# CHARACTERISTIC OF FLUIDS. A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude.

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1 CHARACTERISTIC OF FLUIDS A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude. In a fluid at rest, normal stress is called pressure. 1

2 Dimensions, Dimensional homogeneity and Units Fluid has qualitative and quantitative characteristic. Qualitative : To identify the nature of fluid such as length, time, stress and velocity. Quantitative : Numerical measure of the characteristic. Quantitative requires both a number and a standard. Such standards are called unit. 2

3 Primary quantity : L : Length T : Time M : Mass θ: Temperature Secondary quantity : L 2 : Area LT -1 : Velocity ML -3 : Density 3

4 All theoretically derived equations are dimensionally homogeneous. The dimension of the left side of the equation must be the same as those on the right side, and all additive separate terms must have the same dimensions. Example : V LT = V o = + at LT + LT

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6 UNIT 3 major systems that are commonly used in engineering. 1. British Gravitational (BG) System Length foot (ft) Time second (s) Force pound (lb) Temperature Fahrenheit (ºF) 2. International System (SI) Length meter (m) Time second (s) Mass kilogram (kg) Temperature Kelvin (K) The relation of Kelvin and Celsius is; K = C

7 3. English Engineering (EE) System Length foot (ft) Time second (s) Mass pound mass (lbm) Force pound (lb or lbf) Temperature Rankine (ºR) 7

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9 Density MEASURES OF FLUID MASS AND WEIGHT Designated by the Greek symbol ρ (rho). Defined as its mass per unit volume. ρ = mass = volume kg 3 m Specific volume, is the volume per unit mass. This property is not commonly used in fluid mechanics but is used widely in thermodynamics. volume υ = mass = 1 ρ 9

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12 Specific weight Designated by the Greek symbol γ (gamma). Defined as its weight per unit volume. weight mg kg g γ = = = = ρg 3 volume volume m Specific gravity Designated as SG. Defined as the ratio of the density of the fluid to the density of water at some specified temperature. Usually the specified temperature is taken as 4ºC. SG = ρ H 2 ρ C 12

13 Ideal gas law Gases are highly compressible in comparison to liquids, with changes in gas density directly related to changes in pressure and temperature through the equation ; P = ρrt P : pressure ρ : density R : gas constant T : temperature 13

14 The pressure in the ideal gas law must be expressed as an absolute pressure (abs), which means that it is measured relative to absolute zero pressure (a pressure that would only occur in a perfect vacuum) Standard sea-level atmospheric pressure is 14.7 psi and 101 kpa, respectively. 14

15 VISCOSITY The property of viscosity is described the fluidity of the fluid. To resist the applied force, P, a shearing stress, τ, would be developed at the plate-material interface. The equilibrium is ; P = τa It revealed that as the shearing stress, τ, is increased by increasing P. 15

16 We can say that shear stress, τ, has direct proportion with the velocity gradient that is ; τ du dy The shearing stress and velocity gradient can be related with a relationship of the form ; τ = µ (mu) is dynamic viscosity. µ du dy 16

17 Fluids for which the shearing stress is linearly related to the rate of shearing strain are designated as Newtonian fluids. Fluids for which the shearing stress is not linearly related to the rate of shearing strain are designated as non-newtonian fluids. 17

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20 BULK MODULUS A property that is commonly used to characterize compressibility is the bulk modulus. Defined as ; dp E = = dp υ dv dρ V ρ we conclude that liquids can be considered as incompressible for most practical engineering applications. 20

21 COMPRESSION & EXPANSION OF GAS When gases are compressed (or expanded) the relationship between pressure and density depends on the nature of the process. If the compression or expansion takes place under constant temperature conditions (isothermal process), then ; P = constant ρ If the compression or expansion is frictionless and no heat is exchanged with the surroundings (isentropic process), then ; P k ρ = constant 21

22 k is the ratio of the specific heat at constant pressure, c p, to the specific heat at constant volume, c v. k = c p c v 22

23 SURFACE TENSION The intensity of the molecular attraction per unit length along any line in the surface is called the surface tension. Designated by the Greek symbol, σ (sigma) Unit is N/m. The forces balance of half-cut spherical is shown as ; 2 2πRσ = PπR 23

24 The forces balance of capillary action is shown as ; 2 2π Rσ cosθ = ρghπr 24

25 Chapter 2 - Pressure INTRODUCTION PRESSURE In this chapter we will consider an important class of problems in which the fluid is either at rest or moving in such a manner that there is no relative motion between adjacent particles. In both instances there will be no shearing stresses in the fluid, and the only forces that develop on the surfaces of the particles will be due to the pressure. The absence of shearing stresses greatly simplifies the analysis There are no shearing stresses present in a fluid at rest. 1

