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# Fenômentos de Transporte

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of change per unit volume of kinetic energy as one follows the fluid motion. The terms on the right-hand side represent, respectively, rate of work done per unit volume by pressure forces 198 Chapter 4: The Momentum Balances rate of reversible conversion to internal energy per unit volume rate of work done per unit volume by viscous forces rate of irreversible conversion to internal energy per unit volume rate of work done per unit volume by gravitational forces To illustrate a common application of Equation (4.2-7), for a Newtonian fluid, where x is the position vector dv = = - 5 i Substituting this in Equation (4.2-7) and using the definitions yields a differential equation called the Navier-S tokes equation. (4.2- 1 1) (4.2- 12) (4.2- 13) I p g = - v p + p v 2 v + p g I (4.2-14) The investigation of solutions of this equation for various system where momentum is transferred by viscous motion is a large part of the topic of momentum transport as a transport phenomenon. For example, the axial flow of an incompressible Newtonian fluid in a horizontal pipe is described in cylindrical coordinates by (4.2- 15) Chapter 6 is devoted entirely to the topic of momentum transfer in fluid flow and utilizes the momentum balance in both its integral and differential forms. At this point it is useful to summarize both the macroscopic and microscopic equations of mass, energy, and momentum. Chapter 4: The Momentum Balances 199 4.3 Summary of Balance Equations and Constitutive Relations hips The balance equations in Table 4.3-1 are not applicable to systems where there is interconversion of energy and mass. Table 4.3-1 Tabulation of balance equations MOMENTUM ENERGY - TOTAL, ENERGY6 - THERMAL ENERGY - MECHANICAL MASS - TOTAL MASS - SPECIES - v p + p v 2 v + p g +$ l ;vpdV = ZF I, (A + 6 + R) p (v . n) dA - v . [P v] + P (v . g) = Q - W Do = - ( v . q ) - p ( v . v ) - ( 5 : v v 1 Dt - (v. p v) - p (- v . v) + P (v . 8) = - 1 w - w - (v . [r. v]) - (- [5 : vv]) v . ( p v ) + $ = 0 I = k r i d V For one-dimensional flow of an incompressible fluid. 200 Chapter 4: The Momentum Balances We summarize the common constitutive relationships - those equations that relate material properties to the fluxes - in Table 4.3-2. Table 4.3-2 Tabulation of common constitutive relationships 4.4 The Momentum Equation in Non-Inertial Reference Frames The momentum equation as we have written it applies to an inertial reference frame system. By an inertial frame we mean a frame which is not accelerated.s (We exclude rotation entirely and permit only uniform translation.) These frames are also referred to as Gallilean frames or Newtonian frames. The acceleration of frames attached to the earth's surface, however, is slight enough that our equations give good answers for most engineering problems - the error is of the order of one-third of 1 percent. For problems involving relatively large acceleration of the reference frame, however, Newtonian mechanics must be reformulated. An example of such a problem is the use of a reference frame attached to a rocket as it accelerates from zero velocity on the pad to some orbital velocity. For such a problem, observations from an inertial frame and from the accelerating frame can differ greatly. For example, if an astronaut in zero gravity but in an accelerating rocket drops a toothbrush, he or she sees it apparently accelerate toward the "floor" (wherever the surface opposite to the direction of the acceleration vector of the spacecraft might be at the time). This observation is with respect to an accelerating frame (the spacecraft) within which the astronaut is fixed. To an observer in an inertial frame, however, the toothbrush simply continues at the One-dimensional. Fanger, C. G. (1970). Engineering Mechanics: Statics and Dynamics. Columhus, OH, Charles E. Merrill Books, Inc. Chapter 4: The Momentum Balances 201 constant velocity it had when released, and the spacecraft accelerates past the toothbrush. In other words, observed acceleration depends on the motion of the reference frame in which it is observed. If this is true, Newton's second law must change in form when written with respect to an accelerating reference frame, because the force experienced by a body should not depend on the motion of the observer. Here we will consider only a simple example of this - that of a system of constant mass.g Return to the astronaut's toothbrush. There is no net external force on the toothbrush (we neglect any drag from the atmosphere within the spacecraft). Newton's law becomes = = atoothbrush, observed from inutial frame atoothbrush. observed from iautial frame (4.4- 1) which implies that the acceleration of the toothbrush with respect to an inertial coordinate fiame is zero. With respect to an accelerating frame (one attached to the spacecraft), however, there is an apparent acceleration of the toothbrush. This apparent acceleration is equal to the negative of the difference in acceleration between that of a frame attached to the spacecraft and that of the inertial frame. Since the acceleration of the inertial frame is zero, the relative acceleration is Newton's law will still be operationally valid in the accelerating coordinate frame if we interpret the acceleration on the right-hand side as the acceleration of the object as observed from the accelerating frame and also add an appropriate force (the inertial body force) to the left-hand side. The inertial body force is the negative of the product of the mass and the acceleration of the accelerating frame (the spacecraft)lo We neglect the drag force from the atmosphere in this discussion. l0 Hansen, A. G. (1967). Fluid Mechanics. New York, NY, Wiley. 202 Chapter 4: The Momentum Balances Another example is a camera sitting on the ledge above the rear seat of your car. If you hit a cow which happens to be crossing the road, and therefore decelerate rapidly from a velocity of 60 mph, the camera flies forward with respect to the car as well as to the ground. The sum of forces on the camera is zero (again neglecting friction from the air). This time the coordinate frame is decelerating, not accelerating. To the hitchhiker standing at the side of the road, the acceleration of the camera is zero - it continues at 60 mph until the intervention of a force (perhaps from the windshield - hopefully not the back of your head). In your accelerating (in this case, decelerating) reference frame, there is still no external force on the object, but there is an inertial body force of (- m aautomobilc), where aautomobilc is the relative acceleration between car and ground (inertial frame). This is balanced on the right-hand side of the equation by the same mass times the relative acceleration of the camera to the car, which is the same in magnitude as and opposite in direction to the relative acceleration of the car and the ground. (4,4-6) In the case of rotating frames things become more complex, and the reader is referred to texts such as that by Fanger," which cover this subject in more detail. I 1 Fanger, C. G. (1970). Engineering Mechanics: Statics and Dynamics. Columbus, OH, Charles E. Merrill Books, Inc., p. 562. Chapter 4: The Momentum Balances 203 Chapter 4 Problems 4.1 A fire hose has an inside diameter of 1 1/2 in. and a nozzle diameter of 0.5 in. The inlet pressure to the hose is 40 psig at a water flow of 8 ft3/min. If a flat velocity profile is assumed, what is the value of and the direction of the force exerted by tbe fmman holding the hose? 4.2 Your company is planning to manufacture