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CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Chapter 3: Newtonian Fluids CM4650 Polymer Rheology Michigan Tech Navier-Stokes Equation v vv p 2 v g t 1 © Faith A. Morrison, Michigan Tech U. Chapter 3: Newtonian Fluid Mechanics TWO GOALS •Derive governing equations (mass and momentum balances •Solve governing equations for velocity and stress fields QUICK START V W x First, before we get deep into 2 v (x ) H derivation, let’s do a Navier-Stokes 1 2 x1 problem to get you started in the x3 mechanics of this type of problem solving. 2 © Faith A. Morrison, Michigan Tech U. 1 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics EXAMPLE: Drag flow between infinite parallel plates •Newtonian •steady state •incompressible fluid •very wide, long V •uniform pressure W x2 v1(x2) H x1 x3 3 EXAMPLE: Poiseuille flow between infinite parallel plates •Newtonian •steady state •Incompressible fluid •infinitely wide, long W x2 2H x1 x3 v (x ) x1=0 1 2 x1=L p=Po p=PL 4 2 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Engineering Quantities of In more complex flows, we can use Interest general expressions that work in all cases. (any flow) volumetric ⋅ flow rate ∬ ⋅ | average 〈 〉 velocity ∬ Using the general formulas will Here, is the outwardly pointing unit normal help prevent errors. of ; it points in the direction “through” 5 © Faith A. Morrison, Michigan Tech U. The stress tensor was Total stress tensor, Π: invented to make the calculation of fluid stress easier. Π ≡ b (any flow, small surface) dS nˆ Force on the S ⋅ Π surface V (using the stress convention of Understanding Rheology) Here, is the outwardly pointing unit normal of ; it points in the direction “through” 6 © Faith A. Morrison, Michigan Tech U. 3 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics To get the total force on the macroscopic surface , we integrate over the entire surface of interest. ‐Π Fluid force on ⋅ the surface S ⋅ , and evaluated at the surface 7 (using the stress convention of Understanding Rheology) © Faith A. Morrison, Michigan Tech U. Engineering Using the general formulas will Quantities of help prevent errors (like Interest forgetting the pressure). (any flow) force on the surface, S ⋅ ̳ z‐component of force on ̂ ⋅ ⋅ ̳ the surface, S 8 (using the stress convention of Understanding Rheology) © Faith A. Morrison, Michigan Tech U. 4 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Engineering Quantities of Interest (any flow) Total Fluid Torque on a ⋅ Π surface, S is the vector from the axis of rotation to 9 (using the stress convention of Understanding Rheology) © Faith A. Morrison, Michigan Tech U. Common surface shapes: rectangular: dS dxdy circular top: dS r drd surface of cylinder: dS Rd dz sphere:()sinsin dS Rd r d R2 d d Note: spherical coordinate system in use by fluid mechanics community For more areas, see uses 0as the angle from the Exam 1 formula =axis to the point. handout: pages.mtu.edu/~fmorriso/cm4650/formu la_sheet_for_exam1_2018.pdf 10 © Faith A. Morrison, Michigan Tech U. 5 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Review: Chapter 3: Newtonian Fluid Mechanics TWO GOALS •Derive governing equations (mass and momentum balances •Solve governing equations for velocity and stress fields We got a Quick Start with Newtonian problem QUICK START V W solving… x First, before we get deep into 2 v (x ) H derivation, let’s do a Navier-Stokes 1 2 x1 problem to get you started in the x3 mechanics of this type of problem solving. 