Answer
$W_{e}=804.248kW$
$E_{e,day}=19301.952\frac{kWh}{day}$
$Money_{day}=1737.18\frac{dollars}{day}$
Work Step by Step
Here we are working with kinetic energy:
The kinetic work is: $W_{k}=\frac{1}{2}mV^2$
Where $m=\rho Q=\rho VA$
$m=1.25\frac{kg}{m^3}*8\frac{m}{s}*(\frac{\pi*(100m)^2}{4})=78539.82\frac{kg}{s}$
And then the energy of the fluid is:
$W_{k}=\frac{1}{2}*78539.82\frac{kg}{s}*(8\frac{m}{s})^2=2513.274kW$
The actual electric generation will be:
$W_{e}=\eta W_{k}=0.32*2513.274kW=804.248kW $
The amount of electric energy per day is:
$E_{e,day}=W_{e}*t=804.248kW*24h=19301.952\frac{kWh}{day}$
And its monetary value is:
$Money_{day}=19301.952\frac{kWh}{day}*0.09\frac{dollars}{kWh}=1737.18\frac{dollars}{day}$
