Method of Images
|
|
- Charity Fletcher
- 6 years ago
- Views:
Transcription
1 . - Marine Hdrodnamics, Spring 5 Lecture Marine Hdrodnamics Lecture Method of Images m Potential for single source: φ = ln + π m ( ) Potential for source near a wall: φ = m ln +( ) +ln +( + ) π m d φ d = m Added source for smmetr Note: e sure to verif that the oundar conditions are satisfied smmetr or calculus for φ () = φ ( ). 1
2 Vorte near a wall (ground effect): φ = + tan 1 ( ) tan 1 ( + ) π - Added vorte for smmetr dφ Verif that = on the wall =. d φ a a Circle of radius a near a wall: = ( ) +( + ) This solution satisfies the oundar condition on the wall ( φ = ), and the degree it n satisfies the oundar condition of no flu through the circle oundar increases as the ratio /a >> 1, i.e., the velocit due to the image dipole small on the real circle 1 1 for >>a. For a D dipole, φ d, φ d.
3 More than one wall: Eample 1: Eample : Eample 3: Forces on a od undergoing stead translation D Alemert s parado Fied odies & translating odies - Galilean transformation. z o z o Fied in space Fied in translating od = ` + t 3
4 Reference sstem O: v, φ, p Reference sstem O : v,φ,p O X S O X S φ = v ˆn = φ = ˆn = (,, ) (n n,n,n z ) = n on od v as φ as Galilean transform: v(,, z, t) φ(,, z, t) + φ( = + t,,z,t) Pressure (no gravit) p = 1 ρv + C o = C o = φ = v ˆn = φ n = v (,, ) as φ as v ( = t,, z,t)+(,, ) = φ ( = t,, z,t)+ = φ (,,z,t) = 1 ρv + C o = C o 1 ρ In O: unstead flow C o p s = ρ φ 1 ρ t }{{} v +C o φ =( + )(φ + )= t }{{} t }{{} t p s = ρ 1 ρ + C o = 1 ρ + C o = C o 1 ρ In O : stead flow p s = ρ φ t }{{} 1 ρ }{{} v +C o = C o p s p = 1 ρ stagnation pressure p s p = 1 ρ stagnation pressure 4
5 3.1. Forces nˆ Total fluid force for ideal flow (i.e., no shear stresses): F = pˆ nds For potential flow, sustitute for p from ernoulli: For the hdrostatic case v φ : φ 1 F = ρ + φ + g +c(t) nds ˆ } t {{ }}{{} F s = ( ρgnˆ) ds = ( ) ( ρg) dυ = ρg ĵ where = dυ }{{} We evaluate onl the hdrodnamic force: φ 1 F d = ρ + φ ˆndS t φ For stead motion : t hdrodnamic force 1 F d = ρ hdrostatic force Gauss outward υ Archimedes υ theorem normal principle v nds ˆ 5
6 3.1.3 Eample Hdrodnamic force on D clinder in a stead uniform stream. S nˆ a ( F d = ) ρ φ π ρ ndl ˆ = φ î = ρa F =F dθ φ nˆ î }{{} = ρa π φ Velocit potential for flow past a D clinder: a φ = r cos θ 1+ r Velocit vector on the D clinder surface: φ 1 φ φ =(v r,v θ )=, }{{} r r θ } {{ } φ π cos θdθ cos θ Square of the velocit vector on the D clinder surface: =4 sin θ ˆnadθ sin θ 6
7 Finall, the hdrodnamic force on the D clinder is given π π ρa (1 ) F = dθ 4 sin θ cos θ = ρ (a) dθ sin θ cos θ = }{{}}{{}}{{}}{{} odd diameter even p s p or π (1 ) F = ρ (a) dθ sin θ sin θ = π 3π w.r.t, projection }{{} Therefore, F = no horizontal force ( smmetr fore-aft of the streamlines). Similarl, In fact, in general we find that F, on an D or 3D od. D Alemert s parado : No hdrodnamic force acts on a od moving with stead translational (no circulation) velocit in an infinite, inviscid, irrotational fluid. The moment as measured in a local frame is not necessaril zero. 7
8 3.13 Lift due to Circulation Eample Hdrodnamic force on a vorte in a uniform stream. φ = + θ = rcos θ + θ π π Consider a control surface in the form of a circle of radius r centered at the point vorte. Then according to Newton s law: Where, F d stead flow ΣF = L CV dt (F V + F CS)+ M NET = F F V = F CS + M NET = Hdrodnamic force eerted on the vorte from the fluid. F V = F = Hdrodnamic force eerted on the fluid in the control volume from the vorte. F CS M NET d L CV dt = Surface force (i.e., pressure) on the fluid control surface. = Net linear momentum flu in the control volume through the control surface. = Rate of change of the total linear momentum in the control volume. F F θ Control volume The hdrodnamic force on the vorte is F = F CS + M IN 8
