p0714 - ) dS = Q W s Hence, since we are given W s , there...

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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 w h i + 1 2 u 2 i W ( u i A ) , 1 Upstream + w h f + 1 2 u 2 f W ( u f A ) , 1 Downstream = Q 7.14. CHAPTER 7, PROBLEM 14 767 So, in terms of volume flux, we have 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, c p V [ T f T i ] = Q Therefore, the temperature downstream of the power plant is T f = T i + Q 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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p0714 - ) dS = Q W s Hence, since we are given W s , there...

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