Mcmaster Mech Eng 3o04 - Fluid Mechanics Notes By Dr. Mohamed S. Hamed
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9/3/2008
Outline and Announcements Outline:z z z z z z z z z
Introduction – Course instructor and website. Assessment. Lectures, Class Notes, and Downloads Assignments, Tutorials, and Exams. Role of the Textbook. Teaching Style. Course objectives. Interesting Flows. The “First Question”.
Important announcements:1.
Tutorials start the week of Sept 15. ME3O04 – Chapter 1
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Assessment z Two-term tests (October and November) = 30%. z Assignments = 15% . z Final examination = 55% .
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Lectures z
Class discussions might include points that are not necessarily included in the textbook.
z
All exams will include questions on theory and concepts covered in lectures and class discussions.
z
Assignments might include questions on theory and concepts covered in lectures and class discussions.
z
Attending lectures is very important!!
z
Laptops and Cell Phones are not allowed during lectures, tutorials, and exams. ME3O04 – Chapter 1
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Class Notes and Downloads z Class notes are print out of my PowerPoint presentations. z Notes taken by students during lectures are very important. z Lecture notes and other material will be posted on the course website in a password protected section. z All material is copyright protected and should not be shared with and/or distributed to others. ME3O04 – Chapter 1
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Assignments z Assignments – will be assigned regularly, roughly every week. z Assignments might include problems from the textbook and from lectures and class discussions. z Due dates are posted on the web. z A late penalty of 10% per day will be applied. ME3O04 – Chapter 1
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Tutorials and Exams z Tutorials are provided to:{ Address any unclear points. { Help you solve assignments.
z All exams will include questions on theory and concepts covered in lectures and class discussions.
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Role of the Textbook z The textbook will be used to assign problems. z It supplements class discussions. z It is not a substitute for lectures!!
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Teaching Style z
I love and enjoy teaching. So, my teaching style is dynamic and interactive.
z
Everybody is encouraged to interact (ask questions, inquire, answer questions, etc.)
z
Bonus cards will awarded to individuals based on their level of interaction and understanding of the material.
z
Each card is worth 0.25%. There is no limit on how many cards you can get!!
z
This bonus is added to your total. So, your total could be more than 100%!! ME3O04 – Chapter 1
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Important Announcements z Please check the “Important Announcements” Announcements section on the web on a regular basis. z Tutorials will start the week of Sept 15. 15
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Course Objectives z Introduce special “vocabulary” vocabulary and “basic concepts” used in fluid mechanics. z Develop a good understanding of these concepts. z Use them to analyze and understand fluid flows in “real (practical)” (practical) problems.
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Why Fluid Mechanics is A Core Course? Please try to think of or name: {An industry, {A piece of machinery, or {Any engineering system,
where fluid mechanics does not play an important role!
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Industries where Concepts of Fluid Mechanics Play a Vital Role •Automobile Engineering •Aerospace Engineering •Oil & Gas Engineering •Power Generation •Thermal Management •Environmental Control •Biotechnology •Energy Conversion •Process Engineering ▪ ME3O04 – Chapter 1
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Some Interesting Flows
Airfoil zero angle
Airfoil 25° angle
Air flow over a Car
Wing Vortex
Personal Plume
Reference: Multimedia Fluid Mechanics CD-ROM Cambridge University Press ISBN-10: 0521604761 ME3O04 – Chapter 1
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OK, So what would be the first question we should address in this class? What is
Fluid
Mechanics ?
Forms of Matter: • Solid. • Liquid. • Gas.
Motion and its “Cause” Force ME3O04 – Chapter 1
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Scope of Fluid Mechanics Fluid Mechanics is the study of the behavior of fluids at rest and in motion. motion
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What is the Definition of a “Fluid” ? Definition of A Fluid: Fluid is a substance that deforms continuously under the application (or the effect) of a shear stress. tangential force area ME3O04 – Chapter 1
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Definition of a Fluid – cont’d When a shear stress is applied: z Solids deform or bend, then stop! z Fluids continuously deform ⇒ i.e., they Flow
Fig. 1.1 ME3O04 – Chapter 1
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Q. What makes a fluid deform? A. The “no-slip condition”.
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Two Famous Flows in Fluid Mechanics 1. Couette flow. 2. Poiseuille flow.
Poiseuille Flow ME3O04 – Chapter 1
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Two Famous Flows in Fluid Mechanics 1. Couette Flow 2. Poiseuille Flow
Poiseuille Flow ME3O04 – Chapter 1
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Basic Equations z In the analysis of any fluid mechanics problem, we need to use a set of equations called “Governing Equations”. z These equations can be classified into: 1. Basic or General Laws. 2. Particular or Special Laws.
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Basic or General Laws z Conservation of mass. z Newton’s second law of motion. z The principle of angular momentum.
From Mechanics
z The first law of thermodynamics. z The second law of thermodynamics.
From Thermodynamics ME3O04 – Chapter 1
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Basic or General Laws – Cont’d z Note that these laws do not depend on the type of fluid involved in the problem. That’s why they are called “General Laws”. z Not all these laws are required every time. z In many problems, it is necessary to bring into the analysis additional relations ⇒ “Particular Laws”
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Particular or Special Laws z These laws describe the behavior of physical properties of specific types of fluids. z Example: Ideal Gas Law ⇒
P = ρ RT
¾ This law can be used only for ideal gases. ¾ It relates absolute pressure and temperature to gas density. ¾ P in Pa = N/m2, ρ in kg/m3, and T in K.
¾ R = gas constant, for air = 286.9 J/kg.K ■
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Methods of Description 1. The system approach (the Lagrangian approach), which follows one specific particle or system (e.g., following the motion of a falling object).
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Methods of Description - Continue 2. The control volume approach (the Eurlerian approach), which focuses on a certain region not on a certain particle (e.g., flow in a pipeline).
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Methods of Analysis in Thermodynamics 9 System (or “Closed System”)
9 Control Volume (or “Open System”)
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Methods of Analysis in Fluid Mechanics 1. Infinitesimal or finite control volume. 2. Large control volume.
Differential control volume
Large Control volume
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Methods of Analysis in Fluid Mechanics 1. Infinitesimal or finite system. 2. Control volume.
z In each case the equations will look different. z In (1), the resulting equations are differential equations, which provide details of the flow. z In (2), the resulting equations are integral, which give a global or overall behavior of the flow. ■ ME3O04 – Chapter 1
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Why Use Infinitesimal Control Volume (Differential) Approach? z Differential approach allows us to determine flow details, e.g., velocity distribution:-
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Why Use Control Volume (Integral) Approach? z Integral approach allows the determination of global or overall values, such as, average velocities, forces, etc..
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Units and Dimensions z
A unit is a specific quantitative measure of a physical quantity. Example, the foot and the meter, which are used to measure the physical quantity ‘length’.
z
A physical quantity, such as length, is called a dimension.
z
Dimensions can classified as basic or primary dimensions and secondary dimensions:1. Length and time are basic dimensions. 2. Velocity is a secondary dimension because it can be represented using primary dimensions (velocity = length / time). ME3O04 – Chapter 1
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Systems of Basic Dimensions
1. [M], [L], [t], and [T]. 2. [F], [L], [t], and [T]. 3. [F],[M], [L], [t], and [T].
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Systems of Units 1. MLtT - SI (kg, m, s, K). 2. FLtT - British Gravitational (lbf, ft, s, oR). 3. FMLtT - English Engineering (lbf, lbm, ft, s, oR).
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Preferred Systems of Units SI (kg, m, s, K)
British Gravitational (lbf, ft, s, oR)
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Dimensional Consistency z All equations and formulas must have consistent dimensions. z I.e., all terms in any equation must have the same dimension. z What is the dimension of each term of Bernoulli’s equation?
r2 r2 V1 p1 V2 p2 + g z1 + = + g z2 + 2 ρ 2 ρ ME3O04 – Chapter 1
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Dimensional Consistency – Cont’d z
In your engineering studies and practice, you might deal with two types of equations having inconsistent dimensions:1. Semi-empirical equations, e.g., the Manning equation, used to calculate the velocity of flow in an open channel (such as a canal):2/3 1/ 2
Rn S0 V= n
Where V is the velocity in m/s, R is the hydraulic radius of the channel in m, and S is the channel slope (ratio). For unfinished concrete, n = 0.014.
z
What is the unit of V? Does n have a unit?
z
If we use n =0.014, and R in ft, can we get the correct value of V in ft/s? ME3O04 – Chapter 1
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Dimensional Consistency – Cont’d 2. In the second type the dimensions of an equation are consistent but the use of units is not. z
For example, the commonly used Energy Efficiency ratio (EER) of an air conditioner is:-
cooling rate energy / time EER = = electrical input energy / time z
The equation is dimensionally consistent, with EER being dimensionless (ratio). However, it is used in an inconsistent way.
z
A good A/C has EER = 10, which means 10 Btu/hr for each 1 W of electrical power.
z
One must say 10 (Btu/hr)/W because it is not dimensionless. ME3O04 – Chapter 1
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Exercise A sky diver with a mass of 75 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be FD = kV2, where k = 0.228 N.s2/m2. Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.
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Vocabulary List 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
A Fluid. Mechanics. Shear force. Normal force (pressure). Flow = continuous deformation. Scope of Fluid Mechanics. Basic Equations. Methods of Description. Methods of Analysis. No-slip condition. Lagrangian approach. Eulerian approach. Infinitesimal or finite system = infinitesimal control volume (C.V.). 14. Infinitesimal C.V. = Differential approach. 15. Control volume or integral approach. 16. Basic dimensions. ME3O04 – Chapter 1
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A flow is described by the velocity field,
Problem
V = a y iˆ + b t ˆj Where a = 1 1/s and b = 2 m/s2. t in seconds. (a) Plot the pathline of the particle that passed point (1,2) at t = 2. (b) Streakeline at t = 3 of the particles that passed point (1,2).
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Note the difference between the meaning of t and to ME3O04 Chapter 2
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Note the difference between the meaning of t and to ME3O04 Chapter 2
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Problem – time permit Fluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?
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Chapter 2 - Fundamental Concepts Outline:-
¾ ¾ ¾ ¾ ¾ ¾ ¾
¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾
Fluid as a Continuum. Scalar quantities. Vector quantities - Velocity Field. Steady Vs Unsteady Flow Fields. Uniform Flow Fields. One-, Two-, and Three-Dimensional Flows. Visual Representation of flow fields: Timeline. Streamlines. Pathlines. Streaklines. Types of Forces acting on a fluid element. Stresses (name and sign convention). Stress Field (stress at a point). Shear Stress and rate of deformation of a fluid element. Viscosity. 1. Newtonian and 2. Non-Newtonian Fluids. Surface Tension. Classification of Fluid Motions. Compressibility effect. ME3O04 Chapter 2
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Nature of Fluids (Matter) z Fluids (gases or liquids) are forms of matter and they consist of molecules with atoms and space in between. z So, generally speaking, mass of a fluid is not continuously distributed in space: (mass – space – mass – space – mass – etc…)
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Fluid as a Continuum Continuum Hypothesis: We can assume that fluids are continuous medium. 1.
What does it mean? It means that a fluid regardless of its molecular nature, can be treated as a continuous medium.
2.
What is the result or benefit of this assumption? a)
Each fluid property is assumed to have a definite value at every point in space, thus
b)
Fluid properties are considered to be continuous function of position and time.
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Continuum Hypothesis – Cont’d 3. When it is not possible to make such an assumption? We can not use this assumption if the mean free path of the molecules is of the same order of magnitude as the smallest significant characteristic dimension of the problem. {
Molecules always vibrate in space.
{
Mean free path is the distance the molecules travel as it vibrate.
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Mean Free Path, Lm z Examples, for air:1. At STP (15º C and 101.3 kPa) ⇒ Lm = 6 × 10-8 m = 0.06 μm = 60 nm. 2. At P = 1.33 × 10-7 kPa (Rarefied Gas) ⇒ Lm = 0.9 m = 90 cm.
z
Applications = air flow between parallel plates with a = 10 cm at atmospheric pressure i.e., P = 101.3 kPa.
?
⇐
Can we assume that air is a continuous medium in this case ?
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Quantities of interest In fluid mechanics we are interested in quantities that are: 1. Scalar quantities, such as: density, temperature, pressure, etc. 2. Vector quantities, such as: velocity, acceleration, stress, ect.
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Scalar Quantities – e.g., Density ρ z Since ρ is a scalar quantity, we need only the specification of its magnitude. z The complete or field representation of ρ is given by: (2.1) z Equation (2.1) represents a scalar field.
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Density ρ - Cont’d z Density is sometimes expressed in terms of specific gravity, SG for liquids:ρ SG = ρw max
where, ρwmax = max density of water = 1000 kg/m3 at 4 ºC. zFor gases,
SG =
ρ ρa
where ρa= air density.
zOr, in terms of specific weight, γ:
weight m g γ= = =ρg ∀ volume
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Vector Quantities – e.g., Velocity Field, V z Similar to ρ, we can represent the velocity as:(2.2)
z Eqn. 2.2 represents the velocity field, which is a vector field. z A Vector has magnitude and direction.
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Velocity Field,
V
- Cont’d
z Velocity can be written in terms of its 3 scalar components (u, v, w) in the x, y, and z directions:
where iˆ, ˆj , and kˆ = unit vectors in x, y, and z directions, respectively. z In general each component u, v, and w is a function of x,y, z, and t, e.g., u = u(x,y,z,t)
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Types of Flows - Steady Flow z
If properties at every point in a flow field do not change with time (t), the flow is called steady.
z
If η is any fluid property in a steady flow field, mathematically:
∂η =0, ∂t
or
η =η (x, y, z)
z
Remember : if f = f(x), then rate of change of f w.r.t. x =
z
If f=f(x,y), then rate of change of f w.r.t. x = ∂f ∂x ME3O04 Chapter 2
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■
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Types of Flows - One-, Two-, and ThreeDimensional Flows Depending on the number of space coordinates required to specify the velocity field, a flow can be described as 1D, 2D or 3D. 9
V = V ( x, y , z , t )
⇐
3D, unsteady
9
V = V ( x, y , t )
⇐
2D, unsteady
9
V = V (x)
⇐
1D, steday
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Examples – Steady Flow
1D
2D
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Uniform Flow Sometimes we can neglect the no-slip condition at the wall and assume that the velocity is uniform across the whole cross section, as shown below:
u = u(x,y)
u=u(x)
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Visual Representation of Flow Fields Sometime we want to have a visual representation of flow fields. Such a representation is provided by:
¾ Timelines. ¾ Streamlines. ¾ Path lines. ¾ Streak lines. ME3O04 Chapter 2
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Timelines z A timeline is a line connecting the positions of a set of fluid particles at a given instant. This is a timeline
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Streamlines A streamline is the line drawn in the flow field so that at any instant in time it is tangent to the direction of the flow, i.e., tangent to the velocity vector.
v dy i.e., = tan α = , u dx
⇒ u dy − v dx = 0
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Streamlines – Cont’d
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Pathlines z A path line is the line traced out by a given particle as it flows from one point to another.
z Path lines are useful in studying , for example, the trajectory of a contaminant leaving a smokestack.
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Pathlines – Cont’d
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Streaklines A streakline is the locus of all particles that at an earlier instant in time, passed through a prescribed point in space.
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Streaklines – Cont’d
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What type of Lines are these? z The redlines, and z The white lines?
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Important points to note z A streamline and a timeline are instantaneous lines, e.g., snap shots. z While streaklines and pathlines are generated by the passage of time, e.g., video recording. z In steady flow, the velocity at each point in the flow field remains constant with time and consequently, in a steady flow, pathlines, pathlines streaklines, streaklines and streamlines are identical. ME3O04 Chapter 2
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Example Problem A two-dimensional unsteady velocity field is given by u = x (1 + 2 t), v = y. Find:1. The time-varying streamlines which pass through some reference point (xo,yo). Sketch some for the case of xo=1, yo = 1. 2. Find the equation of the pathline which passes through the point (xo, yo) at t = 0. Sketch this pathline for the case of xo=1, yo = 1.
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Streamlines at different times
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Pathline for t < 0 to t > 0
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Types of Forces Acting on a Fluid Element 1. Surface Forces. 2. Body Forces.
ME3O04 Chapter 2
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Types of Forces Acting on a Fluid Element A. Surfaces Forces:- e.g., Pressure and Friction These are forces generated on the surface of the fluid element by contact with other fluid particles or a solid surface.
B. Body Forces:- e.g., Gravity and Electromagnetic Field These forces are not concentrated at the surface of the fluid element; they are rather experienced throughout the particle. Example: gravitational body force acting on a fluid element of volume, d ∀ =
gravitational force = d m × g = ρ d ∀ g ME3O04 Chapter 2
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Surface Force on a Fluid Element y δAx
δAy
δAx F x
Δy δAz
z
Δz Δx
Fluid Element in Cartesian Coordinates
Surface Forces cause Stresses on the surfaces of the fluid element ME3O04 Chapter 2
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Stress Caused by Surface Forces
Fluid element
σ (sigma) is used to denote normal stresses. τ (Tao) is used to denote tangential stresses. ME3O04 Chapter 2
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Stress Caused by Surface Forces – Cont’d
z Normal stress at point C =
σ xx =
z Tangential stress at point C = τ xy =
δFx lim δAx → 0 δ Ax
δFy δF lim δA Or, τ xz = lim δAz δAx →0 x δAx →0 x
ME3O04 Chapter 2
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Name Convection - Stress needs two
directions to be define (force and area). z First letter on the left refers to the plan on which the stress acts.
δAx
z Second letter refers to the direction in which the stress acts. z Note that the subscript in δAx refers to the direction normal to the surface of interest. ME3O04 Chapter 2
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Sign Convention z Velocity or force is positive if it is in the positive direction of the axis. z The plan is considered positive if the normal to it is in the positive direction of the axis.
δ Ax
z Sign of stress is determined by the sign of its plan and force. z What is the sign of σxx ?
ME3O04 Chapter 2
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Stress Field Stress at a point =
⎡σ xx τ xy τ xz ⎤ ⎢ ⎥ ⎢τ yx σ yy τ yz ⎥ ⎢ ⎥ τ τ σ zx zy zz ⎣ ⎦
Note: all stresses shown here are positive. ME3O04 Chapter 2
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Shear Stress in a Fluid Element Exposed to a Shear Force z Fluid continuously deforms (i.e., flows) under the effect of a shear force. z Shear Stress = τ
yx
δFx dF x = δA y → 0 δ A y dA y
= lim
ME3O04 Chapter 2
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Rate of Deformation of a Fluid Element Exposed to a Shear Force z Rate of deformation =
δα dα = lim δt → 0 δ t dt
ME3O04 Chapter 2
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(1)
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Rate of Deformation of a Fluid Element Exposed to a Shear Force z Because δα is very small we can say that: δl = δα. δy (2)
ME3O04 Chapter 2
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Rate of Deformation of a Fluid Element Exposed to a Shear Force Due to the no slip condition, point M’ will be (3) moving at δu, thus : δl = δu. δt
ME3O04 Chapter 2
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Rate of Deformation of a Fluid Element Exposed to a Shear Force z From (2) and (3), Rate of deformation =
δα δu dα du = , or , = δt δy dt dy
ME3O04 Chapter 2
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Relation Between Shear Stress and Rate of Deformation z Because every fluid element exposed to shear stress will deform, there must be a relationship between the shear stress and the rate of deformation. deformation z This relationship depends on the type of fluid:1. Newtonian, or 2. Non-Newtonian. ME3O04 Chapter 2
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Fluid Viscosity z Newtonian Fluids:{Most of the common fluids (water, air, oil, etc.) {“Linear” fluids
z For Newtonian fluids, the rate of deformation is in direct proportion with the shear stress, i.e.,
z The constant of proportionality is the fluid viscosity, μ, i.e.,
ME3O04 Chapter 2
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Non-Newtonian Fluids {Special fluids (e.g., most biological fluids, toothpaste, some paints, etc.) {“Non-linear” fluids In this case the relation between Shear stress and rate of deformation Takes the form:
Which can also be written as :
In this case, η is called the apparent viscosity ME3O04 Chapter 2
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Non-Newtonian Fluids – Cont’d
ME3O04 Chapter 2
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Dynamic and Kinematic Viscosity
z This equation is called Newton’s law of viscosity for a one-dimensional flow (note only y appears in the equation). equation z μ (mu) is called the Dynamic Viscosity of the fluid and it is a physical property of the fluid. z ν (nu) =μ/ρ is called the Kinematic viscosity and it is another physical property. ME3O04 Chapter 2
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Dynamic and Kinematic Viscosity
z Units of μ : Pa.s = N.S/m2 = kg/(m.s), 1 Poise = 1 gm/(cm.s). z Units of ν: m2/s, 1 Stoke = 1 cm2/s. ME3O04 Chapter 2
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Problem A block of mass M slides on a thin film of oil. The film thickness is h and the area of the block is A. When released, mass m exerts tension on the cord, causing the block to accelerate. Neglect friction in the pulley and air resistance. Develop an expression for the viscous force that acts on the block when it moves at speed V. Obtain an expression for the block speed as a function of time. If mass M = 5 kg, m = 1 kg, A = 25 cm2 , and h = 0.5 mm. If it takes 1 s for the speed of the block to reach 1 m/s, find the oil viscosity μ.
ME3O04 Chapter 2
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Surface Tension,
σ
z Surface tension is a force that appears along any common surface (interface) between two fluids. interface
z Units of σ is force per unit length (length of the interface), e.g., N/m or lbf/ft. ME3O04 Chapter 2
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Surface Tension – Cont’d z As shown below, if the two fluids are in contact with a solid surface, a contact angle, θ, develops. z Both of values of σ and θ depend on the type of fluids in contact.
ME3O04 Chapter 2
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ME3O04 Chapter 2
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ME3O04 Chapter 2
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Capillary Rise (CR) and Capillary Depression (CD). z CR occurs when θ < 90°. z CD occurs when θ > 90°.
ME3O04 Chapter 2
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Calculation of Δh
∑F
z
∑F
z
=0
= σ π D cos θ − ρ g Δ∀ = 0
Q Δ∀ =
π D2 4
Δh
From (2) in (1):-
(1)
(2)
4 σ cos θ ∴ Δh = ρgD
Note the sign of Δh when θ > 90º and θ < 90º. ME3O04 Chapter 2
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Description and Classification of Fluid Motions
ME3O04 Chapter 2
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Viscous and Inviscid Flows Under some special circumstances, the effect of fluid viscosity can be ignored (neglected). Example:- in region of flow away from solid surfaces.
ME3O04 Chapter 2
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Laminar and Turbulent Flows z A laminar flow is a flow in which the fluid particles move in smooth layers, laminas. z A Turbulent Flow is a flow in which the fluid particles rapidly mix as they move due to random velocity fluctuations.
z The flow in a pipe is considered laminar if Re < 2300, where, ρV D Re = = Reynolds number μ z Where, ρ,μ, V, & D are the density, viscosity, velocity, and diameter, respectively. ME3O04 Chapter 2
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Compressible and Incompressible Flows z Incompressible flows are those in which variations in density are negligible. z When variations in density are not negligible, the flow is called compressible. z Variations in density are due to changes in pressure and/or temperature.
ME3O04 Chapter 2
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Compressible and Incompressible Flows z Mostly, liquids can be regarded as incompressible fluids. z Pressure and density changes in liquids are reflected by the bulk compressibility modulus, or modulus of elasticity: elasticity
dp Ev = dρ / ρ z For water at 15°C, Ev = 2010 kPa = 2.92 x 105 psi. ME3O04 Chapter 2
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Can we treat a gas flow as an Incompressible flow? z For Mach number M < 0.3, the maximum density variation is less than 5%. z Thus, gas flows with M < 0.3 can be treated as incompressible. z Mach number = V/c, where V is the flow velosity and c is the speed of sound. z The speed of sound in an ideal gas is given by:
c = k RT
z Where k = Cp/Cv, R = gas constant, and T is the absolute temperature. ME3O04 Chapter 2
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Problem At what minimum speed (in mph) would an automobile have to travel for compressibility effects to be important? Assume the local air temperature is 60°F.
ME3O04 Chapter 2
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Problem – time permit Fluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?
ME3O04 Chapter 2
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Vocabulary List 1. 2. 3. 4. 5. 6. 7.
Continuum hypothesis. Mean free path. Scalar quantity. Vector quantity. Uniform flow. Multi-dimension flow. Flow Visualization.
ME3O04 Chapter 2
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Vocabulary List 1. 2. 3. 4.
Timeline. Streamline. Pathline. Streakline.
ME3O04 Chapter 2
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Vocabulary List 1. 2. 3. 4. 5. 6. 7. 8. 9.
Body force. Surface force. Stress field. Shear stress. Normal stress. Rate of deformation. Viscid and Inviscid flows. Laminar and Turbulent flows. Compressible and Incompressible flow. ME3O04 Chapter 2
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Outline:- Fluid Statics – Chapter 3 ¾ The Basic Equation of Fluid Statics ¾ ¾ ¾ ¾ ¾
Types of Pressures. Pressure Variation in a Static Fluid. Example Problem. Hydrostatic Force on Submerged Surfaces. Example Problems.
ME3O04 – Chapter 3
1
3
1
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What does Static Fluid mean? z Statics means that the fluid is not moving, i.e., its velocity =0; and its acceleration = 0. z Fluid velocity = 0 means that it does not flow. flow z If a static fluid does not flow, how much shear stress the fluid is exposed to? no flow = no deformation = no shear ME3O04 – Chapter 3
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What does Static Fluid mean? z In this case, fluid can be exposed to only normal forces and behaves as
“a rigid body” – no deformation
ME3O04 – Chapter 3
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The Basic Equation of Fluid Statics Consider the following fluid element:-
ME3O04 – Chapter 3
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The Basic Equation of Fluid Statics Consider the following fluid element:-
Newton’s 2nd law:-
r r ∑ dF = dm . a = 0
Divide both sides by d∀ gives:
r r dF ∑ d∀ = ρ .a = 0
ME3O04 – Chapter 3
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The Basic Equation of Fluid Statics – Cont’d z Forces affecting on the fluid element = surface + body forces, i.e.,
r r r ∑ dF = dFs + dFB
z Body Force =
or,
ME3O04 – Chapter 3
1
3
6
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Surface Force in y- direction, dFys
Surface force in y-direction = r ⎡⎛ ∂ p dy d F S y = ⎢ ⎜⎜ p − ∂y 2 ⎣⎝
= −
⎞ ⎛ ∂ p dy ⎞ ⎤ ⎟⎟ − ⎜⎜ p + ⎟⎟ ⎥ dx .dz ∂y 2 ⎠⎦ ⎠ ⎝
∂p dx .dy .dz ∂y ME3O04 – Chapter 3
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Surface Force, dFs z Combining the other two directions, we get:
r ⎛ ∂p ˆ ∂p ˆ ∂p ˆ ⎞ dFS = −⎜⎜ i + j + k ⎟⎟dx.dy.dz ∂y ∂z ⎠ ⎝ ∂x = −∇p.dx.dy.dz z Where, ∇p = gradient of p.
ME3O04 – Chapter 3
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Basic Equation of Fluid Statics Recall – Newton’s 2nd Law:-
r r dF ∑ d∀ = ρ .a = 0 r r
(1)
r ∑ dF = dFs + dFB
r dFS = −∇p.dx.dy.dz
(2)
From (2) in (1):
Basic Equation of Fluid Statics ME3O04 – Chapter 3
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Basic Equation of Fluid Statics
Note that equation above is not really one equation, it is rather three equations in the three directions x, y, and z.
∂p ∂p ∂p − + ρ g x = 0, − + ρ g y = 0, − + ρ g z = 0 ∂x ∂y ∂z ME3O04 – Chapter 3
1
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Basic Equation of Fluid Statics z Knowing that
r g = g x iˆ + g y ˆj + g z kˆ
z Assuming that z is the vertical direction, we can say that gx = gy = 0, gz = - g, the three equations:
−
∂p ∂p ∂p + ρ g x = 0, − + ρ g y = 0, − + ρ g z = 0 ∂x ∂y ∂z
ME3O04 – Chapter 3
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Basic Equation of Fluid Statics Can be written as :-
∂p = 0, ∂x
∂p = 0, ∂y
i.e., pressure is not function of x
∂p = −ρ g ∂z
i.e., pressure is not function of y
z i.e.,
dp = − ρ g = −γ dz
ME3O04 – Chapter 3
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Important Restrictions:In order to use the following equation, three conditions must be satisfied. satisfied
dp = − ρ g = −γ dz 1. Fluid must be static, i.e., velocity = acceleration = 0. 2. Gravity is the only body force. 3. The Z axis is vertical and upward. ME3O04 – Chapter 3
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Types of Pressures
Fig. 3.2
Pabs = Patm + Pgage ME3O04 – Chapter 3
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Pressure Variation in a Static Fluid For Incompressible Fluid: Manometers
dp = −ρ g dz
p
z
po
zo
∫ dp = − ∫ ρ g dz ME3O04 – Chapter 3
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Pressure Variation in a Static Fluid For Incompressible Fluid: Manometers
p
z
po
zo
∫ dp = − ∫ ρ g dz
p − po = − ρ g ( z − zo ) = ρ g ( zo − z ) or, p = po + ρ g h ME3O04 – Chapter 3
1
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Simple Rules to Analyze Multiple-Liquid Manometer Problems 1. Any two points at the same elevation in a continuous volume of the same liquid are at the same pressure. 2. Pressure increases as one goes down a liquid column.
