Reynolds Transport Theorem
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10. Collisions • Use Conservation of Momentum and Energy and The
10. Collisions • Use conservation of momentum and energy and the center of mass to understand collisions between two objects. • During a collision, two or more objects exert a force on one another for a short time: -F(t) F(t) Before During After • It is not necessary for the objects to touch during a collision, e.g. an asteroid flied by the earth is considered a collision because its path is changed due to the gravitational attraction of the earth. One can still use conservation of momentum and energy to analyze the collision. Impulse: During a collision, the objects exert a force on one another. This force may be complicated and change with time. However, from Newton's 3rd Law, the two objects must exert an equal and opposite force on one another. F(t) t ti tf Dt From Newton'sr 2nd Law: dp r = F (t) dt r r dp = F (t)dt r r r r tf p f - pi = Dp = ò F (t)dt ti The change in the momentum is defined as the impulse of the collision. • Impulse is a vector quantity. Impulse-Linear Momentum Theorem: In a collision, the impulse on an object is equal to the change in momentum: r r J = Dp Conservation of Linear Momentum: In a system of two or more particles that are colliding, the forces that these objects exert on one another are internal forces. These internal forces cannot change the momentum of the system. Only an external force can change the momentum. The linear momentum of a closed isolated system is conserved during a collision of objects within the system. -
Modeling Conservation of Mass
Modeling Conservation of Mass How is mass conserved (protected from loss)? Imagine an evening campfire. As the wood burns, you notice that the logs have become a small pile of ashes. What happened? Was the wood destroyed by the fire? A scientific principle called the law of conservation of mass states that matter is neither created nor destroyed. So, what happened to the wood? Think back. Did you observe smoke rising from the fire? When wood burns, atoms in the wood combine with oxygen atoms in the air in a chemical reaction called combustion. The products of this burning reaction are ashes as well as the carbon dioxide and water vapor in smoke. The gases escape into the air. We also know from the law of conservation of mass that the mass of the reactants must equal the mass of all the products. How does that work with the campfire? mass – a measure of how much matter is present in a substance law of conservation of mass – states that the mass of all reactants must equal the mass of all products and that matter is neither created nor destroyed If you could measure the mass of the wood and oxygen before you started the fire and then measure the mass of the smoke and ashes after it burned, what would you find? The total mass of matter after the fire would be the same as the total mass of matter before the fire. Therefore, matter was neither created nor destroyed in the campfire; it just changed form. The same atoms that made up the materials before the reaction were simply rearranged to form the materials left after the reaction. -
Derivation of Fluid Flow Equations
TPG4150 Reservoir Recovery Techniques 2017 1 Fluid Flow Equations DERIVATION OF FLUID FLOW EQUATIONS Review of basic steps Generally speaking, flow equations for flow in porous materials are based on a set of mass, momentum and energy conservation equations, and constitutive equations for the fluids and the porous material involved. For simplicity, we will in the following assume isothermal conditions, so that we not have to involve an energy conservation equation. However, in cases of changing reservoir temperature, such as in the case of cold water injection into a warmer reservoir, this may be of importance. Below, equations are initially described for single phase flow in linear, one- dimensional, horizontal systems, but are later on extended to multi-phase flow in two and three dimensions, and to other coordinate systems. Conservation of mass Consider the following one dimensional rod of porous material: Mass conservation may be formulated across a control element of the slab, with one fluid of density ρ is flowing through it at a velocity u: u ρ Δx The mass balance for the control element is then written as: ⎧Mass into the⎫ ⎧Mass out of the ⎫ ⎧ Rate of change of mass⎫ ⎨ ⎬ − ⎨ ⎬ = ⎨ ⎬ , ⎩element at x ⎭ ⎩element at x + Δx⎭ ⎩ inside the element ⎭ or ∂ {uρA} − {uρA} = {φAΔxρ}. x x+ Δx ∂t Dividing by Δx, and taking the limit as Δx approaches zero, we get the conservation of mass, or continuity equation: ∂ ∂ − (Aρu) = (Aφρ). ∂x ∂t For constant cross sectional area, the continuity equation simplifies to: ∂ ∂ − (ρu) = (φρ) . ∂x ∂t Next, we need to replace the velocity term by an equation relating it to pressure gradient and fluid and rock properties, and the density and porosity terms by appropriate pressure dependent functions. -
Linear-Nonequilibrium Thermodynamics Theory for Coupled Heat and Mass Transport
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Chemical and Biomolecular Research Papers -- Yasar Demirel Publications Faculty Authors Series 2001 Linear-nonequilibrium thermodynamics theory for coupled heat and mass transport Yasar Demirel University of Nebraska - Lincoln, [email protected] Stanley I. Sandler University of Delaware, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/cbmedemirel Part of the Chemical Engineering Commons Demirel, Yasar and Sandler, Stanley I., "Linear-nonequilibrium thermodynamics theory for coupled heat and mass transport" (2001). Yasar Demirel Publications. 8. https://digitalcommons.unl.edu/cbmedemirel/8 This Article is brought to you for free and open access by the Chemical and Biomolecular Research Papers -- Faculty Authors Series at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Yasar Demirel Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Published in International Journal of Heat and Mass Transfer 44 (2001), pp. 2439–2451 Copyright © 2001 Elsevier Science Ltd. Used by permission. www.elsevier.comllocate/ijhmt Submitted November 5, 1999; revised September 6, 2000. Linear-nonequilibrium thermodynamics theory for coupled heat and mass transport Yasar Demirel and S. I. Sandler Center for Molecular and Engineering Thermodynamics, Department of Chemical Engineering, University of Delaware, Newark, DE 19716, USA Corresponding author — Y. Demirel Abstract Linear-nonequilibrium thermodynamics (LNET) has been used to express the entropy generation and dis- sipation functions representing the true forces and flows for heat and mass transport in a multicomponent fluid. These forces and flows are introduced into the phenomenological equations to formulate the cou- pling phenomenon between heat and mass flows. -
Law of Conversation of Energy
Law of Conservation of Mass: "In any kind of physical or chemical process, mass is neither created nor destroyed - the mass before the process equals the mass after the process." - the total mass of the system does not change, the total mass of the products of a chemical reaction is always the same as the total mass of the original materials. "Physics for scientists and engineers," 4th edition, Vol.1, Raymond A. Serway, Saunders College Publishing, 1996. Ex. 1) When wood burns, mass seems to disappear because some of the products of reaction are gases; if the mass of the original wood is added to the mass of the oxygen that combined with it and if the mass of the resulting ash is added to the mass o the gaseous products, the two sums will turn out exactly equal. 2) Iron increases in weight on rusting because it combines with gases from the air, and the increase in weight is exactly equal to the weight of gas consumed. Out of thousands of reactions that have been tested with accurate chemical balances, no deviation from the law has ever been found. Law of Conversation of Energy: The total energy of a closed system is constant. Matter is neither created nor destroyed – total mass of reactants equals total mass of products You can calculate the change of temp by simply understanding that energy and the mass is conserved - it means that we added the two heat quantities together we can calculate the change of temperature by using the law or measure change of temp and show the conservation of energy E1 + E2 = E3 -> E(universe) = E(System) + E(Surroundings) M1 + M2 = M3 Is T1 + T2 = unknown (No, no law of conservation of temperature, so we have to use the concept of conservation of energy) Total amount of thermal energy in beaker of water in absolute terms as opposed to differential terms (reference point is 0 degrees Kelvin) Knowns: M1, M2, T1, T2 (Kelvin) When add the two together, want to know what T3 and M3 are going to be. -
Key Concepts: Conservation of Mass, Momentum, Energy Fluid: a Material
