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 ríocanalizado bajo la presa, según se muestra. Las pérdidasdel sistema por fricción son h, = 3.5V/(2g), donde Ves la velocidad media en el conducto de entrada. ¿Paraqué caudal en metros cúbicos por segundo se extraeráuna potencia de 25 MW? ¿Cuál de las dos solucionestiene un mejor "rendimiento de conversión"?P3.145BES%3DZ1 = 50 m VD=4 m22 = 10 m23 = 0 mTurbina

Given data: Write the expression of velocity in terms of discharge.

Answered: The turbine in the figure uses the flow…

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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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

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