This change of v with respect to x corresponds to an angular deformation rate

# This change of v with respect to x corresponds to an

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in a generally counter-clockwise direction. This change of v with respect to x corresponds to an angular deformation rate ∂v/∂x . (Think of v being momentum per unit mass, so its time rate of change—due to an adjacent fluid element—creates a force which then produces angular deformation of the fluid element.) Similarly, in the right-hand figure we depict the effects of changes in u with respect to y , and hence the angular deformation rate ∂u/∂y (caused by a fluid element below, and
76 CHAPTER 3. THE EQUATIONS OF FLUID MOTION x y v u x face Figure 3.10: Sources of angular deformation of face of fluid element. perpendicular to, the one shown), also producing a distortion generally in the counter-clockwise direction. We note that it is not necessary that ∂v/∂x and ∂u/∂y both produce counter-clockwise distortions (or clockwise ones), but only that these occur in the xy plane. It is clear from Fig. 3.9 that this is the plane on which τ xy acts, and the preceding physical argument shows that both ∂v/∂x and ∂u/∂y contribute to this component of shear stress as we have previously argued on purely mathematical grounds. We leave as an exercise for the reader construction of a physical argument, analogous to that given here, showing that ∂v/∂x and ∂u/∂y are the contributions to τ yx on the y face of the fluid element shown in Fig. 3.9, thus suggesting that τ yx = τ xy and providing analogous arguments justifying the forms of the other components of shear stress appearing in Eq. (3.37). Normal Viscous Stresses and Pressure We remark here that, as depicted in Fig. 3.9, there are actually two separate contributions to the normal force: one involves pressure, as would be expected; but there is a second type of normal force of viscous origin. We can see evidence of this when pouring very viscous fluids such as molasses or cold pancake syrup. Although they flow, they do so very slowly because the normal viscous forces are able to support considerable tensile stresses arising from gravitational force acting on the falling liquid. (We note that surface tension, discussed in Chap. 2, is also a factor here, but not the only one.) This can easily be recognized if we consider dropping a solid object having the same density as molasses, e.g. , at the same time we begin to pour the latter. We would see that the solid object would fall to the floor much more quickly, retarded only by drag from the surrounding air (also of viscous origin, but not nearly as effective in slowing motion as are the internal viscous normal stresses in this case). Finally, we observe that viscous normal stresses also occur in flow of gases, but they must be compressive because gases cannot support tensile stresses. As indicated in Fig. 3.9, the normal stresses are different from pressure (which is always com- pressive, as the figure indicates), and they are denoted τ xx , τ yy and τ zz . We should expect, by analogy with the shear stresses, that, e.g. , τ xx acts in the x direction, and on the x face of the fluid
3.4. MOMENTUM BALANCE—THE NAVIER–STOKES EQUATIONS 77 element. This combination requires that (again, by analogy with the form of shear stress) τ xx = μ

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