Key concepts: Conservation of mass, momentum, energy Fluid: a material that deforms continuously and permanently under the application of a shearing stress, no matter how small. Fluids are either gases or liquids. (Under very specialized conditions, a phase of intermediate properties can be stable, but we won’t consider that possibility.) In liquids, the molecules are relatively closely spaced, allowing the magnitude of their (attractive, electrically-based) interaction energy to be of the same magnitude as their kinetic energy. As a result, they exist as a loose collection of clusters. In gases, the molecules are much more widely separated, so the kinetic energy (at a given temperature, identical to that in the liquid) is far greater than the interaction energy (much less than in the liquid), and molecule do not form clusters. In a liquid, the molecules themselves typically occupy a few percent of the total space available; in a gas, they occupy a few thousandths of a percent. Nevertheless, for our purposes, all fluids are considered to be continua (no voids or holes).The absence of significant intermolecular attraction allows gases to fill whatever volume is available to them, whereas the presence of such attraction in liquids prevents them from doing so. The attractive forces in liquid water are unusually strong, compared to other liquids. Properties of Fluids: m Density is mass/volume: ρ = . The density of liquid water is V 3 o −3 3 ~1.0 kg/m ; that of air at 20 C is ~1.2x10 kg/m . mg W Specific weight is weight/volume: γ = = = ρg V V C:\Adata\CLASNOTE\342\Class Notes\Key concepts_Topic 1.doc 1 γ Specific gravity is density normalized to the density of water: sg..= i γ w V 1 Specific volume is volume/mass: V = = m ρ Bulk modulus or modulus of elasticity is the pressure change per dp dp fractional change in volume or density: E =− = . -
Continuity Equation in Pressure Coordinates
Continuity Equation in Pressure Coordinates Here we will derive the continuity equation from the principle that mass is conserved for a parcel following the fluid motion (i.e., there is no flow across the boundaries of the parcel). This implies that δxδyδp δM = ρ δV = ρ δxδyδz = − g is conserved following the fluid motion: 1 d(δM ) = 0 δM dt 1 d()δM = 0 δM dt g d ⎛ δxδyδp ⎞ ⎜ ⎟ = 0 δxδyδp dt ⎝ g ⎠ 1 ⎛ d(δp) d(δy) d(δx)⎞ ⎜δxδy +δxδp +δyδp ⎟ = 0 δxδyδp ⎝ dt dt dt ⎠ 1 ⎛ dp ⎞ 1 ⎛ dy ⎞ 1 ⎛ dx ⎞ δ ⎜ ⎟ + δ ⎜ ⎟ + δ ⎜ ⎟ = 0 δp ⎝ dt ⎠ δy ⎝ dt ⎠ δx ⎝ dt ⎠ Taking the limit as δx, δy, δp → 0, ∂u ∂v ∂ω Continuity equation + + = 0 in pressure ∂x ∂y ∂p coordinates 1 Determining Vertical Velocities • Typical large-scale vertical motions in the atmosphere are of the order of 0. 01-01m/s0.1 m/s. • Such motions are very difficult, if not impossible, to measure directly. Typical observational errors for wind measurements are ~1 m/s. • Quantitative estimates of vertical velocity must be inferred from quantities that can be directly measured with sufficient accuracy. Vertical Velocity in P-Coordinates The equivalent of the vertical velocity in p-coordinates is: dp ∂p r ∂p ω = = +V ⋅∇p + w dt ∂t ∂z Based on a scaling of the three terms on the r.h.s., the last term is at least an order of magnitude larger than the other two. Making the hydrostatic approximation yields ∂p ω ≈ w = −ρgw ∂z Typical large-scale values: for w, 0.01 m/s = 1 cm/s for ω, 0.1 Pa/s = 1 μbar/s 2 The Kinematic Method By integrating the continuity equation in (x,y,p) coordinates, ω can be obtained from the mean divergence in a layer: ⎛ ∂u ∂v ⎞ ∂ω ⎜ + ⎟ + = 0 continuity equation in (x,y,p) coordinates ⎝ ∂x ∂y ⎠ p ∂p p2 p2 ⎛ ∂u ∂v ⎞ ∂ω = − ⎜ + ⎟ ∂p rearrange and integrate over the layer ∫p ∫ ⎜ ⎟ 1 ∂x ∂y p1⎝ ⎠ p ⎛ ∂u ∂v ⎞ ω(p )−ω(p ) = (p − p )⎜ + ⎟ overbar denotes pressure- 2 1 1 2 ⎜ ⎟ weighted vertical average ⎝ ∂x ∂y ⎠ p To determine vertical motion at a pressure level p2, assume that p1 = surface pressure and there is no vertical motion at the surface. -
Heat and Energy Conservation
1 Lecture notes in Fluid Dynamics (1.63J/2.01J) by Chiang C. Mei, MIT, Spring, 2007 CHAPTER 4. THERMAL EFFECTS IN FLUIDS 4-1-2energy.tex 4.1 Heat and energy conservation Recall the basic equations for a compressible fluid. Mass conservation requires that : ρt + ∇ · ρ~q = 0 (4.1.1) Momentum conservation requires that : = ρ (~qt + ~q∇ · ~q)= −∇p + ∇· τ +ρf~ (4.1.2) = where the viscous stress tensor τ has the components = ∂qi ∂qi ∂qk τ = τij = µ + + λ δij ij ∂xj ∂xi ! ∂xk There are 5 unknowns ρ, p, qi but only 4 equations. One more equation is needed. 