The average wind speed for Kuwait is approximately 4.6 m/s [1]. However, 1.5 m/s was added to this number to compensate for the Weibull Distribution. (At greater speeds, there is a greater power output.) In order to find the power per Unit Area of Land, the velocity of 6.1 m/s and the density of air at a constant value of 1.22 kg/m^3 is plugged into the equation below:
Power per Unit Area of Land= (Pi/400) (rho)(V)^3
where rho is the density and V is the velocity.
The result of power due to wind is 2.175 kg/s^2, which is pretty average for wind power. This is equal to 2.175 W*s/m^2. Now converting to kWh/m^2 per day and then multiplying by 80% of the area of Kuwait land we get 206712 kWh. This gives us 0.074 kWh per person per day.
[1] http://www.sciencedirect.com/science/article/pii/S0960148105000145
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