9 a. Net linear momentum flu in the control volume through the control surfaces, M NET. Recall that the control surface has the form of a circle of radius r centered at the point vorte. a.1 The velocit components on the control surface are u = sin θ πr v = cos θ πr The radial velocit on the control surface is therefore, given u r = r = cos θ = V nˆ v θ = πr θ a. The net horizontal and vertical momentum flues through the control surface are given π π (M NET ) = ρ dθruv r = ρ dθr sin θ cos θ = πr π π (MNET) = ρ dθrvv r = ρ dθr cos θ cos θ πr π ρ ρ = cos θdθ = π 9
10 . Pressure force on the control surface, F CS..1 From ernoulli, the pressure on the control surface is p = 1 ρ v + C. The velocit v on the control surface is given v =u + v = sin θ + cos θ πr πr = sin θ + πr πr.3 Integrate the pressure along the control surface to otain F CS π (F CS ) = dθrp( cos θ) = π π ρ (F CS ) = dθrp( sin θ) = ( r) dθ sin 1 θ = πr ρ c. Finall, the force on the vorte F is given }{{} π F =(F CS ) +(M ) IN = F =(F CS ) +(M ) IN = ρ i.e., the fluid eerts a downward force F = ρ on the vorte. Kutta-Joukowski Law D : F = ρ 3D : F = ρ Generalized Kutta-Joukowski Law: n F = ρ i where F is the total force on a sstem of n vortices in a free stream with speed. 1 i=1
OUTLINE FOR Chapter 3
013/4/ OUTLINE FOR Chapter 3 AERODYNAMICS (W-1-1 BERNOULLI S EQUATION & integration BERNOULLI S EQUATION AERODYNAMICS (W-1-1 013/4/ BERNOULLI S EQUATION FOR AN IRROTATION FLOW AERODYNAMICS (W-1-.1 VENTURI
More informationOffshore Hydromechanics Module 1
Offshore Hydromechanics Module 1 Dr. ir. Pepijn de Jong 4. Potential Flows part 2 Introduction Topics of Module 1 Problems of interest Chapter 1 Hydrostatics Chapter 2 Floating stability Chapter 2 Constant
More informationApply mass and momentum conservation to a differential control volume. Simple classical solutions of NS equations
Module 5: Navier-Stokes Equations: Appl mass and momentum conservation to a differential control volume Derive the Navier-Stokes equations Simple classical solutions of NS equations Use dimensional analsis
More informationFluid Mechanics II. Newton s second law applied to a control volume
Fluid Mechanics II Stead flow momentum equation Newton s second law applied to a control volume Fluids, either in a static or dnamic motion state, impose forces on immersed bodies and confining boundaries.
More informationLifting Airfoils in Incompressible Irrotational Flow. AA210b Lecture 3 January 13, AA210b - Fundamentals of Compressible Flow II 1
Lifting Airfoils in Incompressible Irrotational Flow AA21b Lecture 3 January 13, 28 AA21b - Fundamentals of Compressible Flow II 1 Governing Equations For an incompressible fluid, the continuity equation
More information3.5 Vorticity Equation
.0 - Marine Hydrodynamics, Spring 005 Lecture 9.0 - Marine Hydrodynamics Lecture 9 Lecture 9 is structured as follows: In paragraph 3.5 we return to the full Navier-Stokes equations (unsteady, viscous
More informationy=1/4 x x=4y y=x 3 x=y 1/3 Example: 3.1 (1/2, 1/8) (1/2, 1/8) Find the area in the positive quadrant bounded by y = 1 x and y = x3
Eample: 3.1 Find the area in the positive quadrant bounded b 1 and 3 4 First find the points of intersection of the two curves: clearl the curves intersect at (, ) and at 1 4 3 1, 1 8 Select a strip at
More informationAll that begins... peace be upon you
All that begins... peace be upon you Faculty of Mechanical Engineering Department of Thermo Fluids SKMM 2323 Mechanics of Fluids 2 «An excerpt (mostly) from White (2011)» ibn Abdullah May 2017 Outline
More informationWater is sloshing back and forth between two infinite vertical walls separated by a distance L: h(x,t) Water L
ME9a. SOLUTIONS. Nov., 29. Due Nov. 7 PROBLEM 2 Water is sloshing back and forth between two infinite vertical walls separated by a distance L: y Surface Water L h(x,t x Tank The flow is assumed to be
