convective acceleration must be nonzero and hence the total acceleration given

Convective acceleration must be nonzero and hence the

This preview shows page 57 - 60 out of 164 pages.

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.
Image of page 57
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.
Image of page 58
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 .
Image of page 59
Image of page 60

You've reached the end of your free preview.

Want to read all 164 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors