Answer:
a) 6.93 ×10^8 W
b) 1.01
Explanation:
height of the dam h= 221 m
power output= 1300 MW
power:
P=Fv
also, Q= 655 [tex]m^3/s[/tex]
depth d = 108 m
also, pressure p=F/A= ρdg
ρ= density of fluid, g= acceleration due to gravity
ρdg= 1000×108×9.81=1059400 N/m^2
Power flow= p×Q
now, Power flow= 1059400×655 =693,907000 W = 6.93 ×10^8 W
b) Ratio of this power to the facility = Power flow / average power
=[tex]\frac{693\times10^6}{680\times10^6}[/tex]
Ratio of this power to the facility= 1.01