COURSE NUMBER: ME 321 Fluid Mechanics I Fluid statics. Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET

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1 COURSE NUMBER: ME 321 Fluid Mechanics I Fluid statics Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET 1

2 Fluid statics Fluid statics is the study of fluids in which there is no relative motion between fluid particles. If there is no relative motion; no shearing stresses exist. The only stress that exists is a normal stress, the pressure. So, it is the pressure that is of primary interest in fluid statics. When the fluid velocity is zero, denoted as the hydrostatic condition, the pressure variation is due only to the weight. For a known fluid in a given gravity field, the pressure may easily be calculated by integration. If the fluid is moving in rigid-body motion, such as a tank of liquid which has been spinning for a long time, the pressure can be easily calculated, because the fluid is free of shear stress. 2

3 Fluid statics Three situations are usually investigated in fluid statics. These include (a) fluids at rest, such as liquid in a reservoir, (b) fluids contained in devices that undergo linear acceleration, and (c) fluids contained in rotating cylinders. In addition to the examples shown above, we consider instruments called manometers and investigate the forces of buoyancy and the stability of floating bodies. 3

4 PRESSURE AT A POINT Pressure at a point is defined as being the infinitesimal normal compressive force divided by the infinitesimal area over which it acts. Now does the pressure, at a given point, vary as the normal to the area changes direction? To show that this is not the case, even for fluids in motion, consider the wedge-shaped element of unit depth (in the z- direction) shown in Fig

5 Assume that a pressure p acts on the hypotenuse and that a different pressure acts on each of the other areas, as shown. Since the forces on the two end faces are in the z-direction, we have not included them on the element. Now, let us apply Newton's second law to the element, for both the x- and y-directions: 5

6 Where we have used The pressures shown are due to the surrounding fluid and are the average pressure on the areas. Substituting Eqs take the form As the element shrinks to a point, D x 0 and D y 0. Hence the right-hand sides in the equations above go to zero, even for fluids in motion, providing with the result that, at a point, Since q is arbitrary, this relationship holds for all angles at a point. Thus we conclude that the pressure in a fluid is constant at a point. That is, pressure is a scalar function. It acts equally in all directions at a given point for both a static fluid and a fluid that is in motion. 6

7 Pressure Variation in a Fluid Body To determine the pressure variation in fluids at rest or fluids undergoing an acceleration while the relative position of fluid elements to one another remains the same, consider the infinitesimal element displayed in Fig.2.4, where the z-axis is in the vertical direction. 7

8 Pressure Variation in a Fluid Body If we assume that p is the pressure at the center of this element, the pressures at any other point can be expressed by using a first-order Taylor series expansion with p(x, y, z): lf we move from the center to a face a distance (dx/2) away, we see that the pressure is The pressures at all faces are expressed in this manner, as shown in Fig

9 Pressure Variation in a Fluid Body Newton's second law is written in vector form for a constantmass system as 9

10 Pressure Variation in a Fluid Body 10

11 Fluids at Rest A fluid at rest does not undergo any acceleration. Therefore, Eq reduces to or This equation implies that there is no pressure variation in the x and y directions, that is, in the horizontal plane. The pressure varies in the z direction only. Also note that dp is negative if dz is positive; that is, the pressure decreases as we move up and increases as we move down, 11

12 Fluids at Rest Hydrostatic-pressure distribution. 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. Potnt D has a different pressure from A, B, and C because it is not connected to them by a water Path. 12

13 Pressures in Liquids at Rest If the density can be assumed constant (which is the case with liquids), Eq may be integrated to yield so that pressure increases with depth. Note that z is positive in the upward direction. The quantity (p/g + z) is often referred to as the piezometric head. If the point of interest were a distance h below a free surface Eq would result in This equation is quite useful in converting pressure to an equivalent height of liquid. 13

14 Gage Pressure and Vacuum Pressure Engineers usually specify pressures as (1) the absolute or (2) the gage pressure ( pressure relative to the local ambient atmosphere). The second case occurs because many pressure instruments are of dffirential type and record, not an absolute magnitude, but the difference between the fluid pressure and the atmosphere. 14

15 Linearly Accelerating Containers Consider a fluid at rest relative to a reference frame that is linearly accelerating with a horizontal component a x and a vertical component a z. Then Eq simplifies to 15

16 Linearly Accelerating Containers Note that density or viscosity does not appear in the above equation. 16

17 Linearly Accelerating Containers A coffee mug on a horizontal tray is accelerated at 7 m/s 2.The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest. (a) Assuming rigid body acceleration of the coffee, determine whether it will spill out of the mug,(b) Calculate the gage pressure in the corner at point A if the density of coffee is 1010 kg/m 3. (a) Regardless of the shape of the mug, the free surface tilts at an angle q given by as a z = 0. 17

Linearly Accelerating Containers If the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted face intersects the original rest surface exactly at the centerline,

