Ch1 Introduction

INTRODUCTION

Characteristics of fluids • A fluid may be liquid, vapour or gas. It has no permanent shape but takes up the shape of a containing vessel or channel or is shaped by external forces (e.g. the atmosphere). • A fluid consists of atoms/molecules in random motion (translation) and in continual collision with the surroundings. • Fluids are readily deformable, and flow. • Solids have ‘frozen’ molecules that vibrate and do not translate. Solids resist change of shape. • A fluid is defined as a substance that deforms continuously when acted on by a shearing stress of any magnitude.

Characteristics of fluids • For a solid, application of a shear stress causes a deformation which, if modest, is not permanent and solid regains original position.

Attached plates

Solid

Characteristics of fluids • For a fluid, continuous deformation takes place with an infinite number of layers sliding over each other. Deformation continues until the force is removed.

Fluid

• A fluid is a substance for which a shear stress tends to produce unlimited deformation.

Dimensions and Units • Fluid characteristics are described qualitatively in terms of basic dimensions: length, L, time, T, and mass, M. •

All theoretically derived equations are dimensionally homogeneous.

• For a quantitative description units are required • Two system of units will be used: – International System (SI); m, s, kg, K – British Gravitational (BG) System; ft, s, lb, ºF or ºR

Properties and characteristics of fluids • From a microscopic point of view a fluid is not a continuous and homogeneous substance. • We take the engineering macroscopic view such that we can examine a sufficiently large ‘particle’ of fluid to allow the concept of velocity and density ‘at a point’. • Density is the mass per unit volume, and is a macroscopic concept. Density at a point is:

 m ρ = lim   v →0 V  • Specific volume, specific weight, specific gravity

Properties and characteristics of fluids • Pressure = (normal force) / area. The pressure at a point is:

F  p = lim   A→0  A • In the absence of shear forces (fluid at rest or in uniform motion) pressure at a point is independent of direction

Properties and characteristics of fluids • Perfect gas law – In this course all gases obey the perfect gas law

pV = mRT

or

p = ρRT

Viscosity • For elastic solids shearing strain is proportional to the shearing stress • For fluids shearing stress is proportional to the rate of shearing strain

• For Newtonian fluids shearing stress is linearly proportional to the rate of shearing strain • The study of non-Newtonian fluids is called rheology • Viscosity is very sensitive to temperature

Example 1.5: The velocity distribution for the flow of a Newtonian fluid between two wide, parallel plates is given by the equation

3V u 2



2  y   1     h   

where V is the mean velocity. The fluid has a viscosity of 0.0r lb·s/ft2. When V = 2 ft/s and h = 0.2 in. determine: (a) the shearing stress acting on the bottom wall, and (b) the shearing stress acting on a plane parallel to the walls and passing through the centerline (midplane)

Example 1.5: The velocity distribution for the flow of a Newtonian fluid between two wide, parallel plates is given by the equation

3V u 2



2  y   1     h   

where V is the mean velocity. The fluid has a viscosity of 0.0r lb·s/ft2. When V = 2 ft/s and h = 0.2 in. determine: (a) the shearing stress acting on the bottom wall, and (b) the shearing stress acting on a plane parallel to the walls and passing through the centerline (midplane) Solution.

du dy

Shearing stress

 

Velocity distribution

du 3Vy  2 dy h

(a) Along the bottom wall, y = -h

du 3V  dy h

shearing stress

 bot wall  14.4 lb/ft 2

(b) Along the midplane, y = 0

du 0 dy

shearing stress

 midplane  0

Properties and characteristics of fluids • Compressibility: all fluids are compressible, especially gases. Most liquids can be regarded as incompressible for most purposes. • The bulk modulus of elasticity, Ev , is a property which is used to account for compressive effects:

 p  Ev         • Speed of sound is the velocity at which small disturbances propagate in a fluid. For ideal gases speed of sound:

c  kRT

Vapor Pressure • Vapor pressure is a pressure exerted by a vapour on the fluid when they are in equilibrium in a closed vessel • Vapor pressure is a function of temperature • A liquid boils when the pressure is reduced to vapor pressure • When the liquid pressure is dropped below the vapor pressure due to flow phenomena, we call the process cavitation • Cavitation is the formation and subsequent collapse of vapor bubbles in a flowing fluid

Surface Tension •

Liquid, being unable to expand freely, will form an interface with a second liquid or gas

•

This surface phenomenon is due to unbalanced cohesive forces acting on the liquid molecule on the fluid surface

•

The intensity of molecular attraction per unit length along any line in the surface is called the surface tension coefficient, σ (N/m)

•

The value of surface tension decreases as temperature increases

•

If the interface is curved, then there is a pressure difference across the interface, the pressure being higher on the concave side ( drop of fluid, bubble)

•

Capillary action in small tubes, which involves a liquid-gas solid interface, is also caused by surface tension

Effect of capillary action in small tubes. (a) Rise of column for a liquid that wets the tube. (b) Free-body diagram for calculating column height. (c) Depression of column for a nonwetting liquid. The height h is governed by the value of the surface tension, σ, tube radius, R, specific weight of the liquid, γ, and the angle of contact, θ.

Effect of capillary action in small tubes. (a) Rise of column for a liquid that wets the tube. (b) Free-body diagram for calculating column height. (c) Depression of column for a nonwetting liquid. The height h is governed by the value of the surface tension, σ, tube radius, R, specific weight of the liquid, γ, and the angle of contact, θ.

 R 2 h  2 R cos  h

2 cos  R

Example 1.8: Pressures are sometime determined by measuring the height of a column of liquid in a vertical tube. What diameter of clean glass tubing is required so that the rise of water at 20ºC in a tube due to capillary action (as opposed to pressure in the tube) is less than 1.0 mm?

Example 1.8: Pressures are sometime determined by measuring the height of a column of liquid in a vertical tube. What diameter of clean glass tubing is required so that the rise of water at 20ºC in a tube due to capillary action (as opposed to pressure in the tube) is less than 1.0 mm? Solution

h

2 cos  R

R

2 cos  h

For water at 20ºC (from Table B.2), σ = 0.0728 N/m and γ = 9.789 kN/m3. Since θ ≈ 0º it follows that for h = 1.0 mm,

R

 9.789 10

2  0.0728 N/m   1 3

N/m

3

  1.0 mm   10

3

m/mm

And the minimum required tube diameter, D, is

D  2 R  0.0298 m  29.8 mm



 0.0149 m

Problem 1.65: A 12-in.-diameter circular plate is placed over a fixed bottom plate with a 0.1-in. gap between the two plates filled with glycerin. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible.

Problem 1.65: A 12-in.-diameter circular plate is placed over a fixed bottom plate with a 0.1-in. gap between the two plates filled with glycerin. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible.

Solution Torque due to shearing stress on plate:

dT  r dA where

dA  2 rdr Thus dT  r 2 rdr

and R

T  2  r 2 dr 0

Problem 1.65: A 12-in.-diameter circular plate is placed over a fixed bottom plate with a 0.1-in. gap between the two plates filled with glycerin. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible.

Solution Torque due to shearing stress on plate: R

T  2  r 2 dr 0

Velocity distribution

du V r   dy   Searing stress

 

du r  dy 

Problem 1.65: A 12-in.-diameter circular plate is placed over a fixed bottom plate with a 0.1-in. gap between the two plates filled with glycerin. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible.

Solution Torque due to shearing stress on plate: R

T  2  r 2 dr 0

Searing stress

 

du r  dy 

Torque

2 R 3 2 R 4 T r dr   0   4

Problem 1.65: A 12-in.-diameter circular plate is placed over a fixed bottom plate with a 0.1-in. gap between the two plates filled with glycerin. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible.

Solution Torque

4

lb s   rev   rad   1 min   6  2  0.0313 2 2        2 R 4 ft 2   min   rev   60 s   12 ft   T   0.0772 ft lb  4  0.1  ft  4   12   

END OF LECTURE 

Dimensions Associated with Common Physical Quantities

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Density of water as a function of temperature back

(a) Deformation of material placed between two parallel plates. (b) Forces acting on upper plate.

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Behavior of a fluid placed between two parallel plates

du   dy

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Linear variation of shearing stress with rate of shearing strain for common fluid back

Variation of shearing stress with rate of shearing strain for several types of fluids, including common non-Newtonian fluids. back

Dynamic (absolute) viscosity of some common fluids as a function of temperature

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Forces acting on one-half of a liquid drop Pressure drop across the surface of the droplet

2 R  p R 2 2 p  pi  pe  R

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