DIMENSIONS AND UNITS


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1 DIMENSIONS AND UNITS A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Primary Dimension Dimensions SI Units English Units Mass M Kilogram (Kg) Mass of a 1 lb object is 1/32.2 slug on earth Length L Meter (m) Foot (ft) Time T Second (s) Second (s) Temperature Θ Kevin (K) Rankin ( o R) 1
2 TABLE 1: Some Selected Properties Quantity Dimensions Si Units English Units Gravitational Acceleration [g] = L/T 2 Length/time m/s ft/s 2 Mass M Kg Mass of a 1 lb object is 1/32.2 slug on earth Force F 1N = 1 kg m/s 2 lb (pound) (Newton) Energy or Work = Force x Distance [E] = F L 1 J = N m ftlb (Joule) Power = Work per unit time [P] = F L/T 1 W = Nm/s (Watt) 1 hp = 550 ftlb/sec Pressure [p] = F/L 2 Force/unit area Pa = 1 N/m 2 (Pascal) lb/in 2 = 144 lb/ft 2 2
3 The Principle of Dimensional Homogeneity In engineering and science, all equations must be dimensionally homogeneous, that is, each additive term in an equation must have the same dimensions. Consistent Units Not only must all fluid mechanics equations be dimensionally homogeneous, one must use consistent units; that is, each additive term must have the same unit. 3
4 Specific Weight γ = ρg [ N / m 3 ] or [ lbf / ft 3 ] Weight per unit volume 20 o C, 1 atm) γ water = (998 kg/m 3 )(9.807 m 2 /s) = 9790 N/m 3 [= 62.4 lbf/ft 3 ] γ air = (1.205 kg/m 3 )(9.807 m 2 /s) = 11.8 N/m 3 [= lbf/ft 3 ] 4
5 Specific Gravity Ratio of fluid density to density at STP 20 o C, 1 atm) SG liquid = ρ ρ liquid water ρliquid = 9790kg / m 3 SG gas = ρ ρ gas air = ρ gas 1.205kg / m 3 Water SG water = 1 Mercury SG Hg = 13.6 Air SG air = 1 5
6 Pressure Pressure is the (compression) stress at a point in a static fluid [p] = Force/unit area = F/L 2 Temperature Temperature is related to the internal energy level of a fluid. K o R = o = o C F
7 Ideal Gas Law Equation of state pv = nr T p = ρ RT, R = Rn / M n R n = universal gas constant M = molecular weight of the gas 7
8 Ideal Gas Equation of State Often used by ME s, occasionally used by CE s. Equation of state: p = ρ R T where p = absolute pressure (e.g. p = 0 = total vacuum, minimum) ρ = mass density R = gas constant (see Table A.4, p 817) K = absolute temperature (T = 0 o K is minimum) 8
9 Thermal Energy Specific heat, c: capacity of a substance to store thermal heat or the energy required to raise the temperature of a unit mass of a substance by one degree, (kj/kg o Cor kj/kg K) c v: specific heat under constant specific volume c p: specific heat under constant pressure k : ratio of specific heats (c p / c v ), it is often a constant (e.g. k = 1.4 for diatomic particles such as O 2, N 2 ) 9
10 Viscosity Newton's Law of Viscosity: very important (Eq p. 26) τ = μ (dv/dy) Says τ, the applied shear stress (force per unit area) is directly proportional to velocity gradient. Proportionality constant is μ dynamic viscosity, it is constant for Newtonian fluids (this course) Example: water. If it s not constant it is called a nonnewtonian fluid (not considered in this course). Example: blood 10
11 Viscosity Velocity gradient (dv/dy): how fast does V change for a given change in y (e.g. 1 mm of height): does V changes a little or a lot? y V A fluid deforms continuously Same velocity profile at each point in direction of movement. 11
12 Viscosity (continued) General Linear Velocity Relationship V = ay + b Apply bottom Boundary Condition (BC) V TOP V BOT = 0 V = 0 at y = 0 0 = a 0 + b results in a zero intercept, e.g. b = 0 So our general equation is becoming more specific V = ay The top BC V = V TOP at y = h applied in V = ay is: V TOP = a h a = V TOP / h h 12
13 Reynolds Number The primary parameter correlating the viscous behavior of all Newtonian fluids, a dimensionless parameter Re = ρvl VL = μ ν V and L are characteristic velocity and length scales of the flow. Low Re: Lamina Flow High Re: Turbulent Flow Kinematic viscosity ν = μ ρ 13
14 Surface Tension 14
15 Surface Tension 1. Liquid surface acts like stretched membrane 2. It sags because of higher attraction for glass & Water than air 3. Not many air molecules to attract above 4. no water molecules above to attract water air water Capillary action. For water on glass, water molecules are strongly attracted to glass, creeps up sides. It will creep to some point & stop when water is in equilibrium Look at ΣF z = 0, (forces in vertical direction water Surface tension is typically measured in dynes/cm, the force in dynes required to break a film of length 1 cm 15 F σ
16 F σ,v = σ l cosθ Weight of water Surface Tension (continued) What is holding up water? Surface Tension Force This Force is: F σ = σ l W = mg = (ρv) g = γv = γδha So ΣF v = 0 is: F s W = 0 Set terms equal: σ l cos θ = γ ΔhA (Eq. A) θ is very small so: cos θ ~ 1 σ = tension for water glass air a force per unit length l = length of contact cos θ = gets vertical component of force 16
17 Surface Tension Case #1: Circular Tube For circular tube of diameter d: l = πd A = 1/4 πd 2 in Eq. A: σ l = γδha σπd = γδh (1/4) πd 2 Solve for Δh Δh = (4σ) / (γd) Circular tube only Δh Physical significance: Larger diameter, smaller Δh 17
18 Surface Tension Case #2: Flat Plates Again ΣF v = 0 W + F S = 0 with F s cos θ ~ F s Δh W l Here W = mg = γ V = γ Δh A = γ Δh t l F S = σ length of contact = σ (2 l ) Δh is not a function of length of plates; solve for Δh Δh = 2σ γt 18
19 Chapter 2: Pressure Distribution in a Fluid 19
20 P = lim t Δ A 0 A. Some Basics = Δ F Δ A = df da We use the absolute pressure datum occasionally More often we use gage pressure (see Fig. 2.3 below) Read Chapter 2 to get a good sense of how fluid pressure works. 20
21 The fluid pressure acts equally in all directions at a point for ρ = constant. See proof using a wedge (Fig. 2.1, p. 64). dx, dy, dz 0 All p are same in magnitude. 21
22 Apply Newton s Second Law ΣF = ma For a = 0 ΣF = 0 In vertical ΣF z = 0 Choose fluid volume dx dy dz Three forces to consider: Pressure Forces on bottom & top, & W F B = p B A B = p B dx dy F T = p T A T = (p B + dp) dx dy (by inspection dp < 0) W = γ = γdx dy dz 22
23 Linear variation of pressure in vertical From Newton s Law Substituting these 3 forces into ΣF z = 0 F B F T W = 0 p B dx dy (p B + dp) dx dy  γ dx dy dz = 0 Multiple by 1 & move last term to RHS dp =  γdz dz dp/dz =  γ Eq Version 1 of linear variation For water, each foot up in z, the pressure decreases (hence minus sign) by 62.4 lbs/ft 2. To get another form, multiple by dz, integrate dp =  γ dz Assume γ is constant in z (ok for liquids) carry out integration p =  γ z + const or p 2 p 1 =  γ (z 2 z 1 )= γ (z 1 z 2 ) 23
24 Piezometric Head p1 γ 2 p 2 / γ p + z γ = constant 1 z 2 z 1 z = 0 Open Tank 24
25 Piezometric Head p 1 γ p 2 γ p + z γ = constant 1 z 1 2 p 3 γ z 2 Pressurized Tank 3 z 3 z = 0 25
26 Linear variation of pressure in vertical (Rectangular Volume cont.) p + γz = constant (Eq. 2.15) Version 2 Get third version of this rule move from point LHS stays the same so we can write 1 to 2 p 1 + γz 1 = p 2 + γz 2 (Eq. 2.14) Version 3 26
27 Linear variation of pressure in vertical (cont) Apply rules: (be sure all points connect & the fluid is not flowing. Point 1and 2 The rule is p + γz = constant Lets apply it Version 3:.1. 2 p 1 + γz 1 = p 2 + γz 2 = constant h 13 Because z 1 = z 2, p 1 = p 2 Point 1and 3 Also because p 1 + γz 1 = p 3 + γz 3 Solve for p 3 = p 1 + γ (z 1 z 3 ) Define Δz 13 = z 1 z 3 Define h 13 always as absolute value of Δz 13 to get: p 3 = p 1 + γ h
28 Pressure Variation Equation Version 4 of this expression is obtained by taking the pressure at a starting point, and adding or changing pressures due to heights of fluids: p end = p start + Σ (γh) down  Σ (γh) up 28
29 p end = p start + Σ (γh) down  Σ (γh) up What about gases? Example: Find pressure difference in air 1 meter apart. Apply Ver. 4 equation Density air & water p A = p B + γ air h 1 ρ air = 1.22 kg/m 3 B A h 1 = 1m So γ = air ρ air g = (1.22)(9.81) kg/m 3 m/s 2 = N/m 3 = kn/m 3 29
30 Gas pressure (cont) So Δp = p A p B = γ h 1 1m of air: = [ kn/m 3 ] (1m) = kn/m 2 1m of water: = [ kn/m 3 ](1m) = kn/m 2 Thus the pressure difference for 1m of air compared to water is (.012) / (9.799) =.001 = 0.1% Δp air is only about 1/10 of a percent of Δp w So for problems with liquids we can usually ignore gas weight, Do include when looking at atmosphere only, especially large distances 30
31 Example, What is the pressure of the air in the tank shown in the figure if l 1 = 40 cm, l 2 = 100 cm, l 3 = 80 cm? Main Equation: p end = p start + Σ (γh) down  Σ (γh) up Balance: Oil: except l 1 Mercury (Hg): except l 3 Air: Ignore Start Pt. S p s = 0 on top of Hg End Pt. A on top of oil; (p A unknown).. S 31
32 C. Forces on Flat Plates An arbitrary shape completely submerged in the liquid 32
33 Magnitude of Force 1. Setup coordinate system x, y with origin at the plate s centroid 2. Setup dummy coordinate ξ down from the surface in the plane of the plate An element area da with depth h p = pa + γh F = γ hda pda = ( pa + γ h) da = pa A+ h = ξ sinθ 1 ξ CG = ξda A 33
34 F = pa A+ γ sin θ ξda = pa A+ γ sinθξ CG A h CG = ξcg sinθ F = = p a A + γh A ( p + γh ) A = p A a CG CG CG The force on one side of any plane submerged surface in a uniform fluid equals the pressure at the plate centroid times the plate area, independent of the shape of the plate at the angle θ at which it is slant 34
35 As the pressure forces distribute on the surface of the plane plate, the resultant force F acts not through the centroid but below it toward the highpressure side. It line of action passes through the center of pressure CP of the plate. Find (x CP, y CP )? Sum moments of the elemental force pda about the centroid and equate to the moment of resultant F. Fy CP = ypda = = γ sinθ ( yda ) p a = 0 y yξda ( p + γξ sinθ ) a da (Because the origin of coordinate system is in centroid.) ξ = ξ CG y ( 2 Fy ) CP γ sinθ ξcg yda y da = γ sinθi xx = 0 = ( yda ) I xx  the area moment of inertia of the plate area about its centroidal x axis moment of inertia of area, or second moment of inertia 35
36 Fy CP ( 2 ξ ) CG yda y da = γ sinθi xx = γ sinθ y CP I = γ sinθ p xx CG A Use the sample principle x CP = γ sinθ p I xy CG A I xy = xyda I xy  the product of inertial of the plate When the shape is symmetry, x cp = 0, and the center of pressure lies directly below the centroid on the y axis Using gage pressure p = γ CG h CG F = p CG A = γh CG A y CP = Ixx sinθ h A CG x CP Ixy sinθ = h A CG 36
37 37
