Answer:
![e_(mec)=162J/kg](https://img.qammunity.org/2021/formulas/physics/college/i9zom3b89jvso9rqohuiqv7rypx5xh20od.png)
![P_(potential)=18312480W=18312.5kW](https://img.qammunity.org/2021/formulas/physics/college/kqgcw1f1bnfr02vjgoyayus8atmzj0vsik.png)
![P_(actual)=5493.7kW](https://img.qammunity.org/2021/formulas/physics/college/3u9qeo6r9066o3c1gdnb9b3oeu03yalfsb.png)
Step-by-step explanation:
Given data
Diameter d=80m
Speed v=18 m/s
Efficiency n=30%
Air density p=1.25 kg/m³
For Mechanical energy of air per unit mass
The power potential of wind per unit mass could be defined as follow:
![e_(mec)=(v^2)/(2)\\e_(mec)=((18m/s)^2)/(2) \\e_(mec)=162J/kg](https://img.qammunity.org/2021/formulas/physics/college/rqwlpbx6la4cuhz9l5xexcbm38i47hbo72.png)
For Power generation potential
The generation potential of turbine will determined from the available kinetic energy of air:
![P_(potential)=e_(mec)m\\P_(potential)=e_(mec)pV\\P_(potential)=e_(mec)p(dV)/(dt)\\ P_(potential)=e_(mec)pA(dx)/(dt) \\P_(potential)=e_(mec)pr^2\pi v\\P_(potential)=(162J/kg)(1.25kg/m^3)((80m)/(2) )^2\pi (18m/s)\\P_(potential)=18312480W=18312.5kW](https://img.qammunity.org/2021/formulas/physics/college/uxgw9plvs8vxq7fumslvdpmdu46tu7k0gs.png)
For Actual power
The actual power generation could be defined as follow as: