The turbine in the figure uses the flow of the river channeled under the dam, as shown. The friction losses of the system are hf=3.5V^2/(2g), where V is the average velocity in the inlet duct. For what flow in m^3/s would power of 25 MW be extracted? Which of the two solutions has a better "conversion yield"?

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
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The turbine in the figure uses the flow of the river channeled under the dam, as shown. The friction losses of the system are hf=3.5V^2/(2g), where V is the average velocity in the inlet duct. For what flow in m^3/s would power of 25 MW be extracted? Which of the two solutions has a better "conversion yield"?

La turbina de la Figura P3.145 utiliza el flujo del río
canalizado bajo la presa, según se muestra. Las pérdidas
del sistema por fricción son h, = 3.5V/(2g), donde V
es la velocidad media en el conducto de entrada. ¿Para
qué caudal en metros cúbicos por segundo se extraerá
una potencia de 25 MW? ¿Cuál de las dos soluciones
tiene un mejor "rendimiento de conversión"?
P3.145
BES
%3D
Z1 = 50 m V
D=4 m
22 = 10 m
23 = 0 m
Turbina
Transcribed Image Text:La turbina de la Figura P3.145 utiliza el flujo del río canalizado bajo la presa, según se muestra. Las pérdidas del sistema por fricción son h, = 3.5V/(2g), donde V es la velocidad media en el conducto de entrada. ¿Para qué caudal en metros cúbicos por segundo se extraerá una potencia de 25 MW? ¿Cuál de las dos soluciones tiene un mejor "rendimiento de conversión"? P3.145 BES %3D Z1 = 50 m V D=4 m 22 = 10 m 23 = 0 m Turbina
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