A) 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 330m³/s and...



Thermodynamics Question.


The answer from A is 60 days, I want to see the resolution and at B I don't know the final answer, thanks.


A) The time for a hydroelectric plant reservoir to reach half of its useful height.<br>Knowing that this reservoir is 10km long and 278m wide, which today the inflow<br>into the reservoir is 330m³/s and outflow is 531 m³/s. And that the current useful<br>height is of 750m.<br>b| The minimum mass flow in kg/s required to maintain energy production of<br>152MW per turbine, knowing that the minimum height in the reservoir for the<br>plant's operation is 375m. One time that this system is composed of 8 turbines<br>what will be the volumetric flow (m³/s) that needs to be reached to the reservoir<br>so that the production of this amount of energy per turbine is maintained. Note:<br>disregard the effect of kinetic energy; specific mass of water equal to 1000 kg/m²;<br>g= 9.81 m<br>

Extracted text: A) 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 330m³/s and outflow is 531 m³/s. And that the current useful height is of 750m. b| The minimum mass flow in kg/s required to maintain energy production of 152MW per turbine, knowing that the minimum height in the reservoir for the plant's operation is 375m. One time that this system is composed of 8 turbines what will be the volumetric flow (m³/s) that needs to be reached to the reservoir so 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/m²; g= 9.81 m

Jun 06, 2022
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