Question

2-14 Two sites are being considered for wind power gen- eration. In the first site, the wind blows steadily at 7 m/s for 3000 hours per year, whereas in the second site the wind blows at 10 m/s for 1500 hours per year. Assuming the wind velocity is negligible at other times for simplicity, deter- mine which is a better site for wind power generation. Hint: Note that the mass flow rate of air is proportional to wind velocity
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Answer 1

Answer:

The Kinetic energy possessed by the wind is considered as the mechanical energy.

Express the kinetic energy per unit mass for first site.

Ke1 = v^21/2

Here, kinetic energy of the wind in first site Ke1 and velocity of the wind in first is V1.

Substitute 7m/s for v1.

ke1 = (7m/s)2/2

= 24.5m2/s2

= 24.5m2/s2(IKJ/KG/1,000M2/S2)

= 0.0245KJ/KG

Thus, kinetic energy per unit mass for first site is 0.0245KJ/Kg

Explanation:

Answer 2

The second site is better for wind power generation.

Given:

• Velocity ($V_1$) = 7 m/s
• Velocity ($V_2$) = 10 m/s
• Time ($T_1$) = 3000 hrs/year
• Time ($T_2$) = 1500 hrs/year

To Find:

• The calculation to determine which site is better for wind power generation.

Solution:

• Power generation = rate of change in time with respect to kinetic energy.
• The formula is given by, Power = ΔKE = $\frac{m*v^{2}}{2}$
• From the above formula, we get to know that m ∝ v
• So power ∝ $V^{3}$ , power = $Constant*V^{3}$  
• since mass flow rate is proportional to wind velocity, the area also would be same as turbine used.
• For Site 1,
• Power ($P_{1}$) = const.$V^{3}$  = constant$(7 m/s)^{3}$ = constant (343) W
• For Site 2,
• Power ($P_2$ = const. $V^{3}$ = constant$(10 m/s)^{3}$ = constant(1000) W
• Now let us calculate energy per year for each of the site.
• $E_{year_1}$ = $P_1*T_1 = constant*343*3000 = constant(1029000) W.hr/year$
• $E_{year_2}=P_2*T_2 = constant*1000*1500 = constant(1500000) W.hr/year$
• By the above data, E(year 2) > E(year 1)

∴ The second site is better for wind power generation.