v T Pressure Extra Molecular Stresses Constitutive equations for Stress v t Observation: the stress tensor is symmetric

Similar documents
ME 141. Engineering Mechanics

() t. () t r () t or v. ( t) () () ( ) = ( ) or ( ) () () () t or dv () () Section 10.4 Motion in Space: Velocity and Acceleration

Science Advertisement Intergovernmental Panel on Climate Change: The Physical Science Basis 2/3/2007 Physics 253

Motion on a Curve and Curvature

D zone schemes

LECTURE 5. is defined by the position vectors r, 1. and. The displacement vector (from P 1 to P 2 ) is defined through r and 1.

Circuits 24/08/2010. Question. Question. Practice Questions QV CV. Review Formula s RC R R R V IR ... Charging P IV I R ... E Pt.

Reinforcement learning

Ch.4 Motion in 2D. Ch.4 Motion in 2D

Physics 201, Lecture 5

Technical Vibration - text 2 - forced vibration, rotational vibration

PHYSICS 102. Intro PHYSICS-ELECTROMAGNETISM

Today - Lecture 13. Today s lecture continue with rotations, torque, Note that chapters 11, 12, 13 all involve rotations


Average & instantaneous velocity and acceleration Motion with constant acceleration

1. Kinematics of Particles

f(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2

A Kalman filtering simulation

e t dt e t dt = lim e t dt T (1 e T ) = 1

4.8 Improper Integrals

Ans: In the rectangular loop with the assigned direction for i2: di L dt , (1) where (2) a) At t = 0, i1(t) = I1U(t) is applied and (1) becomes

MEEN 617 Handout #11 MODAL ANALYSIS OF MDOF Systems with VISCOUS DAMPING

September 20 Homework Solutions

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

PHYSICS 1210 Exam 1 University of Wyoming 14 February points

Version 001 test-1 swinney (57010) 1. is constant at m/s.

Contraction Mapping Principle Approach to Differential Equations

Motion. ( (3 dim) ( (1 dim) dt. Equations of Motion (Constant Acceleration) Newton s Law and Weight. Magnitude of the Frictional Force

An object moving with speed v around a point at distance r, has an angular velocity. m/s m

PHYS PRACTICE EXAM 2

Sections 3.1 and 3.4 Exponential Functions (Growth and Decay)

Lecture 3. Electrostatics

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =

INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information.

Orthotropic Materials

6. Gas dynamics. Ideal gases Speed of infinitesimal disturbances in still gas

Chapter Direct Method of Interpolation

Lecture-V Stochastic Processes and the Basic Term-Structure Equation 1 Stochastic Processes Any variable whose value changes over time in an uncertain

0 for t < 0 1 for t > 0

Rotations.

Available online Journal of Scientific and Engineering Research, 2018, 5(10): Research Article

Molecular Dynamics Simulations (Leach )

Addition & Subtraction of Polynomials

156 There are 9 books stacked on a shelf. The thickness of each book is either 1 inch or 2

RESPONSE UNDER A GENERAL PERIODIC FORCE. When the external force F(t) is periodic with periodτ = 2π

ENGR 1990 Engineering Mathematics The Integral of a Function as a Function

Chapter 4 Circular and Curvilinear Motions

KINEMATICS OF RIGID BODIES

vs. Chapter 4: Standard Flows Chapter 4: Standard Flows for Rheology shear elongation 2/1/2016 CM4650 Lectures 1-3: Intro, Mathematical Review

Relative and Circular Motion

ÖRNEK 1: THE LINEAR IMPULSE-MOMENTUM RELATION Calculate the linear momentum of a particle of mass m=10 kg which has a. kg m s

Lecture 22 Electromagnetic Waves

graph of unit step function t

Faraday s Law. To be able to find. motional emf transformer and motional emf. Motional emf

, on the power of the transmitter P t fed to it, and on the distance R between the antenna and the observation point as. r r t

r P + '% 2 r v(r) End pressures P 1 (high) and P 2 (low) P 1 , which must be independent of z, so # dz dz = P 2 " P 1 = " #P L L,

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

Homework 5 for BST 631: Statistical Theory I Solutions, 09/21/2006

Computer Propagation Analysis Tools

Department of Chemical Engineering University of Tennessee Prof. David Keffer. Course Lecture Notes SIXTEEN

