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p0406

# p0406 - a = U 2 L}w 1 x L W i y L j 4.6(b On y = 0 the...

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4.6. CHAPTER 4, PROBLEM 6 375 4.6 Chapter 4, Problem 6 4.6(a): Since the given velocity vector is u = U w 1+ x L W i U y L j the acceleration vector is thus a = u t + u · u = u u x + v u y This is true because the flow is steady ( / t 0 ) and two dimensional ( / z 0 ). So, u x = U L i and u y = U L j Hence, the acceleration is a = U w 1+ x L Ww U L i W U y L w U L j W = U 2 L w 1+ x L W i + U 2 L y L j Upon regrouping terms, the acceleration vector is
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Unformatted text preview: a = U 2 L }w 1 + x L W i + y L j ] 4.6(b): On y = 0 , the acceleration, a = a x i + a y j , is a x = U 2 L w 1 + x L W and a y = 0 So, the ratio of the acceleration at points 2 and 1 is a x 2 a x 1 = 1 + x L = 2 = ⇒ x = L...
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