Wind Energy Handbook. Michael Barton Graham. Читать онлайн. Newlib. NEWLIB.NET

Автор: Michael Barton Graham
Издательство: John Wiley & Sons Limited
Серия:
Жанр произведения: Физика
Год издания: 0
isbn: 9781119451167
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href="#ulink_1a7b97b7-6568-5512-830a-ac739abdab12">(3.52) a relationship between the flow induction factors can be obtained. Dividing Eq. (3.52) by the modified Eq. (3.51), modified to include the additional loss of axial momentum from the pressure drop term Δpd2 in the far wake due to the centrifugal swirl generated radial pressure gradient, leads to:

      The flow angle ϕ is given by

StartStartFraction StartFraction upper C Subscript l Baseline Over upper C Subscript d Baseline EndFraction StartFraction 1 minus a Over lamda mu left-parenthesis 1 plus a prime right-parenthesis EndFraction minus 1 OverOver StartFraction upper C Subscript l Baseline Over upper C Subscript d Baseline EndFraction plus StartFraction 1 minus a Over lamda mu left-parenthesis 1 plus a prime right-parenthesis EndFraction EndEndFraction equals StartFraction lamda mu a prime left-parenthesis 1 minus a right-parenthesis Over a left-parenthesis 1 minus a right-parenthesis plus left-parenthesis a prime lamda mu right-parenthesis squared EndFraction

      Simplifying:

      These are consistent at the rotor tip (where μ = 1) with Eq. (3.2) provided a is sufficiently small compared with unity for terms in a2 to be neglected. This is normally true at the rotor tip, and these results agree exactly with the momentum theory prediction, because no losses such as aerodynamic drag have been included, and the number of blades is assumed to be large. This last assumption means that every fluid particle that passes through the rotor disc interacts strongly with a blade, resulting in the axial velocity being more uniform over the area of the disc. If the same analysis is followed excluding the swirl pressure drop term, then a = 1/3 – a small term ∼2/(9λμ)2, which is negligible except very close to the axis (blade root) or when the rotor tip speed ratio is very low.

delta upper Q equals 4 pi rho upper U Subscript infinity Baseline normal upper Omega italic r a prime left-parenthesis 1 minus a right-parenthesis r squared delta r equals 4 pi rho StartFraction upper U Subscript infinity Baseline Superscript 3 Baseline Over normal upper Omega EndFraction a left-parenthesis 1 minus a right-parenthesis squared r delta r

      The component of the lift per unit span in the tangential direction is therefore

upper L sine phi equals 4 pi rho StartFraction upper U Subscript infinity Baseline Superscript 3 Baseline Over normal upper Omega EndFraction a left-parenthesis 1 minus a right-parenthesis squared

      By the Kutta–Joukowski theorem the lift per unit span is

upper L equals rho upper W normal upper Gamma

      where Γ is the sum of the individual blade circulations and W is the component of incident velocity mutually perpendicular to both Γ and L.

      It