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p0746

# p0746 - H Hence H = h t 1 6 H = ⇒ h t = 5 6 H Thus the...

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7.46. CHAPTER 7, PROBLEM 46 811 7.46 Chapter 7, Problem 46 Figure 7.23: Hydroelectric plant. Applying the approximate energy equation between the reservoir free surfaces (Points 1 and 2), we have p 1 ρ g + α 1 u 2 1 2 g + z 1 = p 2 ρ g + α 2 u 2 2 2 g + z 2 + h t + h L Now, the pressure is atmospheric at both Points 1 and 2 so that p 1 = p 2 . Also, u 1 0 and u 2 0 , while z 1 z 2 = H
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Unformatted text preview: H . Hence, H = h t + 1 6 H = ⇒ h t = 5 6 H Thus, the power generated is ˙ W t = ˙ mgh t = 5 6 ˙ mgH In terms of the given quantities, noting that ˙ m = ρ ˙ V , we find ˙ W t = 5 6 X 1000 kg m 3 ~ X 108 m 3 sec ~ w 9 . 807 m sec 2 W (9 . 5 m) = 8385 kW...
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