ME3O04 – Chapter 3
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Example- Problem Determine the gage pressure in psig at point “a”, if liquid A has SG = 0.75 and liquid B has SG = 1.20. The liquid surrounding point “a” is water and the tank on the left is open to the atmosphere.
ME3O04 – Chapter 3
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Hydrostatic Force on Submerged Surfaces.
p = po + ρ g h
z This equation allows us to determine how pressure varies in a static fluid. z We would like to determine the force due to that pressure on a surface submerged in a liquid.
ME3O04 – Chapter 3
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Hydrostatic Force on Submerged Surfaces – Cont’d In order to fully determine the force on a surface submerged in a liquid, we must determine the following:1. The magnitude of the force; 2. The direction of the force; and 3. The line of action of the force.
ME3O04 – Chapter 3
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1- Direction of the Force on a Plane Submerged Surface z Since fluid is not moving (static), there is no shear, i.e., only normal forces might exist. z Since this force is caused by pressure of fluid, it will always be normal to the surface. surface z This determines the direction of the force.
ME3O04 – Chapter 3
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2- Magnitude of the Force on a Plane Submerged Surface
dF = P. dA P = Po + ρ g h
(1) (2)
From (2) in (1)
∴ FR = ∫ dF = ∫ (Po + ρ g h ) dA A
(3)
A
ME3O04 – Chapter 3
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2- Magnitude of the Force on a Plane Note how θ is measured Submerged Surface
∴ FR = ∫ dF = ∫ (Po + ρ g h ) dA A
(3)
A
but,
h = y sin θ
(4)
From (4) in (3)
FR = Po A + ρ g ∫ y sin θ .dA A
Where, Po is the pressure at the fluid surface, A is the surface area. ME3O04 – Chapter 3
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3- Line of Action of the Force on a Plane Submerged Surface
∴ FR . y ' = ∫ dF . y = ∫ P. dA. y A
A
but,
P = Po + ρ g h
and
h = y sin θ
(
)
∴ FR . y ' = ∫ Po . y + ρ g y 2 sin θ . dA A ME3O04 – Chapter 3
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3- Line of Action of the Force on a Plane Submerged Surface
∴ F . y = ∫ (P . y + ρ g y '
R
o
2
)
sin θ . dA
A
Solving for y’
1 1 ∴y = .∫ Po . y.dA + .∫ ρ g y 2 sin θ . dA FR A FR A '
ME3O04 – Chapter 3
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Hydrostatic Force on Submerged Surfaces – Cont’d 1. The magnitude of the force;
FR = Po A + ρ g ∫ y sin θ .dA A
2. The direction of the force = normal to the surface. 3. The line of action of the force.
∴ y' =
1 1 .∫ Po . y.dA + .∫ ρ g y 2 sin θ . dA FR A FR A ME3O04 – Chapter 3
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Example Problem The pressure in the air gap is 8000 Pa gage. The tank is cylindrical. Calculate the net hydrostatic force (a) On the bottom of the tank; (b) On the cylindrical sidewall CC; (c) On the annular plane panel BB.
ME3O04 – Chapter 3
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Example Problem Gate AB is a homogeneous mass of 180 kg, 1.2 m wide into the paper, resting on smooth bottom B. All fluids are at 20°C. For what water depth h will the force at point B be zero? Assume specific gravity of Glycerin = 1.26.
ME3O04 – Chapter 3
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Vocabulary List 1. 2. 3. 4. 5. 6.
Static fluid Manometer. Hydrostatic pressure. Gauge pressure. Vacuum. Hydrostatic force on a submerged surface.
ME3O04 – Chapter 3
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Outline Chapter 4- Basic Equations in Integral Form for a Control Volume. ¾ ¾ ¾
Definition of a control volume. Dot product of Two Vectors. Volume and Mass Rate of Flow through a C.V. 9 9 9
¾
Conservation of Mass. 9
¾ ¾ ¾ ¾ ¾ ¾ ¾
Volume Flux. Mass Flux. Sign Convention. Special cases.
Example Problem. Extensive and Intensive Fluid Properties. Reynolds Transport Theorem. Momentum Equation. Sign Convention of Terms in the Momentum Equation. Types of forces. Example problem.
ME3O04 – Chapter 4
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¾Conservation of Mass. ¾Conservation of Momentum
Basic Equations in Integral Form for a Control Volume Chapter 4
ME3O04 – Chapter 4
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Why do we need Basic Equations in integral Form ? In some applications we do not need details of the flow field, we rather need some global values, values such as:1. Average velocity at a certain section. 2. Force due to fluid flow. 3. Mass or Volume flow rates.
ME3O04 – Chapter 4
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Examples:Given velocity at the exit section, fine mass flow rate and average velocity at the inlet section.
ME3O04 – Chapter 4
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Examples:Find velocity V3 and force on the scale due to fluid flow.
ME3O04 – Chapter 4
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Examples:Find force on the 90° elbow.
ME3O04 – Chapter 4
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Examples:Find flow rate and force on the gate in the open and closed positions. gate
ME3O04 – Chapter 4
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Basic Equations in Integral Form
Q1
How do we obtain these equations?
A1
By applying conservation laws on a control volume.
Q2 A2
Which conservation laws?
Q3
What is a control volume?
1- Conservation of mass. 2- Conservation of momentum.
ME3O04 – Chapter 4
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Definition of a Control Volume 1. A control volume is an arbitrary volume in space through which fluid flows. The geometric boundary of the control volume is called the control surface. surface 2. The control surface may be real or imaginary. Imaginary surface
Real surface Fig 3.1
ME3O04 – Chapter 4
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Background - Dot Product of Two
Vectors
Dot product of V and dA = V dA cos α = projection of V on dA.
α
α ME3O04 – Chapter 4
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Fig 4.3 1 2 3
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Sign Convention
Fig 4.3
r r r r V . dA = V dA cos 0 = +V dA V . dA = V dA cos180 = −V dA
ME3O04 – Chapter 4
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Volume Flow Rate through a C.V.
z Volume flow rate = volume flux through dA r r = dQ = V cosα . dA = V dA cos α = V . dA z Total volume flow rate through control surface (CS)=
r r & & Q = ∫ dQ = ∫ V . dA CS
CS ME3O04 – Chapter 4
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Mass Flow Rate through a C.V.
z Mass flow rate = Mass flux through dA r r = ρ V . dA = ρ dQ = dm z Total mass flow rate through control surface (CS)=
r r m& = ∫ dm& = ∫ ρ V . dA CS
CS ME3O04 – Chapter 4
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Sign Convention
Fig 4.3
r r volume flux = V . dA = +V dA
r r volume flux = V . dA = −V dA
ME3O04 – Chapter 4
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Conservation of Mass of any System z M = mass of the system = constant. z Or, the time rate of change of the mass of the system = 0
i.e.,
dM ⎞ =0 ⎟ dt ⎠system ME3O04 – Chapter 4
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Conservation of Mass of a Control Volume z Time rate of decrease of mass within the control volume = net mass outflow rate from the control volume. z Time rate of decrease of mass within the control volume =
∂ − ρ d∀ ∫ ∂t CV
z net mass outflow rate from the control volume =
r r dM ⎞ ∂ = ρ d∀ + ∫ ρ V .dA = 0 ⎟ ∫ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
r r ∫ ρ V . dA CS
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Special Cases
r r dM ⎞ ∂ = ⎟ ∫ ρ d∀ + ∫ ρ V .dA = 0 dt ⎠ system ∂t CV CS
1. Unsteady Incompressible Flow, i.e., ρ = c
r r ∂ take ρ outside integrals, ρ d∀ + ρ ∫ V .dA = 0 ∫ ∂t CV CS
divide by ρ , but,
r r ∂ d∀ + ∫ V .dA = 0 ∫ ∂t CV CS
∫ d∀ = ∀ = the volume of the C.V. CV
thus,
r r ∂∀ + ∫ V .dA = 0 ∂t CS ME3O04 – Chapter 4
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Special Cases
1. Unsteady Incompressible Flow, i.e., ρ = c
thus,
r r ∂∀ + ∫ V .dA = 0 ∂t CS
∂∀ for a non - deformable (fixed) C.V., = 0, thus ∂t
r r ∫ V .dA = 0 CS ME3O04 – Chapter 4
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Special Cases
r r dM ⎞ ∂ = ⎟ ∫ ρ d∀ + ∫ ρ V .dA = 0 dt ⎠ system ∂t CV CS
2. Unsteady Incompressible flow through a fixed C.V. ∂∀
ρ = c, ∀ = c, or
=0
∂t r r ∂ ρ d∀ + ρ ∫ V .dA = 0 ∫ ∂t CV CS
devide by ρ ,
r r ∂ d∀ + ∫ V .dA = 0 ∫ ∂t CV CS ME3O04 – Chapter 4
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Special Cases
r r dM ⎞ ∂ = ⎟ ∫ ρ d∀ + ∫ ρ V .dA = 0 dt ⎠ system ∂t CV CS
2. Unsteady Incompressible flow through a fixed C.V.
thus,
r r ∂∀ + ∫ V .dA = 0 ∂t CS
∂∀ but = 0, thus ∂t
r r ∫ V .dA = 0 CS
ME3O04 – Chapter 4
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Special Cases
r r dM ⎞ ∂ ρ d∀ + ∫ ρ V .dA = 0 = ⎟ ∫ dt ⎠ system ∂t CV CS
3. Steady Compressible flow
∂∀ Steady ⇒ = 0, Compressible ⇒ ρ ≠ c ∂t
r r ∫ ρ V .dA = 0
(1)
CS
r r r r Recall, dm& = ρ V .dA, thus, ∫ ρ V .dA = ∑ m& CS
in (1),
r r ∑ m& = ∫ ρ V .dA = 0, i.e., m& in = m& out CS ME3O04 – Chapter 4
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Example Problem Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
⎡ r2 ⎤ u = umax ⎢1 − 2 ⎥ , ⎣ R ⎦
and umax = 10 ft/s.
ME3O04 – Chapter 4
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Solution 1. Water = incompressible fluid → ρ = constant. ∂ ∂t
2. Flowing steadily = steady flow, i.e., = 0. Conservation of mass equation:-
r r ∂ ρ d∀ + ∫ ρ V .dA = 0 ∫ ∂t CV CS
(1)
Under conditions 1 and 2, equation (1) can be written as:
r r ∫ V .dA = 0
CS ME3O04 – Chapter 4
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Extensive Vs Intensive Fluid Properties z An extensive property is one that depends on the size of the C.V. Example:- volume or mass. z An intensive property is one that does not depend on the size of the C.V. Example:- Specific volume, specific enthalpy.
ME3O04 – Chapter 4
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Reynolds Transport Theorem z Let N = any extensive property.
N = the intensive property corresponding zη = m to N.
z For any C.V., Reynolds Transport Theorem is:
r r dN ⎞ ∂ = η ρ d∀ + ∫ η ρ V .dA ⎟ ∫ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
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(I)
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Reynolds Transport Theorem z If N = mass, M
N =1 ⇒ η= m
in (I),
z Conservation of mass equation:
r r dM ⎞ ∂ = ρ d∀ + ∫ ρ V .dA = 0 ⎟ ∫ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
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Interpretation of each term in R.T.T.
r r dN ⎞ ∂ η ρ d∀ + ∫ η ρ V .dA = ⎟ ∫ dt ⎠ system ∂t CV CS
(I)
dN ⎞ z = time rate of change of any extensive ⎟ dt ⎠ system property of the system.
∂ η ρ d∀ = time rate of change of N within the C.V. z ∫ ∂t CV z
ρ d∀ =
z
η ρ d∀ =
mass of an element contained in the C.V. amount of N in that element. ME3O04 – Chapter 4
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Interpretation of each term in R.T.T.
r r dN ⎞ ∂ = η ρ d∀ + ∫ η ρ V .dA ⎟ ∫ dt ⎠ system ∂t CV CS
(I)
r r z ∫ η ρ V .dA = the net rate of flux of N out through the C.S.
r r ρ V .dA =
CS
z
the rate of mass flux exiting dA = mass flow rate.
ME3O04 – Chapter 4
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Momentum Equation for an Inertial (not accelerating) C.V.
r r P = linear momentum = M V
r r Let N = P = M V In R.T.T.
N r ∴η = =V M
r r dN ⎞ ∂ η ρ d∀ + ∫ η ρ V .dA = ⎟ ∫ dt ⎠ system ∂t CV CS
r r r r r dP ⎞ ∂ ⎟ V ρ d∀ + ∫ V ρ V .dA = ∫ ⎟ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
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Momentum Equation “Newton’s Second Law” for an Inertial C.V.
r r dP ∑ F = time rate of change of momentum = dt from R.T.T.,
r r r r r dP ⎞ ∂ ⎟ = V ρ d∀ + ∫ V ρ V .dA ∫ ⎟ dt ⎠ system ∂t CV CS
r ∂ r r r r ∴∑ F = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
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Momentum Equation – meaning of each term
r ∂ r r r r ∴∑ F = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS (A)
=
(B)
+
(C)
z (A) = sum of all forces acting on the fixed C.V. z (B) = time rate of change of momentum inside the C.V. z (C) = net rate of flux of momentum out through the C.S. ME3O04 – Chapter 4
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r r r Types of Forces ∑ F = ∑ FS + ∑ FB r r ∑ FS = surface forces, e.g., pressure ⇒ Fs = ∫ − P dA r r ∑ FB = body forces, e.g., gravity ⇒ FB =
CS
∫ ρ g d∀
CV
sub in
r ∂ r r r r ∑ F = ∂t ∫ V ρ d∀ + ∫ V ρ V .dA CV CS
r r r r r r ∂ FS + FB = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
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Scalar Equations of the Momentum Equation Note
r r r r r r ∂ FS + FB = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS
(I)
Note
z Equation (I) is a vector equation. Therefore, it may be written as three scalar component equations. Note z In Cartesian coordinates (x, y, z), equation (I) can be written as:
Note
r r ∂ x - diretion equation : Fsx + FBx = u ρ d∀ + ∫ u ρ V .dA ∫ ∂t CV CS r r ∂ y - diretion equation : Fsy + FBy = v ρ d∀ + ∫ v ρ V .dA ∫ ∂t CV CS r r ∂ z - diretion equation : Fsz + FBz = w ρ d∀ + ∫ w ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
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Sign Convention of Terms in the Momentum Equation.
r r r r r r ∂ FS + FB = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS
(I)
z F is positive if it is in the positive direction of the coordinate. z V is positive if it is in the positive direction of the coordinate. r r r z Sign of r V .dA depends on the relative directions of V and dA . ME3O04 – Chapter 4
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Problem A jet of water issuing from a stationary nozzle at 15 m/s (Aj = 0.05 m2) strikes a turning vane mounted on a cart as shown. The vane turns the jet through angle θ = 50º. Determine the value of mass, M, required to hold the cart stationary. If the vane angle θ is adjustable, plot the mass, M, needed to hold the cart stationary versus θ for 0 ≤ θ ≤ 180°.
ME3O04 – Chapter 4
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Solution Procedure for This Type of Problems 1. Identify your control volume, C.V. 2. Identify its C.S., and number important sections to consider. 3. Identify your coordinates and draw them. 4. Identify which equation (s) will be used to solve the problem. 5. Identify which assumptions you can make to simplify your equations. ME3O04 – Chapter 4
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Solution Procedure for This Type of Problems 6. Common assumptions are:a. Steady. b. Incompressible, i.e., density, ρ = constant . c. Uniform flow. d. Body force = 0. e. Effect of atmospheric pressure is negligible.
ME3O04 – Chapter 4
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Solution 1. Identify your control volume, C.V. 2. Identify its C.S., and number important sections to consider.
ME3O04 – Chapter 4
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Solution 1. Identify your coordinates and draw them.
ME3O04 – Chapter 4
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Solution – Cont’d 4. Identify which equation (s) will be used to solve the problem.
Rx
Momentum Equation : r r ∂ x - diretion equation : Fsx + FBx = u ρ d∀ + ∫ u ρ V .dA ∫ ∂t CV CS r r ∂ y - diretion equation : Fsy + FBy = v ρ d∀ + ∫ v ρ V .dA ∫ ∂t CV CS r r ∂ z - diretion equation : Fsz + FBz = w ρ d∀ + ∫ w ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
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Solution – Cont’d r r ∂ Fsx + FBx = u ρ d∀ + ∫ u ρ V .dA ∫ ∂t CV CS
Rx
5. Identify which assumptions you can make to simplify your equations. 6. Common assumptions are:9 a. Steady. 9 b. Incompressible, i.e., density, ρ = constant . 9 9 c. Uniform flow. 9 d. Body force = 0. e. Effect of atmospheric pressure is negligible. ME3O04 – Chapter 4
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Solution – Cont’d M=
ρV 2 A g
(1 − cos θ )
Rx
ME3O04 – Chapter 4
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Vocabulary List 1. 2. 3. 4. 5. 6. 7.
Basic equations (conservation laws). Control volume. Dot product of two vectors. Volume flux. Extensive Property. Intensive Property. Reynolds Transport Theorem
ME3O04 – Chapter 4
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Exercise Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
⎡ r2 ⎤ u = umax ⎢1 − 2 ⎥ , ⎣ R ⎦
and umax = 10 ft/s.
ME3O04 – Chapter 4
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Solution 1. Water = incompressible fluid → ρ = constant. ∂ ∂t
2. Flowing steadily = steady flow, i.e., = 0. Conservation of mass equation:-
r r ∂ ρ d∀ + ∫ ρ V .dA = 0 ∫ ∂t CV CS
(1)
Under conditions 1 and 2, equation (1) can be written as:
r r ∫ V .dA = 0
CS
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution – cont’d
r r ∫ V .dA = 0
CS
⇒
r r r r ∫ V .dA + ∫ V .dA = 0 1
(a)
2
r r ⎡ r2 ⎤ at (1) V = U = constant, at (2) V = umax ⎢1- 2 ⎥, dA = 2π r dr ⎣ R ⎦ R
in (a),
R
r2 − ∫ U .2 π rdr + ∫ + umax (1 − 2 ) 2 π r dr = 0 R 0 0
Note the sign of the first and the second integrals. ME3O04 – Chapter 4
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Solution – cont’d
R
in (a),
R
r2 − ∫ U .2 π rdr + ∫ umax (1 − 2 ) 2 π r dr = 0 R 0 0 R
⎡r r ⎤ − U .π R + 2 π umax ⎢ − 2 ⎥ = 0 ⎣ 2 4R ⎦ 0 2
4
2
Solving (b) for U :
U=
(b)
umax = 5.0 ft/s. 2
ME3O04 – Chapter 4
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Outline Chapter 5- Differential Analysis of Fluid Motion. ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾
¾
Why differential. Region of interest. Conservation of Mass for an infinitesimal C.V. (fluid particle) Special cases. Lagrangian and Eulerian descriptions of the motion of a fluid particle. Material, substantial, or particle derivative. Example problem. Particle acceleration. Types of motion of a fluid particle. Rotation and vorticity vectors. Fluid deformation: 1. Angular deformation. 2. Linear deformation. Differential momentum equation. ME3O04 – Chapter 5
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Differential Analysis of Fluid Motion Why differential? To obtain detailed knowledge, we must apply the basic equations of fluid motion in differential form.
ME3O04 – Chapter 5
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Region of interest Since we are interested in formulating differential equations, our analysis will be in terms of infinitesimal systems or infinitesimal control volumes (i.e., differential C.V.)
Fig 5.1
ME3O04 – Chapter 5
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Conservation of Mass for an infinitesimal C.V. (fluid particle)
(I) =Time rate of decrease of mass inside C.V. = (II) = net rate of mass flux (mass flow rate) out through the C.S. ME3O04 – Chapter 5
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Conservation of Mass for an infinitesimal C.V. (fluid particle) (I) = Time rate of decrease of mass inside C.V. =
∂ρ =− d∀ ∂t
∂ρ =− dx dy dz ∂t
ME3O04 – Chapter 5
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(II) = net rate of mass flux (mass flow rate) out through the C.S. in the x-direction =
∂( ρ u) ⎞ ⎛ = ⎜ρu + dx ⎟dy dz − (ρ u ) dy dz ∂x ⎝ ⎠
∂( ρ u) =+ dx dy dz ∂x
ME3O04 – Chapter 5
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Conservation of Mass for an infinitesimal C.V. (fluid particle) (II) = net rate of mass flux (mass flow rate) out through the C.S. in the three directions =
⎡ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) ⎤ + + dx dy dz ( II ) = + ⎢ ⎥ ∂y ∂z ⎦ ⎣ ∂x ∂ρ ∂ρ (I ) = − d∀ = − dx dy dz ∂t ∂t
∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) ∴ + + + =0 ∂t ∂x ∂y ∂z ME3O04 – Chapter 5
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Conservation of Mass - Special Cases ∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + = 0 ⇐ Compressible , Unsteady + ∂x ∂y ∂z ∂t 1. Incompressible Flow:- ρ = constant ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + =0 ∂x ∂y ∂z
2. Steady Flow:-
∂ =0 ∂t
⇒
∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + =0 ∂x ∂y ∂z
⇒
⇒
∂ρ =0 ∂t
∂u ∂v ∂ w + =0 + ∂x ∂y ∂z r i.e., ∇. V = 0
∂ρ =0 ∂t ⇒
r i.e., ∇. ( ρV ) = 0
ME3O04 – Chapter 5
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Example Problem Which of the following sets of equations represent possible two-dimensional incompressible flow cases? a)
u = 2x2 + y2 − x2 y v = x 3 + x( y 2 − 2 y )
b)
u = 2 xy − x + y 2
v = 2 xy − y 2 + x 2 ME3O04 – Chapter 5
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Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle 1. Lagrangian Description:In the Lagrangian description, anyrfluid property, “F”, r is function of the position vector X and time (t), i.e., X
r F = F ( X , t) r r dX ∴ particle velocity, V = dt r r 2 r dV d X and particle acceleration, a = = 2 dt dt ME3O04 – Chapter 5
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Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle 2. Eulerian Description:In the Elurian description, any fluid property, “F”, is function of the space coordinates x, y, z, and time (t), i.e.,
F = F ( x, y , z , t )
∂x ∴ = local velocity in the x - direction ∂t (not a certain particle velocity). ∂F = local change of F (at a certain x, y, z location) ∂t Eulerian description is used in most fluid mechanics problems. ME3O04 – Chapter 5
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Problem The temperature, T, in a long tunnel is known to vary approximately as:
T = To − α e
−x
L
sin( 2 π t / τ )
where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel. A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle. Is the required rate of change equal to
∂T ? ∂t
ME3O04 – Chapter 5
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Substantial Derivative If we want to find the time rate of change of any fluid property,”F”, following a certain particle and still use Eulerian descriction, we have to use what is called a “substantial” substantial or “particle” particle or “material” derivative. derivative
ME3O04 – Chapter 5
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Substantial Derivative In the Eulerian description:-
F = F(x, y, z, t)
∂F ∂F ∂F ∂F dF = dt + dx+ dy+ dz ∂t ∂x ∂y ∂z devide by dt,
dF ∂F ∂F dx ∂F dy ∂F dz = + + + dt ∂t ∂x dt ∂y dt ∂z dt
(1)
If we are following a certain particle:-
dx =u , dt
dy =v , dt
dz =w dt
ME3O04 – Chapter 5
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Substantial Derivative dF ∂F ∂F dx ∂F dy ∂F dz = + + + dt ∂t ∂x dt ∂y dt ∂z dt
dx = u, dt in (1) or,
dy = v, dt
(1)
dz =w dt
∂F ∂F ∂F DF ∂F = +u +w +v Dt ∂t ∂y ∂x ∂z DF ∂F r = + V .∇F Dt ∂t ME3O04 – Chapter 5
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Substantial Derivative – meaning of each term ∂F DF ∂F ∂F ∂F =
+u
+v
+w
∂t ∂x ∂y ∂z DF ∂F r or, + V .∇F = Dt ∂t DF = total time rate of change of F seen or experienced Dt by a certain particle. Dt
∂F = local time rate of change of F. ∂t r V .∇F = convective rate of change = rate of change due to the motion of the particle. ME3O04 – Chapter 5
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Problem The temperature, T, in a long tunnel is known to vary approximately as:
T = To − α e
−x
L
sin( 2 π t / τ )
where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel. A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle.
ME3O04 – Chapter 5
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Particle Acceleration r r V = V ( x, y , z , t ) In the Eulerian description:r r ∂V DV = local acceleration = particle acceleration = a p ∂t Dt r r r r r ∂V ∂V ∂V DV ∂V +v +w ∴ap = = +u (I) ∂z Dt ∂t ∂x ∂y r ∂V r r or, ap = + V .∇V ∂t Note:- Equation (I) is a vector equation, so three equations can be written in the three coordinates. ME3O04 – Chapter 5
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Particle Acceleration
or,
r r r r r ∂V ∂V ∂V DV ∂V ∴ap = = +u +v +w Dt ∂t ∂x ∂y ∂z r ∂V r r + V .∇V ap = ∂t
∴in x - direction : ∴in y - direction :
∴in z - direction :
a px = a py =
a pz
(I)
Du ∂u ∂u ∂u ∂u = +u +v +w Dt ∂t ∂z ∂x ∂y ∂v ∂v Dv ∂v ∂v = +u +v + w Dt ∂t ∂z ∂x ∂y
Dw ∂w ∂w ∂w ∂w = = +u +v +w Dt ∂z ∂t ∂x ∂y
ME3O04 – Chapter 5
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Types of Motion of A Fluid Particle (Kinematics) 1. 2. 3. 4.
Translational. Rotation. Linear Deformation. Angular Deformation.
Fig 5.5 ME3O04 – Chapter 5
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Deformation and Rotation of a Fluid Element
Rate of Deformation =
(Δα + Δβ ) Δt
Rate of rotation = average rotational speed =
ME3O04 – Chapter 5
1 (Δα − Δβ ) 2 Δt 1 2 3
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Rotation of a Fluid Element Δα =
Δη , Δx
Δβ =
Δξ Δy
(1)
⎛ ∂u ∂u ⎞ Δξ = ⎜⎜ u + Δy ⎟⎟.Δt − u.Δt = Δy.Δt ∂y ∂y ⎠ ⎝ ∂v ∂v ⎞ ⎛ Δη = ⎜ v + Δx ⎟.Δt − v.Δt = Δx.Δt ∂x ⎠ ∂x ⎝
in (1)
∴ Δα =
∂v , ∂x
Δβ =
∂u ∂y
(2)
ME3O04 – Chapter 5
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Rotation Vector
1 Δα − Δβ Average rotational velocity = ωz = 2 Δt
1 ⎛ ∂v ∂u ⎞ ∴ωz = ⎜⎜ − ⎟⎟ 2 ⎝ ∂x ∂y ⎠
from (2)
ωz = component of the rotation vector about the z-axis.
1 ⎛ ∂w ∂v ⎞ ωx = ⎜⎜ − ⎟⎟, 2 ⎝ ∂y ∂z ⎠
1 ⎛ ∂u ∂w ⎞ ωy = ⎜ − ⎟ 2 ⎝ ∂z ∂x ⎠
iˆ
r 1 r 1 ∂ ) 1 ˆ ˆ ω = rotation vector = ωx i + ωy j + ωw k = ∇ × V = curl V = 2 2 ∂x 2 u r
ME3O04 – Chapter 5
1 2 3
ˆj
kˆ
∂ ∂y v
∂ ∂z w 23
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Vorticity Vector
r ) 1 ω = rotation vector = ωx iˆ + ωy ˆj + ωw k = ∇ × V = 2 ˆj iˆ kˆ r
r 1 ∂ 1 = curl V = 2 2 ∂x u
∂ ∂y v
∂ ∂z w
iˆ r r r ∂ ζ = vorticity vector = 2ω = ∇ × V = ∂x u ME3O04 – Chapter 5
ˆj ∂ ∂y v
kˆ ∂ ∂z w
1 2 3
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Fluid Deformation: (1) Angular Deformation
Rate of angular deformation in x - y plane = but,
Δα =
∂v , ∂x
Δβ =
∂u ∂y
Δα + Δβ Δt
⎛ ∂v ∂u ⎞ ∴ Rate of angular deformation in x - y plane = ⎜⎜ + ⎟⎟ ⎝ ∂x ∂y ⎠ ⎛ ∂w ∂u ⎞ ∴ Rate of angular deformation in x - z plane = ⎜ + ⎟ x ∂ ∂z ⎠ ⎝ ⎛ ∂w ∂v ⎞ + ⎟⎟ ∴ Rate of angular deformation in y - z plane = ⎜⎜ ⎝ ∂y ∂z ⎠ ME3O04 – Chapter 5
1 2 3
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Types of Motion of A Fluid Particle (Kinematics) 1. 2. 3. 4.