Physics 2A HW #3 Solutions

Magnetostatics Bar Magnet. Magnetostatics Oersted s Experiment

Electric Potential. and Equipotentials

MATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)

Equations from The Four Principal Kinetic States of Material Bodies. Copyright 2005 Joseph A. Rybczyk

General Non-Arbitrage Model. I. Partial Differential Equation for Pricing A. Traded Underlying Security

Computer Aided Geometric Design

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218

Classification of Equations Characteristics

Lecture 5. Chapter 3. Electromagnetic Theory, Photons, and Light

Lecture 17: Kinetics of Phase Growth in a Two-component System:

(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.

Control Volume Derivation

MATHEMATICAL FOUNDATIONS FOR APPROXIMATING PARTICLE BEHAVIOUR AT RADIUS OF THE PLANCK LENGTH

Axis. Axis. Axis. Solid cylinder (or disk) about. Hoop about. Annular cylinder (or ring) about central axis. central axis.

FM Applications of Integration 1.Centroid of Area

Lecture 4 Kinetics of a particle Part 3: Impulse and Momentum

Ch.1. Group Work Units. Continuum Mechanics Course (MMC) - ETSECCPB - UPC

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

The Production of Polarization

1 Using Integration to Find Arc Lengths and Surface Areas

3 Motion with constant acceleration: Linear and projectile motion

Suggested Practice Problems (set #2) for the Physics Placement Test

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle

Diffusion & Viscosity: Navier-Stokes Equation

ENGI 4430 Advanced Calculus for Engineering Faculty of Engineering and Applied Science Problem Set 9 Solutions [Theorems of Gauss and Stokes]

7 Wave Equation in Higher Dimensions

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

r r r r r EE334 Electromagnetic Theory I Todd Kaiser

Physics 101 Lecture 4 Motion in 2D and 3D

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

The order of reaction is defined as the number of atoms or molecules whose concentration change during the chemical reaction.

Fluids & Bernoulli s Equation. Group Problems 9

Mathematics 805 Final Examination Answers

Phys 110. Answers to even numbered problems on Midterm Map

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.

Compressive modulus of adhesive bonded rubber block

Transcription:

Momenum Blnce (coninued Momenum Blnce (coninued Now, wh o do wih Π? Pessue is p of i. bck o ou quesion, Now, wh o do wih? Π Pessue is p of i. Thee e ohe, nonisoopic sesses Pessue E Molecul Sesses definiion: An isoopic foce/e of molecul oigin. Pessue is he sme on ny sufce dwn hough poin nd cs nomlly o he chosen sufce. p pessue = p I = p ee ˆ ˆ + p eˆ eˆ + p eˆ eˆ = Tes: wh is he foce on sufce wih uni noml? ṋ p p definiion: The e sesses e he molecul sesses h e no isoopic E sess enso, Now, wh o do wih? Π p I i.e. eveyhing compliced in molecul defomion This becomes he cenl quesion of heologicl sudy Momenum Blnce (coninued Momenum Blnce (coninued Consiuive equions fo Sess e enso equions ele he velociy field o he sesses geneed by molecul foces e bsed on obsevions (empiicl o e bsed on molecul models (heoeicl e ypiclly found by il-nd-eo e jusified by how well hey wok fo sysem of inees e obseved o be symmeic = f ( v, meil popeies Obsevion: he sess enso is symmeic Micoscopic momenum blnce ρ + v v = Π + ρ g In ems of he e sess enso: ρ + v v = p + ρ g Equion of Moion Equion of Moion Momenum Blnce (coninued Momenum Blnce (coninued Newonin Consiuive equion = µ ( v + ( v T fo incompessible s (see e fo compessible s is empiicl my be jusified fo some sysems wih molecul modeling clculions How is he Newonin Consiuive equion eled o Newon s Lw of iscosiy? = µ ( v + ( v T = µ incompessible s incompessible s eciline flow (sigh lines no viion in -diecion