18 Linearly Accelerating Containers If the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted face intersects the original rest surface exactly at the centerline, as shown. Thus the deflection at the left side of the mug is z = (3 cm)(tan q) = 2.14 cm. This is less than the 3-cm clearance available, so the coffee will not spill. (b) When at rest, the gage pressure at point A is given by p A = rg(z sur - z A ) = (1010 kg/m 3 )(9.81 m/s 2 )(0.07 m) = 694 Pa. During acceleration, the pressure at A becomes p A = rg(z sur - z A ) = (1010 kg/m 3 )(9.81 m/s 2 )( m) = 906 Pa, which is 31% higher than the pressure at rest. 18

19 Linearly Accelerating Containers 19

20 Linearly Accelerating Containers 20

21 Linearly Accelerating Containers 21

22 Rotating Containers 22

23 Rotating Containers 23

24 Rotating Containers Note that density or viscosity does not appear in the above equation. 24

Rotating Containers If r 2 = R = the container radius, then the total rise of water level at the wall, Since the volume of a paraboloid is one-half the base area time height,

25 Rotating Containers If r 2 = R = the container radius, then the total rise of water level at the wall, Since the volume of a paraboloid is one-half the base area time height, the still water level is exactly halfway between the high and low points of the free surface. That is the center of the fluid drops an amount, the edges rise an equal amount. and 25

26 Rotating Containers A 10 cm deep coffee mug contains 7 cm coffee of density 1010 kg/m 3. It is given a rigid body rotation about its central axis. Find (a) the angular velocity which will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this condition. => 0.03 m = 2 (0.03m) 2 /4(9.81m/s 2 ) => = 36.2 rad/s = 345 rpm Pressure at A may be determined by eq or even by P A = gh A = (1010 *9.81 N/m 3 )(0.1m)= 990 Pa 26

27 Rotating Containers 27

28 Rotating Containers 28

29 Rotating Containers 29

30 Manometers Manometers are instruments that use columns of liquids to measure pressures. The figure shows a simple open manometer for measuring p A in a closed chamber relative to atmospheric pressure, p a. The manometric fluid (r 2 ) is chosen different from the chamber fluid (r 1 ) to isolate the chamber fluid from the environment and to suitably scale the length of the open tube. 30

31 Manometers Here, one can begin at A, apply the basic hydrostatic formula down to z 1, jump across fluid 2 to the same pressure p 1, and then use the basic hydrostatic formula up to level z 2. The physical reason that we can jump across" at section I is that a continuous length of the same fluid connects these two equal elevations. Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure. 31

32 First get the specific weights from Tables. Now proceed from A to B, calculating the pressure change in each fluid and adding: 32

33 33

34 Solution First identify the relevant points 1, 2, 3, 4, and 5 as shown in the figure. For simplicity, neglect the weight of the air and assume the pressure at point 3 is equal to the pressure at point 4. 34

35 Forces on Submerged Planes In the design of submerged devices and objects, such as dams, flow barriers, ships, and holding tanks, it is necessary to calculate the magnitudes and locations of forces that act on their surfaces, both plane and curved. Here, we consider only plane surfaces, such as the plane surface of general shape shown in the Fig. Note that a side view is given as well as a view showing the shape of the plane. 35

36 Forces on Submerged Planes The total force of the liquid on the plane surface is found by integrating the pressure over the area, that is, The x and y coordinates are in the plane of the plane surface, as shown. Assuming p = 0 at h = 0, we know that where h is measured vertically down from the free surface to the elemental area da and y is measured from point 0 on the free surface. 36

37 Forces on Submerged Planes The force may then be expressed as The distance to a centroid is defined as The expression for the force then becomes 37

38 Forces on Submerged Planes Where is the vertical distance from the free surface to the centroid of the area and p c is the pressure at the centroid. Thus the magnitude of the force on a submerged plane surface is the pressure at the centroid multiplied by the area. The force does not depend on the angle of inclination, a. 38

39 Forces on Submerged Planes The force does not, in general, act at the centroid. The location of the resultant force is found by taking the sum of the moments of all the infinitesimal pressure forces acting on the area equal to the moment of the resultant force. Let the force F act at the point (x p,y p ), the center of pressure (c.p.). 39

40 Forces on Submerged Planes The value of y p can be obtained by equating moments about the x- axis: where the second moment of the area about the x-axis is 40

41 Forces on Submerged Planes 41

42 Forces on Submerged Planes 42

43 Forces on Submerged Planes 43

44 Forces on Submerged Planes 44

45 Forces on Submerged Planes 45

46 Forces on Submerged Planes 46

47 Forces on Submerged Planes Problem: Find the force P needed to hold the 3 m wide rectangular gate shown. Sol hints: F = = 9810*(5sin40 o /2)*(5x3) N 3 3*5 12 = m (5x3)*2.5 : 7*Psin40 o = (5 - y p )*F 47

