The purpose of this Appendix is to review all the calculations of the Lekakis triple hot wire anemometry (3HWA) measurement method [27] because some errors have been found in the development of the mathematical algorithm. The corrected procedure has been applied also in the present work for 3HWA measurements (see Sec. 4.1). This method belongs to the category of Analytical ones as previously explained in Appendix A because it permits to explicitly solve the non-linear system of equations that describe the response of each sensor of the probe.
Lekakis [27] approach is based on the concept of an effective cooling velocity, Q, that relates the anemometer output voltage, E, to the velocity magnitude and direction: E = f (Q).
Jorgensen [20] directional response law has been chosen by Lekakis [27] to represent the cooling velocity:
Q 2 = u 2 N + k 2 u 2 T + h 2 u 2 B (B.1) where:
u N is the velocity component normal to the sensor and in the plane of its supporting prongs;
u T is the velocity component tangential to the sensor;
u B is the velocity component normal to the sensor and normal to the plane of its supporting prongs.
k and h are the yaw and pitch coefficients, respectively, and they assume different values for each wire.
The geometry of the triple sensor probe is defined in Fig. B.1 in terms of the
to the probe axis and the angles θ i . The coordinate system used to define the instantaneous velocity vector is attached to the probe with the y − axis normal to the plane of sensor 3 and its supporting prongs, and the x − axis parallel to the probe axis.
The effective cooling velocities for each sensor (i = 1, 2, 3) of the probe, in terms
z, w
-x, -u
y, v g
b
2 1
3
d
1d
3d
2d
1q
iFigure B.1. Triple sensor probe geometry and coordinate system
of the instantaneous velocity − → V and the probe geometry are:
Q 2 1 = [u cos θ 1 + (w cos δ 2 + v sin δ 2 ) sin θ 1 ] 2 + +k 1 2 [u sin θ 1 − (w cos δ 2 + v sin δ 2 ) cos θ 1 ] 2 + +h 2 1 [w sin δ 2 − v cos δ 2 ] 2
Q 2 1 = [u cos θ 2 + (w cos δ 1 − v sin δ 1 ) sin θ 2 ] 2 + +k 2 2 [u sin θ 2 − (w cos δ 1 − v sin δ 1 ) cos θ 2 ] 2 + +h 2 2 [w sin δ 1 + v cos δ 1 ] 2
Q 2 1 = (u cos θ 3 − w sin θ 3 ) 2 + +k 3 2 (u sin θ 3 + w cos θ 3 ) 2 + +h 2 3 v 2
(B.2)
where u, v, w are the components of − → V in the x, y and z directions, respectively.
In terms of the polar and azimuthal angles β and γ (Fig. B.1) the velocity com- ponents are:
u = | − → V | cos γ v = | − →
V | sin γ cos β w = | − →
V | sin γ sin β
(B.3)
Substituting the third of Eq. B.3 into Eq. B.2 the equations relating the measured effective cooling velocities to the instantaneous velocity vector become:
Q 2 1 = | − → V | 2 h
sin 2 γ [G 11 sin 2 (β + δ 2 ) + G 21 ]+
+ cos 2 γ G 31 + sin(2γ) sin(β + δ 2 )G 41 i Q 2 2 = | − →
V | 2 h
sin 2 γ [G 12 sin 2 (β − δ 1 ) + G 22 ]+
+ cos 2 γ G 32 + sin(2γ) sin(β − δ 1 )G 42 i Q 2 3 = | − →
V | 2 h
sin 2 γ [G 13 sin 2 β + G 23 ]+
+ cos 2 γ G 33 − sin(2γ) sin β G 43
i
(B.4)
where:
G 1i = k i 2 − h 2 i + (1 − k 2 i ) sin 2 θ i
G 2i = h 2 i
G 3i = 1 − (1 − k i 2 ) sin 2 θ i
G 4i = 1 2 (1 − k 2 i ) sin(2θ i ) i = 1, 2, 3
(B.5)
These coefficient are assumed to have been determined from calibrations (h and
From Eq. B.4, doing the ratios of effective cooling velocities, two coefficients inde- pendent of the magnitude of the velocity vector (A and B) can be introduced as follows:
A = Q
2 1
Q
23= sin
2
γ[G
11sin
2(β+δ
2)+G
21]+G
31cos
2γ+G
41sin(2γ) sin(β+δ
2) sin
2γ[G
13sin
2β+G
23]+G
33cos
2γ−G
43sin(2γ) sin β
B = Q Q
222 3= sin
2γ[G
12sin
2(β−δ
1)+G
22]+G
32cos
2γ+G
42sin(2γ) sin(β−δ
1) sin
2γ[G
13sin
2β+G
23]+G
33cos
2γ−G
43sin(2γ) sin β
(B.6)
Their solution yields β and γ and the third of Eq. B.4 gives | − → V | 2 :
| − →
V | 2 = Q 3
psin
22 γ[G 13 sin 2 β + G 23 ] + G 33 cos 2 γ − G 43 sin(2γ) sin β (B.7)
Assuming that γ 6= π 2 1 , trigonometric relations reduce Eq. B.6 to:
( T 1 tan 2 γ + T 3 tan γ + D 7 = 0
T 2 tan 2 γ + T 4 tan γ + D 8 = 0 (B.8)
where:
T 1 = D 1 sin 2 β + G 51 cos 2 β + 1 2 G 61 sin(2β) + D 3 T 2 = D 2 sin 2 β + G 52 cos 2 β − 1 2 G 62 sin(2β) + D 4
T 3 = 2[D 5 sin β + G 71 cos β]
T 4 = 2[D 6 sin β − G 72 cos β]
(B.9)
and:
G 51 = G 1 1 sin 2 δ 2 G 52 = G 1 2 sin 2 δ 1 G 61 = G 1 1 sin(2δ 2 ) G 62 = G 1 2 sin(2δ 1 ) G 71 = G 4 1 sin δ 2 G 72 = G 4 2 sin δ 1
(B.10)
1