HW_8solS06

# HW_8solS06 - Physics 112 HW#8 Solutions Spring 2006...

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Physics 112 HW #8 — Solutions Spring 2006 6-91 [Hydropower] Assuming a constant flow rate, the amount of water (volume Δ V and mass Δ m = ρΔ V) that enters the dam from above in any time interval Δ t equals the amount of water that leaves the dam at its base ( ρ = the density of water = 1000 kg/m 3 ) during the same time. The total amount of water in between the top and bottom of the dam at any time remains constant, as does the total gravitational PE of this water. In time Δ t the net change in gravitational potential energy of all the water equals the change that would take place simply by moving a mass Δ m directly from top to bottom by height change H: Δ U g = - Δ mgH = - ρΔ VgH The rate of conversion of gravitational PE is: Δ U g Δ t = Δ mgH Δ t = ρ gH Δ V Δ t The electrical power output is 92% of this rate: P = 0.92 ρ gH Δ V Δ t = 2.0 × 10 9 W So the required volume flow rate is: Δ V Δ t = P 0.92 ρ gH = 2.0 × 10 9 W 0.92(1000 kg/m

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## This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

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HW_8solS06 - Physics 112 HW#8 Solutions Spring 2006...

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