CE 213 - Fluid Mechanics
Capillary Effect, Compressibility and Pascal’s Law
Bachu Anilkumar
Department of Civil and Environmental EngineeringIIT Patna
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
1Learning Objectives
Surface tension and few special casesCapillary effectCompressibilityPascal’s law
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
2Surface Tension
Surface tension, σ =FL
(1)
Points to rememberThat surface tension should occur only whenthere is a surface of separation between twofluids
Surface tension of the liquid is not theproperty of the liquid itself, it is a binaryproperty of the liquid and the gas.
Interfacial tension
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
3Surface TensionCase 1: Fluid on one Side and air on another Side of a thin fluid film
pi and po = Pressure on its convex and concave sides respectively
2σr2dθ2 sin(dθ1
2) + 2σr1dθ1 sin(
dθ2
2) = (pi − po)r1r2dθ1dθ2 (2)
pi − po = ∆p = σ
(1r1
+1r2
)(3)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
4Surface TensionCase 2: Air on both sides of a thin liquid film
2x2σr2dθ2 sin(dθ1
2) + 2x2σr1dθ1 sin(
dθ2
2) = (pi − po)r1r2dθ1dθ2 (4)
pi − po = ∆p = 2xσ(
1r1
+1r2
)(5)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
5Surface TensionSpecial Cases
Case i: Spherical Liquid Drop
The radius of curvature is same in all directionsr1 = r2 = r i.e. the radius of the dropInteracting with air from only one side
∆p = σ
(1r
+1r
)=
2σr. (6)
Case ii: Spherical Bubble
The radius of curvature is same in all directionsr1 = r2 = rInteracting with air from both sides (inside the bubble and outside the bubble).
∆p = 2σ(
1r
+1r
)=
4σr. (7)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
6Surface TensionSpecial Cases
Case iii: Cylindrical Liquid Jet
The radius of curvature is r is one direction and∞ in another directionr1 = r and r2 =∞
∆p = σ
(1r
+1∞
)=σ
r. (8)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
7Capillary Effect
The tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube
Examples
Rise of kerosene through a cotton wick inserted into the reservoir of a kerosene lampRise of water to the top of tall trees
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
8Capillary EffectContact angles
Interface Contact angleMercury–glass 140Water–glass 0Water–paraffin 107Water–silver 90Organic liquids (most)–glass 0Ethyl alcohol–glass 0Kerosene–glass 26
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
9Capillary Effect
Capillary effect can be explainedmicroscopically by considering cohesiveforces (the forces between likemolecules, such as water and water)and adhesive forces (the forces betweenunlike molecules, such as water andglass).The relative magnitudes of these forcesdetermine whether a liquid wets a solidsurface or not.
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
10Capillary Effect
W = ~Fsurface (9)
W = mg = ρVg = ρg(πR2h) (10)
W = ~Fsurface → ρg(πR2h) = 2πRσscosφ
Capillary rise,h =2σs
ρgRcosφ (11)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
11Capillary Effect
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
12Compressibility
Volume of a fluid changes with a changein its temperature or pressure.Expansion - Heated up or depressurizedContraction - Cooled down orpressurizedFluids act like elastic solids with respectto pressure
Bulk modulus of elasticity,E = lim∆v→0
(−∆P
∆vv
)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
13Compressibility
Bulk modulus of elasticity,E = lim∆v→0
(−∆P
∆vv
)= −v
dpdv
For a given mass of body, ρv = m (constant)
dρρ
+dvv
= 0→ −dvv
=dρρ
E = −vdpdv
= ρdpdρ
(12)
Interpretation:The fluids for which the k is high, it requires very high pressure to change a definiteamount of density of fluid.For a given change in pressure, the change in volume will be very small, if the valueof k is very high
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
14Compressibility
Ewater = 2x105 KN/m2
Eair = 101 KN/m2
Air is 104 times compressible than water at atmospheric conditions.
All incompressible fluids exhibit incompressible flow in nature ?
All compressible fluids exhibit compressible flow in nature ?
When to consider the compressibility of fluid/ What is the criteria to decide whether thefluid is compressible or incompressible - Based on Mach number of the flow
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
15Compressibility
Criteria: Considering whether the change in pressure brought about by the fluid motioncauses large change in volume or density
Bernoulli equation:
O(∆p) = O(ρv2
2)
From Eq. (12),
E = ρdpdρ→ dρ
ρ=
dpE,
dρρ
=ρv2
2E→ dρ
ρ=
12
v2
Eρ
,
W.K.T. the velocity of sound, a =√
E/ρ
∴dρρ
=12
v2
a2 →12
Ma2
Ma is the ratio of velocity of flow to the acousticvelocity in the flowing medium
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
16Compressibility
Considering a maximum relative change in density of 5% a criterion of an incompressibleflow,
dρρ< 0.05→ 1
2Ma2 < 0.05→ Ma < 0.33
CriterionMa < 0.33, incompressible flow else compressible flow
For air at STP conditions, a = 335m/s
A Mach number of 0.33 corresponds to a velocity of 110 m/s.
∴ Flow of air up to a velocity of 100 m/s under standard condition can be considered asincompressible flow.
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
17Compressibility
FLUID STATICS
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
18Fluid Statics
A fluid element, in isolation from its surroundings, will experience two types of forces:
Body forcesForces act throughout the body of fluid and distributed over entire massCaused by external factors: gravitational, electro magnetic, or electro static fields
Surface forcesForces exerted its surroundings through direct contact at the surface.Two components:
Normal force - component along the normal to the areaShear force - along the plane of the area.
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
19Fluid StaticsNormal Stresses in a Stationary Fluid
Assumption: Fluid is at restNo shear stresses and tensile stressesonly normal forces - compressive in nature
Equations of static equilibrium
ΣFx = σx
(∆y∆z
2
)− σn∆Acosα = 0.
ΣFy = σy
(∆z∆x
2
)− σn∆Acosβ = 0.
ΣFz = σz
(∆x∆y
2
)−σn∆Acosγ−ρg
6(∆x∆y∆z) = 0.
ΣFx ,ΣFy ,ΣFz - net forces acting on fluid elementcosα, cosβ, cosγ - direction cosines of the normal to the inclined plane
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
20Fluid StaticsNormal Stresses in a Stationary Fluid
∆Acosα = ∆y∆z2
∆Acosβ = ∆x∆y2
∆Acosγ = ∆y∆x2
Upon equating and substituting,
σx = σy = σz = σn (13)
Anil (IITP) | CE 213 - Fluid Mechanics | Lecture - 3 | August 7, 2019
21Pascal’s Law
Pascal’s lawNormal stress at any point in a fluid atrest are
Directed towards the point from alldirectionsEqual magnitude
σx = σy = σz = −p (14)
where p is a scalar quantity, defined ashydrostatic/fluid static/thermodynamicpressure
THANK YOU !!