p0722

# p0722 - 782 CHAPTER 7 CONSERVATION OF ENERGY 7.22 Chapter 7...

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782 CHAPTER 7. CONSERVATION OF ENERGY 7.22 Chapter 7, Problem 22 g u 1 Turbine u 2 d 2 d z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ . . . . . . . . . . . . . . . . . . . Figure 7.10: Flow through a turbine. 7.22(a): To analyze this flow we select a stationary control volume that is external to the turbine. 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 Because the flow is steady and incompressible, the instantaneous rate of change of mass vanishes. Also, u rel = u . Hence, the mass-conservation equation simplifies to 8 s 8 S ρ u · n dS = 0 Expanding the closed-surface integral, we have ρ } u 1 π 4 d 2 ] ± 1 Section 1 + ρ } u 2 π 4 (2 d ) 2 ] ± 1 Section 2 = 0 = π 4 ρ u 1 d 2 = πρ u 2 d 2 = ˙ m where u 1 and u 2 denote the magnitude of the velocity at Sections 1 and 2, respectively. Thus, u 1 = 4 ˙ m πρ d 2 and u 2 = 1 4 u 1 = ˙ m πρ d 2 Energy Conservation: To complete the solution, we appeal to energy conservation, i.e., 8 s 8 S ρ } h + 1 2 u · u + gz ] ( u · n ) dS = ˙ Q ˙ W s Hence, since we are given ˙ Q = 0 , there follows (with z = 0 at Section 2) ρ w h 1 + 1 2 u 2 1 + g z W } u 1 π 4 d 2 ] ± 1 Section 1 + ρ w h 2 + 1 2 u 2 2 W } u 2 π 4 (2 d ) 2 ] ± 1 Section 2 = ˙ W s

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