**Please write so it\'s possible to read, had problems in the past with a good a

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**Please write so it's possible to read, had problems in the past with a good answer that wasn't legible at all Appreciate it

***Please only answer if you're very confident in your answer and you want to provide insight into answering the question, answers just thrown together so you can get your answer totals up are not what I'm looking for.

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3.) A homeowner installs a wind turbine with a rotor diameter of 3.7 m to supplement electricity from the public utility and feed into the grid. The cost of the turbine, the associated electronics and a guyed tower of 100 ft is $12,000. If the turbine has an efficiency of 35% and the energy is directly fed into the grid, what is the payback period for the investment? Assume there are minimal maintenance costs, electricity from the public utility costs $0.15 per kWh and the wind velocity is constant at 9 m/s as an average for 100 ft above ground (truly outstanding site)

Explanation / Answer

Air incident on the rotor area per unit time is pi*d^2/4*v= pi*5.7^2/4*9= 229.66m^3/s. So mass per unit time is density of air at 100ft height from ground*volume per unit time= 0.84*229.66= 192.9kg/s.

Now, kinetic energy of this much air, incident on the rotor area is = 0.5*mass per unit time*velocity^2= 0.5*192.9*9^2= 7813 J/s= 7813W

Now, 35% of this is actually converted into electricity= 7813*0.35= 2.73 kW

Now, for 12000$, and electricity sold at 0.15$ per kWh, net kWh that costs 12000$ is 12000/0.15= 80000kWh

Now, 2.73kW is produced and 80000 kWh is required, so time taken = 80000/2.73 = 29304hr= 29304/ 24= 1221 days or, the payback period for the investment is 3 years 4 months 6 days

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