Vatten(byggnad) VVR145 Vatten. 2. Vätskors egenskaper (1.1, 4.1 och 2.8) (Föreläsningsanteckningar)

Size: px

Start display at page:

Download "Vatten(byggnad) VVR145 Vatten. 2. Vätskors egenskaper (1.1, 4.1 och 2.8) (Föreläsningsanteckningar)"

George Harvey
7 years ago
Views:

1 Vatten(byggnad) Vätskors egenskaper (1) Hydrostatik (3) Grundläggande ekvationer (5) Rörströmning (4) 2. Vätskors egenskaper (1.1, 4.1 och 2.8) (Föreläsningsanteckningar) Vätska som kontinuerligt medium Densitet Kompressibilitet Viskositet Ytspänning, kapillaritet Övningstal: B1, B2, B5 och B7 1

8) (Föreläsningsanteckningar) Vätska som kontinuerligt medium Densitet

2 FLUID AS A CONTINUUM A fluid is considered to be a continuum in which h there are no holes or voids velocity, pressure and temperature fields are continuous. Validity criteria: Smallest length scale in a flow >> average spacing between molecules composing the fluid. 2

3 DENSITY (ρ) Mass/ unit volume (kg/m 3 ) Density decreases normally with increasing temperature ρ water = ρ(t,s,p) i.e., dependent on - Temperature - Salt content (ρ S, S in per mille; S = 3.5% in ocean ρ= 1026 kg/m 3 ) - Pressure (but only a small variability) OTHER DEFINITIONS Weight = mass gravity acceleration (W = mg, [N = kg m/s 2 ]) (Eqn. 1.4) Weight density ( tunghet ) )(or specific weight)= density gravity acceleration (w = ρg, [N/m 3 = kg /(m 2 s 2 )]) (Eqn. 1.6) Specific volume ν = reciprocal of density (ν = 1/ρ, [m 3 /kg]) Relative density (or specific gravity), s, is the density normalized with the density of water at a specific temperature and pressure (normally 4 C and atmospheric pressure): s = R.d. = ρ/ρ water (often = ρ/1000) (Eqn. 1.7) 3

1.6) Specific volume ν = reciprocal of density (ν = 1/ρ, [m 3 /kg]) Relative density (or specific gravity), s, is the density normalized with the density of water at a specific temperature and

4 Example density. The specific weight of water at ordinary temperature and pressure is 9.81 kn/m 3. The specific gravity of mercury is Compute the density of water and the specific weight and density of mercury. COMPRESSIBILITY All fluids can be compressed by application of pressure elastic energy being stored Modulus of elasticity ( elastitetsmodul ) describes the compressibility properties of the fluid and is defined on the basis of volume 4

Compute the density of water and the specific weight and density of mercury.

5 Modulus of elasticity: E = -dp/(dv/v 1 ) [Pa] For liquids, region of engineering interest is when V/V 1 1 ΔV Δp V E E water ~ Pa (function of temperature) B1 What pressure must be applied to water to reduce its volume 1 %? 5

ΔV Δp V E E water ~ 2 10 9 Pa (function of temperature)

6 Example compressibility. At a depth of 8 km in the ocean the pressure is 81.8 MPa. Assume that the specific weight of sea water at the surface is kn/m 3 and that the average volume modulus of elasticity is 2.34*10 9 N/m 2 for the pressure range. A) What will be the change in specific volume between that t at the surface and at that t depth? B) What will be the specific volume at that depth? C) What will be the specific weight at that depth? IDEAL FLUID A fluid in which there is no friction REAL FLUID A fluid in which shearing forces always exist whenever motion takes place due to the fluid s inner friction viscosity. 6

A) What will be the change in specific volume between that t at the surface and at that t depth? B) What will be the specific volume at that depth?

7 VISCOSITY Viscosity is a measure of a fluid s inner friction or resistance to shear stress. It arises from the interaction and cohesion of fluid molecules. All fluids posses viscosity, but to a varying degree. For instance, syrup has a considerably higher viscosity than water. DEFINITION OF DYNAMIC VISCOSITY - μ y Shearing of thin fluid film between two plates. The upper plate has an area A. Experiments have shown that for a large number of fluids: F ~ AV/h (if V and h not too large) Linear velocity profile V/h = dv/dy 7

For instance, syrup has a considerably higher viscosity than water.

