I'm missing something easy (have to be)... A turbine produces 200 kilowatts of power. How much energy, in Kilowatt - hours, can such a turbine generate in a year? Given that the average house hold uses about 10,000 kilowatt -hours of energy each year, how many households can be powered by a single turbine. Where do I start with these two questions? please and thank you.

Question

I'm missing something easy (have to be)... A turbine produces 200 kilowatts of power. How much energy, in Kilowatt - hours, can such a turbine generate in a year? Given that the average house hold uses about 10,000 kilowatt -hours of energy each year, how many households can be powered by a single turbine.  Where do I start with these two questions? please and thank you.

anonymous · Answer

Start with the fact that one watt is equal to 1 Joule per second.

anonymous · Answer

figure out how many seconds are in a year and then use dimensional analasys to solve for how many Joules per year.

anonymous · Answer

Ok, lets see here... so, conversion down from 365 days/year to seconds/day?

anonymous · Answer

1year x 7days/1week x 24hr/1day x 60min/1hr x 60s/1min?

anonymous · Answer

you should get about 31536000 seconds per year

anonymous · Answer

Yeah, got that (31,556,926 seconds/year) by googling the answer, but I'm trying to figure out how to start the equation and show all the work.  forgive my ignorance, first math course in awhile.  Thank you for helping!

whpalmer4 · Answer

$$1 year * \frac{365days}{1 year} * \frac{24 hours}{1 day} * \frac{60 min}{1 hour} *\frac{60 second}{1 min} = $$$$365*24*60*60 seconds = 31536000 seconds$$

whpalmer4 · Answer

In a leap year, you'll use 366 days.  Over the years, it averages out to 365.2422 days per year, but for most people's purposes, it's either 365 or 366 days.

anonymous · Answer

I see, Thank you for that!