Translational. Rotation. Linear Deformation. Angular Deformation.
Fig 5.5 ME3O04 – Chapter 5
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Fluid Deformation: (2) Linear Deformation
∂u ⎤ ⎡ + Δx ⎥ Δt − u Δt u ⎢⎣ ∂u ∂x ⎦ = Δt Linear deformation in x - direction = Δx ∂x
∂u Rate of linear deformation in x - direction = ∂x ME3O04 – Chapter 5
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Fluid Deformation: (2) Linear Deformation
∂u Rate of linear deformation in x - direction = ∂x Similarly, rate of linear deformation in y - direction =
∂v ∂y
∂w and rate of linear deformation in z - direction = ∂z ME3O04 – Chapter 5
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Rate of Volume Deformation or Dilation
Rate of volume deformation or dilation = ∂u ∂v ∂w + + = ∂x ∂y ∂z r = ∇.V
ME3O04 – Chapter 5
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Recall - Conservation of Mass - Special Cases
∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + + =0 ∂t ∂x ∂y ∂z Incompressible Flow:- ρ = constant
∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + =0 ∂x ∂y ∂z
⇒
⇐ Compressible ⇐ , Unsteady
⇒
∂ρ =0 ∂t
∂u ∂v ∂ w + + =0 ∂x ∂y ∂z
r i.e., ∇.V = 0 i.e., for an incompressible fluid, rate of volume deformation = 0 ME3O04 – Chapter 5
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Differential Momentum Equation. For an incompressible fluids with constant density and viscosity, momentum equations are:1) x–direction,
2) y-direction,
3) z-direction,
ME3O04 – Chapter 5
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Outline Chapter 6- Incompressible Inviscid Flow ¾ Momentum equation - special cases: 9 Incompressible flow with constant viscosity. 9 Inviscid (frictionless flow), i.e., μ = 0 - Euler’s equation. ¾ Euler’s equations along a streamline. ¾ Bernoulli’s Equation. ¾ Hydrostatic, Static, Dynamic, and Stagnation Pressures. ¾ Applications of Bernoulli’s equation. ¾ Energy Grade Line (EGL) and Hydraulic Grade Line (HGL). ¾ Example problem. ME3O04 – Chapter 6
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Differential Momentum Equation. For an incompressible fluids with constant density and viscosity, momentum equations are:1) x–direction,
2) y-direction,
3) z-direction,
ME3O04 – Chapter 6
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Incompressible Inviscid Flow z Incompressible means ρ = constant.
z Inviscid means that viscosity μ = 0.
ME3O04 – Chapter 6
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Differential Momentum Equation
Or,
For an incompressible fluid (ρ = c) with constant viscosity, momentum equation is:r r Inviscid ⇒ μ = 0 DV 2 ρ = ρ g − ∇p + μ ∇ V Dt
1) x–direction,
2) y-direction,
3) z-direction,
ME3O04 – Chapter 6
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Euler’s Equation For an incompressible (ρ = c) , inviscid flow (μ = 0), 0) momentum equation is:- DVr ρ = ρ g − ∇p Dt Or, 1) x–direction,
2) y-direction,
3) z-direction, ME3O04 – Chapter 6
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Euler’s Equation in Streamline Coordinates z Euler’s equations shown in the previous slide are written using x-y-z coordinates. z In steady flow a fluid particle will move along a streamline because, for a steady flow, pathlines and streamlines coincide. z Thus, in describing the motion of a fluid particle in a steady flow, the distance along a streamline is a logical coordinate to use in writing the equations of motion. ME3O04 – Chapter 6
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Fluid Particle Moving along a Streamline
Fig 6.1 ME3O04 – Chapter 6
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Euler’s Equation in Streamline Coordinates 1. Apply Newton’s 2nd Law in the s-direction
∑ F = FS + FB = m as
∂p ds ⎤ ∂p ds ⎤ ⎡ ⎡ p dn . dx p dn.dx − ρ g sin β d∀ = ρ d∀ as − + − ⎢ ⎥ ⎢⎣ ⎥ ∂s 2 ⎦ ∂s 2 ⎦ ⎣ ME3O04 – Chapter 6
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Euler’s Equation in S-direction ∂p ds ⎤ ∂p ds ⎤ ⎡ ⎡ ⎢⎣ p − ∂s 2 ⎥⎦ dn.dx − ⎢⎣ p + ∂s 2 ⎥⎦ dn.dx − ρ g sin β d∀ = ρ d∀ as (1) dz but, sin β = ds r r r r ∂z DV ∂V ∂V 1 ∂p − g = as = = +V in (1) − ∂s Dt ∂t ρ ∂s ∂s If we neglect the body force and consider only steady flow:-
r 1 ∂p r ∂V − =V ∂s ρ ∂s ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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Euler’s Equation in Streamline Coordinates 2. Apply Newton’s 2nd Law in the n-direction
∑ F = FS + FB = m an
∂p dn ⎤ ∂p dn ⎤ ⎡ ⎡ − − + p ds . dx p ds.dx − ρ g cos β d∀ = ρ d∀ an ⎢⎣ ⎥ ⎥ ⎢ ∂n 2 ⎦ ∂n 2 ⎦ ⎣ ME3O04 – Chapter 6
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Euler’s Equation in n-direction ∂p dn ⎤ ∂p dn ⎤ ⎡ ⎡ ⎢ p − ∂n 2 ⎥ ds.dx − ⎢ p + ∂n 2 ⎥ ds.dx − ρ g cos β d∀ = ρ d∀ an ⎣ ⎦ ⎣ ⎦ (2)
but,
dz cosβ = dn
r2 ∂z V 1 ∂p −g in (2) − = an = − ∂n ρ ∂n R an = centripetal acceleration, R = radius of curvature. If we neglect the body force and consider only steady flow:-
r2 1 ∂p V = ρ ∂n R
ME3O04 – Chapter 6
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Euler’s Equation in n-direction
r2 1 ∂p V = ρ ∂n R
z This equation indicates that pressure increases in the direction outwards from the center of curvature of the streamline.
∂p =0 z In case of flow in a straight line:- R = ∞ ∴ ∂n i.e., there is no variation in pressure in the n-direction. ME3O04 – Chapter 6
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Measurement of Static Pressure
r2 1 ∂p V = ρ ∂n R ∂p R=∞ ∴ =0 ∂n ME3O04 – Chapter 6
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Measurement of Static Pressure
Can we put the pressure gage at the elbow?
r2 1 ∂p V = ρ ∂n R ME3O04 – Chapter 6
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Bernoulli’s Equation z Bernoulli’s equation results from the integration of Euler’s equation along a streamline for a steady flow. z Euler’s equation in s-direction (along a streamline):r r ∂z ∂V r ∂V 1 ∂p − −g = +V ∂s ∂t ∂s ρ ∂s
z For steady flow:-
r r ∂z ∂V 1 ∂p − − g =V ∂s ∂s ρ ∂s ME3O04 – Chapter 6
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Bernoulli’s Equation – Cont’d Since we are moving a long a streamline, thus, p=p(s,t) ∂p ∂p ∂p ∴ dp = ds ds + dt = ∂s ∂t ∂s ∂z ∂z ∂z and, dz = ds + dt = ds ∂s ∂t ∂s r r r r ∂V ∂V ∂V ds + dt = ds and, dV = ∂s ∂t ∂s Recall (3)
r 1 ∂p ∂z r ∂V − − g =V ρ ∂s ∂s ∂s ME3O04 – Chapter 6
(3)
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1
r r dp − g dz = V d V
in (3)
−
or,
r r dp V d V + g dz + =0
ρ
ρ
(4)
z Integration of (4) along S:-
r2 V dp +gz+∫ = constant 2 ρ
(5)
z For an incompressible fluid ρ = c, in (5):
r2 V p + g z + = constant 2 ρ
⇐ Bernoulli’s equation
ME3O04 – Chapter 6
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Restrictions of The Application of r2 Bernoulli’s Equation V p + g z + = constant 2 ρ Bernoulli’s equation is a very powerful tool, but is has to be used very carefully. carefully Because it has very strict applicability limitations. 1. Incompressible flow, i.e., ρ = c, Mach number, Ma < 0.3. 2. Inviscid flow, i.e., μ = 0.
∂ 3. Steady flow, i.e., =0 ∂t 4. Along a streamline. ME3O04 – Chapter 6
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Hydrostatic, Static, Dynamic, and Stagnation Pressures 1. Hydrostatic pressure is pressure resulting from weight of a fluid column. P = ρ g h. 2. Ps=Static pressure is pressure due to the thermodynamic state of the fluid. 3. Pd=Dynamic pressure is pressure due to the velocity of r2 the fluid. ρV Pd =
2
4. Po=Stagnation pressure is pressure exerted by a moving fluid when brought from motion to rest. Po=Ps + Pd ME3O04 – Chapter 6
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Measurement of Static Pressure, Ps
Fig 6.2
ME3O04 – Chapter 6
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Measurement of Stagnation Pressure
Po = Ps + Pd
ME3O04 – Chapter 6
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Simultaneous Measurement of Static and Stagnation Pressures
r2 r2 P V Po Vo From Bernoulli’s + + gzo = + + gz ρ 2 ρ 2 r2 r V ∴ Po = P + ρ Vo = 0 and z o = z, 2 ME3O04 – Chapter 6
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Applications of Bernoulli’s Equation r2 V p + g z + = constant 2 ρ Bernoulli’s equation can be written for any two points along a same stream line as:
r2 r2 V1 p1 V2 p2 + g z1 + = + g z2 + 2 ρ 2 ρ ME3O04 – Chapter 6
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Measurement of Fluid Velocity at a Point Apply Bernoulli’s equation between points A and B: r2 r2 PA VA PB VB + + g zA = + + g zB (1) ρ ρ 2 2
Note :
zA = z B
and VB = 0, thus PB = Po
r2 P P V in (1) A + A = o ρ ρ 2
∴ VA =
2 (Po − PA )
ME3O04 – Chapter 6
ρ
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Applications – Nozzle Flow
One can relate info at section 2 to those at section 1 using:
r2 r2 V1 p1 V2 p2 + g z1 + = + g z2 + 2 ρ 2 ρ ME3O04 – Chapter 6
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Flow through a Siphon One can relate info at sections 1, A, and 2 using:
r2 r2 r2 V1 p1 VA p A V2 p2 + g z1 + = + g zA + = + g z2 + 2 ρ 2 ρ 2 ρ
Note : P1 = P2 = Patm. Since area reservoir >> area pipe
then, V1 ≈ 0 z If one uses Patm =0, this means that pressures are gage. z If one uses Patm = 101.3 kPa, this means that pressures are absolute. ME3O04 – Chapter 6
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Meaning of Each Term in Bernoulli’s Equation r2 p V + + g z = constant r2 ρ 2 p V devide by g : + +z = H ρ g 2g p p∀ = = flow energy per unit weight of the flowing fluid= ρ g mg head due to local static pressure. r r 2 mV 2 V 2 = kinetic energy per unit weight = head due to local = 2g mg dynamic pressure. mg z z= = Potential energy per unit weight = head mg due to elevation. H = total mechanical energy per unit weight = total head of the flow ME3O04 – Chapter 6
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Example Problem A tank with a reentrant orifice called a Borda mouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.
ME3O04 – Chapter 6
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
64 4EGL 7 44 8 r2 p V +z+ =H 2g ρg 123 HGL
z EGL is a line representing the total head. z HGL is a line representing the sum of elevation heads and static pressure head. z The difference EGL-HGL = dynamic head. ME3O04 – Chapter 6
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
64 4EGL 7 44 8 r2 p V +z+ =H 2g ρg 123 HGL
z At point (1) p1 = patm.=0 (gage) and V1 = 0, thus H1 = z1. z At point 4, p4=patm. = 0, thus the height of HGL = z4.
ME3O04 – Chapter 6
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
64 4EGL 7 44 8 r2 p V +z+ =H ρg 2g 123 HGL
z Flow through a constant cross section will have a horizontal HGL (bec. V = c).
ME3O04 – Chapter 6
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Example 1 A tank with a reentrant orifice called a Borda mouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.
ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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+
ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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Aj
1 = ∴ Cc = Ao 2 ME3O04 – Chapter 6
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Example 2 Heavy loads can be moved with relative ease on air cushions by using a load pallet as shown. Air is supplied from the plenum through porous surface AB. It enters the gap vertically at uniform speed, q. Once in the gap, all air flows in the positive x direction (there is no flow in across the plane at x = 0) Assume air flow in the gap is incompressible and uniform at each cross section, with speed u(x) as shown in the enlarged view. Although the gap is narrow (h<
ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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Water is admitted at the center pipe of the platform shown below at a rate of 1 m3/s and discharged into the air around the periphery. The upper circular plate in the figure is horizontal and is fixed in position to the ceiling. The lower annular plate is free to move vertically and is not supported by the pipe. The annular plate weighs 30 N, and the weight of water on it should be considered. If the distance, d, between the two plates is to be maintained at 3.5 cm, what is the total weight W that this platform can support?
ME3O04 –Review
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Water is discharged from a narrow slot in a 150 mm diameter pipe. The resulting horizontal two-dimensional jet is 1 m long and 15 mm thick, but of non-uniform velocity. The pressure at the inlet section is 30 kPag. Calculate (a) the volume flow rate at the inlet section and (b) the forces required at the coupling to hold the spray pipe in place. Neglect the mass of the pipe and the mass of water it contains. Coupling
ME3O04 –Review
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Outline Chapter 7- Dimensional Analysis and Similitude ¾ ¾ ¾
Meaning of Similitude. Dimensionless numbers. Methods of dimensional analysis:1. Nondimensionalizing the basic differential equations. 2. Using Buckingham PI Theorem.
¾ ¾ ¾
Buckingham PI Theorem. Procedure to determine the PI groups (illustrative example - Drag force on a sphere). Significant Dimensionless Groups in Fluid Mechanics.
¾ ¾ ¾ ¾ ¾
Significant Dimensionless Groups in Fluid Mechanics. Flow Similarity and Model Studies. Scaling with Multiple Dependent Parameters. Scaling with Multiple Dependent Parameters. Example Problem 1
ME3O04 - Chapter 7
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Similitude V 2 m2 s 2 1 = 2 gh s m m
??
= dimensionless
One of the important dimensionless 2
numbers in this problem =
V gh 2
ME3O04 - Chapter 7
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Similitude 2V ??
V V = gh1 gh2 2 1
2 2
h2 2 V = V1 h1 2 2
V =2 V 2 2
2 1
i.e., V2 = 2 V1 = 2 V 3
ME3O04 - Chapter 7
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Roasting Time of a Turkey What are the parameters affecting roasting time of a turkey (t) ?
t = constant
ρm
c
3 1. Mass, m p 2. Density, ρ 3. Thermal conductivity, k 4. Specific heat, cp Using dimensional analysis we can show that:-
2
k
Or,
tk
=constant = dimensionless number
cp 3 ρ m2 4
ME3O04 - Chapter 7
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Roasting Time of a Turkey
⎛ tk ⎞ ⎛ tk ⎞ ⎜ ⎟ ⎟ ⎜ = ⎜ c 3 ρ m2 ⎟ ⎜ c 3 ρ m2 ⎟ ⎠bird no. 2 ⎝ p ⎠bird no.1 ⎝ p So, if we know the cooking time of bird no.1, we can calculate the cooking time of bird no. 2 from this equation.
5
ME3O04 - Chapter 7
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Drag Force on a Sphere Drag force on a sphere, F depends on:1. Diameter, D 2. Fluid density, ρ 3. Fluid viscosity, μ 4. Fluid velocity, U Dimensional analysis shows that this problem is governed by two dimensionless numbers :
And that
⎛ ρ VD ⎞ ⎟⎟ = f ⎜⎜ 2 2 ρV D ⎝ μ ⎠ F
6
ME3O04 - Chapter 7
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z Does the size of sphere (i.e., D) matter in using this figure? z Does the type of fluid (i.e., ρ and μ) matter in using this figure? 7
ME3O04 - Chapter 7
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Similitude z So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarly. similarly z The “QUESTION” now is how can we determine important “Dimensionless Numbers” Numbers in any problem of interest? z Methods to do that: 1. Nondimensionalizing basic differential equations. 2. Using Buckingham PI Theorem. 8
ME3O04 - Chapter 7
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Nondimensionalizing the Basic Differential Equations Example: Steady, incompressible, two-dimensional flow, of a Newtonian fluid, with constant viscosity. Assumptions: ∂ ⇒ =0 1. Steady ∂t 2. Incompressible ⇒ ρ =c ∂ ⇒ =w=0 3. Two-dimensional ∂z 4. Newtonian fluid with constant μ. 9
ME3O04 - Chapter 7
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For an incompressible fluid with constant viscosity, momentum equations are:1) x–direction,
2) y-direction,
3) z-direction,
10
ME3O04 - Chapter 7
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Momentum Equations after Applying all Other Assumptions 1) x–direction,
2) y-direction,
3) z-direction,
x − direction
⎡ ∂ 2u ∂ 2u ⎤ ⎡ ∂u ∂u ⎤ ∂p ρ ⎢u + v ⎥ = − + μ ⎢ 2 + 2 ⎥ ∂y ⎦ ∂x ∂y ⎦ ⎣ ∂x ⎣ ∂x
(2)
y − direction
⎡ ∂ 2v ∂ 2v ⎤ ⎡ ∂v ∂v ⎤ ∂p ρ ⎢u + v ⎥ = − − ρg + μ ⎢ 2 + 2 ⎥ ∂y ⎦ ∂y ∂y ⎦ ⎣ ∂x ⎣ ∂x
(3) 11
ME3O04 - Chapter 7
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Non- Dimensionalizing Equations 1 and 2 z Non-dimensionalizing these equations means : 1. To divide all lengths by a reference length, L;
x x = , L *
y y = L *
2. To divide all velocities by a reference velocity, V∞;
u v * u = , v = V∞ V∞ *
z Note: dimensionless quantities are denoted with asterisks: 12
ME3O04 - Chapter 7
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Non- Dimensionalizing Equations 1 and 2
3. To divide pressure by twice the dynamic head
= ρV∞2.
p p = 2 ρ V∞ *
z Note: dimensionless quantities are denoted with asterisks:
13
ME3O04 - Chapter 7
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Equations in Dimensionless Form Momentum, x − direction : * * * 2 * 2 *⎤ ⎡ μ u u p u u ∂ ∂ ∂ ∂ ∂ u * * + v* * = − * + + ⎢ ⎥ ρ V∞ L ⎣ ∂x*2 ∂y *2 ⎦ ∂x ∂y ∂x
(2)
Momentum, y − direction :
μ ⎡ ∂ 2 v* ∂ 2 v* ⎤ ∂v* * ∂v* ∂p* gL u =− * − 2 + +v ⎢ *2 + *2 ⎥ * * ∂x ∂y ∂y V∞ ρ V∞ L ⎣ ∂x ∂y ⎦ *
ρ V∞ L Re = = Reynolds number μ
Fr =
(3)
V∞2 = Froud number gL 14
ME3O04 - Chapter 7
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Recall - Similitude z So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarly. z The “QUESTION” now is how can we determine these important “Dimensionless Numbers” Numbers in any problem of interest? z Methods to do that: 9 1. Nondimensionalizing the basic differential equations. Ö 2. Using Buckingham PI Theorem. 15
ME3O04 - Chapter 7
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Methods to find the Dimensionless Numbers Relevant to a Certain Problem of Interest z
The method of non-dimensionalizing differential equations depends on knowing which equations to use, which is not always the case.
z
If we do not know the equations, we use the second method known as the “Buckingham PI Theorem” .
z
Before we discuss the Buckingham PI Theorem, we need first to discuss two important concepts:1. Fundamental, or Independent, or Primary Dimensions. 2. Dimensional homogeneity.
16
ME3O04 - Chapter 7
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Fundamental, or Independent, or Primary Dimensions z Dimensions of all physical quantities can be expressed in terms of a group of “independent or primary dimensions”. z These primary dimensions are: L 1. Mass, M; Velocity = V = = L. t −1 2. Length, L; t 3. Time, t; V L −2 = = = L . t , Accelerati on 4. Temperature, T. 2 t t z Examples:-
L Force = ma = M . 2 = M .L.t −2 t
17
ME3O04 - Chapter 7
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Dimensional Homogeneity Any valid equation that relates physical quantities must be dimensionally homogeneous, homogeneous i.e., each term in the equation must have the same dimensions. Example: Newton’s Law of Viscosity
∂u τ =μ ∂y L.H.S. of (1)
(1)
force L 1 M τ= = M. 2 . 2 = 2 area t L t .L 18
ME3O04 - Chapter 7
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Dimensional Homogeneity
∂u τ =μ ∂y Qμ =
N.s m2
∴ R.H.S. of (1)
(1) =
Equation (1) can be written as :
M .L t L 1 M = . . t 2 L2 t L t 2 .L
τ ∂u μ ∂y
− 1 = 0,
i.e., f (τ , μ , u, y) = 0 19
ME3O04 - Chapter 7
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Buckingham PI Theorem This theorem states that: Given a relation among n parameters, q1, q2, …, qn, of the form:
g (q1 , q2 ,....., qn ) = 0 The n parameters may be grouped into (n-m) independent dimensionless groups or ratios (π parameters), expressible in functional form by:
G (π 1 , π 2 ,....., π n −m ) = 0 where m = r = the minimum number of independent dimensions required to specify the dimensions of the n parameters. 20
ME3O04 - Chapter 7
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Procedure to Determine the π Groups Illustrative example: Drag force on a sphere. Step 1 List all dimensional parameters involved in the problem and determine n. Where n is the number of dimensional parameters. Drag force on a sphere, FD depends on:1. Diameter, D 2. Fluid density, ρ 3. Fluid viscosity, μ 4. Fluid velocity, U
g ( FD , D, U , μ , ρ ) = 0
∴n=5 21
ME3O04 - Chapter 7
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Procedure to Determine the π Groups – Cont’d Step 2 Select a set of primary dimensions = M, L, t, T Step 3 Construct the “Dimensional Matrix” by listing all parameters in terms of the primary dimensions. M −1 − 2 F = M L t , = 2 Lt FD D U μ ρ D = L, 0 0 1 1 M1 L 1 1 −1 − 3 L 1 U = = L t −1 , t −1 0 t − 2 0 −1 M μ = = M L−1 t −1 0 0 0 T 0 0 Lt M ρ = 3 = M L− 3 “Dimensional Matrix” L 22
ME3O04 - Chapter 7
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Procedure to Determine the π Groups – Cont’d Step 4 Determine the rank of the dimensional matrix, r. where the rank, r, is the order of the largest non-zero determinant in the matrix. D
U
μ
ρ
M 1
0
0
1
1
L 1
1
1
−1 − 3
t −2 0 T 0 0
−1 0
FD
−1 0
0 0
∴ r=3 23
ME3O04 - Chapter 7
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Procedure to Determine the π Groups – Cont’d Step 5 Select a set of r dimensional parameters that includes all the primary dimensions used in step 3 (i.e., M, L, and t). Since r = 3, then select
[ U, D, ρ ]
These parameters are called the repeating parameters. Note: It is common to select a velocity, velocity a dimension, dimension and a fluid property. property
24
ME3O04 - Chapter 7
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Procedure to Determine the π Groups – Cont’d Step 6 Set dimensional equations using the repeating parameters and one of the other parameters, one-at-atime. a c
M ML L π 1 = U a .D b .ρ c .FD = ⎛⎜ ⎞⎟ .Lb .⎛⎜ 3 ⎞⎟ . 2 = M 0 .L0 .t 0 ⎝t⎠
Solving for the exponents: of t: -a -2 = 0 of M: c + 1 = 0 of L: a + b -3c + 1 = 0
⎝L ⎠
t
⇒ a = -2 ⇒ c = -1 ⇒ b = -2
FD = drag coefficient ∴π1 = 2 2 ρU D 25
ME3O04 - Chapter 7
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Procedure to Determine the π Groups – Cont’d Step 6 Set dimensional equations using the repeating parameters and one of the other parameters, one-at-atime.
π 2 = U .D .ρ .μ = a
b
c
a
c
⎛L⎞ b ⎛M ⎞ M ⎜ ⎟ .L .⎜ 3 ⎟ . = M 0 .L0 .t 0 ⎝t⎠ ⎝ L ⎠ L.t
Solving for the exponents: of t: -a -1 = 0 of M: c + 1 = 0 of L: a + b -3c -1 = 0
⇒ a = -1 ⇒ c = -1 ⇒ b = -1
μ
1 ∴π 2 = = ρ U D Re 26
ME3O04 - Chapter 7
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Procedure to Determine the π Groups – Cont’d Step 7 Check that each π group is dimensionless. The use of Buckingham theorem in the problem of drag force on a sphere shows there are two important dimensionless numbers in this problem, which are: drag coefficient =
FD ρ U 2 D2
and, Reynolds number =
ρU D μ
But how these two groups relate, needs to be determined experimentally. 27
ME3O04 - Chapter 7
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Relationship between Drag Coefficient and Reynolds number
28
ME3O04 - Chapter 7
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Buckingham PI Theorem Important points to note:z Buckingham theorem allows us to determine:1. The number of π groups (dimensionless numbers) involved in the problem. 2. The form of each one of these dimensionless numbers (π’s).
z But, it does not allow us to determine the form of the function, G (π 1 , π 2 ,....., π n −m ) = 0 , which has to be determined experimentally. experimentally 29
ME3O04 - Chapter 7
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Forces encountered in Fluid Mechanics Forces encountered in flowing fluids include forces due to: 1. 2. 3. 4. 5. 6.
Inertia Viscosity Pressure Gravity Surface tension Compressibility.
The ratio of any two forces will be dimensionless, and defines a significant dimensionless number in Fluid Mechanics. 30
ME3O04 - Chapter 7
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Dimensions of These Forces
(
)
⎛V ⎞ 1. Inertia force = m.a = ρ L . ⎜ ⎟, ⎝t ⎠ 3
L but t = , V
2 ⎛ ⎞ V 3 Thus, m.a = ρ L . ⎜⎜ ⎟⎟ = ρ L2 V 2 ⎝ L ⎠
(
)
V 2 ∂u . A = μ .L = μ V L 2. Viscous force = τ . A = μ ∂y L 3. Pressure force =
Δp. A = Δp.L2
31
ME3O04 - Chapter 7
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Dimensions of These Forces – Cont’d 4.
(
)
3 m . g = ρ L .g Gravity force =
5. Surface tension force = σ . L , where, σ = surface tension = force per unit length. 6. Compressibility force = Ev . A = Ev . L , where Ev is the compressibility modulus, or modulus of elasticity. 2
Ev =
force dp = = stress ⎛⎜ dρ ⎞⎟ area ⎝ ρ⎠ 32
ME3O04 - Chapter 7
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Significant Dimensionless Groups in Fluid Mechanics inertia force ρ L2V 2 ρ LV = = = Re = Reynolds number 1. 1. μ viscous force μ V L 2 Δ p L Δp pressure force 2. = = = Eu = Euler number 2 2 1 ρ LV inertia force ρV 2 22 2 2 3. inertia force = ρ L V = V = Fr 2 = (Froud number )2 gravity force ρ L3 g gL
inertia force ρ L2V 2 ρ L V 2 4. surface tension force = σ L = σ = We = Weber number 33
ME3O04 - Chapter 7
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Significant Dimensionless Groups in Fluid Mechanics – Cont’d inertia force ρ L2V 2 ρ V 2 V2 2 = = = = Ma 5. compressibility force Ev L2 Ev Ev ρ
where,
V Ev
=
ρ
V V = = Ma = Mach number dp C dρ
and C = speed of sound =
dp dρ
Note:- in case of incompressible flow, ρ = constant, i.e., dρ = 0. Thus, C = ∞ ⇒ Ma = 0. 34
ME3O04 - Chapter 7
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Significant Dimensionless Groups in Fluid Mechanics – Cont’d p − pv pressure force Δp L2 = = = Ca = Cavitation number 2 2 1 6. inertia force ρ L V ρV 2 2
where, p = pressure in liquid stream. pv = vapor pressure of liquid. Note:- As Ca ↓, the more likely cavitation to occur.
