Ae 233 (chapter 1) Fluid Mechanics For Chemical Engineering.ppt

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FLUID MECHANICS FOR
CHEMICAL ENGINEERING
Chapter 1: Fluid Mechanics and
Fluid Properties
SEQUENCE OF CHAPTER 1
Introduction
Objectives
1.1 Definition of A Fluid
Shear stress in moving fluid
Differences between liquid and gases
Newtonian and NonNewtonian Fluid
1.2 Engineering Units
1.3 Fluid Properties
Vapor Pressure
Engineering significance of vapor pressure
Surface Tension
Capillarity Example 1.2
Example 1.3
Summary
Introduction
• Fluid mechanics is a study of the behavior of fluids,
either at rest (fluid statics) or in motion (fluid
dynamics).
• The analysis is based on the fundamental laws of
mechanics, which relate continuity of mass and energy
with force and momentum.
• An understanding of the properties and behavior of
fluids at rest and in motion is of great importance in
engineering.
Define the basic fluid properties. and solve problems using their relationships. Identify the relationships between specific weight. 3. 4.Objectives 1. 2. specific gravity and density. Properly set up equations to ensure consistency of units. . force and mass. Identify the units for the basic quantities of time. length.
1 Definition of Fluid • Fluid mechanics is a division in applied mechanics related to the behaviour of liquid or gas which is either in rest or in motion. • Fluid can be defined as a substance which can deform continuously when being subjected to shear stress at any magnitude. . it can flow continuously as a result of shearing action. • The study related to a fluid in rest or stationary is referred to fluid static. In other words. This includes any liquid or gas. otherwise it is referred to as fluid dynamic.1.
1 Definition of Fluid A fluid is a substance. or flows. when subjected to shearing force In fact if a shear stress is acting on a fluid it will flow and if a fluid is at rest there is no shear stress acting on it. Fluid Flow Shear stress – Yes Fluid Rest Shear stress – No . which deforms continuously.1.
oil and air.1. this concept corresponds to loose or very loose bonding between molecules of liquid or gas. with exception to solids. In microscopic point of view. any other matters can be categorised as fluid. respectively. • Examples of typical fluid used in engineering applications are water. .1 Definition of Fluid • Thus.
On the surface. the cohesion forms a resultant force directed into the liquid region and the combination of cohesion forces between adjacent molecules from a tensioned membrane known as free surface. the molecules can move freely but are constrained through a traction force called cohesion.1. For liquids. . For gases. it is very weak which enables the gas to disintegrate and move away from its container. This force is interchangeable from one molecule to another.1 Fluid Concept In fluid. which is suitable for application as hydraulic fluid such as oil. it is stronger which is sufficient enough to hold the molecule together and can withstand high compression.
Liquids and Gases • For solid.1. imagine that the molecules can be fictitiously linked to each other with springs.1 Comparison Between Solids.1 Definition of Fluid Free surface k k k k (a) Solid (b) Liquid (c) Gas Figure 1. .
Adjacent particles have different velocities. the fluid particles are at rest relative to each other. no shear stress will be produced.Shear stress in moving fluid • If fluid is in motion. Shear force Moving plate Fluid particles New particle position Fixed surface . causing the shape of the fluid to become distorted • On the other hand. the velocity of the fluid is the same at every point. shear stress are developed if the particles of the fluid move relative to each other.
cannot normally be neglected and are related to temperature Occupies a fixed volume and will take the shape of the container No fixed volume. . Completely fill the vessel so that no free surface is formed. it changes volume to expand to fill the containing vessels A free surface is formed if the volume of container is greater than the liquid.Differences between liquid and gases Liquid Gases Difficult to compress and often regarded as incompressible Easily to compress – changes of volume is large.
1) = shear stress = viscosity of fluid du/dy = shear rate. • The magnitude of the velocity gradient (du/dy) has no effect on the magnitude of . du dy (1.Newtonian and NonNewtonian Fluid Fluid obey Newton’s law of viscosity refer Newton’s’ law of viscosity is given by. particularly its temperature. . rate of strain or velocity gradient Newtonian fluids Example: Air Water Oil Gasoline Alcohol Kerosene Benzene Glycerine • The viscosity is a function only of the condition of the fluid.
the slope is constant the viscosity is constant nonNewtonian fluids slope of the curves for nonNewtonian fluids varies .Newtonian fluids • The viscosity of the nonNewtonian fluid is dependent on the velocity gradient as well as the condition of the fluid. Newtonian Fluids a linear relationship between shear stress and the velocity gradient (rate of shear).Newtonian and NonNewtonian Fluid Do not obey Fluid Newton’s law of viscosity Non.
e.Figure 1. and cement. velocity gradient Bingham plastic : resist a small shear stress but flow easily under large shear stresses. and jellies.1 Shear stress vs. sewage sludge. e. Dilatants : viscosity decreases with increasing velocity gradient. Pseudo plastic : most nonNewtonian fluids fall under this group. colloidal substances like clay. Viscosity decreases with increasing velocity gradient.g. milk. e. quicksand.g. . toothpaste.g.
such as L for length. or derived dimensions. such as area (L). T for time. • This dimension system is known as the MLT system where it can be used to provide qualitative description for secondary quantities. the FLT system is also used.2 Units and Dimensions • The primary quantities which are also referred to as basic dimensions. • In some countries. velocity (LT1) and density (ML3). . M for mass and Q for temperature.1. where the quantity F stands for force.
2 Units and Dimensions • An example is a kinematic equation for the velocity V of a uniformly accelerated body. a the acceleration and t the time interval. we can expand that LT1 = LT 1 + LT2 • T . In terms for dimensions of the equation.1. V = V0 + at where V0 is the initial velocity.
t is time and x is displacement. Determine the dimension for the stiffness variable k.Example The free vibration of a particle can be simulated by the following differential equation: du m kx 0 dt where m is mass. u is velocity. .
[m]•[u] [k] = [ t ] • [ x ] = M • LT1 LT = MT2 .Example By making the dimension of the first term equal to the second term: [u] [m] • = [k]•[x] [t] Hence.
2 Engineering Units Primary Units Quantity SI Unit Length Metre. . m Mass Kilogram. kg Time Seconds. K Current Ampere.1. A Luminosity Candela In fluid mechanics we are generally only interested in the top four units from this table. s Temperature Kelvin.
m2/s3 pressure (or stress) Pascal (P) P = N/m2 = kg/m/s2 density kg/m3  specific weight N/m3 = kg/m2/s2 N/m3 = kg/m2/s2 relative density a ratio (no units) dimensionless viscosity N.s/m2 = kg/m/s surface tension N/m N/m = kg/s2 .m/s2 energy (or work) Joule (J) J = N.Derived Units Quantity SI Unit velocity m/s  acceleration m/s2  force Newton (N) N = kg.s/m2 N.m/s = kg.m = kg.m2/s2 power Watt (W) W = N.
. 3. including units. 6. Use correct conversion factors to eliminate unwanted units and obtain the proper units as described in Step 2. Solve the equation algebraically for the desired terms. 5. Cancel units that appear in both the numerator and denominator of any term. Perform the calculations. 2. Substitute known values.Unit Cancellation Procedure 1. 4. Decide on the proper units of the result.
m/s2 F= 800N .Example Given m = 80 kg and a=10 m/s2. Find the force Solution F = ma F = 80 kg x 10 m/s2 = 800 kg.
.2) Units: kg/m3 Typical values: Water = 1000 kg/m3.23 kg/m3 . Air = 1.1. • slightly affected by changes in temperature and pressure. Definition: mass per unit volume.3 Fluid Properties Density Density of a fluid. = mass/volume = m/ (1.
Fluid Properties (Continue) Specific weight Specific weight of a fluid. Air = 12.3) Units: N/m3 Typical values: Water = 9814 N/m3.07 N/m3 . • Definition: weight of the fluid per unit volume • Arising from the existence of a gravitational force • The relationship and g can be found using the following: Since therefore = m/ = g (1.
or Definition 2: A ratio of the specific weight of a substance to the specific weight of water at standard temperature (4C) and atmospheric pressure. s w @ 4C (1. SG s w @ 4C Unit: dimensionless.4) .Fluid Properties (Continue) Specific gravity The specific gravity (or relative density) can be defined in two ways: Definition 1: A ratio of the density of a substance to the density of water at standard temperature (4C) and atmospheric pressure.
The reservoir has a volume of 0. Compute the density.81 8829 N / m 3 volume SGoil oil w @ STP 900 0. specific weight. Solution: oil mass m 825 900kg / m 3 volume 0.Example A reservoir of oil has a mass of 825 kg.9 998 .917 oil weight mg g 900 x 9.917 m3. and specific gravity of the oil.
Fluid with a high viscosity such as syrup deforms more slowly than fluid with a low viscosity such as water. due to cohesion and interaction between molecules.s/m2 or kg/m/s Typical values: Water = 1. The ease with which a fluid pours is an indication of its viscosity.78x105 kg/m/s .14x103 kg/m/s. Units: N. is the property of a fluid.Fluid Properties (Continue) Viscosity • Viscosity. which offers resistance to shear deformation. The viscosity is also known as dynamic viscosity. Air = 1. . • Different fluids deform at different rates under the same shear stress.
In general.14x106 m2/s. Units: m2/s Typical values: Water = 1. .46x105 m2/s. Definition: is the ratio of the viscosity to the density. / • will be found to be important in cases in which significant viscous and gravitational forces exist. Air = 1. viscosity of liquids with temperature. whereas viscosity of gases with in temperature.Kinematic viscosity.