More informationVorticity Equation Marine Hydrodynamics Lecture 9. Return to viscous incompressible flow. N-S equation: v. Now: v = v + = 0 incompressible
13.01 Marine Hydrodynamics, Fall 004 Lecture 9 Copyright c 004 MIT - Department of Ocean Engineering, All rights reserved. Vorticity Equation 13.01 - Marine Hydrodynamics Lecture 9 Return to viscous incompressible
More informationAE301 Aerodynamics I UNIT B: Theory of Aerodynamics
AE301 Aerodynamics I UNIT B: Theory of Aerodynamics ROAD MAP... B-1: Mathematics for Aerodynamics B-: Flow Field Representations B-3: Potential Flow Analysis B-4: Applications of Potential Flow Analysis
More informationMAE 101A. Homework 7 - Solutions 3/12/2018
MAE 101A Homework 7 - Solutions 3/12/2018 Munson 6.31: The stream function for a two-dimensional, nonviscous, incompressible flow field is given by the expression ψ = 2(x y) where the stream function has
More informationragsdale (zdr82) HW7 ditmire (58335) 1 The magnetic force is
ragsdale (zdr8) HW7 ditmire (585) This print-out should have 8 questions. Multiple-choice questions ma continue on the net column or page find all choices efore answering. 00 0.0 points A wire carring
More informationV (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t)
IV. DIFFERENTIAL RELATIONS FOR A FLUID PARTICLE This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common
More informationcalled the potential flow, and function φ is called the velocity potential.
J. Szantr Lectre No. 3 Potential flows 1 If the flid flow is irrotational, i.e. everwhere or almost everwhere in the field of flow there is rot 0 it means that there eists a scalar fnction ϕ,, z), sch
More informationME 321: FLUID MECHANICS-I
6/07/08 ME 3: LUID MECHANI-I Dr. A.B.M. Toufique Hasan Professor Department of Mechanical Engineering Bangladesh Universit of Engineering & Technolog (BUET), Dhaka Lecture- 4/07/08 Momentum Principle teacher.buet.ac.bd/toufiquehasan/
More information- Marine Hydrodynamics. Lecture 14. F, M = [linear function of m ij ] [function of instantaneous U, U, Ω] not of motion history.
2.20 - Marine Hydrodynamics, Spring 2005 ecture 14 2.20 - Marine Hydrodynamics ecture 14 3.20 Some Properties of Added-Mass Coefficients 1. m ij = ρ [function of geometry only] F, M = [linear function
More informationi.e. the conservation of mass, the conservation of linear momentum, the conservation of energy.
04/04/2017 LECTURE 33 Geometric Interpretation of Stream Function: In the last class, you came to know about the different types of boundary conditions that needs to be applied to solve the governing equations
More informationProblems set # 2 Physics 169 February 11, 2015
Prof. Anchordoqui Problems set # 2 Phsics 169 Februar 11, 2015 1. Figure 1 shows the electric field lines for two point charges separated b a small distance. (i) Determine the ratio q 1 /q 2. (ii) What
More informationPART II: 2D Potential Flow
AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page1 PART II: 2D Potential Flow II(a): Euler s Equation& Irrotational Flow We have now completed our tour through the fundamental conservation laws that apply
More information11.1 Introduction Galilean Coordinate Transformations
11.1 Introduction In order to describe physical events that occur in space and time such as the motion of bodies, we introduced a coordinate system. Its spatial and temporal coordinates can now specify
More informationKINEMATIC RELATIONS IN DEFORMATION OF SOLIDS
Chapter 8 KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS Figure 8.1: 195 196 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS 8.1 Motivation In Chapter 3, the conservation of linear momentum for a
More informationVector Calculus. Vector Fields. Reading Trim Vector Fields. Assignment web page assignment #9. Chapter 14 will examine a vector field.
