Wind Energy Handbook. Michael Barton Graham
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Название: Wind Energy Handbook

Автор: Michael Barton Graham

Издательство: John Wiley & Sons Limited

Жанр: Физика

Серия:

isbn: 9781119451167

isbn:

СКАЧАТЬ

      The torque per unit span acting on all the blades is given by the Kutta–Joukowski theorem. The lift per unit radial width L is

upper L equals rho left-parenthesis upper W times normal upper Gamma right-parenthesis

      where (W × Γ) is a vector product, and W is the relative velocity of the air flow past the blade:

      (3.35)delta upper Q equals rho upper W times normal upper Gamma r sine phi Subscript t Baseline delta r equals rho normal upper Gamma italic r upper U Subscript infinity Baseline left-parenthesis 1 minus a right-parenthesis delta r

      Equating the two expressions for δQ gives

a prime equals StartFraction normal upper Gamma Over 4 pi r squared normal upper Omega EndFraction a prime equals StartFraction upper U Subscript infinity Baseline Superscript 2 Baseline a left-parenthesis 1 minus a right-parenthesis Over left-parenthesis upper Omega r right-parenthesis squared EndFraction equals StartFraction a left-parenthesis 1 minus a right-parenthesis Over lamda Subscript r Baseline Superscript 2 Baseline EndFraction

      At the outer edge of the disc the tangential induced velocity is

      If a′ is retained in Eq. (3.32), there is a small inconsistency here between vortex theory and the one‐dimensional actuator disc theory, which ignores rotation effects.

      3.4.5 Torque and power

      The torque on an annulus of radius r and radial width δr (ignoring a′ as actuator disc theory ignores rotation) is

      (3.37)StartFraction italic d upper Q Over italic d r EndFraction delta r equals rho upper W normal upper Gamma r sine phi Subscript t Baseline delta r equals StartFraction rho Baseline 4 pi italic r upper U Subscript infinity Baseline Superscript 3 Baseline a left-parenthesis 1 minus a right-parenthesis squared Over normal upper Omega EndFraction delta r

      The radial distribution of power is

      (3.38)StartFraction italic d upper P Over italic d r EndFraction equals normal upper Omega StartFraction italic d upper Q Over italic d r EndFraction equals one half rho upper U Subscript infinity Baseline Superscript 3 Baseline Baseline 2 pi r 4 a left-parenthesis 1 minus a right-parenthesis squared

      and, therefore, the total power is

      (3.39)upper P equals one half rho upper U Subscript infinity Baseline Superscript 3 Baseline pi upper R squared Baseline 4 a left-parenthesis 1 minus a right-parenthesis squared

      Power coefficient:

      (3.40)upper C Subscript upper P Baseline equals 4 a left-parenthesis 1 minus a right-parenthesis squared equals 4 a prime Subscript t Baseline left-parenthesis 1 minus a right-parenthesis lamda squared

      What is particularly interesting is that the residual rotational flow in the wake makes no apparent reduction in the efficiency of the power extraction.

      3.4.6 Axial flow field

      The axial velocity within the wake in this model falls discontinuously across the wake boundary from the external value and is radially uniform at the disc and in the far wake, just as the momentum theory predicts. There is a small acceleration of the flow around the disc immediately outside of the wake. The induced velocity at the wake cylinder surface itself and hence its convection velocity is −½ a at the disc and −a in the far wake.

      3.4.7 Tangential flow field