4.1.1 Conservation of total energy Consider both mechanical ad thermal energy. Let e be the internal (thermal) energy per unit mass due to microscopic motion, and q2/2 be the kinetic energy per unit mass due to macroscopic motion. Conservation of energy requires D q2 ρ e + dV rate of incr. of energy in V (t) Dt ZZZV 2 ! = − Q~ · ~ndS rate of heat flux into V ZZS + ρf~ · ~qdV rate of work by body force ZZZV + Σ~ · ~qdS rate of work by surface force ZZX Use the kinematic transport theorm, the left hand side becomes D q2 ρ e + dV ZZZV Dt 2 ! 2 Using Gauss theorem the heat flux term becomes ∂Qi − QinidS = − dV ZZS ZZZV ∂xi The work done by surface stress becomes Σjqj dS = (σjini)qj dS ZZS ZZS ∂(σijqj) = (σijqj)ni dS = dV ZZS ZZZV ∂xi Now all terms are expressed as volume integrals over an arbitrary material volume, the following must be true at every point in space, 2 D q ∂Qi ∂(σijqi) ρ e + = − + ρfiqi + (4.1.3) Dt 2 ! ∂xi ∂xj As an alternative form, we differentiate the kinetic energy and get De -
Chapter 3 Newtonian Fluids
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. -
Lecture 5: Magnetic Mirroring
!"#$%&'()%"*#%*+,-./-*+01.2(.*3+456789* !"#$%&"'()'*+,-".#'*/&&0&/-,' Dr. Peter T. Gallagher Astrophysics Research Group Trinity College Dublin :&2-;-)(*!"<-$2-"(=*%>*/-?"=)(*/%/="#* o Gyrating particle constitutes an electric current loop with a dipole moment: 1/2mv2 µ = " B o The dipole moment is conserved, i.e., is invariant. Called the first adiabatic invariant. ! o µ = constant even if B varies spatially or temporally. If B varies, then vperp varies to keep µ = constant => v|| also changes. o Gives rise to magnetic mirroring. Seen in planetary magnetospheres, magnetic bottles, coronal loops, etc. Bz o Right is geometry of mirror Br from Chen, Page 30. @-?"=)(*/2$$%$2"?* o Consider B-field pointed primarily in z-direction and whose magnitude varies in z- direction. If field is axisymmetric, B! = 0 and d/d! = 0. o This has cylindrical symmetry, so write B = Brrˆ + Bzzˆ o How does this configuration give rise to a force that can trap a charged particle? ! o Can obtain Br from " #B = 0 . In cylindrical polar coordinates: 1 " "Bz (rBr )+ = 0 r "r "z ! " "Bz => (rBr ) = #r "r "z o If " B z / " z is given at r = 0 and does not vary much with r, then r $Bz 1 2&$ Bz ) ! rBr = " #0 r dr % " r ( + $z 2 ' $z *r =0 ! 1 &$ Bz ) Br = " r( + (5.1) 2 ' $z *r =0 ! @-?"=)(*/2$$%$2"?* o Now have Br in terms of BZ, which we can use to find Lorentz force on particle. o The components of Lorentz force are: Fr = q(v" Bz # vzB" ) (1) F" = q(#vrBz + vzBr ) (2) (3) Fz = q(vrB" # v" Br ) (4) o As B! = 0, two terms vanish and terms (1) and (2) give rise to Larmor gyration. -
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. -
Fluid Dynamics and Mantle Convection
Subduction II Fundamentals of Mantle Dynamics Thorsten W Becker University of Southern California Short course at Universita di Roma TRE April 18 – 20, 2011 Rheology Elasticity vs. viscous deformation η = O (1021) Pa s = viscosity µ= O (1011) Pa = shear modulus = rigidity τ = η / µ = O(1010) sec = O(103) years = Maxwell time Elastic deformation In general: σ ε ij = Cijkl kl (in 3-D 81 degrees of freedom, in general 21 independent) For isotropic body this reduces to Hooke’s law: σ λε δ µε ij = kk ij + 2 ij λ µ ε ε ε ε with and Lame’s parameters, kk = 11 + 22 + 33 Taking shear components ( i ≠ j ) gives definition of rigidity: σ µε 12 = 2 12 Adding the normal components ( i=j ) for all i=1,2,3 gives: σ λ µ ε κε kk = (3 + 2 ) kk = 3 kk with κ = λ + 2µ/3 = bulk modulus Linear viscous deformation (1) Total stress field = static + dynamic part: σ = − δ +τ ij p ij ij Analogous to elasticity … General case: τ = ε ij C'ijkl kl Isotropic case: τ = λ ε δ + ηε ij ' kk ij 2 ij Linear viscous deformation (2) Split in isotropic and deviatoric part (latter causes deformation): 1 σ ' = σ − σ δ = σ + pδ ij ij 3 kk ij ij ij 1 ε' = ε − ε δ ij ij 3 kk ij which gives the following stress: σ = − δ + ςε δ + ηε 'ij ( p p) ij kk ij 2 'ij ε = With compressibility term assumed 0 (Stokes condition kk 0 ) 2 ( ς = λ ' + η = bulk viscosity) 3 τ Now τ = 2 η ε or η = ij ij ij ε 2 ij η = In general f (T,d, p, H2O) Non-linear (or non-Newtonian) deformation General stress-strain relation: ε = Aτ n n = 1 : Newtonian n > 1 : non-Newtonian n → ∞ : pseudo-brittle 1 −1 1−n 1 −1/ n (1−n) / n Effective viscosity η = A τ = A ε eff 2 2 Application: different viscosities under oceans with different absolute plate motion, anisotropic viscosities by means of superposition (Schmeling, 1987) Microphysical theory and observations Maximum strength of materials (1) ‘Strength’ is maximum stress that material can resist In principle, viscous fluid has zero strength.