Vector alculus Vector Fields Reading Trim 14.1 Vector Fields Assignment web page assignment #9 hapter 14 will eamine a vector field. For eample, if we eamine the temperature conditions in a room, for ever
More information10. The dimensional formula for c) 6% d) 7%
UNIT. One of the combinations from the fundamental phsical constants is hc G. The unit of this epression is a) kg b) m 3 c) s - d) m. If the error in the measurement of radius is %, then the error in the
More informationUNSTEADY LOW REYNOLDS NUMBER FLOW PAST TWO ROTATING CIRCULAR CYLINDERS BY A VORTEX METHOD
Proceedings of the 3rd ASME/JSME Joint Fluids Engineering Conference Jul 8-23, 999, San Francisco, California FEDSM99-8 UNSTEADY LOW REYNOLDS NUMBER FLOW PAST TWO ROTATING CIRCULAR CYLINDERS BY A VORTEX
More informationThe Calculus of Vec- tors
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 1 The Calculus of Vec- Summary: tors 1. Calculus of Vectors: Limits and Derivatives 2. Parametric representation of Curves r(t) = [x(t), y(t),
More informationMechanics Departmental Exam Last updated November 2013
Mechanics Departmental Eam Last updated November 213 1. Two satellites are moving about each other in circular orbits under the influence of their mutual gravitational attractions. The satellites have
More informationConservation of Linear Momentum
Conservation of Linear Momentum Once we have determined the continuit equation in di erential form we proceed to derive the momentum equation in di erential form. We start b writing the integral form of
More information(Jim You have a note for yourself here that reads Fill in full derivation, this is a sloppy treatment ).
Lecture. dministration Collect problem set. Distribute problem set due October 3, 004.. nd law of thermodynamics (Jim You have a note for yourself here that reads Fill in full derivation, this is a sloppy
More informationCONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM
Chapter 4 CONSERVATION OF ANGULAR MOMENTUM FOR A CONTINUUM Figure 4.1: 4.1 Conservation of Angular Momentum Angular momentum is defined as the moment of the linear momentum about some spatial reference
More informationConservation of Linear Momentum for a Differential Control Volume
Conservation of Linear Momentum for a Differential Control Volume When we applied the rate-form of the conservation of mass equation to a differential control volume (open sstem in Cartesian coordinates,
More informationPressure in stationary and moving fluid Lab- Lab On- On Chip: Lecture 2
Pressure in stationary and moving fluid Lab-On-Chip: Lecture Lecture plan what is pressure e and how it s distributed in static fluid water pressure in engineering problems buoyancy y and archimedes law;
More informationPEMP ACD2505. M.S. Ramaiah School of Advanced Studies, Bengaluru
Two-Dimensional Potential Flow Session delivered by: Prof. M. D. Deshpande 1 Session Objectives -- At the end of this session the delegate would have understood PEMP The potential theory and its application
More informationCHAPTER 3 Introduction to Fluids in Motion
CHAPTER 3 Introduction to Fluids in Motion FE-tpe Eam Review Problems: Problems 3- to 3-9 nˆ 0 ( n ˆi+ n ˆj) (3ˆi 4 ˆj) 0 or 3n 4n 0 3. (D) 3. (C) 3.3 (D) 3.4 (C) 3.5 (B) 3.6 (C) Also n n n + since ˆn
More informationMarine Hydrodynamics Prof.TrilochanSahoo Department of Ocean Engineering and Naval Architecture Indian Institute of Technology, Kharagpur
Marine Hydrodynamics Prof.TrilochanSahoo Department of Ocean Engineering and Naval Architecture Indian Institute of Technology, Kharagpur Lecture - 10 Source, Sink and Doublet Today is the tenth lecture
More informationSOLUTIONS TO CONCEPTS CHAPTER 2
SOLUTIONS TO CONCPTS CHAPTR 1. As shown in the figure, The angle between A and B = 11 = 9 A = and B = 4m Resultant R = A B ABcos = 5 m Let be the angle between R and A 4 sin9 = tan 1 = tan 1 (4/) = 5 4cos9
More informationMATH 228: Calculus III (FALL 2016) Sample Problems for FINAL EXAM SOLUTIONS
MATH 228: Calculus III (FALL 216) Sample Problems for FINAL EXAM SOLUTIONS MATH 228 Page 2 Problem 1. (2pts) Evaluate the line integral C xy dx + (x + y) dy along the parabola y x2 from ( 1, 1) to (2,
More informationTHE VORTEX PANEL METHOD
THE VORTEX PANEL METHOD y j m α V 4 3 2 panel 1 a) Approimate the contour of the airfoil by an inscribed polygon with m sides, called panels. Number the panels clockwise with panel #1 starting on the lower
More informationOffshore Hydromechanics
Offshore Hydromechanics Module 1 : Hydrostatics Constant Flows Surface Waves OE4620 Offshore Hydromechanics Ir. W.E. de Vries Offshore Engineering Today First hour: Schedule for remainder of hydromechanics
More information2 4πε ( ) ( r θ. , symmetric about the x-axis, as shown in Figure What is the electric field E at the origin O?
