Chapter 4. The Yawed Hawt
4.1 The Yaw problem
Large upwind Hawts sparingly use grid power to yaw, only in response to long term
persistent windshifts. So the rotor axis often has sizeable yaw angle  to the current wind. The
2D rotortube momentum result 1.2 would imply the best power must decrease at least as the
cos projection factor on swept area. A cos3 power variation results from the high speed ratio
limit (1.5.8) of the naïve vector generalisation of differential momentum theory at the rotor; but
with only the axial component Vcos of the wind reduced, the recomposed velocity in the wake
skews at angle of naively
/ 2a) ≈3to the rotor axis for a=⅓ or 2 to the wind . The BGM assumes no net force on
the rotortube but here with the asymmetry a lift force perpendicular to the wind increases the
thrust and the power. Instead Coleman’s (1945) analysis of a skewed vortex ring wake explains
most simply the power as the typically measured cos2
….
4.2 Skewed Vortex Tube Wake
Consider a rotor of B blades of constant circulation K. So trailing vortices of circulation K
leave the tips at angle to the tangent at a certain spacing
(2Rsin  /B) . The circulation around the axial radial segment subdivided by the vortices is the
same by the curl theorem because the projected normal spacing is the same. Consider the wake
as quickly free of axial vorticity from the blade inner ends by the centrifuging of Chap 1.4.,
breaking the trailing tip vortices into a series of closed vortex rings, circulation  parallel to
the rotor and so at angle to the wake separated in time by / B In this interval the (far)
wake flow translates the circular rings a skewed distance s down the waketube, s = UaB
where Ua is the ultimate average velocity of the rings. The circulation around a path like the one
indicated in Fig 4.1 at any azimuth, shows the induced flow component down this solenoid is
still uniformly/s, despite the skew. Since the ring (rotor image) planes are separated by normal
distance scoshe crosswake component of the velocity jump at the top of the wake cylinder is
sin/ scos which should match the potential flow slip of the crosswake - resolution of the
outer wind over the top of the elliptic wake cylinder from Batchelor 6.6.21 ,
sin/scos Vsin(-)(1+1/cos
≈2Vsin(-) (4.2.1)
So /≈½/Vs.
(4.2.2)
Both jump and slip vary sinusoidally with azimuthal angle to then match everywhere. Ring
elements of length Rd sticking through the horizontal centerplane plane on the left and right
are indistinguishable from unskewed ones so the flow they induce is parallel to the wake
centerline. Top and bottom ring elements the same Rd long induce on the centerline an equal
speed but perpendicular to their skew . This motivates that the net induced flow is horizontal at
angle ‘ ‘ to and at the centerline of ½. As in a normal solenoid , Biot-Savart integration
shows this angle is constant throughout the wake. That allows the net U0 of wind plus uniform
induced velocity to flow down the waketube as by halving (4.2.1) , the crosswake induced
tan / s equals Vsin (-), the crosswake component of the wind. So thus ≈½ or the
induced flow bisects the angle between the rotor axis and the wake cylinder it has the same
component along each, s =2In say 2aVcos as in the BGM, so (4.2.2) gives ≈/a) (4.2.3)
so 3/2 is ½ the naïve value at a=⅓ So the real wake skew to the wind is ≈ a/ a)
(4.2.4) or ½ which is a small ¼ the wild naïve BEM value at a=⅓
The mean of the velocities outside and inside the wake cylinder Ua= Vcos(-)-½/s or Ua /V=
cos(-)-acos translates the rings, so the ring spacing is s = UaB thus
= 2In
Ua /B (4.2.5)
Exercise: Show that H=Ua/s and Ua≈ (1-a)V (1+2/16) where the last term is evaluated at
a=⅓ .
So the ring convection speed is hardly unchanged with yaw at a=⅓ . Now the power P from
the B blades is BQ/ (from Batchelor 6.6.26) (which automatically rejects spanwise flow)
or P=2InUaQ (4.2.6).
Ignoring the near wake and its expansion, every point on the rotor at azimuth  has a mirror
point at  + of equal induced vector on the imaginary upstream half of an infinite vortex
cylinder, so the average induced flow normal to the rotor must be half the full cylinder 2In. This
makes
Q=(1-a) VS cos as in the BGM . But now Ua≈ (1-a)V instead of another (1-a) VS cos in the
BGM gives the optimum P at a=⅓ to vary a bit less than cos2 as 1-152/16. The change is
entirely due to the difference in cos(-) from the BGM coshe component of wind
perpendicular to the vortex plane helps convect vorticity along the skewed wake increasing the
H.
For=0 tangent blades  elements =clW sin  is on azimuthal average cl V cos (1a) almost perfectly matching the above = acos V Ua / B for Ua/V ≈ (1-a)V for any wind
direction to the rotor(The non-dimensional chord Bkl=4a indicates that a changes from
optimum linearly with V which is the speed variation non-robustness with tangent blades as in
Chapters 2 and 6)
Exercise: With the more accurate Ua show the match to second order in a=⅓) at high x would
require Bkl=7a/2 and  =- (1-a)/ 8x compared with =(1-a)/x.