26 Chapter 2 - Pressure PRESSURE 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. The equations of motion (Newton's second law, (F=ma) in the y and z directions are, respectively. y s y y a z y x s x p z x p F 2 sin δ δ δ ρ θ δ δ δ δ = = Σ z s z z a z y x z y x g s x p y x p F 2 2 cos δ δ δ ρ δ δ δ ρ θ δ δ δ δ = = Σ 2

27 Chapter 2 - Pressure where p s, p y, and p z are the average pressures on the faces, γ and ρ are the fluid specific weight and density, respectively, and a y, a z the accelerations. Note that a pressure must be multiplied by an appropriate area to obtain the force generated by the pressure. Since we are really interested in what is happening at a point, we take the limit as δx, δy, and δz approach zero (while maintaining the angle θ), and it follows that p = p = y z p s The pressure at a point in a fluid at rest is independent of direction. We can conclude that 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. This important result is known as Pascal's law named in honor of Blaise Pascal ( ), 3

28 Chapter 2 - Pressure BASIC EQUATION FOR PRESSURE FIELD For liquids or gases at rest the pressure gradient in the vertical direction at any point in a fluid depends only on the specific weight of the fluid at that point. dp dx = 0 dp dy = 0 dp dz = γ 4

29 Chapter 2 - Pressure INCOMPRESSIBLE FLOW h = p 1 ρg p 2 5

30 Chapter 2 - Pressure Pascal s Paradox 6

31 Chapter 2 - Pressure STANDARD ATMOSPHERE 7

32 Chapter 2 - Pressure MEASUREMENT OF PRESSURE The pressure at a point within a fluid mass will be designated as either an absolute pressure or a gage pressure. Absolute pressure is measured relative to a perfect vacuum (absolute zero pressure), whereas gage pressure is measured relative to the local atmospheric pressure. 8

33 Chapter 2 - Pressure A barometer is used to measure atmospheric pressure. mercury barometer p = ρ gh + atm p vapor 9

34 Chapter 2 - Pressure 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. The mercury barometer is an example of one type of manometer, but there are many other configurations possible, depending on the particular application. Three common types of manometers include the piezometer tube, the U-tube manometer, and the inclined-tube manometer. 10

35 Chapter 2 - Pressure PIEZOMETER TUBE p = ρ gh + p o p A = γ = 1h1 ρ1gh1 11

36 Chapter 2 - Pressure U-TUBE MANOMETER p A = ρ 2gh2 ρ1gh1 12

37 Chapter 2 - Pressure INCLINED-TUBE MANOMETER p A p = ρ l B 2g 2 sinθ + ρ3gh3 ρ1gh1 13

38 Chapter 2 - Pressure MECHANICAL AND ELECTRONIC PRESSURE DEVICES A Bourdon tube pressure gage uses a hollow, elastic, and curved tube to measure pressure. 14

39 PASCAL S LAW FOR PRESSURE AT A POINT By considering the equilibrium of a small fluid element in the form of a triangular prism surrounding a point in the fluid, as shown below, a relationship can be established between the pressure p x in the x-direction, p y in the y-direction and p s normal to any plane inclined at any angle to the horizontal at this point. If the fluid is at rest, p x will act at right angles to the plane ABFE, p y at right angles to CDEF and p s at right angle to ABCD. Since the fluid is at rest, there will be no shearing forces on the faces of the element and the element will not be accelerating. The sum of the forces in any direction must, therefore, be zero. 1

40 Considering the x-direction : Force due to p x = p x area ABFE = p x δyδz Component of force due to p s = = ( p areaabcd) sinθ s δy = psδsδz δs = p δyδz s δy sin θ = δs As p y has no compound in the x-direction, the element will be in equilibrium if : p δyδz + ( p δyδz) = 0 p x x = p s s 2

41 similarly in y-direction : Force due to p y = p y area CDEF = p y δxδz Component of force due to p s = = ( p areaabcd) cosθ s δx = psδsδz δs = p δxδz s δx cos θ = δs Weight of element = = specific weight Volume 1 = ρg δ xδyδz 2 As p x has no component in the y-direction, the element will be in equilibrium if ; p δ xδz + p δxδz) + ( ρg 1 δxδyδz ) = 0 y ( s 2 3