35
ME3O04 - Chapter 7
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Flow Similarity and Model Studies z A model test must yield data that can be scaled to obtain information of interest on the full-scale prototype. z To be able to do that, there are conditions that have to be met to ensure similarity of model and prototype flow. z Model and prototype must have:1. Geometrical similarity. 2. Kinematic similarity. 3. Dynamic similarity.
36
ME3O04 - Chapter 7
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Geometrical Similarity Geometrical similarity requires that:1. The model and prototype be of the same shape. 2. All linear dimensions of the model related to corresponding dimensions of the prototype by a constant scale factor. factor
37
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Kinematic Similarity z Two flows are kinematically similar when the velocities at corresponding points are in the same direction and differ by a constant scale factor. z Thus two kinematically similar flows have streamlines related by a constant scale factor. z Since the boundaries form the boundary streamlines of the flow, flows that are kinematically similar must be geometrically similar. i.e., geometrical similarity is a prerequisite for kinematic similarity. 38
ME3O04 - Chapter 7
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Dynamic Similarity z Dynamic similarity requires that Identical types of forces to be :1. Parallel, and 2. Related in magnitude by a constant scale. z For condition 1, dynamic similarity requires kinematic similarity, hence geometric similarity too. z For condition 2, each independent dimensionless group (force ratio) must have the same value in the model and the prototype. 39
ME3O04 - Chapter 7
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Example – Drag Force on a Sphere
F = f ( D, U , μ , ρ ) ⎛ ρV D ⎞ F ⎟⎟ = f 2 (Re) = f 2 ⎜⎜ Drag coefficient = 2 2 ρU D ⎝ μ ⎠ Dynamic similarity is achieved if we use a model sphere (could be smaller or bigger) and keep:⎛ ρV D ⎞ ⎛ ρV D ⎞ ⎟⎟ ⎟⎟ Re model = Re prototype , i.e., ⎜⎜ = ⎜⎜ ⎝ μ ⎠ model ⎝ μ ⎠ prototype and,
⎞ ⎛ ⎞ ⎛ F F ⎟ ⎜ ⎟ ⎜ = ⎜ ρ U 2 D2 ⎟ ⎜ ρ U 2 D2 ⎟ ⎠ prototype ⎠ model ⎝ ⎝ 40
ME3O04 - Chapter 7
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Example – Drag Force on a Sphere – Cont’d ⎛ ρ V D ⎞ ⎛ ρV D ⎞ ⎜⎜ ⎝
μ
⎟⎟ = ⎜⎜ ⎠ model ⎝
μ
⎟⎟ ⎠ prototype
⎛ ⎞ ⎛ ⎞ F F ⎜ ⎟ ⎜ ⎟ =⎜ 2 2 ⎟ ⎜ ρ U 2 D2 ⎟ ⎝ ⎠ model ⎝ ρ U D ⎠ prototype
(1)
(2)
Equations (1) and (2) state that:1. We do not need to use the same fluid to test the model. 2. The resulted drag force in the model will not be equal to the drag force in the prototype. However, the drag coefficients are the same. 41
ME3O04 - Chapter 7
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Scaling with Multiple Dependent Parameters z In the example of drag force on a sphere, we were interested in one dependent parameter, parameter which is the drag force. z In some practical applications there might be more than one dependent parameter of interest. z In such cases, dimensionless groups must be formed separately for each one of those dependent parameters. z Example – performance of a typical centrifugal pump. 42
ME3O04 - Chapter 7
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Centrifugal Pump or Fan
43
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Example for Scaling with Multiple Dependent Parameters Performance of a typical centrifugal pump. z In this case, dependent parameters of interest are:1. Pressure rise or head developed by the pump, h. 2. Power input required to derive the pump, P. z We are interested to know how h and P depend on:1. Volume flow rate, Q. 2. Angular speed, ω. 3. Impeller diameter, D. 4. Fluid properties ρ and μ. 44
ME3O04 - Chapter 7
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Performance of a Typical Centrifugal Pump
∴ h = f1 (Q, ω , ρ , μ , D)
and P = f 2 (Q, ω , ρ , μ , D) If we apply Buckingham Theorem, we can find out that:-
⎛ Q ρ ω D2 ⎞ h ⎟ = f1 ⎜⎜ , 2 2 3 ⎟ μ ω D ω D ⎝ ⎠
⎛ Q ρ ω D2 ⎞ ⎟ and, , = f 2 ⎜⎜ 3 5 3 μ ⎟⎠ ρω D ⎝ω D P
h = head coefficient where, 2 2 ωD P , = power coefficient 3 5 ρω D 45
ME3O04 - Chapter 7
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Performance of a Typical Centrifugal Pump
∴ h = f1 (Q, ω , ρ , μ , D)
and P = f 2 (Q, ω , ρ , μ , D) If we apply Buckingham Theorem, we can find out that:⎛ Q ρ ω D2 ⎞ h ⎟ = f1 ⎜⎜ , 2 2 3 μ ⎟⎠ ω D ⎝ω D
⎛ Q ρ ω D2 ⎞ ⎟ and, , = f 2 ⎜⎜ 3 5 3 μ ⎟⎠ ρω D ⎝ω D P
Q flow coefficient, = 3 ωD
ρωD2 ρ (ωD) D ρ V D and = ≡ μ μ μ
is a form of Re number 46
ME3O04 - Chapter 7
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Performance of a Typical Centrifugal Pump – Cont’d z From practice it has been found that viscous effects are relatively unimportant w.r.t. inertial effects. z So, we can exclude Re number, and thus:
h '⎛ Q ⎞ ⎟ = f1 ⎜⎜ 3⎟ 2 2 ω D ⎝ω D ⎠
Q ⎞ ⎟ and, = f 2 ⎜⎜ 3 5 3⎟ ρω D ⎝ω D ⎠ P
'⎛
47
ME3O04 - Chapter 7
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Performance Curves of a Typical Centrifugal Pump h '⎛ Q ⎞ ⎟ = f1 ⎜⎜ 2 2 3⎟ ω D ⎝ω D ⎠
Q ⎞ ⎜ ⎟ = f2 ⎜ 3⎟ 3 5 ρω D ⎝ω D ⎠ P
'⎛
P Efficiency = Shaft Horsepower
Fig. 7.5 48
ME3O04 - Chapter 7
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Similarity in Pump Performance Complete similarity in pump performance test would require:Q1
ω1D12 h1
ω12 D12 P1
ρ1 ω13 D15
Q2
=
=
ω2 D2 2 h2
ω2 2 D2 2
=
P2
ρ 2 ω23 D25 49
ME3O04 - Chapter 7
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Example Problem The drag of an airfoil at zero angle of attack is a function of density, viscosity, and velocity, in addition to a length parameter. A 1/10-scale model of an airfoil was tested in a wind tunnel at a Reynolds number of 5.5 X 106, based on chord length. Test conditions in the wind tunnel air stream were 15°C and 10 atmospheres absolute pressure. The prototype airfoil has a chord length of 2 m, and it is to be flown in air at standard conditions. Determine the speed at which the wind tunnel model was tested, and the corresponding prototype speed. 50
ME3O04 - Chapter 7
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α = angle of attack.
α=0 51
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Outline Chapter 8- Internal Incompressible Viscous Flow ¾ ¾ ¾ ¾ ¾
Classification of Continuum Fluid Mechanics. Internal Incompressible Viscous Flow. Flow Regimes – Laminar and Turbulent. Boundary Layer and meaning of fully developed flow. Fully developed, Laminar Flow between Infinite Parallel Plates. ¾ Flow in Pipes and Ducts:1. Velocity Profiles in fully developed pipe flow. 2. Turbulent velocity profiles in fully developed pipe flow “Power Law” . 3. Calculation of head loss:a. Major losses. b. Minor losses. ME3O04 – Chapter 8
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Classification of Continuum Fluid Mechanics
9
Chapter 6 To be covered in 4th year
144 42444 3 Chapters 8 and 9
ME3O04 – Chapter 8
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Internal Incompressible Viscous Flow z Internal means that the flow is completely bounded by solid surfaces. Examples:- flow in nozzles, ducts, diffusers, pipes, etc. z Flow Regimes:- internal flows can be classified, based on the flow regime, into:1. Laminar flow: fluid flows in layers or laminas. 2. Turbulent flow: flow is characterized by highfrequency velocity fluctuations. ME3O04 – Chapter 8
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a) Laminar Flow (Re < 2300).
b) Turbulent Flow (Re > 2300).
ME3O04 – Chapter 8
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Osborne Reynolds Experiments
a) Laminar Flow (Re < 2300). b) Turbulent Flow (Re > 2300). ME3O04 – Chapter 8
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Boundary Layer z Viscosity is responsible for what we call the no-slip condition. z Therefore, as fluid approaches a solid surface and due to the no-slip condition, a region of significant deformation, i.e., significant velocity gradient is formed. z This region is called the boundary layer.
ME3O04 – Chapter 8
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Entrance Region thickness of boundary layer
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Fully Developed Flow When the boundary layer reaches its maximum thickness the velocity distribution in the direction of flow does not change anymore, anymore at which case, the flow is said to have been fully developed. developed ∂u ∂u i.e., ≠ 0, i.e., u = u(r, x) i.e., = 0, i.e., u = u(r) ∂x ∂x thickness of boundary layer
ME3O04 – Chapter 8
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Entrance length = Development length, L for Flow in a Pipe.
L = 0.06 Re D
1. For Laminar Flow:
e.g., at Re = 2300, L = 138 D. where D = pipe diameter
2. For Turbulent Flow:
L = 80 D
ME3O04 – Chapter 8
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Fully-Developed Laminar Flows Common cases are:1. Fully developed laminar flow between infinite parallel plates –two cases:a. Both plates stationary. b. Upper plate moving with constant speed , U. 2. Fully developed laminar flow in a pipe.
ME3O04 – Chapter 8
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Fully Developed Laminar Flow Between Infinite Parallel Plates –upper plate moving with constant
speed U.
Assumptions:1. Steady flow, i.e.,
∂ =0 ∂t
2. Incompressible flow, i.e., ρ = constant. 3. Constant viscosity, i.e., μ = constant.
∂u = 0 ≡ u = u(y) 4. If L>> a, then flow is fully developed, i.e., ∂x where, L = length of plates and a = height of gap between plates. 5. Infinite plates means
∂ = 0, i.e., flow is two-dimensional. ∂z
6. Neglect body forces in x and y directions. ME3O04 – Chapter 8
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Fully Developed Laminar Flow Between Infinite Parallel Plates – upper plate moving with constant speed U.
Boundary conditions:1. at y = 0, u = v = 0. 2. at y = a, u = U, v = 0.
ME3O04 – Chapter 8
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Analysis For this type of flow, we would like to determine the following :1. 2. 3. 4. 5. 6.
Velocity distribution, Pressure distribution. Shear stress distribution. Volume flow rate. Average velocity. Point of maximum velocity.
ME3O04 – Chapter 8
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Analysis – Differential Equations 1. Conservation of mass:-
∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + + =0 ∂t ∂x ∂y ∂z Applying assumptions number 1, 5 , and 2:-
∂u ∂v + =0 ∂x ∂y
⇒
∂v =0 ∂y
⇒ v
Since v = 0 at any x ⇒ v ≠ f ( x)
= f ( x) or = constant
∴ v = constant = 0 everywhere. ME3O04 – Chapter 8
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Analysis – Differential Equations – cont’d For an incompressible fluid (ρ = c) with constant viscosity, momentum equation is:-
r r DV 2 ρ = ρ g − ∇p + μ ∇ V Dt Momentum equation in y - direction:-
⎡ ∂ 2v ∂ 2v ∂ 2v ⎤ ⎡ ∂v ∂v ⎤ ∂v ∂p ∂v ρ ⎢ +u + v + w ⎥ = − + ρ gy + μ ⎢ 2 + 2 + 2 ⎥ ∂z ⎦ ∂y ∂y ∂x ∂y ∂z ⎦ ⎣ ∂t ⎣ ∂x
∴
∂p =0 ⇒ ∂y
= p ( x)9 p or = constant
Can not be constant. If it is constant, there will be no flow.
ME3O04 – Chapter 8
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Analysis – Differential Equations – cont’d Momentum equation in x-direction
⎡ ∂ 2u ∂ 2u ∂ 2u ⎤ ⎡ ∂u ∂u ⎤ ∂p ∂u ∂u ρ ⎢ + u + v + w ⎥ = − + ρ gx + μ ⎢ 2 + 2 + 2 ⎥ ∂z ⎦ ∂x ∂y ∂x ∂y ∂z ⎦ ⎣ ∂t ⎣ ∂x
∴
∂p ∂ 2u =μ 2 ∂x ∂y
(1)
Recall ⇒ p = p ( x) f(x) ∴ L.H.S. of equation (1)
and u = u ( y ) f ( x) ≠ f ( y )
f(y)
or
and R.H.S. of equation (1) constant 9 ME3O04 – Chapter 8
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or constant
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Pressure Distribution
∴
∂p = c1 ∂x
∂p ∂ 2u = μ 2 = constant, say = c1 ∂x ∂y
(2)
(3) at x = 0 p = p1 at x = L p = p2 p1 − p2 Δp ∴ c2 = p1 and c1 = − =− L L Δp in (3) ∴p = − x + p1 ⇐ Pressure distribution L
∴
⇒
p = c1.x + c2
Note that p is function of x. ME3O04 – Chapter 8
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Velocity Distribution
but
∂p ∂ 2u =μ 2 Recall ∂x ∂y ∂p Δp in (1) = c1 = − L ∂x ∂u Δp ∴ =y + C3 ∂y μL
Δp 2 and u = y + C3 y + C4 2μ L
(4)
(1) ∂ 2u Δp =2 μL ∂y at y = 0
u=0
at y = a u = U
U U 1 ∂p Δp 2 u = y+ ya− y = u− ya − y 2 a 2μ L a 2 μ ∂x
(
)
ME3O04 – Chapter 8
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Shear Stress Distribution
Qτ yx
∂u =μ ∂y
(
Δp 2 U y −a y and u = y 2μ L a
)
U Δp (a − 2 y ) ∴ τ =μ + a 2L ME3O04 – Chapter 8
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Volume Flow Rate, Q
r r a Q = ∫ V .dA = ∫ u ( y ) .Wdy A
0
(
)
⎡U ⎤ Δp 2 y − a y ⎥ W .dy Q = ∫⎢ y− 2μ L ⎦ 0⎣a ⎡Ua Δp a 3 ⎤ + ∴Q=⎢ ⎥W ⎣ 2 12 μ L ⎦ a
ME3O04 – Chapter 8
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Average Velocity
1 Q ⎡Ua Δp a 3 ⎤ uav = = ⎢ + ⎥W × A ⎣ 2 12 μ L ⎦ Wa
U Δp a ∴ uav = + 2 12 μ L 2
ME3O04 – Chapter 8
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Point of Maximum Velocity Point of maximum velocity is defined by:-
∂u = 0, ∂y
i.e., when τ yx = 0
U Δp (a − 2 y ) = 0 ⇒ τ =μ + a 2L
a μ LU y= + 2 a Δp
Note: - Point of maximum velocity is not at y =a/2. ME3O04 – Chapter 8
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Velocity Distribution Velocity distribution will change as pressure gradient changes.
(
U 1 ∂p 2 u = y+ y −a y a 2 μ ∂x
Pressure decreases to the right.
)
Pressure increases to the right, i.e., decreases to the left. ME3O04 – Chapter 8
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Velocity distribution will change if the upper plate is not moving
Pressure decreases to the right.
(
1 ∂p 2 U u = y+ y −a y a 2 μ ∂x ME3O04 – Chapter 8
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Flow in Pipes and Ducts For an incompressible, Inviscid (i.e., μ = 0) flow, Bernoulli’s equation can be applied between points 1 and 2 as:-
r2 r2 p1 V1 p2 V2 + + g z1 = + + g z 2 = constant ρ 2 ρ 2
(1)
inviscid flow ME3O04 – Chapter 8
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Flow in Pipes and Ducts z Due to friction, i.e., due to shear stress at the wall, we do not expect the right hand side of Bernoulli’s equation to remain constant for an incompressible viscid flow.
r2 r2 p1 V1 p2 V2 + + g z1 = + + g z 2 + hlT 2 2 ρ ρ
(2)
where hlT = total energy loss per unit mass.
viscid flow ME3O04 – Chapter 8
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Velocity Profiles for Fully Developed Pipe Flow
ME3O04 – Chapter 8
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Velocity Profile in Fully Developed Pipe Flow – Laminar Flow 1. Laminar Flow:2 ⎡ R ∂p ⎛r⎞ ⎤ u = u (r ) = − ⎢1 − ⎜ ⎟ ⎥ 4 μ ∂x ⎢⎣ ⎝ R ⎠ ⎥⎦ 2
ME3O04 – Chapter 8
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Velocity Profile in Fully Developed Pipe Flow – Turbulent Flow 2. Turbulent Flow:- velocity vector in case of fully developed turbulent flow in a pipe can be represented by:
r V = (u + u ′) iˆ + v′ ˆj
where, u = time-mean velocity. u’ and v’ are fluctuating velocity components in x- and ydirections, respectively
ME3O04 – Chapter 8
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Velocity Profile in Fully Developed Pipe Flow – Turbulent Flow The velocity profile for turbulent flow through a smooth pipe may be approximated by the empirical power-law 1/ n
u ⎛ r⎞ = ⎜1 − ⎟ U ⎝ R⎠
(1)
where, U = maximum velocity = velocity at the centerline. for ReU > 2x104,
n = -1.7+1.8 log ReU
ReU = Reynolds number calculated using U =
ME3O04 – Chapter 8
ρU D μ 1 2
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Energy Equation for Viscid Flow in Pipes r r V12 p2 V22 + α1 + g z1 = + α2 + g z 2 + hlT ρ 2 ρ 2
p1
(I)
z Equation (I) is not the known Bernoulli’s equation. This equation has two major differences:1. hlT = total energy loss per unit mass, which takes into account energy loss due to friction between points 1 and 2. 2. Kinetic energy coefficients α1 and α2, which allow us to use average velocities V1 and V2. z α = 2 in case of laminar flow, and = 1 in case of turbulent flow.
ME3O04 – Chapter 8
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Solution of Pipe Flow Problems The energy conditions relating conditions at any two points 1 and 2 for a single-path pipe system (i.e., no branching) is: r2 r2 p1 V p V + α1 1 + g z1 + Δh pump = 2 + α 2 2 + g z2 + hlT ρ 2 ρ 2
(II)
where hlT = total energy loss per unit mass. Δp pump Pump power W& pump = = Δhpump= head caused by a pump= ρ Mass flow rate m& α = kinetic energy coefficient = 2.0 for laminar flow, Re < 2300. = 1.0 for turbulent flow Re ≥ 2300. ME3O04 – Chapter 8
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Calculation of Head Loss, hlT hlT = total energy loss per unit mass due to friction, is calculated from:
hlT = ∑ hl + ∑ hlm
where ∑hl = sum of major losses due to frictional effects in fully developed flow in constant-area sections. sections
∑hlm=sum of minor losses due to changes in flow direction and cross section area, area e.g., entrances, elbows, contractions, etc. ME3O04 – Chapter 8
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Calculation of Major Losses, ∑hl For each straight part of the pipe system
L Vav2 hl = f D 2
Where, D = pipe diameter, L = pipe length (length of straight part), f = friction coefficient, which can be calculated from:1. Moody chart, or 2. The following equations:
for laminar flow, i.e., Re < 2300 for Turbulent flow, i.e., Re ≥ 2300
64 f = Re
⎡e / D 1 2.51 ⎤ = −2.0 log ⎢ + 0 .5 ⎥ 3 . 7 f Re f ⎣ ⎦
ME3O04 – Chapter 8
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Moody Chart – Fig 8.12
e = surface roughness
ME3O04 – Chapter 8
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Surface Roughness, e Depends on pipe material – Table 8.1
ME3O04 – Chapter 8
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Calculation of Major Losses, ∑hlm z These are additional losses encountered, primarily as a result of flow separation (due to change in flow direction and/or change in cross section area) in pipe fittings. z Depending on the type of fitting, minor losses are computed in one of two ways:1. Using k = the loss coefficient, which is determined experimentally. Vav2 hlm = k 2 2. Or, using Le = the equivalent length of straight pipe, which is also determined experimentally. Le Vav2 hlm = f D 2 ME3O04 – Chapter 8
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Inlets and Exits – Table 8.2
ME3O04 – Chapter 8
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Sudden Enlargements and Contractions – Fig 8.14
Note the difference in the velocity to be used in each case ME3O04 – Chapter 8
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Gradual Contractions – Nozzles
Vav2 2 hlm = k 2 Value of k given in Table 8.3
ME3O04 – Chapter 8
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Gradual Enlargements - Diffusers Vav2 1 hlm = (Cpi − Cp ) 2 Where, Cp is the pressure recovery coefficient – from Fig 8.15. Cpi is the ideal pressure recovery coefficient.
Cpi = 1 −
1 AR 2
AR = area ratio, to calculated as shown below
ME3O04 – Chapter 8
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Fig 8.15
ME3O04 – Chapter 8
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Bends – Fig 8.16
Le Vav2 hlm = f D 2
ME3O04 – Chapter 8
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Valves and Other Fittings – Table 8.4 Le Vav2 hlm = f D 2
ME3O04 – Chapter 8
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Pumps, Fans, and Blowers
Δh pump =
Δp pump ∴ Δh pump
Δp pump
ρ
W& pump = & Q
W& pump W& pump = = ρ Q& m&
ME3O04 – Chapter 8
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Non-Circular Ducts z Instead of using D, we use the hydraulic diameter, Dh, defined by:-
4A Dh = P
Where, A = cross-section area and P = wetted perimeter. z For a rectangle with two sides a and b, A = a x b, P = 2(a+b), thus:
4ab 2ab = Dh = 2 ( a + b) ( a + b)
z For a square a = b, thus:
Dh = a
ME3O04 – Chapter 8
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Example Problem Water is pumped at the rate of 2 ft3/S from a reservoir 20 ft above a pump to a free discharge 90 ft above the pump. The pressure on the intake side of the pump is 5 psig and the pressure on the discharge side is 50 psig. All pipes are commercial steel of 6 in. diameter. Determine (a) the head supplied by the pump and (b) the total head loss between the pump and point of free discharge.
ME3O04 – Chapter 8
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Outline Chapter 9- External Incompressible Viscous Flow ¾ The Concept of Boundary Layer z Effect of Boundary Layer on a Blunt Body z Effect of Boundary Layer on a Streamlined Body ¾ Boundary Layer Thicknesses: 1. Disturbance Thickness, δ99 2. Displacement Thickness, δ* ¾ Example Problem. ¾ Fluid Flow about Immersed Bodies ¾ Drag and Lift. ¾ Types of Drag. ¾ CD and CL. ME3O04 - Chapter 9
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External Flows z External flows are flows over bodies immersed in an unbounded fluid.
z Examples of external flows are the flow fields around such objects as airfoils, automobiles, and airplanes. ME3O04 - Chapter 9
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Definition of Boundary Layer The boundary layer is the region adjacent to a solid surface in which viscous stresses are present.
ME3O04 - Chapter 9
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Boundary Layer Thicknesses 1. Disturbance Thickness, δ99 2. Displacement Thickness, δ*
ME3O04 - Chapter 9
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Disturbance Thickness, δ99 z δ99 is the boundary layer thickness at which u equals to 99% of the free stream velocity, U. z In other words, it is the distance from the surface at which the velocity is within 1 % of the free stream
ME3O04 - Chapter 9
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Displacement Thicknesses , δ* What do we do to make both mass flow rates equal?
h
r mass flow = ∫ ρ U dA h
h
>
r mass flow = ∫ ρ u dA
0
h 0
ME3O04 - Chapter 9
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Displacement Thicknesses , δ* z Raise the plate a distance δ* so that mass flow rates are equal. z The displacement thickness, δ*, is the distance the plate would be moved so that the loss of mass flux (due to reduction in uniform flow area) is equivalent to the loss the boundary layer causes.
ME3O04 - Chapter 9
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Displacement Thicknesses , δ*
∞
r ∞ mass flux with no B.L = ∫ ρ U dA = ∫ ρ U W.dy ∞
0
∞
0
r mass flux with B.L = ∫ ρ u dA = ∫ ρ u W.dy 0
0 ∞
r ∞ difference in mass flux = ∫ ρ (U − u ) dA = ∫ ρ (U − u ) W.dy 0
(1)
0
ME3O04 - Chapter 9
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Displacement Thicknesses , δ*
difference in mass flux = ρ Uδ * W
(2)
δ u ⎛ ⎞ ⎛ u⎞ from (1) = (2) ∴δ * = ∫ ⎜1 − ⎟ dy = ∫ ⎜1 − ⎟ dy U⎠ U⎠ 0 ⎝ 0 ⎝ ∞
Note: W is depth normal to paper. ME3O04 - Chapter 9
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The Use of Displacement Thicknesses , δ*, in Practical Applications
One can use Bernoulli’s equation to design or determine the pressure drop in a duct by reducing duct dimensions by 2δ* as shown above.
ME3O04 - Chapter 9
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Example Problem Laboratory wind tunnels have test sections 1 ft square and 2 ft long. With nominal air speed U1 = 80 ft/s at the test section inlet, turbulent boundary layers form on the top, bottom, and side walls of the tunnel. The boundary-layer thickness is δ1 = 0.8 in. at the inlet and δ2 = 1.2 in. at the outlet from the test section. The boundary-layer velocity profiles are of power-law form, with: 1/ 7
u ⎛ y⎞ =⎜ ⎟ U ⎝δ ⎠
a) Evaluate the freestream velocity, U2, at the exit from the windtunnel test section. b) Determine the change in static pressure along the test section.
ME3O04 - Chapter 9
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Fluid Flow about Immersed Bodies
Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
ME3O04 - Chapter 9
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Forces from the Surrounding Fluid on a TwoDimensional Object
(a) Pressure force
(b) Viscous (shear or friction) force
Source: Fundamentals of Fluid Mechanics by B. R. Munson et. al., Wiley, 1994. ME3O04 - Chapter 9
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Resultant Force
Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
ME3O04 - Chapter 9
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Drag and Lift
Net force F is resolved into:1. The drag force, FD, defined as the component of the force parallel to the direction of motion, and 2. The lift force, FL, defined as the component of the force perpendicular to the direction of motion
Source of fig: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001. ME3O04 - Chapter 9
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Types of Drag 1. Pressure or Form Drag: This type of drag is due to pressure difference in front of and at the back of the object. The formation of a low pressure wake behind the object depends on the shape or “form” form of the object. This is why this type of drag is called “form” drag.
In case of a streamlined object, the total drag will be due to friction, as in (a). In (b), due to this large wake (region of low pressure), form drag will be significant. ME3O04 - Chapter 9
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Types of Drag 2. Friction or Skin Friction Drag: This type of drag results from viscous (shear) stresses at the surface of the object.
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Drag Coefficient
with
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Flow over a Flat Plate Parallel to the Flow: only friction drag, no pressure drag
Boundary Layer can be 100% laminar, partly laminar and partly turbulent, or essentially 100% turbulent; hence several different drag coefficients are available ME3O04 - Chapter 9
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Flow over a Flat Plate Parallel to the Flow: Only Friction Drag
Laminar BL:
Turbulent BL:
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Drag Coefficient for a Smooth Flat Plate –
Fig 9.8
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Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag, No Friction Drag.
Drag coefficients are usually obtained empirically ME3O04 - Chapter 9
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Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag (Continued)
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Flow over a Sphere: Friction and Pressure Drag
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Drag Coefficient - Typical Values
Source: Principles of Fluid Mechanics by A. Alexandrou, Prentice Hall, 2001. ME3O04 - Chapter 9
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Effect of flow Regime on Form or Pressure Drag - flow over a cylinder
< 1.0
> 1.0
Source: Fluid Mechanics by F.M. White, Wiley, 2003.
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Streamlining Used to Reduce Wake and hence reduce pressure drag
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Important Note Regarding the Area, A
1. In case of pure friction drag, the area, A, is the total surface area in contact with the fluid (i.e., the wetted area). area 2. In case of pure pressure drag, the area A is the frontal area or projected area of the object. 3. For combined cases, drag coefficient for flow over an immersed object usually is based on the frontal area or projected area of the object, object except for airfoils and wings. 4. for airfoils and wings, use the planform area. ME3O04 - Chapter 9
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Airfoils and Wings Planform Area
Planform area is the maximum projected area of the wing. Source: Fluid Mechanics by F.M. White, Wiley, 2003.
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How de calculate Lift? Lift Coefficient, CL
Note: Ap is the planform area = maximum projected area.
