Question

A state-of-the-art Dutch windmill, shown on the left of Figure \(5.10\) was operating in 1937. The height of its axis was \(16[\mathrm{~m}]\), the radius of its sails \(14.35[\mathrm{~m}]\). With a wind velocity of \(7.77\left[\mathrm{~m} \mathrm{~s}^{-1}\right]\) it was able to pump an amount of water \(34.1\left[\mathrm{~m}^{3} \mathrm{~s}^{-1}\right]\) over a height of \(1.67[\mathrm{~m}]\). (a) Calculate the pumping power. (b) Calculate the input wind energy. (c) The wind energy entering into the rotation of the wings was measured and a value \(c_{\mathrm{p}}=0.17\) was found. Calculate the power taken from the wind. Explain the difference with your result in (a).

Short answer

The pumping power is approximately 559,045 W, the input wind energy is 192,172 W, and the extracted power is approximately 32,669 W. The discrepancy suggests differences due to system losses or approximations in the given parameters.

Step-by-step solution

Understand the Problem

We need to calculate the pumping power produced by the windmill, the input wind energy, and the power extracted from the wind using the given data.

Calculate Pumping Power (a)

Pumping power can be calculated using the formula: \[ P = \rho \cdot g \cdot Q \cdot h \]where \( \rho \) is the density of water \( (1000 \mathrm{~kg/m^3}) \),\( g \) is the acceleration due to gravity \( (9.81 \mathrm{~m/s^2}) \),\( Q \) is the flow rate \( (34.1 \mathrm{~m^3/s}) \), and\( h \) is the height \( (1.67 \mathrm{~m}) \).Substituting these values gives:\[ P = 1000 \times 9.81 \times 34.1 \times 1.67 \approx 559,045 \mathrm{~W} \]

Calculate Input Wind Energy (b)

The power available in the wind can be calculated using:\[ P_w = \frac{1}{2} \cdot \rho_{air} \cdot A \cdot v^3 \]where \( \rho_{air} \) is the air density \( (1.225 \mathrm{~kg/m^3}) \),\( A = \pi r^2 \) is the area swept by the windmill's sails \( (\pi \times (14.35)^2 \approx 647.76 \mathrm{~m^2}) \), and\( v \) is the wind velocity \( (7.77 \mathrm{~m/s}) \).Substituting these values gives:\[ P_w = \frac{1}{2} \times 1.225 \times 647.76 \times (7.77)^3 \approx 192,172 \mathrm{~W} \]

Calculate Power Extracted with Given Efficiency (c)

Given the coefficient of performance \( c_p = 0.17 \), the power extracted from the wind is:\[ P_{extracted} = c_p \times P_w = 0.17 \times 192,172 \approx 32,669 \mathrm{~W} \]

Compare the Results and Explain the Difference

The pumping power calculated in (a), \( 559,045 \mathrm{~W} \), was more than the power extracted from the wind calculated here, \( 32,669 \mathrm{~W} \). This discrepancy indicates that the real system losses and the coefficient \( c_p \) given for the wind power conversion were not accounted accurately compared to the theoretical calculations.

Key concepts

Windmill Efficiency

Windmill efficiency refers to how effectively a windmill converts the energy available from the wind into mechanical energy.It is often measured using the coefficient of performance, denoted as \( c_p \).This coefficient represents the fraction of the wind's power that can be extracted by the windmill.In the exercise, the calculated \( c_p \) is 0.17, which means that only 17% of the wind energy is converted into usable mechanical energy.This value is a simplified representation, as real-world conditions can affect efficiency, including wind variability, mechanical losses, and design limitations.To further understand windmill efficiency, consider the formula:\[ c_p = \frac{P_{extracted}}{P_w} \]where \( P_{extracted} \) is the power extracted from the wind and \( P_w \) is the total power available in the wind.Achieving higher efficiency is crucial for better energy conversion, but it is also limited by physical and technological constraints, such as the Betz Limit, which is theoretically about 59%.

Pumping Power

Pumping power indicates how much energy is used to move water or any fluid.It is directly associated with variables such as fluid density, gravity, flow rate, and the height over which the liquid is lifted.The exercise provides the formula:\[ P = \rho \cdot g \cdot Q \cdot h \]where:

• \( \rho \) is the fluid density, typically for water it's \( 1000 \; \text{kg/m}^3 \).
• \( g \) is gravitational acceleration, approximately \( 9.81 \; \text{m/s}^2 \).
• \( Q \) is the volumetric flow rate.
• \( h \) is the lift height.

In our problem, these substituted values yield a pumping power of around \( 559,045 \; \text{W} \).This calculation shows how much energy is utilized to lift the given volume of water to the specified height.The result is notably high, reflecting the practical challenges and energy demand of such operations.

Wind Energy Input

Wind energy input denotes the raw energy available from wind passing through the area swept by the windmill's blades.This is a critical calculation that helps determine how much potential energy can be harnessed by the windmill.The formula used is:\[ P_w = \frac{1}{2} \cdot \rho_{air} \cdot A \cdot v^3 \]where:

• \( \rho_{air} \) is the air density, typically \( 1.225 \; \text{kg/m}^3 \).
• \( A \) is the area swept by the blades, found using \( A = \pi r^2 \).
• \( v \) is the wind velocity.

In the exercise, the swept area calculated was \( 647.76 \; \text{m}^2 \), and with a wind speed of \( 7.77 \; \text{m/s} \), the wind energy input was approximately \( 192,172 \; \text{W} \).This figure indicates the immense power potential coursing through the windmill's area before any conversion losses or efficiency considerations.

Power Extraction

Power extraction refers to the actual amount of power obtained from the wind, after considering the windmill's efficiency.This power is the practical output after accounting for all losses and conversion factors inherent in energy transformation.The exercise calculated the power extracted as:\[ P_{extracted} = c_p \times P_w \]Given the coefficient of performance \( c_p = 0.17 \) and the wind energy input \( P_w \approx 192,172 \; \text{W} \), the power extracted was about \( 32,669 \; \text{W} \).This discrepancy from the theoretical maximum highlights the significant impact of losses due to friction, air resistance, and other inefficiencies.By understanding power extraction, students can better grasp the real-world challenges of harnessing wind energy.It illustrates the gap between the theoretical potential energy and the actual usable power after processing through a windmill system.