With an imaginary liquid above a surface, the pressure at a point on the curved surface is equal on both sides, but the elemental force components in the vertical direction are opposite in sign. Hence, the direction of the vertical force component is reversed when an imaginary fluid is above the surface. In some cases a confined liquid may be above the curved surface, and an imaginary liquid must be added (or subtracted) to determine the free surface. 08/22/2014

• The line of action of the vertical component is determined by equating moments of the elemental vertical components about a convenient axis with the moment of the resultant force. • With the axis at O (Fig. 13. ), in which is the distance from O to the line of action. • Since Fv = γV the distance to the centroid of the volume. • Therefore, the line of action of the vertical force passes through the centroid of the volume, real or imaginary, that extends above the curved surface up to the real or imaginary free surface. 08/22/2014