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CL is Function of Re and Angle of Attack, α z Examples: NACA 23015; NACA 662-215
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Minimum Flight Speed, Vmin z At steady-state flight conditions, lift force, FL, must be equal to aircraft weight, W, thus: 1 FL = W = C L ρ V 2 A 2
z V = Vmin = Minimum flight speed when CL=CLmax Vmin =
2W ρ C Lmax A
z The question is how to maximize CL? ME3O04 - Chapter 9
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Increase of CL using Winglets (Flaps)
Figure 9.23 ME3O04 - Chapter 9
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Outline and Announcements Outline:z z z z z z z z z
Introduction – Course instructor and website. Assessment. Lectures, Class Notes, and Downloads Assignments, Tutorials, and Exams. Role of the Textbook. Teaching Style. Course objectives. Interesting Flows. The “First Question”.
Important announcements:1.
Tutorials start the week of Sept 15. ME3O04 – Chapter 1
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Assessment z Two-term tests (October and November) = 30%. z Assignments = 15% . z Final examination = 55% .
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Lectures z
Class discussions might include points that are not necessarily included in the textbook.
z
All exams will include questions on theory and concepts covered in lectures and class discussions.
z
Assignments might include questions on theory and concepts covered in lectures and class discussions.
z
Attending lectures is very important!!
z
Laptops and Cell Phones are not allowed during lectures, tutorials, and exams. ME3O04 – Chapter 1
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Class Notes and Downloads z Class notes are print out of my PowerPoint presentations. z Notes taken by students during lectures are very important. z Lecture notes and other material will be posted on the course website in a password protected section. z All material is copyright protected and should not be shared with and/or distributed to others. ME3O04 – Chapter 1
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Assignments z Assignments – will be assigned regularly, roughly every week. z Assignments might include problems from the textbook and from lectures and class discussions. z Due dates are posted on the web. z A late penalty of 10% per day will be applied. ME3O04 – Chapter 1
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Tutorials and Exams z Tutorials are provided to:{ Address any unclear points. { Help you solve assignments.
z All exams will include questions on theory and concepts covered in lectures and class discussions.
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Role of the Textbook z The textbook will be used to assign problems. z It supplements class discussions. z It is not a substitute for lectures!!
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Teaching Style z
I love and enjoy teaching. So, my teaching style is dynamic and interactive.
z
Everybody is encouraged to interact (ask questions, inquire, answer questions, etc.)
z
Bonus cards will awarded to individuals based on their level of interaction and understanding of the material.
z
Each card is worth 0.25%. There is no limit on how many cards you can get!!
z
This bonus is added to your total. So, your total could be more than 100%!! ME3O04 – Chapter 1
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Important Announcements z Please check the “Important Announcements” Announcements section on the web on a regular basis. z Tutorials will start the week of Sept 15. 15
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Course Objectives z Introduce special “vocabulary” vocabulary and “basic concepts” used in fluid mechanics. z Develop a good understanding of these concepts. z Use them to analyze and understand fluid flows in “real (practical)” (practical) problems.
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Why Fluid Mechanics is A Core Course? Please try to think of or name: {An industry, {A piece of machinery, or {Any engineering system,
where fluid mechanics does not play an important role!
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Industries where Concepts of Fluid Mechanics Play a Vital Role •Automobile Engineering •Aerospace Engineering •Oil & Gas Engineering •Power Generation •Thermal Management •Environmental Control •Biotechnology •Energy Conversion •Process Engineering ▪ ME3O04 – Chapter 1
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Some Interesting Flows
Airfoil zero angle
Airfoil 25° angle
Air flow over a Car
Wing Vortex
Personal Plume
Reference: Multimedia Fluid Mechanics CD-ROM Cambridge University Press ISBN-10: 0521604761 ME3O04 – Chapter 1
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OK, So what would be the first question we should address in this class? What is
Fluid
Mechanics ?
Forms of Matter: • Solid. • Liquid. • Gas.
Motion and its “Cause” Force ME3O04 – Chapter 1
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Scope of Fluid Mechanics Fluid Mechanics is the study of the behavior of fluids at rest and in motion. motion
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What is the Definition of a “Fluid” ? Definition of A Fluid: Fluid is a substance that deforms continuously under the application (or the effect) of a shear stress. tangential force area ME3O04 – Chapter 1
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Definition of a Fluid – cont’d When a shear stress is applied: z Solids deform or bend, then stop! z Fluids continuously deform ⇒ i.e., they Flow
Fig. 1.1 ME3O04 – Chapter 1
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Q. What makes a fluid deform? A. The “no-slip condition”.
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Two Famous Flows in Fluid Mechanics 1. Couette flow. 2. Poiseuille flow.
Poiseuille Flow ME3O04 – Chapter 1
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ME3O04 – Chapter 1
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Two Famous Flows in Fluid Mechanics 1. Couette Flow 2. Poiseuille Flow
Poiseuille Flow ME3O04 – Chapter 1
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Basic Equations z In the analysis of any fluid mechanics problem, we need to use a set of equations called “Governing Equations”. z These equations can be classified into: 1. Basic or General Laws. 2. Particular or Special Laws.
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Basic or General Laws z Conservation of mass. z Newton’s second law of motion. z The principle of angular momentum.
From Mechanics
z The first law of thermodynamics. z The second law of thermodynamics.
From Thermodynamics ME3O04 – Chapter 1
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Basic or General Laws – Cont’d z Note that these laws do not depend on the type of fluid involved in the problem. That’s why they are called “General Laws”. z Not all these laws are required every time. z In many problems, it is necessary to bring into the analysis additional relations ⇒ “Particular Laws”
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Particular or Special Laws z These laws describe the behavior of physical properties of specific types of fluids. z Example: Ideal Gas Law ⇒
P = ρ RT
¾ This law can be used only for ideal gases. ¾ It relates absolute pressure and temperature to gas density. ¾ P in Pa = N/m2, ρ in kg/m3, and T in K.
¾ R = gas constant, for air = 286.9 J/kg.K ■
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Methods of Description 1. The system approach (the Lagrangian approach), which follows one specific particle or system (e.g., following the motion of a falling object).
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Methods of Description - Continue 2. The control volume approach (the Eurlerian approach), which focuses on a certain region not on a certain particle (e.g., flow in a pipeline).
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Methods of Analysis in Thermodynamics 9 System (or “Closed System”)
9 Control Volume (or “Open System”)
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Methods of Analysis in Fluid Mechanics 1. Infinitesimal or finite control volume. 2. Large control volume.
Differential control volume
Large Control volume
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Methods of Analysis in Fluid Mechanics 1. Infinitesimal or finite system. 2. Control volume.
z In each case the equations will look different. z In (1), the resulting equations are differential equations, which provide details of the flow. z In (2), the resulting equations are integral, which give a global or overall behavior of the flow. ■ ME3O04 – Chapter 1
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Why Use Infinitesimal Control Volume (Differential) Approach? z Differential approach allows us to determine flow details, e.g., velocity distribution:-
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Why Use Control Volume (Integral) Approach? z Integral approach allows the determination of global or overall values, such as, average velocities, forces, etc..
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Units and Dimensions z
A unit is a specific quantitative measure of a physical quantity. Example, the foot and the meter, which are used to measure the physical quantity ‘length’.
z
A physical quantity, such as length, is called a dimension.
z
Dimensions can classified as basic or primary dimensions and secondary dimensions:1. Length and time are basic dimensions. 2. Velocity is a secondary dimension because it can be represented using primary dimensions (velocity = length / time). ME3O04 – Chapter 1
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Systems of Basic Dimensions
1. [M], [L], [t], and [T]. 2. [F], [L], [t], and [T]. 3. [F],[M], [L], [t], and [T].
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Systems of Units 1. MLtT - SI (kg, m, s, K). 2. FLtT - British Gravitational (lbf, ft, s, oR). 3. FMLtT - English Engineering (lbf, lbm, ft, s, oR).
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Preferred Systems of Units SI (kg, m, s, K)
British Gravitational (lbf, ft, s, oR)
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Dimensional Consistency z All equations and formulas must have consistent dimensions. z I.e., all terms in any equation must have the same dimension. z What is the dimension of each term of Bernoulli’s equation?
r2 r2 V1 p1 V2 p2 + g z1 + = + g z2 + 2 ρ 2 ρ ME3O04 – Chapter 1
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Dimensional Consistency – Cont’d z
In your engineering studies and practice, you might deal with two types of equations having inconsistent dimensions:1. Semi-empirical equations, e.g., the Manning equation, used to calculate the velocity of flow in an open channel (such as a canal):2/3 1/ 2
Rn S0 V= n
Where V is the velocity in m/s, R is the hydraulic radius of the channel in m, and S is the channel slope (ratio). For unfinished concrete, n = 0.014.
z
What is the unit of V? Does n have a unit?
z
If we use n =0.014, and R in ft, can we get the correct value of V in ft/s? ME3O04 – Chapter 1
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Dimensional Consistency – Cont’d 2. In the second type the dimensions of an equation are consistent but the use of units is not. z
For example, the commonly used Energy Efficiency ratio (EER) of an air conditioner is:-
cooling rate energy / time EER = = electrical input energy / time z
The equation is dimensionally consistent, with EER being dimensionless (ratio). However, it is used in an inconsistent way.
z
A good A/C has EER = 10, which means 10 Btu/hr for each 1 W of electrical power.
z
One must say 10 (Btu/hr)/W because it is not dimensionless. ME3O04 – Chapter 1
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Exercise A sky diver with a mass of 75 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be FD = kV2, where k = 0.228 N.s2/m2. Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.
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Vocabulary List 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
A Fluid. Mechanics. Shear force. Normal force (pressure). Flow = continuous deformation. Scope of Fluid Mechanics. Basic Equations. Methods of Description. Methods of Analysis. No-slip condition. Lagrangian approach. Eulerian approach. Infinitesimal or finite system = infinitesimal control volume (C.V.). 14. Infinitesimal C.V. = Differential approach. 15. Control volume or integral approach. 16. Basic dimensions. ME3O04 – Chapter 1
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A flow is described by the velocity field,
Problem
V = a y iˆ + b t ˆj Where a = 1 1/s and b = 2 m/s2. t in seconds. (a) Plot the pathline of the particle that passed point (1,2) at t = 2. (b) Streakeline at t = 3 of the particles that passed point (1,2).
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Note the difference between the meaning of t and to ME3O04 Chapter 2
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Note the difference between the meaning of t and to ME3O04 Chapter 2
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Problem – time permit Fluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?
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ME3O04 Chapter 2
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Chapter 2 - Fundamental Concepts Outline:-
¾ ¾ ¾ ¾ ¾ ¾ ¾
¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾
Fluid as a Continuum. Scalar quantities. Vector quantities - Velocity Field. Steady Vs Unsteady Flow Fields. Uniform Flow Fields. One-, Two-, and Three-Dimensional Flows. Visual Representation of flow fields: Timeline. Streamlines. Pathlines. Streaklines. Types of Forces acting on a fluid element. Stresses (name and sign convention). Stress Field (stress at a point). Shear Stress and rate of deformation of a fluid element. Viscosity. 1. Newtonian and 2. Non-Newtonian Fluids. Surface Tension. Classification of Fluid Motions. Compressibility effect. ME3O04 Chapter 2
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Nature of Fluids (Matter) z Fluids (gases or liquids) are forms of matter and they consist of molecules with atoms and space in between. z So, generally speaking, mass of a fluid is not continuously distributed in space: (mass – space – mass – space – mass – etc…)
ME3O04 Chapter 2
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Fluid as a Continuum Continuum Hypothesis: We can assume that fluids are continuous medium. 1.
What does it mean? It means that a fluid regardless of its molecular nature, can be treated as a continuous medium.
2.
What is the result or benefit of this assumption? a)
Each fluid property is assumed to have a definite value at every point in space, thus
b)
Fluid properties are considered to be continuous function of position and time.
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Continuum Hypothesis – Cont’d 3. When it is not possible to make such an assumption? We can not use this assumption if the mean free path of the molecules is of the same order of magnitude as the smallest significant characteristic dimension of the problem. {
Molecules always vibrate in space.
{
Mean free path is the distance the molecules travel as it vibrate.
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Mean Free Path, Lm z Examples, for air:1. At STP (15º C and 101.3 kPa) ⇒ Lm = 6 × 10-8 m = 0.06 μm = 60 nm. 2. At P = 1.33 × 10-7 kPa (Rarefied Gas) ⇒ Lm = 0.9 m = 90 cm.
z
Applications = air flow between parallel plates with a = 10 cm at atmospheric pressure i.e., P = 101.3 kPa.
?
⇐
Can we assume that air is a continuous medium in this case ?
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Quantities of interest In fluid mechanics we are interested in quantities that are: 1. Scalar quantities, such as: density, temperature, pressure, etc. 2. Vector quantities, such as: velocity, acceleration, stress, ect.
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Scalar Quantities – e.g., Density ρ z Since ρ is a scalar quantity, we need only the specification of its magnitude. z The complete or field representation of ρ is given by: (2.1) z Equation (2.1) represents a scalar field.
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Density ρ - Cont’d z Density is sometimes expressed in terms of specific gravity, SG for liquids:ρ SG = ρw max
where, ρwmax = max density of water = 1000 kg/m3 at 4 ºC. zFor gases,
SG =
ρ ρa
where ρa= air density.
zOr, in terms of specific weight, γ:
weight m g γ= = =ρg ∀ volume
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Vector Quantities – e.g., Velocity Field, V z Similar to ρ, we can represent the velocity as:(2.2)
z Eqn. 2.2 represents the velocity field, which is a vector field. z A Vector has magnitude and direction.
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Velocity Field,
V
- Cont’d
z Velocity can be written in terms of its 3 scalar components (u, v, w) in the x, y, and z directions:
where iˆ, ˆj , and kˆ = unit vectors in x, y, and z directions, respectively. z In general each component u, v, and w is a function of x,y, z, and t, e.g., u = u(x,y,z,t)
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Types of Flows - Steady Flow z
If properties at every point in a flow field do not change with time (t), the flow is called steady.
z
If η is any fluid property in a steady flow field, mathematically:
∂η =0, ∂t
or
η =η (x, y, z)
z
Remember : if f = f(x), then rate of change of f w.r.t. x =
z
If f=f(x,y), then rate of change of f w.r.t. x = ∂f ∂x ME3O04 Chapter 2
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■
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Types of Flows - One-, Two-, and ThreeDimensional Flows Depending on the number of space coordinates required to specify the velocity field, a flow can be described as 1D, 2D or 3D. 9
V = V ( x, y , z , t )
⇐
3D, unsteady
9
V = V ( x, y , t )
⇐
2D, unsteady
9
V = V (x)
⇐
1D, steday
ME3O04 Chapter 2
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Examples – Steady Flow
1D
2D
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Uniform Flow Sometimes we can neglect the no-slip condition at the wall and assume that the velocity is uniform across the whole cross section, as shown below:
u = u(x,y)
u=u(x)
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Visual Representation of Flow Fields Sometime we want to have a visual representation of flow fields. Such a representation is provided by:
¾ Timelines. ¾ Streamlines. ¾ Path lines. ¾ Streak lines. ME3O04 Chapter 2
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Timelines z A timeline is a line connecting the positions of a set of fluid particles at a given instant. This is a timeline
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Streamlines A streamline is the line drawn in the flow field so that at any instant in time it is tangent to the direction of the flow, i.e., tangent to the velocity vector.
v dy i.e., = tan α = , u dx
⇒ u dy − v dx = 0
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Streamlines – Cont’d
ME3O04 Chapter 2
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Pathlines z A path line is the line traced out by a given particle as it flows from one point to another.
z Path lines are useful in studying , for example, the trajectory of a contaminant leaving a smokestack.
ME3O04 Chapter 2
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Pathlines – Cont’d
ME3O04 Chapter 2
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Streaklines A streakline is the locus of all particles that at an earlier instant in time, passed through a prescribed point in space.
ME3O04 Chapter 2
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Streaklines – Cont’d
ME3O04 Chapter 2
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What type of Lines are these? z The redlines, and z The white lines?
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Important points to note z A streamline and a timeline are instantaneous lines, e.g., snap shots. z While streaklines and pathlines are generated by the passage of time, e.g., video recording. z In steady flow, the velocity at each point in the flow field remains constant with time and consequently, in a steady flow, pathlines, pathlines streaklines, streaklines and streamlines are identical. ME3O04 Chapter 2
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Example Problem A two-dimensional unsteady velocity field is given by u = x (1 + 2 t), v = y. Find:1. The time-varying streamlines which pass through some reference point (xo,yo). Sketch some for the case of xo=1, yo = 1. 2. Find the equation of the pathline which passes through the point (xo, yo) at t = 0. Sketch this pathline for the case of xo=1, yo = 1.
ME3O04 Chapter 2
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Streamlines at different times
ME3O04 Chapter 2
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Pathline for t < 0 to t > 0
ME3O04 Chapter 2
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Types of Forces Acting on a Fluid Element 1. Surface Forces. 2. Body Forces.
ME3O04 Chapter 2
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Types of Forces Acting on a Fluid Element A. Surfaces Forces:- e.g., Pressure and Friction These are forces generated on the surface of the fluid element by contact with other fluid particles or a solid surface.
B. Body Forces:- e.g., Gravity and Electromagnetic Field These forces are not concentrated at the surface of the fluid element; they are rather experienced throughout the particle. Example: gravitational body force acting on a fluid element of volume, d ∀ =
gravitational force = d m × g = ρ d ∀ g ME3O04 Chapter 2
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Surface Force on a Fluid Element y δAx
δAy
δAx F x
Δy δAz
z
Δz Δx
Fluid Element in Cartesian Coordinates
Surface Forces cause Stresses on the surfaces of the fluid element ME3O04 Chapter 2
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Stress Caused by Surface Forces
Fluid element
σ (sigma) is used to denote normal stresses. τ (Tao) is used to denote tangential stresses. ME3O04 Chapter 2
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Stress Caused by Surface Forces – Cont’d
z Normal stress at point C =
σ xx =
z Tangential stress at point C = τ xy =
δFx lim δAx → 0 δ Ax
δFy δF lim δA Or, τ xz = lim δAz δAx →0 x δAx →0 x
ME3O04 Chapter 2
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Name Convection - Stress needs two
directions to be define (force and area). z First letter on the left refers to the plan on which the stress acts.
δAx
z Second letter refers to the direction in which the stress acts. z Note that the subscript in δAx refers to the direction normal to the surface of interest. ME3O04 Chapter 2
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Sign Convention z Velocity or force is positive if it is in the positive direction of the axis. z The plan is considered positive if the normal to it is in the positive direction of the axis.
δ Ax
z Sign of stress is determined by the sign of its plan and force. z What is the sign of σxx ?
ME3O04 Chapter 2
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Stress Field Stress at a point =
⎡σ xx τ xy τ xz ⎤ ⎢ ⎥ ⎢τ yx σ yy τ yz ⎥ ⎢ ⎥ τ τ σ zx zy zz ⎣ ⎦
Note: all stresses shown here are positive. ME3O04 Chapter 2
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Shear Stress in a Fluid Element Exposed to a Shear Force z Fluid continuously deforms (i.e., flows) under the effect of a shear force. z Shear Stress = τ
yx
δFx dF x = δA y → 0 δ A y dA y
= lim
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Rate of Deformation of a Fluid Element Exposed to a Shear Force z Rate of deformation =
δα dα = lim δt → 0 δ t dt
ME3O04 Chapter 2
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(1)
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Rate of Deformation of a Fluid Element Exposed to a Shear Force z Because δα is very small we can say that: δl = δα. δy (2)
ME3O04 Chapter 2
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Rate of Deformation of a Fluid Element Exposed to a Shear Force Due to the no slip condition, point M’ will be (3) moving at δu, thus : δl = δu. δt
ME3O04 Chapter 2
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Rate of Deformation of a Fluid Element Exposed to a Shear Force z From (2) and (3), Rate of deformation =
δα δu dα du = , or , = δt δy dt dy
ME3O04 Chapter 2
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Relation Between Shear Stress and Rate of Deformation z Because every fluid element exposed to shear stress will deform, there must be a relationship between the shear stress and the rate of deformation. deformation z This relationship depends on the type of fluid:1. Newtonian, or 2. Non-Newtonian. ME3O04 Chapter 2
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Fluid Viscosity z Newtonian Fluids:{Most of the common fluids (water, air, oil, etc.) {“Linear” fluids
z For Newtonian fluids, the rate of deformation is in direct proportion with the shear stress, i.e.,
z The constant of proportionality is the fluid viscosity, μ, i.e.,
ME3O04 Chapter 2
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Non-Newtonian Fluids {Special fluids (e.g., most biological fluids, toothpaste, some paints, etc.) {“Non-linear” fluids In this case the relation between Shear stress and rate of deformation Takes the form:
Which can also be written as :
In this case, η is called the apparent viscosity ME3O04 Chapter 2
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Non-Newtonian Fluids – Cont’d
ME3O04 Chapter 2
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Dynamic and Kinematic Viscosity
z This equation is called Newton’s law of viscosity for a one-dimensional flow (note only y appears in the equation). equation z μ (mu) is called the Dynamic Viscosity of the fluid and it is a physical property of the fluid. z ν (nu) =μ/ρ is called the Kinematic viscosity and it is another physical property. ME3O04 Chapter 2
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Dynamic and Kinematic Viscosity
z Units of μ : Pa.s = N.S/m2 = kg/(m.s), 1 Poise = 1 gm/(cm.s). z Units of ν: m2/s, 1 Stoke = 1 cm2/s. ME3O04 Chapter 2
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Problem A block of mass M slides on a thin film of oil. The film thickness is h and the area of the block is A. When released, mass m exerts tension on the cord, causing the block to accelerate. Neglect friction in the pulley and air resistance. Develop an expression for the viscous force that acts on the block when it moves at speed V. Obtain an expression for the block speed as a function of time. If mass M = 5 kg, m = 1 kg, A = 25 cm2 , and h = 0.5 mm. If it takes 1 s for the speed of the block to reach 1 m/s, find the oil viscosity μ.
ME3O04 Chapter 2
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Surface Tension,
σ
z Surface tension is a force that appears along any common surface (interface) between two fluids. interface
z Units of σ is force per unit length (length of the interface), e.g., N/m or lbf/ft. ME3O04 Chapter 2
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Surface Tension – Cont’d z As shown below, if the two fluids are in contact with a solid surface, a contact angle, θ, develops. z Both of values of σ and θ depend on the type of fluids in contact.
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ME3O04 Chapter 2
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ME3O04 Chapter 2
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Capillary Rise (CR) and Capillary Depression (CD). z CR occurs when θ < 90°. z CD occurs when θ > 90°.
ME3O04 Chapter 2
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Calculation of Δh
∑F
z
∑F
z
=0
= σ π D cos θ − ρ g Δ∀ = 0
Q Δ∀ =
π D2 4
Δh
From (2) in (1):-
(1)
(2)
4 σ cos θ ∴ Δh = ρgD
Note the sign of Δh when θ > 90º and θ < 90º. ME3O04 Chapter 2
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Description and Classification of Fluid Motions
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Viscous and Inviscid Flows Under some special circumstances, the effect of fluid viscosity can be ignored (neglected). Example:- in region of flow away from solid surfaces.
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Laminar and Turbulent Flows z A laminar flow is a flow in which the fluid particles move in smooth layers, laminas. z A Turbulent Flow is a flow in which the fluid particles rapidly mix as they move due to random velocity fluctuations.
z The flow in a pipe is considered laminar if Re < 2300, where, ρV D Re = = Reynolds number μ z Where, ρ,μ, V, & D are the density, viscosity, velocity, and diameter, respectively. ME3O04 Chapter 2
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Compressible and Incompressible Flows z Incompressible flows are those in which variations in density are negligible. z When variations in density are not negligible, the flow is called compressible. z Variations in density are due to changes in pressure and/or temperature.
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Compressible and Incompressible Flows z Mostly, liquids can be regarded as incompressible fluids. z Pressure and density changes in liquids are reflected by the bulk compressibility modulus, or modulus of elasticity: elasticity
dp Ev = dρ / ρ z For water at 15°C, Ev = 2010 kPa = 2.92 x 105 psi. ME3O04 Chapter 2
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Can we treat a gas flow as an Incompressible flow? z For Mach number M < 0.3, the maximum density variation is less than 5%. z Thus, gas flows with M < 0.3 can be treated as incompressible. z Mach number = V/c, where V is the flow velosity and c is the speed of sound. z The speed of sound in an ideal gas is given by:
c = k RT
z Where k = Cp/Cv, R = gas constant, and T is the absolute temperature. ME3O04 Chapter 2
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Problem At what minimum speed (in mph) would an automobile have to travel for compressibility effects to be important? Assume the local air temperature is 60°F.
ME3O04 Chapter 2
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Problem – time permit Fluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?
ME3O04 Chapter 2
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Vocabulary List 1. 2. 3. 4. 5. 6. 7.
Continuum hypothesis. Mean free path. Scalar quantity. Vector quantity. Uniform flow. Multi-dimension flow. Flow Visualization.
ME3O04 Chapter 2
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Vocabulary List 1. 2. 3. 4.
Timeline. Streamline. Pathline. Streakline.
ME3O04 Chapter 2
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Vocabulary List 1. 2. 3. 4. 5. 6. 7. 8. 9.
Body force. Surface force. Stress field. Shear stress. Normal stress. Rate of deformation. Viscid and Inviscid flows. Laminar and Turbulent flows. Compressible and Incompressible flow. ME3O04 Chapter 2
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Outline:- Fluid Statics – Chapter 3 ¾ The Basic Equation of Fluid Statics ¾ ¾ ¾ ¾ ¾
Types of Pressures. Pressure Variation in a Static Fluid. Example Problem. Hydrostatic Force on Submerged Surfaces. Example Problems.
ME3O04 – Chapter 3
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What does Static Fluid mean? z Statics means that the fluid is not moving, i.e., its velocity =0; and its acceleration = 0. z Fluid velocity = 0 means that it does not flow. flow z If a static fluid does not flow, how much shear stress the fluid is exposed to? no flow = no deformation = no shear ME3O04 – Chapter 3
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What does Static Fluid mean? z In this case, fluid can be exposed to only normal forces and behaves as
“a rigid body” – no deformation
ME3O04 – Chapter 3
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The Basic Equation of Fluid Statics Consider the following fluid element:-
ME3O04 – Chapter 3
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The Basic Equation of Fluid Statics Consider the following fluid element:-
Newton’s 2nd law:-
r r ∑ dF = dm . a = 0
Divide both sides by d∀ gives:
r r dF ∑ d∀ = ρ .a = 0
ME3O04 – Chapter 3
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The Basic Equation of Fluid Statics – Cont’d z Forces affecting on the fluid element = surface + body forces, i.e.,
r r r ∑ dF = dFs + dFB
z Body Force =
or,
ME3O04 – Chapter 3
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Surface Force in y- direction, dFys
Surface force in y-direction = r ⎡⎛ ∂ p dy d F S y = ⎢ ⎜⎜ p − ∂y 2 ⎣⎝
= −
⎞ ⎛ ∂ p dy ⎞ ⎤ ⎟⎟ − ⎜⎜ p + ⎟⎟ ⎥ dx .dz ∂y 2 ⎠⎦ ⎠ ⎝
∂p dx .dy .dz ∂y ME3O04 – Chapter 3
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Surface Force, dFs z Combining the other two directions, we get:
r ⎛ ∂p ˆ ∂p ˆ ∂p ˆ ⎞ dFS = −⎜⎜ i + j + k ⎟⎟dx.dy.dz ∂y ∂z ⎠ ⎝ ∂x = −∇p.dx.dy.dz z Where, ∇p = gradient of p.
ME3O04 – Chapter 3
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Basic Equation of Fluid Statics Recall – Newton’s 2nd Law:-
r r dF ∑ d∀ = ρ .a = 0 r r
(1)
r ∑ dF = dFs + dFB
r dFS = −∇p.dx.dy.dz
(2)
From (2) in (1):
Basic Equation of Fluid Statics ME3O04 – Chapter 3
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Basic Equation of Fluid Statics
Note that equation above is not really one equation, it is rather three equations in the three directions x, y, and z.
