Answer
The second side is better.
Work Step by Step
In this case we are working with kinetic energy:
$e_{k}=e_{mec}=\frac{V^2}{2}$
$e_{k,1}=\frac{(7\frac{m}{s})^2}{2}=24.5\frac{J}{kg}$
$e_{k,2}=\frac{(10\frac{m}{s})^2}{2}=50\frac{J}{kg}$
And the flux of mass:
$m=\rho VA$
Assuming $A=1m^2$ to simplify calculus
$m_{1}=(1.25\frac{kg}{m^3})*(7\frac{m}{s})*(1m^2)=8.75kg/s$
$m_{2}=(1.25\frac{kg}{m^3})*(10\frac{m}{s})*(1m^2)=12.5kg/s$
The power generation potential is:
$W_{g}=E_{k}=m*e_{k}$
$W_{g,1}=8.75\frac{kg}{s}*24.5\frac{J}{kg}=214.38\frac{J}{s}=214.38W$
$W_{g,2}=12.5\frac{kg}{s}*50\frac{J}{kg}=625\frac{J}{s}=625W$
And finally the electric power generations per year is:
$E_{e}=W_{g}*t$
$E_{e}=W_{g,1}*t_{1}=0.21438kW*3000\frac{h}{year}=643.14\frac{kWh}{year}$
$E_{e}=W_{g,2}*t_{2}=0.625kW*1500\frac{h}{year}=937.5\frac{kWh}{year}$
As you can see, the second side is better.
