Chapter:1 Fluids & Properties Fluid Statics Fluid Dynamics.

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Transcript of Chapter:1 Fluids & Properties Fluid Statics Fluid Dynamics.

Page 1: Chapter:1 Fluids & Properties Fluid Statics Fluid Dynamics.

Chapter:1 Fluids & Properties

Fluid Statics Fluid Dynamics

Page 2: Chapter:1 Fluids & Properties Fluid Statics Fluid Dynamics.

Fluid Mechanics

Statics Fluid

Dynamic fluid

Kinetic fluid Kinematic fluid

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Density or Mass density: It is define as the ration of mass of fluid to its volume Unit: kg/m3

Water= 1000 kg/m3

Specific Weight or Weight density: It is the ration of Weight of fluid to its volume

Unit: 9.81 * 1000 N/ m3

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Specific Volume: It is the ration of Volume of fluid to its MassUnit: m3/kg

Specific Gravity: Specific gravity is defined as ration of Weight density of fluid to the weight density of standard fluid.

Density (ς) = S x 1000 kg/m3

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Ex 1 Calculate the Specific weight, density and Specific gravity of One liter of a liquid which weighs 7 N

Specific Weight (W) = Weight/Volume

Density(ς) = W/g

Specific Gravity = Density of Liquid/ Density of water

Ans. 7000N/m3, 731.5 kg/m3, 0.7135

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Ex 2 Calculate the density, specific weight and weight of one liter of Petrol of specific gravity 0.7

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Viscosity: Viscosity is defined as the property of a fluid which offers resistance to movement of one layer of fluid over another adjacent (adjoining) layer of the fluid.

dy

u + du

duu

u

yVelocity Profile

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Kinematic Viscosity: It is define as the ration between the dynamic viscosity and density of fluid

1 stoke = 1 cm2/s = 10-4 m2/s1 Centistoke = 1/100 stoke

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Types of Fluid:1) Ideal Fluid: A fluid, which is incompressible and is having no

viscosity, is known as an ideal fluid2) Real Fluid: A fluid, which having a viscosity, is known as real

fluid3) Newtonian fluid: A real fluid, in which the shear stress is

directly proportional to the rate of shear strain .4) Non-Newtonian Fluid: A real fluid, in which the shear stress is

not proportional to the rate of shear strain.5) Ideal Plastic Fluid: A fluid, in which shear stress is more than

the yield value and shear stress is proportional to the rate of shear strain

I F

N F

N N F

I P F

Shear Stress

Velocity Gradient

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Ex 2 A Plate 0.0025 mm distant from a fixed plate, moves at 60 cm/sAnd requires a force of 2N per unit area i.e. 2 N/m2 to maintain thisSpeed. Determine the fluid viscosity between the plates.

Ans: 8.33 x 10-4 poise

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Ex 3 Two horizontal plates are placed 1.25 cm apart, the space Between them being filled with oil of viscosity 14 poises. CalculateThe shear stress in oil if upper plate is moved with a velocity of2.5 m/s.

Ans: 280 N/m2

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Ex 4 Determine the specific gravity of a fluid having viscosity 0.05 Poise and kinematic viscosity 0.035 stokes.

Ans: ς=1428.5 kg/m3, S =1.43

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Ex 5 Determine the viscosity of a liquid having kinematic viscosity 6Stokes and specific gravity 1.9

Ans: μ =1.14 NS/m2

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Surface tension and Capillarity:Surface tension is defined as the tensile force acting on the surface of A liquid in contact with a gas or on the surface between two immiscibleLiquids such that the contact surface behaves like a membrane Under tension.

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Surface Tension on liquid Droplet:Let σ = Surface tension of the liquid p = Pressure intensity inside the droplet d = Dia. Of droplet.

i)Tensile force due to surface tension acting around the Circumference of the cut portion = σ x Circumference = σ x П d (1)ii) Pressure force on the Area = p x П/4 d2 (2)

Under the Equilibrium Condition 1 =2 p = 4 σ /d

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ii) Surface Tension on a Hollow Bubble:

iii) Surface Tension on a Liquid jet: p = Pressure intensity inside the liquid jet above the outside pressure σ = Surface tension of the lquid.

Force due to pressure = p x area of semi jet = p L dForce due to surface tension = σ 2LEquating the firce , we have

P L d = σ 2 L

p x П/4 d2 = 2 (σ x Пd) p = 8 σ/ d

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Ex 6 The surface tension of water in contact with air at 20o C is 0.00725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm2 greater than the outside pressure.Calculate the Dia. of the droplet of water.

p = 4 σ /d

Ans. 1.45 mm

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Ex 6 Find the surface tension in a soap bubble of 40 mm Dia. whenThe inside pressure is 2.5 N/m2 above atmospheric pressure.

p x П/4 d2 = 2 (σ x Пd) p = 8 σ/ d

Ans. σ = 0.0125 N/m

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Capillarity:Capillarity is defined as a phenomenon of rise or fall of a liquid Surface in a small tube relative to the adjacent general level of Liquid when the tube is held vertically in the liquid.

Expression for Capillary Rise:Let d = diameter of small tube h = height of the liquid in the tube σ = Surface tension of liquid θ = Angle of contact bet. Liquid & Gass

Weight of liquid of height h in tube

(1)

Vertical Component Sur. Tensile For. (2)

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(1) = (2)

Value of θ water = 0

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Expression for Capillary Fall:If the glass tube is dipped in mercury, the levelOf mercury in the tube will be lower than theGeneral level of the outside liquid.

1) Surface tension downward=

2) Hydrostatic force upward =

(1) = (2)

Value of θ = 128o-130o (Mercury & Glass tube)

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Ex Calculate the capillary rise in a glass tube of 2.5 mm dia. When Immersed vertically ina) Waterb) Mercury Take surface tension σ = 0.0725 N/m for water and σ = 0.52 N/m for mercury in contact with airThe specific gravity for mercury is given as 13.6 and angle =130o

Ans. A) h = 1.18 cm b) h = - 0.4 cm

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Ex The capillary rise in the glass tube is not exceed 0.2 mm of waterDetermine its minimum size, given that surface tension for water inContact with air = 0.0725 N/m

Ans. D = 14.8 cm

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Ex Calculate the capillary effect in millimeter in a glass tube of 4mmDia., when immersed in 1) water 2) mercuryThe temperature of the liquid is 20o C and the values of the surfaceTension of water and mercury at 20oC in contact with air 0.073 N/mAnd 0.51 N/m respectively. The angle of contact for water is zeroThat for mercury 130o. Take density of water at 20oC as equal to 998 kg/m3.

Ans. A) h = 7.51mm b) h = - 2.46 mm

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Vapour Pressure:Vapor pressure is the pressure at which the liquid is converted intoVapours, at given temperature.

Cavitation:The pressure developed by the collapsing bubbles is so high that theMaterial from the adjoining boundaries gets eroded and cavities areFormed on them. This phenomenon is known as cavitations

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Prepared by, Dr Dhruvesh Patel

Prepared by, Dr Dhruvesh Patel