Acoplamiento mutuo en líneas de transmisión

As the inviscid flow of interest here is driven solely by matching with the entrainment velocities into boundary layer we expand the velocities and pressure as

U{r,z) = 0 + eu(r, z) + . . . (2.4)

V{r,z) = 0 P e v { r , z ) . (2.5)

u dv H--- f r dz = 0, du du dp dr dv dv dp dz

C H A P T E R 2. A DISC RO TATIN G N E A R H O RIZO N TAL GROUND 31

where e = Re~^!'^. Substituting these into (2.1) - (2.3) along with the assumption th a t R e ^ l leaves us with the axisymmetric Euler equations for u, v and p:

dr ' r ' dz

i i

(2,8)

(2.9) It is reasonable to assume initially th at the flow is irrotational and by the principle of conservation of vorticity it will then remain so for all time. It is therefore possible to pose the full problem in terms of the velocity potential 0 where u = d ^ / d r and

V = d ^ j d z . The velocity potential, continuous through all space outside of the body H, satisfles 1 a # ^2^ V ^^ = ^ + - ^ + ^ = 0 in r > 0, z > 0, (2.10) with — = 0 on z = 0, r > 0, (2.11) —— = 0 on H, (2.12) on ^ = -n + (s ) on ABD+, (2.13) — = Ug (s) on ABD . (2.14) These conditions are the inviscid tangential flow condition on the ground, on the body, and the entrainment into the boundary layer along either side of the disc and the layer. In the full problem there may be some inner-outer interaction, causing the entrainment velocities v f to depend upon the layer shape Z(r) but we are going to model the entrainment velocities as known and the boundary layer flow to be

C H A P T E R 2. A DISC RO TATIN G N E A R H O RIZO N TAL GROUND 32

independent of the shape of the viscous shear layer. We also have a relation on the pressures at the wake. We first make the approximation th a t the pressure is continuous across the layer and later turn to the more accurate case where the pressure jum p across the layer is proportional to its curvature. So, for the time being, we also require

— p~(s) — 0 across the wake BD, (2.15) as a model. The model is found to yield a significant result for the far-field re­ sponse and to generate an accurate computational approach for more general cases considered in section 2.4. For a rotating disc in the absence of ground effect the entrainm ent velocities are given by

7 on A B

'^e ='^e = \ (2.16)

I

where 7 is a constant. The first relation in (2.16) comes from the Von Karman solution and the second is from Smith and Timoshin (1996a) which shows the decay in entrainment of fluid into the boundary layer going a s l / r . This is not the complete story however as, in reality, the entrainment velocities vary smoothly, with a small region over the disc rim smoothly connecting the entrainment from the constant velocities along the disc to the 1/r decay into the layer beyond the rim. As this is very local it is thought not to have an impact on the overall solution. As the problem is linear, the choice of the constant 7 is unim portant and is, for convenience, set as

C H APTE R 2. A D ISC R O TA TIN G N E A R H O RIZO N TAL GROUND 33

0 _r

Figure 2.2: The far-field problem

2.2.2

T he far field response

From the far field the disc and the ground can be viewed as effectively coincident, leaving two distinct regions: one above and one below the entrainment layer which is assumed to asymptote to a straight line through the origin of an unknown slope (given by an angle a). The problem to be solved, in terms of the velocity potential

0, but in spherical coordinates {s,6), is given by (2.10) subject to when 6 = 0, 7T when ^ , when 6 = (2.17) (2.18) (2.19) which are a symmetry condition, the no penetration ground condition and the re­ quirement of entrainment velocities into the layer, respectively (see figure 2.2). The

C H A P TE R 2. A DISC RO TATIN G N E A R H O R IZO N TAL GROUND 34

angle a is the inclination of the layer to the vertical and is to be determined. We also require the pressure to be continuous across the layer. Solving these equations yields the velocity potential 0 as

a

^ = { " (2.20)

—j42 In(ssin^) if 6 > a

and Stokes’s streamfunction ^ as

(

respectively, where ^1,^42 are constants adjusted to satisfy the entrainment con­ dition (2.19). The deflection angle a is now determined from applying Bernoulli’s theorem along the layer. This gives us the requirement th a t ( |f ) ^ + (“ | | ) ^ needs, in order to give pressure continuity across the layer, to be the same on either side, i.e. àt 9 = a^. Applying this it becomes apparent th a t this can only be satisfied for « = I (2.22) These solutions form the far field solution for any bounded rotating blade system in the proximity of the ground. It holds for the rotating disc considered here but, to repeat, it also holds when there are genuine blades and wakes present as in a rotor blade, because the azimuthal dependence erodes away as r increases leaving the same far-field form regardless of the initial rotor set up.

C H A P TE R 2. A D ISC RO TATIN G N E A R H O RIZO N TAL GROUND 35

2.2.3

Flow induced by an infinite disc rotatin g above th e

ground

We temporarily consider an unbounded rotating disc near the ground. This is of interest as it is gives an idea of what is happening beneath a finite disc near to the ground and is useful in the small h analysis near the disc rim considered in subsection 2.3.3. Above the disc we have the Von Karman solution with fiuid constantly fiowing in axially. However, beneath the disc the no penetration condition at the ground prevents this from happening; so what form does the fiow take there?

We have to solve the Euler equations in the region between the disc and the ground subject to no penetration on the ground and constant entrainment into the layer on the disc. We consider r )$> 1 and, assuming v is independent of r at large distances, we write

v { r , z ) = v { z ) , (2.23) which, along with the Euler equations, imply the forms

u{r,z) = ru{z), (2.24)

p{r,z) = r'^p + Po{z),

where p is a constant. Substituting these into the Euler equations reduces them to

2u + v' = 0, (2.25)

a^ + vü' = - 2 p , (2.26)

vv' = -p'o, (2.27)

subject to

C H A P TE R 2. A D ISC RO TATIN G N E A R H O RIZO N TAL GROUND 36

%(0) = 0, (2.29) where 7 is the constant entrainment into the discs boundary layer and ' denotes differentiation with respect to z. Integrating (2.27) immediately yields

Po = “ 2^^’ (2.30)

Differentiating (2.26) and substituting for v' from (2.25) gives

vu” = 0. (2.31)

Solving this, and applying a zero vorticity condition so th a t u = 0, yields the simple form for the velocities and pressure as

u = - g , (2.32)

V = (2.33)

7*^7^ 7^2;^ " = - w - k -

So this determines the flow between an inflnite disc and the ground, or the flow beneath a disc close to the ground but away from the disc rim. See also Debuchy et

al (1998) who consider a similar limit for two inflnite co-rotating discs at small h.

2.3

D eterm ining th e layer shape