p0660 - 6.60. CHAPTER 6, PROBLEM 60 621 6.60 Chapter 6,...

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6.60. CHAPTER 6, PROBLEM 60 621 6.60 Chapter 6, Problem 60 Use a stationary control volume whose boundary lies outside the pipe as shown in Fig- ure 6.60. U p a + p V,p a 1 2 U, p a 4 5 A 1 2 A Area = A y x F = F o ( i + j ) ............................................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...... ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ . . . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . Figure 6.60: Strongly-rotational pipe flow. Mass Conservation: In general, the mass-conservation principle tells us that d dt 888 V ρ dV + 8 s 8 S ρ u rel · n dS =0 The flow is incompressible and the control volume’s size is constant. Thus, the instanta- neous rate of change of the mass in the control volume vanishes. Then, since u rel = u , we have 8 s 8 S ρ u · n dS Expanding the closed-surface integral yields ρ ( UA ) ± 1 Inlet + ρ w 1 2 U 4 5 A W ± 1 Lower outlet + ρ w V 1 2 A W ± 1 Upper outlet = ⇒− 3 5 U + 1 2 V Therefore, the velocity V is V = 6 5 U x -Momentum Conservation: For this control volume, the force F is a reaction force. Thus, 8 s 8 S ρ u ( u · n ) dS = i · 8 s 8 S ( p p a ) n dS F o
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622 CHAPTER 6. CONTROL-VOLUME METHOD First, the net flux of x
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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.

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p0660 - 6.60. CHAPTER 6, PROBLEM 60 621 6.60 Chapter 6,...

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