p0714

# p0714 - dS = ˙ Q − ˙ W s Hence since we are given ˙ W...

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766 CHAPTER7 . CONSERVATIONOFENERGY 7.14 Chapter 7, Problem 14 To analyze this flow we select the stationary control volume shown in Figure 7.3. Figure 7.3: Power plant next to a river. 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 i A ) ± 1 Upstream + ρ ( u f A ) ± 1 Downstream =0 = u i = u f Also, note that this means the volume flux, ˙ V , is constant and equal to u i A = u f A = ˙ V Energy Conservation: From energy conservation, we have 8 s 8 S ρ } h + 1 2 u · u ] ( u ·

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Unformatted text preview: ) dS = ˙ Q − ˙ W s Hence, since we are given ˙ W s ≈ , there follows ρ w h i + 1 2 u 2 i W ( − u i A ) ± ,² 1 Upstream + ρ w h f + 1 2 u 2 f W ( u f A ) ± ,² 1 Downstream = ˙ Q 7.14. CHAPTER 7, PROBLEM 14 767 So, in terms of volume flux, we have ρ ˙ V } h f − h i + 1 2 p u 2 f − u 2 i Q ] = ˙ Q But, u i = u f and h = c p T . Hence, ρ c p ˙ V [ T f − T i ] = ˙ Q Therefore, the temperature downstream of the power plant is T f = T i + ˙ Q ρ c p ˙ V Finally, for the given values, T f = 18 o C + 5 . 6 · 10 7 Watts p 998 kg / m 3 Q (4200 J / kg / K) (8 . 9 m 3 / sec) = 19 . 5 o C...
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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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p0714 - dS = ˙ Q − ˙ W s Hence since we are given ˙ W...

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