Question

Use dimensional analysis to show that the electric power output of a wind turbine is proportional to the cube of the wind speed. The relevant quantities on which the power can depend are the length \(L\) of the rotor blades, the density \(\rho\) of air (SI units \(\mathrm{kg} / \mathrm{m}^{3}\) ), and the wind speed \(v\).

Short answer

In this exercise, we demonstrated that the electric power output of a wind turbine is proportional to the cube of the wind speed by using dimensional analysis. We analyzed the dimensions of the given quantities, calculated the dimensions of electric power, applied Buckingham's Pi theorem, and determined the proportionality between power and wind speed. The analysis showed that the power output of a wind turbine is indeed proportional to the cube of the wind speed, confirming our initial objective.

Step-by-step solution

Determine the dimensions of the given quantities

First, we need to determine the dimensions of the given quantities. We'll be using the following notation for dimensions: [M] for mass, [L] for length, and [T] for time. 1. Length of rotor blades, \(L\): This is a distance, so its dimension is [L] 2. Density of air, \(\rho\): The units for density are \(\mathrm{kg} / \mathrm{m}^{3}\), so its dimension is [M][L]^{-3} 3. Wind speed, \(v\): The units for wind speed are \(\mathrm{m} / \mathrm{s}\) , so its dimension is [L][T]^{-1}

Calculate the dimensions of electric power

The electric power output, \(P\), is the energy transferred over time. In SI units, the unit of power is watt (W), which is equal to \(\mathrm{J} \cdot \mathrm{s} ^{-1}\). Energy has units of \(\mathrm{J}\), which is equal to \(\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2}\). Thus, the dimensions of power are [M][L]^{2}[T]^{-3}.

Apply Buckingham's Pi theorem

We can apply Buckingham's Pi theorem to find the relationships between the dimensions of the given quantities and the electric power output. According to this, we need to form a dimensionless product using our three given quantities and electric power. Let's denote the dimensionless product as \(\Pi\). We can write it as: \(\Pi = P ^{a} L^{b} \rho^{c} v^{d}\) Now substituting the dimensions of each variable in this equation, we have: \([M ^{0} L ^{0} T ^{0}] = [M^a L^{2a} T^{-3a}] [L^b] [M^c L^{-3c}] [L^d T^{-d}]\)

Equate dimensions and solve for exponents

Since the product is dimensionless, we can equate the dimensions of mass, length, and time: For mass: \(0 = a + c\) For length: \(0 = 2a + b -3c + d\) For time: \(0 = -3a - d\) We have three equations and four unknowns. However, we are only interested in the relationship between power and wind speed, which corresponds to the exponents \(a\) and \(d\). We can solve the first two equations to get \(b\) and \(c\) in terms of \(a\) and \(d\): From the equation for mass: \(c=-a\) From the equation for length: \(b=-2a+3c+d\) We can now express the dimensionless product as: \(\Pi = P ^{a} L^{-2a+3(-a)+d} \rho^{-a} v^{d}\) Since the product is dimensionless and equals to one, we can rewrite it as: \(P = k \rho^a L^{-2a+3(-a)+d} v^{-d}\) Here, \(k\) is a constant of proportionality.

Determine proportionality

We want to show that power is proportional to the cube of wind speed. This means we need to have \(a=1\) and \(d=-3\): \(P = k \rho^1 L^{-2(1)+3(-1)+(-3)} v^{-(-3)}\) \(P = k \rho L^{-8} v^3\) So, the power output of a wind turbine \(P\) is indeed proportional to the cube of the wind speed \(v\), as we set out to demonstrate.