Linear Momentum and Force

I NTERNATIONAL J OURNAL OF E NERGY AND E NVIRONMENT Volume 4, Issue 3, 2013 pp.357-368 Journal homepage: www.IJEE.IEEFoundation.org ISSN 2076-2895 (Print), ISSN 2076-2909 (Online) ©2013 International Energy & Environment Foundation. All rights reserved. CFD model of air movement in ventilated façade: comparison between natural and forced air flow Miguel Mora Pérez, Gonzalo López Patiño, P. Amparo López Jiménez Hydraulic and Environmental Engineering Department, Universitat Politècnica de Valencia, Spain. Abstract This study describes computational fluid dynamics (CFD) modeling of ventilated façade. Ventilated façades are normal façade but it has an extra channel between the concrete wall and the (double skin) façade. Several studies found in the literature are carried out with CFD simulations about the behavior of the thermodynamic phenomena of the double skin façades systems. These studies conclude that the presence of the air gap in the ventilated façade affects the temperature in the building skin, causing a cooling effect, at least in low-rise buildings. One of the most important factors affecting the thermal effects of ventilated façades is the wind velocity. In this contribution, a CFD analysis applied on two different velocity assumptions for air movement in the air gap of a ventilated façade is presented. A comparison is proposed considering natural wind induced velocity with forced fan induced velocity in the gap. Finally, comparing temperatures in the building skin, the differences between both solutions are described determining that, related to the considered boundary conditions, there is a maximum height in which the thermal effect of the induced flow is significantly observed. Copyright © 2013 International Energy and Environment Foundation - All rights reserved. Keywords: Ventilated Façade; Natural ventilation; Computational Fluid Dynamics (CFD); Architectural design; Wind energy. 1. Introduction Nowadays new strategies in buildings are investigated by architects and engineers to improve the buildings energy performance. Designers commitment to green buildings should involve both, new sustainable buildings design and rehabilitation in the existing ones by installing new systems to make day to day operations more energy efficient and environmentally sensitive. The envelope of a building is the main element responsible for its energy demand. The building skin ought to be a very susceptible part to be modified to improve the whole building energy performance. In this sense, the use of ventilated façades can often have a positive contribution to this objective. The implementation of ventilated façades in buildings has been an object of broad applications especially in recent years. Ventilated façades are a powerful tool when applied to building design, especially in bioclimatic building design. In some countries with high levels of solar radiation, summer over-heating is a big problem in building energy balances. A ventilated façade is a double envelope composed of two skins and a ventilated cavity air gap located between them. Ventilated façade and wall coverings were developed to protect buildings against the combined action of rain and wind by counterbalancing the effects of water beating on walls and keeping the building dry, with high-level aesthetic characteristics and good heat insulation and Linear Momentum and Force Linear Momentum and Force Bởi: OpenStaxCollege Linear Momentum The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object Linear momentum is defined as the product of a system’s mass multiplied by its velocity In symbols, linear momentum is expressed as p = mv Momentum is directly proportional to the object’s mass and also its velocity Thus the greater an object’s mass or the greater its velocity, the greater its momentum Momentum p is a vector having the same direction as the velocity v The SI unit for momentum is kg·m/s Linear Momentum Linear momentum is defined as the product of a system’s mass multiplied by its velocity: p = mv Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110-kg football player running at 8.00 m/s (b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s Strategy No information is given regarding direction, and so we can calculate only the magnitude of the momentum, p (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this 1/6 Linear Momentum and Force example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes p = mv when only magnitudes are considered Solution for (a) To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation pplayer = (110 kg)(8.00 m/s) = 880 kg · m/s Solution for (b) To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation pball = (0.410 kg)(25.0 m/s) = 10.3 kg · m/s The ratio of the player’s momentum to that of the ball is pplayer pball = 880 10.3 = 85.9 Discussion Although the ball has greater velocity, the player has a much greater mass Thus the momentum of the player is much greater than the momentum of the football, as you might guess As a result, the player’s motion is only slightly affected if he catches the ball We shall quantify what happens in such collisions in terms of momentum in later sections Momentum and Newton’s Second Law The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes Using symbols, this law is Fnet = Δp Δt , 2/6 Linear Momentum and Force where Fnet is the net external force, Δp is the change in momentum, and Δt is the change in time Newton’s Second Law of Motion in Terms of Momentum The net external force equals the change in momentum of a system divided by the time over which it changes Δp Fnet = Δt Making Connections: Force and Momentum Force and momentum