∂p ∂p ∂p − + ρ g x = 0, − + ρ g y = 0, − + ρ g z = 0 ∂x ∂y ∂z ME3O04 – Chapter 3
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Basic Equation of Fluid Statics z Knowing that
r g = g x iˆ + g y ˆj + g z kˆ
z Assuming that z is the vertical direction, we can say that gx = gy = 0, gz = - g, the three equations:
−
∂p ∂p ∂p + ρ g x = 0, − + ρ g y = 0, − + ρ g z = 0 ∂x ∂y ∂z
ME3O04 – Chapter 3
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Basic Equation of Fluid Statics Can be written as :-
∂p = 0, ∂x
∂p = 0, ∂y
i.e., pressure is not function of x
∂p = −ρ g ∂z
i.e., pressure is not function of y
z i.e.,
dp = − ρ g = −γ dz
ME3O04 – Chapter 3
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Important Restrictions:In order to use the following equation, three conditions must be satisfied. satisfied
dp = − ρ g = −γ dz 1. Fluid must be static, i.e., velocity = acceleration = 0. 2. Gravity is the only body force. 3. The Z axis is vertical and upward. ME3O04 – Chapter 3
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Types of Pressures
Fig. 3.2
Pabs = Patm + Pgage ME3O04 – Chapter 3
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Pressure Variation in a Static Fluid For Incompressible Fluid: Manometers
dp = −ρ g dz
p
z
po
zo
∫ dp = − ∫ ρ g dz ME3O04 – Chapter 3
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Pressure Variation in a Static Fluid For Incompressible Fluid: Manometers
p
z
po
zo
∫ dp = − ∫ ρ g dz
p − po = − ρ g ( z − zo ) = ρ g ( zo − z ) or, p = po + ρ g h ME3O04 – Chapter 3
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Simple Rules to Analyze Multiple-Liquid Manometer Problems 1. Any two points at the same elevation in a continuous volume of the same liquid are at the same pressure. 2. Pressure increases as one goes down a liquid column.
ME3O04 – Chapter 3
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Example- Problem Determine the gage pressure in psig at point “a”, if liquid A has SG = 0.75 and liquid B has SG = 1.20. The liquid surrounding point “a” is water and the tank on the left is open to the atmosphere.
ME3O04 – Chapter 3
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Hydrostatic Force on Submerged Surfaces.
p = po + ρ g h
z This equation allows us to determine how pressure varies in a static fluid. z We would like to determine the force due to that pressure on a surface submerged in a liquid.
ME3O04 – Chapter 3
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Hydrostatic Force on Submerged Surfaces – Cont’d In order to fully determine the force on a surface submerged in a liquid, we must determine the following:1. The magnitude of the force; 2. The direction of the force; and 3. The line of action of the force.
ME3O04 – Chapter 3
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1- Direction of the Force on a Plane Submerged Surface z Since fluid is not moving (static), there is no shear, i.e., only normal forces might exist. z Since this force is caused by pressure of fluid, it will always be normal to the surface. surface z This determines the direction of the force.
ME3O04 – Chapter 3
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2- Magnitude of the Force on a Plane Submerged Surface
dF = P. dA P = Po + ρ g h
(1) (2)
From (2) in (1)
∴ FR = ∫ dF = ∫ (Po + ρ g h ) dA A
(3)
A
ME3O04 – Chapter 3
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2- Magnitude of the Force on a Plane Note how θ is measured Submerged Surface
∴ FR = ∫ dF = ∫ (Po + ρ g h ) dA A
(3)
A
but,
h = y sin θ
(4)
From (4) in (3)
FR = Po A + ρ g ∫ y sin θ .dA A
Where, Po is the pressure at the fluid surface, A is the surface area. ME3O04 – Chapter 3
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3- Line of Action of the Force on a Plane Submerged Surface
∴ FR . y ' = ∫ dF . y = ∫ P. dA. y A
A
but,
P = Po + ρ g h
and
h = y sin θ
(
)
∴ FR . y ' = ∫ Po . y + ρ g y 2 sin θ . dA A ME3O04 – Chapter 3
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3- Line of Action of the Force on a Plane Submerged Surface
∴ F . y = ∫ (P . y + ρ g y '
R
o
2
)
sin θ . dA
A
Solving for y’
1 1 ∴y = .∫ Po . y.dA + .∫ ρ g y 2 sin θ . dA FR A FR A '
ME3O04 – Chapter 3
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Hydrostatic Force on Submerged Surfaces – Cont’d 1. The magnitude of the force;
FR = Po A + ρ g ∫ y sin θ .dA A
2. The direction of the force = normal to the surface. 3. The line of action of the force.
∴ y' =
1 1 .∫ Po . y.dA + .∫ ρ g y 2 sin θ . dA FR A FR A ME3O04 – Chapter 3
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Example Problem The pressure in the air gap is 8000 Pa gage. The tank is cylindrical. Calculate the net hydrostatic force (a) On the bottom of the tank; (b) On the cylindrical sidewall CC; (c) On the annular plane panel BB.
ME3O04 – Chapter 3
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Example Problem Gate AB is a homogeneous mass of 180 kg, 1.2 m wide into the paper, resting on smooth bottom B. All fluids are at 20°C. For what water depth h will the force at point B be zero? Assume specific gravity of Glycerin = 1.26.
ME3O04 – Chapter 3
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Vocabulary List 1. 2. 3. 4. 5. 6.
Static fluid Manometer. Hydrostatic pressure. Gauge pressure. Vacuum. Hydrostatic force on a submerged surface.
ME3O04 – Chapter 3
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Outline Chapter 4- Basic Equations in Integral Form for a Control Volume. ¾ ¾ ¾
Definition of a control volume. Dot product of Two Vectors. Volume and Mass Rate of Flow through a C.V. 9 9 9
¾
Conservation of Mass. 9
¾ ¾ ¾ ¾ ¾ ¾ ¾
Volume Flux. Mass Flux. Sign Convention. Special cases.
Example Problem. Extensive and Intensive Fluid Properties. Reynolds Transport Theorem. Momentum Equation. Sign Convention of Terms in the Momentum Equation. Types of forces. Example problem.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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1
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¾Conservation of Mass. ¾Conservation of Momentum
Basic Equations in Integral Form for a Control Volume Chapter 4
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Why do we need Basic Equations in integral Form ? In some applications we do not need details of the flow field, we rather need some global values, values such as:1. Average velocity at a certain section. 2. Force due to fluid flow. 3. Mass or Volume flow rates.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
1 2 3
3
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Examples:Given velocity at the exit section, fine mass flow rate and average velocity at the inlet section.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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4
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Examples:Find velocity V3 and force on the scale due to fluid flow.
ME3O04 – Chapter 4
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1 2 3
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Examples:Find force on the 90° elbow.
ME3O04 – Chapter 4
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1 2 3
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Examples:Find flow rate and force on the gate in the open and closed positions. gate
ME3O04 – Chapter 4
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Basic Equations in Integral Form
Q1
How do we obtain these equations?
A1
By applying conservation laws on a control volume.
Q2 A2
Which conservation laws?
Q3
What is a control volume?
1- Conservation of mass. 2- Conservation of momentum.
ME3O04 – Chapter 4
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Definition of a Control Volume 1. A control volume is an arbitrary volume in space through which fluid flows. The geometric boundary of the control volume is called the control surface. surface 2. The control surface may be real or imaginary. Imaginary surface
Real surface Fig 3.1
ME3O04 – Chapter 4
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Background - Dot Product of Two
Vectors
Dot product of V and dA = V dA cos α = projection of V on dA.
α
α ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
Fig 4.3 1 2 3
10
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Sign Convention
Fig 4.3
r r r r V . dA = V dA cos 0 = +V dA V . dA = V dA cos180 = −V dA
ME3O04 – Chapter 4
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Volume Flow Rate through a C.V.
z Volume flow rate = volume flux through dA r r = dQ = V cosα . dA = V dA cos α = V . dA z Total volume flow rate through control surface (CS)=
r r & & Q = ∫ dQ = ∫ V . dA CS
CS ME3O04 – Chapter 4
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Mass Flow Rate through a C.V.
z Mass flow rate = Mass flux through dA r r = ρ V . dA = ρ dQ = dm z Total mass flow rate through control surface (CS)=
r r m& = ∫ dm& = ∫ ρ V . dA CS
CS ME3O04 – Chapter 4
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Sign Convention
Fig 4.3
r r volume flux = V . dA = +V dA
r r volume flux = V . dA = −V dA
ME3O04 – Chapter 4
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Conservation of Mass of any System z M = mass of the system = constant. z Or, the time rate of change of the mass of the system = 0
i.e.,
dM ⎞ =0 ⎟ dt ⎠system ME3O04 – Chapter 4
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Conservation of Mass of a Control Volume z Time rate of decrease of mass within the control volume = net mass outflow rate from the control volume. z Time rate of decrease of mass within the control volume =
∂ − ρ d∀ ∫ ∂t CV
z net mass outflow rate from the control volume =
r r dM ⎞ ∂ = ρ d∀ + ∫ ρ V .dA = 0 ⎟ ∫ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
r r ∫ ρ V . dA CS
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Special Cases
r r dM ⎞ ∂ = ⎟ ∫ ρ d∀ + ∫ ρ V .dA = 0 dt ⎠ system ∂t CV CS
1. Unsteady Incompressible Flow, i.e., ρ = c
r r ∂ take ρ outside integrals, ρ d∀ + ρ ∫ V .dA = 0 ∫ ∂t CV CS
divide by ρ , but,
r r ∂ d∀ + ∫ V .dA = 0 ∫ ∂t CV CS
∫ d∀ = ∀ = the volume of the C.V. CV
thus,
r r ∂∀ + ∫ V .dA = 0 ∂t CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Special Cases
1. Unsteady Incompressible Flow, i.e., ρ = c
thus,
r r ∂∀ + ∫ V .dA = 0 ∂t CS
∂∀ for a non - deformable (fixed) C.V., = 0, thus ∂t
r r ∫ V .dA = 0 CS ME3O04 – Chapter 4
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Special Cases
r r dM ⎞ ∂ = ⎟ ∫ ρ d∀ + ∫ ρ V .dA = 0 dt ⎠ system ∂t CV CS
2. Unsteady Incompressible flow through a fixed C.V. ∂∀
ρ = c, ∀ = c, or
=0
∂t r r ∂ ρ d∀ + ρ ∫ V .dA = 0 ∫ ∂t CV CS
devide by ρ ,
r r ∂ d∀ + ∫ V .dA = 0 ∫ ∂t CV CS ME3O04 – Chapter 4
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Special Cases
r r dM ⎞ ∂ = ⎟ ∫ ρ d∀ + ∫ ρ V .dA = 0 dt ⎠ system ∂t CV CS
2. Unsteady Incompressible flow through a fixed C.V.
thus,
r r ∂∀ + ∫ V .dA = 0 ∂t CS
∂∀ but = 0, thus ∂t
r r ∫ V .dA = 0 CS
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Special Cases
r r dM ⎞ ∂ ρ d∀ + ∫ ρ V .dA = 0 = ⎟ ∫ dt ⎠ system ∂t CV CS
3. Steady Compressible flow
∂∀ Steady ⇒ = 0, Compressible ⇒ ρ ≠ c ∂t
r r ∫ ρ V .dA = 0
(1)
CS
r r r r Recall, dm& = ρ V .dA, thus, ∫ ρ V .dA = ∑ m& CS
in (1),
r r ∑ m& = ∫ ρ V .dA = 0, i.e., m& in = m& out CS ME3O04 – Chapter 4
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Example Problem Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
⎡ r2 ⎤ u = umax ⎢1 − 2 ⎥ , ⎣ R ⎦
and umax = 10 ft/s.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution 1. Water = incompressible fluid → ρ = constant. ∂ ∂t
2. Flowing steadily = steady flow, i.e., = 0. Conservation of mass equation:-
r r ∂ ρ d∀ + ∫ ρ V .dA = 0 ∫ ∂t CV CS
(1)
Under conditions 1 and 2, equation (1) can be written as:
r r ∫ V .dA = 0
CS ME3O04 – Chapter 4
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Extensive Vs Intensive Fluid Properties z An extensive property is one that depends on the size of the C.V. Example:- volume or mass. z An intensive property is one that does not depend on the size of the C.V. Example:- Specific volume, specific enthalpy.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Reynolds Transport Theorem z Let N = any extensive property.
N = the intensive property corresponding zη = m to N.
z For any C.V., Reynolds Transport Theorem is:
r r dN ⎞ ∂ = η ρ d∀ + ∫ η ρ V .dA ⎟ ∫ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
1 2 3
(I)
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Reynolds Transport Theorem z If N = mass, M
N =1 ⇒ η= m
in (I),
z Conservation of mass equation:
r r dM ⎞ ∂ = ρ d∀ + ∫ ρ V .dA = 0 ⎟ ∫ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Interpretation of each term in R.T.T.
r r dN ⎞ ∂ η ρ d∀ + ∫ η ρ V .dA = ⎟ ∫ dt ⎠ system ∂t CV CS
(I)
dN ⎞ z = time rate of change of any extensive ⎟ dt ⎠ system property of the system.
∂ η ρ d∀ = time rate of change of N within the C.V. z ∫ ∂t CV z
ρ d∀ =
z
η ρ d∀ =
mass of an element contained in the C.V. amount of N in that element. ME3O04 – Chapter 4
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Interpretation of each term in R.T.T.
r r dN ⎞ ∂ = η ρ d∀ + ∫ η ρ V .dA ⎟ ∫ dt ⎠ system ∂t CV CS
(I)
r r z ∫ η ρ V .dA = the net rate of flux of N out through the C.S.
r r ρ V .dA =
CS
z
the rate of mass flux exiting dA = mass flow rate.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Momentum Equation for an Inertial (not accelerating) C.V.
r r P = linear momentum = M V
r r Let N = P = M V In R.T.T.
N r ∴η = =V M
r r dN ⎞ ∂ η ρ d∀ + ∫ η ρ V .dA = ⎟ ∫ dt ⎠ system ∂t CV CS
r r r r r dP ⎞ ∂ ⎟ V ρ d∀ + ∫ V ρ V .dA = ∫ ⎟ dt ⎠ system ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Momentum Equation “Newton’s Second Law” for an Inertial C.V.
r r dP ∑ F = time rate of change of momentum = dt from R.T.T.,
r r r r r dP ⎞ ∂ ⎟ = V ρ d∀ + ∫ V ρ V .dA ∫ ⎟ dt ⎠ system ∂t CV CS
r ∂ r r r r ∴∑ F = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Momentum Equation – meaning of each term
r ∂ r r r r ∴∑ F = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS (A)
=
(B)
+
(C)
z (A) = sum of all forces acting on the fixed C.V. z (B) = time rate of change of momentum inside the C.V. z (C) = net rate of flux of momentum out through the C.S. ME3O04 – Chapter 4
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r r r Types of Forces ∑ F = ∑ FS + ∑ FB r r ∑ FS = surface forces, e.g., pressure ⇒ Fs = ∫ − P dA r r ∑ FB = body forces, e.g., gravity ⇒ FB =
CS
∫ ρ g d∀
CV
sub in
r ∂ r r r r ∑ F = ∂t ∫ V ρ d∀ + ∫ V ρ V .dA CV CS
r r r r r r ∂ FS + FB = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
(I) 1 2 3
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Scalar Equations of the Momentum Equation Note
r r r r r r ∂ FS + FB = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS
(I)
Note
z Equation (I) is a vector equation. Therefore, it may be written as three scalar component equations. Note z In Cartesian coordinates (x, y, z), equation (I) can be written as:
Note
r r ∂ x - diretion equation : Fsx + FBx = u ρ d∀ + ∫ u ρ V .dA ∫ ∂t CV CS r r ∂ y - diretion equation : Fsy + FBy = v ρ d∀ + ∫ v ρ V .dA ∫ ∂t CV CS r r ∂ z - diretion equation : Fsz + FBz = w ρ d∀ + ∫ w ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Sign Convention of Terms in the Momentum Equation.
r r r r r r ∂ FS + FB = V ρ d∀ + ∫ V ρ V .dA ∫ ∂t CV CS
(I)
z F is positive if it is in the positive direction of the coordinate. z V is positive if it is in the positive direction of the coordinate. r r r z Sign of r V .dA depends on the relative directions of V and dA . ME3O04 – Chapter 4
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Problem A jet of water issuing from a stationary nozzle at 15 m/s (Aj = 0.05 m2) strikes a turning vane mounted on a cart as shown. The vane turns the jet through angle θ = 50º. Determine the value of mass, M, required to hold the cart stationary. If the vane angle θ is adjustable, plot the mass, M, needed to hold the cart stationary versus θ for 0 ≤ θ ≤ 180°.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution Procedure for This Type of Problems 1. Identify your control volume, C.V. 2. Identify its C.S., and number important sections to consider. 3. Identify your coordinates and draw them. 4. Identify which equation (s) will be used to solve the problem. 5. Identify which assumptions you can make to simplify your equations. ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution Procedure for This Type of Problems 6. Common assumptions are:a. Steady. b. Incompressible, i.e., density, ρ = constant . c. Uniform flow. d. Body force = 0. e. Effect of atmospheric pressure is negligible.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution 1. Identify your control volume, C.V. 2. Identify its C.S., and number important sections to consider.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution 1. Identify your coordinates and draw them.
ME3O04 – Chapter 4
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Solution – Cont’d 4. Identify which equation (s) will be used to solve the problem.
Rx
Momentum Equation : r r ∂ x - diretion equation : Fsx + FBx = u ρ d∀ + ∫ u ρ V .dA ∫ ∂t CV CS r r ∂ y - diretion equation : Fsy + FBy = v ρ d∀ + ∫ v ρ V .dA ∫ ∂t CV CS r r ∂ z - diretion equation : Fsz + FBz = w ρ d∀ + ∫ w ρ V .dA ∫ ∂t CV CS ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution – Cont’d r r ∂ Fsx + FBx = u ρ d∀ + ∫ u ρ V .dA ∫ ∂t CV CS
Rx
5. Identify which assumptions you can make to simplify your equations. 6. Common assumptions are:9 a. Steady. 9 b. Incompressible, i.e., density, ρ = constant . 9 9 c. Uniform flow. 9 d. Body force = 0. e. Effect of atmospheric pressure is negligible. ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution – Cont’d M=
ρV 2 A g
(1 − cos θ )
Rx
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Vocabulary List 1. 2. 3. 4. 5. 6. 7.
Basic equations (conservation laws). Control volume. Dot product of two vectors. Volume flux. Extensive Property. Intensive Property. Reynolds Transport Theorem
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Exercise Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
⎡ r2 ⎤ u = umax ⎢1 − 2 ⎥ , ⎣ R ⎦
and umax = 10 ft/s.
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution 1. Water = incompressible fluid → ρ = constant. ∂ ∂t
2. Flowing steadily = steady flow, i.e., = 0. Conservation of mass equation:-
r r ∂ ρ d∀ + ∫ ρ V .dA = 0 ∫ ∂t CV CS
(1)
Under conditions 1 and 2, equation (1) can be written as:
r r ∫ V .dA = 0
CS
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution – cont’d
r r ∫ V .dA = 0
CS
⇒
r r r r ∫ V .dA + ∫ V .dA = 0 1
(a)
2
r r ⎡ r2 ⎤ at (1) V = U = constant, at (2) V = umax ⎢1- 2 ⎥, dA = 2π r dr ⎣ R ⎦ R
in (a),
R
r2 − ∫ U .2 π rdr + ∫ + umax (1 − 2 ) 2 π r dr = 0 R 0 0
Note the sign of the first and the second integrals. ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Solution – cont’d
R
in (a),
R
r2 − ∫ U .2 π rdr + ∫ umax (1 − 2 ) 2 π r dr = 0 R 0 0 R
⎡r r ⎤ − U .π R + 2 π umax ⎢ − 2 ⎥ = 0 ⎣ 2 4R ⎦ 0 2
4
2
Solving (b) for U :
U=
(b)
umax = 5.0 ft/s. 2
ME3O04 – Chapter 4
http://mech.mcmaster.ca/~hamedm/me3o04/
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Outline Chapter 5- Differential Analysis of Fluid Motion. ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾
¾
Why differential. Region of interest. Conservation of Mass for an infinitesimal C.V. (fluid particle) Special cases. Lagrangian and Eulerian descriptions of the motion of a fluid particle. Material, substantial, or particle derivative. Example problem. Particle acceleration. Types of motion of a fluid particle. Rotation and vorticity vectors. Fluid deformation: 1. Angular deformation. 2. Linear deformation. Differential momentum equation. ME3O04 – Chapter 5
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Differential Analysis of Fluid Motion Why differential? To obtain detailed knowledge, we must apply the basic equations of fluid motion in differential form.
ME3O04 – Chapter 5
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Region of interest Since we are interested in formulating differential equations, our analysis will be in terms of infinitesimal systems or infinitesimal control volumes (i.e., differential C.V.)
Fig 5.1
ME3O04 – Chapter 5
1 2 3
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Conservation of Mass for an infinitesimal C.V. (fluid particle)
(I) =Time rate of decrease of mass inside C.V. = (II) = net rate of mass flux (mass flow rate) out through the C.S. ME3O04 – Chapter 5
1 2 3
4
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4
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Conservation of Mass for an infinitesimal C.V. (fluid particle) (I) = Time rate of decrease of mass inside C.V. =
∂ρ =− d∀ ∂t
∂ρ =− dx dy dz ∂t
ME3O04 – Chapter 5
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5
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(II) = net rate of mass flux (mass flow rate) out through the C.S. in the x-direction =
∂( ρ u) ⎞ ⎛ = ⎜ρu + dx ⎟dy dz − (ρ u ) dy dz ∂x ⎝ ⎠
∂( ρ u) =+ dx dy dz ∂x
ME3O04 – Chapter 5
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6
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Conservation of Mass for an infinitesimal C.V. (fluid particle) (II) = net rate of mass flux (mass flow rate) out through the C.S. in the three directions =
⎡ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) ⎤ + + dx dy dz ( II ) = + ⎢ ⎥ ∂y ∂z ⎦ ⎣ ∂x ∂ρ ∂ρ (I ) = − d∀ = − dx dy dz ∂t ∂t
∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) ∴ + + + =0 ∂t ∂x ∂y ∂z ME3O04 – Chapter 5
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Conservation of Mass - Special Cases ∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + = 0 ⇐ Compressible , Unsteady + ∂x ∂y ∂z ∂t 1. Incompressible Flow:- ρ = constant ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + =0 ∂x ∂y ∂z
2. Steady Flow:-
∂ =0 ∂t
⇒
∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + =0 ∂x ∂y ∂z
⇒
⇒
∂ρ =0 ∂t
∂u ∂v ∂ w + =0 + ∂x ∂y ∂z r i.e., ∇. V = 0
∂ρ =0 ∂t ⇒
r i.e., ∇. ( ρV ) = 0
ME3O04 – Chapter 5
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Example Problem Which of the following sets of equations represent possible two-dimensional incompressible flow cases? a)
u = 2x2 + y2 − x2 y v = x 3 + x( y 2 − 2 y )
b)
u = 2 xy − x + y 2
v = 2 xy − y 2 + x 2 ME3O04 – Chapter 5
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Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle 1. Lagrangian Description:In the Lagrangian description, anyrfluid property, “F”, r is function of the position vector X and time (t), i.e., X
r F = F ( X , t) r r dX ∴ particle velocity, V = dt r r 2 r dV d X and particle acceleration, a = = 2 dt dt ME3O04 – Chapter 5
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Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle 2. Eulerian Description:In the Elurian description, any fluid property, “F”, is function of the space coordinates x, y, z, and time (t), i.e.,
F = F ( x, y , z , t )
∂x ∴ = local velocity in the x - direction ∂t (not a certain particle velocity). ∂F = local change of F (at a certain x, y, z location) ∂t Eulerian description is used in most fluid mechanics problems. ME3O04 – Chapter 5
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Problem The temperature, T, in a long tunnel is known to vary approximately as:
T = To − α e
−x
L
sin( 2 π t / τ )
where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel. A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle. Is the required rate of change equal to
∂T ? ∂t
ME3O04 – Chapter 5
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Substantial Derivative If we want to find the time rate of change of any fluid property,”F”, following a certain particle and still use Eulerian descriction, we have to use what is called a “substantial” substantial or “particle” particle or “material” derivative. derivative
ME3O04 – Chapter 5
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Substantial Derivative In the Eulerian description:-
F = F(x, y, z, t)
∂F ∂F ∂F ∂F dF = dt + dx+ dy+ dz ∂t ∂x ∂y ∂z devide by dt,
dF ∂F ∂F dx ∂F dy ∂F dz = + + + dt ∂t ∂x dt ∂y dt ∂z dt
(1)
If we are following a certain particle:-
dx =u , dt
dy =v , dt
dz =w dt
ME3O04 – Chapter 5
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Substantial Derivative dF ∂F ∂F dx ∂F dy ∂F dz = + + + dt ∂t ∂x dt ∂y dt ∂z dt
dx = u, dt in (1) or,
dy = v, dt
(1)
dz =w dt
∂F ∂F ∂F DF ∂F = +u +w +v Dt ∂t ∂y ∂x ∂z DF ∂F r = + V .∇F Dt ∂t ME3O04 – Chapter 5
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Substantial Derivative – meaning of each term ∂F DF ∂F ∂F ∂F =
+u
+v
+w
∂t ∂x ∂y ∂z DF ∂F r or, + V .∇F = Dt ∂t DF = total time rate of change of F seen or experienced Dt by a certain particle. Dt
∂F = local time rate of change of F. ∂t r V .∇F = convective rate of change = rate of change due to the motion of the particle. ME3O04 – Chapter 5
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Problem The temperature, T, in a long tunnel is known to vary approximately as:
T = To − α e
−x
L
sin( 2 π t / τ )
where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel. A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle.
ME3O04 – Chapter 5
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Particle Acceleration r r V = V ( x, y , z , t ) In the Eulerian description:r r ∂V DV = local acceleration = particle acceleration = a p ∂t Dt r r r r r ∂V ∂V ∂V DV ∂V +v +w ∴ap = = +u (I) ∂z Dt ∂t ∂x ∂y r ∂V r r or, ap = + V .∇V ∂t Note:- Equation (I) is a vector equation, so three equations can be written in the three coordinates. ME3O04 – Chapter 5
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Particle Acceleration
or,
r r r r r ∂V ∂V ∂V DV ∂V ∴ap = = +u +v +w Dt ∂t ∂x ∂y ∂z r ∂V r r + V .∇V ap = ∂t
∴in x - direction : ∴in y - direction :
∴in z - direction :
a px = a py =
a pz
(I)
Du ∂u ∂u ∂u ∂u = +u +v +w Dt ∂t ∂z ∂x ∂y ∂v ∂v Dv ∂v ∂v = +u +v + w Dt ∂t ∂z ∂x ∂y
Dw ∂w ∂w ∂w ∂w = = +u +v +w Dt ∂z ∂t ∂x ∂y
ME3O04 – Chapter 5
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Types of Motion of A Fluid Particle (Kinematics) 1. 2. 3. 4.
Translational. Rotation. Linear Deformation. Angular Deformation.
Fig 5.5 ME3O04 – Chapter 5
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Deformation and Rotation of a Fluid Element
Rate of Deformation =
(Δα + Δβ ) Δt
Rate of rotation = average rotational speed =
ME3O04 – Chapter 5
1 (Δα − Δβ ) 2 Δt 1 2 3
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Rotation of a Fluid Element Δα =
Δη , Δx
Δβ =
Δξ Δy
(1)
⎛ ∂u ∂u ⎞ Δξ = ⎜⎜ u + Δy ⎟⎟.Δt − u.Δt = Δy.Δt ∂y ∂y ⎠ ⎝ ∂v ∂v ⎞ ⎛ Δη = ⎜ v + Δx ⎟.Δt − v.Δt = Δx.Δt ∂x ⎠ ∂x ⎝
in (1)
∴ Δα =
∂v , ∂x
Δβ =
∂u ∂y
(2)
ME3O04 – Chapter 5
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Rotation Vector
1 Δα − Δβ Average rotational velocity = ωz = 2 Δt
1 ⎛ ∂v ∂u ⎞ ∴ωz = ⎜⎜ − ⎟⎟ 2 ⎝ ∂x ∂y ⎠
from (2)
ωz = component of the rotation vector about the z-axis.