Momenum Blnce (coninued Momenum Blnce (coninued Bck o he momenum blnce... ρ + v v = p + ρ g We cn incopoe he Newonin consiuive equion ino he momenum blnce o obin momenum-blnce equion h is specific o incompessible, Newonin s = µ Equion of Moion ( v + ( v T Nvie-Sokes Equion v ρ + v v = p + µ v + ρ g incompessible s Newonin s EXAMPLE: Dg flow beween infinie pllel ples sedy se incompessible infiniely wide, long W EXAMPLE: Poiseuille flow beween infinie pllel ples sedy se incompessible infiniely wide, long W v ( H H = v ( =L p=p o p=p L EXAMPLE: Poiseuille flow in ube Sedy se incompessible long ube A z coss-secion A: z EXAMPLE: Tosionl flow beween pllel ples Sedy se incompessible v = zf ( coss-secionl view: Ω Ω z v z ( L H R R

Newonin s: = µγ& vs. non-newonin s: µγ& Simple She Flow velociy field v ( H = = γ& H γ& = consn How cn we invesige non-newonin behvio? CONSTANT TORQUE MOTOR H v ( ( ( + ς& ( v ph lines Ne solid sufces, he flow is she flow. Epeimenl She Geomeies (z-plne y secion (z-plne secion y H (-plne secion φ ο (φ plne secion (z-plne secion (-plne secion (z-plne secion (-plne secion Sndd Nomenclue fo She Flow gdien diecion Why is she sndd flow? simple velociy field epesens ll sliding flows simple sess enso neul diecion flow diecion

How do picles move p in she flow? Conside wo picles in he sme - plne, iniilly long he is. = > ( l o v ( P (, l,, P (,, l, l v lo lo γ& o P ( γ& oll,l,, l P ( γ& ol,l,, l l loγ& o long imes How do picles move p in she flow? Conside wo picles in he sme - plne, iniilly long he is ( =. γ& v = Ech picle hs diffeen velociy depending on is posiion: v = γ& P : P : v = & γ l v = & γ l The iniil posiion of ech picle is =. Afe seconds, he wo picles e he following posiions: P ( : P ( : = & γ l = & γ l lengh locion = iniil + ime ( ime Wh is he sepion of he picles fe ime? γ& l Uniil Elongionl Flow l γ& ( l l l = l + = l l = l [& γ ( l l ] = l + & γ l ( + & γ + & γ l & γ negligible s γ& l l lγ& l In she he disnce beween poins is diecly popoionl o ime velociy field, & ε( ( & ε v ( & ε ε& ( > Uniil Elongionl Flow Elongionl flow occus when hee is seching - die ei, flow hough concions, ph lines & ε( ( & ε v ( & ε ε& ( > 4

Epeimenl Elongionl Geomeies i-bed o suppo smple Why is elongion sndd flow? simple velociy field epesens ll seching flows simple sess enso o o+ o+ h(o R( h( hin, lubicing lye on ech ple R(o How do picles move p in elongionl flow? Conside wo picles in he sme - plne, iniilly long he is. l o o l = loe ε& = > P P P P l P,, o l P,, o lo ε& P o,, e lo ε& o P,, e A second ype of she-fee flow: Biil Seching befoe fe & ε( ( & ε v ( & ε P i unde pessue ε& ( < fe befoe How do uniil nd biil defomions diffe? How do uniil nd biil defomions diffe? Conside uniil flow in which picle is doubled in lengh in he flow diecion. Conside biil flow in which picle is doubled in lengh in he flow diecion. /4 5

& ε ( ε& ( > v ( & ε A hid ype of she-fee flow: Pln Elongionl Flow All hee she-fee flows cn be wien ogehe s: & ε( ( + b v = & ε( ( b & ε( / Elongionl flow: b=, ε& ( > Biil seching: b=, ε& ( < Pln elongion: b=, ε& ( > Why hve we chosen hese flows? ANSWER: Becuse hese simple flows hve symmey. And symmey llows us o dw conclusions bou he sess enso h is ssocied wih hese flows fo ny subjeced o h flow. In genel: = Bu he sess enso is symmeic leving 6 independen sess componens. Cn we choose flow o use in which hee e fewe hn 6 independen sess componens? Yes we cn symmeic flows How does he sess enso simplify fo she (nd le, elongionl flow? P (,, e ê e ê P(,, Wh would he velociy funcion be fo Newonin in his coodine sysem? H v v = 6

