{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

View Full Document Right Arrow Icon
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
Background image of page 1

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

View Full Document Right Arrow Icon
622 CHAPTER 6. CONTROL-VOLUME METHOD First, the net flux of x
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}