48 Forces on Curved Surfaces 48

49 Forces on Curved Surfaces 49

50 Forces on Curved Surfaces 50

51 Forces on Curved Surfaces 51

52 Forces on Curved Surfaces 52

53 Forces on Curved Surfaces 53

54 Forces on Curved Surfaces 54

55 Forces on Curved Surfaces 55

56 Buoyancy Two laws of buoyancy discovered by Archimedes in the third century B.C.: 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. These two laws are easily derived by referring to the Fig. The body lies between an upper curved surface 1 and a lower curved surface 2. The body experiences a net upward force 56

57 Buoyancy Alternatively, we can sum the vertical forces on elemental slices through the immersed body as shown in the Fig.: This result is identical to the previous one and equivalent to law I above. Here, it is assumed that the fluid has uniform specific weight. 57

58 Buoyancy The line of action of the buoyant force passes through the centroid of the displaced liquid volume only if it has uniform density. This point through which F B acts is called the center of buoyancy. Of course, the center of buoyancy may or may not correspond to the actual center of mass of the body's own material, which may have variable density. 58

59 Buoyancy Gases also exert buoyancy on any body immersed in them. For example, human beings have an average specific weight of about 60 lbf/ft 3. If the weight of a person is 180 lbf, the person's total volume will be 3.0 ft 3. However, in so doing we are neglecting the buoyant force of the air surrounding the person. At standard conditions, the specific weight of air is lbf/ft 3 ; hence the buoyant force is approximately 0.23 lbf. If measured in vacuo, the person would weigh about 0.23 lbf more. For balloons, the buoyant force of air, instead of being negligible, is the controlling factor in the design. 59

60 Buoyancy of Floating Bodies Floating bodies are a special case; only a portion of the body is submerged, with the remainder poking up out of the free surface. This is illustrated in the Fig. From a static force balance, it may be derived that 60

61 Buoyancy of Floating Bodies Not only does the buoyant force equal the body weight but also they are collinear since there can be no net moments for static equilibrium. The above equation is the mathematical equivalent of Archimedes' law 2. Occasionally, a body will have exactly the right weight and volume for its ratio to equal the specific weight of the fluid. If so, the body will be neutrally buoyant and remain at rest at any point where it is immersed in the fluid. Small neutrally buoyant particles are sometime used for flow visualization. A submarine can achieve positive, neutral, or negative buoyancy by pumping water in or out of its ballast 61

62 Stability of Floating Bodies If a floating object is raised a small distance, the buoyant force decreases and the object's weight returns the object to its original position. Conversely, if a floating object is lowered slightly, the buoyant force increases and the larger buoyant force returns the object to its original position. Thus a floating object has vertical stability since a small departure from equilibrium results in a restoring force. 62

63 Rotational Stability of Submerged Bodies Let us now consider the rotational stability of a submerged body. If the center of gravity G of the body is above the centroid C (also referred to as the center of buoyancy) of the displaced volume and a small angular rotation results in a moment that will continue to increase the rotation; the body is unstable and overturning would result. Engineers must design to avoid floating instability. 63

64 Rotational Stability of Submerged Bodies If the center of gravity G is below the centroid C, a small angular rotation provides a restoring moment and the body is stable. If the center of gravity and the centroid coincide, the body is said to be neutrally stable, a situation that is encountered whenever the density is constant throughout the floating body. 64

65 Stability of Floating Bodies If the center of gravity is below the centroid, the body is always stable, as with submerged bodies. The body may be stable, though, even if the center of gravity is above the centroid, as sketched. 65

66 Stability of Floating Bodies When the body rotates the centroid of the volume of displaced liquid moves to the new location C'. If the centroid C' moves sufficiently far, a restoring moment develops and the body is stable. This is determined by the metacentric height GM. Metacentre M is the point of intersection of the buoyant force before rotation with the buoyant force after rotation. If GM is positive, as shown, the body is stable; if GM is negative (M lies below G), the body is unstable. 66

67 Stability of Floating Bodies To determine a quantitative relationship for the distance GM consider the sketch, which shows the uniform cross section of the floating body in rotated condition. 67

68 Stability of Floating Bodies An expression for, the x- coordinate of the centroid of the displaced volume can be found by considering the volume to be the original volume plus the added wedge with cross-sectional area DOE minus the subtracted wedge with cross-sectional area AOB. To locate the centroid of the composite volume, we take moments as follows: 68

69 Stability of Floating Bodies 69

70 Stability of Floating Bodies 70

71 Stability of Floating Bodies 71

72 Stability of Floating Bodies 72

Objectives deals with forces applied by fluids at rest or in rigid-body motion.

Objectives deals with forces applied by fluids at rest or in rigid-body motion. The fluid property responsible for those forces is pressure, which is a normal force exerted by a fluid per unit area. discussion