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Unformatted text preview: U o ) (1 x/x o ) 3 ( 1 /x o ) = 2 U 2 o x o (1 x/x o ) 5 Therefore, the acceleration vector for this flow is a = 2 U 2 o x o (1 x/x o ) 5 i (b) First, note that (1 x/x o ) 1 = u U o Thus, the acceleration can be rewritten as a = 2 U 2 o x o w u U o W 5 / 2 So, we can solve for the velocity ratio as follows. w u 2 u 1 W 5 / 2 = a 2 a 1 = u 2 u 1 = w a 2 a 1 W 2 / 5 Thus, if a 2 /a 1 = 1 / 32 , the velocity ratio is u 2 u 1 = w 1 32 W 2 / 5 = 1 4...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.
 Spring '06
 Phares

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