38 D. Forces on Curved Surfaces (Section 2.6 p. 84) Get formulas by observation. Then prove them by derivation. Horizontal Component of Force, F H What if I lowered a boatshaped piece of steel into water on string? Do you expect it to move suddenly sideways? No Therefore the horizontal force on left = horizontal force on right. 38
39 General rules 1. The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the component. 2. The vertical component of pressure force on a curved surface equals in magnitude and direct the weight of the entire column of fluid, both liquid and atmosphere, above the curved surface. F v = W 1 + W 2 + W 3 39
40 Example, 2.8 A dam has a parabolic shape z/z 0 = (x/x 0 ) 2 as shown in fig, with x 0 = 10 ft and z 0 = 24 ft. The fluid is water, γ = 62.4 lbf/ft 3, and atmospheric pressure may be omitted. Compute the forces F H and F V on the dam and their line of action. The width of the dam is 50 ft. 40
41 41
42 E. Hydrostatic Forces in Layered Fluids A single formula cannot solve the problem because the slop of the linear pressure distribution changes between layers. The total force on the plate does not equal the pressure at the centroid times the plate area, but the plate portion in each layer does satisfy the formula. F = ΣF i = Σp CGi A i The centroid of the plate portion in each layer can be used to locate the center of pressure on that portion y x CP CP ρig sinθii = p A CG ρig sinθii = p A CG i i i i xx xy i i 42
43 Example, 2.10 A Tank 20 ft deep and 7 ft wide is layered with 8 ft of oil, 6 ft of water, and 4 ft of mercury. Compute (a) the total hydrostatic force and (b) the resultant center of pressure of the fluid on the righthand side of the tank. 43
44 E. Buoyancy and Stability Archimedes principle 1. A body immersed in a fluid experiences a vertical buoyant force equation to the weight of the fluid it displaces. 2. A floating body displaces its own weight in the fluid in which it floats F B = F v (2) F v (1) = (fluid weight above 2) (fluid weight above 1) = weight of fluid equivalent to body volume F B F B = body( p2 p1) dah = γ ( z2 z1) dah = γ ( )( displaced volume) = floatingbody weight = γ ( )( body volume ) 44 (floating body)
45 F B ( )( displaced volume) = floating body weight = γ (floating body) The buoyant force and gravitational force are collinear since there are no net moments for a static equilibrium. 45
46 Example 1 Test the dimensional homogeneity of the boundarylayer x momentum equation: 46
47 Example 2 The belt moves at a steady velocity V and skims the top of a tank of oil of viscosity μ, as shown in the fig. Assuming a linear velocity profile in the oil, develop a simple formula for the required beltdrive power P as a function of (h, L, V, b, μ). What beltdrive power P, in watts, is required if the belt moves at 2.5 m/s over SAE 30W oil at 20 o C, with L = 2 m, b = 60 cm, and h= 3 cm? 47
48 Example 3 A solid cylindrical needle of diameter d, length L, and density ρ n may float in liquid of surface tension Y. Neglect buoyancy and assume a contact angle of 0 o. Derive a formula for the maximum diameter d max able to float in the liquid. Calculate d max for a steel needle (SG = 7.84) in water at 20 o C 48
49 Example 4 In the figure below, the tank contains water and immiscible oil at 20 o C. What is h in cm if the density of the oil is 898 kg/m 3? 49
50 Example 5 Gate AB in the figure is 15 ft long and 8 ft wide into the paper and is hinged at B with a stop at A. The water is at 20 o C. The gate is 1inthick steel, SG = Compute the water level h for which the gate will start to fall. 50
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