The compressibility of a fluid is expressed by its bulk modulus of elasticity. is increased by Δp and the volume is changed by Δ. i.05x109 N/m 2. change in pressure K volumetric strain Thus.e.62x109 N/m2 . Oil = 1. which describes the variation of volume with change of pressure.Bulk Modulus All fluids are compressible under the application of an external force and when the force is removed they expand back to their original volume. if the pressure intensity of a volume of fluid. then p K / p K Typical values:Water = 2. K. .
Since this depends upon molecular activity. which is a function of temperature. It reaches a stage of equilibrium when this pressure reaches saturated vapor pressure. for example if the pressure is reduced sufficiently boiling may occur at room temperature. If the pressure above a liquid reaches the vapor pressure of the liquid. the vapor pressure of a fluid also depends on its temperature and increases with it.Vapor Pressure A liquid in a closed container is subjected to a partial vapor pressure in the space above the liquid due to the escaping molecules from the surface. boiling occurs. .
very serious damage can result due to the very large force with which the liquid hits the surface. If this should occur in contact with a solid surface. Cavitations in a closed hydraulic system can be avoided by maintaining the pressure above the vapor pressure everywhere in the system.Engineering significance of vapor pressure In a closed hydraulic system. water vaporizes rapidly in regions where the pressure drops below the vapor pressure. since the flow of fluid can sweep this cloud of bubbles on into an area of higher pressure where the bubbles will collapse suddenly. . There will be local boiling and a cloud of vapor bubbles will form. and can cause serious problems. Cavitations can affect the performance of hydraulic machinery such as pumps. turbines and propellers. Ex. and the impact of collapsing bubbles can cause local erosion of metal surface. This phenomenon is known as cavitations. in pipelines or pumps.
liquids can resist small tensile forces at the interface between the liquid and air. known as surface tension. The reason for the existence of this force arises from intermolecular attraction. This causes a slight deformation at the surface of the liquid (the meniscus effect).Surface Tension Liquids possess the properties of cohesion and adhesion due to molecular attraction.2: Surface Tension . 1. 1. . At the surface of the liquid (Fig. and they act rather like a very thin membrane under tension. a molecule is surrounded by other molecules and intermolecular forces are symmetrical and in equilibrium.2a). In the body of the liquid (Fig. Figure 1. Due to the property of cohesion. This imbalance forces means that the molecules at the surface tend to be drawn together. a molecule has this force acting only through 180. Surface tension is defined as force per unit length. and its unit is N/m.2b).
. and the rise or fall of liquid in capillary tubes is the results of the surface tension. increases the pressure within a droplet of liquid. surface tension for water at 20C is 0.7) . A steel needle floating on water. balancing the surface tensional force of a spherical droplet of radius r. The internal pressure. Surface tension. the spherical shape of dewdrops.0728 N/m). Surface tension is usually very small compared with other forces in fluid flows (e.g. P. is given by 2R = pR2 2 P r (1.
• For tube larger than 12 mm (1/2 in. such as mercury.) capillarity effects are negligible. such as water.Capillarity • The surface tension leads to the phenomenon known as capillarity • where a column of liquid in a tube is supported in the absence of an externally applied pressure. and fall in tubes that do not wet (cohesion > adhesion).). • Rise or fall of a liquid in a capillary tube is caused by surface tension and depends on the relative magnitude of cohesion of the liquid and the adhesion of the liquid to the walls of the containing vessels. . • Liquid rise in tubes if they wet a surface (adhesion > cohesion). • Capillarity is important when using tubes smaller than 10 mm (3/8 in.
3 Capillary actions 2 cos h r (1.8) where h = height of capillary rise (or depression) = surface tension = wetting (contact) angle = specific weight of liquid r = radius of tube .Figure 1.
81 8829 N / m 3 volume SG oil oil w @ 4 C 900 0. and specific gravity of the oil.9 1000 . Compute the density. Solution: oil mass m 825 900kg / m 3 volume 0. specific weight.917 m3. The reservoir has a volume of 0.Example A reservoir of oil has a mass of 825 kg.917 oil weight mg g 900 x 9.
In a 3mm diameter vertical tube.4 = 83.006 2 2 x 0 .Example Water has a surface tension of 0. if the liquid rises 6 mm above the liquid outside the tube. calculate the wetting angle. 2 cos h r cos rh 9810 x 0. Solution Capillary rise due to surface tension is given by.7 .4 N/m.0015x 0.
Discussion on the vapor pressure of the liquid Surface tension Capillarity phenomena .Summary This chapter has summarized on the aspect below: Understanding of a fluid The differences between the behaviours of liquid and gases Newtonian and nonNewtonian fluid were identified Engineering unit of SI unit were discussed Fluid properties of density. specific gravity. viscosity and bulk modulus were outlined and taken up. specific weight.