Wind Energy Handbook. Michael Barton Graham

Wind Energy Handbook

W squared italic upper B c left-parenthesis upper C Subscript l Baseline cosine phi plus upper C Subscript d Baseline sine phi right-parenthesis delta r"/>

The torque on an annular ring is

(3.47) delta upper Q equals left-parenthesis delta upper L sine phi minus delta upper D cosine phi right-parenthesis r equals one half rho upper W squared italic upper B c r left-parenthesis upper C Subscript l Baseline sine phi minus upper C Subscript d Baseline cosine phi right-parenthesis delta r

where B is the number of blades.

3.5.3 The BEM theory

The basic assumption of the BEM theory is that the force of a blade element is solely responsible for the change of axial momentum of the air that passes through the annulus swept by the element. It is therefore to be assumed that there is no radial interaction between the flows through contiguous annuli: a condition that is, strictly, only true if pressure gradients acting axially on the curved streamlines can be neglected if the axial flow induction factor does not vary radially. In practice, the axial flow induction factor is seldom uniform, but experimental examination of flow through propeller discs by Lock (1924) shows that the assumption of radial independence is acceptable.

Equating the axial thrust on all blade elements, given by Eq. (3.46), with the rate of change of axial momentum of the air that passes through the annulus swept out by the elements, given by Eq. (3.9), with A_D = 2πrδr

(3.48) delta upper T equals one half rho upper W squared italic upper B c left-parenthesis upper C Subscript l Baseline cosine phi plus upper C Subscript d Baseline sine phi right-parenthesis delta r equals 2 italic pi r delta r rho upper U Subscript infinity Baseline left-parenthesis 1 minus a right-parenthesis 2 italic a upper U Subscript infinity Baseline

It should be noted here that the right hand side of Eq. (3.48) ignores the effect of the swirl velocity (2a'ΩR) on the axial momentum balance through generating a centrifugal pressure gradient in the far wake from the axis to the wake boundary. The resulting pressure reduction that generates an additional pressure drop across the disc was termed Δp_d2 when considered previously in Eq. (3.22).

Equating the torque on the elements, given by Eq. (3.47), with the rate of change of angular momentum of the air passing through the swept annulus, given by Eq. (3.34),

(3.49) delta upper Q equals one half rho upper W squared italic upper B c r left-parenthesis upper C Subscript upper L Baseline sine phi minus upper C Subscript upper D Baseline cosine phi right-parenthesis delta r equals 2 italic pi r delta r rho upper U Subscript infinity Baseline left-parenthesis 1 minus a right-parenthesis 2 a prime r squared normal upper Omega

If drag is eliminated from the above two equations, to make a comparison with the results of the vortex theory of Section 3.4, the flow angle ϕ can be determined:

tangent phi equals StartFraction a prime r normal upper Omega Over italic a upper U Subscript infinity Baseline EndFraction equals StartFraction a prime Over a EndFraction StartFraction r Over upper R EndFraction lamda

However, from the velocity triangle at a blade element given by Eq. (3.44), the flow angle is also

tangent phi equals StartFraction 1 minus a Over lamda Subscript r Baseline left-parenthesis 1 plus a prime right-parenthesis EndFraction

Equating the two above expressions for tanϕ

StartFraction a prime Over a EndFraction StartFraction r Over upper R EndFraction lamda equals StartFraction 1 minus a Over lamda Subscript r Baseline left-parenthesis 1 plus a prime right-parenthesis EndFraction

(3.50a) a left-parenthesis 1 minus a right-parenthesis equals lamda Subscript r Baseline Superscript 2 Baseline a prime left-parenthesis 1 plus a prime right-parenthesis

At the outer edge of the rotor μ = 1 and a^′ = a^′_t, so

(3.50b) a left-parenthesis 1 minus a right-parenthesis equals lamda squared a prime Subscript t Baseline left-parenthesis 1 plus a prime Subscript t right-parenthesis

Equation (3.2) is consistent with the earlier Eqs. (3.32) and (3.33).

With drag included the thrust Eq. (3.48) can be reduced to

(3.51) StartFraction upper W squared Over upper U Subscript infinity Baseline Superscript 2 Baseline EndFraction upper B StartFraction c Over upper R EndFraction left-parenthesis upper C Subscript l Baseline cosine phi plus upper C Subscript d Baseline sine phi right-parenthesis equals 8 pi a left-parenthesis 1 minus a right-parenthesis mu

where the parameter mu equals StartFraction r Over upper R EndFraction _.

If the pressure drop term Δp_d2 is not ignored, the right hand side of Eq. (3.51) becomes 8πμ{a(1 − a) + (a^′λμ)²}. The additional term (a^′λμ)² is small and usually negligible except very close to the rotor axis or at low tip speed ratios.

The torque Eq. (3.49) simplifies to

(3.52) StartFraction upper W squared Over upper U Subscript infinity Baseline Superscript 2 Baseline EndFraction upper B StartFraction c Over upper R EndFraction left-parenthesis upper C Subscript l Baseline sine phi minus upper C Subscript d Baseline cosine phi right-parenthesis equals 8 pi lamda mu squared a prime left-parenthesis 1 minus a right-parenthesis

It is convenient to put

(3.53a)

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