convective acceleration must be nonzero and hence the total acceleration given

# Convective acceleration must be nonzero and hence the

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convective acceleration must be nonzero, and hence the total acceleration given by the substantial derivative is also nonzero: Du Dt = uu x + vu y negationslash = 0 , and similarly, Dv Dt = uv x + vv y negationslash = 0 , for this 2-D case. That is, we have demonstrated non-zero acceleration in a steady flow for which the local acceleration is zero.
52 CHAPTER 3. THE EQUATIONS OF FLUID MOTION We now consider a simple example to demonstrate calculation of Lagrangian, or total, acceler- ation for a given velocity field. EXAMPLE 3.1 Let the velocity field U have the components u = x + y + z + t , v = x 2 y 3 zt , w = exp( xyzt ) . Find the components of the Lagrangian acceleration. The first thing to note when starting this calculation is that since the velocity field has three components, we should expect the acceleration to also be a vector with three components. In particular, although we can concisely express Lagrangian acceleration in the form D U Dt = U ∂t + U · ∇ U , it is generally far simpler to work with the individual components. For example, the x -direction acceleration, which we shall denote as a x , is given by a x Du Dt = ∂u ∂t + U · ∇ u , or a x = u t + uu x + vu y + wu z , with analogous expressions holding for the other two components: a y = v t + uv x + vv y + wv z , a z = w t + uw x + vw y + ww z . At this point, all that is required is carry out the indicated partial differentiations and substitute the results into the formulas. We have u t = 1 , u x = 1 , u y = 1 , u z = 1 , v t = x 2 y 3 z , v x = 2 xy 3 zt , v y = 3 x 2 y 2 zt , v z = x 2 y 3 t , and w t = xyz exp( xyzt ) , w x = yzt exp( xyzt ) , w y = xzt exp( xyzt ) , w z = xyt exp( xyzt ) . It then follows from the above equations that a x = 1 + ( x + y + z + t ) · (1) + x 2 y 3 zt · (1) + exp( xyzt ) · (1) , a y = x 2 y 3 z + ( x + y + z + t ) · (2 xy 3 zt ) + ( x 2 y 3 zt ) · (2 xy 3 zt ) + exp( xyzt ) · ( x 2 y 3 t ) , a z = xyz exp( xyzt ) + ( x + y + z + t ) · ( yzt exp( xyzt )) + ( x 2 y 3 zt ) · ( xyt exp( xyzt )) . 3.2 Review of Pertinent Vector Calculus In this section we will briefly review the parts of vector calculus that will be needed for deriving the equations of fluid motion. There are two main theorems from which essentially everything else we will need can be derived: Gauss’s theorem and the general transport theorem. The first of these is usually encountered in elementary physics classes, but we will provide a fairly detailed (but non-rigorous) treatment here. The second is rather obscure and occurs mainly only in fluid dynamics; but it is very important, and it is directly related to an elementary integration formula due to Leibnitz.
3.2. REVIEW OF PERTINENT VECTOR CALCULUS 53 3.2.1 Gauss’s theorem Gauss’s theorem corresponds to a quite simple and rather intuitive idea: the integral of a derivative equals the net value of the function (whose derivative is being integrated) over the boundary of the domain of integration. In one space dimension this is precisely the fundamental theorem of calculus : integraldisplay b a f ( x ) dx = F ( b ) F ( a ) , (3.3) where F ( x ) is a function such that F ( x ) = f ( x ); i.e. , F is the antiderivative or primitive of f .

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