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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The actuator disc concept

      The mechanism described above accounts for the extraction of kinetic energy but in no way explains what happens to that energy: it may well be put to useful work, but some may be spilled back into the wind as turbulence and eventually be dissipated as heat.

      Upstream of the disc, the streamtube has a cross‐sectional area smaller than that of the disc and an area larger than the disc downstream. The expansion of the streamtube is because the mass flow rate must be the same everywhere. The mass of air that passes through a given cross‐section of the streamtube in a unit length of time is ρ AU, where ρ is the air density, A is the cross‐sectional area, and U is the flow velocity. The mass flow rate must be the same everywhere along the streamtube, and so

      (3.1)rho upper A Subscript infinity Baseline upper U Subscript infinity Baseline equals rho upper A Subscript upper D Baseline upper U Subscript upper D Baseline equals rho upper A Subscript upper W Baseline upper U Subscript upper W

      The symbol ∞ refers to conditions far upstream, D refers to conditions at the disc, and W refers to conditions in the far wake.

Schematic illustration of an energy extracting actuator disc and streamtube.

      3.2.1 Simple momentum theory

      The air that passes through the disc undergoes an overall change in velocity, UUW, and a rate of change of momentum equal to the overall change of velocity times the mass flow rate:

      (3.3)italic Rate of change of momentum equals left-parenthesis upper U Subscript infinity Baseline minus upper U Subscript upper W Baseline right-parenthesis rho upper A Subscript upper D Baseline upper U Subscript upper D Baseline

      The force causing this change in momentum comes entirely from the pressure difference across the actuator disc and the axial component of the pressure acting on the curved surface of the streamtube. This latter pressure is usually assumed to be ambient and therefore to give zero contribution, without further explanation. In fact this pressure is different from ambient due to the axial variation of velocity along the streamtube, but the integral of its axial component from far upstream to far downstream can be shown to be exactly equal to zero, the streamwise contribution upstream of the actuator disc exactly balancing the downstream contribution that opposes the stream (Jamieson 2018).

      Therefore,

      To obtain the pressure difference (pD+pD), Bernoulli's equation is applied separately to the upstream and downstream sections of the streamtube: separate equations are necessary because the total energy is different upstream and downstream. Bernoulli's equation states that, under steady conditions, the total energy in the flow, comprising kinetic energy, static pressure energy, and gravitational potential energy, remains constant provided no work is done on or by the fluid. Thus, for a unit volume of air,

      Upstream, therefore, we have

      (3.5b)one half rho Subscript infinity Baseline upper U Subscript infinity Superscript 2 Baseline plus p Subscript infinity Baseline plus rho Subscript infinity Baseline italic g h Subscript infinity Baseline equals one half rho Subscript upper D Baseline upper U Subscript upper D Superscript 2 Baseline plus p Subscript upper D Superscript plus Baseline plus rho Subscript upper D Baseline italic g h Subscript upper D

      Assuming the flow speed to be at low Mach number M (typically M < 0.3 is sufficient), it may be treated as incompressible (ρ = ρD) and to be independent of buoyancy effects (ρgh = ρghD) then,

      (3.5c)one half rho upper U Subscript infinity Superscript 2 Baseline plus p Subscript infinity Baseline equals one half rho upper U Subscript upper D Superscript 2 Baseline plus p Subscript upper D Superscript plus

      Similarly, downstream,

      (3.5d)one half rho upper U Subscript upper W Superscript 2 Baseline plus p Subscript infinity Baseline equals one half rho upper U Subscript upper D Superscript 2 Baseline plus p Subscript upper D Superscript minus

      Subtracting these equations, we obtain

      (3.6)left-parenthesis p Subscript upper D Baseline Superscript plus Baseline minus p Subscript upper D Baseline Superscript minus Baseline right-parenthesis equals one half rho left-parenthesis upper U Subscript infinity Superscript 2 Baseline minus upper U Subscript upper W Superscript 2 Baseline right-parenthesis

      (3.7)one half rho left-parenthesis upper U Subscript infinity Superscript 2 Baseline minus upper U Subscript upper W Superscript 2 Baseline right-parenthesis upper A Subscript upper D Baseline equals left-parenthesis upper U Subscript infinity Baseline minus upper U Subscript upper W Baseline right-parenthesis rho upper A Subscript upper D Baseline upper U Subscript infinity Baseline left-parenthesis 1 minus a right-parenthesis

      and so,

      (3.8)upper U Subscript upper W Baseline equals left-parenthesis 


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