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# Fenômentos de Transporte

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Chapter 4: l%e Momentum Balances 173 I, v, p (v cos a) dA + 2 Sv v, p dV = I; Fx (4.1-12) ~ v , p ( v c o s a ) d A + $ S V v , p d V = LCF, (4.1-13) (4.1 - 14) In general, the convective terms as well as the sum of the forces may have a non-zero component in any direction, although, if possible, we usually try to choose a coordinate system which minimizes the number of non-zero components. Also notice that the forces exerted by the surroundings on the control volume (the forces in the momentum balance) are equal and opposite in sign to the forces exerted by the control volume on the surroundings. le 4.1-1 . . m e n t m flux of fLrud rn lgmmw -flow UL circular p - The (axially symmetric) velocity profile for fully developed laminar flow in a smooth tube of constant circular cross-section is v = v,, [l -(;)'I (4.1-15) Calculate the momentum flux. Choose the flow direction to be the x-direction. Solution a) The momentum flux across any arbitrary pipe cross-section is Momentum flux = V I , ( v . 4 = ~(v-[l-(~)2](i)}(v~[l-(~)2](i*n)}27crLh 0 (4.1 - 16) Noting that for this choice of coordinates, the outward normal is in the x- direction 174 Chapter 4: The Momentum Balances ( i - n ) = 1 Then (4.1- 17) Momentum flux = 2 ~t vkx i (4.1-18) = 2 n v L i l ( r - ~ + $ ) c i r (4.1-19) = ~ ~ V ~ ~ [ C - L + Y - ] I R 2 2R2 6R4 (4.1-20) x R2 vLx (4.1-21) Momentum flux = i 4.1.1 Types of forces There are two types of forces with which we must contend: body forces surface forces. Body forces arise from the same types of fields that we considered when discussing potential energy - gravitational, electromagnetic, etc. The change in energy from position changes in a potential field can be accomplished by a force acting on a body during its displacement. Such a force - for example, gravity - does not necessarily disappear when the body is at rest. All potential fields do not produce what we call body forces, but all body forces are produced by potential fields. Body forces are proportional to the amount of mass and act throughout the interior of the system or control volume. The second type of force is the surface force. Surface forces are exerted on the system by the surroundings via components of the traction on the surface of the system. Traction (force) acting at a point on the surface is a vector since it has both magnitude and direction, and it can therefore be resolved into two Chapter 4: The Momentum Balances I75 components: one parallel to the surface (shear), and one component normal to the surface. The normal component in many problems is constituted of only the negative of the pressure multiplied by the area on which it acts, although non- Newtonian fluids can also have viscous forces acting in the normal direction. At a point in a fluid at rest, the normal force is the Same in all directions and is called the hydrostatic pressure. Hydrostatic pressure is the same as the pressure used in a thermodynamic equation of state. There is no shear in static classical fluid systems because by definition such a fluid would deform continuously under shear. (This classic definition of a fluid becomes somewhat fuzzy when we later discuss non-Newtonian fluids, such a Bingham plastics, that exhibit characteristics of both fluids and solids.) When a general fluid is flowing, the normal stresses are not always equal. If the fluid is incompressible, however, the pressure can be defined satisfactorily as the mean of the normal stresses in the three coordinate directions. The pressure field in either a static or a moving fluid exhibits a gradient caused by gravity. This gradient can often be neglected except in special circumstances; for example, where buoyancy forces enter the problem. 4.1.2 entire surface of irregular objects Influence of uniform pressure over To obtain the net force resulting from pressure on a control volume, we must integrate over the entire outside surface net force produced by on surface of control volume (4.1.2-1) For irregularly shaped volumes, this can lead to cumbersome integrals. We often can avoid integrating the force produced by pressure over the non-flow surface of irregular volumes by performing the integration in gauge pressure. This artifice removes a constant pressure over the entire surface of the control volume from the integral. Adding or subtracting a constant normal stress over the entire outside surface of a control volume, no matter how irregularly shaped, does not give a net contribution to the external force term. 176 Chapter 4: The Momentum Balances This can be seen by dividing an irregular control volume into elementary prisms as shown in Figure 4.1.2- 1, adding the forces on the left- and right-hand faces, and taking the limit as the face areas of the prisms approach zero to obtain the area integral. Xk xi Figure 4.1.2-1 Approximation of solid by prisms (7" . n) dA + normal stress x area )left facc I 1 rime faa (normal stress x area (4.1.2-2) Consider the typical rectangular prism shown in Figure 4.1.2-2 of area dXj by dxk (the coordinate xk is normal to the page) on its left face, a, and whose right face, b, is sliced at some angle p as indicated. The angle y is a right angle, and the outward normal to b makes an angle a with the horizontal. The normal stress is assumed to arise only from the pressure. Notice that if the magnitude of the normal stress (4.1-2-3) Chapter 4: The Momentum Balances I77 Figure 4.1.2-2 Detail of prism is the same on each face, one obtains no net force in the xi-direction, because the increase in area over which the normal force acts on the right face is always just compensated by the decrease in the component of the force acting in that direction Fxi on left face component of normal stress in area on which it acts] (4.1-24) In this case the nonnal stress is just the pressure, which acts normal to the surface and in a direction opposed to that of the outward normal. Fxi on left face = p dx, dx, (4.1-2-5) Considering the right face and noting that the angle a and the angle p are equal because their sides are perpendicular Fxi on right face 1 = component of normal stress in area on which it acts I L dxj dx, = -pdxjdxk (4.1-26) The same conclusion would be reached if the left-hand face were sliced at some different angle. I78 Chapter 4: The Momentum Balances Therefore, the imposition of a uniform pressure (normal force) over the outer surface of an irregularly shaped object contributes no net force tbe surface of the object. 4.1.3 Averages and the momentum equation If we examine the fifst term in the macroscopic momentum balance in cylindrical coordinates as applied to constant-density steady axial flow in a uniform-diameter pipe between upstream cross-section 1 and down-stream cross- section 2, with the positive z-coordinate in the direction of flow, we see that U = 0, that the magnitude of the z-component of v is the same as the magnitude of v itself, and that the cosine is -1 at cross-section 1 and +1 at cross-section 2, so the term reduces to /Av,pvcosadA = l j A 2 v2 M - j A l v2 = P [ (4) - (41 A We often make the assumption in evaluating such a term that (v\u2019) = (v)2 (4.1.3-1) (4.1.3-2) In the following Sections we examine this assumption for a particular turbulent flow profile and for laminar flow.5 Momentum balance approximation - turbulent flow The velocity distribution for turbulent flow in pipes is fairly well represented by the following empirical profile: v = v,(l-*)+ (4.1.3-3) In using the momentum balance for such flows, we often make the further approximation