1 ⎛ ∂w ∂v ⎞ ωx = ⎜⎜ − ⎟⎟, 2 ⎝ ∂y ∂z ⎠
1 ⎛ ∂u ∂w ⎞ ωy = ⎜ − ⎟ 2 ⎝ ∂z ∂x ⎠
iˆ
r 1 r 1 ∂ ) 1 ˆ ˆ ω = rotation vector = ωx i + ωy j + ωw k = ∇ × V = curl V = 2 2 ∂x 2 u r
ME3O04 – Chapter 5
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ˆj
kˆ
∂ ∂y v
∂ ∂z w 23
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Vorticity Vector
r ) 1 ω = rotation vector = ωx iˆ + ωy ˆj + ωw k = ∇ × V = 2 ˆj iˆ kˆ r
r 1 ∂ 1 = curl V = 2 2 ∂x u
∂ ∂y v
∂ ∂z w
iˆ r r r ∂ ζ = vorticity vector = 2ω = ∇ × V = ∂x u ME3O04 – Chapter 5
ˆj ∂ ∂y v
kˆ ∂ ∂z w
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Fluid Deformation: (1) Angular Deformation
Rate of angular deformation in x - y plane = but,
Δα =
∂v , ∂x
Δβ =
∂u ∂y
Δα + Δβ Δt
⎛ ∂v ∂u ⎞ ∴ Rate of angular deformation in x - y plane = ⎜⎜ + ⎟⎟ ⎝ ∂x ∂y ⎠ ⎛ ∂w ∂u ⎞ ∴ Rate of angular deformation in x - z plane = ⎜ + ⎟ x ∂ ∂z ⎠ ⎝ ⎛ ∂w ∂v ⎞ + ⎟⎟ ∴ Rate of angular deformation in y - z plane = ⎜⎜ ⎝ ∂y ∂z ⎠ ME3O04 – Chapter 5
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Types of Motion of A Fluid Particle (Kinematics) 1. 2. 3. 4.
Translational. Rotation. Linear Deformation. Angular Deformation.
Fig 5.5 ME3O04 – Chapter 5
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Fluid Deformation: (2) Linear Deformation
∂u ⎤ ⎡ + Δx ⎥ Δt − u Δt u ⎢⎣ ∂u ∂x ⎦ = Δt Linear deformation in x - direction = Δx ∂x
∂u Rate of linear deformation in x - direction = ∂x ME3O04 – Chapter 5
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Fluid Deformation: (2) Linear Deformation
∂u Rate of linear deformation in x - direction = ∂x Similarly, rate of linear deformation in y - direction =
∂v ∂y
∂w and rate of linear deformation in z - direction = ∂z ME3O04 – Chapter 5
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Rate of Volume Deformation or Dilation
Rate of volume deformation or dilation = ∂u ∂v ∂w + + = ∂x ∂y ∂z r = ∇.V
ME3O04 – Chapter 5
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Recall - Conservation of Mass - Special Cases
∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + + =0 ∂t ∂x ∂y ∂z Incompressible Flow:- ρ = constant
∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + =0 ∂x ∂y ∂z
⇒
⇐ Compressible ⇐ , Unsteady
⇒
∂ρ =0 ∂t
∂u ∂v ∂ w + + =0 ∂x ∂y ∂z
r i.e., ∇.V = 0 i.e., for an incompressible fluid, rate of volume deformation = 0 ME3O04 – Chapter 5
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Differential Momentum Equation. For an incompressible fluids with constant density and viscosity, momentum equations are:1) x–direction,
2) y-direction,
3) z-direction,
ME3O04 – Chapter 5
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Outline Chapter 6- Incompressible Inviscid Flow ¾ Momentum equation - special cases: 9 Incompressible flow with constant viscosity. 9 Inviscid (frictionless flow), i.e., μ = 0 - Euler’s equation. ¾ Euler’s equations along a streamline. ¾ Bernoulli’s Equation. ¾ Hydrostatic, Static, Dynamic, and Stagnation Pressures. ¾ Applications of Bernoulli’s equation. ¾ Energy Grade Line (EGL) and Hydraulic Grade Line (HGL). ¾ Example problem. ME3O04 – Chapter 6
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Differential Momentum Equation. For an incompressible fluids with constant density and viscosity, momentum equations are:1) x–direction,
2) y-direction,
3) z-direction,
ME3O04 – Chapter 6
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Incompressible Inviscid Flow z Incompressible means ρ = constant.
z Inviscid means that viscosity μ = 0.
ME3O04 – Chapter 6
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Differential Momentum Equation
Or,
For an incompressible fluid (ρ = c) with constant viscosity, momentum equation is:r r Inviscid ⇒ μ = 0 DV 2 ρ = ρ g − ∇p + μ ∇ V Dt
1) x–direction,
2) y-direction,
3) z-direction,
ME3O04 – Chapter 6
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Euler’s Equation For an incompressible (ρ = c) , inviscid flow (μ = 0), 0) momentum equation is:- DVr ρ = ρ g − ∇p Dt Or, 1) x–direction,
2) y-direction,
3) z-direction, ME3O04 – Chapter 6
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Euler’s Equation in Streamline Coordinates z Euler’s equations shown in the previous slide are written using x-y-z coordinates. z In steady flow a fluid particle will move along a streamline because, for a steady flow, pathlines and streamlines coincide. z Thus, in describing the motion of a fluid particle in a steady flow, the distance along a streamline is a logical coordinate to use in writing the equations of motion. ME3O04 – Chapter 6
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Fluid Particle Moving along a Streamline
Fig 6.1 ME3O04 – Chapter 6
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Euler’s Equation in Streamline Coordinates 1. Apply Newton’s 2nd Law in the s-direction
∑ F = FS + FB = m as
∂p ds ⎤ ∂p ds ⎤ ⎡ ⎡ p dn . dx p dn.dx − ρ g sin β d∀ = ρ d∀ as − + − ⎢ ⎥ ⎢⎣ ⎥ ∂s 2 ⎦ ∂s 2 ⎦ ⎣ ME3O04 – Chapter 6
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Euler’s Equation in S-direction ∂p ds ⎤ ∂p ds ⎤ ⎡ ⎡ ⎢⎣ p − ∂s 2 ⎥⎦ dn.dx − ⎢⎣ p + ∂s 2 ⎥⎦ dn.dx − ρ g sin β d∀ = ρ d∀ as (1) dz but, sin β = ds r r r r ∂z DV ∂V ∂V 1 ∂p − g = as = = +V in (1) − ∂s Dt ∂t ρ ∂s ∂s If we neglect the body force and consider only steady flow:-
r 1 ∂p r ∂V − =V ∂s ρ ∂s ME3O04 – Chapter 6
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ME3O04 – Chapter 6
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Euler’s Equation in Streamline Coordinates 2. Apply Newton’s 2nd Law in the n-direction
∑ F = FS + FB = m an
∂p dn ⎤ ∂p dn ⎤ ⎡ ⎡ − − + p ds . dx p ds.dx − ρ g cos β d∀ = ρ d∀ an ⎢⎣ ⎥ ⎥ ⎢ ∂n 2 ⎦ ∂n 2 ⎦ ⎣ ME3O04 – Chapter 6
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Euler’s Equation in n-direction ∂p dn ⎤ ∂p dn ⎤ ⎡ ⎡ ⎢ p − ∂n 2 ⎥ ds.dx − ⎢ p + ∂n 2 ⎥ ds.dx − ρ g cos β d∀ = ρ d∀ an ⎣ ⎦ ⎣ ⎦ (2)
but,
dz cosβ = dn
r2 ∂z V 1 ∂p −g in (2) − = an = − ∂n ρ ∂n R an = centripetal acceleration, R = radius of curvature. If we neglect the body force and consider only steady flow:-
r2 1 ∂p V = ρ ∂n R
ME3O04 – Chapter 6
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Euler’s Equation in n-direction
r2 1 ∂p V = ρ ∂n R
z This equation indicates that pressure increases in the direction outwards from the center of curvature of the streamline.
∂p =0 z In case of flow in a straight line:- R = ∞ ∴ ∂n i.e., there is no variation in pressure in the n-direction. ME3O04 – Chapter 6
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Measurement of Static Pressure
r2 1 ∂p V = ρ ∂n R ∂p R=∞ ∴ =0 ∂n ME3O04 – Chapter 6
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Measurement of Static Pressure
Can we put the pressure gage at the elbow?
r2 1 ∂p V = ρ ∂n R ME3O04 – Chapter 6
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Bernoulli’s Equation z Bernoulli’s equation results from the integration of Euler’s equation along a streamline for a steady flow. z Euler’s equation in s-direction (along a streamline):r r ∂z ∂V r ∂V 1 ∂p − −g = +V ∂s ∂t ∂s ρ ∂s
z For steady flow:-
r r ∂z ∂V 1 ∂p − − g =V ∂s ∂s ρ ∂s ME3O04 – Chapter 6
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Bernoulli’s Equation – Cont’d Since we are moving a long a streamline, thus, p=p(s,t) ∂p ∂p ∂p ∴ dp = ds ds + dt = ∂s ∂t ∂s ∂z ∂z ∂z and, dz = ds + dt = ds ∂s ∂t ∂s r r r r ∂V ∂V ∂V ds + dt = ds and, dV = ∂s ∂t ∂s Recall (3)
r 1 ∂p ∂z r ∂V − − g =V ρ ∂s ∂s ∂s ME3O04 – Chapter 6
(3)
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1
r r dp − g dz = V d V
in (3)
−
or,
r r dp V d V + g dz + =0
ρ
ρ
(4)
z Integration of (4) along S:-
r2 V dp +gz+∫ = constant 2 ρ
(5)
z For an incompressible fluid ρ = c, in (5):
r2 V p + g z + = constant 2 ρ
⇐ Bernoulli’s equation
ME3O04 – Chapter 6
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Restrictions of The Application of r2 Bernoulli’s Equation V p + g z + = constant 2 ρ Bernoulli’s equation is a very powerful tool, but is has to be used very carefully. carefully Because it has very strict applicability limitations. 1. Incompressible flow, i.e., ρ = c, Mach number, Ma < 0.3. 2. Inviscid flow, i.e., μ = 0.
∂ 3. Steady flow, i.e., =0 ∂t 4. Along a streamline. ME3O04 – Chapter 6
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Hydrostatic, Static, Dynamic, and Stagnation Pressures 1. Hydrostatic pressure is pressure resulting from weight of a fluid column. P = ρ g h. 2. Ps=Static pressure is pressure due to the thermodynamic state of the fluid. 3. Pd=Dynamic pressure is pressure due to the velocity of r2 the fluid. ρV Pd =
2
4. Po=Stagnation pressure is pressure exerted by a moving fluid when brought from motion to rest. Po=Ps + Pd ME3O04 – Chapter 6
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Measurement of Static Pressure, Ps
Fig 6.2
ME3O04 – Chapter 6
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Measurement of Stagnation Pressure
Po = Ps + Pd
ME3O04 – Chapter 6
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Simultaneous Measurement of Static and Stagnation Pressures
r2 r2 P V Po Vo From Bernoulli’s + + gzo = + + gz ρ 2 ρ 2 r2 r V ∴ Po = P + ρ Vo = 0 and z o = z, 2 ME3O04 – Chapter 6
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Applications of Bernoulli’s Equation r2 V p + g z + = constant 2 ρ Bernoulli’s equation can be written for any two points along a same stream line as:
r2 r2 V1 p1 V2 p2 + g z1 + = + g z2 + 2 ρ 2 ρ ME3O04 – Chapter 6
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Measurement of Fluid Velocity at a Point Apply Bernoulli’s equation between points A and B: r2 r2 PA VA PB VB + + g zA = + + g zB (1) ρ ρ 2 2
Note :
zA = z B
and VB = 0, thus PB = Po
r2 P P V in (1) A + A = o ρ ρ 2
∴ VA =
2 (Po − PA )
ME3O04 – Chapter 6
ρ
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Applications – Nozzle Flow
One can relate info at section 2 to those at section 1 using:
r2 r2 V1 p1 V2 p2 + g z1 + = + g z2 + 2 ρ 2 ρ ME3O04 – Chapter 6
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Flow through a Siphon One can relate info at sections 1, A, and 2 using:
r2 r2 r2 V1 p1 VA p A V2 p2 + g z1 + = + g zA + = + g z2 + 2 ρ 2 ρ 2 ρ
Note : P1 = P2 = Patm. Since area reservoir >> area pipe
then, V1 ≈ 0 z If one uses Patm =0, this means that pressures are gage. z If one uses Patm = 101.3 kPa, this means that pressures are absolute. ME3O04 – Chapter 6
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Meaning of Each Term in Bernoulli’s Equation r2 p V + + g z = constant r2 ρ 2 p V devide by g : + +z = H ρ g 2g p p∀ = = flow energy per unit weight of the flowing fluid= ρ g mg head due to local static pressure. r r 2 mV 2 V 2 = kinetic energy per unit weight = head due to local = 2g mg dynamic pressure. mg z z= = Potential energy per unit weight = head mg due to elevation. H = total mechanical energy per unit weight = total head of the flow ME3O04 – Chapter 6
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Example Problem A tank with a reentrant orifice called a Borda mouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.
ME3O04 – Chapter 6
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
64 4EGL 7 44 8 r2 p V +z+ =H 2g ρg 123 HGL
z EGL is a line representing the total head. z HGL is a line representing the sum of elevation heads and static pressure head. z The difference EGL-HGL = dynamic head. ME3O04 – Chapter 6
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
64 4EGL 7 44 8 r2 p V +z+ =H 2g ρg 123 HGL
z At point (1) p1 = patm.=0 (gage) and V1 = 0, thus H1 = z1. z At point 4, p4=patm. = 0, thus the height of HGL = z4.
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
64 4EGL 7 44 8 r2 p V +z+ =H ρg 2g 123 HGL
z Flow through a constant cross section will have a horizontal HGL (bec. V = c).
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Example 1 A tank with a reentrant orifice called a Borda mouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.
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+
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Aj
1 = ∴ Cc = Ao 2 ME3O04 – Chapter 6
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Example 2 Heavy loads can be moved with relative ease on air cushions by using a load pallet as shown. Air is supplied from the plenum through porous surface AB. It enters the gap vertically at uniform speed, q. Once in the gap, all air flows in the positive x direction (there is no flow in across the plane at x = 0) Assume air flow in the gap is incompressible and uniform at each cross section, with speed u(x) as shown in the enlarged view. Although the gap is narrow (h<
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Water is admitted at the center pipe of the platform shown below at a rate of 1 m3/s and discharged into the air around the periphery. The upper circular plate in the figure is horizontal and is fixed in position to the ceiling. The lower annular plate is free to move vertically and is not supported by the pipe. The annular plate weighs 30 N, and the weight of water on it should be considered. If the distance, d, between the two plates is to be maintained at 3.5 cm, what is the total weight W that this platform can support?
ME3O04 –Review
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Water is discharged from a narrow slot in a 150 mm diameter pipe. The resulting horizontal two-dimensional jet is 1 m long and 15 mm thick, but of non-uniform velocity. The pressure at the inlet section is 30 kPag. Calculate (a) the volume flow rate at the inlet section and (b) the forces required at the coupling to hold the spray pipe in place. Neglect the mass of the pipe and the mass of water it contains. Coupling
ME3O04 –Review
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Outline Chapter 7- Dimensional Analysis and Similitude ¾ ¾ ¾
Meaning of Similitude. Dimensionless numbers. Methods of dimensional analysis:1. Nondimensionalizing the basic differential equations. 2. Using Buckingham PI Theorem.
¾ ¾ ¾
Buckingham PI Theorem. Procedure to determine the PI groups (illustrative example - Drag force on a sphere). Significant Dimensionless Groups in Fluid Mechanics.
¾ ¾ ¾ ¾ ¾
Significant Dimensionless Groups in Fluid Mechanics. Flow Similarity and Model Studies. Scaling with Multiple Dependent Parameters. Scaling with Multiple Dependent Parameters. Example Problem 1
ME3O04 - Chapter 7
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Similitude V 2 m2 s 2 1 = 2 gh s m m
??
= dimensionless
One of the important dimensionless 2
numbers in this problem =
V gh 2
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Similitude 2V ??
V V = gh1 gh2 2 1
2 2
h2 2 V = V1 h1 2 2
V =2 V 2 2
2 1
i.e., V2 = 2 V1 = 2 V 3
ME3O04 - Chapter 7
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Roasting Time of a Turkey What are the parameters affecting roasting time of a turkey (t) ?
t = constant
ρm
c
3 1. Mass, m p 2. Density, ρ 3. Thermal conductivity, k 4. Specific heat, cp Using dimensional analysis we can show that:-
2
k
Or,
tk
=constant = dimensionless number
cp 3 ρ m2 4
ME3O04 - Chapter 7
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Roasting Time of a Turkey
⎛ tk ⎞ ⎛ tk ⎞ ⎜ ⎟ ⎟ ⎜ = ⎜ c 3 ρ m2 ⎟ ⎜ c 3 ρ m2 ⎟ ⎠bird no. 2 ⎝ p ⎠bird no.1 ⎝ p So, if we know the cooking time of bird no.1, we can calculate the cooking time of bird no. 2 from this equation.
5
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Drag Force on a Sphere Drag force on a sphere, F depends on:1. Diameter, D 2. Fluid density, ρ 3. Fluid viscosity, μ 4. Fluid velocity, U Dimensional analysis shows that this problem is governed by two dimensionless numbers :
And that
⎛ ρ VD ⎞ ⎟⎟ = f ⎜⎜ 2 2 ρV D ⎝ μ ⎠ F
6
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z Does the size of sphere (i.e., D) matter in using this figure? z Does the type of fluid (i.e., ρ and μ) matter in using this figure? 7
ME3O04 - Chapter 7
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Similitude z So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarly. similarly z The “QUESTION” now is how can we determine important “Dimensionless Numbers” Numbers in any problem of interest? z Methods to do that: 1. Nondimensionalizing basic differential equations. 2. Using Buckingham PI Theorem. 8
ME3O04 - Chapter 7
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Nondimensionalizing the Basic Differential Equations Example: Steady, incompressible, two-dimensional flow, of a Newtonian fluid, with constant viscosity. Assumptions: ∂ ⇒ =0 1. Steady ∂t 2. Incompressible ⇒ ρ =c ∂ ⇒ =w=0 3. Two-dimensional ∂z 4. Newtonian fluid with constant μ. 9
ME3O04 - Chapter 7
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For an incompressible fluid with constant viscosity, momentum equations are:1) x–direction,
2) y-direction,
3) z-direction,
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Momentum Equations after Applying all Other Assumptions 1) x–direction,
2) y-direction,
3) z-direction,
x − direction
⎡ ∂ 2u ∂ 2u ⎤ ⎡ ∂u ∂u ⎤ ∂p ρ ⎢u + v ⎥ = − + μ ⎢ 2 + 2 ⎥ ∂y ⎦ ∂x ∂y ⎦ ⎣ ∂x ⎣ ∂x
(2)
y − direction
⎡ ∂ 2v ∂ 2v ⎤ ⎡ ∂v ∂v ⎤ ∂p ρ ⎢u + v ⎥ = − − ρg + μ ⎢ 2 + 2 ⎥ ∂y ⎦ ∂y ∂y ⎦ ⎣ ∂x ⎣ ∂x
(3) 11
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Non- Dimensionalizing Equations 1 and 2 z Non-dimensionalizing these equations means : 1. To divide all lengths by a reference length, L;
x x = , L *
y y = L *
2. To divide all velocities by a reference velocity, V∞;
u v * u = , v = V∞ V∞ *
z Note: dimensionless quantities are denoted with asterisks: 12
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Non- Dimensionalizing Equations 1 and 2
3. To divide pressure by twice the dynamic head
= ρV∞2.
p p = 2 ρ V∞ *
z Note: dimensionless quantities are denoted with asterisks:
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Equations in Dimensionless Form Momentum, x − direction : * * * 2 * 2 *⎤ ⎡ μ u u p u u ∂ ∂ ∂ ∂ ∂ u * * + v* * = − * + + ⎢ ⎥ ρ V∞ L ⎣ ∂x*2 ∂y *2 ⎦ ∂x ∂y ∂x
(2)
Momentum, y − direction :
μ ⎡ ∂ 2 v* ∂ 2 v* ⎤ ∂v* * ∂v* ∂p* gL u =− * − 2 + +v ⎢ *2 + *2 ⎥ * * ∂x ∂y ∂y V∞ ρ V∞ L ⎣ ∂x ∂y ⎦ *
ρ V∞ L Re = = Reynolds number μ
Fr =
(3)
V∞2 = Froud number gL 14
ME3O04 - Chapter 7
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Recall - Similitude z So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarly. z The “QUESTION” now is how can we determine these important “Dimensionless Numbers” Numbers in any problem of interest? z Methods to do that: 9 1. Nondimensionalizing the basic differential equations. Ö 2. Using Buckingham PI Theorem. 15
ME3O04 - Chapter 7
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Methods to find the Dimensionless Numbers Relevant to a Certain Problem of Interest z
The method of non-dimensionalizing differential equations depends on knowing which equations to use, which is not always the case.
z
If we do not know the equations, we use the second method known as the “Buckingham PI Theorem” .
z
Before we discuss the Buckingham PI Theorem, we need first to discuss two important concepts:1. Fundamental, or Independent, or Primary Dimensions. 2. Dimensional homogeneity.
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Fundamental, or Independent, or Primary Dimensions z Dimensions of all physical quantities can be expressed in terms of a group of “independent or primary dimensions”. z These primary dimensions are: L 1. Mass, M; Velocity = V = = L. t −1 2. Length, L; t 3. Time, t; V L −2 = = = L . t , Accelerati on 4. Temperature, T. 2 t t z Examples:-
L Force = ma = M . 2 = M .L.t −2 t
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Dimensional Homogeneity Any valid equation that relates physical quantities must be dimensionally homogeneous, homogeneous i.e., each term in the equation must have the same dimensions. Example: Newton’s Law of Viscosity
∂u τ =μ ∂y L.H.S. of (1)
(1)
force L 1 M τ= = M. 2 . 2 = 2 area t L t .L 18
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Dimensional Homogeneity
∂u τ =μ ∂y Qμ =
N.s m2
∴ R.H.S. of (1)
(1) =
Equation (1) can be written as :
M .L t L 1 M = . . t 2 L2 t L t 2 .L
τ ∂u μ ∂y
− 1 = 0,
i.e., f (τ , μ , u, y) = 0 19
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Buckingham PI Theorem This theorem states that: Given a relation among n parameters, q1, q2, …, qn, of the form:
g (q1 , q2 ,....., qn ) = 0 The n parameters may be grouped into (n-m) independent dimensionless groups or ratios (π parameters), expressible in functional form by:
G (π 1 , π 2 ,....., π n −m ) = 0 where m = r = the minimum number of independent dimensions required to specify the dimensions of the n parameters. 20
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Procedure to Determine the π Groups Illustrative example: Drag force on a sphere. Step 1 List all dimensional parameters involved in the problem and determine n. Where n is the number of dimensional parameters. Drag force on a sphere, FD depends on:1. Diameter, D 2. Fluid density, ρ 3. Fluid viscosity, μ 4. Fluid velocity, U
g ( FD , D, U , μ , ρ ) = 0
∴n=5 21
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Procedure to Determine the π Groups – Cont’d Step 2 Select a set of primary dimensions = M, L, t, T Step 3 Construct the “Dimensional Matrix” by listing all parameters in terms of the primary dimensions. M −1 − 2 F = M L t , = 2 Lt FD D U μ ρ D = L, 0 0 1 1 M1 L 1 1 −1 − 3 L 1 U = = L t −1 , t −1 0 t − 2 0 −1 M μ = = M L−1 t −1 0 0 0 T 0 0 Lt M ρ = 3 = M L− 3 “Dimensional Matrix” L 22
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Procedure to Determine the π Groups – Cont’d Step 4 Determine the rank of the dimensional matrix, r. where the rank, r, is the order of the largest non-zero determinant in the matrix. D
U
μ
ρ
M 1
0
0
1
1
L 1
1
1
−1 − 3
t −2 0 T 0 0
−1 0
FD
−1 0
0 0
∴ r=3 23
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Procedure to Determine the π Groups – Cont’d Step 5 Select a set of r dimensional parameters that includes all the primary dimensions used in step 3 (i.e., M, L, and t). Since r = 3, then select
[ U, D, ρ ]
These parameters are called the repeating parameters. Note: It is common to select a velocity, velocity a dimension, dimension and a fluid property. property
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Procedure to Determine the π Groups – Cont’d Step 6 Set dimensional equations using the repeating parameters and one of the other parameters, one-at-atime. a c
M ML L π 1 = U a .D b .ρ c .FD = ⎛⎜ ⎞⎟ .Lb .⎛⎜ 3 ⎞⎟ . 2 = M 0 .L0 .t 0 ⎝t⎠
Solving for the exponents: of t: -a -2 = 0 of M: c + 1 = 0 of L: a + b -3c + 1 = 0
⎝L ⎠
t
⇒ a = -2 ⇒ c = -1 ⇒ b = -2
FD = drag coefficient ∴π1 = 2 2 ρU D 25
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Procedure to Determine the π Groups – Cont’d Step 6 Set dimensional equations using the repeating parameters and one of the other parameters, one-at-atime.
π 2 = U .D .ρ .μ = a
b
c
a
c
⎛L⎞ b ⎛M ⎞ M ⎜ ⎟ .L .⎜ 3 ⎟ . = M 0 .L0 .t 0 ⎝t⎠ ⎝ L ⎠ L.t
Solving for the exponents: of t: -a -1 = 0 of M: c + 1 = 0 of L: a + b -3c -1 = 0
⇒ a = -1 ⇒ c = -1 ⇒ b = -1
μ
1 ∴π 2 = = ρ U D Re 26
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Procedure to Determine the π Groups – Cont’d Step 7 Check that each π group is dimensionless. The use of Buckingham theorem in the problem of drag force on a sphere shows there are two important dimensionless numbers in this problem, which are: drag coefficient =
FD ρ U 2 D2
and, Reynolds number =
ρU D μ
But how these two groups relate, needs to be determined experimentally. 27
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Relationship between Drag Coefficient and Reynolds number
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Buckingham PI Theorem Important points to note:z Buckingham theorem allows us to determine:1. The number of π groups (dimensionless numbers) involved in the problem. 2. The form of each one of these dimensionless numbers (π’s).
z But, it does not allow us to determine the form of the function, G (π 1 , π 2 ,....., π n −m ) = 0 , which has to be determined experimentally. experimentally 29
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Forces encountered in Fluid Mechanics Forces encountered in flowing fluids include forces due to: 1. 2. 3. 4. 5. 6.
Inertia Viscosity Pressure Gravity Surface tension Compressibility.
The ratio of any two forces will be dimensionless, and defines a significant dimensionless number in Fluid Mechanics. 30
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Dimensions of These Forces
(
)
⎛V ⎞ 1. Inertia force = m.a = ρ L . ⎜ ⎟, ⎝t ⎠ 3
L but t = , V
2 ⎛ ⎞ V 3 Thus, m.a = ρ L . ⎜⎜ ⎟⎟ = ρ L2 V 2 ⎝ L ⎠
(
)
V 2 ∂u . A = μ .L = μ V L 2. Viscous force = τ . A = μ ∂y L 3. Pressure force =
Δp. A = Δp.L2
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Dimensions of These Forces – Cont’d 4.
(
)
3 m . g = ρ L .g Gravity force =
5. Surface tension force = σ . L , where, σ = surface tension = force per unit length. 6. Compressibility force = Ev . A = Ev . L , where Ev is the compressibility modulus, or modulus of elasticity. 2
Ev =
force dp = = stress ⎛⎜ dρ ⎞⎟ area ⎝ ρ⎠ 32
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Significant Dimensionless Groups in Fluid Mechanics inertia force ρ L2V 2 ρ LV = = = Re = Reynolds number 1. 1. μ viscous force μ V L 2 Δ p L Δp pressure force 2. = = = Eu = Euler number 2 2 1 ρ LV inertia force ρV 2 22 2 2 3. inertia force = ρ L V = V = Fr 2 = (Froud number )2 gravity force ρ L3 g gL
inertia force ρ L2V 2 ρ L V 2 4. surface tension force = σ L = σ = We = Weber number 33
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Significant Dimensionless Groups in Fluid Mechanics – Cont’d inertia force ρ L2V 2 ρ V 2 V2 2 = = = = Ma 5. compressibility force Ev L2 Ev Ev ρ
where,
V Ev
=
ρ
V V = = Ma = Mach number dp C dρ
and C = speed of sound =
dp dρ
Note:- in case of incompressible flow, ρ = constant, i.e., dρ = 0. Thus, C = ∞ ⇒ Ma = 0. 34
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Significant Dimensionless Groups in Fluid Mechanics – Cont’d p − pv pressure force Δp L2 = = = Ca = Cavitation number 2 2 1 6. inertia force ρ L V ρV 2 2
where, p = pressure in liquid stream. pv = vapor pressure of liquid. Note:- As Ca ↓, the more likely cavitation to occur.
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Flow Similarity and Model Studies z A model test must yield data that can be scaled to obtain information of interest on the full-scale prototype. z To be able to do that, there are conditions that have to be met to ensure similarity of model and prototype flow. z Model and prototype must have:1. Geometrical similarity. 2. Kinematic similarity. 3. Dynamic similarity.