p E( r, θ) = cosθ 3 ( sinθ ˆi + cosθ ˆj ) + sinθ cosθ ˆi + ( cos θ 1) ˆj r ( ) ( p = cosθ sinθ ˆi + cosθ ˆj + sinθ cosθ ˆi sinθ ˆj 3 r where the trigonometric identit ( θ ) vectors ˆr and cos 1 = sin θ
More informationPressure in stationary and moving fluid. Lab-On-Chip: Lecture 2
Pressure in stationary and moving fluid Lab-On-Chip: Lecture Fluid Statics No shearing stress.no relative movement between adjacent fluid particles, i.e. static or moving as a single block Pressure at
More informationMAE 3130: Fluid Mechanics Lecture 7: Differential Analysis/Part 1 Spring Dr. Jason Roney Mechanical and Aerospace Engineering
MAE 3130: Fluid Mechanics Lecture 7: Differential Analysis/Part 1 Spring 2003 Dr. Jason Roney Mechanical and Aerospace Engineering Outline Introduction Kinematics Review Conservation of Mass Stream Function
More informationCONSERVATION OF ENERGY FOR ACONTINUUM
Chapter 6 CONSERVATION OF ENERGY FOR ACONTINUUM Figure 6.1: 6.1 Conservation of Energ In order to define conservation of energ, we will follow a derivation similar to those in previous chapters, using
More information7 EQUATIONS OF MOTION FOR AN INVISCID FLUID
7 EQUATIONS OF MOTION FOR AN INISCID FLUID iscosity is a measure of the thickness of a fluid, and its resistance to shearing motions. Honey is difficult to stir because of its high viscosity, whereas water
More informationIncompressible Flow Over Airfoils
Chapter 7 Incompressible Flow Over Airfoils Aerodynamics of wings: -D sectional characteristics of the airfoil; Finite wing characteristics (How to relate -D characteristics to 3-D characteristics) How
More information= C. on q 1 to the left. Using Coulomb s law, on q 2 to the right, and the charge q 2 exerts a force F 2 on 1 ( )
Phsics Solutions to Chapter 5 5.. Model: Use the charge model. Solve: (a) In the process of charging b rubbing, electrons are removed from one material and transferred to the other because the are relativel
More informationSome Basic Plane Potential Flows
Some Basic Plane Potential Flows Uniform Stream in the x Direction A uniform stream V = iu, as in the Fig. (Solid lines are streamlines and dashed lines are potential lines), possesses both a stream function
More informationMath 221 Examination 2 Several Variable Calculus
Math Examination Spring Instructions These problems should be viewed as essa questions. Before making a calculation, ou should explain in words what our strateg is. Please write our solutions on our own
More informationFluid Mechanics Prof. T. I. Eldho Department of Civil Engineering Indian Institute of Technology, Bombay
Fluid Mechanics Prof. T. I. Eldho Department of Civil Engineering Indian Institute of Technology, Bombay Lecture No. # 35 Boundary Layer Theory and Applications Welcome back to the video course on fluid
More informationChapter 27 Sources of Magnetic Field
Chapter 27 Sources of Magnetic Field In this chapter we investigate the sources of magnetic of magnetic field, in particular, the magnetic field produced by moving charges (i.e., currents). Ampere s Law
More informationMath 1B Calculus Test 3 Spring 10 Name Write all responses on separate paper. Show your work for credit.
ath B Calculus Test Spring Name Write all responses on separate paper. Show our work for credit.. Determine whether each integral is convergent or divergent. Evaluate those that are convergent. a. / 4
More informationSolutions to Math 41 Final Exam December 9, 2013
Solutions to Math 4 Final Eam December 9,. points In each part below, use the method of your choice, but show the steps in your computations. a Find f if: f = arctane csc 5 + log 5 points Using the Chain
More informationChapter 11 Reference Frames
Chapter 11 Reference Frames Chapter 11 Reference Frames... 2 11.1 Introduction... 2 11.2 Galilean Coordinate Transformations... 2 11.2.1 Relatively Inertial Reference Frames and the Principle of Relativity...
