BFC FLUID MECHANICS BFC NOOR ALIZA AHMAD


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1 BFC FLUID MECHANICS
2 CHAPTER 1.0: Principles of Fluid 1.1 Introduction to Fluid Mechanics 1.2 Thermodynamic Properties of a Fluid: Density, specific weight, specific gravity, viscocity (kelikatan)berat tentu, compressibility (kemampatan), Bulk modulus (modulus pukal), dynamic & kinematic viscosity (kelikatan dinamik dan kinematik),surface tension (ketegangan permukaan) and capillarity (kererambutan). 2
3 CHAPTER 1.0: Principles of Fluid 1.1 Introduction to Fluid Mechanics Mechanics is the oldest physical science that deals with both stationary and moving bodies under the influence of forces. The branch of mechanics that deals with bodies at rest is called statics, while the branch that deals with bodies in motion is called dynamics. The subcategory fluid mechanics is defined as the science that deals with the behavior of fluids at rest (fluid statics)or in motion (fluid dynamics), and the interaction of fluids with solids or other fluids at the boundaries. 3
4 1.1 Introduction to Fluid Mechanics  Fluid engineering applications is enormous: breathing, blood flow, swimming, pumps, fans, turbines, airplanes, ships, rivers, windmills, pipes, icebergs, engines, filters, jets, and sprinklers, to name a few.. 4
5 1.1 Introduction to Fluid Mechanics (Cont d)  From the point of view of fluid mechanics, all matter consists of only two states, fluid and solid. Distinction between a solid and a fluid is made on the basis of the substance s ability to resist an applied shear (or tangential) stress that tends to change its shape.  Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid. The fluid moves and deforms continuously as long as the shear stress is applied. 5
6 1.1 Introduction to Fluid Mechanics (Cont d) Figure 1 illustrates a solid block resting on a rigid plane and stressed by its own weight. The solid sags into a static deflection, shown as a highly exaggerated dashed line, resisting shear without flow. Figure 1 6
7 1.1 Introduction to Fluid Mechanics (Cont d)  The liquid and gas at rest in Figure 2 require the supporting walls to eliminate shear stress. Figure 2 7
8 1.1 Introduction to Fluid Mechanics  The liquid retains its volume and forms a free surface in the container.  If the walls are removed, shear develops in the liquid and a big splash results.  If the container is tilted, shear again develops, waves form, and the free surface seeks a horizontal configuration, pouring out over the lip if necessary. 8
9 1.2 Thermodynamic Properties of a Fluid: a) Density, (ketumpatan) m V ( kg / 3 m )  Density is highly variable in gases and increases nearly proportionally to the pressure level. 9
10 1.2 Thermodynamic Properties of a Fluid: (Cont d)  Density in liquids is nearly constant; the density of water (about 1000 kg/m 3 ) increases only 1 percent if the pressure is increased by a factor of 220. Thus most liquid flows are treated analytically as nearly incompressible.  Compare their densities at 20 C and 1 atm: Mercury: = 13,580 kg/m 3 Hydrogen: = kg/m 3 10
11 The density of liquids and solids depends more strongly on temperature than it does on pressure. At 1 atm, for example, the density of water changes from 998 kg/m3 at 20 C to 975 kg/m3 at 75 C, a change of 2.3 percent, which can still be neglected in many engineering analyses. 0 C Density (kg/m 3 ) Nota : 1000 kg/m 3 = 1.94 slugs/ft 3 11
12 b) Specific volume, (Isipadu tentu) V m 1 c) Relative density (ketumpatan relatif) and is defined as the ratio of the density of a substance to the density ofsome standard substance at a specified temperature (usually water at 4 C, for which H 2 O 1000 kg/m 3 ). 12
13 d) Specific gravity, SG (graviti tentu) the density of a substance is given relative to the density of a wellknown substance. SG H 2Oat4 e) specific weight, s (berat tentu) s = g ( N/m 3 ) 0 C Substances Water Blood Seawater Gasoline Mercury Wood Gold Ice Air SG
14 Example 1: Given Specific Gravity of Mercury is (20 0 C). Calculate mercury s density. Solution: merkuri merkuri 1000kg/ m x10 kg/ m 3 14
15 Example 2: Calculate the gasoline s density at 20 0 C. The mass and volume are 60 kg dan 0.5 m 3 respectively Solution: m V ( kg/ 3 m ) 120kg / m 3 15
16 f) Compressibility, (kebolehmampatan) refers to the change in volume (V) of a substance that is subjected to a change in pressure on it. Added Volume Bar the usual quantity used to measure this phenomenon is the bulk modulus of elasticity or simply bulk modulus, E 16
17 g) Bulk Modulus, E (Modulas Pukal) Pressure changes needed for changing the volume Initial Volume Showing the pressure increment with volume reduction Volume changes E, harder to compress 17
18 h) Viscosity, (Kelikatan) is a property that represents the internal resistance of a fluid to motion or the fluidity, and that property is the viscosity. The force a flowing fluid exerts on a body in the flow direction is called the drag force, and the magnitude of this force depends, in part, on viscosity 18
19 19
20 To obtain a relation for viscosity, consider a fluid layer between two very large parallel plates (or equivalently, two parallel plates immersed in a large body of a fluid) separated by a distance l. Now a constant parallel force F is applied to the upper plate while the lower plate is held fixed. After the initial transients, it is observed that the upper plate moves continuously under the influence of this force at a constant velocity V. 20
21 The fluid in contact with the upper plate sticks to the plate surface and moves with it at the same velocity, and the shear stress acting on this fluid layer is ( = F/A) where A is the contact area between the plate and the fluid. Note that the fluid layer deforms continuously under the influence of shear stress. Details : See Cengel( 2005). Fluid Mechanics. Mc Graw Hill 21
22 Fluids for which the rate of deformation is proportional to the shear stress are called Newtonian fluids Water, air, gasoline, and oils (Newtonian fluids) Blood and liquid plastics (nonnewtonian fluids) In onedimensional shear flow of Newtonian fluids, shear stress can be expressed by the linear relationship Shear stress: 22
23 Example 3: SAE 30 Oil at 20 0 C of in is placed in between two layer. The bottom layer is fixed while upper layer moves with acceleration 13 ft/s. Calculate shear stress for the oil. Solution: = (9.20 x 103 )[ 13/(0.005/2)] = Ib/ ft 2 23
24 Example 4: Benzene at 20 0 C has a coefficient of viscocity, Pa.s. Calculate the shear stress to deform this fluid at velocity gradient of 4900 s1? Solution: = x 4900 = 3.19Pa 24
25 i) Dynamic Viscosity, (Kelikatan dinamik) defined as shear force per unit area Units: N s/m 2, kgm 1s1,Poise P Typically Water =1.14 kgm 1s1, Air =1.78 kgm 1s1, Mercury =1.552 kgm 1s1, Paraffin Oil=1.9 kgm 1s1, 25
26 Dynamic viscocities of some fluids at 1 atm and 20 0 C (unless otherwise stated) 26
27 j) Kinematic Viscosity, Kelikatan kinematik, the ratio of dynamic viscosity to density Two common units of kinematic viscosity are m 2 /s and stoke (1 stoke 1 cm 2 /s m 2 /s). Dynamic viscosity, in general, does not depend on pressure, but kinematic viscosity does. 27
28 Typically Water =1.14 x 106, m 2 /s Air =1.46 x 105 m 2 /s, Mercury =1.145 x 104 m 2 /s, Paraffin Oil =2.375 x 103 m 2 /s 28
29 k) Surface Tension, s (Ketegangan permukaan) a drop of blood forms a hump on a horizontal glass! a drop of mercury forms a nearperfect sphere and can be rolled just like a steel ball over a smooth surface! water droplets from rain or dew hang from branches or leaves of trees! 29
30 In these and other observances, liquid droplets behave like small spherical balloons filled with the liquid, and the surface of the liquid acts like a stretched elastic membrane under tension. The pulling force that causes this tension acts parallel to the surface and is due to the attractive forces Between the molecules of the liquid. The magnitude of this force per unit length is called surface tension s and is usually expressed in the unit N/m s F 2b 30
31 31
32 How surface tension arises?? Let see at microscopic view in Figure. By considering two liquid molecules, one at the surface and one deep within the liquid body. The attractive forces applied on the interior molecule by the surrounding molecules balance each other because of symmetry. 32
33 But the attractive forces acting on the surface molecule are not symmetric,and the attractive forces applied by the gas molecules above are usually very small. Therefore, there is a net attractive force acting on the molecule at the surface of the liquid, which tends to pull the molecules on the surface toward the interior of the liquid. This force is balanced by the repulsive forces from the molecules below the surface that are being compressed. 33
34 The resulting compression effect causes the liquid to minimize its surface area. This is the reason for the tendency of the liquid droplets to attain a spherical shape, which has the minimum surface area for a given volume. 34
35 That why with amusement, that some insects can land on water or even walk on water and that small steel needles can float on water. These phenomena are again made possible by surface tension that balances the weights of these objects. 35
36 l) Capillary effect (Kererambutan) Another interesting consequence of surface tension is the capillary effect,which is the rise or fall of a liquid in a smalldiameter tube inserted into the liquid. This effect is usually expressed by saying that water wets the glass (by sticking to it) while mercury does not 36
37 The contact angle for wetting and nonwetting fluids. A liquid is said to wet the surface when < 90 and not to wet the surface when >
38 The capillary rise of water and the capillary fall of mercury in a smalldiameter glass tube. The phenomenon of capillary effect can be explained microscopically by considering cohesive forces (the forces between like molecules, such as water and water) and adhesive forces (the forces between unlike molecules, such as water and glass). The liquid molecules at the solid liquid interface are subjected to both cohesive forces by other liquid molecules and adhesive forces by the molecules of the solid. 38
39 The magnitude of the capillary rise in a circular tube can be determined from a force balance on the cylindrical liquid column of height h in the tube (see figure) The bottom of the liquid column is at the same level as the free surface of the reservoir, and thus the pressure there must be atmospheric pressure. 39
40 This balances the atmospheric pressure acting at the top surface, and thus these two effects cancel each other. The weight of the liquid column is approximately Equating the vertical component of the surface tension force to the weight gives 40
41 Solving for h gives the capillary rise to be h 2 s cos gr R = constant Example 5: A 0.6mmdiameter glass tube is inserted into water at 20 C in a cup. Determine the capillary rise of water in the tube 41
42 Solution: With assumption, 1 There are no impurities in the water and no contamination on the surfaces of the glass tube. 2 The experiment is conducted in atmospheric air. 42
43 h = 2(0.073 N/m) 1000 kg/m 3 ( 9.81m/s 2 )(0.3 x 103 ) (cos 0 0 ) 1 kg. m/s 2 1 N = m = 5.0cm 43
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