Wh would he velociy funcion be fo Newonin in his coodine sysem? H v v = ecos e independen of coodine sysem, bu in genel he coefficiens will be diffeen when he sme veco is wien in wo diffeen coodine sysems: v v = v v v = v v Fo she flow nd he wo picul coodine sysems we hve jus emined, H v = H = H v = H = Wh do we len if we fomlly nsfom v fom one coodine sysem o he ohe? If we plug in he sme numbe in fo nd, we will NOT be sking bou he sme poin in spce, bu we WILL ge he sme ec velociy veco. Since sess is clculed fom he velociy field, we will ge he sme ec sess enso when we clcule i fom eihe veco epesenion lso. v p = v p This is n unusul cicumsnce only ue fo pk = he picul coodine pk sysems chosen. eˆ eˆ eˆ = = e = e e = = = Becuse of symmey, hee e only 5 nonzeo componens of he e sess enso in she flow. Conclusion: SHEAR: = This gely simplifies he epeimenliss sks s only fou sess componens ( = mus be mesued insed of 6. We hve found coodine sysem (he she coodine sysem in which hee e only 5 non-zeo coefficiens of he sess enso. In ddiion, =. This leves only fou sess componens o be mesued fo his flow, epessed in his coodine sysem. 7

How does he sess enso simplify fo elongionl flow? Becuse of symmey, hee e only nonzeo componens of he e sess enso in elongionl flows. ELONGATION:, = Thee is 8 o of symmey ound ll hee coodine es. This gely simplifies he epeimenliss sks s only hee sess componens mus be mesued insed of 6. One finl commen on mesuing sesses... Densiy does no vy (much wih pessue fo polymeic s. Wh is mesued is he ol sess, Π : p + Π = p + p + 4 ρ g / cm gs densiy polyme densiy incompessible M ρ = P RT Fo he noml sesses we e fced wih he difficuly of seping p fom ii. Compessible s: Incompessible s: Ge p fom nrt p = mesuemens of T nd.? 5 5 5 Pessue (MP Fo incompessible s i is no possible o sepe p fom ii. Noml Sess Diffeences Luckily, his is no poblem since we only need Π = p + Equion of moion + v v = Π + ρ g = P + ρ g Soluion? Noml sess diffeences We do no need ii diecly o solve fo velociies Fis noml sess diffeence Second noml sess diffeence N N Π Π In she flow, hee sess quniies e mesued Π Π N, =, N = In elongionl flow, wo sess quniies e mesued, 8

Fis noml sess effecs: od climbing < Second noml sess effecs: inclined openchnnel flow > E ension in he -diecion pulls down he fee sufce whee dv /d is gees (see DPL p65. E ension in he -diecion pulls zimuhlly nd upwd (see DPL p65. Newonin - glycein iscoelsic - % soln of polyehylene oide in we Newonin - glycein iscoelsic - soluion of N ~ -N / polycylmide in glycein Bid, e l., Dynmics of Polymeic Fluids, vol., Wiley, 987, Figue.- pge 6. (DPL Wh s ne? Emple: Cn he equion of moion pedic od climbing fo ypicl vlues of N, N? z coss-secion A: Ω v = v z κr R A dπ zz Wh is? d Bid e l. p64 We seek o qunify he behvio of nonnewonin s R. I. Tnne, Engineeing Rheology, Ofod 985, Figue.6 pge 4 She ς&( v Even wih jus hese (o 4 sndd flows, we cn sill genee n infinie numbe of flows by vying ς&( nd ε&(. She-fee (elongionl, eensionl & ε ( ( + b v = ε& ( ( b ε&( Elongionl flow: b=, Biil seching: b=, Pln elongion: b=, ε&( > ε& ( < ε&( > Pocedue:. Choose flow ype (she o ype of elongion.. Specify ς& ( o ε&( s ppopie.. Impose he flow on of inees. 4. Mesue sesses. she, N, N elongion, 5. Repo sesses in ems of meil funcions. 6. Compe mesued meil funcions wih pedicions of hese meil funcions (fom poposed consiuive equions. Fih A. Moison, Michign Tech U. 7. Choose he mos ppopie consiuive equion fo use in numeicl modeling. 6b. Compe mesued meil funcions wih hose mesued on ohe meils. 7. Dw conclusions on he likely popeies of he unknown meil bsed on he compison. 9