Answer
$D_{2}=7.77m$
$P_{k}=3.98kW$
Work Step by Step
The diameter of the wind channel dowstream, choosing $1$ as the point before the windmill and $2$ as the point after the windmill, is :
$A_{1}V_{1}=A_{2}V_{2}$
$\frac{\pi D_{1}^2}{4}V_{1}=\frac{\pi D_{2}^2}{4}V_{2}$
$D_{2}=D_{1}\sqrt {\frac{V_{1}}{V_{2}}}=7m*\sqrt {\frac{8\frac{m}{s}}{6.5\frac{m}{s}}}=7.77m$
Now we calculate the specific volume of air:
$\upsilon=\frac{RT}{P}=\frac{0.287\frac{kPam^3}{kgK}*(20+273.15)K}{100kPa}=0.841\frac{m^3}{kg}$
Then the mass flow rate will be:
$m=\frac{Q}{\upsilon}=\frac{A_{1}V_{1}}{\upsilon}=\frac{\frac{\pi (7m)^2}{4}*8\frac{m}{s}}{0.841\frac{m^3}{kg}}=366.08\frac{kg}{s}$
Since we are working with kinetic energy:
$P_{k}=\frac{1}{2}m\Delta V^2=\frac{1}{2}*366.08\frac{kg}{s}*[(8\frac{m}{s})^2-(6.5\frac{m}{s})^2]$
$P_{k}=3.98kW$
