What is the theoretically available wind power density from this area?
Assuming the wind power is about 18 knots or 20.71 mph, we can find the possible number of off shore wind turbines power potential [1]. We can also assume that average wind turbine blade rotor diameter is about 110m [2]. Wind turbines must also be placed 5 times the rotor diameter apart. We also know the coastline of Kuwait is 499 km, or 499,000m [3]. So if we divide the coastline by the rotor diameter with correct spacing, we can find the number of wind turbines, which results into about 907 wind turbines.
In order to find the wind power density, we use the following equation:
Power per Unit Area of Land= (Pi/400) (rho)(V)^3
where V is the velocity. We can then plug in our value of 20.71 mph or 9.26 m/s into our equation given that the density is still 1.222 kg/m^3. Using those given values, the power per unit area of land is equal to 7.608 W/m^2. If we multiply this by the area which is 10000m by 499000m, it will be equal to the available amount of power.
Thus the power is approximately 37.9 GW or 15.16 kWh/d per person for shallow or 30.32 kWh/d per person for deep shore. But it is assumed that power unit area of 3 W/m^2 for offshore wind farms, which would mean 5.05 kWh/d per person for shallow or 10.1 kWh/d per person for deep.
Following McKay, how many tons of concrete and steel will it take to pull this off?
According to McKay, to create 48 kWh/d per person in the UK it would require 60 million tons of concrete and steel, so performing a ratio between the UK and Kuwait it will yield the amount of concrete and steel needed in Kuwait. The amount of concrete and steel for Kuwait is about .03832 million tons.
[1] http://www.windfinder.com/forecasts/wind_persian_gulf_akt.htm
[2] http://en.wikipedia.org/wiki/Wind_turbine
[3] http://www.indexmundi.com/kuwait/coastline.html
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