2 © Faith A. Morrison, Michigan Tech U. Now… Back to exploring the origin of the equations (so we can adapt to non‐Newtonian) © Faith A. Morrison, Michigan Tech U. Chapter 3: Newtonian Fluid Mechanics TWO GOALS •Derive governing equations (mass and momentum balances •Solve governing equations for velocity and stress fields Mass Balance Consider an arbitrary control volume V enclosed by a surface S rate of increase net flux of of mass in CV mass into CV 12 © Faith A. Morrison, Michigan Tech U. 6 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Mathematics Review Polymer Rheology b dS nˆ S V 13 © Faith A. Morrison, Michigan Tech U. Chapter 3: Newtonian Fluid Mechanics Polymer Rheology Mass Balance (continued) Consider an arbitrary volume V enclosed by a surface S rate of increase d dV of mass in V dt V outwardly pointing unit net flux of normal mass into V nˆ v dS through surface S S 14 © Faith A. Morrison, Michigan Tech U. 7 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Chapter 3: Newtonian Fluid Mechanics Polymer Rheology Mass Balance (continued) d dV nˆ v dS Leibnitz dt rule V S dV nˆ v dS Gauss Divergence V t S Theorem v dV V v dV 0 V t 15 © Faith A. Morrison, Michigan Tech U. Chapter 3: Newtonian Fluid Mechanics Polymer Rheology Mass Balance (continued) v dV 0 Since V is t arbitrary, V Continuity equation: microscopic mass balance v 0 t 16 © Faith A. Morrison, Michigan Tech U. 8 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Chapter 3: Newtonian Fluid Mechanics Polymer Rheology Mass Balance (continued) Continuity equation (general fluids) v 0 t v v 0 t D v 0 Dt For =constant (incompressible fluids): v 0 17 © Faith A. Morrison, Michigan Tech U. Chapter 3: Newtonian Fluid Mechanics Polymer Rheology Momentum Balance Consider an arbitrary control volume Momentum is conserved. V enclosed by a surface S rate of increase net flux of sum of of momentum in CV momentum into CV forces on CV resembles the resembles the Forces: rate term in the flux term in the body (gravity) mass balance mass balance molecular forces 18 © Faith A. Morrison, Michigan Tech U. 9 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Momentum Balance Polymer Rheology b dS nˆ S V 19 © Faith A. Morrison, Michigan Tech U. Momentum Balance (continued) Polymer Rheology rate of increase d v dV of momentum in V dt Leibnitz V rule v dV V t net flux of nˆ vv dS Gauss momentum into V S Divergence Theorem vv dV V 20 © Faith A. Morrison, Michigan Tech U. 10 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Momentum Balance (continued) Polymer Rheology Forces on V Body Forces (non-contact) force on V g dV dueto g V 21 © Faith A. Morrison, Michigan Tech U. Chapter 3: Newtonian Fluid Mechanics Polymer Rheology Molecular Forces (contact) – this is the tough one choose a surface stress through P f at P dS on dS the force on P that surface We need an expression for the state of stress at an arbitrary point P in a flow. 22 © Faith A. Morrison, Michigan Tech U. 11 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Molecular Forces (continued) Think back to the molecular picture from chemistry: The specifics of these forces, connections, and interactions must be captured by the molecular forces term that we seek. 23 © Faith A. Morrison, Michigan Tech U. Molecular Forces (continued) •We will concentrate on expressing the molecular forces mathematically; •We leave to later the task of relating the resulting mathematical expression to experimental observations. First, choose a surface: nˆ •arbitrary shape •small f dS stress at P dS f What is f ? on dS 24 © Faith A. Morrison, Michigan Tech U. 12 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Consider the forces on three mutually perpendicular surfaces through point P: x2 P x1 eˆ1 x3 eˆ 2 a eˆ3 b c 25 © Faith A. Morrison, Michigan Tech U. Molecular Forces (continued) a is stress on a “1” surface at P a surface with unit normal eˆ1 b is stress on a “2” surface