Respuesta :
Solution :
Given :
[tex]$V_1 = 7 \ m/s$[/tex]
Operation time, [tex]$T_1$[/tex] = 3000 hours per year
[tex]$V_2 = 10 \ m/s$[/tex]
Operation time, [tex]$T_2$[/tex] = 2000 hours per year
The density, ρ = [tex]$1.25 \ kg/m^3$[/tex]
The wind blows steadily. So, the K.E. = [tex]$(0.5 \dot{m} V^2)$[/tex]
[tex]$= \dot{m} \times 0.5 V^2$[/tex]
The power generation is the time rate of the kinetic energy which can be calculated as follows:
Power = [tex]$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$[/tex]
Regarding that [tex]$\dot m \propto V$[/tex]. Then,
Power [tex]$ \propto V^3$[/tex] → Power = constant x [tex]$V^3$[/tex]
Since, [tex]$\rho_a$[/tex] is constant for both the sites and the area is the same as same winf turbine is used.
For the first site,
Power, [tex]$P_1= \text{const.} \times V_1^3$[/tex]
[tex]$P_1 = \text{const.} \times 343 \ W$[/tex]
For the second site,
Power, [tex]$P_2 = \text{const.} \times V_2^3 \ W$[/tex]
[tex]$P_2 = \text{const.} \times 1000 \ W$[/tex]
