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Vatten(byggnad)

Vätskors egenskaper (1)Hydrostatik (3)Grundläggande ekvationer (5)Rörströmning (4)

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

Vätska som kontinuerligt medium

Densitet

Kompressibilitet

Viskositet

Ytspänning, kapillaritet

Ö

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Övningstal: B1, B2, B5 och B7

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FLUID AS A CONTINUUM

A fluid is considered to be a continuum in hi h h h l id which there are no holes or voids ⇒

velocity, pressure and temperature fields are continuous.

Validity criteria: Smallest length scale in a

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Validity criteria: Smallest length scale in a flow >> average spacing between molecules composing the fluid.

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DENSITY (ρ)Mass/ unit volume (kg/m3)

Density decreases normally with increasing temperatureρwater = ρ(T,S,p)

i.e., dependent on- Temperature

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- Salt content (ρ ≈ 1000 + 0.741⋅S, S in per mille;

S = 3.5% in ocean ⇒ ρ = 1026 kg/m3)

- Pressure (but only a small variability)

OTHER DEFINITIONS Weight = mass × gravity acceleration

(W = mg, [N = kg⋅m/s2]) (Eqn. 1.4)

Weight density (“tunghet”) (or specific weight)= g y ( g ) ( p g )density × gravity acceleration

(w = ρg, [N/m3 = kg⋅/(m2s2)]) (Eqn. 1.6)

Specific volume ν = reciprocal of density(ν = 1/ρ, [m3/kg])

Relative density (or specific gravity) s is the density

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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/ρwater (often = ρ/1000) (Eqn. 1.7)

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Example – density.

The specific weight of water at ordinary temperature and pressure is 9.81 kN/m3. The specific gravity of mercury is 13.56. Compute the density of water and the specific weight and density of mercury

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specific weight and density of mercury.

COMPRESSIBILITY

All fluids can be compressed by application of pressure ⇒ elastic energy being storedpressure ⇒ elastic energy being stored

Modulus of elasticity (“elastitetsmodul”) describes the compressibility properties of the fluid and is defined on the basis of volume

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volume

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Modulus of elasticity:

E = -dp/(dV/V1) [Pa]

For liquids, region of engineering interest is when V/V1 ∼ 1 ⇒pV Δ

−≈Δ

Ewater ~ 2⋅109 Pa (function of temperature)EV

≈

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B1 What pressure must be applied to water to reduce its volume 1 % ?

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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 10.05 kN/m3 and that the average volume modulus of elasticity is 2.34*109 N/m2 for the pressure range.

A) What will be the change in specific volume b t th t t th f d t th t d th?

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between that at the surface and at that depth?B) What will be the specific volume at that depth?C) What will be the specific weight at that depth?

IDEAL FLUIDA fluid in which there is no friction

REAL FLUIDA fluid in which shearing forces always exist whenever motion takes place due to the fluid’s inner friction – viscosity.

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VISCOSITYViscosity is a measure of a fluid’s “inner friction” or resistance to shear stress. friction or resistance to shear stress.

It arises from the interaction and cohesion of fluid molecules.

All fluids posses viscosity, but to a varying

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

DEFINITION OF DYNAMIC VISCOSITY - μ

yy

Shearing of thin fluid film between two plates. The upper plate has an area A.

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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

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Introduction of the proportionality constant μ, named dynamic viscosity, gives Newton’s viscosity law shear force (“skjuvspänning”):

dvVF (Eqn 4 1 4 2) N/m2

μ [Pa⋅s or kg/ms] - Dynamic viscosity

ν = μ/ρ [m2/s] - Kinematic viscosity

dyhAμμτ === (Eqn. 4.1-4.2) N/m2

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ν μ/ρ [m /s] Kinematic viscosity

No-slip condition – water particles adjacent to solid boundary has zero velocity (observational fact)

μ (Pa·s)μ (Pa s)

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Implication of viscosity: a fluid cannot sustain a shear stress without deformation

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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

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in which viscous action is strong

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Laminar flow: smooth, orderly motion in which fluid elements appears to slide over each other in layers (little exchange b t l )between layers).

Turbulent flow: random or chaotic motion of individual fluid particles, and rapid mixing and exchange of these particles through the flow

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Turbulent flow is most common in nature.

Newtonian – non-Newtonian fluidsExamples non-Newtonian fluids:

Plastics, blood, suspensions, paints, foods

Shear vs. rate of strain re-lations for non-Newtonian

ii dy

duττμττ >=− ,

fluids:

Bingham plastic

n>1: Shear-thickening fluid,

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n>1: Shear thickening fluid, n<1: Shear-thinning fluid

n

dy

du)(μτ =

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Example – viscosity.

A 3 id b t t l 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 m2 area through the oil at a velocity of 0 1 m/s if the plate is 1 cm from one

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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

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sides is 29 N. Assuming viscous flow, and neglecting all end effects, calculate the viscosities of the oils.

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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.

(T t ti l f T F

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(Torque = rotational force, T = F · rangular velocity ω = V/r)

SURFACE TENSION (“ytspänning”), CAPILLARITY

Surface tension effects occur at liquid surfaces (interfaces of liquid-liquid, liquid-gas, liquid-solid).

Surface tension effects are often negligible in engineering problems. Exceptions:1. Bubble formation2. Capillary rise of liquids in narrow spaces

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3. Break-up of liquid drops4. Formation of liquid drops5. Investigations using small physical models

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Surface tension, σ [N/m], is thought of as a molecular force in the liquid surface. Surface tension decreases with temperature and is

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pdependent 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 tension1) Capillary rise/drop

Contact angle θ between glassand water Vertical force balanceand water. Vertical force balancebetween surface tension forceand weight of water column gives

σ · 2π r · cosθ = ρ g · h · π r2 →

h = (2 σ · cosθ)/ (ρ g · r)

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(valid if r<2.5 mm)

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Angle of contact θ

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θ depends on relation between cohesive and adhesive forces.

2) For spherical droplet

Balance between internal pressure force and surfacepressure force and surfacetension force:

p · πR2 = 2πR · σ →

p = 2 · σ /R

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3) For bubble

Balance between internalpressure force and surfacetension force:

p · πR2 =2 · 2πR · σ →

p = 4 · σ /R

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Measurement of surface tension

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F = 2σπD ⇒ σ = F/(2πD)

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B7 Calculate the maximum capillary B7 Calculate the maximum capillary rise of water (T = 20°C) to be expected in a vertical glass tube, diameter 1 mm.

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Tal – Delprov 1, 010921

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

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att sjunka? Ytspänningen för dammens vatten är 0.0728 N/m.