at P c is stress on a “3” surface at P We can write these vectors in a Cartesian coordinate system: a a1eˆ1 a2eˆ2 a3eˆ3 11eˆ1 12eˆ2 13eˆ3 stress on a “1” surface in the 1- direction 26 © Faith A. Morrison, Michigan Tech U. 13 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics Molecular Forces (continued) a a1eˆ1 a2eˆ2 a3eˆ3 a is stress on a “1” surface at P 11eˆ1 12eˆ2 13eˆ3 b is stress on a “2” surface at P b b1eˆ1 b2eˆ2 b3eˆ3 c is stress on a “3” surface at P 21eˆ1 22eˆ2 23eˆ3 c c1eˆ1 c2eˆ2 c3eˆ3 31eˆ1 32eˆ2 33eˆ3 So far, this is pk nomenclature; next we relate these Stress on a “p” expressions to force surface in the k-direction on an arbitrary surface. 27 © Faith A. Morrison, Michigan Tech U. Molecular Forces (continued) How can we write f (the force on nˆ an arbitrary surface dS) in terms of the Ppk? f dS f f1eˆ1 f2eˆ2 f3eˆ3 f3 is force on dS in f1 is force on dS in 3-direction 1-direction f2 is force on dS in 2-direction There are three Ppk that relate to forces in the 1-direction: 11, 21, 31 28 © Faith A. Morrison, Michigan Tech U. 14 CM4650 Chapter 3 Newtonian Fluid 2/5/2018 Mechanics nˆ Molecular Forces (continued) f dS How can we write f (the force on an arbitrary surface dS) in terms of the f f1eˆ1 f2eˆ2 f3eˆ3 quantities Ppk? f1 , the force on dS in 1-direction, can be broken into three parts associated with the three stress components: . 11, 21, 31 nˆ eˆ1 dS projection of first part: 11 dA onto the 11nˆ eˆ1 dS 1 surface force area area 29 © Faith A. Morrison, Michigan Tech U. Molecular Forces (continued) f1 , the force on dS in 1-direction, is composed of THREE parts: projection of first part: 11 dA onto the 11nˆ eˆ1 dS 1 surface projection of second part: 21 dA onto the 21nˆ eˆ2 dS 2 surface projection of stress on a 2 -surface third part: 31 dA onto the 31nˆ eˆ3 dS in the 1- direction 3 surface the sum of these three = f1 30 © Faith A.
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  • Chapter 15 - Fluid Mechanics Thursday, March 24Th

    Chapter 15 - Fluid Mechanics Thursday, March 24Th

    Chapter 15 - Fluid Mechanics Thursday, March 24th •Fluids – Static properties • Density and pressure • Hydrostatic equilibrium • Archimedes principle and buoyancy •Fluid Motion • The continuity equation • Bernoulli’s effect •Demonstration, iClicker and example problems Reading: pages 243 to 255 in text book (Chapter 15) Definitions: Density Pressure, ρ , is defined as force per unit area: Mass M ρ = = [Units – kg.m-3] Volume V Definition of mass – 1 kg is the mass of 1 liter (10-3 m3) of pure water. Therefore, density of water given by: Mass 1 kg 3 −3 ρH O = = 3 3 = 10 kg ⋅m 2 Volume 10− m Definitions: Pressure (p ) Pressure, p, is defined as force per unit area: Force F p = = [Units – N.m-2, or Pascal (Pa)] Area A Atmospheric pressure (1 atm.) is equal to 101325 N.m-2. 1 pound per square inch (1 psi) is equal to: 1 psi = 6944 Pa = 0.068 atm 1atm = 14.7 psi Definitions: Pressure (p ) Pressure, p, is defined as force per unit area: Force F p = = [Units – N.m-2, or Pascal (Pa)] Area A Pressure in Fluids Pressure, " p, is defined as force per unit area: # Force F p = = [Units – N.m-2, or Pascal (Pa)] " A8" rea A + $ In the presence of gravity, pressure in a static+ 8" fluid increases with depth. " – This allows an upward pressure force " to balance the downward gravitational force. + " $ – This condition is hydrostatic equilibrium. – Incompressible fluids like liquids have constant density; for them, pressure as a function of depth h is p p gh = 0+ρ p0 = pressure at surface " + Pressure in Fluids Pressure, p, is defined as force per unit area: Force F p = = [Units – N.m-2, or Pascal (Pa)] Area A In the presence of gravity, pressure in a static fluid increases with depth.