are intimately related Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics This statement of Newton’s second law of motion includes the more familiar Fnet=ma as a special case We can derive this form as follows First, note that the change in momentum Δp is given by Δp = Δ(mv) If the mass of the system is constant, then Δ(mv) = mΔv So that for constant mass, Newton’s second law of motion becomes Fnet = Δp Δt Because = Δv Δt mΔv Δt = a, we get the familiar equation Fnet=ma when the mass of the system is constant Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass We will consider systems with varying mass in some detail; however, the relationship between momentum and force remains useful when mass is constant, such as in the following example Calculating Force: Venus Williams’ Racquet 3/6 Linear Momentum and Force During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h) What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? Strategy This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact Newton’s second law stated in terms of momentum is then written as Fnet = Δp Δt As noted ... Chapter 9 Center of Mass and Linear Momentum In this chapter we will introduce the following new concepts: -Center of mass (com) for a system of particles -The velocity and acceleration of the center of mass -Linear momentum for a single particle and a system of particles We will derive the equation of motion for the center of mass, and discuss the principle of conservation of linear momentum Finally we will use the conservation of linear momentum to study collisions in one and two dimensions and derive the equation of motion for rockets (9-1) 1 2 1 1 1 2 2 1 2 2 Consider a system of two particles of masses and at positions and , respectively. We define the position of the center of mass (com) as follow The Center s: of Mass: com m x m x x m m m m x x + = + 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 3 . We can generalize the above definition for a system of particles as follows: Here is the total mass of all the pa . 1 rt . ic n n n n n com i i i n m x m x m x m x m x m x m x m x x m x m m m m n M M M = + + + + + + + + = = = + + + + ∑ 1 2 3 les . We can further generalize the definition for the center of mass of a system of particles in three dimensional space. We assume that the the -th particle ( mass ) has posi n i M m m m m i m = + + + + 1 tion vector 1 n com i i i i r r m r M = = ∑ r r r (9-2) 1 1 The position vector for the center of mass is given by the equation: ˆ ˆ ˆ The position vector can be written as: The components of are given by the n com i i i com com com com com r m r M r x i y j z k r = = = + + ∑ r r r r 1 1 1 1 1 1 equations: n n n com i i com i i com i i i i i x m x y m y z m z M M M = = = = = = ∑ ∑ ∑ The center of mass has been defined using the quations given above so that it has the following prope The center of mass of a system of particles moves as though all the system's mass were conc rty: etra The abolve statement will be proved lat ted there, and that the vector sum of a er. An example is given in the figure. ll t A b he external forces were aseball bat is flipped applied th into the a ere ir and moves under the influence of the gravitation force. The center of mass is indicated by the black dot. It follows a parabolic path as discussed in Chapter 4 (projectile motion) All the other points of the bat follow more complicated paths (9-3) Solid bodies can be considered as systems with continuous distribution of matter The sums that are used for the calculation of the c The Center of Mass for Solid Bod enter of mass of systems with d ies iscrete distribution of mass become integrals: The integrals above are rather complicated. 1 1 1 A simpler special case is that of uniform objects in wh ic com com com x xdm y ydm z zdm M M M = = = ∫ ∫ ∫ h the mass density is constant and equal to In objects with symetry elements (symmetry point, symmetry line, symmetry 1 1 1 plane) it is not com com com x xdV y ydV z zdV V V dm M dV V V ρ = = = = ∫ ∫ ∫ necessary to eveluate the integrals. The center of mass lies on the symmetry element. For example the com of a uniform sphere coincides with the sphere center In a uniform rectanglular object the com lies at the intersection of the diagonals (9-4) . C C O m 1 m 3 m 2 F 1 F 2 F 2 x Lathi, B.P. “Ordinary Linear Differential and Difference Equations” Digital Signal Processing Handbook Ed. Vijay K. Madisetti and Douglas B. Williams Boca Raton: CRC Press LLC, 1999 c  1999byCRCPressLLC 2 Ordinary Linear Differential and Difference Equations B.P. Lathi California State University, Sacramento 2.1 Differential Equations Classical Solution • MethodofConvolution 2.2 Difference Equations Initial Conditions andIterativeSolution • Classical Solution • MethodofConvolution References 2.1 Differential Equations Afunctioncontainingvariablesandtheirderivativesiscalledadifferentialex pression,andanequation involvingdifferentialexpressionsiscalledadifferentialequation. Adifferentialequationisanordinary differential equation if it contains only one independent variable; it is a partial differential equation if it contains more than one independentvariable. Weshall deal here only withordinary differential equations. In the mathematical texts, the independent variable is generally x, which can be anything such as time, distance, velocity, pressure, and so on. In most of the applications in control systems, the independent variable is time. For this reason we shall use here independent variable t for time, although it canstand for any other variable as well. The following equation  d 2 y dt 2  4 + 3 dy dt + 5y 2 (t) = sint is an ordinary differential equation of