Question

Two sites are being considered for wind power generation. In the first site, the wind blows steadily at \(7 \mathrm{m} / \mathrm{s}\) for 3000 hours per year, whereas in the second site the wind blows at \(10 \mathrm{m} / \mathrm{s}\) for 1500 hours per year. Assuming the wind velocity is negligible at other times for simplicity, determine which is a better site for wind power generation. Hint: Note that the mass flow rate of air is proportional to wind velocity.

Short answer

Answer: Site 2 is a better site for wind power generation, as it generates more energy compared to Site 1.

Step-by-step solution

Calculate the total number of hours for both sites

We are given the time, the wind blows for Site 1 and Site 2 as - Site 1: 3000 hours per year - Site 2: 1500 hours per year We will calculate the total number of hours of steady wind for both sites as: Total number of hours = Hours for Site 1 + Hours for Site 2 Total number of hours = 3000 + 1500 = 4500 hours per year

Calculate the energy generated by the wind blowing at each site

We can now use the given velocities and the power formula to calculate the energy generated by the wind at both sites. - Site 1: Wind velocity is 7 m/s - Site 2: Wind velocity is 10 m/s Since mass flow rate is proportional to wind velocity, we can denote \(m_1\) and \(m_2\) as the mass flow rates for Site 1 and Site 2. For Site 1, the energy generated E1 is given by: E1 = (Hours for Site 1) x (Power for Site 1) E1 = 3000 x (0.5 x \(m_1\) x (7)^2) For Site 2, the energy generated E2 is given by: E2 = (Hours for Site 2) x (Power for Site 2) E2 = 1500 x (0.5 x \(m_2\) x (10)^2)

Find the ratio of mass flow rates

Since the mass flow rate of air is proportional to the wind velocity, we have: \(m_1\) : \(m_2\) = 7 : 10

Compare the energy generated by both sites

Now we need to compare E1 and E2 to determine which site generates more energy: E1 = 3000 x (0.5 x \(m_1\) x (7)^2) E2 = 1500 x (0.5 x \(m_2\) x (10)^2) Using the ratio of mass flow rates, we can rewrite \(m_2\) in terms of \(m_1\): \(m_2\) = \(\frac{10}{7} m_1\) Substitute back into the equation for E2: E2 = 1500 x (0.5 x \(\frac{10}{7} m_1\) x (10)^2) Now compare E1 and E2: E1 = 3000 x (0.5 x \(m_1\) x (7)^2) E2 = 1500 x (0.5 x \(\frac{10}{7} m_1\) x (10)^2) Calculate the energies: E1 = 102900\(m_1\) E2 = 107142.9\(m_1\) Since E2 > E1, the second site is a better site for wind power generation.