Question

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 80-m-diameter blades at that location. Also, determine the actual electric power generation, assuming an overall efficiency of 30 percent. Take the air density to be 1.25 kg/m3.

Answer 1

Answer:

1. The specific mechanical energy of the air in the specific location is 40.5 J/kg.
2. The power generation potential of the wind turbine at such place is of 2290 kW
3. The actual electric power generation is 687 kW

Explanation:

1. The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
2. The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$.
3. The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
4. To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
5. Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
6. Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
7. The actual electric power generation is calculated using the definition of efficiency:$\eta=\frac{E_P}{E_I}}$, where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$