second order because the highest derivative is of the second order. An nth-order differential equation is linear ifit is of the form a n (t) d n y dt n + a n−1 (t) d n−1 y dt n−1 +···+a 1 (t) dy dt + a 0 (t)y(t) = r(t) (2.1) where the coefficients a i (t) are not functions of y(t). If these coefficients (a i ) are constants, the equation is linear with constant coefficients. Many engineering (as well as nonengineering) systems can be modeled by these equations. Systems modeled by these equations are known as linear time- invariant (LTI) systems. In this chapter we shall deal exclusively with linear differential equations with constant coefficients. Certain other forms of differential equations are dealt with elsewhere in this volume. c  1999 by CRC Press LLC Role of Auxiliary Conditions in Solution of Differential Equations We now show that a differential equation does not, in general, have a unique solution unless some additional constraints (or conditions) on the solution are known. This fact should not come as a surprise. A function y(t) has a unique derivative dy/dt, but for a given derivative dy/dt there are infinite possible functions y(t).Ifwearegivendy/dt , it is impossible to determine y(t) uniquely unless an additional piece of information about y(t) is given. For example, the solution of a differential equation dy dt = 2 (2.2) obtained by integrating both sides of the equation is y(t) = 2t + c (2.3) for any value of c. Equation 2.2 specifies a function whose slope is 2 for all t. Any straight line with a slope of 2 satisfies this equation. Clearly the solution is not unique, but if we place an additional constraint on the solution y(t), then we specify a unique solution. For example, suppose we require that y(0) = 5; then out of all the possible solutions available, only one function has a slope of 2 and an intercept with the vertical axis at 5. By setting t = 0 in Equation 2.3 and substituting y(0) = 5 in the same equation, we obtain y(0) = 5 = c and y(t) = 2t + 5 which is the unique solution satisfying both Equation 2.2 and the constraint y(0) = 5. Inconclusion, differentiation isanirreversibleoperationduringwhichcertain informationislost. Toreversethisoperation,onepieceofinformationabouty(t)mustbeprovidedtorestoretheoriginal y(t). Usingasimilarargument,wecanshowthat,givend 2 y/dt 2 ,wecandeterminey(t)uniquelyonly if two additional Telemark University College Department of Electrical Engineering, Information Technology and Cybernetics Faculty of Technology, Postboks 203, Kjølnes ring 56, N-3901 Porsgrunn, Norway. Tel: +47 35 57 50 00 Fax: +47 35 57 54 01 Introduction to Visual Studio and C# HANS-PETTER HALVORSEN, 2012.08.17 2 Table of Contents 1 Introduction 5 1.1 Visual Studio 5 1.2 C# . 6 1.3 .NET Framework 6 1.4 Object-Oriented Programming (OOP) . 7 2 Visual Studio . 8 2.1 Introduction . Introduction to Linear Momentum and Collisions Introduction to Linear Momentum and Collisions Bởi: OpenStaxCollege Each rugby player has great momentum, which will affect the outcome of their collisions with each other and the ground (credit: ozzzie, Flickr) We use the term momentum in various ways in everyday language, and most of these ways are consistent with its precise scientific definition We speak of sports teams or politicians gaining and maintaining the momentum to win We also recognize that momentum has something to with collisions For example, looking at the rugby players in the photograph colliding and falling to the ground, we expect their momenta to have great effects in the resulting collisions Generally, momentum implies 1/2 Introduction to Linear Momentum and Collisions a tendency to continue on course—to move in the same direction—and is associated with great mass and speed Momentum, like energy, is important because it is conserved Only a few physical quantities are conserved in nature, and studying them yields fundamental insight into how nature works, as we shall see in our study of momentum 2/2 Embedded Systems Design: An Introduction to Processes, Tools, and Techniques by Arnold S. Berger ISBN: 1578200733 CMP Books © 2002 (237 pages) An easy-to-understand guidebook for those embarking upon an embedded processor development project. Table of Contents Embedded Systems Design—An Introduction to Processes, Tools, and Techniques Preface Introduction Chapter 1 - The Embedded Design Life Cycle Chapter 2 - The Selection Process Chapter 3 - The Partitioning Decision Chapter 4 - The Development Environment Chapter 5 - Special Software Techniques Chapter 6 - A Basic Toolset Chapter 7 - BDM, JTAG, and Nexus Chapter 8 - The ICE — An Integrated Solution Chapter 9 - Testing Chapter 10 - The Future Index List of Figures List of Tables List of Listings List of Sidebars TEAMFLY Team-Fly ® Embedded Systems Design—An Introduction to Processes, Tools, and Techniques Arnold Berger CMP Books CMP Media LLC 1601 West 23rd Street, Suite 200 Lawrence, Kansas 66046 USA www.cmpbooks.com Designations used by companies to distinguish their products are often claimed as trademarks. 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Copyright © 2002 by CMP Books, except where noted otherwise. ... it changes Δp Fnet = Δt Making Connections: Force and Momentum Force and momentum are intimately related Force acting over time can change momentum, and Newton’s second law of motion, can be stated... relationship between momentum and force remains useful when mass is constant, such as in the following example Calculating Force: Venus Williams’ Racquet 3/6 Linear Momentum and Force During the... terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes Using symbols, this law is Fnet = Δp Δt , 2/6 Linear Momentum and Force