Wind Energy Harvesting

Micro-to-Small Scale Turbines

Ravi Kishore, Shashank Priya, Colin Stewart

• Publisher: De Gruyter
• Year: 2018
• eBook ISBN: 9781614519799
• Edition: 1
• Topic: Physical Sciences
• Subtopic: Energy

This book provides the fundamental concepts required for the development of an efficient small-scale wind turbine. For centuries, engineers and scientists have used wind turbines of all shapes and sizes to harvest wind energy. Large-scale wind turbines have been successful at producing great amounts of power when deployed in sites with vast, open space, such as in fi elds or in offshore waters. For environments with limited space, such as dense urban environments, small-scale wind turbines are an attractive alternative for taking advantage of the ubiquity of wind. However, many of today's tools for aerodynamic design and analysis were originally developed for large-scale turbines and do not scale down to these smaller devices. Arranged in a systematic and comprehensive manner, complete with supporting examples, Wind Energy Harvesting: Micro- To Small-Scale Turbines is a useful reference for undergraduate and graduate level classes on energy harvesting, sustainable energy, and fl uid dynamics, and an introduction to the field for non-technical readers.

1Introduction

1.1Wind energy

Wind is the large-scale movement of air caused by differences in atmospheric pressure. The uneven heating of the earth by the sun, primarily between the equator and the poles, leads to these pressure differences and causes air to rush from high-pressure regions to low-pressure regions. However, wind does not flow in a straight path; the effects of Coriolis and centrifugal forces due to rotation of the earth cause wind to deflect. The local wind condition is finally determined by the geographical features in and around the given region, such as plains, mountains, forests, oceans, rivers, and buildings. It should also be noted that although the rotational forces and regional geography remain more or less constant over small periods of time, the solar heating changes throughout the year, resulting in the seasonal change of wind flow.

Intuitively, one can expect that wind power is a function of wind speed. Mathematically, the power P of wind flowing at a speed u∞ through a region of cross-sectional area A is given by

$P=\frac{1}{2}\rho A{u}_{\infty }^3,\left(1.1\right)$

where ρ is the density of air. It is interesting to note from Eq. (1.1) that wind power is proportional to the cube of wind speed. To derive this equation, let us consider a disc with cross-sectional area A through which wind flows with a constant velocity u∞. The air that flows through the disc in an infinitesimal time step dt travels a distance dx = u∞dt, creating a cylinder of volume dV = Adx as shown in Fig. 1.1. The mass dm of air inside the cylinder can be calculated using

$\text{d}m=\rho \text{d}V=\rho A{u}_{\infty }\text{d}t.\left(1.2\right)$

The kinetic energy of this air is equal to

$\text{KE}=\frac{1}{2}\text{d}m{u}_{\infty }^2=\frac{1}{2}\left(\rho A{u}_{\infty }\text{d}t\right)u_{\infty }^2=\frac{1}{2}\rho A{u}_{\infty }^3\text{d}t.\left(1.3\right)$

Assuming that the change in internal and potential energies of the air are negligible, the power of the wind can be calculated as the rate of change of its kinetic energy:

$P=\frac{\text{dKE}}{\text{d}t}.\left(1.4\right)$

Fig. 1.1: Wind flowing through a cylindrical differential volume.

We arrive at the desired power equation by substituting Eq. (1.3) into Eq. (1.4):

$P=\frac{1}{2}\rho A{u}_{\infty }^3.\left(1.5\right)$

Example 1.1. What is the power of the wind flowing through a circular cross-section of diameter 5 m with velocity 10m/s normal to the cross-section? Assume the density of air is 1.225 kg/m3.

Solution. The area of the circular cross-section is given by A = π (D/2)2, where D is the diameter. Setting diameter D = 5 myields A = π (5/2)2 m2 = 19.64 m2. The power is then calculated using Eq. (1.5):

$P=\left(1/2\right)\left(1.225{\text{kg/m}}^{\text{3}}\right)\left(19.64{\text{m}}^{\text{2}}\right){\left(10\text{m/s}\right)}^3=12\text{kW}\text{.}$

Example 1.2. What is the average power output of a wind turbine with a 5 m rotor diameter, deployed on a location with average wind velocity 10 m/s? Assume the net efficiency of the wind turbine is 25%.

Solution. The total available wind power is again given by Eq. (1.5). Following Example (1.1), we find P = 12 kW. The power output of the wind turbine is then given by Pout = ηP, where η denotes the wind turbine efficiency. Thus,

$P_{\text{out}}=\eta P=\left(0.25\right)(...$

Table of contents

• Cover
• Title Page
• Copyright
• Preface
• Contents
• List of Figures
• List of Tables
• 1 Introduction
• 2 Wind turbines
• 3 Components of a small-scale wind turbine
• 4 Aerodynamics of a wind turbine
• 5 Applying BEM to small-scale wind turbine blade design
• 6 CFD analysis of wind turbines: Fundamentals
• 7 CFD analysis of wind turbines: Practical guidelines
• 8 Diffuser-Augmented Small-Scale Wind Turbine
• 9 Unconventional wind energy harvesters
• References
• Index