1) The time for a hydroelectric plant reservoir to reach half of its useful height.Knowing that this reservoir is 10km long and 278m wide, which today the inflowinto the reservoir is 330m /s and outflow is 531 m/s. And that the current usefulheight is of 750m.2) The minimum mass flow in kg/s required to maintain energy production of152MW per turbine, knowing that the minimum height in the reservoir for theplant's operation is 375m. One time that this system is composed of 8 turbineswhat will be the volumetric flow (m³/s) that needs to be reached to the reservoirso that the production of this amount of energy per turbine is maintained. Note:disregard the effect of kinetic energy; specific mass of water equal to 1000 kg/m³;g = 9.81 m

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1) The time for a hydroelectric plant reservoir to reach half of its useful height. Knowing that this reservoir is 10km long and 278m wide, which today the inflow into the reservoir is 330m/s and outflow is 531 m³/s. And that the current useful height is of 750m.

1) The time for a hydroelectric plant reservoir to reach half of its useful height. Knowing that this reservoir is 10km long and 278m wide, which today the inflow into the reservoir is 330m/s and outflow is 531 m³/s. And that the current useful height is of 750m.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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