1 million kWh = 1 GWh, so 927.7 million kWh = 927.7 GWh.
With 2.8 million people and an average Canadian load of 2.5 kW per person (one of the highest in the world), the average Toronto load is about 7 Gigawatts, which is about 160 GWh each day, or 160,000,000 kWh per day.
It would be about 5.88 barrels. 10,000 kWh / 1,700kWh per boe = 5.882352941176471.... Source: http://bioenergy.ornl.gov/papers/misc/energy_conv.html Formula: Barrel of oil equivalent (boe) = approx. 6.1 GJ (5.8 million Btu), equivalent to 1,700 kWh.
1 mwh = 1000 kwh hence 4 mwh = 4000 kwh
The answer to this question is zero. There is no kWh given.
To find kWh a time frame has to be given. That is what the h in kWh stands for.
Installed capacity is 1189 MW, annual generation is 4466 GWH, which is 4,466 million kWh.
With 2.8 million people and an average Canadian load of 2.5 kW per person (one of the highest in the world), the average Toronto load is about 7 Gigawatts, which is about 160 GWh each day, or 160,000,000 kWh per day.
Your question should read, "How many kW.h are there in 0.07 GW.h?" It's important to use the correct symbols. There are 1000 000 000 kilowatt hours in one gigawatt hour -so you can now work it out for yourself.
1 GWh = 1 000 MWh
1 gigawatt = 1000 megawatts 1 MW of electricical Power means different GWh in a year according to how long the power can be used through the year. 1 year = 8760 hours. So at 100%, a power plant of 1 MW will produce 8760 MWh, ie 8,76 GWh For instance, with a nuclear power plant that usually runs 80% of the time you have: 1 MW -> 7 GWh in a year You have approximately the same ratio (a bit more) with a thermal power plant (fossil fuel) With wind power plant, they usually run 23% of the time in windy areas (due to meterological conditions). You then have in such a case : 1 MW -> 2 GWh in a year For solar power it is less and up to areas. In California, 1 MW -> 1,4 GWh in a year In Germany, 1 MW -> 0,6 GWh in a year Electrical power of plants are up to types of plants: Nuclear & Fuel : around 1000 MW (up to power plant size...) 1 Wind turbine : up to 5 MW Solar plant (photovoltaic) : up to 20 MW Solar plant (other, experimental) : up to 500 MW so far
9,000
about 13698630.13 kWh per dayPower production is the primary function of the Bonneville Dam. The two Bonneville powerhouses generate about 5 billion kWh of electricity each year; or 13.7 million kWh /day.
1000
Divide the kWh by 1 million
Divide by 1000.
General Hospital - 1963 1-9277 was released on: USA: 21 June 1999
One thousand cubic feet of gas (Mcf) -> 1.027 million BTU = 1.083 billion J = 301 kWh by Lyon