8 Introduction of the proportionality constant μ, named dynamic viscosity, gives Newton s viscosity law shear force ( skjuvspänning ): F V dv τ = = μ = μ (Eqn. 4142) ) N/m 2 A h dy μ [Pa s or kg/ms] - Dynamic viscosity ν = μ/ρ [m 2 /s] - Kinematic viscosity No-slip condition water particles adjacent to solid boundary has zero velocity (observational fact) μ (Pa s) 8

2) N/m 2 A h dy μ [Pa s or kg/ms] - Dynamic viscosity ν = μ/ρ [m 2 /s] - Kinematic viscosity

9 Implication of viscosity: a fluid cannot sustain a shear stress without deformation Implications of Newton s law: τ, μ independent of pressure (in contrast to solids) no velocity gradient no shear stress Restriction of Newton s law: law only valid if the fluid flow is laminar in which viscous action is strong 9

contrast to solids) no velocity gradient no shear stress Restriction of

10 Laminar flow: smooth, orderly motion in which fluid elements appears to slide over each other in layers (little exchange between layers). Turbulent flow: random or chaotic motion of individual fluid particles, and rapid mixing and exchange of these particles through the flow Turbulent flow is most common in nature. Newtonian non-newtonian fluids Examples non-newtonian fluids: Plastics, blood, suspensions, paints, foods Shear vs. rate of strain relations for non-newtonian fluids: Bingham plastic du τ τ = μ, τ > τ i dy i n>1: Shear-thickening fluid, n<1: Shear-thinning fluid du n τ = μ( ) dy 10

flow is most common in nature. Newtonian non-newtonian fluids Examples non-newtonian fluids: Plastics, blood, suspensions, paints, foods Shear vs.

11 Example viscosity. A space, 3 cm wide, between two plane horizontal surfaces is filled with SAE 30 Western lubricating oil at 20 C. What force is required to drag a very thin plate of 1 m 2 area through the oil at a velocity of 0.1 m/s if the plate is 1 cm from one surface? B2 A very large thin plate is centred in a gap of width 0.06 m with different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow, and neglecting all end effects, calculate the viscosities of the oils. 11

B2 A very large thin plate is centred in a gap of width 0.06 m with different oils of unknown viscosities above and below; one viscosity is twice the other.

12 B5* A circular disc of diameter d is slowly rotated in a liquid of large viscosity m at a small distance h from a fixed surface. Derive an expression for the torque T necessary to maintain an angular velocity ω. Neglect centrifugal effects. (Torque = rotational ti force, T = F r angular velocity ω = V/r) SURFACE TENSION ( ytspänning ), CAPILLARITY Surface tension effects occur at liquid surfaces (interfaces of liquid-liquid, liquid-gas, liquidsolid). Surface tension effects are often negligible in engineering problems. Exceptions: 1. Bubble formation 2. Capillary rise of liquids in narrow spaces 3. Break-up of liquid drops 4. Formation of liquid drops 5. Investigations using small physical models 12

(Torque = rotational ti force, T = F r angular velocity ω = V/r) SURFACE TENSION ( ytspänning ), CAPILLARITY Surface tension effects occur at liquid surfaces (interfaces of

13 Surface tension, σ [N/m], is thought of as a molecular force in the liquid surface. Surface tension decreases with temperature and is dependent on the contact fluid (surface tension usually quoted in contact with air). The surface tension force will support small loads if liquid surface is curved. Implications of surface tension 1) Capillary rise/drop Contact angle θ between glass and water. Vertical force balance between surface tension force and weight of water column gives σ 2π r cosθ = ρ g h π r 2 h = (2 σ cosθ)/ (ρ g r) (valid if r<2.5 mm) 13

The surface tension force will support small loads if liquid surface is curved.

14 Angle of contact θ θ depends on relation between cohesive and adhesive forces. 2) For spherical droplet Balance between internal pressure force and surface tension force: p πr 2 = 2πR σ p = 2 σ /R 14

15 3) For bubble Balance between internal pressure force and surface tension force: p πr 2 =2 2πR σ p = 4 σ /R Measurement of surface tension F = 2σπD σ = F/(2πD) 15

16 B7 Calculate the maximum capillary rise of water (T = 20 C) to be expected in a vertical glass tube, diameter 1 mm. Tal Delprov 1, En insekt hålls uppe på ytan av en damm av ytspänningen (vattnet väter inte insektens ben), se Figur. Insekten har sex ben och varje ben är i kontakt med vattnet över en längd av 5 mm. Vad är den maximala massan (i gram) för insekten om den skall undvika att sjunka? Ytspänningen för dammens vatten är N/m. 16

ben), se Figur. Insekten har sex ben och varje ben är i kontakt med vattnet över en längd av 5 mm.

Fluid Mechanics: Static s Kinematics Dynamics Fluid

Fluid Mechanics: Static s Kinematics Dynamics Fluid Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three