In particular it is seen to separate recirculate and then reattach at a

# In particular it is seen to separate recirculate and

• 164

This preview shows page 124 - 127 out of 164 pages.

In particular, it is seen to separate, recirculate and then reattach at a downstream location. This recirculation region is actually quite prominant in some flows, and as represented here would form a donut-shaped region around the inside of the expansion section of the pipe adjacent to the point of expansion. It is important to recognize that these are caused mainly by viscous effects, so our present analysis will not be able to account for all details.
4.3. CONTROL-VOLUME MOMENTUM EQUATION 119 D 3 n x r n 3 2 1 n D n n n n 1 n Figure 4.8: Flow through a rapidly-expanding pipe. Figure 4.8 also displays the control volume we will employ in the present analysis. This extends from location 1 upstream of the point of expansion, denoted as location 2, on downstream to location 3 and is fixed. In carrying out this calculation we will assume the flow is uniform at each cross section; i.e. , we take the pressure and velocity to be constant across each cross section. Then the flow field can vary only in the x direction, and this leads to only one nonzero velocity component, namely, the one in the x direction. We will denote this as u in what follows. The goal of this analysis is to predict the change in pressure of the flow resulting from rapid expansion of the pipe, assuming the upstream flow speed is known. From the continuity equation we have u 1 A 1 + u 3 A 3 = u 1 πD 2 1 4 + u 3 πD 2 3 4 = 0 , or u 1 πD 2 1 4 = u 3 πD 2 3 4 u 3 = u 1 parenleftbigg D 1 D 3 parenrightbigg 2 . Observe that the minus sign appearing in the first term of the first of these equations occurs because the flow direction at location 1 is positive, but the outward unit normal vector to the control volume at this location is in the negative x direction. The integral momentum equation can now be written as integraldisplay S e ρu 2 n x dA = integraldisplay S p n dA . where n x denotes the x -direction component of the general control surface outward unit normal vector. Observe that this is the only nonzero component of this vector acting at the entrance and exit of the control volume. Also note that the control surface on which the pressure acts is the surface of the entire control volume—not simply that of entrances and exits. This is one of the major differences between working with the control-volume momentum equation and the corresponding continuity equation.
120 CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS We first evaluate the left-hand side integral, which contains contributions only from entrances and exits, to obtain integraldisplay S e ρu 2 n x dA = ρu 2 1 πD 2 1 4 + ρu 2 3 πD 2 3 4 . Similarly, evaluation of the right-hand side integral yields integraldisplay S p n dA = p 1 πD 2 1 4 + p 2 π 4 ( D 2 3 D 2 1 ) p 3 πD 2 3 4 . It should be noted that although pressure is acting on the lateral boundaries of the pipe, there is no contribution from this because such contributions cancel on opposite sides of the pipe across the diameter, i.e. , the outward normal directions are opposite, and since pressure is assumed constant in each cross section cancellation occurs.

#### You've reached the end of your free preview.

Want to read all 164 pages?

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern