p0440

# p0440 - 528 CHAPTER 4 KINEMATICS 4.40 Chapter 4 Problem 40...

This preview shows pages 1–2. Sign up to view the full content.

528 CHAPTER 4. KINEMATICS 4.40 Chapter 4, Problem 40 Problem: The velocity for a two-dimensional flow is u = Uxy 2 /H 3 i Ux 2 y/H 3 j , where U and H are constants of dimensions L/T and L , respectively. Compute the circulation, Γ = ± C u · d s , on the rectangular contour shown. Verify that your result is consistent with Equation (4.33). 0 Lx 0 H y Contour C d s = i dx d s = i dx d s = j dy d s = j dy ............................................................................................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............... . . . . . . . . . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . . . . . . ...... ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution: In general, to evaluate a line integral, we treat the integral as the sum of conventional integrals on each segment of the contour, with the sign determined by the differential distance vector, d s . Hence, referring to the figure, we see that Γ = 8 L 0 u ( x, 0) · ( i dx )+ 8 H 0 u ( L,y ) · ( j dy ) + 8 L 0 u ( x,H ) · ( i dx 8 H 0 u (0 ,y ) · ( j dy ) = 8 L 0 [ u ( x, 0) u ( )] dx + 8 H 0 [ v ( ) v (0 )] dy Thus, for the given velocity vector,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

### Page1 / 2

p0440 - 528 CHAPTER 4 KINEMATICS 4.40 Chapter 4 Problem 40...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online