42 Since δx, δy and δz are all very small quantities, δxδyδz is negligible in comparison with the other two terms, and the equation reduces to p y = p s Now, we can conclude that ; p = p = x y p s 4

43 THE INFLUENCE OF HEIGHT IN PRESSURE For static equilibrium the sum of the horizontal forces must be zero : p1 A = p2a In mathematical term, if (x, y) is the horizontal plane: p p = 0 x and = 0 y 5

44 The sum of all vertical forces must be zero : ) ( z z g A p A p mg A p A p = = ρ We can conclude as : ( 1) z z g p p = ρ Thus, in any fluid under gravitational attraction, pressure decrease with increase of height z. 6

45 Chapter 3 - Hydrostatic force on a submerged plane surface HYDROSTATIC FORCE ON A SUBMERGED 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. We also know that the pressure will vary linearly with depth as shown in Figure 1 if the fluid is incompressible. Figure 1 1

46 Chapter 3 - Hydrostatic force on a submerged plane surface The magnitude of the resultant fluid force is equal to the pressure acting at the centroid of the area multiplied by the total area. F R = ρ gh A (equation 1) c Figure 2 2

47 Chapter 3 - Hydrostatic force on a submerged plane surface Note that the magnitude of the force is independent of the angle θ and depends only on the specific weight of the fluid, the total area, and the depth of the centroid of the area below the surface. In effect, Equation 1 indicates that the magnitude of the resultant force is equal to the pressure at the centroid of the area multiplied by the total area. Since all the differential forces that were summed to obtain F R are perpendicular to the surface, the resultant F R must also be perpendicular to the surface. The point through which the resultant force acts is called the center of pressure. 3

48 Chapter 3 - Hydrostatic force on a submerged plane surface Coordinate for center of pressure (y R, x R ) : I xc y R = + yc A y c I xyc x R = + y A c x c 4

49 Chapter 3 - Hydrostatic force on a submerged plane surface Centroidal coordinates and moments of inertia for some common areas are given in Figure 3. Figure 3 5

50 Chapter 3 Pressure prism for rectangular shape PRESSURE PRISM An informative and useful graphical interpretation can be made for the force developed by a fluid acting on a plane area. Consider the pressure distribution along a vertical wall of a tank of width b, which contains a liquid having a specific weight γ(=ρg). Since the pressure must vary linearly with depth, we can represent the variation as is shown in Figure 4, where the pressure is equal to zero at the upper surface and equal to γh(=ρgh) at the bottom. Figure 4 1

51 Chapter 3 Pressure prism for rectangular shape The base of this volume in pressure-area space is the plane surface of interest, and its altitude at each point is the pressure. This volume is called the pressure prism, and it is clear that the magnitude of the resultant force acting on the surface is equal to the volume of the pressure prism. The magnitude of the resultant fluid force is equal to the volume of the pressure prism and passes through its centroid 2

52 Chapter 3 Pressure prism for rectangular shape Specific values can be obtained by decomposing the pressure prism into two parts, ABDE and BCD, as shown in Figure 5. Thus, F R = F 1 + F 2 1 h FR = volume = ρ gh)( bh) = ρg( ) A 2 ( 2 The location of F R can be determined by summing moments about some convenient axis, such as one passing through A. In this instance F R y A = F1 y1 + F2 y2 Figure 5 3

53 Chapter 3 Pressure prism for rectangular shape Figure 6 For inclined plane surfaces the pressure prism can still be developed, and the cross section of the prism will generally be trapezoidal as is shown in Figure 6. The use of pressure prisms for determining the force on submerged plane areas is convenient if the area is rectangular so the volume and centroid can be easily determined. 4

54 Chapter 3 Pressure prism for rectangular shape However, for other nonrectangular shapes, integration would generally be needed to determine the volume and centroid. In these circumstances it is more convenient to use the equations developed in the previous section, in which the necessary integrations have been made and the results presented in a convenient and compact form that is applicable to submerged plane areas of any shape. 5

55 Chapter 3 Pressure prism for rectangular shape We note that in this case the force on one side of the wall now consists of F R as a result of the hydrostatic pressure distribution, plus the contribution of the atmospheric pressure, p atm A, where A is the area of the surface. However, if we are going to include the effect of atmospheric pressure on one side of the wall we must realize that this same pressure acts on the outside surface (assuming it is exposed to the atmosphere), so that an equal and opposite force will be developed as illustrated in the figure 7. Thus, we conclude that the resultant fluid force on the surface is that due only to the gage pressure contribution of the liquid in contact with the surface the atmospheric pressure does not contribute to this resultant. Figure 7 6

56 HYDROSTATIC FORCES ON A SUBMERGED CURVED PLANE The equations developed in previous lesson for the magnitude and location of the resultant force acting on a submerged surface only apply to plane surfaces. However, many surfaces of interest (such as those associated with dams, pipes, and tanks) are non-planar. We will consider the equilibrium of the fluid volume enclosed by the curved surface of interest and the horizontal and vertical projections of this surface. 1

57 Horizontal Force ; F H = ρ gh c A Vertical Force ; F V = ρgv Resultant Force ; F = F + R 2 H F 2 V 2

58 Example 1 The 6-m-diameter drainage conduit of Figure 1 is half full of water at rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1-m length of the curved section BC of the conduit wall. Figure 1 3

59 Example 2 A 4-m-long curved gate is located in the side of a reservoir containing water as shown in Figure 2. Determine the magnitude of the horizontal and vertical components of the force of the water on the gate. Will this force pass through point A? Explain. Figure 2 4

60 Example 3 Determine the magnitude of the horizontal and vertical components of the force (per unit length) of the water on the concrete seawall of Figure 3. Figure 3 5

61 Chapter 4 Buoyancy, Floatation and Stability BUOYANCY, FLOATATION AND STABILITY Archimedes Principle When a stationary body is completely submerged in a fluid, or floating so that it is only partially submerged, the resultant fluid force acting on the body is called the buoyant force. Note that the forces F 1, F 2, F 3, and F 4 are simply the forces exerted on the plane surfaces, W(=mg) is the weight of the shaded fluid volume, and F B is the force the body is exerting on the fluid. The forces on the vertical surfaces, such as F 3 and F 4, are all equal and cancel, so the equilibrium equation of interest is in the z direction and can be expressed as F F mg F B = 2 1 If the specific weight of the fluid is constant, then ; F F = ρ g h h ) A (1) 2 1 ( 2 1 where A is the horizontal area of the upper (or lower) surface, and Equation (1) can be written as ; = ρ g h h ) A g ( h h ) A V F B [ ] ( 2 1 ρ 2 1 1

62 Chapter 4 Buoyancy, Floatation and Stability Simplifying, we arrive at the desired expression for the buoyant force ; F B = ρgv Figure 1 2

63 Chapter 4 Buoyancy, Floatation and Stability Archimedes' principle states that the buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward. Thus, we conclude that the buoyant force passes through the centroid of the displaced volume as shown in Figure 1(c). The point through which the buoyant force acts is called the center of buoyancy. 3

64 Chapter 4 Buoyancy, Floatation and Stability Example A spherical buoy has a diameter of 1.5 m, weighs 8.50 kn, and is anchored to the seafloor with a cable as is shown in Figure 2(a). Although the buoy normally floats on the surface, at certain times the water depth increases so that the buoy is completely immersed as illustrated. For this condition what is the tension of the cable? Figure 2 4

65 Chapter 4 Buoyancy, Floatation and Stability Stability As illustrated by the Figure 3, a body is said to be in a stable equilibrium position if, when displaced, it returns to its equilibrium position. Conversely, it is in an unstable equilibrium position if, when displaced (even slightly), it moves to a new equilibrium position. Stability considerations are particularly important for submerged or floating bodies since the centers of buoyancy and gravity do not necessarily coincide. Figure 3 5

66 Chapter 4 Buoyancy, Floatation and Stability A small rotation can result in either a restoring or overturning couple. For example, for the completely submerged body shown in Figure 4, 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, F B, 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 as long as the center of gravity falls below the center of buoyancy, this will always be true; that is, the body is in a stable equilibrium position with respect to small rotations. Figure 4 6

67 Chapter 4 Buoyancy, Floatation and Stability However, as is illustrated in Figure 5, 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. Figure 5 7

68 Chapter 4 Buoyancy, Floatation and Stability For floating bodies the stability problem is more complicated, since as the body rotates the location of the center of buoyancy (which passes through the centroid of the displaced volume) may change. As is shown in Figure 6, 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 the buoyant force, F B, shifts to pass through the centroid of the newly formed displaced volume and, as illustrated, combines with the weight, W, to form a couple which will cause the body to return to its original equilibrium position. Figure 6 8

69 Chapter 4 Buoyancy, Floatation and Stability However, for the relatively tall, slender body shown in Figure 7, a small rotational displacement can cause the buoyant force and the weight to form an overturning couple as illustrated. Figure 7 9

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