More informationFundamentals of Fluid Dynamics: Ideal Flow Theory & Basic Aerodynamics
Fundamentals of Fluid Dynamics: Ideal Flow Theory & Basic Aerodynamics Introductory Course on Multiphysics Modelling TOMASZ G. ZIELIŃSKI (after: D.J. ACHESON s Elementary Fluid Dynamics ) bluebox.ippt.pan.pl/
More information2.5 Stokes flow past a sphere
Lecture Notes on Fluid Dynamics.63J/.J) by Chiang C. Mei, MIT 007 Spring -5Stokes.tex.5 Stokes flow past a sphere Refs] Lamb: Hydrodynamics Acheson : Elementary Fluid Dynamics, p. 3 ff One of the fundamental
More informationFluid Dynamics: Theory, Computation, and Numerical Simulation Second Edition
Fluid Dynamics: Theory, Computation, and Numerical Simulation Second Edition C. Pozrikidis m Springer Contents Preface v 1 Introduction to Kinematics 1 1.1 Fluids and solids 1 1.2 Fluid parcels and flow
More informationImpact of a Jet. Experiment 4. Purpose. Apparatus. Theory. Symmetric Jet
Experiment 4 Impact of a Jet Purpose The purpose of this experiment is to demonstrate and verify the integral momentum equation. The force generated by a jet of water deflected by an impact surface is
More informationHomework Two. Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu
Homework Two Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu Contents 1 BT Problem 13.15 (8 points) (by Nick Hunter-Jones) 1 2 BT Problem 14.2 (12 points: 3+3+3+3)
More informationFundamentals of Applied Electromagnetics. Chapter 2 - Vector Analysis
Fundamentals of pplied Electromagnetics Chapter - Vector nalsis Chapter Objectives Operations of vector algebra Dot product of two vectors Differential functions in vector calculus Divergence of a vector
More informationHurricane Modeling E XPANDING C ALCULUS H ORIZON
Februar 5, 2009 :4 Hurricane Modeling E XPANDING THE Sheet number Page number can magenta ellow black C ALCULUS H ORIZON Hurricane Modeling... Each ear population centers throughout the world are ravaged
More informationComplex functions in the theory of 2D flow
Complex functions in the theory of D flow Martin Scholtz Institute of Theoretical Physics Charles University in Prague scholtz@utf.mff.cuni.cz Faculty of Transportation Sciences Czech Technical University
More informationControl Volume. Dynamics and Kinematics. Basic Conservation Laws. Lecture 1: Introduction and Review 1/24/2017
Lecture 1: Introduction and Review Dynamics and Kinematics Kinematics: The term kinematics means motion. Kinematics is the study of motion without regard for the cause. Dynamics: On the other hand, dynamics
More informationLecture 1: Introduction and Review
Lecture 1: Introduction and Review Review of fundamental mathematical tools Fundamental and apparent forces Dynamics and Kinematics Kinematics: The term kinematics means motion. Kinematics is the study
More informationSummary: Applications of Gauss Law
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 1 Summary: Applications of Gauss Law 1. Field outside of a uniformly charged sphere of radius a: 2. An infinite, uniformly charged plane
More informationChapter 9 Flow over Immersed Bodies
57:00 Mechanics of Fluids and Transport Processes Chapter 9 Professor Fred Stern Fall 009 1 Chapter 9 Flow over Immersed Bodies Fluid flows are broadly categorized: 1. Internal flows such as ducts/pipes,
More informationLecture 04. Curl and Divergence
Lecture 04 Curl and Divergence UCF Curl of Vector Field (1) F c d l F C Curl (or rotor) of a vector field a n curlf F d l lim c s s 0 F s a n C a n : normal direction of s follow right-hand rule UCF Curl
More informationYell if you have any questions
Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P36-1 Before Starting All of your grades should now be posted
More informationFluid Mechanics for International Engineers HW #4: Conservation of Linear Momentum and Conservation of Energy
2141-365 Fluid Mechanics for International Engineers 1 Problem 1 RTT and Time Rate of Change of Linear Momentum and The Corresponding Eternal Force Notation: Here a material volume (MV) is referred to
More informationContinuum Mechanics Lecture 7 Theory of 2D potential flows
Continuum Mechanics ecture 7 Theory of 2D potential flows Prof. http://www.itp.uzh.ch/~teyssier Outline - velocity potential and stream function - complex potential - elementary solutions - flow past a
More informationAerodynamics. Basic Aerodynamics. Continuity equation (mass conserved) Some thermodynamics. Energy equation (energy conserved)