Negative pitch is unusual and with normal positive pitch, the match can’t be sustained, and a
would decrease with yaw and the performance would drift below optimum 1-.942 with ,
such as to the cos2 measured in the Figure 3.12 for a fixed pitch rotor.
There must be a tradeoff with robust for
well as power average windspeed. The yaw
best fixed pitch according to the statistics of
statistics would be specific to each design
wind speed and direction and yaw
and control strategy. Unfortunately even
response. Of course the designed optimum
wind direction statistics are little studied.
should be at the w(V) power average yaw as
Exercise: Show that with Ua≈ (1-a)V , designing for robustness at a mean value of yaw o, still
gives the design pitch to bisect the design =(1-a)cos o /x at large x but for the vortex theory
=2a/lx (whereas for the resolved flow BM =2a cos o /lx)
4.3 Momentum balances, Induced flow variations and Yaw moment
The blades axial A thrust BK r per unit span totaling to 2InUaS, (or SH) is higher by
the cos  factor than the rate of change 2InQ of rotortube momentum along the rotor axis.
Also why at high x, is there a component of I in the rotorplane normal to the pressure jumpH
across the rotor? The induced flow J and so dQ at the rotor should be symmetric about the page
in Fig 4.1 so there should no net blade drive to left or right. These discrepancies indicate a
greater pressure on windward side of the rotortube as it passes the rotor than on the leeward side.
One force the outer irrotational fluid can exert on the rotortube inside it is a lift 2L perpendicular
to V and away from the skew, so it would have a rotor axis component, (half of it momentum
flux Batchelor 407). The rotortube momentum balance is then 2Q I=L -A . So taking
crossrotor or rotorplane components P≈2InQ tan. Then resp. vortex A and momentum rotor
thrust components are 2In times UaS vs Q(1+ tan tan ) =V(1-a)S(1+¼) so L overcorrects
the axial momentum balance by a factor of 4/3 at a=⅓ but agreeing as a and the expansion goes
to 0 taking  to ½.he 3/16 might be taken as an expansion correction to A and the power
P=Au for a net 1-¾2 at a=⅓, though this would require even more stall-defying negative pitch
to maintain with . At any rate (resolved) momentum theories err in yaw because they generalise
from windwise with no drag (Chapter 1.2) to no rotor tube force neglecting the possibility of
skewed rotortube lift L normal to the wind.
Such a lift requires net vertical vorticity inside the wake tube and the unbalanced leading lip
of the vortex rings from the upwind (right) edge of the rotor has the right orientation of K and
with its area scaling as R2 its lift scales correctly as O(InUaS The lift also implies net
pressure on the rotortube walls and indeed the low pressure backside of the rotor is particularly
close to the upwind leading wedge of the waketube. As the pressure varies, the velocity must
vary for the uniform H across the rotortube tube indicated by the uniform velocity far upstream
and downstream. Indeed the induced flow J on downwind side of the rotor mouth of the vortex
tube closer to the rings should be greater than that on the upwind side, but symmetrical about the
horizontal. So the leading term in a Taylor series would a linear variation of J from I of J≈
vyR where y is distance to the left of a blade element from the rotor center . Say it spans about
say a diameter flow-wise. Then there would be a net Bernoulli lateral force of 2R∫u vsin2
Rd or 2u vR2 on the rotortube to equal 2uS aVcostan thus v≈aVtan  so J≈IyR
agreeing with Coleman’s direct integration at the rotor for the induction J of unexpanded wake
vortex rings.
This J variation implies just a variation in local power with no change to the net power,
which depends on the uniform 
But for a physical blade element to have the constant
despite the J variation in  requires azimuthally varying pitch to maintain constant  . Many
have violated the constant K in calculating the tangent blade circulation as (c times) V cos (1a) - aVyR with relative variation ayR(1-a) It is more reasonable to generalise the above
picture (implicitly with vortex rays shed by the blades) to say that the induced flow J at the
rotor is aVy/R more than I , half the induced flow downstream on the same streamline. Then
for a tangent blade the circulation varies as (c times) V cos (1-a) -I-aVy/R whilst the now
‘local” resulting induced flow in the wake would vary as 2aV +2I . For these to be
proportional requires I to be -a2Vy/R/(1-2a) for relative variation as ay/R at small a, or
quibbling  aof the naive value. More importantly such variation may still be sufficient to
trip a blade from lift enhanced by Coriolos secondary flow (appendix 2) into a deep stall.
Exercise: Show this simultaneous variation of K and u reduces the tangent blade power
fractionally as the average square of ay/R over the area or /64 at a=⅓ . Show it causes a
yaw restoring torque equal to the thrust times the arm ⅜aRabout the rotor center, as if it were
directed downwind at a virtual center ⅜aR behind the rotor.
How a Hawt actually passively yaws requires analysis of the larger yaw moments of the
tailvane with particular attention to any aerodynamic damping. Structurally rapid yaw can
impose large gyroscopic moments on the spinning rotor so additional strengthening or damping
can be needed.
An implication would be that other tangent blade crosswind mills such as the oscillating wing at
low wind tolerate yaw as cos2 also. Of course our next major subject the (tangent blade) Vawt
is already perfect in yaw.