Hello, I hope this answers your question.
From reading, here are the useful facts to solve this question:
1-A windmill is a device for extracting kinetic energy from the moving air.
2- kinetic energy in the wind is turned into electric energy. Denote kinetic energy: \(E_k\) and electrical energy: \(E_e\)
3-Suppose that when the wind blows at 5.0 m/s the kinetic energy per unit time that blows past a windmill is 0.50 MJ/s
Let the initial speed: \(v_{initial}=5.0 m/s\), and given the initial kinetic energy, \(E_k=0.50*10^6 J\)
Hint: (Although the text is talking about power and energy, they are the same here due to "1 second" (Since P=E/t -->P=E if t=1))
By the Kinetic Energy equation:
\(E_k=\frac{1}{2}mv^2\), we can find the mass, m of the air. \(m=\frac{2E_k}{v^2}=\frac{2*0.50*10^6}{(5.0)^2}=40 000 kg\) (1)
4-Windmills cannot capture all the kinetic energy from the air passing.
From this, we deduce that windmills are not 100% efficient (Thus, we will use the equation of Efficiency), and the text explained this further:
5-The maximum possible efficiency of a windmill is called the Betz limit. Given, the theoretical Betz limit: 16/27 of the kinetic energy of air passing through it.
Now to the question:
Suppose that the wind speed increases to 10.0 m/s (Denote, \(v_{final}=10.0m/s\))
If the windmill we considered before can operate at 80% of the Betz limit, how many megawatts of power does it produce?
It is crucial to note from this, that it is the same windmill, thus the following is an accurate assumption in solving: mass of air passing through the windmill blades is constant, as found in (1) -->40000kg.
Applying the efficiency equation: \(Efficiency=\frac{E_e}{E_k}\)
Given, the efficiency is 80% of the betz limit -->\(0.80*\frac{16}{27}=0.47407407\) is the efficiency.
\(E_k=\frac{1}{2}(40000)(10^2)\) \(=2.0*10^6 J\)
Thus, \(E_e=0.47407407*2.0*10^6=948148.14 J\)
Recall, this is the energy per second hence the power.
So the question wans the final answer to be in megawatts:
\(948148.14J/s=948148.14W=0.94814814MW\)
Which can be approximated to be: \(0.95MW\)
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