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Unformatted text preview: 41 Chapter 4 POWER FROM THE WIND 4.1 Introduction Discussion presented in the previous chapter dealt with variations in the availability of wind power, and the quantification of these variations. Once we can quantify the wind resource in terms of the likelihood that the wind speed will remain within a certain range, and its long term characteristics, the next questions naturally are, how much power is contained in the wind, and what fraction of that power can be extracted. Unlike power from fossil fuel plants, the output of a wind turbine generator (WTG) varies with the season, timeofday and surrounding physical conditions. There are also other causes of variations including turbulence, gusts, inversion layers and changing weather patterns in a season as discussed in the previous chapter. The approach presented in this chapter deals with both mechanical shaft power and electrical power outputs from wind turbine generators (WTG). At first we develop a model to quantify the extractable power from the wind. This is followed by a discussion on the causes of variations in the wind power output. Then we discuss coefficient of performance and how it is affected by the blade pitch angle, and the operating speed of the WTG. The blade pitch angle reflects the orientation of the blade as it faces the incoming wind. The chapter is closed by a discussion of the capacity factor and the long term average of wind power availability. The capacity factor is a comprehensive index useful for comparing different WTG designs, and their performance under various prevailing wind conditions. 4.2 Power Extracted by a Wind Machine The power in the wind can be computed using the concept of kinetics. Power is the energy delivered per unit time. The available energy is the kinetic energy of wind in motion. The kinetic energy (K.E.) of any particle is onehalf of the product of its mass (M) and velocity (v) squared. 42 2M(v)21K.E.=Mass, M can be replaced by the product of density ρand volume (Vol). The volume in turn, can be replaced by the distance covered by a vertical layer of wind and its crosssectional area. If we assume ‘dist’ to be the thickness of the layer of wind passing through the WTG, K.E.12(Vol)(v)12(dist)(area)(v)==ρρ22(4.1) K.E. for blade area A = 12(dist)(v)ρA2(4.2) 32A(v)21A(v)Tdist21TimeK.E.Powerρρ===watts(4.3) Here we consider the velocity v to be the distance covered in time T. Where, ρ =1.2 kg / m (density of air)3A = wind turbine blade area in m2v = wind speed in m/sec 43 Let us take a simple example to check how significant is the effect of velocity on the WTG output. Example 4.1 Let us compute the power available to a WTG under the following conditions....
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This note was uploaded on 03/16/2012 for the course ECE 5374G taught by Professor Srahman during the Spring '12 term at Virginia Tech.
 Spring '12
 SRahman
 The Land

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