p1184 - GCI = 1 . 25 | 6 h | 2 1 . 78 1 = 0 . 51 | 6 h |...

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

View Full Document Right Arrow Icon
11.84. CHAPTER 11, PROBLEM 84 123 11.84 Chapter 11, Problem 84 Figure 11.41 shows the opening screen for Program POTFLOW DATA . Figure 11.41: Opening screen of Program POTFLOW DATA . Executing Program POTFLOW and examining the program output confirms that for 101, 201 and 401 grid points the velocity at x =0 and y =2 a is u (0 , 2 a ) = 1.236173, 1.231433 and 1.232812, respectively. 11.84(a): Since the number of finite-difference cells is 100, 200 and 400 for the three computations, the grid-refinement ratio is r =2 . So, identifying φ = u (0 , 2 a ) ,wehave φ r 2 h =1 . 236173 , φ rh =1 . 231433 , φ h =1 . 232812 The order of accuracy is p = f n | ( φ rh φ r 2 h ) / ( φ rh φ h ) | f nr = f n | (1 . 231433 1 . 236173) / (1 . 231433 1 . 232812) | f n 2 =1 . 78 11.84(b): Using the fact that the order of accuracy is 1.78, the grid convergence index is
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: GCI = 1 . 25 | 6 h | 2 1 . 78 1 = 0 . 51 | 6 h | 124 CHAPTER 11. POTENTIAL FLOW For the 401-point grid, the GCI expressed in percent is thus GCI = 51 e e e e e h rh h e e e e e = 0 . 06% 11.84(c): Based on Richardson extrapolation, the error for the 401-point grid is E h = 1 3 ( h rh ) = 1 3 (1 . 232812 1 . 231433) = 0 . 000460 Hence, as a percentage, there follows 100 E h h = 0 . 04% The error estimates are similar because the numerical solution is nearly second-order accurate at this point in the flowfield....
View Full Document

Page1 / 2

p1184 - GCI = 1 . 25 | 6 h | 2 1 . 78 1 = 0 . 51 | 6 h |...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online