Substitution of 332 and analogous results for y and z momentum into Eq 329

Substitution of 332 and analogous results for y and z

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Substitution of (3.32) (and analogous results for y and z momentum) into Eq. (3.29) permits us to write the latter equation as integraldisplay R ( t ) ρ D U Dt dV = integraldisplay R ( t ) F B dV + integraldisplay S ( t ) F S dA . (3.33) There are two main steps needed to complete the derivation. The first is analogous to what was done in the case of the continuity equation; namely, we need to convert the integral equation to a differential equation. The second is treatment of the surface forces. We will first provide a preliminary mathematical characterization of these forces that will lead to the differential equation. Details of the surface forces will be given in a separate section. Preliminaries on Surface Forces It is important to first understand the mathematical structure of the surface forces. This will not only aid in required manipulations of the equation, but it will also provide further insight into how these forces should be represented. We begin by noting that F S must be a vector (because ρ U and F B are vectors), and this suggests that there must be a matrix, say T , such that F S = T · n , where, as usual, n is the outward unit normal vector to the surface S ( t ). We remark that the “dot” notation for the matrix-vector product is used to emphasize that each component of F S is the (vector) dot product of the corresponding row of T with the column vector n . This is actually a completely trivial observation; but as will become more clear as we proceed, it is very important. Furthermore, it will also be important to recognize that the physics represented by the vector F S must somehow be incorporated into the elements of the matrix T since n is purely geometric. Basic Form of Differential Momentum Equation The above expression for F S allows us to write Eq. (3.33) as integraldisplay R ( t ) ρ D U Dt dV = integraldisplay R ( t ) F B dV + integraldisplay S ( t ) T · n dA ,
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3.4. MOMENTUM BALANCE—THE NAVIER–STOKES EQUATIONS 73 and application of Gauss’s theorem to the surface integral, followed by rearrangement, yields integraldisplay R ( t ) ρ D U Dt F B − ∇ · T dV = 0 . (3.34) Now recall that R ( t ) is an arbitrary fluid element that can be chosen to be arbitrarily small; hence, it follows from arguments used in an analogous situation while deriving the continuity equation that ρ D U Dt F B − ∇ · T = 0 . (3.35) This provides a fundamental, and very general, momentum balance that is valid at all points of any fluid flow (within the confines of the continuum hypothesis). 3.4.2 Treatment of surface forces We can now complete the derivation of the Navier–Stokes equations by considering the details of T which must contain the information associated with surface forces. We again recall that T is a matrix, and F S is a 3-D vector. Thus, T must be a 3 × 3 matrix having a total of nine elements. Since, as already noted, these must carry the same information found in the components of the surface force vector F S , we know from earlier discussions of shear stress and pressure that both of these must be represented in the elements of T . That is, these elements must be associated with two types of forces: i
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