p0410

# p0410 - 4.10 CHAPTER 4 PROBLEM 10 379 4.10 Chapter 4...

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4.10. CHAPTER 4, PROBLEM 10 379 4.10 Chapter 4, Problem 10 U y x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................... ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................... . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... Figure 4.2: Stagnation-point flow. 4.10(a): In general, we know that the acceleration is a = d u dt ±,²1 Total = u t ±,²1 Unsteady + u · u ± 1 Convective First, compute the unsteady acceleration, i.e., u t = 2 R dU dt ( x i y j ) Next, evaluate the differential operator u · .W eh av e u · = 2 U R X x x y y ~ Thus, the convective acceleration is u · u = 2 U R X x x y y ~ 2 U R ( x i y j ) = 4 U 2 R 2 ( x i + y j ) Therefore the total acceleration is a = d u dt = 2 R X dU dt +2 U 2 R ~ x i 2 R X dU

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p0410 - 4.10 CHAPTER 4 PROBLEM 10 379 4.10 Chapter 4...

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