Flow with no friction (inviscid) Aerodynamics Basic Aerodynamics Continuity equation (mass conserved) Flow with friction (viscous) Momentum equation (F = ma) 1. Euler s equation 2. Bernoulli s equation
More informationGeneral Solution of the Incompressible, Potential Flow Equations
CHAPTER 3 General Solution of the Incompressible, Potential Flow Equations Developing the basic methodology for obtaining the elementary solutions to potential flow problem. Linear nature of the potential
More informationIn this section, mathematical description of the motion of fluid elements moving in a flow field is
Jun. 05, 015 Chapter 6. Differential Analysis of Fluid Flow 6.1 Fluid Element Kinematics In this section, mathematical description of the motion of fluid elements moving in a flow field is given. A small
More information2.25 Advanced Fluid Mechanics
MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Problem 10.3 This problem is from Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin Consider the three different, steady,
More informationFLUID MECHANICS. 1. Division of Fluid Mechanics. Hydrostatics Aerostatics Hydrodynamics Gasdynamics. v velocity p pressure ρ density
FLUID MECHANICS. Diision of Fluid Mechanics elocit p pressure densit Hdrostatics Aerostatics Hdrodnamics asdnamics. Properties of fluids Comparison of solid substances and fluids solid fluid τ F A [Pa]
More informationFinal Review of AerE 243 Class
Final Review of AeE 4 Class Content of Aeodynamics I I Chapte : Review of Multivaiable Calculus Chapte : Review of Vectos Chapte : Review of Fluid Mechanics Chapte 4: Consevation Equations Chapte 5: Simplifications
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring 2013 Exam 3 Equation Sheet. closed fixed path. ! = I ind.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 013 Exam 3 Equation Sheet Force Law: F q = q( E ext + v q B ext ) Force on Current Carrying Wire: F = Id s " B # wire ext Magnetic
More informationSimplifications to Conservation Equations
Chater 5 Simlifications to Conservation Equations 5.1 Steady Flow If fluid roerties at a oint in a field do not change with time, then they are a function of sace only. They are reresented by: ϕ = ϕq 1,
More informationIntegrals along a curve in space. (Sect. 16.1)
Integrals along a curve in space. (Sect. 6.) Line integrals in space. The addition of line integrals. ass and center of mass of wires. Line integrals in space Definition The line integral of a function
More informationMAE 222 Mechanics of Fluids Final Exam with Answers January 13, Give succinct answers to the following word questions.
MAE 222 Mechanics of Fluids Final Exam with Answers January 13, 1994 Closed Book Only, three hours: 1:30PM to 4:30PM 1. Give succinct answers to the following word questions. (a) Why is dimensional analysis
More informationCandidacy Exam Department of Physics February 6, 2010 Part I
Candidacy Exam Department of Physics February 6, 2010 Part I Instructions: ˆ The following problems are intended to probe your understanding of basic physical principles. When answering each question,
More informationPHY 2049 FALL 2000 EXAM 1
PHY 09 FALL 000 EXAM 1 1. Figure below shows three arrangements of electric field lines. A proton is released from point X. It is accelerated by the electric field toward point Y. Points X and Y have eual
More information2.25 Advanced Fluid Mechanics
MIT Department of Mechanical Engineering.5 Advanced Fluid Mechanics Problem 4.05 This problem is from Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin Consider the frictionless, steady
More informationEMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 4 Pure Bending
EA 3702 echanics & aterials Science (echanics of aterials) Chapter 4 Pure Bending Pure Bending Ch 2 Aial Loading & Parallel Loading: uniform normal stress and shearing stress distribution Ch 3 Torsion:
More informationComputational Fluid Dynamics (CFD, CHD)*
1 / 1 Computational Fluid Dnamics (CFD, CHD)* PDE (Shocks 1st); Part I: Basics, Part II: Vorticit Fields Rubin H Landau Sall Haerer, Producer-Director Based on A Surve of Computational Phsics b Landau,
More informationBSL Transport Phenomena 2e Revised: Chapter 2 - Problem 2B.11 Page 1 of 5
BS Transport Phenomena 2e Revised: Chapter 2 - Problem 2B11 Page 1 of 5 Problem 2B11 The cone-and-plate viscometer (see Fig 2B11 A cone-and-plate viscometer consists of a flat plate and an inverted cone,
More informationENGI 4430 Advanced Calculus for Engineering Faculty of Engineering and Applied Science Problem Set 3 Solutions [Multiple Integration; Lines of Force]
ENGI 44 Advanced Calculus for Engineering Facult of Engineering and Applied Science Problem Set Solutions [Multiple Integration; Lines of Force]. Evaluate D da over the triangular region D that is bounded
More information1. Fluid Dynamics Around Airfoils
1. Fluid Dynamics Around Airfoils Two-dimensional flow around a streamlined shape Foces on an airfoil Distribution of pressue coefficient over an airfoil The variation of the lift coefficient with the
More informationE&M. 1 Capacitors. January 2009
E&M January 2009 1 Capacitors Consider a spherical capacitor which has the space between its plates filled with a dielectric of permittivity ɛ. The inner sphere has radius r 1 and the outer sphere has
More informationS12.1 SOLUTIONS TO PROBLEMS 12 (ODD NUMBERS)
OLUTION TO PROBLEM 2 (ODD NUMBER) 2. The electric field is E = φ = 2xi + 2y j and at (2, ) E = 4i + 2j. Thus E = 2 5 and its direction is 2i + j. At ( 3, 2), φ = 6i + 4 j. Thus the direction of most rapid
More informationAA210A Fundamentals of Compressible Flow. Chapter 1 - Introduction to fluid flow
AA210A Fundamentals of Compressible Flow Chapter 1 - Introduction to fluid flow 1 1.2 Conservation of mass Mass flux in the x-direction [ ρu ] = M L 3 L T = M L 2 T Momentum per unit volume Mass per unit
More informationMATHEMATICS 200 December 2011 Final Exam Solutions
MATHEMATICS December 11 Final Eam Solutions 1. Consider the function f(, ) e +4. (a) Draw a contour map of f, showing all tpes of level curves that occur. (b) Find the equation of the tangent plane to
More informationChapter 4 MOTION IN TWO AND THREE DIMENSIONS
Chapter 4 MTIN IN TW AND THREE DIMENSINS Section 4-5, 4-6 Projectile Motion Projectile Motion Analzed Important skills from this lecture: 1. Identif the projectile motion and its velocit and acceleration
More informationMTHE 227 Problem Set 10 Solutions. (1 y2 +z 2., 0, 0), y 2 + z 2 < 4 0, Otherwise.
MTHE 7 Problem Set Solutions. (a) Sketch the cross-section of the (hollow) clinder + = in the -plane, as well as the vector field in this cross-section. ( +,, ), + < F(,, ) =, Otherwise. This is a simple
More informationPhysics 2212 K Quiz #1 Solutions Summer q in = ρv = ρah = ρa 4
Physics 2212 K Quiz #1 Solutions Summer 2016 I. (18 points A uniform infinite insulating slab of charge has a positive volume charge density ρ, and a thickness 2t, extending from t to +t in the z direction.
More informationPHYS152 Lecture 8. Eunil Won Korea University. Ch 30 Magnetic Fields Due to Currents. Fundamentals of Physics by Eunil Won, Korea University
PHYS152 Lecture 8 Ch 3 Magnetic Fields Due to Currents Eunil Won Korea University Calculating the Magnetic Field Due to a Current Recall that we had the formula for the electrostatic force: d E = 1 ɛ dq
More informationSingle Particle Motion
Single Particle Motion C ontents Uniform E and B E = - guiding centers Definition of guiding center E gravitation Non Uniform B 'grad B' drift, B B Curvature drift Grad -B drift, B B invariance of µ. Magnetic
More informationPhysics 1 Second Midterm Exam (AM) 2/25/2010
Physics Second Midterm Eam (AM) /5/00. (This problem is worth 40 points.) A roller coaster car of m travels around a vertical loop of radius R. There is no friction and no air resistance. At the top of
More information1.060 Engineering Mechanics II Spring Problem Set 3
1.060 Engineering Mechanics II Spring 2006 Due on Monday, March 6th Problem Set 3 Important note: Please start a new sheet of paper for each problem in the problem set. Write the names of the group members
More information1.1 The Equations of Motion
1.1 The Equations of Motion In Book I, balance of forces and moments acting on an component was enforced in order to ensure that the component was in equilibrium. Here, allowance is made for stresses which
More informationMath 1B Final Exam, Solution. Prof. Mina Aganagic Lecture 2, Spring (6 points) Use substitution and integration by parts to find:
Math B Final Eam, Solution Prof. Mina Aganagic Lecture 2, Spring 20 The eam is closed book, apart from a sheet of notes 8. Calculators are not allowed. It is your responsibility to write your answers clearly..
More information