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Geometrical Similarity Geometrical similarity requires that:1. The model and prototype be of the same shape. 2. All linear dimensions of the model related to corresponding dimensions of the prototype by a constant scale factor. factor
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Kinematic Similarity z Two flows are kinematically similar when the velocities at corresponding points are in the same direction and differ by a constant scale factor. z Thus two kinematically similar flows have streamlines related by a constant scale factor. z Since the boundaries form the boundary streamlines of the flow, flows that are kinematically similar must be geometrically similar. i.e., geometrical similarity is a prerequisite for kinematic similarity. 38
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Dynamic Similarity z Dynamic similarity requires that Identical types of forces to be :1. Parallel, and 2. Related in magnitude by a constant scale. z For condition 1, dynamic similarity requires kinematic similarity, hence geometric similarity too. z For condition 2, each independent dimensionless group (force ratio) must have the same value in the model and the prototype. 39
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Example – Drag Force on a Sphere
F = f ( D, U , μ , ρ ) ⎛ ρV D ⎞ F ⎟⎟ = f 2 (Re) = f 2 ⎜⎜ Drag coefficient = 2 2 ρU D ⎝ μ ⎠ Dynamic similarity is achieved if we use a model sphere (could be smaller or bigger) and keep:⎛ ρV D ⎞ ⎛ ρV D ⎞ ⎟⎟ ⎟⎟ Re model = Re prototype , i.e., ⎜⎜ = ⎜⎜ ⎝ μ ⎠ model ⎝ μ ⎠ prototype and,
⎞ ⎛ ⎞ ⎛ F F ⎟ ⎜ ⎟ ⎜ = ⎜ ρ U 2 D2 ⎟ ⎜ ρ U 2 D2 ⎟ ⎠ prototype ⎠ model ⎝ ⎝ 40
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Example – Drag Force on a Sphere – Cont’d ⎛ ρ V D ⎞ ⎛ ρV D ⎞ ⎜⎜ ⎝
μ
⎟⎟ = ⎜⎜ ⎠ model ⎝
μ
⎟⎟ ⎠ prototype
⎛ ⎞ ⎛ ⎞ F F ⎜ ⎟ ⎜ ⎟ =⎜ 2 2 ⎟ ⎜ ρ U 2 D2 ⎟ ⎝ ⎠ model ⎝ ρ U D ⎠ prototype
(1)
(2)
Equations (1) and (2) state that:1. We do not need to use the same fluid to test the model. 2. The resulted drag force in the model will not be equal to the drag force in the prototype. However, the drag coefficients are the same. 41
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Scaling with Multiple Dependent Parameters z In the example of drag force on a sphere, we were interested in one dependent parameter, parameter which is the drag force. z In some practical applications there might be more than one dependent parameter of interest. z In such cases, dimensionless groups must be formed separately for each one of those dependent parameters. z Example – performance of a typical centrifugal pump. 42
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Centrifugal Pump or Fan
43
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Example for Scaling with Multiple Dependent Parameters Performance of a typical centrifugal pump. z In this case, dependent parameters of interest are:1. Pressure rise or head developed by the pump, h. 2. Power input required to derive the pump, P. z We are interested to know how h and P depend on:1. Volume flow rate, Q. 2. Angular speed, ω. 3. Impeller diameter, D. 4. Fluid properties ρ and μ. 44
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Performance of a Typical Centrifugal Pump
∴ h = f1 (Q, ω , ρ , μ , D)
and P = f 2 (Q, ω , ρ , μ , D) If we apply Buckingham Theorem, we can find out that:-
⎛ Q ρ ω D2 ⎞ h ⎟ = f1 ⎜⎜ , 2 2 3 ⎟ μ ω D ω D ⎝ ⎠
⎛ Q ρ ω D2 ⎞ ⎟ and, , = f 2 ⎜⎜ 3 5 3 μ ⎟⎠ ρω D ⎝ω D P
h = head coefficient where, 2 2 ωD P , = power coefficient 3 5 ρω D 45
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Performance of a Typical Centrifugal Pump
∴ h = f1 (Q, ω , ρ , μ , D)
and P = f 2 (Q, ω , ρ , μ , D) If we apply Buckingham Theorem, we can find out that:⎛ Q ρ ω D2 ⎞ h ⎟ = f1 ⎜⎜ , 2 2 3 μ ⎟⎠ ω D ⎝ω D
⎛ Q ρ ω D2 ⎞ ⎟ and, , = f 2 ⎜⎜ 3 5 3 μ ⎟⎠ ρω D ⎝ω D P
Q flow coefficient, = 3 ωD
ρωD2 ρ (ωD) D ρ V D and = ≡ μ μ μ
is a form of Re number 46
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Performance of a Typical Centrifugal Pump – Cont’d z From practice it has been found that viscous effects are relatively unimportant w.r.t. inertial effects. z So, we can exclude Re number, and thus:
h '⎛ Q ⎞ ⎟ = f1 ⎜⎜ 3⎟ 2 2 ω D ⎝ω D ⎠
Q ⎞ ⎟ and, = f 2 ⎜⎜ 3 5 3⎟ ρω D ⎝ω D ⎠ P
'⎛
47
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Performance Curves of a Typical Centrifugal Pump h '⎛ Q ⎞ ⎟ = f1 ⎜⎜ 2 2 3⎟ ω D ⎝ω D ⎠
Q ⎞ ⎜ ⎟ = f2 ⎜ 3⎟ 3 5 ρω D ⎝ω D ⎠ P
'⎛
P Efficiency = Shaft Horsepower
Fig. 7.5 48
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Similarity in Pump Performance Complete similarity in pump performance test would require:Q1
ω1D12 h1
ω12 D12 P1
ρ1 ω13 D15
Q2
=
=
ω2 D2 2 h2
ω2 2 D2 2
=
P2
ρ 2 ω23 D25 49
ME3O04 - Chapter 7
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Example Problem The drag of an airfoil at zero angle of attack is a function of density, viscosity, and velocity, in addition to a length parameter. A 1/10-scale model of an airfoil was tested in a wind tunnel at a Reynolds number of 5.5 X 106, based on chord length. Test conditions in the wind tunnel air stream were 15°C and 10 atmospheres absolute pressure. The prototype airfoil has a chord length of 2 m, and it is to be flown in air at standard conditions. Determine the speed at which the wind tunnel model was tested, and the corresponding prototype speed. 50
ME3O04 - Chapter 7
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α = angle of attack.
α=0 51
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Outline Chapter 8- Internal Incompressible Viscous Flow ¾ ¾ ¾ ¾ ¾
Classification of Continuum Fluid Mechanics. Internal Incompressible Viscous Flow. Flow Regimes – Laminar and Turbulent. Boundary Layer and meaning of fully developed flow. Fully developed, Laminar Flow between Infinite Parallel Plates. ¾ Flow in Pipes and Ducts:1. Velocity Profiles in fully developed pipe flow. 2. Turbulent velocity profiles in fully developed pipe flow “Power Law” . 3. Calculation of head loss:a. Major losses. b. Minor losses. ME3O04 – Chapter 8
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Classification of Continuum Fluid Mechanics
9
Chapter 6 To be covered in 4th year
144 42444 3 Chapters 8 and 9
ME3O04 – Chapter 8
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Internal Incompressible Viscous Flow z Internal means that the flow is completely bounded by solid surfaces. Examples:- flow in nozzles, ducts, diffusers, pipes, etc. z Flow Regimes:- internal flows can be classified, based on the flow regime, into:1. Laminar flow: fluid flows in layers or laminas. 2. Turbulent flow: flow is characterized by highfrequency velocity fluctuations. ME3O04 – Chapter 8
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a) Laminar Flow (Re < 2300).
b) Turbulent Flow (Re > 2300).
ME3O04 – Chapter 8
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Osborne Reynolds Experiments
a) Laminar Flow (Re < 2300). b) Turbulent Flow (Re > 2300). ME3O04 – Chapter 8
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Boundary Layer z Viscosity is responsible for what we call the no-slip condition. z Therefore, as fluid approaches a solid surface and due to the no-slip condition, a region of significant deformation, i.e., significant velocity gradient is formed. z This region is called the boundary layer.
ME3O04 – Chapter 8
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Entrance Region thickness of boundary layer
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Fully Developed Flow When the boundary layer reaches its maximum thickness the velocity distribution in the direction of flow does not change anymore, anymore at which case, the flow is said to have been fully developed. developed ∂u ∂u i.e., ≠ 0, i.e., u = u(r, x) i.e., = 0, i.e., u = u(r) ∂x ∂x thickness of boundary layer
ME3O04 – Chapter 8
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Entrance length = Development length, L for Flow in a Pipe.
L = 0.06 Re D
1. For Laminar Flow:
e.g., at Re = 2300, L = 138 D. where D = pipe diameter
2. For Turbulent Flow:
L = 80 D
ME3O04 – Chapter 8
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Fully-Developed Laminar Flows Common cases are:1. Fully developed laminar flow between infinite parallel plates –two cases:a. Both plates stationary. b. Upper plate moving with constant speed , U. 2. Fully developed laminar flow in a pipe.
ME3O04 – Chapter 8
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Fully Developed Laminar Flow Between Infinite Parallel Plates –upper plate moving with constant
speed U.
Assumptions:1. Steady flow, i.e.,
∂ =0 ∂t
2. Incompressible flow, i.e., ρ = constant. 3. Constant viscosity, i.e., μ = constant.
∂u = 0 ≡ u = u(y) 4. If L>> a, then flow is fully developed, i.e., ∂x where, L = length of plates and a = height of gap between plates. 5. Infinite plates means
∂ = 0, i.e., flow is two-dimensional. ∂z
6. Neglect body forces in x and y directions. ME3O04 – Chapter 8
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Fully Developed Laminar Flow Between Infinite Parallel Plates – upper plate moving with constant speed U.
Boundary conditions:1. at y = 0, u = v = 0. 2. at y = a, u = U, v = 0.
ME3O04 – Chapter 8
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Analysis For this type of flow, we would like to determine the following :1. 2. 3. 4. 5. 6.
Velocity distribution, Pressure distribution. Shear stress distribution. Volume flow rate. Average velocity. Point of maximum velocity.
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Analysis – Differential Equations 1. Conservation of mass:-
∂ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w) + + + =0 ∂t ∂x ∂y ∂z Applying assumptions number 1, 5 , and 2:-
∂u ∂v + =0 ∂x ∂y
⇒
∂v =0 ∂y
⇒ v
Since v = 0 at any x ⇒ v ≠ f ( x)
= f ( x) or = constant
∴ v = constant = 0 everywhere. ME3O04 – Chapter 8
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Analysis – Differential Equations – cont’d For an incompressible fluid (ρ = c) with constant viscosity, momentum equation is:-
r r DV 2 ρ = ρ g − ∇p + μ ∇ V Dt Momentum equation in y - direction:-
⎡ ∂ 2v ∂ 2v ∂ 2v ⎤ ⎡ ∂v ∂v ⎤ ∂v ∂p ∂v ρ ⎢ +u + v + w ⎥ = − + ρ gy + μ ⎢ 2 + 2 + 2 ⎥ ∂z ⎦ ∂y ∂y ∂x ∂y ∂z ⎦ ⎣ ∂t ⎣ ∂x
∴
∂p =0 ⇒ ∂y
= p ( x)9 p or = constant
Can not be constant. If it is constant, there will be no flow.
ME3O04 – Chapter 8
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Analysis – Differential Equations – cont’d Momentum equation in x-direction
⎡ ∂ 2u ∂ 2u ∂ 2u ⎤ ⎡ ∂u ∂u ⎤ ∂p ∂u ∂u ρ ⎢ + u + v + w ⎥ = − + ρ gx + μ ⎢ 2 + 2 + 2 ⎥ ∂z ⎦ ∂x ∂y ∂x ∂y ∂z ⎦ ⎣ ∂t ⎣ ∂x
∴
∂p ∂ 2u =μ 2 ∂x ∂y
(1)
Recall ⇒ p = p ( x) f(x) ∴ L.H.S. of equation (1)
and u = u ( y ) f ( x) ≠ f ( y )
f(y)
or
and R.H.S. of equation (1) constant 9 ME3O04 – Chapter 8
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or constant
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Pressure Distribution
∴
∂p = c1 ∂x
∂p ∂ 2u = μ 2 = constant, say = c1 ∂x ∂y
(2)
(3) at x = 0 p = p1 at x = L p = p2 p1 − p2 Δp ∴ c2 = p1 and c1 = − =− L L Δp in (3) ∴p = − x + p1 ⇐ Pressure distribution L
∴
⇒
p = c1.x + c2
Note that p is function of x. ME3O04 – Chapter 8
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Velocity Distribution
but
∂p ∂ 2u =μ 2 Recall ∂x ∂y ∂p Δp in (1) = c1 = − L ∂x ∂u Δp ∴ =y + C3 ∂y μL
Δp 2 and u = y + C3 y + C4 2μ L
(4)
(1) ∂ 2u Δp =2 μL ∂y at y = 0
u=0
at y = a u = U
U U 1 ∂p Δp 2 u = y+ ya− y = u− ya − y 2 a 2μ L a 2 μ ∂x
(
)
ME3O04 – Chapter 8
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Shear Stress Distribution
Qτ yx
∂u =μ ∂y
(
Δp 2 U y −a y and u = y 2μ L a
)
U Δp (a − 2 y ) ∴ τ =μ + a 2L ME3O04 – Chapter 8
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Volume Flow Rate, Q
r r a Q = ∫ V .dA = ∫ u ( y ) .Wdy A
0
(
)
⎡U ⎤ Δp 2 y − a y ⎥ W .dy Q = ∫⎢ y− 2μ L ⎦ 0⎣a ⎡Ua Δp a 3 ⎤ + ∴Q=⎢ ⎥W ⎣ 2 12 μ L ⎦ a
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Average Velocity
1 Q ⎡Ua Δp a 3 ⎤ uav = = ⎢ + ⎥W × A ⎣ 2 12 μ L ⎦ Wa
U Δp a ∴ uav = + 2 12 μ L 2
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Point of Maximum Velocity Point of maximum velocity is defined by:-
∂u = 0, ∂y
i.e., when τ yx = 0
U Δp (a − 2 y ) = 0 ⇒ τ =μ + a 2L
a μ LU y= + 2 a Δp
Note: - Point of maximum velocity is not at y =a/2. ME3O04 – Chapter 8
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Velocity Distribution Velocity distribution will change as pressure gradient changes.
(
U 1 ∂p 2 u = y+ y −a y a 2 μ ∂x
Pressure decreases to the right.
)
Pressure increases to the right, i.e., decreases to the left. ME3O04 – Chapter 8
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Velocity distribution will change if the upper plate is not moving
Pressure decreases to the right.
(
1 ∂p 2 U u = y+ y −a y a 2 μ ∂x ME3O04 – Chapter 8
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Flow in Pipes and Ducts For an incompressible, Inviscid (i.e., μ = 0) flow, Bernoulli’s equation can be applied between points 1 and 2 as:-
r2 r2 p1 V1 p2 V2 + + g z1 = + + g z 2 = constant ρ 2 ρ 2
(1)
inviscid flow ME3O04 – Chapter 8
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Flow in Pipes and Ducts z Due to friction, i.e., due to shear stress at the wall, we do not expect the right hand side of Bernoulli’s equation to remain constant for an incompressible viscid flow.
r2 r2 p1 V1 p2 V2 + + g z1 = + + g z 2 + hlT 2 2 ρ ρ
(2)
where hlT = total energy loss per unit mass.
viscid flow ME3O04 – Chapter 8
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Velocity Profiles for Fully Developed Pipe Flow
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Velocity Profile in Fully Developed Pipe Flow – Laminar Flow 1. Laminar Flow:2 ⎡ R ∂p ⎛r⎞ ⎤ u = u (r ) = − ⎢1 − ⎜ ⎟ ⎥ 4 μ ∂x ⎢⎣ ⎝ R ⎠ ⎥⎦ 2
ME3O04 – Chapter 8
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Velocity Profile in Fully Developed Pipe Flow – Turbulent Flow 2. Turbulent Flow:- velocity vector in case of fully developed turbulent flow in a pipe can be represented by:
r V = (u + u ′) iˆ + v′ ˆj
where, u = time-mean velocity. u’ and v’ are fluctuating velocity components in x- and ydirections, respectively
ME3O04 – Chapter 8
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Velocity Profile in Fully Developed Pipe Flow – Turbulent Flow The velocity profile for turbulent flow through a smooth pipe may be approximated by the empirical power-law 1/ n
u ⎛ r⎞ = ⎜1 − ⎟ U ⎝ R⎠
(1)
where, U = maximum velocity = velocity at the centerline. for ReU > 2x104,
n = -1.7+1.8 log ReU
ReU = Reynolds number calculated using U =
ME3O04 – Chapter 8
ρU D μ 1 2
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Energy Equation for Viscid Flow in Pipes r r V12 p2 V22 + α1 + g z1 = + α2 + g z 2 + hlT ρ 2 ρ 2
p1
(I)
z Equation (I) is not the known Bernoulli’s equation. This equation has two major differences:1. hlT = total energy loss per unit mass, which takes into account energy loss due to friction between points 1 and 2. 2. Kinetic energy coefficients α1 and α2, which allow us to use average velocities V1 and V2. z α = 2 in case of laminar flow, and = 1 in case of turbulent flow.
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Solution of Pipe Flow Problems The energy conditions relating conditions at any two points 1 and 2 for a single-path pipe system (i.e., no branching) is: r2 r2 p1 V p V + α1 1 + g z1 + Δh pump = 2 + α 2 2 + g z2 + hlT ρ 2 ρ 2
(II)
where hlT = total energy loss per unit mass. Δp pump Pump power W& pump = = Δhpump= head caused by a pump= ρ Mass flow rate m& α = kinetic energy coefficient = 2.0 for laminar flow, Re < 2300. = 1.0 for turbulent flow Re ≥ 2300. ME3O04 – Chapter 8
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Calculation of Head Loss, hlT hlT = total energy loss per unit mass due to friction, is calculated from:
hlT = ∑ hl + ∑ hlm
where ∑hl = sum of major losses due to frictional effects in fully developed flow in constant-area sections. sections
∑hlm=sum of minor losses due to changes in flow direction and cross section area, area e.g., entrances, elbows, contractions, etc. ME3O04 – Chapter 8
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Calculation of Major Losses, ∑hl For each straight part of the pipe system
L Vav2 hl = f D 2
Where, D = pipe diameter, L = pipe length (length of straight part), f = friction coefficient, which can be calculated from:1. Moody chart, or 2. The following equations:
for laminar flow, i.e., Re < 2300 for Turbulent flow, i.e., Re ≥ 2300
64 f = Re
⎡e / D 1 2.51 ⎤ = −2.0 log ⎢ + 0 .5 ⎥ 3 . 7 f Re f ⎣ ⎦
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Moody Chart – Fig 8.12
e = surface roughness
ME3O04 – Chapter 8
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Surface Roughness, e Depends on pipe material – Table 8.1
ME3O04 – Chapter 8
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Calculation of Major Losses, ∑hlm z These are additional losses encountered, primarily as a result of flow separation (due to change in flow direction and/or change in cross section area) in pipe fittings. z Depending on the type of fitting, minor losses are computed in one of two ways:1. Using k = the loss coefficient, which is determined experimentally. Vav2 hlm = k 2 2. Or, using Le = the equivalent length of straight pipe, which is also determined experimentally. Le Vav2 hlm = f D 2 ME3O04 – Chapter 8
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Inlets and Exits – Table 8.2
ME3O04 – Chapter 8
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Sudden Enlargements and Contractions – Fig 8.14
Note the difference in the velocity to be used in each case ME3O04 – Chapter 8
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Gradual Contractions – Nozzles
Vav2 2 hlm = k 2 Value of k given in Table 8.3
ME3O04 – Chapter 8
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Gradual Enlargements - Diffusers Vav2 1 hlm = (Cpi − Cp ) 2 Where, Cp is the pressure recovery coefficient – from Fig 8.15. Cpi is the ideal pressure recovery coefficient.
Cpi = 1 −
1 AR 2
AR = area ratio, to calculated as shown below
ME3O04 – Chapter 8
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Fig 8.15
ME3O04 – Chapter 8
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Bends – Fig 8.16
Le Vav2 hlm = f D 2
ME3O04 – Chapter 8
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Valves and Other Fittings – Table 8.4 Le Vav2 hlm = f D 2
ME3O04 – Chapter 8
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Pumps, Fans, and Blowers
Δh pump =
Δp pump ∴ Δh pump
Δp pump
ρ
W& pump = & Q
W& pump W& pump = = ρ Q& m&
ME3O04 – Chapter 8
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Non-Circular Ducts z Instead of using D, we use the hydraulic diameter, Dh, defined by:-
4A Dh = P
Where, A = cross-section area and P = wetted perimeter. z For a rectangle with two sides a and b, A = a x b, P = 2(a+b), thus:
4ab 2ab = Dh = 2 ( a + b) ( a + b)
z For a square a = b, thus:
Dh = a
ME3O04 – Chapter 8
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Example Problem Water is pumped at the rate of 2 ft3/S from a reservoir 20 ft above a pump to a free discharge 90 ft above the pump. The pressure on the intake side of the pump is 5 psig and the pressure on the discharge side is 50 psig. All pipes are commercial steel of 6 in. diameter. Determine (a) the head supplied by the pump and (b) the total head loss between the pump and point of free discharge.
ME3O04 – Chapter 8
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Outline Chapter 9- External Incompressible Viscous Flow ¾ The Concept of Boundary Layer z Effect of Boundary Layer on a Blunt Body z Effect of Boundary Layer on a Streamlined Body ¾ Boundary Layer Thicknesses: 1. Disturbance Thickness, δ99 2. Displacement Thickness, δ* ¾ Example Problem. ¾ Fluid Flow about Immersed Bodies ¾ Drag and Lift. ¾ Types of Drag. ¾ CD and CL. ME3O04 - Chapter 9
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External Flows z External flows are flows over bodies immersed in an unbounded fluid.
z Examples of external flows are the flow fields around such objects as airfoils, automobiles, and airplanes. ME3O04 - Chapter 9
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Definition of Boundary Layer The boundary layer is the region adjacent to a solid surface in which viscous stresses are present.
ME3O04 - Chapter 9
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Boundary Layer Thicknesses 1. Disturbance Thickness, δ99 2. Displacement Thickness, δ*
ME3O04 - Chapter 9
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Disturbance Thickness, δ99 z δ99 is the boundary layer thickness at which u equals to 99% of the free stream velocity, U. z In other words, it is the distance from the surface at which the velocity is within 1 % of the free stream
ME3O04 - Chapter 9
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Displacement Thicknesses , δ* What do we do to make both mass flow rates equal?
h
r mass flow = ∫ ρ U dA h
h
>
r mass flow = ∫ ρ u dA
0
h 0
ME3O04 - Chapter 9
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Displacement Thicknesses , δ* z Raise the plate a distance δ* so that mass flow rates are equal. z The displacement thickness, δ*, is the distance the plate would be moved so that the loss of mass flux (due to reduction in uniform flow area) is equivalent to the loss the boundary layer causes.
ME3O04 - Chapter 9
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Displacement Thicknesses , δ*
∞
r ∞ mass flux with no B.L = ∫ ρ U dA = ∫ ρ U W.dy ∞
0
∞
0
r mass flux with B.L = ∫ ρ u dA = ∫ ρ u W.dy 0
0 ∞
r ∞ difference in mass flux = ∫ ρ (U − u ) dA = ∫ ρ (U − u ) W.dy 0
(1)
0
ME3O04 - Chapter 9
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Displacement Thicknesses , δ*
difference in mass flux = ρ Uδ * W
(2)
δ u ⎛ ⎞ ⎛ u⎞ from (1) = (2) ∴δ * = ∫ ⎜1 − ⎟ dy = ∫ ⎜1 − ⎟ dy U⎠ U⎠ 0 ⎝ 0 ⎝ ∞
Note: W is depth normal to paper. ME3O04 - Chapter 9
http://mech.mcmaster.ca/~hamedm/me3o04/
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The Use of Displacement Thicknesses , δ*, in Practical Applications
One can use Bernoulli’s equation to design or determine the pressure drop in a duct by reducing duct dimensions by 2δ* as shown above.
ME3O04 - Chapter 9
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Example Problem Laboratory wind tunnels have test sections 1 ft square and 2 ft long. With nominal air speed U1 = 80 ft/s at the test section inlet, turbulent boundary layers form on the top, bottom, and side walls of the tunnel. The boundary-layer thickness is δ1 = 0.8 in. at the inlet and δ2 = 1.2 in. at the outlet from the test section. The boundary-layer velocity profiles are of power-law form, with: 1/ 7
u ⎛ y⎞ =⎜ ⎟ U ⎝δ ⎠
a) Evaluate the freestream velocity, U2, at the exit from the windtunnel test section. b) Determine the change in static pressure along the test section.
ME3O04 - Chapter 9
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Fluid Flow about Immersed Bodies
Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
ME3O04 - Chapter 9
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Forces from the Surrounding Fluid on a TwoDimensional Object
(a) Pressure force
(b) Viscous (shear or friction) force
Source: Fundamentals of Fluid Mechanics by B. R. Munson et. al., Wiley, 1994. ME3O04 - Chapter 9
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Resultant Force
Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
ME3O04 - Chapter 9
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Drag and Lift
Net force F is resolved into:1. The drag force, FD, defined as the component of the force parallel to the direction of motion, and 2. The lift force, FL, defined as the component of the force perpendicular to the direction of motion
Source of fig: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001. ME3O04 - Chapter 9
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Types of Drag 1. Pressure or Form Drag: This type of drag is due to pressure difference in front of and at the back of the object. The formation of a low pressure wake behind the object depends on the shape or “form” form of the object. This is why this type of drag is called “form” drag.
In case of a streamlined object, the total drag will be due to friction, as in (a). In (b), due to this large wake (region of low pressure), form drag will be significant. ME3O04 - Chapter 9
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Types of Drag 2. Friction or Skin Friction Drag: This type of drag results from viscous (shear) stresses at the surface of the object.
ME3O04 - Chapter 9
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Drag Coefficient
with
ME3O04 - Chapter 9
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Flow over a Flat Plate Parallel to the Flow: only friction drag, no pressure drag
Boundary Layer can be 100% laminar, partly laminar and partly turbulent, or essentially 100% turbulent; hence several different drag coefficients are available ME3O04 - Chapter 9
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Flow over a Flat Plate Parallel to the Flow: Only Friction Drag
Laminar BL:
Turbulent BL:
ME3O04 - Chapter 9
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Drag Coefficient for a Smooth Flat Plate –
Fig 9.8
ME3O04 - Chapter 9
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Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag, No Friction Drag.
Drag coefficients are usually obtained empirically ME3O04 - Chapter 9
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Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag (Continued)
ME3O04 - Chapter 9
http://mech.mcmaster.ca/~hamedm/me3o04/
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Flow over a Sphere: Friction and Pressure Drag
ME3O04 - Chapter 9
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Drag Coefficient - Typical Values
Source: Principles of Fluid Mechanics by A. Alexandrou, Prentice Hall, 2001. ME3O04 - Chapter 9
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Effect of flow Regime on Form or Pressure Drag - flow over a cylinder
< 1.0
> 1.0
Source: Fluid Mechanics by F.M. White, Wiley, 2003.
ME3O04 - Chapter 9
http://mech.mcmaster.ca/~hamedm/me3o04/
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Streamlining Used to Reduce Wake and hence reduce pressure drag
ME3O04 - Chapter 9
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Important Note Regarding the Area, A
1. In case of pure friction drag, the area, A, is the total surface area in contact with the fluid (i.e., the wetted area). area 2. In case of pure pressure drag, the area A is the frontal area or projected area of the object. 3. For combined cases, drag coefficient for flow over an immersed object usually is based on the frontal area or projected area of the object, object except for airfoils and wings. 4. for airfoils and wings, use the planform area. ME3O04 - Chapter 9
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Airfoils and Wings Planform Area
Planform area is the maximum projected area of the wing. Source: Fluid Mechanics by F.M. White, Wiley, 2003.
ME3O04 - Chapter 9
http://mech.mcmaster.ca/~hamedm/me3o04/
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How de calculate Lift? Lift Coefficient, CL
Note: Ap is the planform area = maximum projected area.
ME3O04 - Chapter 9
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CL is Function of Re and Angle of Attack, α z Examples: NACA 23015; NACA 662-215
ME3O04 - Chapter 9
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Minimum Flight Speed, Vmin z At steady-state flight conditions, lift force, FL, must be equal to aircraft weight, W, thus: 1 FL = W = C L ρ V 2 A 2
z V = Vmin = Minimum flight speed when CL=CLmax Vmin =
2W ρ C Lmax A
z The question is how to maximize CL? ME3O04 - Chapter 9
http://mech.mcmaster.ca/~hamedm/me3o04/
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Increase of CL using Winglets (Flaps)
Figure 9.23 ME3O04 - Chapter 9
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